3

In many textbooks, the interval

$$ I = -(c\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 $$

is taken for granted as the same for two events in any reference frame.

Is it possible to derive this just from the two postulates,

  1. That the laws of physics are the same in all reference frames
  2. That the speed of light c is constant for all reference frames?
joshphysics
  • 57,120
rupertonline
  • 397
  • 2
    possible duplicate of Einstein's postulates <==> Minkowski space. (In layman's terms) –  Sep 09 '13 at 01:54
  • Actually, the above is related, but on a second read-through the OP never got what he asked for, which was a simple, algebraic-manipulation explanation. The work of Bacry & Lévy-Leblond is more than a bit technical. –  Sep 09 '13 at 01:57
  • 1
    Yes. See, for example, An Introduction to Mechanics by Kleppner and Kolenkow for a short, algebraic derivation if the Lorentz Transformation along these lines. Otherwise, you can find such derivations all over the web by googling "derivation of Lorentz transformation". Invariance of the interval then follows from the form of the transformation. – joshphysics Sep 09 '13 at 02:16
  • The postulates-to-Lorentz-transformation development can be done several ways. None of them are trivial but many involve only easy steps. Once you have the transformation the invariance of the interval can be proved by tedious but simple direct substitution. I gave this to my modern physics class, and they complained but did it without difficulty. – dmckee --- ex-moderator kitten Sep 09 '13 at 02:38
  • So if I understand correctly, it's possible to get to this via the Lorentz transformations (which are a direct result of the postulates?), but not directly from the postulates with simplicity? – rupertonline Sep 09 '13 at 02:41
  • Hmmm...now that I think about it, the interval for a light ray must be zero (by definition of $c$), and one of the postulates is that these intervals are the same in all frames. So part of the proof is by assumption. I don't know a short way to show the rest, however. – dmckee --- ex-moderator kitten Sep 09 '13 at 02:44
  • That's as far as I got. Reading here on Wikipedia, it states that the second postulate can be used to give a stronger version of this, but I couldn't find any reason this should be so. – rupertonline Sep 09 '13 at 02:49
  • If $I_a = 0$ in frame $a$, then $I_b = 0$ in any other frame $b$ (this follows from 2.). So you have $I_a = f(v_{ab}) I_b$ where $v$ is relative speed, $f$ is unknown function. $f$ cannot depend on time, space or direction of $v$, otherwise laws would be different in places, times, directions. Now consider the third frame $c$: $I_a = f(v_{ac}) I_c$, $I_b = f(v_{bc}) I_c$. On the other hand $I_b = f(v_{bc}) I_c$. Combining them you get $f(v_{bc}) = f(v_{ba})f(v_{ac})$. LHS depends on $v_{ba}$ and $v_{ac}$. However, it also depends on angle between them. But RHS does not – xaxa Sep 10 '13 at 15:13
  • So $f(v)=const$ and $const = const \times const$ thus $const=1$. – xaxa Sep 10 '13 at 15:15

0 Answers0