At the risk of adding my confusion to the mix, I'm probably missing something stupidly obvious here, but...
zzzoak, your condition (1) seems to me to say that the coordinate matrix in O of the metric tensor M=\eta = diag(1,-1,-1,-1), in units such that c = 1, which means that O is an orthonormal coordinate system. Together with (1), your condition (3) says the transformation from O to O' is a Lorentz transformation, and that therefore O' is also orthonormal. Is that right? If so, why doesn't the problem become a trivial matter of expressing 1 and -1 in terms of any combination of those variables we like?
fys iks, could you explain (for my curiosity!) how you got those answers?
The transformation is, in matrix form:
T = \begin{bmatrix}<br />
\alpha & \beta & 0 & 0\\ <br />
\mu & \nu & 0 & 0\\ <br />
0 & 0 & a & 0\\ <br />
0 & 0 & 0 & b<br />
\end{bmatrix}
Where mu, nu, alpha and beta are functions of velocity, making T a boost or, synonymously, a hyperbolic rotation.
A general boost along the x-axis is given by
\begin{bmatrix}<br />
\frac{1}{\sqrt{1-v^2}} & \frac{-v}{\sqrt{1-v^2}} & 0 & 0\\ <br />
\frac{-v}{\sqrt{1-v^2}} & \frac{1}{\sqrt{1-v^2}} & 0 & 0\\ <br />
0 & 0 & 1 & 0\\ <br />
0 & 0 & 0 & 1<br />
\end{bmatrix} = \begin{bmatrix}<br />
\cosh(\phi) & -\sinh(\phi) & 0 & 0\\ <br />
-\sinh(\phi) & \cosh(\phi) & 0 & 0\\ <br />
0 & 0 & 1 & 0\\ <br />
0 & 0 & 0 & 1<br />
\end{bmatrix}
where \phi = \text{artanh}(v), the inverse hyperbolic tangent of v. Then \mu\nu - \alpha\beta = 0, and a^2=b^2=1^2=1. But \mu^2-\alpha^2=-\sinh^2(\phi)-\cosh^2(\phi) is never equal to 1 for any real value of phi. And \nu^2-\beta^2=\cosh^2(\phi)+\sinh^2(\phi) is never equal to -1 for any real value of phi.
http://www.wolframalpha.com/input/?i=-sinh^2%28x%29-cosh^2%28x%29%3D1
Whereas:
\mu^2+\alpha^2=-\sinh^2(\phi)+\cosh^2(\phi)=1
-\nu^2-\beta^2=-\cosh^2(\phi)+\sinh^2(\phi)=-1
Even if conditions (1) and (3) weren't requirements of the original problem, shouldn't the answers be consistent with the case where (1) and (3) are true, at least?
On the assumption that
ds^2=x^TMx=(Tx)^TMTx=x^TT^TMTx=(ds')^2
\Rightarrow M=T^TMT
I tried working back from the given answers to M, but this resulted in a horrible mess! Zeros in all the right places, but I had
M_{33}=a^4
M_{44}=b^4
and the top left 2x2 square full of stuff like this:
M_{00}=\alpha^2\mu^2-\alpha^4+\alpha\mu^2\nu+\alpha^2\beta+\mu^2\nu^2-\mu^2\beta^2
Argh.
(By the way, the LaTeX problem you mention in your edit is a bug in PF that appeared a few months ago. I get around it, in Firefox, by copying the message I'm working on, closing the tab I'm typing my message in, and selecting "Clear Recent History" from the Tools menu, then opening a new tab, going back to PF, starting a new message and pasting in where I'd got up to. Bit of a faff, but you get stoical after a while, except late at night, if you keep noticing mistakes to correct...)
