11

Relativity has always been explained to me (in books I've read, etc) as a superset of newton's laws - that is; it encapsulates all of Newton's mechanics in addition to other effects (observer effect, time dilation, space-time geometry, etc).

I can kind of imagine 2 of Newton's 3 laws of motion to be incapsulated within Einstein's Special Relativity, but the one about "for every action there is an equal and opposite reaction", I'm struggling to find a place for that within Special Relativity.

Does this sit outside Special Relativity's explanatory power or is it inferred within somehow?

Qmechanic
  • 201,751
Mike S
  • 942
  • In addition to @BrianBi's answer, I would like to point out that there is something inherently non-relativistic in the third law so that one can straightaway realize that it cannot be incorporated into the relativistic framework, namely, it refers to the simultaneity of spatially separated events--a notion that is not covariant in relativity. Of course, that is a trouble for the third law and not for relativity because the deeper and the more general physical content behind the third law, namely, the conservation of momentum, can be easily expressed in relativity. –  Oct 22 '18 at 15:37

2 Answers2

12

Newton's third law is really a special case of the conservation of momentum. Suppose you have two rigid bodies with momenta $\mathbf{p}_1$ and $\mathbf{p}_2$. If they only interact with each other, then $\mathbf{p}_1 + \mathbf{p}_2$ is constant, since total momentum is conserved. Differentiating this gives $\frac{d\mathbf{p}_1}{dt} + \frac{d\mathbf{p}_2}{dt} = 0$. But this is just $\mathbf{F}_{21} + \mathbf{F}_{12}$ (i.e., the external force exerted on body 1 plus the external force exerted on body 2).

In relativity we usually don't use the concept of force, preferring to deal with momentum instead. Momentum is conserved in relativity, just like it is in Newtonian mechanics.

Brian Bi
  • 6,491
  • 27
  • 62
  • Postulate versus derivation does not distinguish Newtonian mechanics and relativity at all. Noether's theorem allows you to derive conservation of momentum from translational symmetry just as well in Newtonian mechanics as in relativity. In either theory, you can start out by postulating symmetries or postulating conservation laws. – Keshav Srinivasan Feb 26 '14 at 03:16
  • @KeshavSrinivasan I agree. – Brian Bi Feb 26 '14 at 03:17
  • @BrianBi I think your answer might be exaggerating the power of symmetry in relativity. One needs to specify an action to which one can apply Noether's theorem, and that requires extra physical input, namely some information about the interactions involved. I don't see how one get's conservation of momentum for free simply from knowing that whatever interactions one uses, they must be Poincare-invariant. In other words, conservation of momentum is not somehow contained in the principle of relativity alone. – joshphysics Feb 26 '14 at 03:37
  • @joshphysics All right, I get that, but I'll edit the answer to make it less confusing. – Brian Bi Feb 26 '14 at 03:41
  • 1
    @joshphysics, what you mean you don't see how you get conservation of momentum if you stipulate that all interactions are Poincare-invariant? Poincare-invariant implies translation-invariant implies momentum conservation. –  Feb 26 '14 at 03:53
  • 1
    @dgh I think he's trying to say that, while you know there is some conserved quantity that arises from translational invariance, you won't be able to compute that quantity in terms of observables until you've written down the action. – Brian Bi Feb 26 '14 at 03:55
  • @BrianBi That's precisely what I mean. – joshphysics Feb 26 '14 at 03:59
  • @BrianBi Well, Landau and Lifshitz are able to prove from first principles that the Lagrangian of a free particle in classical mechanics has to be proportional to the square of the velocity, and they're able to use that to show that the conserved quantity that arises from translational invariance must be proportional to the velocity. Can't you do a similar derivation in the relativistic case? – Keshav Srinivasan Feb 26 '14 at 04:01
  • @KeshavSrinivasan That's still the free particle. The statement of momentum conservation we'd like to make presumably refers to complicated interactions (i.e. electromagnetic), in which case you need some more physical data. – joshphysics Feb 26 '14 at 04:09
  • @joshphysics Can't we simply say that there is this conserved quantity (namely, $\frac{\partial\mathcal{L}}{\partial (\partial_\mu{q})}$--and stating this shouldn't require an explicit expression for the Lagrangian. Knowing that $\mathcal{L}$ is translationally invariant should suffice, right?) arising out of the translational invariance and that is what we would like to name momentum? Isn't it what we do anyway? –  Oct 22 '18 at 15:31
0

Newton's first law (really Galileo's law of inertia):

This works as well in Special Relativity as it does for Newton. If an object has a constant relative speed near the speed of light and no force acts on it, it keeps moving at a constant relative speed. Check. Can't go faster than light, though.

Newton's second law: F=ma Acceleration depends on time, and time is relative to the state of motion of the observer. For someone doing physics in an inertial reference frame, of course F=ma still works like it always did, at low speeds at least. Trying to push something from rest to relativistic speeds is a whole different story. It appears to get more massive (and harder to accelerate). And the speed of light is still the top speed limit.

Newton's third law: Action - Reaction (force pairs) Works until you are applying force to something already moving near the speed of light. Energy still gets pumped in, but it doesn't move much faster, and can't exceed the speed of light under any circumstances.