0
$\begingroup$

The two postulates of special relativity are:

The choice of what inertial frame to use is arbitrary: all laws of physics are invariant. (the principle of relativity)

The metric $$(\Delta s)^2 = (\displaystyle\sum_{\mu=1}^3 \Delta x^\mu)^2 - c^2 (\Delta x^0)^2 = 0$$ for any light-like geodesic. (the invariance of the speed of light)

Now, my question is whether the Lorentz transformation necessary follow from these two postulates, or if the Lorentz transformation is just one possible solution that satisfies the postulates. Why does it has to be of the form $\gamma(x^1-v x^0)$?

You could, of course, have an empirical argument that the Lorentz group is preserved in experiments but that's not really what I'm searching for .

$\endgroup$
  • 1
    $\begingroup$ You should probably be aware that there are many other statements of "the" two postulates of relativity. In fact, this is a new one to me though they seem at first blush like they might be equivalent to the more common ways of writing them. $\endgroup$ – dmckee Feb 12 '16 at 3:58
  • $\begingroup$ Relativity doesn't consider "all laws of physics". The laws of Newtonian mechanics are, for instance, not invariant under a Lorentz transformation. That was exactly the point of Einstein's papers: it's more important to rescue Maxwell's equations than it is to rescue classical mechanics. As for the last part: all of physics stems from empiricism. There are no "postulates", those things all live in the mathematics department. $\endgroup$ – CuriousOne Feb 12 '16 at 4:31
  • $\begingroup$ The Lorentz transformation follows from the assumption that specifically Maxwell's laws are identical in any inertial frame along with the invariance of the speed of light. $\endgroup$ – Chris Feb 12 '16 at 7:35
  • $\begingroup$ Related: physics.stackexchange.com/q/12664/2451 and links therein. Note that the Lorentz group $O(3,1):= \{ \Lambda\in {\rm Mat}_{4\times 4}(\mathbb{R}) \mid \Lambda^T\eta \Lambda = \eta \}$ is defined as the $4\times 4$ matrices that preserve the Minkowski metric $\eta$. $\endgroup$ – Qmechanic Feb 15 '16 at 15:58
0
$\begingroup$

In some sense it is not the only possibility. And that's ignoring the obvious typos like missing signs or inconsistent use of superscripts and subscripts, and the fact that your equations require velocities to be dimensionless to even be dimensionally correct.

In particular the first principle tells you next to nothing if you haven't specified which particular statements you label as laws of physics.

And the second one all by itself is pretty weak as well. Basically you want to say that if two events can be traversed by an object going at speed $c$ then $(\Delta (ct))^2-(\Delta x^1)^2=0$ but that's pretty much a tautology all by itself. The real content should be that different frames agree on whether the object is going at $c$ but you haven't actually stated that $x^0=ct$ or that $c$ is actually a constant that doesn't depend on frame. So even if you find that $ct$ transforms a certain way, this might not give us Lorentz transformations. If $c$ can depend on frame then $t$ can change differently so long as $ct$ transforms correctly.

And your laws of physics might be the same, for instance Maxwell in Gaussian units really only has $ct$ appear.

As you can tell, a lot of this boils down to your statements being vague about what $c$ is. And there are ways to derive the Lorentz transformations by assuming a linear transformation, a group structure, an isotropy, a homogeneity, and then getting a frame invariant speed that is determined empirically (an infinite invariant speed giving Galilean Relativity and an invariant speed equal to $c$ giving Einsteinian Relativity). When you just set it equal to one, you exclude Galilean Relativity, and you make it hard to distinguish all the other Relativities that were possible.

$\endgroup$
0
$\begingroup$

Have you ever heard Taiji relativity ? See by Hsu & Hsu.

They derive the Lorentz transformations and all the results of Special Theory of Relativity just based the principle of relativity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.