We're coming into a new era with the internet education. Susskind is one of the greatest teachers available, togehter with a handful others, and I'm very grateful that his lectures are freely available online. I'm 40 yo, regrettably didn't find physics back in the school days, but I've had an exploding interest in physics the last 3-4 years. Now there's the math barrier - luckily, I've taken all the required maths during my education, so some refreshment is needed.
All they know the Einstein's stuff is just a HUGE HOAX that has been devealed at: numbermusicrevolution.com/srt/ Indeed, Einstein's lies allowed many institutions and scholars to rise so much funds rising, so they are not willing to kill that Golden Chicken. Notwithstanding, all this hoax has to come to and end, sooner or later, and many people will have to face justice.
[Dark energy - not related to the course]; Del operator 15:00; Gravitational field and Gauss law 21:00; Gauss theorem 32:00; Equivalence Principle 59:30; Galilean transformation 1:09:00; Accelerating frame of reference 1:12:00; Light motion in accelerating frame 1:18:00; Tidal forces 1:32:00; Curvature 1:37:00
The finishing sentence of this lecture gave me a real breakthrough in understanding. In all of these series of lectures, how can anyone thank Prof. Susskind enough for his generosity of spirit?
@Andrew Martin Not sure what Robin Betts meant, but yeah, that was an enlightening insight right there for me. Or at least it makes me feel like I understand something important which I previously did not. I admire the clarity with which Prof. Susskind can gauge the right moment to feed such illustrative central points of insight to the audience so they can make the connection. Even when he's getting tired. A great teacher!
This is some hardcore stuff. I feel so humbled watching this haha. I think I can barely understand it but man its a mindfuck. Its very warming to see that people like the professor not only understand it, but can teach it. And this is only an introduction to relativity!
The depth of questions that Professor Susskind has to bare and answer is.. nerve-challenging. Hats up for this remarkable professor for his beautiful mind and personality.
This is an advanced undergraduate or early graduate course, to be able to grasp the concepts you need a foundation of college level calculus of several variables.
This professor is SUPREMELY patient. The student asking the questions over and over again is justified in asking. He paid for the course. He has every right. But he also needs to be getting his notes out and trying to solve on his own. Otherwise, he's just asking for the prof to do his work for him.
The student was looking at his notes but must have not paid attention when the professor canceled the 4pi, the when he looked at his notes to catch up.he was like, but wait professor ☝
Alternative form of the calculation of gravitational acceleration at a point inside a spherical mass. A = m G / r^2 which in terms of whole sphere is: A = n M G / R^2 R : Radius to point on surface of sphere r : Radius to point inside radius of sphere n : Ratio of r to R
For all of you who say these students ask stupid questions: "We must know. We will know." - David Hilbert Asking questions is never stupid. Not asking questions for fear of sounding stupid. Now, that's really stupid.
I think they approach it from an opposite direction in perceiving the concept'sas myself . I see a concept in my mind and attempt to put the math to it. Others with a more mathematics background I think may see the equation and from that derived the image in the mind. I thought the same at first but I was a 4.0 physics major, not mathematics. I can't expect others to see the concept's presented as quickly. That doesn't lesson their questions however. Koodos to the students. It a challenging concept to truly grasp.
The case where red sphere is inside A question in the class is that a 4*pi is missing from the equation. I think the reason for this would be like this: by dissolving guass theorem we get- 4 * pi *G * integral(rho * cube(dx))=Aperpendicular*- 4* pi * R2. Now as explained by the sir this case has a unit mass density uniformly distributed. so the equation can be written, => 4 * pi *G* rho* integral(cube(dx))=Aperpendicular* -4* pi * R2. As you can see that rho in the integral is taken out of the integral. the remaining integral would be volume of the red sphere. so, => 4 * pi *G* rho* 4/3 * pi * R3=Aperpendicular* -4* pi * R2 => 4* pi * G* rho * R/3 = -Aperpendicular. Now if you check this equation with the one on the right board these would match and no extra pi is missing.
AHHHHHHHHH! God I’m dumb. I figured that he was right and that the students were wrong because he’s saying that the conditions are different. I didn’t understand how he got his answer until you explained it. A uniform ρ means that it’s constant and can be left out of the integral. So it’s just -4πGρ on the left side. So I’m assuming then, that means the integral of d^3x is actually the volume of the sphere, right? So if you evaluated it, you’d get (-4πGρ)(4πR^3/3) = 4πR^2. You’d cancel the 4π on either side and be left with (-4πGρ)R^3/3 = R^2. Then you cancel again which gives you -4πGpR/3. That is equivalent to Newton’s equation. He got the wrong answer because he left out the initial -4π.
I recommend to read Professor Susskind's books as well (the Theoretical Minimum Series), in my opnion before looking the lectures. I t is a big luck that a first class, premium, five star scientist with such a reputation obviously has fun and an excellent ability to teach his stuff. Thank you for making this publicly available. Kind regards from Germany!
the presentation is well organised and the lecturer too, i think he deals well with his students. i like the way he handle question, is been enlightening lecture.on relativity.theory.
for the field to be inside on the left hand side it should be-- (-4/3 πR^3 4πρG)=4πR^2 A(perpendicular) then the missing pi can be located and the equation will become appropriate
+deadmeat1471 p = m/v m = pv v_sphere = (4/3)πr^3 m = p((4/3)πr^3) For the equations to be consistent with each other, the "m" on the right board should be able to be replaced with the "pv" equivalent expression above, resulting in "A_perp = -(4π*r^3*p*G)/(3*r^2) = -(4π*p*G/3)*r" but instead the left board only shows "A_perp = -(p*G/3)*r". That's where the 4π is missing on the left (assuming the right was derived correctly)
The error was in the right side of the equation on the left board. He equated A_perp*4pi*R^2 with the total mass inside the sphere (multiplied with G) and then expressed that as the volume of the sphere times the mass density. The correct way to apply Gauss' theorem would have been to equate the left side of this equation with the volume of the sphere V times the divergence inside, which is according to Gauss' law -4pi*rho*G. Thr final equation looks like this: A_perp*4pi*R^2=-4pi*rho*G*V and using V=(4/3)*pi*R^3: A_perp*4pi*R^2=-4pi*rho*G*(4/3)*pi*R^3 which has an additional factor of 4pi on the right side.
deadmeat1417 use "dimensional analysis" to know you are wrong here. 4pi is dimensionless, so 4pi r^3 has dimension (length)^3. So its a volume, not an area.
I'm not a student, I am just fascinated with physics and I love watching these to try to learn anything I can. Very good lectures! Is cosmological constant sort of like the inverse of gravity or something? Gravity is very weak but gets very strong with mass, cosmological constant is very weak but gets very strong with distance. So, in order to live in a universe (or to find ourselves in one from the anthropic principle) in which matter is being held together due to gravity, you would ultimately have to have the repulsion, otherwise it'd collapse on itself, and as the universe gets bigger and bigger, faster and faster, the surface area and distances of it is larger therefore enacting the cosmological constant to grow larger? Does that have anything to do with the formation of voids like Bootes void? To me it seems like that's why you get a filament or sponge like structure if you could zoom out on the universe as a whole.. all of the mass coalescing into valleys of spacetime, and voids bubbling up in relatively empty regions, which pushes matter out toward the valleys? Sorry, I'm sure these are really dumb questions, just trying to figure it out!
@@itchyandscratchy7350 The cosmological constant would get stronger as the time passes, I think Prof. Susskind explained it pretty clearly in the beginning. Also, it's not really getting stronger by the distance but due to its extremely small value, the equation would only sums up into a considerable number when it's multiplied by a huge distance, hence it only matter when you accounted the entire universe for it, but not so much in an astronomical scale or even in global scale. Calling it an inverse gravity is pretty much inaccurate because it's not exactly a force. Let alone a force that repulse each matter or particle from each other, it's something that drive the particle grow apart from each other.
@@Vandalgia 6:20 Professor Susskind tells that the theory on the "Big Rip" describes the cosmological constant to be time-dependant, therefore growing with time. As Professor Susskind reminds, however, that is not true ; the theory on the "Big Rip" violates all principles of physics, and so it is not true based on the given principles of physics. I hope that clarifies Professor Susskind's point about the constant.
I tend to believe that the Felix Bloch professor of theoretical physics at Stanford University and his considerable contribution to string theory probably deserves a little more respect then that.. it is a privilege to watch this and thanks to him for his dedication to education.
To be onest he is real good teacher. Turning complex things to real simple (making it dumb) is high skill. Explaining something complex in a complex way is easy. What hard is turning complex things into a "cosmological constant" dumb level,requires high amount of experience and capeability. Btw : the students having problem with perpendicular A is the mistake that the both examples are different and the perpendicular A is example specific. there are two different A not equal :)
It is the sin(x) = 2MG/Rc^2 with x=2MG/Rc^2 that is used for the aproximation, also there is other aproximation, that for the elapsed time delta t, it is taken as the horizontal component of speed of light is the same c, that can be appear as a logic circle (that is not), since for this is used the a posteriori known fact that the angle is too small, and in the former the use of values of G and c are used to aproximate sinx =x.
The 4pi is lost when he subs into Gauss's Theorem at 44:43. The right-hand side was previously calculated to be -4piGM (at 38:55), but he only subs into -GM, losing the 4pi.
The reason he messed up at 0:45:24 is that he is missing an additional factor of 4pi on the right hand side from Gauss' law: Del.A = G rho 4pi. Integrating this expression over the volume of a sphere leads to 4/3 pi r^3 * G rho 4 pi
Take note that in the discussion of the gravitational field at some radius R inside the earth taken from the center of the earth, that Leonard had dropped a 4pi. Perhaps you may have caught this. So, the g field at radius R inside the earth is actually: G*p*pi*R*4/3
compare the equations at 45:21 and 40:29. They each have a term A*4*pi*R**2 but they are equal to two different things: one is -4*pi*G*M and the other is just M*G. What happened to the 4*pi that was in the equation from 40:29? Am I missing something?
It just make me smile every time Prof. Susskind call the white board "black board". Other than that he's such a great educator. I wish i can attend his classes one day.
Thanks to Google, TH-cam and the internet technology researchers who made this possible to watch these videos. Otherwise I had no chance to see this lecture series.
That disagreement from 45:00 to 52:00, the student was 100% right, the professor got it wrong. He left out a factor of 4pi from the right hand side. Gauss's Law&Theorem state that Aperp*surf. area == -4piGM. So, at 45:00, the surface area is 4piR^2. And the mass of the sphere is the volume of the sphere * rho. The volume of the sphere is 4pi/3 * R^3. Plug all that in, you get: Aperp * 4piR^2 = -4pi * G * 4pi/3 * R^3 * rho. See the problem? When he wrote it down at 45:00, he forgot the first 4pi. He wrote down simply -GM rather than -4piGM. The student was right. After cancelling the 4pis from each side, you end up with: Aperp * R^2 = -G * 4pi/3 * R^3 * rho. So, Aperp = -4pi/3 * rho * G * R. If you changed that back to mass instead of rho, you'd get back to the original Gauss equation: Aperp = -GM/R^2.
greg55666 That is correct. The student was right to be confused by the result Dr. Susskind came to on the left board because M=V*rho. For A perp = -MG/R^2 as on the right board the equation should read -{[(4pi/3)*R^3]*rho*G}/R^2. The (4pi/3)*R^3 is the volume of the sphere so when you multiply the volume by rho you get Mass or M=V*rho. Now all you need to do is substitute V*rho for M in the equation A perp = -MG/R^2 and cancel out the R^2 in the denominator with the R^3 in the numerator. That leaves A perp = (4pi/3)*rho*R*G. I think the confusion here is that the student asking the question had no idea how to articulate the correction he was attempting to point out. Had he been able to say "I can't seem to make the 4pi cancel out like you did due to the 2 equations you set up not being equivalent between the left and right boards. I think you unnecessarily added a 4pi to the L.H.S of the equation on the left board." then I think Dr. Susskind would have immediately realized the error and corrected it. The fact that there were so many conflicting voices in the room and nobody was making any sense led him to just leave it as it was.
@@dylanhaymore608 in student’s defense, he was constantly interrupted by sycophants. I’ve watched this maybe 10x trying to find the error. These posts helped Ty
@@paulodonnell753 I'm glad I was able to help you understand. This material is already difficult to learn so it's even harder to notice when there are errors from the instructor. Good luck on your journey for knowledge!
Yeah. Well put. Spherical space about the center of mass is the reasonable conclusion. I forgot that the mandatory requirement that all mass be contained within R of the center of mass.
I accedentally fell asleep when watching youtube, woke up and had watched the whole lecture one and was at the end of this one. Am I gonna be a genius now?
the description has: "discusses dark energy, the tendency of it to tear atoms apart" Susskind would be so pissed. He only spent about 15 minutes of the class explaining how unlikely it is for dark energy to rip apart atoms.
At 1:13:13, it givs x = (1/2)g*t^2 as a formula for position of a (uniformly) accelerated object. That formula is only strictly applicable if the object started out at rest.
Unbelievable. I have had very little formal education, a lifetime of professional experience traveling and living around the world. Mind blowing knowledge - FOR FREE - presented by a world class professor. It's troubling TH-cam algorithm does not push this content and knowledge More; esp the younger generation. I remember visiting a tiny all girls(K-8th) school started by Irish nuns over a century ago in Peshawar a few years ago; an entire classroom of 6th grade girls were doing Calculus on a true blackboard. I didn't understand a lick of it. Mean while in US, my 5th grade son has not come near Algibra much less Calculus. I'm not complaining, just comparing. Kids living in extreme poverty in a extremely dense ancient city, non stop extreme violence all around, yet they can focus their young minds to unbelievable levels academically. Our kids are distracted by countless electronic devices, social media, video games. Yet it is on those devices and media that you can find a world class education 🤷♂️.
@memyself1125 No. Satelites use it and Particle accelerators use it but until you get to speeds of (lets say 10,000 miles per hour) you do not use it. That being said there is relative motion, which is helpful to the layman, but not a necessity.
Well, actually when the gravity field equation is derived, we assume a test particle with mass m different than zero. When the m term appears on both sides of the equation, we divide by that (thus it must be different than zero). Therefore it is counter-intuitive to suggest that gravity has the same effect on a zero mass particle. On the contrary according to Newton the gravitational force exerted on a zero mass particle is zero.
Yes people are asking question like they don't know what they are talking about. And his style is like he is teaching for those who already have some knowledge.
Well, rho = m/V would be the case if the mass if uniformly distributed in a given volume. If the value of rho depends of the chosen volume we have to write rho = Delta m/Delta V. I think this becomes more obvious if you take the limit Delta V -> 0. Then, the differences become differentials. rho = dm/dV => dm = rho dV Integrate over a given volume to get the mass included in this volume. M(in the chosen volume) = Integral{rho dV} . and rho in general could be a function of V.
I thought this was the final part of the explanation as I had simply opened this video, not the playlist, and TH-cam hadn't recommended any part 3 or bigger on the suggestions. :(
Thank you professor..... One question was : why the effect of mass outside the Gauss surface inside the spherical mass be neglected or canceled ....my answer is I think that the mass is not neglected as you see the density of the field (the gravitation density g) be reduced by reducing the (r) to becomes zero in the centre of sphere which means that the effect of the outside mass which also cause (g) can be considered as superposition summation for the (g) vector at the center and that summation becomes zero when (r) become zero so the disappear of the (g) in the center due to the opposite directions of (g) , from the general relativity point of view also we can understand that the distortion of the metric tensor curvature outside the mass (density of energy) is totally different than when it passes through the it ...thank you
At 1.18 Susskind describes how a light beam would cross an accelerating elevator -- noting that the observer in the elevator would see a downward curving light beam (and an external ,stationary , observer , presumably seeing the light beam travelling straight and horizontally ). And yet had the elevator been travelling at a constant speed (SR), the observer in the elevator would see the beam travelling horizontally within the elevator, and the external observer would see it travelling diagonally upwards. Why does the external observer ,say, see the light beam being affected by the elevator's frame of reference in the SR case , but not in the GR case. ?.
+Metallurgist47 If I understand correctly the difference is in the original vector of the light. With the accelerating frame the assumption is the beam is emitted at the same time the acceleration starts and so it's direction is horizontal. In the constant velocity case, since the velocity is constant, you have to assume that the elevator is in motion at the beginning of the experiment and the beam is emitted from a source inside the elevator and therefore it starts with the upward component already. Also, the curvature is only seen by the observer inside the elevator. To someone outside it travels perfectly horizontally.
Is it a coincidence that the operation of applying an operator to a function looks the same as the operation of multiplying two values? Both are written with simple concatenation. Maybe a number is identified with the operation of multiplying by that number? It could also be identified with the operation of adding that number, and I don't see a principle that says to pick one over the other.
I've always wondered at the assertion that the elevator acceleration is indistinguishable from gravity. Namely, the elevator acceleration field has no divergence, it is everywhere (in the elevator) parallel. However, a gravitational acceleration field has slight convergence, due to the Earth's mass acting as a point mass, with the gravity field radially divergent from there. Thus dropping 2 balls from opposite sides of the elevator, maintain their separation in an elevator acceleration field, but will be slightly closer together at the floor in a gravitational field. Therefore, there is a difference between the two fields.
The _principle_ _of_ _equivalence_ and the modern theory of gravity from Einstein does *NOT* postulate that acceleration and gravitation are physically indistinguishable and universally applicable.
Question about the Einstein’s elevator @ 01:00? Since the direction of the weight is always in the direction of "g" or towards the centre of gravitating mass therefore in which direction a person would feel his weight in elevator ? Similarly, gravity bends light towards gravitating mass therefore why the direction of the bent light is against the direction of "g" in elevator? Just wondering: Shouldn't a ball or person move in the direction of acceleration "g" or towards the centre of elevator due the universal law of gravitation [F= GMm/r^2]. @about 50:00: Would the inside person or object feel "g" if the elevator is consider as a shell?
The reason the student is getting confused in this lecture is because Susskind uses big R on both the left and right board, but they are not the same, R is < than the one on the right because the shell is inside the earth, when you break the rho for density down, that brings out the actual earth radius that is used on the right board.
So there is a newtonian aquivalence Principle( if an object gets pulled by a bigger mass the acceleration is not dependent of the mass of the smaller object) and Einsteins aquivalence princple( when you accelerate a rocket with 9,81m/s^2 it Feels like earth‘s Gravity Field for someone in that rocket )?
How do you justify that? It will be that if rho does not not depend on the volume. Otherwise, you have to calculate the integral to find a relation between M and V.
We heard in the last video that the gravitational field is just a field of acceleration, because the acceleration is independent of the mass of the test particle. The test particle can be any mass at all, a planet or a single molecule, and the acceleration it undergoes will be the same. If I accept that (Newtonian) view, then why should I be surprised that gravity bends light? Surely it is just a particle of zero mass, being affected by the field just as any other test particle is?
Maybe particles are more pushed together than pulled to each other and that's why it increases as you pull them apart, space "pressure" so to speak pushing them together.
I suppose that this is second year subject, but Student are asking question like they did not complete any university maths, or physics before. Gauss's Law is and Gauss theorem is usually first year mathematics, and they usually had to overcome difficulties with new concepts.
Time = vector quantity (foward) Clockwise = could be moving to the right like of the normal clock we have but some people do sell clock that move to the left. As long as it is 360 degrees. So I think there is nothing like 'anti' Ant = se-mut A-rah = direction Jam = clock What is wise? Page 1375 -wise adv combining form. 1 indicating direction or manner: clockwise; likewise. 2 with reference to: businesswise. [OE -wisan (italic); see WISE2] wise1 1 2 prudent; sensible. 3 4 5 6 7 Arch. possessing power or magic. 8 9 10 wise2 n Arch. way, manner, fashion, or respect (esp. in any wise, in no wise) [OE wise (italic) manner]
@pi05pm3 It is! It's extremely helpful to have free open-courseware like this available, especially since I'm planning on doing my PhD in theoretical physics next year ;)
@@mdnazmusshahid4827 if you're confused about what you want to do for your PhD and where you want to do it, you should speak with someone at your university to get some guidance. Probably start with a lecturer who is in the field or close to the field you want to study. My topic was solutions of the self-interacting Maxwell-Dirac equations.
When I look at the shadow of an electrical pole at about mid- afternoon I see a well defined shadow edge near the base but when I follow the shadow away from the pole the edges start to blur. This suggests that either light is attracted to dark or repelled by other light. No gravity effects here? Moon shadows are much crisper
That's just caused by the fact that the light source has dimensions. Light from the edge of the light source is able to "see" further behind the object than light from the center of the light source. No repulsion or attraction necessary
@@gamerdio2503 when I walk at night the dividing line of shaded to unshaded is sharper, more crisp than in the day light. I can see a very distinct shadow line at night compared to the shadow line I get from the sun. It’s like the photons in sunlight are repulsed from each other where the moonlight is running more parallel
@@abberepair8288 Well, moonlight *is* sunlight, so that idea is thrown out the window. I think the answer might come from the fact that there's less light reflecting from the moon, making it harder to make a distinction on the edge. In addition, the fact that moonlight is reflected light and not a light source itself might have an effect
For Gauss' Theorem, i don't get how the example on the left board(calculating A within the earth) is using gauss' theorem. why do we take A4piR^2= Mass within sphere x G ? Shouldn't it be = -4piGM instead?! Please help me someone! Thanks.
Because that isn't Gauss' theorem. You can't just stick a factor of 4pi into it for no reason. Gauss' theorem simply states that the flow of 'something' (anything) in or out of a region is equal to the amount passing through the boundary. That's actually a pretty obvious statement really when you think about it. In mathematical terms the same thing reads 'the volume integral of divergence within a region is equal to the integrated surface flux' That's maybe not so obvious! What adds a little confusion is the 4pi introduced in the equation for gravitation. Essentially it's a modification of 'G' to allow for the fact that G was originally defined by Newton in terms of one dimension but we're now working in 3 dimensions. We get similar factors of 4pi in electromagnetism for the same reason. That factor of 4pi cancels out when you use Gauss' theorem (it HAS to or nothing makes sense - you'd have more water running down the drain then you put into the bath :)
Hi, wonderful video. but a question: there is a book for follow this lectures? i want to have also a book for Einstein relativity Special and general. Thanks all and regards from Italy :-)
there is one called Special Relativity and Classical Field Theory: The Theoretical Minimum . It is coming out in September. Susskind wrote other books such as 'Quantum Mechaniics'
Frame of reference connected in time.... Electron jumps..position superposition... Parallel so ....electrons are paralle to each other...not disappear gband appearing another place.. Photons bend because of mass down?
not sure, but if you have a isotropic distribution of mass (convergence of gravity), the only shape the system can take is a sphere. Simply because at every point on the edge of the sphere it's the same distance to the center.
To say that the "father" of string theory (read: modern particle physics) knows nothing about dark energy is quite a statement.. However, from what he's saying the expansion is increasing, as the repulsive force grows with distance.
Well, yes, you'll get the average density of the particle, but you will not get the same rho which is a function of position. Also, without integrating rho over space you don't know the total mass M (unless it's given), and you can't calculate M over V. If I told you there is a spherical object at the origin of radius 1 and rho of r equal to |r|, what is its mass and average density? You can't really solve that without integrating.
@stenniswood if space itself is converging into point masses then where is it going to? Another universe? As raw material for the expanding edges of our universe?
We're coming into a new era with the internet education. Susskind is one of the greatest teachers available, togehter with a handful others, and I'm very grateful that his lectures are freely available online.
I'm 40 yo, regrettably didn't find physics back in the school days, but I've had an exploding interest in physics the last 3-4 years. Now there's the math barrier - luckily, I've taken all the required maths during my education, so some refreshment is needed.
Thanks Stanford University for posting these lectures
I know.....students pay a lot to be at this seat of learning....so to give this away for free is generous.
All this to disprove the flat earth
10 year old comment jesus christ
@@mattwooten830 you are on the good drugs
All they know the Einstein's stuff is just a HUGE HOAX that has been devealed at: numbermusicrevolution.com/srt/ Indeed, Einstein's lies allowed many institutions and scholars to rise so much funds rising, so they are not willing to kill that Golden Chicken. Notwithstanding, all this hoax has to come to and end, sooner or later, and many people will have to face justice.
Heartfelt thanks to Professor Susskind and Stanford for making these beautiful ideas available to me. I am following all of the series.
[Dark energy - not related to the course]; Del operator 15:00; Gravitational field and Gauss law 21:00; Gauss theorem 32:00; Equivalence Principle 59:30; Galilean transformation 1:09:00; Accelerating frame of reference 1:12:00; Light motion in accelerating frame 1:18:00; Tidal forces 1:32:00; Curvature 1:37:00
❤thank you
he is so patient and calm while giving lecture
The finishing sentence of this lecture gave me a real breakthrough in understanding. In all of these series of lectures, how can anyone thank Prof. Susskind enough for his generosity of spirit?
Robin Betts One of his books (forget which) mentions the great pleasure he gets from teaching physics , for which I'm eternally grateful.
Seriously this guy is great!
Do you mean his line that curvatures in spacetime are equivalent to tidal forces? Or a different line?
@@andrewmartin3671 1:47:16
@Andrew Martin
Not sure what Robin Betts meant, but yeah, that was an enlightening insight right there for me. Or at least it makes me feel like I understand something important which I previously did not.
I admire the clarity with which Prof. Susskind can gauge the right moment to feed such illustrative central points of insight to the audience so they can make the connection. Even when he's getting tired.
A great teacher!
I can’t get enough, this is my idea of finding the bright side of life. Free class at a university I couldn’t afford. I’m loving it🤭😍🥳
This is some hardcore stuff. I feel so humbled watching this haha. I think I can barely understand it but man its a mindfuck. Its very warming to see that people like the professor not only understand it, but can teach it. And this is only an introduction to relativity!
The depth of questions that Professor Susskind has to bare and answer is.. nerve-challenging. Hats up for this remarkable professor for his beautiful mind and personality.
life passes by as i try to understand this 'introduction'
This is an advanced undergraduate or early graduate course, to be able to grasp the concepts you need a foundation of college level calculus of several variables.
This professor is SUPREMELY patient. The student asking the questions over and over again is justified in asking. He paid for the course. He has every right. But he also needs to be getting his notes out and trying to solve on his own. Otherwise, he's just asking for the prof to do his work for him.
Point well taken....I must agree with you here.
Totally agree... but I'm thinking of the other tuition paying students; it does get annoying.
It just disappoints me that they are paying for this class and aren't even enjoying his great physics puns.
visnevskiscom "Information presented in a particular way" is, by definition, a service. But also, it was a joke.
The student was looking at his notes but must have not paid attention when the professor canceled the 4pi, the when he looked at his notes to catch up.he was like, but wait professor ☝
Alternative form of the calculation of gravitational acceleration at a point inside a spherical mass.
A = m G / r^2 which in terms of whole sphere is: A = n M G / R^2
R : Radius to point on surface of sphere
r : Radius to point inside radius of sphere
n : Ratio of r to R
For all of you who say these students ask stupid questions:
"We must know. We will know." - David Hilbert
Asking questions is never stupid. Not asking questions for fear of sounding stupid. Now, that's really stupid.
morani789 that’s such a moronic position, most questions are extremely stupid and shouldn’t be entertained in a serious environment
morani789, those students are not prepared for this class, and the level of this class is not high, but their level is low.
I think they approach it from an opposite direction in perceiving the concept'sas myself . I see a concept in my mind and attempt to put the math to it. Others with a more mathematics background I think may see the equation and from that derived the image in the mind.
I thought the same at first but I was a 4.0 physics major, not mathematics. I can't expect others to see the concept's presented as quickly. That doesn't lesson their questions however. Koodos to the students. It a challenging concept to truly grasp.
I think if we look at it as people just discussing things instead of as a “physics lecture”, then we won’t be so annoyed by the questions :)
Stop defending the indefensible. Some of these questions are atrocious. Really dumb.
Thank yo u to Leonard Susskind for these lectures. I wish that my physics lecturers 40 years ago at Oxford had covered this!
Leonard is my absolutely favorite contemporary physicist.
Thanks for posting these lectures by Dr. Susskind
It is an honor to be taught by this gentleman. He has full command over what he is talking about. Thanks a lot Prof. Susskind.
Susskind is a great Magister. I appreciate this gift. Danke.
The case where red sphere is inside A question in the class is that a 4*pi is missing from the equation. I think the reason for this would be like this:
by dissolving guass theorem we get- 4 * pi *G * integral(rho * cube(dx))=Aperpendicular*- 4* pi * R2.
Now as explained by the sir this case has a unit mass density uniformly distributed. so the equation can be written,
=> 4 * pi *G* rho* integral(cube(dx))=Aperpendicular* -4* pi * R2. As you can see that rho in the integral is taken out of the integral.
the remaining integral would be volume of the red sphere. so,
=> 4 * pi *G* rho* 4/3 * pi * R3=Aperpendicular* -4* pi * R2
=> 4* pi * G* rho * R/3 = -Aperpendicular.
Now if you check this equation with the one on the right board these would match and no extra pi is missing.
Outside sphere
AHHHHHHHHH! God I’m dumb. I figured that he was right and that the students were wrong because he’s saying that the conditions are different. I didn’t understand how he got his answer until you explained it. A uniform ρ means that it’s constant and can be left out of the integral. So it’s just -4πGρ on the left side. So I’m assuming then, that means the integral of d^3x is actually the volume of the sphere, right? So if you evaluated it, you’d get (-4πGρ)(4πR^3/3) = 4πR^2. You’d cancel the 4π on either side and be left with (-4πGρ)R^3/3 = R^2. Then you cancel again which gives you -4πGpR/3. That is equivalent to Newton’s equation.
He got the wrong answer because he left out the initial -4π.
I recommend to read Professor Susskind's books as well (the Theoretical Minimum Series), in my opnion before looking the lectures. I t is a big luck that a first class, premium, five star scientist with such a reputation obviously has fun and an excellent ability to teach his stuff. Thank you for making this publicly available. Kind regards from Germany!
Bro I am read
Bro I am read
it is my first mathematical lecture on general theory.... its very use full to me..😊
PRANK GONE WRONGPRANK GONE WRONGPRANK GONE WRONGPRANK GONE WRONGPRANK GONE WRONGPRANK GONE WRONGPRANK GONE WRONGPRANK GONE WRONGPRANK GONE WRONGPRANK GONE WRONGPRANK GONE WRONGPRANK GONE WRONGPRANK GONE WRONGPRANK GONE WRONGPRANK GONE WRONGPRANK GONE WRONGPRANK GONE WRONGPRANK GONE WRONGPRANK GONE WRONG
I love this stuff I could listen to it all day
PRANK GONE WRONG
So could i. But i still wouldnt understand
I want this guy as my professor.
Haha... I'm a MSc student now and I'm planning to apply for my PhD. But I'm not a physicist, so he won't be my professor either way I look at it.
not a problem - 75 000 USD per year and you in Stanford.
If you continue watching this series... Than he is your prof.
Is he still teach at Stanford, I really want to meet him, just once time =)
You're not alone.
Thanks to TH-cam and Stanford for posting these awesome videos :-)..........really worthwhile!.....oh and thanks to Prof Susskind for his lectures
the presentation is well organised and the lecturer too, i think he deals well with his students. i like the way he handle question, is been enlightening lecture.on relativity.theory.
for the field to be inside on the left hand side it should be--
(-4/3 πR^3 4πρG)=4πR^2 A(perpendicular)
then the missing pi can be located and the equation will become appropriate
+Tarun Kumar the field is worked out by volume? I thought surface area;4pir^3(as he explained it)
+deadmeat1471
p = m/v
m = pv
v_sphere = (4/3)πr^3
m = p((4/3)πr^3)
For the equations to be consistent with each other, the "m" on the right board should be able to be replaced with the "pv" equivalent expression above, resulting in "A_perp = -(4π*r^3*p*G)/(3*r^2) = -(4π*p*G/3)*r" but instead the left board only shows "A_perp = -(p*G/3)*r". That's where the 4π is missing on the left (assuming the right was derived correctly)
The error was in the right side of the equation on the left board. He equated A_perp*4pi*R^2 with the total mass inside the sphere (multiplied with G) and then expressed that as the volume of the sphere times the mass density.
The correct way to apply Gauss' theorem would have been to equate the left side of this equation with the volume of the sphere V times the divergence inside, which is according to Gauss' law -4pi*rho*G. Thr final equation looks like this:
A_perp*4pi*R^2=-4pi*rho*G*V
and using V=(4/3)*pi*R^3:
A_perp*4pi*R^2=-4pi*rho*G*(4/3)*pi*R^3
which has an additional factor of 4pi on the right side.
deadmeat1417 use "dimensional analysis" to know you are wrong here. 4pi is dimensionless, so 4pi r^3 has dimension (length)^3. So its a volume, not an area.
I'm not a student, I am just fascinated with physics and I love watching these to try to learn anything I can. Very good lectures!
Is cosmological constant sort of like the inverse of gravity or something? Gravity is very weak but gets very strong with mass, cosmological constant is very weak but gets very strong with distance. So, in order to live in a universe (or to find ourselves in one from the anthropic principle) in which matter is being held together due to gravity, you would ultimately have to have the repulsion, otherwise it'd collapse on itself, and as the universe gets bigger and bigger, faster and faster, the surface area and distances of it is larger therefore enacting the cosmological constant to grow larger?
Does that have anything to do with the formation of voids like Bootes void? To me it seems like that's why you get a filament or sponge like structure if you could zoom out on the universe as a whole.. all of the mass coalescing into valleys of spacetime, and voids bubbling up in relatively empty regions, which pushes matter out toward the valleys?
Sorry, I'm sure these are really dumb questions, just trying to figure it out!
I don't know, but I think that he said dark energy is a constant, so it doesn't get stronger.
@@itchyandscratchy7350 The cosmological constant would get stronger as the time passes, I think Prof. Susskind explained it pretty clearly in the beginning.
Also, it's not really getting stronger by the distance but due to its extremely small value, the equation would only sums up into a considerable number when it's multiplied by a huge distance, hence it only matter when you accounted the entire universe for it, but not so much in an astronomical scale or even in global scale.
Calling it an inverse gravity is pretty much inaccurate because it's not exactly a force. Let alone a force that repulse each matter or particle from each other, it's something that drive the particle grow apart from each other.
@@Vandalgia
6:20
Professor Susskind tells that the theory on the "Big Rip" describes the cosmological constant to be time-dependant, therefore growing with time. As Professor Susskind reminds, however, that is not true ; the theory on the "Big Rip" violates all principles of physics, and so it is not true based on the given principles of physics.
I hope that clarifies Professor Susskind's point about the constant.
@@powahfulgameplayer Yeah, "stronger" might not be the correct word to describe this. As you stated it, it's more like, it changes as the time passes.
Dr Susskind is a national treasure!
My maths is so damned rusty..Yet, Professor Susskind makes lecture's substance sound very easy... Thoroughly enjoyable and interesting stuff.
٩ة٩٩خخت٩نت٩ةنك٩٩٩و٩٩٩خ٩٩٩٩٩٩ة٩و٩٩٩٩٩٩٩٩ة٩٩ت٩و٩٩٩ وننتظرن وون تت٩ ت٩ ن تت ٠ت ٩ ههههه ٩تو ت٩ خ٩ةتة٩ تت تتت تنخ٩ اه ٩٠نحت اه ٠ون٠٠ تت نن ٠٩٩ت٠٩ت نن نت٩ت٠تن٩٠٠ ٠ن٩٩٠ ٠ون٩
I tend to believe that the Felix Bloch professor of theoretical physics at Stanford University and his considerable contribution to string theory probably deserves a little more respect then that.. it is a privilege to watch this and thanks to him for his dedication to education.
To be onest he is real good teacher. Turning complex things to real simple (making it dumb) is high skill. Explaining something complex in a complex way is easy. What hard is turning complex things into a "cosmological constant" dumb level,requires high amount of experience and capeability.
Btw : the students having problem with perpendicular A is the mistake that the both examples are different and the perpendicular A is example specific. there are two different A not equal :)
It is the sin(x) = 2MG/Rc^2 with x=2MG/Rc^2 that is used for the aproximation, also there is other aproximation, that for the elapsed time delta t, it is taken as the horizontal component of speed of light is the same c, that can be appear as a logic circle (that is not), since for this is used the a posteriori known fact that the angle is too small, and in the former the use of values of G and c are used to aproximate sinx =x.
The 4pi is lost when he subs into Gauss's Theorem at 44:43. The right-hand side was previously calculated to be -4piGM (at 38:55), but he only subs into -GM, losing the 4pi.
The reason he messed up at 0:45:24 is that he is missing an additional factor of 4pi on the right hand side from Gauss' law: Del.A = G rho 4pi. Integrating this expression over the volume of a sphere leads to 4/3 pi r^3 * G rho 4 pi
Thank you!! The good professor goofed a few times in this lecture.
It takes an incredibly well prepared lecturer to throw out derivations on demand.
Austin Fahrenheit
or a very smart and passionate one
not discredit Pr Susskind here, but anyone with basic school knowledge about differentiating could do derivations
To follow them and understand them yes, but to remember where the equations came from takes an incredibly well prepared lecturer.
I suspect Austin is talking about derivations of phyical formulae not about derivations of mathematical functions a.k.a. derivatives.
Take note that in the discussion of the gravitational field at some radius R inside the earth taken from the center of the earth, that Leonard had dropped a 4pi. Perhaps you may have caught this. So, the g field at radius R inside the earth is actually: G*p*pi*R*4/3
Must be quite an honor to actually be on lectures of that guy.
compare the equations at 45:21 and 40:29. They each have a term A*4*pi*R**2 but they are equal to two different things: one is -4*pi*G*M and the other is just M*G. What happened to the 4*pi that was in the equation from 40:29? Am I missing something?
It just make me smile every time Prof. Susskind call the white board "black board". Other than that he's such a great educator. I wish i can attend his classes one day.
Thanks to Google, TH-cam and the internet technology researchers who made this possible to watch these videos. Otherwise I had no chance to see this lecture series.
good to see math kids all disagreeing with each other and one of the worlds best physicists, Gives innumerate biology students like me, hope!
the correct formula for the acceleration inside a body is a(R)=4*π*ρ*R/3 , the 4*π should exist on the formula @46.27
That disagreement from 45:00 to 52:00, the student was 100% right, the professor got it wrong. He left out a factor of 4pi from the right hand side. Gauss's Law&Theorem state that Aperp*surf. area == -4piGM.
So, at 45:00, the surface area is 4piR^2.
And the mass of the sphere is the volume of the sphere * rho.
The volume of the sphere is 4pi/3 * R^3.
Plug all that in, you get:
Aperp * 4piR^2 = -4pi * G * 4pi/3 * R^3 * rho. See the problem? When he wrote it down at 45:00, he forgot the first 4pi. He wrote down simply -GM rather than -4piGM. The student was right. After cancelling the 4pis from each side, you end up with:
Aperp * R^2 = -G * 4pi/3 * R^3 * rho.
So, Aperp = -4pi/3 * rho * G * R.
If you changed that back to mass instead of rho, you'd get back to the original Gauss equation:
Aperp = -GM/R^2.
greg55666 That is correct. The student was right to be confused by the result Dr. Susskind came to on the left board because M=V*rho. For A perp = -MG/R^2 as on the right board the equation should read -{[(4pi/3)*R^3]*rho*G}/R^2. The (4pi/3)*R^3 is the volume of the sphere so when you multiply the volume by rho you get Mass or M=V*rho. Now all you need to do is substitute V*rho for M in the equation A perp = -MG/R^2 and cancel out the R^2 in the denominator with the R^3 in the numerator. That leaves A perp = (4pi/3)*rho*R*G. I think the confusion here is that the student asking the question had no idea how to articulate the correction he was attempting to point out. Had he been able to say "I can't seem to make the 4pi cancel out like you did due to the 2 equations you set up not being equivalent between the left and right boards. I think you unnecessarily added a 4pi to the L.H.S of the equation on the left board." then I think Dr. Susskind would have immediately realized the error and corrected it. The fact that there were so many conflicting voices in the room and nobody was making any sense led him to just leave it as it was.
@@dylanhaymore608 in student’s defense, he was constantly interrupted by sycophants. I’ve watched this maybe 10x trying to find the error. These posts helped Ty
@@paulodonnell753 I'm glad I was able to help you understand. This material is already difficult to learn so it's even harder to notice when there are errors from the instructor. Good luck on your journey for knowledge!
Just love this guys voice and accent
Time 50:00 min
4π is also there in the formula on right side when particle is inside sphere
58:00 I don't understand the 4 pi missing thing either. can someone explain?
Dr. Leonard is enjoying himself when students question him.
Yeah. Well put. Spherical space about the center of mass is the reasonable conclusion. I forgot that the mandatory requirement that all mass be contained within R of the center of mass.
❤Thank you very much Professor Susskind.
I accedentally fell asleep when watching youtube, woke up and had watched the whole lecture one and was at the end of this one. Am I gonna be a genius now?
the description has: "discusses dark energy, the tendency of it to tear atoms apart" Susskind would be so pissed. He only spent about 15 minutes of the class explaining how unlikely it is for dark energy to rip apart atoms.
Well tendency can mean tendency not to. Still, kind of misleading.
About 26:35 into the video, he mentions unit vectors. Good. A very useful concept.
😂😂😂😂😂
At 1:13:13, it givs x = (1/2)g*t^2 as a formula for position of a (uniformly) accelerated object. That formula is only strictly applicable if the object started out at rest.
Unbelievable. I have had very little formal education, a lifetime of professional experience traveling and living around the world. Mind blowing knowledge - FOR FREE - presented by a world class professor. It's troubling TH-cam algorithm does not push this content and knowledge More; esp the younger generation. I remember visiting a tiny all girls(K-8th) school started by Irish nuns over a century ago in Peshawar a few years ago; an entire classroom of 6th grade girls were doing Calculus on a true blackboard. I didn't understand a lick of it. Mean while in US, my 5th grade son has not come near Algibra much less Calculus. I'm not complaining, just comparing. Kids living in extreme poverty in a extremely dense ancient city, non stop extreme violence all around, yet they can focus their young minds to unbelievable levels academically. Our kids are distracted by countless electronic devices, social media, video games. Yet it is on those devices and media that you can find a world class education 🤷♂️.
Due to the gravity of the subject matter in this lecture, I started this video 12 years ago, and just finished it today
@memyself1125 No. Satelites use it and Particle accelerators use it but until you get to speeds of (lets say 10,000 miles per hour) you do not use it.
That being said there is relative motion, which is helpful to the layman, but not a necessity.
These lectures are incredible.
Well, actually when the gravity field equation is derived, we assume a test particle with mass m different than zero. When the m term appears on both sides of the equation, we divide by that (thus it must be different than zero).
Therefore it is counter-intuitive to suggest that gravity has the same effect on a zero mass particle. On the contrary according to Newton the gravitational force exerted on a zero mass particle is zero.
Yes people are asking question like they don't know what they are talking about. And his style is like he is teaching for those who already have some knowledge.
Well, rho = m/V would be the case if the mass if uniformly distributed in a given volume. If the value of rho depends of the chosen volume we have to write rho = Delta m/Delta V.
I think this becomes more obvious if you take the limit Delta V -> 0. Then, the differences become differentials.
rho = dm/dV => dm = rho dV
Integrate over a given volume to get the mass included in this volume.
M(in the chosen volume) = Integral{rho dV} .
and rho in general could be a function of V.
19:55 divergence of a vector is scalar, not vector
He did say that the divergence was a scalar. I think he just misspoke at that instant.
I thought this was the final part of the explanation as I had simply opened this video, not the playlist, and TH-cam hadn't recommended any part 3 or bigger on the suggestions. :(
Since the divergance of the field equation is calculated in terms of Pi does that mean that the mass density can only account for spherical space?
Thank you professor..... One question was : why the effect of mass outside the Gauss surface inside the spherical mass be neglected or canceled ....my answer is I think that the mass is not neglected as you see the density of the field (the gravitation density g) be reduced by reducing the (r) to becomes zero in the centre of sphere which means that the effect of the outside mass which also cause (g) can be considered as superposition summation for the (g) vector at the center and that summation becomes zero when (r) become zero so the disappear of the (g) in the center due to the opposite directions of (g) , from the general relativity point of view also we can understand that the distortion of the metric tensor curvature outside the mass (density of energy) is totally different than when it passes through the it ...thank you
At 1.18 Susskind describes how a light beam would cross an accelerating elevator -- noting that the observer in the elevator would see a downward curving light beam (and an external ,stationary , observer , presumably seeing the light beam travelling straight and horizontally ).
And yet had the elevator been travelling at a constant speed (SR), the observer in the elevator would see the beam travelling horizontally within the elevator, and the external observer would see it travelling diagonally upwards.
Why does the external observer ,say, see the light beam being affected by the elevator's frame of reference in the SR case , but not in the GR case. ?.
+Metallurgist47 If I understand correctly the difference is in the original vector of the light. With the accelerating frame the assumption is the beam is emitted at the same time the acceleration starts and so it's direction is horizontal. In the constant velocity case, since the velocity is constant, you have to assume that the elevator is in motion at the beginning of the experiment and the beam is emitted from a source inside the elevator and therefore it starts with the upward component already. Also, the curvature is only seen by the observer inside the elevator. To someone outside it travels perfectly horizontally.
Thanks a lot to Standford University, it's dificult to find the RG theory...
thanks from Peru. Greetings
Is it a coincidence that the operation of applying an operator to a function looks the same as the operation of multiplying two values? Both are written with simple concatenation.
Maybe a number is identified with the operation of multiplying by that number? It could also be identified with the operation of adding that number, and I don't see a principle that says to pick one over the other.
I've always wondered at the assertion that the elevator acceleration is indistinguishable from gravity. Namely, the elevator acceleration field has no divergence, it is everywhere (in the elevator) parallel. However, a gravitational acceleration field has slight convergence, due to the Earth's mass acting as a point mass, with the gravity field radially divergent from there. Thus dropping 2 balls from opposite sides of the elevator, maintain their separation in an elevator acceleration field, but will be slightly closer together at the floor in a gravitational field. Therefore, there is a difference between the two fields.
The _principle_ _of_ _equivalence_ and the modern theory of gravity from Einstein does *NOT* postulate that acceleration and gravitation are physically indistinguishable and universally applicable.
Nope, there was a 4pi missing. Should be for
R>R_0: A=-MG/R²
R
00 lol you can 00 lol 00 00 loll 00 like 00000000000000000000000)00 0000lolllll
Thank you Stanford!!
Session starts at 14.20
PRANK GONE WRONG
14:20
It is impossible to overrate the quality of this lecture.
Question about the Einstein’s elevator @ 01:00?
Since the direction of the weight is always in the direction of "g" or towards the centre of gravitating mass therefore in which direction a person would feel his weight in elevator ?
Similarly, gravity bends light towards gravitating mass therefore why the direction of the bent light is against the direction of "g" in elevator?
Just wondering: Shouldn't a ball or person move in the direction of acceleration "g" or towards the centre of elevator due the universal law of gravitation [F= GMm/r^2].
@about 50:00: Would the inside person or object feel "g" if the elevator is consider as a shell?
Einstein's work is riddled with errors.
That was an analogy thought up for Special Relativity, on the subject of 'time dilation, not General Relativity.
But does that suggest that the divergence is a relative scalar?
good lecture
The reason the student is getting confused in this lecture is because Susskind uses big R on both the left and right board, but they are not the same, R is < than the one on the right because the shell is inside the earth, when you break the rho for density down, that brings out the actual earth radius that is used on the right board.
So there is a newtonian aquivalence Principle( if an object gets pulled by a bigger mass the acceleration is not dependent of the mass of the smaller object) and Einsteins aquivalence princple( when you accelerate a rocket with 9,81m/s^2 it Feels like earth‘s Gravity Field for someone in that rocket )?
How do you justify that? It will be that if rho does not not depend on the volume. Otherwise, you have to calculate the integral to find a relation between M and V.
We heard in the last video that the gravitational field is just a field of acceleration, because the acceleration is independent of the mass of the test particle. The test particle can be any mass at all, a planet or a single molecule, and the acceleration it undergoes will be the same. If I accept that (Newtonian) view, then why should I be surprised that gravity bends light? Surely it is just a particle of zero mass, being affected by the field just as any other test particle is?
Maybe particles are more pushed together than pulled to each other and that's why it increases as you pull them apart, space "pressure" so to speak pushing them together.
I suppose that this is second year subject, but Student are asking question like they did not complete any university maths, or physics before. Gauss's Law is and Gauss theorem is usually first year mathematics, and they usually had to overcome difficulties with new concepts.
52:42 The equations in the 2 cases don't match coz a 4pi was accidentally left out which comes from the Gauss's Law
Time = vector quantity (foward)
Clockwise = could be moving to the right like of the normal clock we have but some people do sell clock that move to the left. As long as it is 360 degrees.
So I think there is nothing like 'anti'
Ant = se-mut
A-rah = direction
Jam = clock
What is wise?
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-wise adv combining form. 1 indicating direction or manner: clockwise; likewise. 2 with reference to: businesswise. [OE -wisan (italic); see WISE2]
wise1
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2 prudent; sensible.
3
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7 Arch. possessing power or magic.
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wise2
n Arch. way, manner, fashion, or respect (esp. in any wise, in no wise) [OE wise (italic) manner]
i absolutely appreciate your videos they are amazingly useful
@pi05pm3 It is! It's extremely helpful to have free open-courseware like this available, especially since I'm planning on doing my PhD in theoretical physics next year ;)
Have u completed your PhD?
@@mdnazmusshahid4827 yes, I finished in 2015. :)
Would u mind to let me know ur topic and workplace please.
me also interested in doing PhD but little confused about topic and outcome of doing PhD.
@@mdnazmusshahid4827 if you're confused about what you want to do for your PhD and where you want to do it, you should speak with someone at your university to get some guidance. Probably start with a lecturer who is in the field or close to the field you want to study. My topic was solutions of the self-interacting Maxwell-Dirac equations.
@@fermista Thanks for ur suggestions.
What an amazing lecture!
When I look at the shadow of an electrical pole at about mid- afternoon I see a well defined shadow edge near the base but when I follow the shadow away from the pole the edges start to blur. This suggests that either light is attracted to dark or repelled by other light. No gravity effects here? Moon shadows are much crisper
That's just caused by the fact that the light source has dimensions. Light from the edge of the light source is able to "see" further behind the object than light from the center of the light source. No repulsion or attraction necessary
@@gamerdio2503 I understand what you are saying but noonlight doesn’t dissipate the way unreflected sunlight does
@@abberepair8288 What do you mean moonlight? You mean the shadows on the moon itself?
@@gamerdio2503 when I walk at night the dividing line of shaded to unshaded is sharper, more crisp than in the day light. I can see a very distinct shadow line at night compared to the shadow line I get from the sun. It’s like the photons in sunlight are repulsed from each other where the moonlight is running more parallel
@@abberepair8288 Well, moonlight *is* sunlight, so that idea is thrown out the window. I think the answer might come from the fact that there's less light reflecting from the moon, making it harder to make a distinction on the edge. In addition, the fact that moonlight is reflected light and not a light source itself might have an effect
For Gauss' Theorem, i don't get how the example on the left board(calculating A within the earth) is using gauss' theorem. why do we take A4piR^2= Mass within sphere x G ? Shouldn't it be = -4piGM instead?! Please help me someone! Thanks.
Because that isn't Gauss' theorem. You can't just stick a factor of 4pi into it for no reason.
Gauss' theorem simply states that the flow of 'something' (anything) in or out of a region is equal to the amount passing through the boundary. That's actually a pretty obvious statement really when you think about it.
In mathematical terms the same thing reads 'the volume integral of divergence within a region is equal to the integrated surface flux' That's maybe not so obvious!
What adds a little confusion is the 4pi introduced in the equation for gravitation. Essentially it's a modification of 'G' to allow for the fact that G was originally defined by Newton in terms of one dimension but we're now working in 3 dimensions. We get similar factors of 4pi in electromagnetism for the same reason.
That factor of 4pi cancels out when you use Gauss' theorem (it HAS to or nothing makes sense - you'd have more water running down the drain then you put into the bath :)
Best course ever
the student who was confused was right susskind was off by a factor of 4. because -4piMG=A4piR^2. susskind started his equation as A4piR^2=-MG
Hi, wonderful video. but a question: there is a book for follow this lectures?
i want to have also a book for Einstein relativity Special and general.
Thanks all and regards from Italy :-)
there is one called Special Relativity and Classical Field Theory: The Theoretical Minimum . It is coming out in September. Susskind wrote other books such as 'Quantum Mechaniics'
Frame of reference connected in time....
Electron jumps..position superposition...
Parallel so ....electrons are paralle to each other...not disappear gband appearing another place..
Photons bend because of mass down?
not sure, but if you have a isotropic distribution of mass (convergence of gravity), the only shape the system can take is a sphere. Simply because at every point on the edge of the sphere it's the same distance to the center.
as of this writing
Lecture 1: 5,504,000 views
Lecture 2: 938,000 views
To say that the "father" of string theory (read: modern particle physics) knows nothing about dark energy is quite a statement.. However, from what he's saying the expansion is increasing, as the repulsive force grows with distance.
Well, yes, you'll get the average density of the particle, but you will not get the same rho which is a function of position.
Also, without integrating rho over space you don't know the total mass M (unless it's given), and you can't calculate M over V.
If I told you there is a spherical object at the origin of radius 1 and rho of r equal to |r|, what is its mass and average density? You can't really solve that without integrating.
37:09 - a piece of art work: a little sitting bear he has candidly drawn 😊
@stenniswood if space itself is converging into point masses then where is it going to? Another universe? As raw material for the expanding edges of our universe?
45:20 He remembers the G but forgets the original 4pi from Gauss' Law.
45:20