0
$\begingroup$

I want to derive the Lorentz equations without the postulate that the speed of light is constant in all reference frames. Let $L$ denote the matrix that transforms spacetime coordinates $$ \begin{bmatrix}x' \\ t'\end{bmatrix} = L(v) \begin{bmatrix}x \\ t\end{bmatrix} $$ depending on the velocity $v$. So far I have found that $L$ is of the form $$ L(v) = \begin{bmatrix} \gamma & - \gamma v \\ \varepsilon & \gamma \end{bmatrix}, $$ where $\gamma$ and $\varepsilon$ themselves depend on $v$. Since $L$ is an element of the special orthogonal group, $L^{-1} = L^\mathsf{T}$ implies $\varepsilon = \gamma v$ and $\det L = 1$ implies $\gamma = (1 + v^2)^{-1/2}$, completing the proof. But how can I argue that $L$ must be orthogonal in the first place? Why must it preserve the spacetime norm? How is this fact related to the conservation of energy?

$\endgroup$
1
$\begingroup$

There's a mistake in your proof: $L^{-1}\neq L^T$. The correct version is $$L^T \eta L =\eta, \tag1$$ where $\eta$ is the Minkowski metric. That's why you have the wrong $\gamma$. The reason for this is because the Lorentz transformation is not part of the orthogonal group, it's part of the Lorentz group. The upshot is that we need the Lorentz transformations to preserve the quadratic form $$\sum_{\mu,\nu=0}^3 x^\mu\eta_{\mu\nu}y^\nu \tag2$$ for every possible pair of 4-vectors $x^\mu$ and $y^\nu$. From (2) requirement you can derive the requirement in (1).

As for why it must preserve the form in (2), it's a sufficient (and possibly necessary) condition that any linear transformation has to obey if \begin{align} 0 &= \sum_{\mu,\nu=0}^3 \eta_{\mu\nu} \Delta x^\mu \Delta x^\nu \\ & = \sum_{\mu,\nu,\alpha\beta=0}^3 \eta_{\mu\nu} L^\mu_{\hphantom{\mu}\alpha'}\Delta x^{\alpha'} L^\mu_{\hphantom{\nu}\beta'}\Delta x^{\beta'} \end{align} for every $\Delta x^\nu$ that a light ray moves along (fancy way of saying "all observers see light moving with speed $c$").

$\endgroup$
  • $\begingroup$ Thanks alot! I vaguely recalled Lorentz transformations to be rotations but missed that they are not Euclidean. $\endgroup$ – za893411 Feb 16 '18 at 23:53

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.