You’re the best!

This is so underrated. Great job, Noah!

now finally i have video of Feynman lecture too !!! nicee

17:25 the rastafari equation. Thank you for taking the time to make this video.

Excellent!!! Thank you 🙏🏼

Wow! I never really understood why the integration by parts is done so. We used to memorize this as students!

Incredibly well explained.

@ Michał Prządka : The Taylor expansion seems correct to me. First a comment on the equation from Noah's comment : In f(x + E) = f(x) + E*f'(x)+.., the ' can mean deriving by x or by E, which amounts to the same thing I think. Define g(eps) = S(x + eps*eta) Hence g(0) = S(x) To find the value of g(eps), we can use the Taylor expansion ('meaning deriving by eps) : g(0+eps) = g(0) + eps*g'(0) = S(x) + eps*S'(x) = S(x) + eps*(S'(x+eps*eta) at eps=0) That said, I have many issues with the derivation. I have a question too : That principle of least action looks false. It is easly to find unrealistic paths that have a lower action than the real path. For example, in a gravitational field, have a particle follow a higher path at slower speed. Hence the kinetic energy will be lower, the potential energy higher and the travel time longer. In order to meaningfully talk about a minimum, one must consider a set of paths, all of which violate the laws of physics, except for the real one. In what way may the laws of physics be varied to form that set ?

@knarfamoranemix6030

4 жыл бұрын

An answer to my own question : My proposed path with lower action is not allowed, for it would violate the time constraint, as the time interval is fixed. The paths under consideration are world lines, which are paths in spacetime and have I suspect very few limitations beyond the constraints and being continuous. Maybe speed needs to remain finate too.

I'm confused at the particular step demonstrated at 20:53. The first order approximation uses the derivative evaluated at the x(t)+EN(t). Why isn't it just x(t)? I thought the expansion was around x(t), so the derivative should be evaluated at just x(t). For instance, you wrote above that, f(x+E)=f(x)+Ef'(x)+.... Here f' is evaluated at x, rather than x+E. Also what is f' wrt? Is it df/dx? Or df/d(x+E)? I'm following the formula, f(x)=f(a)+(df(a)/dx)(x-a)+... where in the former case, x=x(t)+EN(t), and a=x(t). As for the latter case, x=x+E and a=x.

@noahexplainsphysics

5 жыл бұрын

The formula for a Taylor expansion is f(x) = f(a) + (x-a)f'(a)+... If we replace "x" with "x + E" and "a" with "x", this becomes f(x + E) = f(x) + E*f'(x)+...

@przadka

4 жыл бұрын

NoahExplainsPhysics to me this part is also confusing. If we follow your answer in the comment above, in the video I think we should replace E with epsilon*eta(t) but this is not what you do in the video - in the second term on the right hand side of the equation you have just epsilon and not full E(=epsilon*eta(t)). Am I missing something? The problem I see is that in your Taylor’s formula f(x+epsilon), the epsilon has a different meaning that the epsilon in the formula for S. I the formula for S epsilon is simply a variable that S(ie action) depends (like plain old x in a single variable calculus. And you want to evaluate S(epsilon) knowing S(epsilon) at point epsilon=0. And now you can use Taylor’s formula f(epsilon) = f(0)+f’(0)*(epsilon-0)+... which is pretty much exactly what you have on the right hand side of your equation in your video. PS. Thanks for the videos, they are great!!!

@knarfamoranemix6030

4 жыл бұрын

A way you can see that using eps*eta as a coefficient before the second term of the Taylor expansion would be wrong, is checking the dimensions of the terms. In a sum all terms must have the same dimension. That way one can see that the coefficient must have the dimension of eps (which is dimensionless). Eta has the dimension of meter.

@gytoser801

2 жыл бұрын

Because it is a property of functions. Considering f'(x) = 0 you get with Taylor f(x+e) = f'(x)

39:07 there should be eta dot not just eta. Great video thank you.

What if the bracketed part in 51:55 (claimed to be 0) were a trigonometric function which has the same function values at t1 and t2, divided by eta(t). Wouldn't that also leave the integral to be 0? On what grounds could we reject this?

@GlockenspielHero

Жыл бұрын

The premise of the proof is that dS = 0 for any eta(t). Otherwise, xbar + episilon*eta(t) or xbar - epsilon*eta(t) will be lower, and xbar will not be the global minimum of S. Thus, the bracketed part must depend on a constant solution xbar. If xbar were a function of one eta(t), the argument doesn't hold up - what works for one choice of eta(t) won't work for another!

@nupursarkar8397

Жыл бұрын

@@GlockenspielHero can you pls explain 52:18 that part in detail I'm a little confused as eta(t1)=eta(t2)=0 and everything =0 if we take eta(t) common in the third line just as the same we took in 2nd line . kindly explain.

tahnkyou

@masgar2008

4 жыл бұрын

Indeed love to watch your videos. Fabulously explained. Great sir. I follow your explanation to make my student understand GRA8

46:66