A question about conservation of momentum (Lagrangian)

[Coffee_]

http://imgur.com/QhYG54l

The image seems to be not showing here is the link : http://imgur.com/QhYG54lWhat does Landau mean here by the Lagrangian remaining ''unchanged''. Is it the value of the lagrangian as a function that may not change or is it the form that may not change?

Also how does this curly delta differ from the normal differential symbol?

I'm confused because what follows seems like a derivation as if the value of the Lagrangian as a function can't change, but then why all those curly delta's from variational calculus?

Thanks

 

[UltrafastPED]

The notation is from the calculus of variations, and represents a "virtual displacement"; this is all usually explained in an advanced course in analytical mechanics. Laundau is like a summary of an advanced text - great for reference once you have studied it previously, but perhaps not so good as an introduction.

Here he is making statements about Noether's theorem. I suggest that you look for a fuller presentation, then his rather cryptic style will become clearer - he just cuts out all of the steps between A and Z.

 

[Radarithm]

Basically what he's saying is that if you shift all of the particles in the system to a new position:
$$r_{\alpha}\to r_{\alpha}+\epsilon$$
and find that the Lagrangian stays the same, then it is demonstrating translational invariance. As a consequence of this translational invariance, momentum is conserved. Like what the UltrafastPED said, this is part of Noether's Theorem.

Here is an excerpt from Taylor's Classical Mechanics:

In particular, the potential energy must be unaffected by this displacement, so that
$$U(r_1+\epsilon,...,r_N+\epsilon,t)=U(r_1,...,r_N,t)$$
or, more briefly,
$$\delta U=0$$
where $\delta U$ denotes the change U under [the translation stated above]. Clearly, the velocities are unchanged by [the above translation]. (Adding a constant $\epsilon$ to all the $r_{\alpha}$ doesn't change the $\dot{r_{\alpha}}$.) Therefore $\delta T=0$, and hence
$$\delta \mathcal{L}=0$$
Under the [above translation]. This result is true for any displacement $\epsilon$. If we choose $\epsilon$ to be an infinitesimal displacement in the x direction, then all of the x coordinates $x_1,...,x_N$ increase by $\epsilon$, while the y and z coordinates are unchanged. For this translation, the change in $\mathcal{L}$ is
$$\delta \mathcal{L}=\epsilon\frac{\partial\mathcal{L}}{\partial x_1}+...\epsilon\frac{\partial\mathcal{L}}{\partial x_N}=0.$$
This implies that
$$\sum_{\alpha=1}^N \frac{\partial\mathcal{L}}{\partial x_{\alpha}}=0$$
Now using Lagrange's equations we can rewrite each derivative as
$$\frac{\partial\mathcal{L}}{\partial x_{\alpha}}=\frac{\mathrm{d}}{\mathrm{dt}}\frac{\partial\mathcal{L}}{\partial\dot{x_{\alpha}}}=\frac{\mathrm{d}}{\mathrm{dt}}p_{\alpha x}$$
where $p_{\alpha x}$ is the x component of the momentum of particle $\alpha$. This, [the above] becomes
$$\sum_{\alpha=1}^N\frac{\mathrm{d}}{\mathrm{dt}}p_{\alpha x}=\frac{\mathrm{d}}{\mathrm{dt}}P_x=0$$
Where $P_x$ is the x component of the total momentum

Now if the rate of change of the total momentum (eg. the external forces on a system) is 0, then momentum is conserved. This is a consequence of Noether's Theorem.

 

[Coffee_]

Basically what he's saying is that if you shift all of the particles in the system to a new position:
$$r_{\alpha}\to r_{\alpha}+\epsilon$$
and find that the Lagrangian stays the same, then it is demonstrating translational invariance. As a consequence of this translational invariance, momentum is conserved. Like what the UltrafastPED said, this is part of Noether's Theorem.

Here is an excerpt from Taylor's Classical Mechanics:


Now if the rate of change of the total momentum (eg. the external forces on a system) is 0, then momentum is conserved. This is a consequence of Noether's Theorem.

So it really is the ''VALUE'' of the Lagrangian that has to remain the same in the context of conservation of momentum + angular mom?

 

[Radarithm]

So it really is the ''VALUE'' of the Lagrangian that has to remain the same in the context of conservation of momentum + angular mom?

Yes, conserved quantities arise from invariances within the Lagrangian. For example, if the Lagrangian does not depend on time, the Hamiltonian (total energy of the system) is conserved.

 

[Radarithm]

If you haven't read an intermediate text before L&L (eg. Taylor, Marion and Thornton, etc) then I highly recommend going through https://www.amazon.com/dp/189138922X/?tag=pfamazon01-20 before venturing into Landau or Goldstein or anything at that level.

 

[Coffee_]

If you haven't read an intermediate text before L&L (eg. Taylor, Marion and Thornton, etc) then I highly recommend going through https://www.amazon.com/dp/189138922X/?tag=pfamazon01-20 before venturing into Landau or Goldstein or anything at that level.

Yeah I know but we have a slightly different system. We don't quite follow a certain book in class but the prof just makes a ''course text'' that is based on different standard books. This is intro to analytical mechanics class but the prof based the first quarter of the course on Landau and Lifschitz. So I don't really have a choice to learn it this way I guess.

 

[Coffee_]

Yes, conserved quantities arise from invariances within the Lagrangian. For example, if the Lagrangian does not depend on time, the Hamiltonian (total energy of the system) is conserved.

May I ask then, what about Galilean invariance for the Lagrangian?

For homogeneous and isotropic space we assume the VALUE of the lagrangian doesn't change for these symmetric changes in coordinates.

Do you mind elaborating on how it is different from the argument that uses the fact that it has to be invariant under a constant velocity difference with the situation before? Then it isn't the value that has to remain constant anymore right?

 

[Radarithm]

Do you mind elaborating on how it is different from the argument that uses the fact that it has to be invariant under a constant velocity difference with the situation before? Then it isn't the value that has to remain constant anymore right?

Do you mind elaborating a bit more on that?

 

[Coffee_]

Do you mind elaborating a bit more on that?

Basically I'm just interested in some explanation/clarification on the following thing.

Once you know that the lagrangian for a free particle is a function of the magnitude of vector v or just v² then in L&L they do the following:

L=L((v)²).

Galilean invariance must hold so:

L((v)²) and L((v+e)²) must be ''equivalent'' they say.

To show this equivalence they go to the action and say it has to be minimal for both cases so they look for the total time derivative in the second largangian etc...

Anyway I'm interested in why in the latter case they go to the action and in the other cases (conservations) they go look at the value of the Lagrangian.

 

[Radarithm]

Basically I'm just interested in some explanation/clarification on the following thing.

Once you know that the lagrangian for a free particle is a function of the magnitude of vector v or just v² then in L&L they do the following:

L=L((v)²).

Galilean invariance must hold so:

L((v)²) and L((v+e)²) must be ''equivalent'' they say.

To show this equivalence they go to the action and say it has to be minimal for both cases so they look for the total time derivative in the second largangian etc...

Anyway I'm interested in why in the latter case they go to the action and in the other cases (conservations) they go look at the value of the Lagrangian.

I think its just another way of proving Galilean invariance (I'm not very familiar with the Galilean invariance of the Lagrangian). Do you mind showing me how they go about proving this (the part where they go to the action)?
I might be able to use that information and relate it to the value of the Lagrangian.

 

[Coffee_]

I think its just another way of proving Galilean invariance (I'm not very familiar with the Galilean invariance of the Lagrangian). Do you mind showing me how they go about proving this (the part where they go to the action)?
I might be able to use that information and relate it to the value of the Lagrangian.

Here: http://imgur.com/x6qyGSR

 

[Radarithm]

I'm interested in why in the latter case they go to the action and in the other cases (conservations) they go look at the value of the Lagrangian.

Apparently it seems like a convention.
Correct me if I'm wrong but Landau shows that $L=L'$ under Galilean transformations. Did you want me to explain how? Unless that's the case, then I guess him going to the action is probably a convention, a way to prove this invariance. I think if you took a look at Noether's theorem you'd probably have more appreciation over how scattered the symmetries are in the Lagrangian.

 

[Coffee_]

Apparently it seems like a convention.
Correct me if I'm wrong but Landau shows that $L=L'$ under Galilean transformations. Did you want me to explain how? Unless that's the case, then I guess him going to the action is probably a convention, a way to prove this invariance. I think if you took a look at Noether's theorem you'd probably have more appreciation over how scattered the symmetries are in the Lagrangian.

Oh no I just realize that you can't use the lagrangian ''value'' reasoning in this case. Obviously when you change your reference frame to another velocity the value of the Lagrangian will change.