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Thank you very much.

We don't drive it. We define it. Just like we have defined the energy operator, right in the Schrödinger equation. The easiest way to get to the definition is to take a wave functions ψ=Ae ^i(xp-Εt) and differentiate it wrt x. Then, you'll immediately see how the momentum operator pops out.

Physics is fun😊

I wish you reply p,x’s poisson relation with ih.

C^k means defined and continuous, doesn't it? Not just defined

1:00

Yes! Thank you very much. I did look at a few videos on the momentum operator but none were as clear as this one. A bit cranky and untidy but that’s part of the fun 🎉 . Dr-J

Thank you !

When it comes to faithful homomorphisms (including group actions) somehow it clicked for me when my professor said - that homomorphism has no kernel (meaning the only element in the kernel is just a neutral element).

"I happen to have a linear map here." - pulls out toilet paper tube 🤣 (I might try that during my thesis defense)

Thank you so much 😭 This video was really helpful

based

Really appreciate you taking the over pedantic description out of your explanation, super easy to follow along

Please exercises

I could not find any other source or reference that describes the maximum or net force of the impact in terms or Fi + Fg... rather all of them say the F = m*a*(h/d)... Can you provide another source that talks about this inertial frame of reference with respect to a falling object?

Newton's 2nd law can also be derived from the Euler-Lagrange equations where IBP is also involved.

Bro why did you stop making videos for this playlist it is a life saver please continue the playlist please 🙏🥺

I DO NOT BELIEVE THAT, the RATIO of. " h / d " is the. REAL CORRELATION OF NET IMPACT .. . THAT. IS. ONLY. A. FANCY. EQUATION, that gives unTRUE RESULTS .. . I MADE MY OWN RESEARCH ABOUT. THE " CODES OF ENERGY TRANSFORMATION ", TO ACCURATELY DERIVE THE TRUE IMPACT ,, , coz everytime I drop the ball I get different values of " d " from the same height and same amount of Mass.. . That is why using the REBOUND DISTANCE ( d ) is not an APPROPRIATE CORRELATION TO HAVE EXACT AMOUNT OF ENERGY. THE ENERGY OF POTENTIAL WORK AT REBOUND DISTANCE ( d ) , ,, IS NOT A COMPLETE TASK BECAUSE THAT ENERGY OF WORK in that TRANSFORMATION HAS EXPERIENCING ENERGY DISSIPATION coz some Shocks of IMPACT WAS DIRECTLY ABSORBED BY THE GROUND OR FLEBILITY OR RIGIDITY OF OBJECT BEING DROP ... TAKE NOTE THAT EVERYTHING WAS CREATED BY GOD WAS CLOTHE WITH THE " ENERGY ", AND THE ENERGY IS THE STRENGTH AND SHIELD OF PROTECTION OF THE MATTER TO SUSTAIN ITSELF OF ITS EXISTENCE, AS LONG AS THE STRENGTH OF THE OBJECT DID NOT MATCH THE EXTERNAL FORCES THAT CAN DESTROY IT, IT WILL NOT FACE ITS EXTINCTION .. . AND PERFECT WAY TO MEASURE THE ENERGY OF IMPACT THRU POTENTIAL WORK, IS TO MEASURE THE DEFORMATION OF THE GROUND .. . AND REMODEL THE SCALE OF IMPACT USING WEIGHING SCALE THAT WILL YIELD SAME SETTLEMENT OR DEFORMATION THAT WILL BE LOADED BY EQUIVALENT MASS OF GROSS IMPACT... OR REMODEL THE DEFORMATION OF OBJECT BEING DROPPED.. . IF YOU DRAW THE ACTUAL CURVE OF IMPACT DONE BY TEST EXPERIMENTS, IT IS NOT A LINEAR EQUATION, BUT IT IS SOMETHING LIKE A PARABOLIC OR 2ND DEGREE EXPONENTIAL CURVE ... AND I ALREADY DERIVE THAT PERFECT FORMULA FOR CURVE OF IMPACT FORCE .. .

to much hand waving then again your catering to physicist

Amazing video. You got a new subscriber.

Sir upload more video

Awesome series. Really enjoying it so far. I don't know if I'll be able to keep the smirk off my face the next time my prof says unfaithful representation.

thank you but i think you messed the thing more

Thank you for this. Deriving this was a homework question on our first ever QM hw assignment (it wasn't even hinted at in the book). I never would have figured this out on my own.

I have never heard of stopping distance = 1ft. This seems incredibly facile, at best. To me, the whole equation falls to pieces when you make generalizations like this. SO many factors are playing around with each other when a body (be it yours or your friend's or a length of wood being rigged) comes to a stop, I can't see working with numbers that make such recklessly static assumptions. Where's rope stretch? Diameter? Systemic friction? Materials present? This topic is interesting, but not at all useful pragmatically for a practicing arborist. Theory versus praxis.

i think am gonna pass manifolds course after watching this video

What is the difference between velocity and speed in this context?

How the hell does the integral of psi*psi goes from negative infinity to positive infinity goes to zero when for a normalised psi when it's literally defined as equal to 1 for a normalised function as the probability of finding a particle from positive to negative infinity is one .. this makes no sense and its driving me crazy ughh 🤬🤬

Thanks it really helped me out :D

So cool. New climber but I've worked under my father, who's been a climber for 30+ years, for about 4 or 5 years total now. I've been roping all that time. He's a master arborist and has always stressed that physics is vital to a safe take down. Thanks for the video!

I spent some time this week trying to derive the 20% rule for dismantling a spar. ie when an arborist cuts down a spar, each piece should be 20% of the total height of the spar for it to land flat after a 3/4 rotation. I got height of last cut off the ground (s) is 4.45X length of the log(l), which is pretty close to the ideal 4.0, but I wonder if I did it correctly. I used total energy available in the first 90 deg, turning that into the rotational energy of a log spinning from one end PLUS the kinetic energy of the center of mass moving downward. I got w=sqrt((12g)/7l). That gave me angular velocity (w in radians), and downward velocity was wl/2 when the log was at 90 degrees, where l is length of log. From there I calculated how high the arborist has to be to allow another pi radians of rotation, which worked out to s=l*((7pi^2/24)+pi/2). I'm wondering whether at liftoff I should have made note that the moment of inertia changes from (ml^2)/3 to (ml^2)/12, but when I tried that I got a completely wrong answer. Reason? When the log leaves the stump, it should instantly begin to rotate about its center axis, so angular velocity should increase to conserve momentum, like a skater who pulls her arms in. Why doesn't that work out?

this is such a great vid, thanks so much!

Which sources did you use as research material?

What a great playlist, I can't believe it has so little views. I feel like I found the holy grail since I need to prepare a 90 minute talk about this. Thanks!

One comment: the Eg = mgh more accurately is Eg=mg(h+d). And you are assuming objects are rigid. I think the force you feel on impact if Fi, not Fnet. Otherwise when you place an object on top of your head and this is in static state you wouldn't feel any force (Fnet is 0).

I believe there is a mistake at the equation at time 03:47 Instead of phi inverse on the LHS it should be psi inverse. The idea of this is to show that smoothness is independent of chart.

dang I know next to nothing about topology and even I almost understood this, nicely done!

This is simply a consistency check.

instantly subbed, thankkks!!

Love how entertaining you've made this (on top of representation theory already being a blast, obviously). Thanks!

You did it, you son of a b****😂

Thank u

Me: "Why is that definition so over the top and complicated?" Him: "and we know how maps work from R^n to R^k" Me: "Ahhhhh" Well thank you Sir! :D

Excellent

Wow, for only ~350 views, this video is extremely well done. Nice work for the vid, and apparently nice work that night too. ;) P.S. When you said "When someone gives me this, I'm like... what?" I felt that.

Dude, wonderful video. Thanks so much

You sir are a hero :) thank you so much <3

Super short videos are super good compared with the half an hour videos

The extra explanations (which textbooks lack) is good for understanding.

Yes! Physics for Arborists that's great! when using this formula for negative rigging - would "h" be the distance of free fall or the total distance until completely stopped? I've been thinking a bit about physics and negative rigging lately, since the distance is measured from the piece's center of gravity - a long piece would fall a longer distance before the rope can stop it, but I don't really have a good understanding of how to calculate forces in a tipping motion, have you ever thought of this?