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I'm trying to learn special relativity right now, and for various reasons I'd like to start by getting the invariance of proper time, rather than starting with the Lorentz transformations - if you can get the former, you can almost immediately recover the latter by exponentiating basis vectors of the Lie algebra associated with the group of symmetries preserving proper time.

Is there a nice proof out there of the invariance of proper time, starting solely from the postulate of invariance of the speed of light among observers in different inertial frames? I can do this in the case where the proper time is zero, but can't figure out what to do otherwise.


Edit: if anyone is curious about how to recover the Lorentz transformations from the invariance of proper time, let $O(3,1)$ be the group of linear transformations $M$ which leave the quadratic form $$Q = \begin{pmatrix}1 & 0 & 0 &0 \\0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}$$ the same. Then let $\varphi$ be a curve through the identity in $O(3,1)$. Differentiating the identity $\varphi^T Q \varphi = Q$ and evaluating at zero gives the identity $\dot\varphi^T Q = - Q \dot\varphi$. Computing this by coordinates lets you find a basis of $T_{\text{Id}}O(3,1)$ consisting of of skew-symmetric matrices in the upper-left $3 \times 3$ block and matrices whose last row and last column are symmetric. Taking the exponential of each of these gives you the simplest rotations and Lorentz transformations, which then generate the connected component of the Lie group $O(3,1)$.

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    $\begingroup$ Related: physics.stackexchange.com/q/175266/50583 $\endgroup$ – ACuriousMind Nov 1 '15 at 14:15
  • $\begingroup$ I have very little idea about Lie algebra, but I wanted to ask, the procedure you've shown i.e. differentiating a curve in a group where it's value is identity at 0 is equivalent to Lie brackets and pretty much the definition of Lie algebras? $\endgroup$ – Cheeku Nov 1 '15 at 15:04
  • $\begingroup$ @Cheeku: It is quite hard to parse your question. Differentiating curves at $0$ has precisely nothing to do with the definition of a Lie algebra per se, but everything to do with the definition of the Lie algebra associated to a particular Lie group. $\endgroup$ – WillO Nov 1 '15 at 15:12
  • $\begingroup$ @WillO That is what I needed to hear. $\endgroup$ – Cheeku Nov 1 '15 at 15:15
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    $\begingroup$ If you start with the invariance of the speed of light you immediately recover the entire Lorentz symmetry in a few lines. It's hard for me to see where one could squeeze another shortcut to proper time in there or why that may have physical significance. All of this can be done with complete ignorance about Lie groups and algebras. $\endgroup$ – CuriousOne Nov 1 '15 at 15:32
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Do you know Bernard Schutz's book: A First Course in General Relativity? Check out the first chapter of that book. There is a derivation of invariance of proper time using first principles in section 1.6.

Basically, the idea is to start from expressing $\Delta \bar s ^2$ (interval in the barred frame) as a linear combination of $\Delta x_i$'s (vector components in the unbarred frame). Then use tricks in linear algebra and invoke the two postulates of special relativity to eventually arrive at the invariance result. The argument is quite interesting, and should be a fun exercise if you want to challenge yourself.

Good luck!

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