# Instantaneous force propagation in classical mechanics (1 Viewer)

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#### aloyisus

I'm working my way (slowly) through Landau & Lif***z Classical Mechanics. I'm finally nearing the end of chapter one, and although I hit another stumbling block, I think I've got it now. If anyone has the time to check my reasoning, I'd be grateful.

I will quote the passage that was confusing me in full:

"The fact that the potential energy depends only on the positions of the particles at a given instant shows that a change in the position of any particle instantaneously affects all the other particles. We may say that the interactions are instantaneously propagated. The necessity for interactions in classical mechanics to be of this type is closely related to the premises upon which the subject is based, namely the absolute nature of time and Galileo's relativity principle. If the propagation of interactions were not instantaneous, but took place with a finite velocity, then that velocity would be different in different frames of reference in relative motion, since the absoluteness of time necessarily implies that the ordinary law of composition of velocities is applicable to all phenomena. The laws of motion for interacting bodies would then be different in different inertial frames, a result which would contradict the relativity principle."

And here's my attempt to understand this:

Consider a 2 particle system. Particle 2 is acted upon by a force due to particle 1, but because it does not act instantaneously, the strength of force depends on the relative positions of particle 1 at the current time t, and particle 2 at an earlier time $t - \Delta t$, where $\Delta t$ represents the time taken for the interaction to propagate through space.

i.e. $U = U(\left |\mathbf{r}_{1}(t) - \mathbf{r}_{2}(t-\Delta t)\right |)$

Clearly $\Delta t$ is a function of the state of the system, i.e. it will vary depending on relative positions and velocities of the particles. It is also the time interval between two events, an interaction (or could I instead say 'a signal' here?) leaving particle 1 and reaching particle 2. Moreover, because of the absoluteness of time in classical mechanics, $\Delta t$ is the same in all reference frames.

For the laws of physics to be the same in a different inertial frame travelling with constant velocity -k relative to the original frame, the potential U must be invariant under a Galilean transformation r' = r + kt. So we have $\mathbf{r}'(t-\Delta t) = \mathbf{r}(t-\Delta t) + \mathbf{k}t - \mathbf{k}\Delta t$, and by substituting into the potential, we can see that while the kt terms cancel, there is a $\mathbf{k}\Delta t$ term left over.

i.e. $U' = U(\left |\mathbf{r}_{1}(t) - \mathbf{r}_{2}(t-\Delta t) + \mathbf{k}\Delta t\right |)$

So now the potential is dependent on the reference frame, which it can't be, and the only way around this is to make $\Delta t = 0$.

Does this explanation make sense? Also, and perhaps most confusingly for me, if we consider that the Lagrangian is only defined to within a total time derivative of a function f(r,t), could that invalidate the final part of my argument (i.e. the potential might still be invariant under Galilean transformation if the difference between U' and U is a total time derivative)?

If you made it this far, thanks for reading!

Last edited:

#### Naty1

All looks good...I don't know about the last question regarding the Langrangian.

#### aloyisus

Thanks for looking over it, Naty1. I'm glad it makes sense.

I'll restate my question about the Lagrangian.

In order for the laws of physics must be the same in all inertial reference frames, Lagrange's equations of motion must keep the same form in all inertial reference frames, which places restrictions on the Lagrangian. More specifically, for the Lagrangian to give identical equations of motion when it undergoes a Galilean transormation (r' = r + kt, v' = v + k), the transformed Lagrangian L' must take this form:

L' = L + d/dt f(r,t)

So we're allowed to have this extra term, and if it's present, the laws of physics remain unchanged under the transformation.

The question is, going back to my original post where I said:

$U' = U(\left |\mathbf{r}_{1}(t) - \mathbf{r}_{2}(t-\Delta t) + \mathbf{k}\Delta t\right |)$

How can I satisfy myself that this transformed potential isn't just the original potential with an extra d/dt f(r,t), which would leave the physics invariant?

Anybody got any ideas? And thanks again for reading.

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