Amazing, as a theoretical physicist myself, I'm looking forward to see more mathematical videos about this topic 😁 Edit: makes me wanna sit down with numerical methods and analyze more reallistic example of shrinking and expanding core (true R(t))
@@pressaltf4forfreevbucks179 Hong Kong $ ... so only 64 US$ I'm poor too 😭😉 (and given I'm now in Philippines it shows 2500PHP... not sure why as my visa was charged 500HKD that I've selected)
WOW, thank you so much for your kind support, I am grateful for the appreciation. This drives me to keep making these videos. I enjoy making them (despite the consumption of my free time) so it is very gratifying to know that the content is appreciated. Thanks again!
Man, I am a high schooler and I am trying my best to learn Physics, a subject that I love with all of my heart, I am self studing calculus to, one day, contribute a little to this wonderful field, the things that I find more interesting are the possibilities that the Island of stability can give us, and I want to dedicate my life to that. Take my money (earned by surveys so no problem for me) and buy a cup of something
Sounds like your more proactive with your education than I was at your age. Good for you. I did take calculus in high school but not until the second half of senior year after taking pre calculus over the summer. I wish I did it sooner or tried testing into some more advanced classes because I could of gotten the entire calc chain including differential equations out of the way before college if I wasn’t afraid to take AP courses. I could of also gotten physics 1 and 2 as transfer credits I think if I took the calc based versions. It would have been intense but easier than doing it in college and put me 2 years ahead for my Bachelor’s. It may have lowered my GPA but I didn’t go to an elite University so it wouldn’t have mattered anyway. And if your aiming for an elite school, AP courses are a good way to stand out if you do well in your classes. In hindsight, I think it’s worthwhile to take some risks and not play it too safe in high school because if you have big goals your going to have to do that at some point.
@@jacobharris5894 hello, thank you for your words and your experiences. I'll surely try all I can, here we have many good universities but to attend the best one you need to pass a very difficult exam with at least 65-70% (only 32 people) else you are out. It is needed to start studing at least 1-2 years before. I still have more than 2 years, if my will is enough I’ll make it *fingers crossed*, if no there are many others very good.
Absolutely fascinating. I have always been interested how this was calculated ever since I did a school assignment on the Manhattan project when I was 14 in 1974.
glad you liked it; similar story here: I read about the Manhattan Project when I was a kid and got obsessed and decided to be a physicist. That was a few decades back and here I am.
This is the first time that I am excited for a homework assignment and I can't believe it came from TH-cam, I really want to try to do this! Brilliant video and thanks for educating us on this fascinating topic!
A little mistake, min 10:17. On radial term you have inside the brackets (1/r^2 df/dt) but it is ( r^2 df/dt ). After you use the function f=u/r all come back to be good.
Thank for the notification, you are right and another viewer also let me know about this, I messed up the r² term already at 10:10 so I have included an erratum in the video description. Thanks so much for pointing this out and in such a courteous way (not very common on the internet), it is a good catch. Fortunately, as you proved, it doesn't affect the calculation because it is correctly applied, this is just a typo on the video but not an error in the calculation. I do my best to avoid these typos but after watching everything many times some minor details slip through. Thanks again.
I graduated in physics in 1971 in Bristol, UK. When we were shown the neutron diffusion equation, our lecturer said "you all have all of the mathematical tools that you need to solve this equation, but I'm not allowed to show you how to do it, for reasons of national security." How times have changed!
maybe your lecturer was joking or trying to make it more interesting because the use of diffusion theory for weapons design was rendered obsolete pretty soon, it is of little use to someone attempting to build a nuclear weapon, and even the reports including these calculations were declassified in 1965
@@jkzero many thanks for your reply. Klaus Fuchs gained his PhD at Bristol and worked on the theoretical calculations for nuclear weapons. Many years later was imprisoned in the UK for passing nuclear secrets to the Russians. This may have led to the staff at Bristol becoming over cautious in protecting "secrets."
glad to see that people appreciate the puns... there is a Taylor Swift joke hidden in a previous video than maybe was too subtle because only one person noticed it
Thanks for the endorsement; I wish I could get to wider audiences but the TH-cam algorithm is driven by engagement so you can actively help the channel by liking, subscribing, and sharing. This type of support is highly appreciated so I can continue making videos.
Thank you so much for your continuous support, I will make sure to enjoy a coffee while being grateful for the appreciation of strangers on the other side of the planet that find this content of interest. Thanks again!
Not going to lie, the maths is a bit beyond me, but I love how you worked out the critical masses. All of your videos have been amazing and I always look forward to the next one.
thanks for the feedback, this video was math heavy; people seem to like these so I will make more but I will also continue with the standard video of some history sprinkled with a less-intense calculation on interesting topics, even beyond nuclear weapons; I don't want the FBI knocking on my door :D
Man, this theory was taught to me by a professor the last semester, amazing the application and the smooth way this works to explain such a beauty phenomenon
Awesome! I am curious to know what brings viewers to the channel, were you searching for something in particular or did the 'mighty algorithm' find you?
You could always look at TH-cam channel "Nuclear Engineering Lectures" NE560, here: th-cam.com/video/iSbLUBuaT78/w-d-xo.html - but remember transfer function analyses are perturbative, and the full range of control rod operation is not, so you have to hope things stay linear. Also remember transfer function theory often looks at the Zero-Power state (to avoid pesky thermal feedback considerations) of a reactor at criticality. It's a useful tool but extending it to more general cases is not trivial.
I think N=0 at boundary of the ball is not "no neutrons escape" boundary condition, it is more like "all neutrons trying to escape are vanishing" because neutron current J is nonzero through boundary
I think you are totally right. Honestly, the "elementary" boundary condition is quite artificial, in fact, it is only used because it allows an analytical solution for R_c in terms of the relevant parameters ( u, cross-sections, density, etc.), it gives a glimpse of what drives R_c and the critical mass. However, this is the only reason this condition is introduced, mostly for academic purposes. The "advanced" boundary condition is what has to be used in a practical calculation.
I cannot thank you enough for your continuous support. The only way, I guess, is by keeping the videos coming and the quality standards. Thank you so much!
Actually, I am amazed by your work. I don’t know what to say. I thought a lot about the effort you made to make this video. It must have been hard. I don’t think anyone can thank you as you deserve. In fact, I am traveling to Russia, where there are no means of support or Communication with the outside world, even TH-cam ads are prohibited, so I turned on a VPN so that ads would appear on your videos and I could watch them. Indeed, I watched 14 ads and visited 11 advertiser’s websites. This took some time, but it is nothing compared to the time that you spent making this video, and perhaps this is worth a few cents to the channel, but I feel that it is more moral support than material support. Thank you from the bottom of my heart, you brilliant mind.❤
Thanks for the endorsement; I wish I could get to wider audiences but the TH-cam algorithm is driven by engagement so you can actively help the channel by liking, subscribing, and sharing. This type of support is highly appreciated so I can continue making videos. Thanks also for watching the ads, I do the same with small channels that I watch, if there is an ad I let it go until the end because I know I am contributing with a few cents to the creator.
I think that there is a small error at 13:35. At this point in the video it is stated that the third boundary condition is that no neutrons escape the sphere and this is done by requiring that N(r=R, t)=0. Then there is a quick disclaimer that this condition is not completely true and the more advanced solution is briefly mentioned. I may just be misunderstanding it, but I understood this to mean that the condition N(r=R, t)=0 does imply that no neutrons will escape the core however this boundary condition is not practical or optimal. Using the solution that satisfies the condition N(r=R, t)=0, I calculated the total neutron flux at the boundary from the neutron density and found that it is not zero but proportional to R*N(r=0, t). I am not criticizing the use of the boundary condition N(r=R, t)=0, but I think it is incorrect to say that this condition is equivalent to requiring that no neutrons escape the core.
Thanks for your exposition. Honestly, the "elementary" boundary condition is quite artificial, in fact, it is only used because it allows an analytical solution for R_c in terms of the relevant parameters ( u, cross-sections, density, etc.), it gives a glimpse on what is driving R_c and the critical mass. However, this is the only reason this condition is introduced, mostly for academic purposes. The "advanced" boundary condition is what has to be used in a practical calculation.
thanks for that, you can support the channel by liking, commenting, and sharing the videos, that engagement drives the algorithm that gives the channel exposure. Thanks in advance.
oh no... you are totally right, I messed up the subscripts at 15:58, they should be reversed, it should say: lambda_f = 16.89 cm and lambda_t = 3.596 cm. I will add an errata in the description. Thanks so much for pointing this out, it is a good catch. Fortunately, it doesn't affect most of the elementary calculation because the two mean-free paths are multiplied but it does make as difference in the 'advanced' solution, where lambda_t appears isolated. Thanks again. I do my best to avoid these typos but after watching everything many times some minor details slip through. Thanks also for asking in such a nice way contrary to patronizing mode of writing comments favored on the internet.
Excellent! Could you make a video on the implosion physics ? For example, it is often said that although it was known that hollow cores would have a much better yield, the original bomb used a "conservative" design based on a solid core ("Christy core") because the maths of hollow core implosion was too difficult to solve at the time (but it was finally implemented in all designs a few years later). I could not find any more detailed info on that on the web.
did you check my "Nuclear Weapons Q&A" video? I don't go into the details of the solid vs. hollow core but I do calculate how much the core is compressed because somebody asked about it.
My kind of YT video, just at the right intersection of math and physics, delivered at the right cadence, not too fast, not too slow, loved it. One question, if sin(kR) = 0, wouldn't that give us multiple solutions? ie, kR = n * pi? Any particular physical constraint that we must consider only the lowest numbered "harmonic", n = 1?
If you used solutions with multiples of pi you would have areas with negative neutron density which I guess would not make much physical sense, but I am kinda guessing. The reason why you can use multiple solutions in QM is that you multiply the function with it’s complex conjugate before you get the probability, which means that it will always be positive:)
nice catch, that is a very nice observation; you are right, the general solution is kappa*R=n*pi; however, the higher "harmonics" will only make the bomb core bigger, we are searching for the minimum radius. Also, a solution for n>1 can lead to a negative density, which is also a non-physical solution. Mathematically, your point is 100% correct; however, here is when physics guides the final solution. Just like setting B=0 to remove the non-physical cos(kappa*r) term.
Thank you Prof, that makes sense. You did proactively point out all other, non-valid, solutions and explained why they are not applicable. So this stuck with me. This was the only point, everything else i was able to follow-up rather clearly. These are not easy thing to communicate, you struck absolutely the right balance between explaining the points in sufficient detail, without getting bogged down. That balance is hard to get right, you hit the ball out of the park. Thank you!
@@ytashu33 in the video at 14:20 I refer to the solution used as "the simplest solution" this is precisely to address your question. I could have explained more details but the video was already 25 min long. Thanks for your appreciation, I love teaching and created this channel mostly because I wanted to share fun and education stuff, I just never thought that a video only about solving a hard equation would produce this level of interest, which I find great. Welcome to the channel and make sure to check the other videos.
How is the boundary condition expression for the advanced solution derived (20:59)? I'd really be thrilled to read the derivation, can you provide something? Is it something like Neumann vs Dirichlet boundary condition? Thank you for all of your work!
Thanks fro the feedback, I am so excited that his video got so many people interested. The elementary boundary condition (BC) is a Dirichlet BC but it is quite artificial, the correct way of doing the calculation is by using a Neumann BC. I am glad someone got hooked on that detail, it is a lengthy derivation presented for the first time by Robert Serber, one of Oppenheimer's protegés, on his annotated version of The Los Alamos Primer. He only did it in Cartesian coordinates but it can be generalized to spherical coordinates chooser.crossref.org/?doi=10.2307%2Fj.ctvw1d5pf.6
Thanks for the endorsement; I wish I could get to wider audiences but the TH-cam algorithm is driven by engagement so you can actively help the channel by liking, subscribing, and sharing. This type of support is highly appreciated so I can continue making videos.
I believe that the elementary solution does not correspond to neutrons not escaping the core, but rather that neutrons that leave the core are permanently lost, so the function is 0 outside the sphere. I think an actual condition for neutrons not leaving the center would be the spatial derivative wrt to r being 0 at the center, as expected from Fick's law, instead of the actual value. Otherwise sub-criticality is not achievable because the number of neutrons would not only be conserved but also continuously multiplying. Other than that, loved the video, think your calm and collected teaching style is great.
I am glad you liked the video, thanks for the feedback. Honestly, "elementary" boundary condition is quite artificial, in fact, it is only used because it allows an analytical solution for R_c in terms of the relevant parameters ( u, cross-sections, density, etc.), it gives a glimpse on what is driving R_c and the critical mass. However, this is the only reason this condition is introduced, mostly for academic purposes. The "advanced" boundary condition is what has to be used in a practical calculation.
Thanks for producing this excellent video. It answered two long-standing questions I have had: what is the uranium critical mass, and how does implosion reduce it. I have 2 comments: 1. Being too lazy to calculate, I'm guessing that putting a boundary condition on the neutron flux at the surface instead of the density means that there is a higher density of neutrons out near the boundary, producing more fissions there. Thus a lower critical mass. 2. At the end of WWII, German atomic scientists, including Werner Heisenberg, were interned at an English estate known as Farm Hall. Their conversations were bugged and translated. When they learned of the Hiroshima explosion, Heisenberg gave this lecture to the other guests. As the physicist Jeremy Bernstein pointed out, his inept presentation made it clear that he had never tried to solve this diffusion problem before. This indicates the primary reason why the Germans never got very far in making an atomic bomb: utter incompetence at the very top of a rigid, hierarchical research/development system. As Bernstein put it, Heisenberg was a great physicist (invented quantum mechanics) but not a very good one (couldn't get the units right in a simple calculation). The Manhattan project had the likes of Fermi, who was both great and good.
Thanks, I am glad you liked the video and that it answer many of your questions. Regarding your follow-up questions: 1. yes, relaxing the boundary condition gives you a lower critical mass. 2. Your words echo very much Berstein's sentiment. I do not buy the narrative of the scientists' boycotting the bomb program but at the same time I disagree on Heisenberg being so inept, the evidence suggests that they simply worked on other things. His lecture at Farm Hall shows that it took him just a couple of days to correctly determine the critical mass, although the transcripts also reveal that he knew little to nothing about bomb physics. I have a full video on Heisenberg and the German bomb th-cam.com/video/6zIJTwQ2blU/w-d-xo.html
It's ridiculous to say that Heisenberg was not a very good physicist. Heisenberg made a mistake in his first attempt to calculate the critical mass, relying on "random walk" model that resulted in need for dozens of tons of fissile material for nuclear detonation. Edward Teller admitted in an interview he made the same mistake first as well. The difference between them (Heisenberg vs Teller, or more generally, most German physicist remained in their Motherland vs Manhattan project participants) is not greatness as scientists. The reason is simply the motivation. Teller and his fellows continued relentlessly with a different approach because of the fear of the Nazis. In contrast to them, Heisenberg didn't have such phobia as a driving force against Americans. He wasn't particularly eager for building the bomb and could abandon this goal easily on one hand. On the other hand (perhaps more importantly), he could set his heart at rest, seemingly not being have to worry anymore about Germany being bombed with nuclear weapon. Teller praised Heisenberg as a "hero" for this. In a 2000 paper that summarizes and discusses Heisenberg's Farm Hall lecture, author M.S El Nachie also concluded that Heisenberg has never thought seriously about the bomb before he learned of Hiroshima bombing. Talking about scientific greatness El Nachie praises Heisenberg for his razor sharp intellect for deriving the correct result alone in couple of days, without tables, calculator, his notes etc, being de facto in captivity, knowing nothing of his family. He wrote that Heisenberg was doing what theoretical physicists do best. (Note that in the "Los Alamos Primer" -lectures based on the work of dozens of well supported scientists in the Manhattan Project- also only presented the "elementary solution" for critical mass.)
@@ukornel77 Your comment, "Talking about scientific greatness El Nachie praises Heisenberg for his razor sharp intellect for deriving the correct result alone in couple of days, without tables, calculator, his notes etc, being de facto in captivity, knowing nothing of his family." is puzzling. When I was an undergraduate physics major in the 1970's, I would have been expected to be able to do this calculation off the top of my head. The mathematical techniques used therein have been part of the standard literature since the 19th century. Notes, tables, calculator are not required, although the fission cross section is required to convert the analytical result into an actual critical mass. Heisenberg most certainly should have been able to do this calculation at the beginning of the German bomb program. The physicist Jeremy Bernstein suggested that, since Heisenberg got it wrong early on, the rigid, hierarchical nature of the German research system made it very difficult for others to correct the lapse. So they all dropped the subject.
@@paulkolodner2445 You are right, calculators, tables are not needed for the correct mathematical solution. But in the Farm Hall Heisenberg wanted to answer the question: how the Americans were able to build an air transportable bomb with a reported amount of fissile material of 4 kg. For this he needed cross section data, neutron multiplication data, that were measured by a different group within the German research program and the measurement had large uncertainty. About the rigidity of the German research system: I believe Bernstein got it wrong. There were other capable scientists in the program, and they were not prohibited from reconsidering the critical mass problem. They chose freely not to reconsider it because they were not motivated enough to build the bomb. They lacked the motivation because they didn't like the Nazis, and they wear not afraid of being bombed by American nuclear weapons. As soon they learned that they were wrong, Heisenberg got his motivation (curiosity) and provided the correct solution. I'm pretty sure that if they had known what progress the Americans (and even the Soviets!) had achieved, far ahead of them, not only Heisenberg, but his colleagues also would have repeated their calculations, doubling down their efforts to counter the Allied nuclear weapon(s) and they would have derived the right conclusion years before the Farm Hall lecture. One addition: trusting in Heisenberg's false calculation by his less prestigious colleagues is not a nature of the German research system, but might be a more general, cultural thing, a kind of basic German national characteristic preferring orderly collaboration over competition. There is a story about German pocket battleship Graf Spee got suck in the port of Montevideo. Her captain Lagendorff believed that he spotted a British aircraft carrier through his telescope, right in front of the port. Lagendorff actually saw a transport ship for the Ark Royal, there were no British carriers in several thousands of kilometers. Not a single officer of the ship questioned the captain's decision to destroy the ship.
I really appreciate that in every formula you describe what each variable represents and how it relates to the physical interpretation. Even if I don't completely understand the math, the implications are not completely lost and I am still learning. Thanks for making these.
Thanks, I am glad you liked the video. I am curious to know what brings viewers to the channel, were you searching for something in particular or did the 'mighty algorithm' find you?
@@jkzero Your video about Planck caught my interest initially and then I started looking through your other videos. However I initially saw it on my feed.
@@spencerwenzel7381 Thanks for sharing and I am glad the algorithm has been showing my video to so many people, I hope you find the other videos of interest too and welcome to the channel.
Yes, the cylindrical case is a bit more complicated than the spherical and cubic core, and yeah, if you encounter a Bessel equation then you are on the right track
Hi @jkzero, here is my solution to the homework problem of the cube of uranium: I used the Laplacian operator in Cartesian coordinates, solved the equations for all three axes, and found the "critical side" a_c=19 cm, which gives me a critical mass of 538 kg. Is this correct? How can using a cube instead of a sphere increase the critical mass by a factor 4?
first of all: congrats on accepting the challenge, you got the critical side of the uranium cube correct but the critical mass seems to be off. Could you share how you got the mass from the size of the "critical cube"?
this is great, I have been told of others using my videos in their classes, this is such an honor; curious to know how it is received, please report back on how this interesting applied problem is received by your students
thanks, I appreciate the feedback. I hope to keep making videos, I really enjoy this and the interaction with the community. I am curious to know what brings viewers to the channel, were you searching for something in particular or did the 'mighty algorithm' find you?
@@jkzero i really love physic and math, i study engineering so i really like to Watch videos about what i like, i found you about 5 months ago when you did the video about the Enola Gay, today i opened TH-cam and in the home i found your video, i really enjoyed It!😊
@@danieleambrosini1681 the Enola Gay video was my official debut with the channel and having a community of nerds interested in physics stories and that don't shy away from some math is just great. Thanks for returning for more!
Hmm when we set the boundary condition N(r,t) =0 at the critical radius R, is it equivalent to the Uranium ball being placed inside a shell of perfect neutron mirror? I still haven't entirely grasp how allowing neutrons to leak out (advanced solution) end up reducing the critical mass compared to the case of neutrons being trapped inside (basic solution) while in the basic case fission reaction also outputs more neutron compared to the input.... That being said, hopefully your channel gets enough support so us audience can get more cool stuff like this.
the "elementary" boundary condition is quite artificial, in fact, it is only used because it allows an analytical solution for R_c in terms of the relevant parameters ( u, cross-sections, density, etc.), it gives a glimpse on what is driving R_c and the critical mass. However, this is the only reason this condition is introduced, mostly for academic purposes. The "advanced" boundary condition is what has to be used in a practical calculation. Thanks for watching and the positive feedback. Welcome to the channel.
First of all, excellent video. It's refreshing to see a subject that is often *talked* about get the mathematical treatment. I was wondering if there's somewhere I could find a derivation of your expression for the diffusion constant D = 1/3*lambda_f*. Most explanations I've seen are very handwave-y and not particularly rigorous.
Thanks, I am glad you found the content of interest. I agree with you, most derivation of the diffusion equation are quite fishy. Regarding the explicit form of the diffusion constant the only legit derivation that I have found is in the appendix of the book The Physics of the Manhattan Project by B. Cameron Reed. It is a long a quite tedious calculation involving several angular integrations.
Thanks; I hope to get to wider audiences but the TH-cam algorithm is driven by engagement so you can actively help the channel by liking, subscribing, and sharing. This type of support is highly appreciated so I can continue making videos.
Thanks, I am glad you like the video. Regarding your question, we need to take the limit r→0 of sin(κr)/r. It is tempting to say that this is 0/0 but the limit has to be taken with care. First, you expand the sin(κr) in the numerator as a Taylor series sin(κr) = κr - (κr)³/6 + ..., then divide this by r and you get sin(κr)/r ≈ κ - κ³r²/6 + ... and now you take the limit r→0 and only the first term survives, all the others contain r so they vanish, and you get lim r→0 sin(κr)/r ≈ κ. I hope this helps.
clicked on your video just as the cider i'd had earlier informed me that actually, it has not yet fully diffused through my system and that perhaps i'd get more from your video if i delayed watching it until tomorrow. til then! thank you.
I love your video. It is very clear, simultaneously advanced and illustrative. But there is a lapsus calami when you wrote the sphaerical laplacian. You put 1/r2 inside braket, but it is r2. Thank you very much for your video. I repeat, I love it.
You are totally right, I messed up the r² term already at 10:10. Thanks so much for pointing this out and for going through the calculation, it is a good catch. Other viewers also pointed this out and I added an erratum in the description. Fortunately, it doesn't affect the calculation because it is correctly applied, this is just a typo on the video but not an error in the calculation; however, I do my best to avoid these typos but after watching everything many times some minor details slip through. Thanks again.
Sorry to sound stupid, but at ~12:25 it is said that as r approaches 0 the limit is C times kappa. I am still a high schooler and I have very surface level experience with calculus, but why does this limit approach C times kappa and not just C? When (sin(x))/(x) as lim x->0 approaches 1? edit: oh and great video, even though I am not yet versed in advanced mathematics I would love to nonetheless and I enjoyed the more in-deptj analysis!
This is not stupid at all, it is a great question, I remember how many times my brain exploded when learning limits. Intuitively, the ratio sin(x)/x should go to infinity when x→0 because the denominator approaches zero; however, this is not the case and why we study limits in early calculus. There are several ways to see that sin(x)/x→1 when x→0: 1. make a plot of sin(x) for small values of x, you will see that for small x the function sin(x) looks like a straight line with slope 1, in other words, sin(x)→x when x→0; from here you get sin(x)/x→1 when x→0. 2. Another way to see this is using Taylor series, this is the idea of approximating any function as an infinite series. You can look it up, the Taylor series for sin(x) = x - x³/6 + ... (this is an infinite sum of positive odd powers of x), from here again, the ratio sin(x)/x = 1 - x²/6 + ... and now when you take the limit x→0 only the first term survives (all the positive powers of x→0). 3. A final way is using the so-called L'Hôpital's rule, this requires that are familiar with derivatives, but I give you the name in case you want to look it up. I hope this helps. Thanks for watching such a high-level video being in high school.
A very nice video. You explain very clearly the basic physics concepts of a chain reaction. Laplacian, Ficks diffusion law, Continuity, conservation of Mass. All these concepts are within reach of any undergraduate engineering student. Many of these concepts were taught to me while as an undergrad in Chemical Engineering. Very well done!
Thanks you, I am glad you liked the video. You have a good point, there are many concepts packed but I did my best make the calculation flow. I warned the general viewer about the math because this video is in fact a response to the request from viewers after I shared the historical aspects of the calculation of the critical mass in a video some months back (th-cam.com/video/LduH7613QXw/w-d-xo.html). Please check the other videos, in my latest I began a series on quantum mechanics that with your background you might enjoy and also be surprised (many of the physicists viewers were surprised), check it out here th-cam.com/video/gXeAp_lyj9s/w-d-xo.html
@@jkzero I will definitely check out the video on quantum mechanics. There are also other very good TH-cam channels which very nicely explain complex topics such as what you do. For example, a couple of days ago, the Veritasium channel showed a very nice explanation of various concepts in solid state physics, including: 1. band gap 2. the various bands, (conduction and valence) 3. A very nice animation showing the Poisson distribution and how it creates the PN junction discontinuity. Lots of good stuff. th-cam.com/video/AF8d72mA41M/w-d-xo.html I am now a subscriber to your very fine channel!
Yes, Veritasium offers excellent content. Thanks for watching and the positive feedback. Also thanks for supporting the channel with your subscription and welcome to the channel!
thanks fro the sub, I am glad you found the content of interest. I am already working on the next video and more juicy calculations are coming. I want to balance the content with some historical context but always including some high-level calculation to complement the story, and more importantly, for viewers to follow along and reproduce themselves if desired. I am curious to know what brings viewers to the channel, were you searching for something in particular or did the 'mighty algorithm' find you?
@@jkzero Thanks for the reply, recently I was searching on the topic detonation wave hydrodynamics. I found that topic fascinating because of welch lab channel. Now TH-cam recommended this video and absolutely love it...thanks again 👍🙏🙏👌👌👏👏😍
@@Unique-Concepts thanks for sharing and great that the searching algorithm brought you here and thanks again for your support, welcome to the channel! I might not know much about the topic that you were searching for but you might find my seminar on blast waves of interest (link below). It is about a paper that I published in 2022 extending the result from Sedov-Taylor-von Neumann but written for undergraduate-level students of physics and engineering. It required solving hydrodynamic equations for properly describing an expanding blast wave and I include all the steps in the seminar, check it out and let me know if this helps th-cam.com/video/JySY4bkW5wY/w-d-xo.html
What a great video! i truly want to thank you for going through the effort of laying out the math in the latex format you were talking about, it looks clean, the math is a bit higher than my level but your explanation makes total sense, i will see if i can try the questions you put in the end, your content is really enjoyable!
I am glad you liked the video and thanks for appreciating the effort, writing that code was a lot of work. I hope you give the "assignments" a try and let me know if any help is needed; I am happy that others have also showed interest in attempting it
What would be the equations for critical mass if you had a perfectly spherical neutron reflector and a neutron half-life of 11 minutes? Thanks again for the video. I loved the math derivation.
Okay, I am a little stumped at the laplacian part. So I already specifically got the form down pretty well: (nabla)^(2)*F + (kappa)^(2)*F = 0 but now how exactly do I plug in the "auxiliary function" to this figure in order to get the positional function. At least just give me a hint or point me in a general direction that I could learn how to do this (note: I am familiar with Calculus, however, I am not entirely versed in doing these equations on my own).
Replace u(r) = r*f(f) in the Laplacian ∇²f= (1/r²)*d/dr(r² df/dr), expand all the terms, simplify what can be simplified, and you will get now get the harmonic-motion equation for u(r). Please note that there is a typo on the video: at 10:10 the Laplacian of f should say: ∇²f= (1/r²)*d/dr(r² df/dr) (not 1/r² in the derivative bracket), this is a typo and not an error in the calculation. I hope this helps. If not just let me know.
Great video, Jorge. Thank you! In the method of separating variables you assume that N(x,t) can be factorized. Is there a reason why this makes sense from a physics point of view?
The reason is mathematical: Given certain conditions, you can prove uniqueness of the solution and thus the solution he found (obeying the boundary conditions) can be the only solution. On the other hand, if he would have chosen more complicated geometries and boundary conditions, the separation ansatz would probably not have worked.
Physicists are quite informal in general but specially when solving equations; instead of trying to prove that a solution exists and that it is unique, we just take an ansatz, work it out, and if a solution is found then that is the solution. Period. The factorization of partial solutions is just an ansatz that works. It is the first thing we always try. If it fails, then we try more formal methods.
@@jkzero yeah, that is what you have mathematicians for who will prove that everything you have done was right ;). I was just wondering wether another ansatz would have yield another solution but that was answered by the reply of @digxx.
Chiarissimo dottor Jorge, ho notato, con grande sorpresa e ammirazione, che ha vergato i suoi appunti in data 25 dicembre. La stimo ancor più per questo: chi può lavorare il giorno di Natale? Solo un appassionato, un entusiasta (etimo: en + theos = un dio che ti parla dentro) può fare una cosa del genere, perciò, c'è ne fosse bisogno, rinnovo e attesto la mia perspicua ammirazione per la sua persona e il suo lavoro. Se lei permette, appena avrò un poco di tempo, riporterò su un testo la sua disanima corredandola di ulteriori semplificazioni e ampliamenti per renderla ancor più comprensibile per i meno competenti. E naturalmente citerò la fonte, cioè: lei. Grazie. Forte nello spirito, tenero nel cuore. G.
I am glad you liked it, I am curious to know what brings viewers to the channel, were you searching for something in particular or did the 'mighty algorithm' find you?
@@jkzeroyour video appeared on my feed randomly yeah, wasn't looking for anything related. I guess that being a new chanel plus the "oppenheimer" related theme pushed your videos to many people. I hope more people are able to watch your videos!
@@JazzBerri thanks fro sharing, I am glad the algorithm brought you here. Make sure to check the other videos, there is one dedicated to the real meaning of the famous words of Oppenheimer0 Welcome to the channel, I hope to have back for more videos.
At 10:10, the expression for the Laplacian operator in spherical coordinates is wrong. Inside the parenthesis, there should be r squared, not 1 over r squared. Apart from this, very interesting video!
Thanks, I am glad you liked the video. You are totally right, I messed up the r² term. Thanks so much for pointing this out, it is a good catch. Fortunately, it doesn't affect the calculation because it is correctly applied, this is just a typo on the video but not an error in the calculation. Others have also notified it and I included an erratum in the video description. I do my best to avoid these typos but after watching everything many times some minor details slip through. Thanks again.
When doing problems like this, there are an infinite number of solutions that satisfy the boundary condition (just keep adding pi for the Neumann BC). The coefficients for these solutions depend on the starting conditions over the whole domain. However, the higher harmonics change the value of kappa which changes the exponential growth rate. For heat diffusion problems, the exponent gets more negative which causes fine structures to quickly damp out leaving the first few harmonics. Here, it looks like the negative in front of nu' also means that it is negative in the exponent (or less positive) for higher harmonics. However, it should still contribute somewhat to the overall reaction. For bombs that have a neutron generator in the pit, how important is the initial distribution to the critical mass?
You are right: the general solution is kappa*R=n*pi; however, the higher "harmonics" will only make the bomb core bigger, we are searching for the minimum radius. Also, a solution for n>1 can lead to a negative density, which is also a non-physical solution. Mathematically, your point is 100% correct; however, here is when physics constrains the final solution, just like setting B=0 to remove the non-physical cos(kappa*r) term.
soy físico teórico, todo esto lo aprendí por mi cuenta por gusto, simplemente me fascina el tema y creé este canal porque pensé que era hora de compartir la pasión, afortunadamente la recepción ha sido fenomenal lo que me anima a seguir creando este tipo de videos. Cuéntame que te trajo por acá, ¿te encontró el algoritmo?
"the mighty algorithm", me alegro que te haya traído por acá. ¿En tipo de física trabajas? ¿Eres estudiante, postdoc, académico o estás en la industria?
using primes (') to denote the derivative with respect to r, the Laplacian of f is ∇²f = 1/r^2 (1/r^2 f')' if you replace f=u/r, then using the derivative of a ratio f' = (u'r - u)/r^2, then the Laplacian of f becomes ∇²f = 1/r^2 (u'r - u)' because the terms r^2 cancel out inside the parenthesis, but there is still the derivative outside the parenthesis to be applied; here be careful because there are two terms (u'r and u) but the first term (u'r) is a product so you will end up with three terms (two of which will cancel each other out): the full Laplacian becomes ∇²f = 1/r^2 (u'r - u)' = 1/r^2 (u''r + u' - u') = 1/r^2 (u''r) = u''/r, as shown at 10:37. If this is still unclear just let me know.
@@KBIMT thanks for getting back on this, the reason is a typo on the video. I messed up the term inside the parenthesis: it should be r² (and not 1/r²). Now the r² will cancel out. Note that the typo doesn't affect the calculation because it is correctly applied, this is just a typo on the video but not an error in the calculation. I do my best to avoid these typos but after watching everything many times some minor details slip through, sorry about that.
@@KBIMT of course, I am making these videos to interact with this community, so happy to get back to you on this issue. Plus I am the one who screwed up one the key points in the calculation with a typo, sorry about that.
I have not touched a graphing calculator in decades (when I spent all my savings in a TI-92+ and barely used it). I would expect that you can define custom functions but I guess the easiest way would be explicitly replacing nu' (criticality factor) in the definition of g(t). Note that you will need to define a radius, say 30 cm, that is bigger than the critical value; otherwise, your solution will be constant (because nu'=0 for R=R_c)
@@AlbertCamus-r6i I honestly don't know how to code this in a graphing calculator but I have created a Python code to generate the plots shown in the video you can find in this link github.com/jsdiazpo/JK0/blob/main/11-Neutron-Diffusion-Eq.ipynb Please let me know if this helps
this is a fantastic video. the math is beautiful. no doubt that this is also the same equations used for the generation of nuclear power for thorium reactors. so for nuclear reactors, nu prime is not allowed to reach zero but always kept negative. amazing work
thanks, I am glad you liked the video. Honestly, I am not an expert on reactors, I just made one video showing that they cannot explode like a bomb but I have a good friend who is an expert on thorium reactors, if you have questions on this topic make sure to write them in this post, I will try to have him in the future th-cam.com/users/postUgkxkfK_xvCYFftK_Wsw6wh0L6ECtL87-Dcv
I love the video @jkzero! But may I ask you a question? I know that the video is not supposed to extremely technical and formal, but as a matter of curiosity: when you say that sin(kR)=0 implies kR=\pi (because it is simpler), would the solution be more "complete" if you consider every possible solution for sin(kR)=0? I don't wanna get into technicalities (eigenfunctions, Fourier expansion, etc.), but it got me curious. Thanks in advance! Edit: Extremely underrated content. I will definitely subscribe!
thanks for the feedback and the sub, I am glad the video has been of interest for many people. You have a good point: the general solution is kappa*R=n*pi; however, the higher "harmonics" will only make the bomb core bigger, we are searching for the minimum radius. Also, a solution for n>1 can lead to a negative density, which is also a non-physical solution. Mathematically, your point is 100% correct; however, here is when physics constrains the final solution, just like setting B=0 to remove the non-physical cos(kappa*r) term.
to be honest, i liked the way you present the video with formulas and graphs but it is very hard to grasp and track. at least i as a non-physicist had a hard time to understand and could not finish the video. but i liked your type and really love to see chemistry version of you (or not this much of (unheard) definition and formulas). but I know that some people really like this kind of videos so dont worry about my comment.
thanks for your honest comment, one thing missing at the beginning is that this video is the response to the request of the step-by-step calculation of the critical mass in a previous video. If you want a more detailed description of the story behind this video please check "Critical Mass: when the atomic bomb got real" (th-cam.com/video/LduH7613QXw/w-d-xo.html). The present video by its own is pure mathematical physics but the other video provides the context and historical relevance of this calculation. I hope you like that video better.
every physicist knows that when it comes to optimization the answer will be a sphere for which you would never use Cartesian coordinates. You can solve the problem in Cartesian coordinates for a cubic bomb core, that is in fact mentioned at 23:38
@@jkzero Tbh at first glance, I wasn't a fan of this idea because the neutrons can pass through the center or 'r=0' in this case. Using spherical coordinates to describe particle motion passing through center? I really thought what in the crazy world we live in. Yeah anyways, the concept and video is real good. Would recommend to my classmates. and friends.
@@overlordprincekhan Thanks, I am glad you liked the video and thanks fro sharing it. I don't understand why it bothers you that neutrons pass through r=0, they are in fact all over the place and mostly at the center because the initial burst is produced there
There is one thing that I do not understand about implosion. Well, there are many things that I do not understand about it, but I have this particular question. How are the atoms in a metal sphere of Pu-239 compressed? Are there little voids in the sphere due to the casting process, or is the chemical energy of the explosive lenses enough to overcome the repulsive force between atoms, thereby bringing them closer to each other? Many thanks.
those a really good questions and I can safely say that I do not know the answers, I would have to do some digging. I cannot guarantee to be able to fulfill all the requests but I always open to collecting suggestions, thanks.
small correction on language: The Laplacian operator does not depend on the coordinate system; however it's representation certainly does. Laplacian, and the physics its describes, don't care about our coordinate systems.
Love this. I am pedant myself, and this is the kind the pedantic comment that I really enjoy. Yes, you are totally right, I misused the word dependence when I should have referred to representation. I appreciate the correction.
@@jkzero is it pedantic to say I wasn't being pedantic? No, I just hope one of your viewers makes the jump to _"coordinate free physics"_, which was a revelatory moment in my education, received directly from the Guru of relativity himself (Kip).
@@DrDeuteron I did my PhD on the study of Lorentz invariance, I recall the exact moment when the "Nature doesn't care about coordinates" really clicked in my head, I was reading a paper by my supervisor in a bus on my way to Walmart, I remember that moment vividly.
@@jkzero oh good, because I recently posted something to the effect: you don't understand GR until you accept that the Ptolemaic model of the solar system is just as "right" as the Copernican version, but I get written off as a crackpot before explaining Coriolis and centrifugal, like "gee" , are all a result of non-geodesic coordinates. Stand by for a thought problem...
@@jkzero so the push back is that "rotation is real", hence the standard fictitious forces. Here's the question: A mass m=1 is "at rest" at a radius "R"--what's the Coriolis/Centrifugal forces on it? Every one says 0/0, ofc. But that's frame dependent. In a frame rotating at w, you get a centrifugal: w^2R, and here's the kicker, for Corilois: 2v X w = -2wRw = -2w^2R, so the total is a apparent inward w^2R keeping the mass in orbit in the rotating frame--even though there are no forces at all.
At 5:33, it is written: lambda_f = A / (rho * N_A * sigma_f) lambda_t = A / (rho * N_A * sigma_t) sigma_t = sigma_f + sigma_e So, if sigma_e >= 0, lambda_t
Thanks for your comment. I see that you spotted the errata. For the advanced solution, you can check the appendix on the Los Alamos Primer, available as a book but also for free online.
I am interested in taking up the challenge and following through the procedure to attempt to calculate the critical mass for a cylinder. I do have a background in electrical engineering, and as part of that I studied quantum mechanics, and nuclear physics, and did extensive training in calculus.... However I am more interested in trying to calculate the critical mass that relates to a nuclear reactor instead of a nuclear weapon, would you be able to do a video on that or point to where one could find information how to make such a calculation? I am thinking something like calculating the critical mass for something like a Chicago pile, CP-1... minus the bottle of Chianti Bertolli wine... and Enrico fermi!
I have the suspicion that the procedure followed here might serve as a guide for solving the neutron diffusion equation, but the controlled conditions in a reaction can be quite different. I am not a nuclear engineer but maybe the lectures on the MIT course Introduction to Nuclear Engineering and Ionizing Radiation might be more useful (lectures are on TH-cam).
@@jkzero Thank you Dr Diaz, I do appreciate your suggestion, and I have searched for the course, and bookmarked it, as it is something that I will attempt at the end of the year, as it will be an interesting and cool challenge, as I am currently completing a Masters of IT, and wont have the bandwidth until then.
I was using formula shown by you to calculate laplacian and keep getting wrong answer and the suddenly realises it's wrong (later verified with wiki page shown by u just some seconds ago) and then got it finally. BTW amazing video keep doing this. serious physics stuff of this kind is much needed.
you are totally right, I messed up the r² term already at 10:10. Thanks so much for pointing this out and for going through the calculation, it is a good catch. Other viewers also pointed this out and I added an erratum in the description. Fortunately, it doesn't affect the calculation because it is correctly applied, this is just a typo on the video but not an error in the calculation; however, I do my best to avoid these typos but after watching everything many times some minor details slip through. Thanks again.
Another interesting condition involves uranium dissolved in aqueous solution. In solution, the uranium is not nearly so dense as is the case with uranium metal, but the water slows the neutrons down. It turns out that the fission cross-section for U-235 is higher for slower neutrons. Now, we have to take neutron scattering off of water molecules into account, and how that scattering slows down the neutrons. There have been a number of serious nuclear criticality accidents involving inadvertently exceeding the maximum uranium concentration limits for a given solution geometry. That has led to the imposition of critically safe geometries for places where fissile materials are dissolved in aqueous solutions.
Great Video! In the continuity equation dN/dt describes the change in neutron density due to neutron absorption and emission, while del J describes the change in neutron density due to neutrons diffusing away. So why is the entire expression (dN/dt+ del J) equal to the neutron sink and source term (S+ + S-)? Shouldn't there be an additional term on the right side, accounting for the neutrons being lost due to their motion out of the volume that is being considered?
the term ∂N/∂t describes the change on neutron density in a infinitesimal volume with time, that is all, the reasons are given by other terms in the equation. The term ∇·J describes the differential number of neutrons moving in and out of the infinitesimal volume (they can move in and out by any reason, in our case by diffusion when we impose Fick's law). If the number of particles were fixed then the RHS of the equation would be zero, like in the case of electric currents or fluid motion, but here we need to tell the equation that inside the infinitesimal volume the neutron number is not conserved and in fact neutrons are created and destroyed, this is why the two terms on the RHS are needed.
This equation is local, that is, it describes what happens in an infinitesimally small volume around a particular point (of coordinates x). Sphere boundary is far away.
Amazing channel and videos. A question: I'm very puzzled by the various prompt critical excursions with the demon core and friends that occured in history. I understand nuclear reactors rely on delayed neutrons to keep a reaction in balance. I understand a prompt supercritical state will have a neutron population grow extremely fast. I understand an unconstrained core will tear itself apart before full combustion. What I don't understand is how can a core balance in prompt critical state (blue flash and all) and neither fizzle nor explode. I've been given explanations like Doppler broadening and thermal expansions which certainly could fit the bill but, without knowing how quickly temperature changes vs how quickly the neutron population grows, I find unconvincing. Is there a way to do back of the envelope calculations to think about these prompt critical balanced reactions?
my most sincere and honest answer is: I do not know. I am pretty sure a nuclear engineer would know the answer, they study these reactions in non-ideal scenarios in detail. As a physicist myself, we mostly do calculations in ideal conditions, prove the feasibility based on the physics, conservation laws, etc. and then leave the dirty (but crucial) dirty work to engineers, they really make things happen.
Great video! However, I think that your explanation of he trivial b.c. is a bit misleading. We are not 'disallowing the neuterons to escape' the sphere, however we set the radius of the sphere where N=0. Hence a larger Rc, we artificially require the spere to end only where the neuterons would have stopped, assuming the same λ_t. But I'm nitpicking
thanks, I am glad you liked the video. Honestly, "elementary" boundary condition is quite artificial, in fact, it is only used because it allows an analytical solution for R_c in terms of the relevant parameters ( u, cross-sections, density, etc.), it gives a glimpse on what is driving R_c and the critical mass. However, this is the only reason this condition is introduced, mostly for academic purposes. The "advanced" boundary condition is what has to be used in a practical calculation.
Can you cover any usage of Markov Chains? I believe they were used/developed during the Manhattan Project for neutron diffusion. Thanks. Great video bro.
...excellent video as always. One question immedietly pops up to my mind. If criticality condition equals exactly 0 does this theoretically mean that the fuel won't explode but "burn" as a self sustaining nuclear reactor until whole mass is "consumed"? On th side note by knowing critical mass we can now calculate minimal theoretical yield of this device. Since critical mass is about 50 kg and energy denistiy of uranium/plutonium is 17 kt/kg minimal theoretical yield would be around 850 kt of TNT. Realistic result is unfortunately much lower as the Trinity test (similar device design) produced yiel of only 25 kt. Mass/energy convertion factor was much lower since majority of the mass was blown up (unused). Btw, I always wondered what is "cost/benefit" analysis of uranium enrichment in terms of energy invested and energy gained. Is there any information as to how much energy was used to produce first atom bomb since enriching facility is generally massive. So massive in fact that even with this secret known, one needs a facility the size of a small town to produce a bomb. It would be interesting to know if one needs to invest more energy to enrich the uranium than the bomb would produce...:)
IIRC fuel for nuclear power plants needed as much as 4% of energy released for the enrichment. There are equations on Wikipedia saying how much more energy it takes to enrich to 90% than to just 4%. However, this also depend on how you enrich. Zippe type centrifuges are far more efficient than gasous diffusion.
@@kwzieleniewski ...you mean to say that 4% of (thermal) energy released from the nuclear fuel is needed to enrich that exact ammount of uranium ore to be graded as nuclear fuel? Thing is that 4% (3-5%) is basically average ammount of the enrichemnt that ore needs to be processed to to be classified as nuclear fuel so I don't want to be confused with this number similarity. Beside the obvious difference in enrichemnt methods there is probably some efficiency drop as the enrichment goes to the higher levels because extracting the remaining mass is that much harder since the saturation of U235 is pretty high and mass difference between U235 and U238 almost insignificant...:/
Using the random walk, Heisenberg "forced" a single neutron to fission every uranium nuclei, this is the key mistake: one neutron fissions one uranium nuclei, then moves until it encounters another nuclei and produces a second fission, and so on. In a real chain reaction, the neutron density grows exponentially but all the new neutrons also produce fission, this makes the reaction much faster and there is no need of a single neutron doing all the work. These new neutrons diffuse in the uranium while fissioning. The original neutron does not need to travel far, it can even be absorbed, that's irrelevant, there are so many new neutrons diffusing in the material fissioning it, that in the end the chain reaction is much more efficient than assuming a simple random walk.
Thanks, I am glad you liked it. Honestly, I made this video during Christmas time, I really enjoyed making it, although it took many hours to animate all the calculation. When I finished it I thought "nobody will watch this, but I love the result." It is the video that I which I had watched years ago and that it why I made it. The surprise was that this video made my channel explode (no pun intended). Thanks for watching.
Pheonomenal content. Subscribed! as a side note, I am a highschool student aiming to pursue a career in theoretical physics. How do you suggest I proceed with that? All and any help would be appreciated
I'd love a little more on the theoretical basis (assumptions) that underlies these equations. Why is the diffusion model different from the random walk that Heisenberg assumed? I thought diffusion occurs from many particles doing a random walk.
Using the random walk, Heisenberg "forced" a single neutron to fission every uranium nuclei, this is the key mistake: one neutron fissions one uranium nuclei, then moves until it encounters another nuclei and produces a second fission, and so on. In a real chain reaction, the neutron density grows exponentially but all the new neutrons also produce fission, this makes the reaction much faster and there is no need of a single neutron doing all the work. These new neutrons diffuse in the uranium while fissioning. The original neutron does not need to travel far, it can even be absorbed, that's irrelevant, there are so many new neutrons diffusing in the material fissioning it, that in the end the chain reaction is much more efficient than assuming a simple random walk.
@@jkzero Thank you so much for the explanation! Although tbh I'm still a little confused. It sounds like you're saying he modeled a single neutron having to meet and fission every single uranium nucleus. But that wouldn't ever work, right? I thought the exponential chain reaction is what's necessary for an explosion.
@@jkzero That's confounding. Everyone knew the exponential multiplication was the key to the chain reaction. Why would he reduce the problem to something he knew was wildly unrealistic?
you are totally right, this is one of the things what I meant when I said that the calculation shown in the videos involves many approximations. But it could be argued that diffusion theory is not even applicable because the size of the material is barely bigger than the transport mean free path. Careful calculations show that many effects ignored here cancel each other out but this is just a lucky coincidence.
in any case yours is an excellent video, I have a degree in nuclear engineering and in the nuclear reactor physics courses I had followed, the first approach to use for calculating the distribution of neutrons was to use the diffusion equation, we then had the most precise calculations with the transport equation @@jkzero
you are right that the standard density of plutonium is closer to 20 g/cm^3; however, depending on temperature plutonium exhibits different crystalline phases. The relevant phase used for weapons is the so-called “delta” phase and that is the density that I have provided.
@@AlbertCamus-r6i exactly; I mean you can have plutonium metal in other phases too but for the bomb they used the delta phase due to stability and for machining purposes
A proton is a collection of 1836 expanding electrons and add a bouncing expanding electron makes a hydrogen atom. “G” calculated from first principles- the hydrogen atom- in 2002. No energy, charge, photons, waves, spin, fields, potential, quantum,space, time, space-time etc. All Standard Theory/Model was replaced by Expansion Theory in 2002. “The Final Theory: Rethinking Our Scientific Legacy “, Mark McCutcheon for proper physics.
Thanks for watching; you can actively help the channel by liking, subscribing, and sharing. This type of support is highly appreciated so I can continue making videos.
This is so interesting. I bet a similar equation can be written for a biological cell producing some biomolecule in its 'core' (cytoplasm or nucleus or some organelle) and then it is flowing out. What do you think Jorge?
This topic goes beyond my expertise so whatever I say should be taken with a grain of salt. Moreover, I do not like it when people comment too confidently on topics beyond their expertise. With that said, you can search for modeling of so-called "reaction-diffusion systems" that include many biological and chemical systems. There you will encounter Fick's law again and the diffusion equation for modeling, for example, patterns in ecological systems or the diffusion of ions in living systems. There is a whole branch that makes use of this type of equation called mathematical biology.
@jkzero The math should be equivalent. I do research on biochemical systems and can vouch for the biology side. Recently I've been teaching myself some modelling which is why I came across your videos. Perhaps I can pick your brain some time to validate my equations. I work at the max planck in dresden. I like the idea where you can set the boundary flux to be zero or nonzero. It's just like a cell's receptors not letting stuff out for zero flux or secreting something, for nonzero.
Glad you think so! I am curious to know what brings viewers to the channel, were you searching for something in particular or did the 'mighty algorithm' find you?
@@cyrilguillouard1266 Thanks for sharing and I am glad the algorithm is working, I hope you find the other videos of interest too and welcome to the channel
yes, I use Manim in all my videos. I know that Grant recommends to not use Manim for just showing equations but I can't help it, I like how LaTeX formulas are shown too much. Thanks for stopping by and make sure to check the other videos.
You are right: the general solution is kappa*R=n*pi; however, the higher "harmonics" will only make the bomb core bigger, we are searching for the minimum radius. The solution kappa*R=0 is useless because it requires kappa=0, which is mathematically correct but physically useless leading to N(r,t)=0, which of course solves the equation but it provides no information. Also, a solution for n>1 can lead to a negative density, which is also a non-physical solution. Mathematically, your point is 100% correct; however, here is when physics constrains the final solution, just like setting B=0 to remove the non-physical cos(kappa*r) term.
So, because bismuth-209 is technically radioactive it has a critical mass too. Can you calculate it? It would be nice to know even if there isn't nearly enough bismuth in the entire universe to create it.
Not really, you are confusing radioactive with fissile, they are very different properties. If any radioactive material could be made to undergo a nuclear chain reactions we would be making bombs with bananas (from their radioactive potassium content). Instead only a few isotopes like U-235 and Pu-239 can be used because they are fissile.
in this calculation the critical mass is defined as the minimum amount of fissile material to trigger a self-sustained nuclear chain reaction. Of course as soon as the reaction starts, the material will evaporate and the whole assembly will self-destroy. This is not accounted for in the calculation shown in the video.
Hi! Is there a way to predict theoretically the fission and capture cross-sections of our processes? You can get more or less the order of magnitude of this treating the U-235 as an hard sphere and using the Bohr-Wheeler theory that tells you that the higher Z²/A, the higher the fission probability. I tried to evaluate the U-235 fission cross section and I get something like 6.20 time 10^(-23) cm². But I'm not able to see how to improve this little model in order to study capture cross section. Also I have no idea how to distinguish slow and fast neutrons! By the way, your videos are beautiful.
Thanks for the feedback and I am glad you liked the videos. Yes, you can motivate the order of magnitude for the cross-sections but at the scale of neutrons and nuclei, quantum-mechanical effects can become relevant and classical intuition falls short. Plus a nucleus with 235 nucleons is really hard to model and you can only use approximation methods, in the end the best way is just go a measure it, all you need is a few microgram of uranium-235, which is easy to "make" with a cyclotron
tau can be determined in terms of the fission mean-free-path and the average neutron speed. I showed the detail of this in the video about the Physics of a Nuclear Explosion at the 4:04 mark th-cam.com/video/6VSrGDOrWXc/w-d-xo.htmlsi=5uq4tpjHFlhVPBah&t=244
@@kotowaty3386 excellent! I'm glad that helped. Thanks to you for watching and for following the calculations along. There are some typos later in the video, please check the errata in the video description.
You are totally right, I messed up the r² term. Thanks so much for pointing this out, it is a good catch. Fortunately, it doesn't affect the calculation because it is correctly applied, this is just a typo on the video but not an error in the calculation. Others have also notified it and I have included an erratum in the video description. I do my best to avoid these typos but after watching everything many times some minor details slip through. Thanks again
In the remarkable story two flaws are negating each others. First, Diffusion is over estimated for fast neutrons (of bomb core) and for thermal ones as well. Second, the source neutron term due to fission must be decreased for capturing neutrons. But at criticality threshold these mistakes happly are zeroesed. Good luck!
it is interesting how the simple calculation in this video was enough despite the many limitations of the formalism. The most remarkable is the fact that two non-negligible effects have been ignored: 1. neutrons can also been captured by U235 without fissioning, this was completed ignored; and 2. the elastic-scattering cross section is not isotropic so care must be taken when calculating the corresponding cross-section. Nonetheless, when these two effects are incorporated they happen to cancel each other out. Just a lucky coincidence.
Amazing, as a theoretical physicist myself, I'm looking forward to see more mathematical videos about this topic 😁
Edit: makes me wanna sit down with numerical methods and analyze more reallistic example of shrinking and expanding core (true R(t))
Is that actually 2500 dollars?!?!? I should become a theoritical physicist too💀💀💀
@@pressaltf4forfreevbucks179 Hong Kong $ ... so only 64 US$ I'm poor too 😭😉 (and given I'm now in Philippines it shows 2500PHP... not sure why as my visa was charged 500HKD that I've selected)
WOW, thank you so much for your kind support, I am grateful for the appreciation. This drives me to keep making these videos. I enjoy making them (despite the consumption of my free time) so it is very gratifying to know that the content is appreciated. Thanks again!
its pesos lol 2500 is $44
still great@@pressaltf4forfreevbucks179
@@pressaltf4forfreevbucks179 no, its about 40 bucks.
Man, I am a high schooler and I am trying my best to learn Physics, a subject that I love with all of my heart, I am self studing calculus to, one day, contribute a little to this wonderful field, the things that I find more interesting are the possibilities that the Island of stability can give us, and I want to dedicate my life to that.
Take my money (earned by surveys so no problem for me) and buy a cup of something
thanks so much for your support, the appreciation from strangers watching my videos really drives me to keep making them. Thank you!
Sounds like your more proactive with your education than I was at your age. Good for you. I did take calculus in high school but not until the second half of senior year after taking pre calculus over the summer. I wish I did it sooner or tried testing into some more advanced classes because I could of gotten the entire calc chain including differential equations out of the way before college if I wasn’t afraid to take AP courses. I could of also gotten physics 1 and 2 as transfer credits I think if I took the calc based versions. It would have been intense but easier than doing it in college and put me 2 years ahead for my Bachelor’s. It may have lowered my GPA but I didn’t go to an elite University so it wouldn’t have mattered anyway. And if your aiming for an elite school, AP courses are a good way to stand out if you do well in your classes. In hindsight, I think it’s worthwhile to take some risks and not play it too safe in high school because if you have big goals your going to have to do that at some point.
@@jacobharris5894 hello, thank you for your words and your experiences.
I'll surely try all I can, here we have many good universities but to attend the best one you need to pass a very difficult exam with at least 65-70% (only 32 people) else you are out.
It is needed to start studing at least 1-2 years before.
I still have more than 2 years, if my will is enough I’ll make it *fingers crossed*, if no there are many others very good.
Absolutely fascinating. I have always been interested how this was calculated ever since I did a school assignment on the Manhattan project when I was 14 in 1974.
glad you liked it; similar story here: I read about the Manhattan Project when I was a kid and got obsessed and decided to be a physicist. That was a few decades back and here I am.
@@jkzero congratulations
This is the first time that I am excited for a homework assignment and I can't believe it came from TH-cam, I really want to try to do this! Brilliant video and thanks for educating us on this fascinating topic!
happy to see that some viewers have accepted the challenge, let me know what you get
A little mistake, min 10:17. On radial term you have inside the brackets (1/r^2 df/dt) but it is ( r^2 df/dt ). After you use the function f=u/r all come back to be good.
Thank for the notification, you are right and another viewer also let me know about this, I messed up the r² term already at 10:10 so I have included an erratum in the video description. Thanks so much for pointing this out and in such a courteous way (not very common on the internet), it is a good catch. Fortunately, as you proved, it doesn't affect the calculation because it is correctly applied, this is just a typo on the video but not an error in the calculation. I do my best to avoid these typos but after watching everything many times some minor details slip through. Thanks again.
Yeah, I tried the expansion with the form written in the video and it was generating a googolplex of terms 🙂
I graduated in physics in 1971 in Bristol, UK. When we were shown the neutron diffusion equation, our lecturer said "you all have all of the mathematical tools that you need to solve this equation, but I'm not allowed to show you how to do it, for reasons of national security." How times have changed!
maybe your lecturer was joking or trying to make it more interesting because the use of diffusion theory for weapons design was rendered obsolete pretty soon, it is of little use to someone attempting to build a nuclear weapon, and even the reports including these calculations were declassified in 1965
@@jkzero many thanks for your reply. Klaus Fuchs gained his PhD at Bristol and worked on the theoretical calculations for nuclear weapons. Many years later was imprisoned in the UK for passing nuclear secrets to the Russians. This may have led to the staff at Bristol becoming over cautious in protecting "secrets."
@@stevebeal73 oh... yeah, I can see that. Fuchs' case was a big thing. It was quite sad from the Peierls family side
A coffee/beer from a nuclear colleague from Poland, with kind regards!⚛
Thank you so much for your kind support, I am grateful for the appreciation, which drives me to keep making these videos. Thanks again!
17:15 _This will very quickly start blowing up_ LOL, I've often heard of exponentials "blowing up" but here, never a more true definition 😂
glad to see that people appreciate the puns... there is a Taylor Swift joke hidden in a previous video than maybe was too subtle because only one person noticed it
This channel deserves more views and subs, one of the best Physics YT channels out there keep it up!
Thanks for the endorsement; I wish I could get to wider audiences but the TH-cam algorithm is driven by engagement so you can actively help the channel by liking, subscribing, and sharing. This type of support is highly appreciated so I can continue making videos.
That's worth a coffee☕
Thank you so much for your continuous support, I will make sure to enjoy a coffee while being grateful for the appreciation of strangers on the other side of the planet that find this content of interest. Thanks again!
Not going to lie, the maths is a bit beyond me, but I love how you worked out the critical masses. All of your videos have been amazing and I always look forward to the next one.
thanks for the feedback, this video was math heavy; people seem to like these so I will make more but I will also continue with the standard video of some history sprinkled with a less-intense calculation on interesting topics, even beyond nuclear weapons; I don't want the FBI knocking on my door :D
Thank you for this video, I got stuck while working on building my atomic bomb and this was very helpful.
I am glad you liked the video; however, I cannot wish you well in your endeavors :)
Man, this theory was taught to me by a professor the last semester, amazing the application and the smooth way this works to explain such a beauty phenomenon
Awesome! I am curious to know what brings viewers to the channel, were you searching for something in particular or did the 'mighty algorithm' find you?
Can you please do a video on deriving the transfer function for a nuclear reactor in terms of control rods vs output thermal power?
You could always look at TH-cam channel "Nuclear Engineering Lectures" NE560, here:
th-cam.com/video/iSbLUBuaT78/w-d-xo.html
- but remember transfer function analyses are perturbative, and the full range of control rod operation is not, so you have to hope things stay linear. Also remember transfer function theory often looks at the Zero-Power state (to avoid pesky thermal feedback considerations) of a reactor at criticality. It's a useful tool but extending it to more general cases is not trivial.
I think N=0 at boundary of the ball is not "no neutrons escape" boundary condition, it is more like "all neutrons trying to escape are vanishing" because neutron current J is nonzero through boundary
I think you are totally right. Honestly, the "elementary" boundary condition is quite artificial, in fact, it is only used because it allows an analytical solution for R_c in terms of the relevant parameters (
u, cross-sections, density, etc.), it gives a glimpse of what drives R_c and the critical mass. However, this is the only reason this condition is introduced, mostly for academic purposes. The "advanced" boundary condition is what has to be used in a practical calculation.
Thanks!
Thanks to you for supporting me to keep making these videos!
🚀Thanks!
I cannot thank you enough for your continuous support. The only way, I guess, is by keeping the videos coming and the quality standards. Thank you so much!
Actually, I am amazed by your work. I don’t know what to say. I thought a lot about the effort you made to make this video. It must have been hard. I don’t think anyone can thank you as you deserve. In fact, I am traveling to Russia, where there are no means of support or Communication with the outside world, even TH-cam ads are prohibited, so I turned on a VPN so that ads would appear on your videos and I could watch them. Indeed, I watched 14 ads and visited 11 advertiser’s websites. This took some time, but it is nothing compared to the time that you spent making this video, and perhaps this is worth a few cents to the channel, but I feel that it is more moral support than material support. Thank you from the bottom of my heart, you brilliant mind.❤
Thanks for the endorsement; I wish I could get to wider audiences but the TH-cam algorithm is driven by engagement so you can actively help the channel by liking, subscribing, and sharing. This type of support is highly appreciated so I can continue making videos. Thanks also for watching the ads, I do the same with small channels that I watch, if there is an ad I let it go until the end because I know I am contributing with a few cents to the creator.
@@jkzeroDone❤
I think that there is a small error at 13:35. At this point in the video it is stated that the third boundary condition is that no neutrons escape the sphere and this is done by requiring that N(r=R, t)=0. Then there is a quick disclaimer that this condition is not completely true and the more advanced solution is briefly mentioned. I may just be misunderstanding it, but I understood this to mean that the condition N(r=R, t)=0 does imply that no neutrons will escape the core however this boundary condition is not practical or optimal. Using the solution that satisfies the condition N(r=R, t)=0, I calculated the total neutron flux at the boundary from the neutron density and found that it is not zero but proportional to R*N(r=0, t). I am not criticizing the use of the boundary condition N(r=R, t)=0, but I think it is incorrect to say that this condition is equivalent to requiring that no neutrons escape the core.
Thanks for your exposition. Honestly, the "elementary" boundary condition is quite artificial, in fact, it is only used because it allows an analytical solution for R_c in terms of the relevant parameters (
u, cross-sections, density, etc.), it gives a glimpse on what is driving R_c and the critical mass. However, this is the only reason this condition is introduced, mostly for academic purposes. The "advanced" boundary condition is what has to be used in a practical calculation.
Shame how underrated this channel is😢
thanks for that, you can support the channel by liking, commenting, and sharing the videos, that engagement drives the algorithm that gives the channel exposure. Thanks in advance.
Being sigma_t greater then sigma_f (min 5:31), how is it possible that lambda_t is greater than lambda_f (min 15:58)?
oh no... you are totally right, I messed up the subscripts at 15:58, they should be reversed, it should say: lambda_f = 16.89 cm and lambda_t = 3.596 cm. I will add an errata in the description. Thanks so much for pointing this out, it is a good catch. Fortunately, it doesn't affect most of the elementary calculation because the two mean-free paths are multiplied but it does make as difference in the 'advanced' solution, where lambda_t appears isolated. Thanks again. I do my best to avoid these typos but after watching everything many times some minor details slip through. Thanks also for asking in such a nice way contrary to patronizing mode of writing comments favored on the internet.
@@jkzero You're welcome! Anyway, thank you for the amazing video and the explanation!
Excellent! Could you make a video on the implosion physics ? For example, it is often said that although it was known that hollow cores would have a much better yield, the original bomb used a "conservative" design based on a solid core ("Christy core") because the maths of hollow core implosion was too difficult to solve at the time (but it was finally implemented in all designs a few years later). I could not find any more detailed info on that on the web.
did you check my "Nuclear Weapons Q&A" video? I don't go into the details of the solid vs. hollow core but I do calculate how much the core is compressed because somebody asked about it.
Read the NWFAQ by Carey Sublette.
My kind of YT video, just at the right intersection of math and physics, delivered at the right cadence, not too fast, not too slow, loved it. One question, if sin(kR) = 0, wouldn't that give us multiple solutions? ie, kR = n * pi? Any particular physical constraint that we must consider only the lowest numbered "harmonic", n = 1?
If you used solutions with multiples of pi you would have areas with negative neutron density which I guess would not make much physical sense, but I am kinda guessing. The reason why you can use multiple solutions in QM is that you multiply the function with it’s complex conjugate before you get the probability, which means that it will always be positive:)
nice catch, that is a very nice observation; you are right, the general solution is kappa*R=n*pi; however, the higher "harmonics" will only make the bomb core bigger, we are searching for the minimum radius. Also, a solution for n>1 can lead to a negative density, which is also a non-physical solution. Mathematically, your point is 100% correct; however, here is when physics guides the final solution. Just like setting B=0 to remove the non-physical cos(kappa*r) term.
Thank you Prof, that makes sense. You did proactively point out all other, non-valid, solutions and explained why they are not applicable. So this stuck with me. This was the only point, everything else i was able to follow-up rather clearly. These are not easy thing to communicate, you struck absolutely the right balance between explaining the points in sufficient detail, without getting bogged down. That balance is hard to get right, you hit the ball out of the park. Thank you!
@@ytashu33 in the video at 14:20 I refer to the solution used as "the simplest solution" this is precisely to address your question. I could have explained more details but the video was already 25 min long.
Thanks for your appreciation, I love teaching and created this channel mostly because I wanted to share fun and education stuff, I just never thought that a video only about solving a hard equation would produce this level of interest, which I find great. Welcome to the channel and make sure to check the other videos.
How is the boundary condition expression for the advanced solution derived (20:59)? I'd really be thrilled to read the derivation, can you provide something? Is it something like Neumann vs Dirichlet boundary condition? Thank you for all of your work!
Thanks fro the feedback, I am so excited that his video got so many people interested. The elementary boundary condition (BC) is a Dirichlet BC but it is quite artificial, the correct way of doing the calculation is by using a Neumann BC. I am glad someone got hooked on that detail, it is a lengthy derivation presented for the first time by Robert Serber, one of Oppenheimer's protegés, on his annotated version of The Los Alamos Primer. He only did it in Cartesian coordinates but it can be generalized to spherical coordinates chooser.crossref.org/?doi=10.2307%2Fj.ctvw1d5pf.6
Very underrated TH-cam channel.
Thanks for the endorsement; I wish I could get to wider audiences but the TH-cam algorithm is driven by engagement so you can actively help the channel by liking, subscribing, and sharing. This type of support is highly appreciated so I can continue making videos.
I believe that the elementary solution does not correspond to neutrons not escaping the core, but rather that neutrons that leave the core are permanently lost, so the function is 0 outside the sphere.
I think an actual condition for neutrons not leaving the center would be the spatial derivative wrt to r being 0 at the center, as expected from Fick's law, instead of the actual value. Otherwise sub-criticality is not achievable because the number of neutrons would not only be conserved but also continuously multiplying.
Other than that, loved the video, think your calm and collected teaching style is great.
I am glad you liked the video, thanks for the feedback. Honestly, "elementary" boundary condition is quite artificial, in fact, it is only used because it allows an analytical solution for R_c in terms of the relevant parameters (
u, cross-sections, density, etc.), it gives a glimpse on what is driving R_c and the critical mass. However, this is the only reason this condition is introduced, mostly for academic purposes. The "advanced" boundary condition is what has to be used in a practical calculation.
Thanks for producing this excellent video. It answered two long-standing questions I have had: what is the uranium critical mass, and how does implosion reduce it.
I have 2 comments:
1. Being too lazy to calculate, I'm guessing that putting a boundary condition on the neutron flux at the surface instead of the density means that there is a higher density of neutrons out near the boundary, producing more fissions there. Thus a lower critical mass.
2. At the end of WWII, German atomic scientists, including Werner Heisenberg, were interned at an English estate known as Farm Hall. Their conversations were bugged and translated. When they learned of the Hiroshima explosion, Heisenberg gave this lecture to the other guests. As the physicist Jeremy Bernstein pointed out, his inept presentation made it clear that he had never tried to solve this diffusion problem before. This indicates the primary reason why the Germans never got very far in making an atomic bomb: utter incompetence at the very top of a rigid, hierarchical research/development system. As Bernstein put it, Heisenberg was a great physicist (invented quantum mechanics) but not a very good one (couldn't get the units right in a simple calculation). The Manhattan project had the likes of Fermi, who was both great and good.
Thanks, I am glad you liked the video and that it answer many of your questions. Regarding your follow-up questions: 1. yes, relaxing the boundary condition gives you a lower critical mass. 2. Your words echo very much Berstein's sentiment. I do not buy the narrative of the scientists' boycotting the bomb program but at the same time I disagree on Heisenberg being so inept, the evidence suggests that they simply worked on other things. His lecture at Farm Hall shows that it took him just a couple of days to correctly determine the critical mass, although the transcripts also reveal that he knew little to nothing about bomb physics. I have a full video on Heisenberg and the German bomb th-cam.com/video/6zIJTwQ2blU/w-d-xo.html
It's ridiculous to say that Heisenberg was not a very good physicist.
Heisenberg made a mistake in his first attempt to calculate the critical mass, relying on "random walk" model that resulted in need for dozens of tons of fissile material for nuclear detonation. Edward Teller admitted in an interview he made the same mistake first as well. The difference between them (Heisenberg vs Teller, or more generally, most German physicist remained in their Motherland vs Manhattan project participants) is not greatness as scientists.
The reason is simply the motivation. Teller and his fellows continued relentlessly with a different approach because of the fear of the Nazis.
In contrast to them, Heisenberg didn't have such phobia as a driving force against Americans. He wasn't particularly eager for building the bomb and could abandon this goal easily on one hand. On the other hand (perhaps more importantly), he could set his heart at rest, seemingly not being have to worry anymore about Germany being bombed with nuclear weapon.
Teller praised Heisenberg as a "hero" for this.
In a 2000 paper that summarizes and discusses Heisenberg's Farm Hall lecture, author M.S El Nachie also concluded that Heisenberg has never thought seriously about the bomb before he learned of Hiroshima bombing.
Talking about scientific greatness El Nachie praises Heisenberg for his razor sharp intellect for deriving the correct result alone in couple of days, without tables, calculator, his notes etc, being de facto in captivity, knowing nothing of his family. He wrote that Heisenberg was doing what theoretical physicists do best.
(Note that in the "Los Alamos Primer" -lectures based on the work of dozens of well supported scientists in the Manhattan Project- also only presented the "elementary solution" for critical mass.)
@@ukornel77 Your comment, "Talking about scientific greatness El Nachie praises Heisenberg for his razor sharp intellect for deriving the correct result alone in couple of days, without tables, calculator, his notes etc, being de facto in captivity, knowing nothing of his family." is puzzling. When I was an undergraduate physics major in the 1970's, I would have been expected to be able to do this calculation off the top of my head. The mathematical techniques used therein have been part of the standard literature since the 19th century. Notes, tables, calculator are not required, although the fission cross section is required to convert the analytical result into an actual critical mass. Heisenberg most certainly should have been able to do this calculation at the beginning of the German bomb program. The physicist Jeremy Bernstein suggested that, since Heisenberg got it wrong early on, the rigid, hierarchical nature of the German research system made it very difficult for others to correct the lapse. So they all dropped the subject.
@@paulkolodner2445
You are right, calculators, tables are not needed for the correct mathematical solution. But in the Farm Hall Heisenberg wanted to answer the question: how the Americans were able to build an air transportable bomb with a reported amount of fissile material of 4 kg. For this he needed cross section data, neutron multiplication data, that were measured by a different group within the German research program and the measurement had large uncertainty.
About the rigidity of the German research system: I believe Bernstein got it wrong. There were other capable scientists in the program, and they were not prohibited from reconsidering the critical mass problem. They chose freely not to reconsider it because they were not motivated enough to build the bomb. They lacked the motivation because they didn't like the Nazis, and they wear not afraid of being bombed by American nuclear weapons. As soon they learned that they were wrong, Heisenberg got his motivation (curiosity) and provided the correct solution. I'm pretty sure that if they had known what progress the Americans (and even the Soviets!) had achieved, far ahead of them, not only Heisenberg, but his colleagues also would have repeated their calculations, doubling down their efforts to counter the Allied nuclear weapon(s) and they would have derived the right conclusion years before the Farm Hall lecture.
One addition: trusting in Heisenberg's false calculation by his less prestigious colleagues is not a nature of the German research system, but might be a more general, cultural thing, a kind of basic German national characteristic preferring orderly collaboration over competition. There is a story about German pocket battleship Graf Spee got suck in the port of Montevideo. Her captain Lagendorff believed that he spotted a British aircraft carrier through his telescope, right in front of the port. Lagendorff actually saw a transport ship for the Ark Royal, there were no British carriers in several thousands of kilometers. Not a single officer of the ship questioned the captain's decision to destroy the ship.
I really appreciate that in every formula you describe what each variable represents and how it relates to the physical interpretation. Even if I don't completely understand the math, the implications are not completely lost and I am still learning. Thanks for making these.
Thanks, I am glad you liked the video. I am curious to know what brings viewers to the channel, were you searching for something in particular or did the 'mighty algorithm' find you?
@@jkzero Your video about Planck caught my interest initially and then I started looking through your other videos. However I initially saw it on my feed.
@@spencerwenzel7381 Thanks for sharing and I am glad the algorithm has been showing my video to so many people, I hope you find the other videos of interest too and welcome to the channel.
This is an amazing video! Made me even more excited to learn DPEs next semester!
If anything in my videos can make people excited about physics and math, I can say: mission accomplished! Thanks for sharing
You know it’s a good day when you get a notification for Dr. Jorge S. Diaz 🔥
Thank you I was trying to find this for a long long time
Great Vidéo! In cylindrical coordinates we are using the Bessel Function right?
Yes, the cylindrical case is a bit more complicated than the spherical and cubic core, and yeah, if you encounter a Bessel equation then you are on the right track
Hi @jkzero, here is my solution to the homework problem of the cube of uranium: I used the Laplacian operator in Cartesian coordinates, solved the equations for all three axes, and found the "critical side" a_c=19 cm, which gives me a critical mass of 538 kg. Is this correct? How can using a cube instead of a sphere increase the critical mass by a factor 4?
first of all: congrats on accepting the challenge, you got the critical side of the uranium cube correct but the critical mass seems to be off. Could you share how you got the mass from the size of the "critical cube"?
Very nice! I will use some of this as an exercise for my PDE course.!
this is great, I have been told of others using my videos in their classes, this is such an honor; curious to know how it is received, please report back on how this interesting applied problem is received by your students
Awesome! Make a lot of videos on physics because you are really talented! :) Love from Italy
thanks, I appreciate the feedback. I hope to keep making videos, I really enjoy this and the interaction with the community. I am curious to know what brings viewers to the channel, were you searching for something in particular or did the 'mighty algorithm' find you?
@@jkzeroI searched relaxing physics video and found you😂
@@ion8264 nothing more relaxing than taking time off and spending free time solving partial differential equations 😃
@@jkzero i really love physic and math, i study engineering so i really like to Watch videos about what i like, i found you about 5 months ago when you did the video about the Enola Gay, today i opened TH-cam and in the home i found your video, i really enjoyed It!😊
@@danieleambrosini1681 the Enola Gay video was my official debut with the channel and having a community of nerds interested in physics stories and that don't shy away from some math is just great. Thanks for returning for more!
Hmm when we set the boundary condition N(r,t) =0 at the critical radius R, is it equivalent to the Uranium ball being placed inside a shell of perfect neutron mirror?
I still haven't entirely grasp how allowing neutrons to leak out (advanced solution) end up reducing the critical mass compared to the case of neutrons being trapped inside (basic solution) while in the basic case fission reaction also outputs more neutron compared to the input....
That being said, hopefully your channel gets enough support so us audience can get more cool stuff like this.
the "elementary" boundary condition is quite artificial, in fact, it is only used because it allows an analytical solution for R_c in terms of the relevant parameters (
u, cross-sections, density, etc.), it gives a glimpse on what is driving R_c and the critical mass. However, this is the only reason this condition is introduced, mostly for academic purposes. The "advanced" boundary condition is what has to be used in a practical calculation.
Thanks for watching and the positive feedback. Welcome to the channel.
First of all, excellent video. It's refreshing to see a subject that is often *talked* about get the mathematical treatment. I was wondering if there's somewhere I could find a derivation of your expression for the diffusion constant D = 1/3*lambda_f*. Most explanations I've seen are very handwave-y and not particularly rigorous.
Thanks, I am glad you found the content of interest. I agree with you, most derivation of the diffusion equation are quite fishy. Regarding the explicit form of the diffusion constant the only legit derivation that I have found is in the appendix of the book The Physics of the Manhattan Project by B. Cameron Reed. It is a long a quite tedious calculation involving several angular integrations.
@@jkzero Alright, thanks a whole bunch for the reference.
Wow, so well explained. Such advanced physics explained step by step. I hope more people checks your wonderful channel
Thanks; I hope to get to wider audiences but the TH-cam algorithm is driven by engagement so you can actively help the channel by liking, subscribing, and sharing. This type of support is highly appreciated so I can continue making videos.
Hi, love the video so far, but could you explain why at 12:35 the r’s cancel
Thanks, I am glad you like the video. Regarding your question, we need to take the limit r→0 of sin(κr)/r. It is tempting to say that this is 0/0 but the limit has to be taken with care. First, you expand the sin(κr) in the numerator as a Taylor series sin(κr) = κr - (κr)³/6 + ..., then divide this by r and you get sin(κr)/r ≈ κ - κ³r²/6 + ... and now you take the limit r→0 and only the first term survives, all the others contain r so they vanish, and you get lim r→0 sin(κr)/r ≈ κ. I hope this helps.
clicked on your video just as the cider i'd had earlier informed me that actually, it has not yet fully diffused through my system and that perhaps i'd get more from your video if i delayed watching it until tomorrow. til then! thank you.
stay safe and enjoy the video later
I love your video. It is very clear, simultaneously advanced and illustrative. But there is a lapsus calami when you wrote the sphaerical laplacian. You put 1/r2 inside braket, but it is r2. Thank you very much for your video. I repeat, I love it.
You are totally right, I messed up the r² term already at 10:10. Thanks so much for pointing this out and for going through the calculation, it is a good catch. Other viewers also pointed this out and I added an erratum in the description. Fortunately, it doesn't affect the calculation because it is correctly applied, this is just a typo on the video but not an error in the calculation; however, I do my best to avoid these typos but after watching everything many times some minor details slip through. Thanks again.
Sorry to sound stupid, but at ~12:25 it is said that as r approaches 0 the limit is C times kappa. I am still a high schooler and I have very surface level experience with calculus, but why does this limit approach C times kappa and not just C? When (sin(x))/(x) as lim x->0 approaches 1?
edit: oh and great video, even though I am not yet versed in advanced mathematics I would love to nonetheless and I enjoyed the more in-deptj analysis!
This is not stupid at all, it is a great question, I remember how many times my brain exploded when learning limits. Intuitively, the ratio sin(x)/x should go to infinity when x→0 because the denominator approaches zero; however, this is not the case and why we study limits in early calculus. There are several ways to see that sin(x)/x→1 when x→0:
1. make a plot of sin(x) for small values of x, you will see that for small x the function sin(x) looks like a straight line with slope 1, in other words, sin(x)→x when x→0; from here you get sin(x)/x→1 when x→0.
2. Another way to see this is using Taylor series, this is the idea of approximating any function as an infinite series. You can look it up, the Taylor series for sin(x) = x - x³/6 + ... (this is an infinite sum of positive odd powers of x), from here again, the ratio sin(x)/x = 1 - x²/6 + ... and now when you take the limit x→0 only the first term survives (all the positive powers of x→0).
3. A final way is using the so-called L'Hôpital's rule, this requires that are familiar with derivatives, but I give you the name in case you want to look it up.
I hope this helps. Thanks for watching such a high-level video being in high school.
Either:
lim_x->0 [sin(k*x)]/[x] = k * lim_x->0 [sin(k*x)]/[k*x] = k
Or, with De l'hospital theorem:
lim_x->0 [sin(k*x)]/[x] = lim_x->0 [diff sin(k*x)]/[diff x] = lim_x->0 [k * cos(kx)]/[1] = k
A very nice video.
You explain very clearly the basic physics concepts of a chain reaction.
Laplacian, Ficks diffusion law, Continuity, conservation of Mass. All these concepts are within reach of any undergraduate engineering student.
Many of these concepts were taught to me while as an undergrad in Chemical Engineering.
Very well done!
Thanks you, I am glad you liked the video. You have a good point, there are many concepts packed but I did my best make the calculation flow. I warned the general viewer about the math because this video is in fact a response to the request from viewers after I shared the historical aspects of the calculation of the critical mass in a video some months back (th-cam.com/video/LduH7613QXw/w-d-xo.html). Please check the other videos, in my latest I began a series on quantum mechanics that with your background you might enjoy and also be surprised (many of the physicists viewers were surprised), check it out here th-cam.com/video/gXeAp_lyj9s/w-d-xo.html
@@jkzero I will definitely check out the video on quantum mechanics.
There are also other very good TH-cam channels which very nicely explain complex topics such as what you do.
For example, a couple of days ago, the Veritasium channel showed a very nice explanation of various concepts in solid state physics, including:
1. band gap
2. the various bands, (conduction and valence)
3. A very nice animation showing the Poisson distribution and how it creates the PN junction discontinuity.
Lots of good stuff.
th-cam.com/video/AF8d72mA41M/w-d-xo.html
I am now a subscriber to your very fine channel!
Yes, Veritasium offers excellent content. Thanks for watching and the positive feedback. Also thanks for supporting the channel with your subscription and welcome to the channel!
Subscribed!Thank you for this video.Please do more videos on advanced mathematical modelling.
thanks fro the sub, I am glad you found the content of interest. I am already working on the next video and more juicy calculations are coming. I want to balance the content with some historical context but always including some high-level calculation to complement the story, and more importantly, for viewers to follow along and reproduce themselves if desired. I am curious to know what brings viewers to the channel, were you searching for something in particular or did the 'mighty algorithm' find you?
@@jkzero Thanks for the reply, recently I was searching on the topic detonation wave hydrodynamics. I found that topic fascinating because of welch lab channel. Now TH-cam recommended this video and absolutely love it...thanks again 👍🙏🙏👌👌👏👏😍
@@Unique-Concepts thanks for sharing and great that the searching algorithm brought you here and thanks again for your support, welcome to the channel!
I might not know much about the topic that you were searching for but you might find my seminar on blast waves of interest (link below). It is about a paper that I published in 2022 extending the result from Sedov-Taylor-von Neumann but written for undergraduate-level students of physics and engineering. It required solving hydrodynamic equations for properly describing an expanding blast wave and I include all the steps in the seminar, check it out and let me know if this helps th-cam.com/video/JySY4bkW5wY/w-d-xo.html
i would love to see a complete playlist about the calculus for nuclear reactions
thanks for the request; I cannot guarantee to be able to fulfill all the requests but I am collecting suggestions, thanks
What a great video! i truly want to thank you for going through the effort of laying out the math in the latex format you were talking about, it looks clean, the math is a bit higher than my level but your explanation makes total sense, i will see if i can try the questions you put in the end, your content is really enjoyable!
I am glad you liked the video and thanks for appreciating the effort, writing that code was a lot of work. I hope you give the "assignments" a try and let me know if any help is needed; I am happy that others have also showed interest in attempting it
What would be the equations for critical mass if you had a perfectly spherical neutron reflector and a neutron half-life of 11 minutes? Thanks again for the video. I loved the math derivation.
Okay, I am a little stumped at the laplacian part. So I already specifically got the form down pretty well: (nabla)^(2)*F + (kappa)^(2)*F = 0 but now how exactly do I plug in the "auxiliary function" to this figure in order to get the positional function. At least just give me a hint or point me in a general direction that I could learn how to do this (note: I am familiar with Calculus, however, I am not entirely versed in doing these equations on my own).
Replace u(r) = r*f(f) in the Laplacian ∇²f= (1/r²)*d/dr(r² df/dr), expand all the terms, simplify what can be simplified, and you will get now get the harmonic-motion equation for u(r). Please note that there is a typo on the video: at 10:10 the Laplacian of f should say: ∇²f= (1/r²)*d/dr(r² df/dr) (not 1/r² in the derivative bracket), this is a typo and not an error in the calculation. I hope this helps. If not just let me know.
@@jkzero Major thanks.
Great video, Jorge. Thank you! In the method of separating variables you assume that N(x,t) can be factorized. Is there a reason why this makes sense from a physics point of view?
The reason is mathematical: Given certain conditions, you can prove uniqueness of the solution and thus the solution he found (obeying the boundary conditions) can be the only solution. On the other hand, if he would have chosen more complicated geometries and boundary conditions, the separation ansatz would probably not have worked.
Makes sense, thank you!@@digxx
Physicists are quite informal in general but specially when solving equations; instead of trying to prove that a solution exists and that it is unique, we just take an ansatz, work it out, and if a solution is found then that is the solution. Period. The factorization of partial solutions is just an ansatz that works. It is the first thing we always try. If it fails, then we try more formal methods.
@@jkzero yeah, that is what you have mathematicians for who will prove that everything you have done was right ;). I was just wondering wether another ansatz would have yield another solution but that was answered by the reply of @digxx.
@@nathansudermann-merx4586 we do many things in physics that would drive mathematicians mad, like "multiplying" or "dividing" by dx when integrating
Chiarissimo dottor Jorge, ho notato, con grande sorpresa e ammirazione, che ha vergato i suoi appunti in data 25 dicembre. La stimo ancor più per questo: chi può lavorare il giorno di Natale? Solo un appassionato, un entusiasta (etimo: en + theos = un dio che ti parla dentro) può fare una cosa del genere, perciò, c'è ne fosse bisogno, rinnovo e attesto la mia perspicua ammirazione per la sua persona e il suo lavoro. Se lei permette, appena avrò un poco di tempo, riporterò su un testo la sua disanima corredandola di ulteriori semplificazioni e ampliamenti per renderla ancor più comprensibile per i meno competenti. E naturalmente citerò la fonte, cioè: lei.
Grazie.
Forte nello spirito, tenero nel cuore.
G.
Didn’t expect to find such an interesting video on youtube!
I am glad you liked it, I am curious to know what brings viewers to the channel, were you searching for something in particular or did the 'mighty algorithm' find you?
@@jkzeroyour video appeared on my feed randomly yeah, wasn't looking for anything related.
I guess that being a new chanel plus the "oppenheimer" related theme pushed your videos to many people.
I hope more people are able to watch your videos!
@@JazzBerri thanks fro sharing, I am glad the algorithm brought you here. Make sure to check the other videos, there is one dedicated to the real meaning of the famous words of Oppenheimer0 Welcome to the channel, I hope to have back for more videos.
@@jkzero yeah thanks! I liked your chanel so im staying hahah
At 10:10, the expression for the Laplacian operator in spherical coordinates is wrong. Inside the parenthesis, there should be r squared, not 1 over r squared. Apart from this, very interesting video!
Thanks, I am glad you liked the video. You are totally right, I messed up the r² term. Thanks so much for pointing this out, it is a good catch. Fortunately, it doesn't affect the calculation because it is correctly applied, this is just a typo on the video but not an error in the calculation. Others have also notified it and I included an erratum in the video description. I do my best to avoid these typos but after watching everything many times some minor details slip through. Thanks again.
When doing problems like this, there are an infinite number of solutions that satisfy the boundary condition (just keep adding pi for the Neumann BC). The coefficients for these solutions depend on the starting conditions over the whole domain. However, the higher harmonics change the value of kappa which changes the exponential growth rate. For heat diffusion problems, the exponent gets more negative which causes fine structures to quickly damp out leaving the first few harmonics. Here, it looks like the negative in front of nu' also means that it is negative in the exponent (or less positive) for higher harmonics. However, it should still contribute somewhat to the overall reaction. For bombs that have a neutron generator in the pit, how important is the initial distribution to the critical mass?
You are right: the general solution is kappa*R=n*pi; however, the higher "harmonics" will only make the bomb core bigger, we are searching for the minimum radius. Also, a solution for n>1 can lead to a negative density, which is also a non-physical solution. Mathematically, your point is 100% correct; however, here is when physics constrains the final solution, just like setting B=0 to remove the non-physical cos(kappa*r) term.
donde trabajeste con neutrones. Central nuclear? que gran trabajo
soy físico teórico, todo esto lo aprendí por mi cuenta por gusto, simplemente me fascina el tema y creé este canal porque pensé que era hora de compartir la pasión, afortunadamente la recepción ha sido fenomenal lo que me anima a seguir creando este tipo de videos. Cuéntame que te trajo por acá, ¿te encontró el algoritmo?
Algo parecido al algoritmo. eres el 3blue1brown de la fisica. Yo soy físico también. Un saludo@@jkzero
"the mighty algorithm", me alegro que te haya traído por acá. ¿En tipo de física trabajas? ¿Eres estudiante, postdoc, académico o estás en la industria?
10:31 I don't understand how auxiliary function substitution can produce the function below, how to approach this?
using primes (') to denote the derivative with respect to r, the Laplacian of f is ∇²f = 1/r^2 (1/r^2 f')' if you replace f=u/r, then using the derivative of a ratio f' = (u'r - u)/r^2, then the Laplacian of f becomes ∇²f = 1/r^2 (u'r - u)' because the terms r^2 cancel out inside the parenthesis, but there is still the derivative outside the parenthesis to be applied; here be careful because there are two terms (u'r and u) but the first term (u'r) is a product so you will end up with three terms (two of which will cancel each other out): the full Laplacian becomes ∇²f = 1/r^2 (u'r - u)' = 1/r^2 (u''r + u' - u') = 1/r^2 (u''r) = u''/r, as shown at 10:37. If this is still unclear just let me know.
How do these terms r^2 cancel out exacly? If f'(r)=(u'(r)*r-u(r))/r^2 then multiplied by 1/r^2 should give (numerator)/r^4
@@KBIMT thanks for getting back on this, the reason is a typo on the video. I messed up the term inside the parenthesis: it should be r² (and not 1/r²). Now the r² will cancel out. Note that the typo doesn't affect the calculation because it is correctly applied, this is just a typo on the video but not an error in the calculation. I do my best to avoid these typos but after watching everything many times some minor details slip through, sorry about that.
Thanks for replying and explaining this
@@KBIMT of course, I am making these videos to interact with this community, so happy to get back to you on this issue. Plus I am the one who screwed up one the key points in the calculation with a typo, sorry about that.
Have a question: How would you input the N(r , t) function into a Graphing Calculator alongside the Criticality Factor equation?
I have not touched a graphing calculator in decades (when I spent all my savings in a TI-92+ and barely used it). I would expect that you can define custom functions but I guess the easiest way would be explicitly replacing nu' (criticality factor) in the definition of g(t). Note that you will need to define a radius, say 30 cm, that is bigger than the critical value; otherwise, your solution will be constant (because nu'=0 for R=R_c)
Can you simplify what you mean a bit? Still kind of new to this. @jkzero
@@AlbertCamus-r6i I honestly don't know how to code this in a graphing calculator but I have created a Python code to generate the plots shown in the video you can find in this link github.com/jsdiazpo/JK0/blob/main/11-Neutron-Diffusion-Eq.ipynb Please let me know if this helps
this is a fantastic video. the math is beautiful. no doubt that this is also the same equations used for the generation of nuclear power for thorium reactors. so for nuclear reactors, nu prime is not allowed to reach zero but always kept negative.
amazing work
thanks, I am glad you liked the video. Honestly, I am not an expert on reactors, I just made one video showing that they cannot explode like a bomb but I have a good friend who is an expert on thorium reactors, if you have questions on this topic make sure to write them in this post, I will try to have him in the future th-cam.com/users/postUgkxkfK_xvCYFftK_Wsw6wh0L6ECtL87-Dcv
Molto interessante e ben fatto. Complimenti vivissimi dottor Jorge.
Grazie mille! Thanks for watching and make sure to check the other videos
Farollo sicuramente.
Grazie.
I love the video @jkzero! But may I ask you a question? I know that the video is not supposed to extremely technical and formal, but as a matter of curiosity: when you say that sin(kR)=0 implies kR=\pi (because it is simpler), would the solution be more "complete" if you consider every possible solution for sin(kR)=0? I don't wanna get into technicalities (eigenfunctions, Fourier expansion, etc.), but it got me curious. Thanks in advance!
Edit: Extremely underrated content. I will definitely subscribe!
thanks for the feedback and the sub, I am glad the video has been of interest for many people. You have a good point: the general solution is kappa*R=n*pi; however, the higher "harmonics" will only make the bomb core bigger, we are searching for the minimum radius. Also, a solution for n>1 can lead to a negative density, which is also a non-physical solution. Mathematically, your point is 100% correct; however, here is when physics constrains the final solution, just like setting B=0 to remove the non-physical cos(kappa*r) term.
@@jkzeroperfect. Thanks a lot and keep up with the amazing work!
@@CaixetaArthur thanks to you for watching and following up with your question. Welcome to the channel!
to be honest, i liked the way you present the video with formulas and graphs but it is very hard to grasp and track. at least i as a non-physicist had a hard time to understand and could not finish the video. but i liked your type and really love to see chemistry version of you (or not this much of (unheard) definition and formulas). but I know that some people really like this kind of videos so dont worry about my comment.
thanks for your honest comment, one thing missing at the beginning is that this video is the response to the request of the step-by-step calculation of the critical mass in a previous video. If you want a more detailed description of the story behind this video please check "Critical Mass: when the atomic bomb got real" (th-cam.com/video/LduH7613QXw/w-d-xo.html). The present video by its own is pure mathematical physics but the other video provides the context and historical relevance of this calculation. I hope you like that video better.
9:20
Why don't just stick with caartesian coordinates?
every physicist knows that when it comes to optimization the answer will be a sphere for which you would never use Cartesian coordinates. You can solve the problem in Cartesian coordinates for a cubic bomb core, that is in fact mentioned at 23:38
@@jkzero Tbh at first glance, I wasn't a fan of this idea because the neutrons can pass through the center or 'r=0' in this case. Using spherical coordinates to describe particle motion passing through center? I really thought what in the crazy world we live in.
Yeah anyways, the concept and video is real good. Would recommend to my classmates. and friends.
@@overlordprincekhan Thanks, I am glad you liked the video and thanks fro sharing it. I don't understand why it bothers you that neutrons pass through r=0, they are in fact all over the place and mostly at the center because the initial burst is produced there
very very clear explanation! thank you
Thanks, I hope you also check the other videos.
There is one thing that I do not understand about implosion. Well, there are many things that I do not understand about it, but I have this particular question.
How are the atoms in a metal sphere of Pu-239 compressed? Are there little voids in the sphere due to the casting process, or is the chemical energy of the explosive lenses enough to overcome the repulsive force between atoms, thereby bringing them closer to each other?
Many thanks.
those a really good questions and I can safely say that I do not know the answers, I would have to do some digging. I cannot guarantee to be able to fulfill all the requests but I always open to collecting suggestions, thanks.
small correction on language: The Laplacian operator does not depend on the coordinate system; however it's representation certainly does. Laplacian, and the physics its describes, don't care about our coordinate systems.
Love this. I am pedant myself, and this is the kind the pedantic comment that I really enjoy. Yes, you are totally right, I misused the word dependence when I should have referred to representation. I appreciate the correction.
@@jkzero is it pedantic to say I wasn't being pedantic?
No, I just hope one of your viewers makes the jump to _"coordinate free physics"_, which was a revelatory moment in my education, received directly from the Guru of relativity himself (Kip).
@@DrDeuteron I did my PhD on the study of Lorentz invariance, I recall the exact moment when the "Nature doesn't care about coordinates" really clicked in my head, I was reading a paper by my supervisor in a bus on my way to Walmart, I remember that moment vividly.
@@jkzero oh good, because I recently posted something to the effect: you don't understand GR until you accept that the Ptolemaic model of the solar system is just as "right" as the Copernican version, but I get written off as a crackpot before explaining Coriolis and centrifugal, like "gee" , are all a result of non-geodesic coordinates. Stand by for a thought problem...
@@jkzero so the push back is that "rotation is real", hence the standard fictitious forces. Here's the question:
A mass m=1 is "at rest" at a radius "R"--what's the Coriolis/Centrifugal forces on it? Every one says 0/0, ofc.
But that's frame dependent. In a frame rotating at w, you get a centrifugal: w^2R, and here's the kicker, for Corilois:
2v X w = -2wRw = -2w^2R, so the total is a apparent inward w^2R keeping the mass in orbit in the rotating frame--even though there are no forces at all.
At 5:33, it is written:
lambda_f = A / (rho * N_A * sigma_f)
lambda_t = A / (rho * N_A * sigma_t)
sigma_t = sigma_f + sigma_e
So, if sigma_e >= 0, lambda_t
Sorry, I didn't see the errata. But, for the 2nd question ?
Is there a link where we can understand this please ?
Thanks for your comment. I see that you spotted the errata. For the advanced solution, you can check the appendix on the Los Alamos Primer, available as a book but also for free online.
@@jkzero Thanks a lot. I've found it. I'll read it.
Do you know please why there is a density of 15.6 g.cm^-3 (22:45) instead of 19.8 g.cm^-3?
I am interested in taking up the challenge and following through the procedure to attempt to calculate the critical mass for a cylinder. I do have a background in electrical engineering, and as part of that I studied quantum mechanics, and nuclear physics, and did extensive training in calculus.... However I am more interested in trying to calculate the critical mass that relates to a nuclear reactor instead of a nuclear weapon, would you be able to do a video on that or point to where one could find information how to make such a calculation? I am thinking something like calculating the critical mass for something like a Chicago pile, CP-1... minus the bottle of Chianti Bertolli wine... and Enrico fermi!
I have the suspicion that the procedure followed here might serve as a guide for solving the neutron diffusion equation, but the controlled conditions in a reaction can be quite different. I am not a nuclear engineer but maybe the lectures on the MIT course Introduction to Nuclear Engineering and Ionizing Radiation might be more useful (lectures are on TH-cam).
@@jkzero Thank you Dr Diaz, I do appreciate your suggestion, and I have searched for the course, and bookmarked it, as it is something that I will attempt at the end of the year, as it will be an interesting and cool challenge, as I am currently completing a Masters of IT, and wont have the bandwidth until then.
I was using formula shown by you to calculate laplacian and keep getting wrong answer and the suddenly realises it's wrong (later verified with wiki page shown by u just some seconds ago) and then got it finally.
BTW amazing video keep doing this. serious physics stuff of this kind is much needed.
you are totally right, I messed up the r² term already at 10:10. Thanks so much for pointing this out and for going through the calculation, it is a good catch. Other viewers also pointed this out and I added an erratum in the description. Fortunately, it doesn't affect the calculation because it is correctly applied, this is just a typo on the video but not an error in the calculation; however, I do my best to avoid these typos but after watching everything many times some minor details slip through. Thanks again.
@@jkzero sorry i didn't notice😅
I really enjoyed this nuclear bomb series and is enjoying the quantum series. eagerly waiting for next episode🤞
@@MayankSharma-mh5bt thanks for following along
Another interesting condition involves uranium dissolved in aqueous solution. In solution, the uranium is not nearly so dense as is the case with uranium metal, but the water slows the neutrons down. It turns out that the fission cross-section for U-235 is higher for slower neutrons. Now, we have to take neutron scattering off of water molecules into account, and how that scattering slows down the neutrons. There have been a number of serious nuclear criticality accidents involving inadvertently exceeding the maximum uranium concentration limits for a given solution geometry. That has led to the imposition of critically safe geometries for places where fissile materials are dissolved in aqueous solutions.
Great Video!
In the continuity equation dN/dt describes the change in neutron density due to neutron absorption and emission, while del J describes the change in neutron density due to neutrons diffusing away. So why is the entire expression (dN/dt+ del J) equal to the neutron sink and source term (S+ + S-)? Shouldn't there be an additional term on the right side, accounting for the neutrons being lost due to their motion out of the volume that is being considered?
the term ∂N/∂t describes the change on neutron density in a infinitesimal volume with time, that is all, the reasons are given by other terms in the equation. The term ∇·J describes the differential number of neutrons moving in and out of the infinitesimal volume (they can move in and out by any reason, in our case by diffusion when we impose Fick's law). If the number of particles were fixed then the RHS of the equation would be zero, like in the case of electric currents or fluid motion, but here we need to tell the equation that inside the infinitesimal volume the neutron number is not conserved and in fact neutrons are created and destroyed, this is why the two terms on the RHS are needed.
This equation is local, that is, it describes what happens in an infinitesimally small volume around a particular point (of coordinates x). Sphere boundary is far away.
@@kwzieleniewski not sure if you are trying to make a particular point or asking for something in particular
This is why neutron cross section value is so important. Was measured in Argonne or was it Fermi lab ?
Amazing channel and videos. A question:
I'm very puzzled by the various prompt critical excursions with the demon core and friends that occured in history.
I understand nuclear reactors rely on delayed neutrons to keep a reaction in balance.
I understand a prompt supercritical state will have a neutron population grow extremely fast.
I understand an unconstrained core will tear itself apart before full combustion.
What I don't understand is how can a core balance in prompt critical state (blue flash and all) and neither fizzle nor explode.
I've been given explanations like Doppler broadening and thermal expansions which certainly could fit the bill but, without knowing how quickly temperature changes vs how quickly the neutron population grows, I find unconvincing.
Is there a way to do back of the envelope calculations to think about these prompt critical balanced reactions?
my most sincere and honest answer is: I do not know. I am pretty sure a nuclear engineer would know the answer, they study these reactions in non-ideal scenarios in detail. As a physicist myself, we mostly do calculations in ideal conditions, prove the feasibility based on the physics, conservation laws, etc. and then leave the dirty (but crucial) dirty work to engineers, they really make things happen.
A nurtron walks into a bar and has a couple of drinks. He asks the bartender what he owes. The bartender says " for you no charg."
Great video! However, I think that your explanation of he trivial b.c. is a bit misleading. We are not 'disallowing the neuterons to escape' the sphere, however we set the radius of the sphere where N=0. Hence a larger Rc, we artificially require the spere to end only where the neuterons would have stopped, assuming the same λ_t. But I'm nitpicking
thanks, I am glad you liked the video. Honestly, "elementary" boundary condition is quite artificial, in fact, it is only used because it allows an analytical solution for R_c in terms of the relevant parameters (
u, cross-sections, density, etc.), it gives a glimpse on what is driving R_c and the critical mass. However, this is the only reason this condition is introduced, mostly for academic purposes. The "advanced" boundary condition is what has to be used in a practical calculation.
Incredible video!
Thanks, I am glad you liked it. make sure to check the other videos on the channel.
Amazing video and amazing channel!
Thanks for the visit and the positive feedback
Thanks. I'll make sure to employ this new information... wisely...
Can you cover any usage of Markov Chains? I believe they were used/developed during the Manhattan Project for neutron diffusion. Thanks. Great video bro.
...excellent video as always. One question immedietly pops up to my mind.
If criticality condition equals exactly 0 does this theoretically mean that the fuel won't explode but "burn" as a self sustaining nuclear reactor until whole mass is "consumed"?
On th side note by knowing critical mass we can now calculate minimal theoretical yield of this device. Since critical mass is about 50 kg and energy denistiy of uranium/plutonium is 17 kt/kg minimal theoretical yield would be around 850 kt of TNT.
Realistic result is unfortunately much lower as the Trinity test (similar device design) produced yiel of only 25 kt. Mass/energy convertion factor was much lower since majority of the mass was blown up (unused).
Btw, I always wondered what is "cost/benefit" analysis of uranium enrichment in terms of energy invested and energy gained. Is there any information as to how much energy was used to produce first atom bomb since enriching facility is generally massive. So massive in fact that even with this secret known, one needs a facility the size of a small town to produce a bomb. It would be interesting to know if one needs to invest more energy to enrich the uranium than the bomb would produce...:)
IIRC fuel for nuclear power plants needed as much as 4% of energy released for the enrichment. There are equations on Wikipedia saying how much more energy it takes to enrich to 90% than to just 4%.
However, this also depend on how you enrich. Zippe type centrifuges are far more efficient than gasous diffusion.
@@kwzieleniewski ...you mean to say that 4% of (thermal) energy released from the nuclear fuel is needed to enrich that exact ammount of uranium ore to be graded as nuclear fuel? Thing is that 4% (3-5%) is basically average ammount of the enrichemnt that ore needs to be processed to to be classified as nuclear fuel so I don't want to be confused with this number similarity.
Beside the obvious difference in enrichemnt methods there is probably some efficiency drop as the enrichment goes to the higher levels because extracting the remaining mass is that much harder since the saturation of U235 is pretty high and mass difference between U235 and U238 almost insignificant...:/
The Differential Equation of DOOM :)
I read this as "The Differential Equation of BOOM" but yeah, both work, I guess
What was conceptually wrong with Heisenberg's initial "random walk" estimate that you mentioned in the previous video?
Using the random walk, Heisenberg "forced" a single neutron to fission every uranium nuclei, this is the key mistake: one neutron fissions one uranium nuclei, then moves until it encounters another nuclei and produces a second fission, and so on. In a real chain reaction, the neutron density grows exponentially but all the new neutrons also produce fission, this makes the reaction much faster and there is no need of a single neutron doing all the work. These new neutrons diffuse in the uranium while fissioning. The original neutron does not need to travel far, it can even be absorbed, that's irrelevant, there are so many new neutrons diffusing in the material fissioning it, that in the end the chain reaction is much more efficient than assuming a simple random walk.
Beautiful work ❤
Thanks, I am glad you liked it. Honestly, I made this video during Christmas time, I really enjoyed making it, although it took many hours to animate all the calculation. When I finished it I thought "nobody will watch this, but I love the result." It is the video that I which I had watched years ago and that it why I made it. The surprise was that this video made my channel explode (no pun intended). Thanks for watching.
Pheonomenal content. Subscribed! as a side note, I am a highschool student aiming to pursue a career in theoretical physics. How do you suggest I proceed with that? All and any help would be appreciated
Go to college for physics, pursue a Masters (i.e. get good grades in undergrad), maybe pursue a PhD (you’ll figure that out when you get closer to it)
Thank you!@@okkam7078
I'd love a little more on the theoretical basis (assumptions) that underlies these equations. Why is the diffusion model different from the random walk that Heisenberg assumed? I thought diffusion occurs from many particles doing a random walk.
Using the random walk, Heisenberg "forced" a single neutron to fission every uranium nuclei, this is the key mistake: one neutron fissions one uranium nuclei, then moves until it encounters another nuclei and produces a second fission, and so on. In a real chain reaction, the neutron density grows exponentially but all the new neutrons also produce fission, this makes the reaction much faster and there is no need of a single neutron doing all the work. These new neutrons diffuse in the uranium while fissioning. The original neutron does not need to travel far, it can even be absorbed, that's irrelevant, there are so many new neutrons diffusing in the material fissioning it, that in the end the chain reaction is much more efficient than assuming a simple random walk.
@@jkzero Thank you so much for the explanation! Although tbh I'm still a little confused. It sounds like you're saying he modeled a single neutron having to meet and fission every single uranium nucleus. But that wouldn't ever work, right? I thought the exponential chain reaction is what's necessary for an explosion.
@@Norsilca you are right, this shows you how incorrect Heisenberg's conceptualization of the problem was
@@jkzero That's confounding. Everyone knew the exponential multiplication was the key to the chain reaction. Why would he reduce the problem to something he knew was wildly unrealistic?
for a more precise calculation, however, the transport equation should be used
you are totally right, this is one of the things what I meant when I said that the calculation shown in the videos involves many approximations. But it could be argued that diffusion theory is not even applicable because the size of the material is barely bigger than the transport mean free path. Careful calculations show that many effects ignored here cancel each other out but this is just a lucky coincidence.
in any case yours is an excellent video, I have a degree in nuclear engineering and in the nuclear reactor physics courses I had followed, the first approach to use for calculating the distribution of neutrons was to use the diffusion equation, we then had the most precise calculations with the transport equation
@@jkzero
For the Pu239 density in g/cm^(3) - isn't it greater than the figure you provide?
you are right that the standard density of plutonium is closer to 20 g/cm^3; however, depending on temperature plutonium exhibits different crystalline phases. The relevant phase used for weapons is the so-called “delta” phase and that is the density that I have provided.
@jkzero Huh. Thanks for that - so basically the standard density is not in that phase at all. Interesting.
@@AlbertCamus-r6i exactly; I mean you can have plutonium metal in other phases too but for the bomb they used the delta phase due to stability and for machining purposes
A proton is a collection of 1836 expanding electrons and add a bouncing expanding electron makes a hydrogen atom. “G” calculated from first principles- the hydrogen atom- in 2002. No energy, charge, photons, waves, spin, fields, potential, quantum,space, time, space-time etc. All Standard Theory/Model was replaced by Expansion Theory in 2002. “The Final Theory: Rethinking Our Scientific Legacy “, Mark McCutcheon for proper physics.
I love this channel
Thanks for watching; you can actively help the channel by liking, subscribing, and sharing. This type of support is highly appreciated so I can continue making videos.
Incredible this video😍
This is so interesting. I bet a similar equation can be written for a biological cell producing some biomolecule in its 'core' (cytoplasm or nucleus or some organelle) and then it is flowing out. What do you think Jorge?
This topic goes beyond my expertise so whatever I say should be taken with a grain of salt. Moreover, I do not like it when people comment too confidently on topics beyond their expertise. With that said, you can search for modeling of so-called "reaction-diffusion systems" that include many biological and chemical systems. There you will encounter Fick's law again and the diffusion equation for modeling, for example, patterns in ecological systems or the diffusion of ions in living systems. There is a whole branch that makes use of this type of equation called mathematical biology.
@jkzero The math should be equivalent. I do research on biochemical systems and can vouch for the biology side. Recently I've been teaching myself some modelling which is why I came across your videos. Perhaps I can pick your brain some time to validate my equations. I work at the max planck in dresden. I like the idea where you can set the boundary flux to be zero or nonzero. It's just like a cell's receptors not letting stuff out for zero flux or secreting something, for nonzero.
@@Archiekunst you can find my email on the "about" of the channel
@ 21:03 REMARK: A gradient is not the same as a derivative!
Beautiful video on a nice subject❤
Glad you think so! I am curious to know what brings viewers to the channel, were you searching for something in particular or did the 'mighty algorithm' find you?
@@jkzero ah the mighty algorithm
@@cyrilguillouard1266 Thanks for sharing and I am glad the algorithm is working, I hope you find the other videos of interest too and welcome to the channel
Is this made on manim? Very nice animation
yes, I use Manim in all my videos. I know that Grant recommends to not use Manim for just showing equations but I can't help it, I like how LaTeX formulas are shown too much. Thanks for stopping by and make sure to check the other videos.
at 15:08 the only solution won't be pie but 0,pie,2pie,3pie ...npie so that if the kappa*R = 0 => kappa=0
You are right: the general solution is kappa*R=n*pi; however, the higher "harmonics" will only make the bomb core bigger, we are searching for the minimum radius. The solution kappa*R=0 is useless because it requires kappa=0, which is mathematically correct but physically useless leading to N(r,t)=0, which of course solves the equation but it provides no information. Also, a solution for n>1 can lead to a negative density, which is also a non-physical solution. Mathematically, your point is 100% correct; however, here is when physics constrains the final solution, just like setting B=0 to remove the non-physical cos(kappa*r) term.
@@jkzero got it thanks for your response. I will try to go over all the cases with pen and paper to be more clear :)
@@PalashTiwari-d3g good luck and please get back letting me know what you get or if you encounter any road-block
So, because bismuth-209 is technically radioactive it has a critical mass too. Can you calculate it? It would be nice to know even if there isn't nearly enough bismuth in the entire universe to create it.
Not really, you are confusing radioactive with fissile, they are very different properties. If any radioactive material could be made to undergo a nuclear chain reactions we would be making bombs with bananas (from their radioactive potassium content). Instead only a few isotopes like U-235 and Pu-239 can be used because they are fissile.
Very good job! I think I will have one question but maybe later ;)
questions are more than welcome, I hope that I can help
How is the mass being destroyed or heavily damaged by it partially fissioning accounted for?
in this calculation the critical mass is defined as the minimum amount of fissile material to trigger a self-sustained nuclear chain reaction. Of course as soon as the reaction starts, the material will evaporate and the whole assembly will self-destroy. This is not accounted for in the calculation shown in the video.
Hi! Is there a way to predict theoretically the fission and capture cross-sections of our processes? You can get more or less the order of magnitude of this treating the U-235 as an hard sphere and using the Bohr-Wheeler theory that tells you that the higher Z²/A, the higher the fission probability. I tried to evaluate the U-235 fission cross section and I get something like 6.20 time 10^(-23) cm². But I'm not able to see how to improve this little model in order to study capture cross section. Also I have no idea how to distinguish slow and fast neutrons! By the way, your videos are beautiful.
Thanks for the feedback and I am glad you liked the videos. Yes, you can motivate the order of magnitude for the cross-sections but at the scale of neutrons and nuclei, quantum-mechanical effects can become relevant and classical intuition falls short. Plus a nucleus with 235 nucleons is really hard to model and you can only use approximation methods, in the end the best way is just go a measure it, all you need is a few microgram of uranium-235, which is easy to "make" with a cyclotron
Hey. I like this explanation however I don't understand how to obtain tau term. I feel like one can get a formula for this however I'm stuck
Should I just calculate an average speed from average neutron energy? What value is that average energy?
tau can be determined in terms of the fission mean-free-path and the average neutron speed. I showed the detail of this in the video about the Physics of a Nuclear Explosion at the 4:04 mark th-cam.com/video/6VSrGDOrWXc/w-d-xo.htmlsi=5uq4tpjHFlhVPBah&t=244
@@jkzero Everything clear now. Thank you for the answer
@@kotowaty3386 excellent! I'm glad that helped. Thanks to you for watching and for following the calculations along. There are some typos later in the video, please check the errata in the video description.
Warning: typo in laplacian of f at 10:10.
You are totally right, I messed up the r² term. Thanks so much for pointing this out, it is a good catch. Fortunately, it doesn't affect the calculation because it is correctly applied, this is just a typo on the video but not an error in the calculation. Others have also notified it and I have included an erratum in the video description. I do my best to avoid these typos but after watching everything many times some minor details slip through. Thanks again
In the remarkable story two flaws are negating each others. First, Diffusion is over estimated for fast neutrons (of bomb core) and for thermal ones as well. Second, the source neutron term due to fission must be decreased for capturing neutrons. But at criticality threshold these mistakes happly are zeroesed. Good luck!
it is interesting how the simple calculation in this video was enough despite the many limitations of the formalism. The most remarkable is the fact that two non-negligible effects have been ignored: 1. neutrons can also been captured by U235 without fissioning, this was completed ignored; and 2. the elastic-scattering cross section is not isotropic so care must be taken when calculating the corresponding cross-section. Nonetheless, when these two effects are incorporated they happen to cancel each other out. Just a lucky coincidence.