Momentum in special relativity

[DrStupid]

Then define, what you mean by "third Law".

Just take Newton's wording and replace "bodies" with non specified entities.

It's clear that in special relativity momentum conservation strictly holds true, because momentum is the "Noether charge" of spatial translation invariance, but for me there's no way to accommodate this with the "third law".

And in classical mechanics it strictly holds due to lex 2 and 3. This formalism fits better to the usual application in classical mechanics (long range interactions between closed systems). But is is still the same conservation.

 

[DrStupid]

On the other hand, if they are point-masses, then in the absence of fields or action-at-a-distance, then the probability of a collision should be zero.

If you don't reduce the problem to one dimension it actually is zero, unless the mass points are neighbours in a continuum.

 

[vanhees71]

Just take Newton's wording and replace "bodies" with non specified entities.

This is too vague. I think the third law only belongs to Newtonian mechanics. In a modern approach via symmetries it's not even needed as a postulate, because then it's substituted by the derivation of momentum conservation from spatial translation invariance (holding in both Newtonian and special-relativistic physics).

 

[DrStupid]

This is too vague.

This response applies to itself. What exactly is the problem?

In a modern approach [...]

There are more than one way to skin a cat. Some of them are modern others are not. You prefer postulating symmetries others are fine with postulated conservations. So what? As long as the results comply with experimental observations it is just a matter of taste.

 

[vanhees71]

You still have not defined, what you mean by "third law" in the context of relativity. My understanding is that the "third law" (i.e., Newton's Lex Tertia) is invalid in relativistic physics, and that's the very reason, why one uses interactions via fields rather than Newtonian forces, which are because of the "third law" necessarily actions at a distance in contradiction to the causality structure of relativistic spacetime descriptions.

 

[DrStupid]

You still have not defined, what you mean by "third law" in the context of relativity.

I did. See #36.

My understanding is that the "third law" (i.e., Newton's Lex Tertia) is invalid in relativistic physics

Is that just gut feeling or can you actually show that it is invalid in relativity?

 

[vanhees71]

In #36 is no clear statement about what you mean by "the third law". That's why I asked for you definition in the first place...

It's obvious that the 3rd law cannot hold in the usual Newtonian action-at-a-distance form. Take Newtonian gravity, according to which $\vec{F}_{12}=-\vec{F}_{21}$ instantaneously, i.e., when the moon moves a bit away from the Earth the force on the Earth instantaneously changes according to Newton's law of force. This cannot happen within a relativistic theory, and that's why one rather uses field theories. There the third law doesn't hold, but the interactions are all local via the field, and that's why momentum conservation still holds, because the field carries momentum changes as do the interacting bodies.

 

[Mister T]

The third law also holds for local interactions.

And it doesn't hold for nonlocal interactions. That's why it has no place in the structure of modern physics. It's part of the Newtonian approximation, outside of that it has little to no utility. It has been replaced by conservation of momentum.

 

[DrStupid]

In #36 is no clear statement about what you mean by "the third law".

I can't help you if you don't tell me what you are missing.

It's obvious that the 3rd law cannot hold in the usual Newtonian action-at-a-distance form.

That's not a problem because there is no "action-at-a-distance" in relativity.

Take Newtonian gravity, according to which $\vec{F}_{12}=-\vec{F}_{21}$ instantaneously

That's not a problem of the third law but of the law of gravitation.

 

[neilparker62]

Just wondering: is it too simplistic to ask if relativistic mass is $γ m_0$, why should relativistic momentum be anything other than $γm_0v$ ?

 

[SiennaTheGr8]

Just wondering: is it too simplistic to ask if relativistic mass is $γ m_0$, why should relativistic momentum be anything other than $γm_0v$ ?

It's a natural guess that happens to be right (and more than a guess if you pose the problem the right way—see the Tolman and Lewis thought experiment), but its "rightness" ultimately hinges on whether $\gamma m_0 \vec v$ is indeed conserved, an empirical question.

 

[Mister T]

Just wondering: is it too simplistic to ask if relativistic mass is $γ m_0$, why should relativistic momentum be anything other than $γm_0v$ ?

Here are two ways to look at it: (Note: My $m$ is the ordinary mass, what some people call the rest mass).

1. Momentum is $mv$. Relativistic momentum is $\gamma mv$. We can call $\gamma m$ the relativistic mass and rename it the mass if we are interested in retaining the definition mass times velocity.

2. Momentum is $\gamma mv$. At low speeds this is approximately $mv$. We can call $\gamma m$ the relativistic mass and rename it the mass if we are interested in retaining the low-speed approximation.

When you look at it the second way it seems silly to introduce relativistic mass, but the term was invented for reasons that are consistent with the first way. That was done a long time ago when researchers hoped that what we now call relativistic physics could be a generalization of Newtonian physics.

 

[neilparker62]

Thanks for the insights - appreciated.

 

[DrStupid]

Here are two ways to look at it: (Note: My $m$ is the ordinary mass, what some people call the rest mass).

1. Momentum is $mv$. Relativistic momentum is $\gamma mv$. We can call $\gamma m$ the relativistic mass and rename it the mass if we are interested in retaining the definition mass times velocity.

2. Momentum is $\gamma mv$. At low speeds this is approximately $mv$. We can call $\gamma m$ the relativistic mass and rename it the mass if we are interested in retaining the low-speed approximation.

Or

3. Momentum is $qv$. We derive $q(v)$ for conserved momentum and call it relativistic mass in relativity. Because we do not like frame-dependent properties in relativity we than define a new frame-independent property $m$ with $q=\gamma m$ and call it mass.

 

[pervect]

It's worth noting that most if not all of the discussion applies only to point particles. The momentum of a non-point particle, for instance a system of finite volume, is more complicated.

In tensor notation, given the stress energy tensor of a system $T$, an observer with some 4-velocity u, the energy and momentum of a unit volume (specified via the form of the energy-momentum 4-vector) are given by $T_{ab} u^a$. The concept of "unit volume" is observer dependent, Lorentz contraction and other relativistic effects make the "unit volume" depend on the observer. The specified observer is the one with the specified 4-velocity $u$.

There are some interesting predicted effects. If one applies a force at one point of a (born) rigid object, the ratio of acceleration/force (F/a) will not be constant as the force is varied. Large forces will induce stresses in the body, stresses that will slightly affect the magnitude of the acceleration. The effects are minor - any realistic body will break apart before the stress can change the force/acceleration ratio much. While the effects are tiny for physically reasonable materials, they are necessary to have a mathematically self-consistent presentation of relativistic dynamics for any system more complicated than a point particle. This result is in contrast to Newtonian mechanics where stresses never affect the ratio of force and acceleration. It's a new effect that appears in special relativity that did not exist in Newtonian mechanics.