Spatial homogeneity condition for a free particle Lagrangian

[renormalize]

What are you basing this on? The Euler-Lagrange equations are not the same.

Are you sure? $4x^{3}v_{x}=4x^{3}\dot{x}=d\left(x^{4}\right)/dt$ is a total time derivative.

 

[PeterDonis]

Are you sure? $4x^{3}v_{x}=4x^{3}\dot{x}=d\left(x^{4}\right)/dt$ is a total time derivative.

Ah, I see. In one dimension, yes, this is true--but then it's confusing why $v_x$ is written in the second term but $v^2$ in the first.

As far as I can tell, L&L do not discuss the fact that you can add a total time derivative to a Lagrangian without changing the equations of motion.

 

[renormalize]

As far as I can tell, L&L do not discuss the fact that you can add a total time derivative to a Lagrangian without changing the equations of motion.

From Landau & Lifshitz, Mechanics (3rd ed.), pg. 4:

 

[cianfa72]

What are you basing this on? The Euler-Lagrange equations are not the same.

$4x^3v_x = 4x^3\dot x$. It is the $\frac {d} {dt} \left (x^4(t) \right )$ therefore it is the "standard" free particle Lagrangian written in an inertial frame + a total derivative term.

 

[PeterDonis]

From Landau & Lifshitz, Mechanics (3rd ed.), pg. 4:

Ah, missed it. Thanks!

Their further discussion appears to assume that no total time derivative term is added and should be read with that qualifier in mind.

 

[PeterDonis]

$4x^3v_x = 4x^3\dot x$. It is the $\frac {d} {dt} \left (x^4(t) \right )$ therefore it is the "standard" free particle Lagrangian written in an inertial frame + a total derivative term.

Yes, @renormalize already pointed that out. And as I noted in post #40 just now, their further discussion appears to assume that no total time derivative is added, so, for example, when they say that the Lagrangian of a free particle in an inertial frame cannot depend on position, what they really mean is "except for a possible total time derivative term that depends on position but does not change the equations of motion". And similarly for other statements they make about Lagrangians, such as when they derive $L = \frac{1}{2} m v^2$ as the only possible free particle Lagrangian in an inertial frame.

 

[cianfa72]

How? $x^3$ is not translation invariant.

If one translates by the same amount $c$ the $x$ coordinates of the start and end events, one gets the same (translated) solution path (as stationary action path).

 

[PeterDonis]

If one translates by the same amount $c$ the $x$ coordinates of the start and end events, one gets the same (translated) solution path (as stationary action path).

Yes, we've resolved that based on @renormalize pointing out that the $x^3 v_x$ term is a total time derivative, so the equation of motion is the same.

 

[cianfa72]

Yes, we've resolved that based on @renormalize pointing out that the $x^3 v_x$ term is a total time derivative, so the equation of motion is the same.

$L = \frac{1}{2} m v^2 + x$ is "translation invariant" as well. Does it "count" as spatially homogeneous ?

 

[PeterDonis]

On the same ground $L = \frac{1}{2} m v^2 + x$ is "translation invariant".

No, it isn't. $x$ is not a total time derivative.

 

[cianfa72]

No, it isn't. $x$ is not a total time derivative.

Yes, it isn't a total time derivative. However: fix a pair of start and end events. Suppose $\alpha(t)$ is a stationary action path for the Lagrangian in #44. Then translates the start and end $x$ coordinate of the same amount $c$. Then $\alpha (t) + c$ will be a stationary action path as well.

 

[renormalize]

Yes, it isn't a total time derivative. However: fix a pair of start and end events. Suppose $\alpha(t)$ is a stationary action path for the Lagrangian in #44. Then translates the start and end $x$ coordinate of the same amount $c$. Then $\alpha (t) + c$ will be a stationary action path as well.

You keep making this statement but it is not true. Please do the actual calculation! Write your Lagrangian as $L\left(x,\dot{x}\right)=\frac{1}{2}m\dot{x}^{2}+kx$ (where I've inserted the coupling constant $k$ to allow us to examine the force-free case by setting $k=0$). Then $L$ and the resulting equations-of-motion are simple enough that you can easily find the stationary solution $x(t)$. Do so and then evaluate $L$ on the stationary path (i.e., put $x$ "on-shell") and then integrate from $t_0$ to $t_1$ to evaluate the value of the stationary action $S[t_0,t_1]$. Use the same technique to calculate $S[t_0+c,t_1+c]$ and you'll find that the difference $S[t_0+c,t_1+c]-S[t_0,t_1]$ is non-zero and proportional to the coupling $k$. The two stationary paths are not the same when $k\neq 0$.

 

[PeterDonis]

Please do the actual calculation!

And on that note, this thread is closed since the OP question has been thoroughly answered.

 

[PeterDonis]

Use the same technique to calculate $S[t_0+c,t_1+c]$

Two notes for clarification:

First, this is a translation in time, but the OP was talking about a translation in space. The point that the action is what needs to be evaluated is still valid.

Second, in any case, this Lagrangian is irrelevant to the claim made in L&L that was asked about in the OP, since this Lagrangian is not the Lagrangian of a free particle in an inertial frame.