Lagrangian coordinate transformation (Noether)

[gionole]

This question has been bugging me and not sure how to ask it in an understandable way, but I will try.

Imagine we have $L = \frac{1}{2}m\dot x^2$.

If we do: $x(t) -> x(t) + \epsilon(t)$, we transform the whole trajectory and We will arrive at $\frac{dP}{dt} = 0$ (momentum conserved).
Even though particle won't move on the transformed trajectory, we still do the transformation. The way we arrived to momentum conservation is by the same way as Euler lagrangian derivation as such $x(t)$ is a true path, so its action must be minimum, so action of $x(t) + \epsilon(t)$ must be less. The only way this is proved is by "action of true path" must be minimum. Is there other type of proof ? The reason I'm asking is with this proof, it's like we're saying, "yeah, we know particle moves on $x(t)$ path, but what if it used a different path($x(t) + \epsilon(t)$), and then we show it can't and arrive at momentum conservation. So it's NOT the kind of proof where particle was at $A$ and then moved to $B$ and momentum didn't change and then moved to C and momentum still didn't change, because since the whole trajectory was transformed, particle wouldn't appear on $C$ as $C$ is on the old path.

So to me, this doesn't seem like understandable for some reason. The better way would be that we know particle moves on the true path $x(t)$ and on that path, there we got $A, B, C$ points and we show when particle moves between them, momentum is conserved. That would be the real, more understandable way, but I guess, we can't do that.

So why are we saying: "when we do transformation, momentum is conserved ?" - particle doesn't actually start using new $x(t) + \epsilon(t)$ trajectory at all, so what's the point of saying the "transformation" at all ? I understand the derivation but not the intuitive idea of considering transformation. I think this "transformation" is the same thing as using the "action must be minimum" over the true path and nothing else - so it seems like the mathematical trick and not the intuitive idea of "coordinate transformation".

Hope I explained well enough, if not, let me know.

 

[topsquark]

This question has been bugging me and not sure how to ask it in an understandable way, but I will try.

Imagine we have $L = \frac{1}{2}m\dot x^2$.

If we do: $x(t) -> x(t) + \epsilon(t)$, we transform the whole trajectory and We will arrive at $\frac{dP}{dt} = 0$ (momentum conserved).
Even though particle won't move on the transformed trajectory, we still do the transformation. The way we arrived to momentum conservation is by the same way as Euler lagrangian derivation as such $x(t)$ is a true path, so its action must be minimum, so action of $x(t) + \epsilon(t)$ must be less. The only way this is proved is by "action of true path" must be minimum. Is there other type of proof ? The reason I'm asking is with this proof, it's like we're saying, "yeah, we know particle moves on $x(t)$ path, but what if it used a different path($x(t) + \epsilon(t)$), and then we show it can't and arrive at momentum conservation. So it's NOT the kind of proof where particle was at $A$ and then moved to $B$ and momentum didn't change and then moved to C and momentum still didn't change, because since the whole trajectory was transformed, particle wouldn't appear on $C$ as $C$ is on the old path.

So to me, this doesn't seem like understandable for some reason. The better way would be that we know particle moves on the true path $x(t)$ and on that path, there we got $A, B, C$ points and we show when particle moves between them, momentum is conserved. That would be the real, more understandable way, but I guess, we can't do that.

So why are we saying: "when we do transformation, momentum is conserved ?" - particle doesn't actually start using new $x(t) + \epsilon(t)$ trajectory at all, so what's the point of saying the "transformation" at all ? I understand the derivation but not the intuitive idea of considering transformation. I think this "transformation" is the same thing as using the "action must be minimum" over the true path and nothing else - so it seems like the mathematical trick and not the intuitive idea of "coordinate transformation".

Hope I explained well enough, if not, let me know.

The whole point of this method is to analyze symmetries and find expressions for conserved currents, so which path is taken is not always particularly important.

However, the method was derived using the concept of the Euler-Lagrange equations of motion, from which we may derive the path x(t) if we like. It is not usually known beforehand.

-Dan

 

[gionole]

```

The whole point of this method is to analyze symmetries and find expressions for conserved currents, so which path is taken is not always particularly important.

However, the method was derived using the concept of the Euler-Lagrange equations of motion, from which we may derive the path x(t) if we like. It is not usually known beforehand.

-Dan

1. To be honest, Unfortunately That doesn't answer my confusions/worries.
2. I derived it using "action" must be minimum(not by euler lagrange).
3. To repeat my queetion:

So why are we saying: "when we do transformation, momentum is conserved ?" - particle doesn't actually start using new $x(t) + \epsilon(t)$ trajectory at all, so what's the point of saying the "transformation" at all ? I understand the derivation but not the intuitive idea of considering transformation. I think this "transformation" is the same thing as using the "action must be minimum" over the true path and nothing else - so it seems like the mathematical trick and not the intuitive idea of "coordinate transformation".

 

[topsquark]

```

1. To be honest, Unfortunately That doesn't answer my confusions/worries.
2. I derived it using "action" must be minimum(not by euler lagrange).
3. To repeat my queetion:

Sorry about that. I missed that you were using the action principle.

Still, the idea is the same. We assume that there is a well-defined path, x(t) the describes the equation of motion, vary that path by adding $\epsilon (t)$, and the action principle tells you how to minimize this and find x(t). It is, in every way, equivalent to Euler-Lagrange, but it's slightly easier to generalize. So the principle is the same, even if I was talking about the wrong method.

-Dan

 

[gionole]

Sorry about that. I missed that you were using the action principle.

Still, the idea is the same. We assume that there is a well-defined path, x(t) the describes the equation of motion, vary that path by adding $\epsilon (t)$, and the action principle tells you how to minimize this and find x(t). It is, in every way, equivalent to Euler-Lagrange, but it's slightly easier to generalize. So the principle is the same, even if I was talking about the wrong method.

-Dan

I agree but what I'm emphesizing is that the word "coordinate transformation" is kind of confusing, because in reality, particle is still moving in the same path, not transformed path. Coordinate transformation to me is the same thing as trying to find equation(with the same philosophy of finding the minimal action) that describes the equation of motion and while doing that, you just end up with conservation.

Question 1: do you find everything I just said true ?

Question 2: Wouldn't just using euler lagrange get you the conservation of law as well ? why are we in need to just talk about "coordinate transformation" ? Maybe there is a better proof instead of using "action minimal" rule ?

Question 3: I noticed that in the video when we do $x(t) + \epsilon(t)$, we do it such that $\epsilon(t_1) =\epsilon(t_2) = 0$. So it's not fully the transformation as we resrict starting and ending point be the same hence another reason why I don't like the "coordinate transformation".

 

[topsquark]

I agree but what I'm emphesizing is that the word "coordinate transformation" is kind of confusing, because in reality, particle is still moving in the same path, not transformed path. Coordinate transformation to me is the same thing as trying to find equation(with the same philosophy of finding the minimal action) that describes the equation of motion and while doing that, you just end up with conservation.

Question 1: do you find everything I just said true ?

Question 2: Wouldn't just using euler lagrange get you the conservation of law as well ? why are we in need to just talk about "coordinate transformation" ? Maybe there is a better proof instead of using "action minimal" rule ?

Question 3: I noticed that in the video when we do $x(t) + \epsilon(t)$, we do it such that $\epsilon(t_1) =\epsilon(t_2) = 0$. So it's not fully the transformation as we resrict starting and ending point be the same hence another reason why I don't like the "coordinate transformation".

Oh! I see what you are getting at.

Frankly, I have never used the term "coordinate transformation" to describe what we are doing. What we are (Mathematically) doing is "varying the path" in order to find an extremum of the path, which is where x(t) is. Does that make a bit more sense?

As to why we use the Action Principle instead of Euler-Lagrange... Well, depending on what we want to do, we often do use Euler-Lagrange. Yes, you can use Euler-Lagrange to derive Noether's Theorem. But the further you get into Physics, the more general the Action Principle can be made. For example, when you get to Quantum Field Theory, we minimize the action defined as the integral of a Lagrangian density. The Euler-Lagrange field equations are, in fact, present in the background, but it is difficult to directly use those to generate a series expansion to approximate a solution. (And the solution "path" represents how the field behaves, it is not simply a path.) The action is very easy to turn into a series. (Solving the terms in the series... not so easy!) Not to mention, the Euler-Lagrange equations are essentially a Mathematical statement; the Principle of Least Action is a Physical one.

-Dan

 

[gionole]

Yes, I think it makes sense. Noether basically says to use least of action principle(using $\delta S = 0$) and she notes that doing that sometimes gives conservation law.

I had a confusion that due to the "coordinate translation/transformation" wording, we were transforming the particle'r location. I'm saying this because on one of Leonard Susskind's lecture, I heard the following: "and we shift all of particle's positions by the same amount as in shift the system in space" -and this got me confusing. In the real world, particle doesn't get shifted. I think all Leonard means is if we got 2 particles with their true paths($x_1(t), x_2(t)$), then we can write the action for the modified paths and use action be minimal rule and this will "sometimes" get us conservation law. Then Noether moves forward and proves that this happens when L doesn't change when we insert modified paths.

In reality, I think this whole thing about noether is only mathematical and not physical, because in physics, why would we decide to choose different/modified path.

Question 1: Thoughts ? do you think I understand it ?

Question 2: I believe Noether's proof is focusing only for such cases of $x(t) -> x(t) + \epsilon(t)$, when $\epsilon(t)$ is 0 at $t_1$ and $t_2$, right ? if not, we wouldn't arrive at conservation of momentum (unless we got special case when $\epsilon$ is a number instead of time dependent ). Correct ?

 

[topsquark]

Yes, I think it makes sense. Noether basically says to use least of action principle(using $\delta S = 0$) and she notes that doing that sometimes gives conservation law.

I had a confusion that due to the "coordinate translation/transformation" wording, we were transforming the particle'r location. I'm saying this because on one of Leonard Susskind's lecture, I heard the following: "and we shift all of particle's positions by the same amount as in shift the system in space" -and this got me confusing. In the real world, particle doesn't get shifted. I think all Leonard means is if we got 2 particles with their true paths($x_1(t), x_2(t)$), then we can write the action for the modified paths and use action be minimal rule and this will "sometimes" get us conservation law. Then Noether moves forward and proves that this happens when L doesn't change when we insert modified paths.

In reality, I think this whole thing about noether is only mathematical and not physical, because in physics, why would we decide to choose different/modified path.

Question 1: Thoughts ? do you think I understand it ?

Question 2: I believe Noether's proof is focusing only for such cases of $x(t) -> x(t) + \epsilon(t)$, when $\epsilon(t)$ is 0 at $t_1$ and $t_2$, right ? if not, we wouldn't arrive at conservation of momentum (unless we got special case when $\epsilon$ is a number instead of time dependent ). Correct ?

Well, we aren't really doing anything to the particle, it's true. We are hypothesizing that there is, in fact, an extremum path that the particle will follow in the end, and that the particle will follow a path $x(t) + \epsilon (t)$. We then insist this path is an extremum and that sets up the equations to find x(t). In that sense it is purely Mathematical. The Physics comes in when we define the action S and say that it must be a minimum for the Physics to happen.

Yes, we need the endpoints of the varied path to only go between the same $x(t_0)$ and $x(t_1)$ for every possible we are considering. This is part of the statement of the Physical problem: How does the particle get from point A to point B? The endpoints themselves are a part of the problem.

Perhaps I should try to clarify this: Where the coordinate transformations come in is when we use Noether's theorem to analyze symmetries. For example, we know that if we change the coordinate system X to X' by a constant translation, then the momentum of the particle (in the primed coordinate system) does not depend on the change of coordinate system. (This is not true if you rotate the coordinate system. "Fictitious" forces show up in the equations to hold the particle's hat on its head.) The particle's path through space stays the same. What we have done is change its coordinate system: we are not doing anything to the particle's actual path through space when we do this.

-Dan

 

[gionole]

For example, we know that if we change the coordinate system X to X' by a constant translation, then the momentum of the particle (in the primed coordinate system) does not depend on the change of coordinate system.

I get that, the question is why would you need to hypothesize to change the coordinate system in that case ? To me, it makes sense what we do with principle of least action and how we find $x(t)$. In between, we can arrive to momentum conservation, but to hypothesize to change X into X' doesn't make sense to me. That was my total confusion into starting this thread.

 

[topsquark]

I get that, the question is why would you need to hypothesize to change the coordinate system in that case ? To me, it makes sense what we do with principle of least action and how we find $x(t)$. In between, we can arrive to momentum conservation, but to hypothesize to change X into X' doesn't make sense to me. That was my total confusion into starting this thread.

For no other reason than it gives us a nice result: the law of conservation of momentum is related to the space translation invariance of the system. Once you know that, you start looking for other invariances and you see how powerful Noether's theorem really is: any symmetry of the Lagrangian implies a conservation law. You don't even need to know the equation of motion, Noether's theorem tells you that straight off and you can use that information in solving for x(t).

-Dan

 

[vanhees71]

This question has been bugging me and not sure how to ask it in an understandable way, but I will try.

Imagine we have $L = \frac{1}{2}m\dot x^2$.

If we do: $x(t) -> x(t) + \epsilon(t)$, we transform the whole trajectory and We will arrive at $\frac{dP}{dt} = 0$ (momentum conserved).

I haven't read the entire thread, but this claim is wrong, because the Lagrangian is not invariant under this very general transformation. If this were true, Newtonian mechanics of a free point particle were forminvariant under arbitrary coordinate transformations, i.e., it would look the same in any non-inertial frame, which for sure is not correct.

Momentum conservation is due to spatial translation invariance, i.e., (in infinitesimal form, which is sufficient for Noether)
$$t'=t, \quad x'=x+\delta a, \quad \delta a=\text{const}.$$
Then for
$$L=\frac{m}{2} \dot{x}^2$$
you get
$$L(\dot{x}')=L(\dot{x}),$$
i.e., you have a symmetry a la Noether (even the most simple case that the Lagrangian itself is invariant).

The conserved quantity is the conjugate momentum,
$$p=\frac{\partial L}{\partial{\dot{x}}}.$$
Indeed the Euler-Lagrange equation tells you
$$\dot{p}=\frac{\partial L}{\partial x}=0.$$

 

[gionole]

I haven't read the entire thread, but this claim is wrong, because the Lagrangian is not invariant under this very general transformation. If this were true, Newtonian mechanics of a free point particle were forminvariant under arbitrary coordinate transformations, i.e., it would look the same in any non-inertial frame, which for sure is not correct.

Momentum conservation is due to spatial translation invariance, i.e., (in infinitesimal form, which is sufficient for Noether)
$$t'=t, \quad x'=x+\delta a, \quad \delta a=\text{const}.$$
Then for
$$L=\frac{m}{2} \dot{x}^2$$
you get
$$L(\dot{x}')=L(\dot{x}),$$
i.e., you have a symmetry a la Noether (even the most simple case that the Lagrangian itself is invariant).

The conserved quantity is the conjugate momentum,
$$p=\frac{\partial L}{\partial{\dot{x}}}.$$
Indeed the Euler-Lagrange equation tells you
$$\dot{p}=\frac{\partial L}{\partial x}=0.$$

Thanks for the proof. Yes, I understood that, but then I watched this:



and he uses $\epsilon(t)$ in the momentum conservation video of noether which is confusing. He shouldn't have done that in my opinion.