Didn’t you miss out a - sign near the end of the derivation of Newton’s equation? Force is the negative of the gradient of the potential.

I just would like to say thank you to you. This is literally the best online materials I could found on motion and dynamics

@drmitchellsphysicschannel2955

2 жыл бұрын

Thanks! I'm really glad the lectures are useful

Interesting to see the E/L 'derived' in terms of Lagrangian as opposed to the purely mathematical derivation. Enjoying this series very much. I went back to Euler's first geometric derivation and split it into position / gradient videos then derived it analytically as Lagrange did in a third video.

thank you i have atrophied all points great refresher and more clarification still a lot of work to do practices inertia

52:19 How do you differentiate L wrt vector quantities here?

at 47:44 shouldn't there be a minus in front of mr''? dS=int(-mr'' ... like this?

@drmitchellsphysicschannel2955

2 жыл бұрын

Yes, thanks for the typo. Both terms in the square brackets have the same sign. Since the terms in the square brackets are together equal to zero, we recover Newton's second law.

Really nice video! Thank you very much. But what I'm missing and what I initially searched for, is an intuitive explanation, on WHY the Lagrangian is T-V. Maybe I'm missing something, but it's not clear for me, what the difference in the Energies has to do with the Action of the Path. Does anyone have a link to an explanation?

@drmitchellsphysicschannel2955

3 ай бұрын

Well, using L=T-V is equivalent to F=ma. One could similarly ask WHY is it F=ma? There are infinitely many logically consistent theories of the universe... but only one of them is the correct one that we observe in Nature. WHY is this the right one? I think there is no answer to that. I don't think an intuitive answer really helps, philosophically, since in the end your intuition is just a learned expectation based on observation. Indeed, as I discuss later in this course, when you take into account special relativity, the Lagrangian is NOT just L=T-V, so anything based on intuition to explain this result must fail at a more precise level of theory. Special relativity is not understood intuitively. Similarly with quantum mechanics etc.

Could you explain why L=T-V

@ziyodullaziyodulla8096

9 ай бұрын

Great question that I have

This is really excellent explanation. I have a basic question: where does L = T - V come from? I mean, how did Lagrange comes up with this?

@drmitchellsphysicschannel2955

8 ай бұрын

It's a very good question! The principle of least action is quite general... but the actual expression for the Lagrangian is not set by this. L=T-V is chosen to reproduce Newton's second law. You just have to reverse engineer it to see that, as we do in this lecture. But if you want to put in Einstein's special relativity, you have to change the form of L a bit. We do this in one of the later lectures and explain a little more deeply where L comes from. You get different Lagrangians when you try to describe different kinds of physics. But there are some guiding principles to help us, such as relativistic invariance, gauge invariance, locality, causality...

I got very confused at 55m0s where you say "this equation is great because it only involves scalar quantities, not vector quantities" - because the Euler-Lagrange equation you derive has "delta r ->" and delta r dotted ->" in it. Even a small change in a vector, is still a vector, isn't it? BTW, I'm not a mathematician or physicist, but I still really enjoyed this, so thanks!

so, ds in definition of action integral is function of all the points in a straight line defined by (z,dz/dt,t)?

@drmitchellsphysicschannel2955

2 жыл бұрын

Yes, it is a functional of the whole path. For a system with one degree of freedom z, the Lagrangian L is a function of z, dz/dt and t. One could plot the particle trajectory in the (z,t) plane (for example a mass being thrown upwards under gravity follows a straight line but has a parabolic trajectory in the (z,t) plane) and the gradient is dz/dt. The Lagrangian is obtained along the path. The action S is obtained by integrating along the path from start time to end time. The infinitesimal dS is obtained along a very short segment of the path.

I am a french woman I like your confernces and your point of vue philosophical but in that demonstration of POLA at 47.20 i have the same question than cesar Prodan one year ago you said " move the dot from here to here at the expense of this minus sign " But why? What's the reason of that ? Thancks for your answer

@user-wi3lg1rz8f

20 күн бұрын

Because of Integration by parts.

When you say locality in space, do you just mean that only the fields at the particle's current location can affect its motion? I have no problem with that, as I can imagine standing at some location far away from the particle and pulling apart two charges which are on top of each other to create an electric field, whose presence will not be immediately felt by the original particle. It all just seems like a truism though.

@drmitchellsphysicschannel2955

3 жыл бұрын

Locality means that only the fields at a particle's position and those at nearby positions can affect its behaviour. This is because information travels at a finite speed through space (the maximum speed is the speed of light). More precisely, when we say "nearby", we mean fields an infinitesimal distance away can affect the particle after an infinitesimal time. Put another way, at a given instant in time, only the fields and their first spatial derivatives can matter. So it's a bit more subtle. In some sense it is a truism... I would rather say that locality is an *axiom*.

12:25 why did you consider at=0 when t=T?

@josuelima5033

3 ай бұрын

unidimensional movement in z direction, like a vertical throw

47:11 can somebody plz explain me how r*(deltar) vanish ?

@felixgep2608

3 ай бұрын

At 34:05 we observe, that delta r (t_1) and delta r (t_2) must be equal to zero, because we start at the same point in space and we end at the same point in space for any path we take. Since the formula of partial integration says, that we have have to evaluate the boundary term at the boundaries of the integration, we have to evaluate delta r at t_1 and at t_2, therefore these terms are 0.

You have missed a MINUS sign.

You lost me at 11:56 when said b = 0 means z = -½gt² + at + b can be written as a completely different equation: z = ½gt ( a' - t ) please explain how got this new equation? It seems to be a complete non-sequitur.

@gibbogle

8 ай бұрын

Since the constant a is arbitrary - unknown until it is fixed by the boundary conditions - you can write a = ga'/2, where a' is just a scaled by a constant factor 2/g: a' = (2/g)a. It is not necessary to change the constant of integration like this, you can follow through the solution using a, and the result will be the same. He did it like this for brevity.

Tough to take

You are assuming that 𝛿r(t) is a very small perturbation. It looks like you are also assuming that d(𝛿r(t))/dt is also a very small quantity for all t from t1 to t2 so that (d(𝛿r(t))/dt)^2 can be ignored. I am a bit worried about the higher order terms, as limits are not being taken to calculate a derivative at the end of the derivation.

@drmitchellsphysicschannel2955

2 жыл бұрын

In the end we are always taking the limit of small deviations becoming infinitesimals, either in a derivative or under the integral. The principle of least action basically means dS=0, a stationary point in the action. So the first order variation vanishes, that's all. There is still a finite second order variation around the minimum, but we don't need this to obtain the Euler-Lagrange equations of motion.

@eamon_concannon

2 жыл бұрын

@@drmitchellsphysicschannel2955 I think a clearer way for me to understand this derivation is to consider ΔS(λ) = S(r(t)+λ𝛿r(t)) - S(r(t)) where ΔS is treated as a function of the scalar λ only as we assume that r(t) is a given extremum of the action S. We can say S(λ) = S(r(t)+λ𝛿r(t)) so S(0) = S(r(t)). Hence, limit ΔS(λ)/λ as λ approaches 0 is 0. The derivation is very similar but it turns out that there is no need to assume anything about the size of 𝛿r(t) or d(𝛿r(t))/dt or to deal with higher order terms.

@yousufo.ramahi126

2 жыл бұрын

I believe the parameterization in terms of the scalar is how perturbation theory in quantum is usually taught (perturbation in the Hamiltonian).