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There’s a postulate in special relativity as following: Physics laws are identical in all inertial reference frames.

I’m a math student, recently when I reviewed special relativity before learning general relativity, I ran into this question which I cast aside long time ago. Here’s my question: what does the physics law mean in the statement? For example, relativity is a physics law as well, would it fail in a non-inertial reference frame as the statement suggests? There’s another example, Newton’s Law $F=ma$, it can be applied in both inertial and non-inertial frame in elementary physics. What’s wrong? What exactly does the statement suggest? Is there any precise mathematical description?

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  • $\begingroup$ FYI - I've updated my answer with more precise definitions of the SR postulates. $\endgroup$
    – safesphere
    Oct 26, 2017 at 22:08

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You are right that SR holds for inertial observers only. As for $F=ma$, standing alone it is not a complete law. Non-inertial observers can also apply it by adding fictitious forces. To distinguish real and fictitious forces, you need Newton's third law. In fact Newton's three laws of motion are equivalent to the law of conservation of momentum and angular momentum. So non-inertial observers may see momentum and/or angular momentum not being conserved.

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Because it is a postulate, it doesn't have a mathematical proof. It is sort of like an axiom in mathematics. Few examples of laws of physics are the conservation of energy and momentum. Energy and momentum are conserved in every inertial reference frame.

The idea is that you can't conduct an experiment from an inertial reference frame in order to find out if you're stationary or moving with constant speed in a straight line. In this sense, there is no difference between a stationary frame and one that is moving with constant speed. So, all inertial frames are equivalent.

Special relativity can handle acceleration. As long as the gravitational effects are not important and spacetime can be considered flat, special relativity can describe accelerated motion. I wouldn't say that special relativity is a law of physics, rather it is a theory.

Newton's second law can work in non-inertial frames only if one considers fictitious forces (pseudo-forces). E.g. centrifugal forces, coriolis forces.

So, the fact that the laws of physics are the same in every inertial reference frame means that the equations that describe those laws have the same form.

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"Laws of physics are the same in all inertial coordinate systems", in relation to deriving the Lorentz transformations, mathematically translates to isotropy of space and homogeneity of space and time.

There are many rigorous derivations out there with this point clarified. This is just one of mamy:

Derivation of the Lorentz Transformations

EDIT: More presicely, the first poastulate of Special Relativity requires only homogeneity of space and time. Then the second postulate can be expressed as either the constancy of the speed of light or isotropy of space.

Interestingly, the constancy of the speed of light (in all inertial frames) is an excessive requirement that is not necessary for Special Relativity. The sufficient requirement is only that light moves with the same speed forward and back (isotropy of space).

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  • $\begingroup$ Do you know a proof (if the fact is true) that the axiom of constancy of speed of light when switching between IRFs implies the isotropy of space? $\endgroup$
    – DanielC
    Oct 27, 2017 at 0:32
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    $\begingroup$ @DanielC Sorry, I don't have the explicit proof, because commonly people prove the opposite, specifically that isotropy of space results in the constant speed of light. It is easy to take this argument and reverse it. For example, here: mathpages.com/home/kmath307/kmath307.htm Kevin argues, "If we replace $v$ with $-v$ these two transformations are exchanged, except for the factor $\mu$, so if we are to have spatial isotropy we must have $\mu$ equal to 1." This further leads to the constant c. You can follow his argument in reverse, postulate constant c and derive $v\rightarrow -v$. $\endgroup$
    – safesphere
    Oct 27, 2017 at 6:46

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