Instantaneous force propagation in classical mechanics

In summary, the author is working their way through Landau and Lif***z Classical Mechanics, and is having difficulty understanding a confusing passage. They state that if 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, but they eventually come to a solution that satisfies the laws of physics in all reference frames.
  • #1
aloyisus
7
0
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 [itex]t - \Delta t[/itex], where [itex]\Delta t[/itex] represents the time taken for the interaction to propagate through space.

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

Clearly [itex]\Delta t[/itex] 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, [itex]\Delta t[/itex] is the same in all reference frames.

For the laws of physics to be the same in a different inertial frame traveling 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 [itex]\mathbf{r}'(t-\Delta t) = \mathbf{r}(t-\Delta t) + \mathbf{k}t - \mathbf{k}\Delta t[/itex], and by substituting into the potential, we can see that while the kt terms cancel, there is a [itex]\mathbf{k}\Delta t[/itex] term left over.

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

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

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:
Physics news on Phys.org
  • #2
All looks good...I don't know about the last question regarding the Langrangian.
 
  • #3
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:

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

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.
 

Related to Instantaneous force propagation in classical mechanics

1. What is instantaneous force propagation in classical mechanics?

Instantaneous force propagation in classical mechanics refers to the concept of how forces act on objects in motion. It is based on the idea that forces act instantaneously, meaning they have an immediate effect on the motion of an object.

2. How does instantaneous force propagation differ from continuous force propagation?

Instantaneous force propagation differs from continuous force propagation in that it assumes forces act instantaneously, while continuous force propagation takes into account the time it takes for the force to have an effect on the motion of an object. In classical mechanics, instantaneous force propagation is used to simplify calculations and assumptions about the motion of objects.

3. What are some examples of instantaneous force propagation in everyday life?

Examples of instantaneous force propagation in everyday life include pushing a door to open it, throwing a ball, or hitting a ping pong ball with a paddle. In each of these cases, the force applied has an immediate effect on the motion of the object.

4. How does Newton's first law of motion relate to instantaneous force propagation?

Newton's first law of motion states that an object at rest will remain at rest and an object in motion will remain in motion at a constant velocity unless acted upon by an external force. Instantaneous force propagation is based on the idea that forces act instantaneously, meaning they have an immediate effect on the motion of an object. Therefore, Newton's first law of motion is related to instantaneous force propagation in that it explains how an object's motion can be changed by the application of a force.

5. What are some limitations of using instantaneous force propagation in classical mechanics?

One limitation of using instantaneous force propagation in classical mechanics is that it does not take into account the time it takes for forces to act on an object. In reality, forces do not act instantaneously, and there may be a delay between the application of a force and its effect on an object's motion. Additionally, instantaneous force propagation is based on idealized conditions and may not accurately reflect real-world situations where friction and other external factors can affect an object's motion.

Similar threads

  • Mechanics
Replies
15
Views
1K
  • Mechanics
Replies
13
Views
1K
Replies
2
Views
1K
Replies
6
Views
1K
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
418
Replies
3
Views
1K
Replies
9
Views
1K
Back
Top