2
$\begingroup$

Given the two-dimensional metric $$ds^2=-r^2dt^2+dr^2$$ How can I find a coordinate transformation such that this metric reduces to the two-dimensional Minkowski metric?

I know that $g_{\mu\nu}=\begin{pmatrix}-r^2&0\\0&1\end{pmatrix}$ (this metric) and $\eta_{\mu\nu}=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ (Minkowski). Obviously, the matrix transformation is $\begin{pmatrix}1/r^2&0\\0&1\end{pmatrix}g_{\mu\nu}=\eta_{\mu\nu}$, but how is that related to the coordinate transformation itself?


EDIT: would the following transformation be acceptable? $$r'=r\cosh t$$ $$t'=r\sinh t$$ Such that: $dr'=\cosh t\ dr+r\sinh t\ dt,\quad dt'=\sinh t\ dr+r\cosh t\ dt$

And: $ds'^2=-dt'^2+dr'^2=-r^2dt^2+dr^2=ds^2$

Where we have: $ds'^2=\eta_{\mu\nu}dx^{\mu}dx^{\nu}$ as requested.

Is that correct? Also, is there a formal way of "deriving" the proper change of coordinates (since mine is more of an educated guess)?

$\endgroup$
2
$\begingroup$

If you were to Wick rotate $t \rightarrow i \theta$, the metric would be $ds^2 = dr^2 + r^2 d\theta^2$, which is just flat space in polar coordinates. The standard cartesian coordinates can be obtained by $x=r\cos\theta$, $y=r\sin\theta$. The same procedure works in the original Lorentzian signature metric, but with hyperbolic trig functions instead of sines and cosines.

By the way, this is two-dimensional Rindler space, which is just a patch of two-dimensional Minkowski space: http://en.wikipedia.org/wiki/Rindler_coordinates.

| cite | improve this answer | |
$\endgroup$
12
$\begingroup$

In the general case you want the Cartan-Karlhede algorithm. It is an algorithm for producing a complete set of classifying invariants for a metric, expressed as functions of the coordinates. Given the components of the metric $g$ in the coordinates $x_1, x_2, \ldots$, the algorithm produces a list \begin{align} \Lambda & = \Lambda(x_i) \\ \Psi_k & = \Psi_k(x_i) \quad k = 0,\ldots, 4 \\ R_{kj} & = R_{kj}(x_i) \quad k,j = 0,\ldots, 2 \\ \Lambda_{00} & = \Lambda_{00}(x_i)\\ & \;\; \vdots \end{align} where each quantity is defined in a way that is coordinate independent. (The names here are standard notation, but what each of them is, is a little beyond the scope of this answer.) This is in contrast to a quantity like $g_{00}$ whose value at a point depends on your coordinates. Of course, expressed as a function of coordinates, $\Lambda$ and the others may look very different in various coordinate systems, but at corresponding points, the value is the same.

Then if we have two metrics given, we can run the algorithm on both. If the metrics are really the same, but different coordinates, the invariants must agree. This gives a system of equations, \begin{align} \Lambda(x_i) & = \Lambda'(y_j) \\ \Psi_k(x_i) & = \Psi_k'(y_j) \\ & \;\; \vdots \end{align} which may or may not have a solution, $x_i = x_i(y_j)$. (For example if you do this with two Schwarzschild metrics in the standard coordinates, you find that it is necessary that $m = m'$.) If there is a solution, this is your change of coordinates.

There is a caveat to the preceding. It may be that not all the equations are independent. In $n$ dimensions we need $n$ equations but the algorithm may produce fewer independent equations. This happens precisely when there is a symmetry in the spacetime. Then there cannot a unique change of coordinates, because at least one coordinate is superfluous. Indeed for the case of flat metrics the entire system is just $0 = 0$.

In this case the algorithm only establishes that there exists a change of coordinates, but you have to look at some other invariant information to find a coordinate change. (You will not be able to find a unique change of coordinates because there are many.) One piece of such information is the Killing vectors.

(This particular case is amenable to the brute force method demonstrated in the other answers, but a more complicated metric in more than two dimensions is not.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This is a very interesting answer, although I would say that it's an answer to a different question. In the case at hand, the space is 2d Rindler space, which is maximally symmetric, and as you point out, this algorithm is useless. In any event, I'm now curious about the algorithm. Is the list of invariant quantities finite? $\endgroup$ – Surgical Commander Mar 2 '15 at 0:23
  • 1
    $\begingroup$ Yes. The invariants are formed from the Riemann tensor and its derivatives. In each iteration another derivative is taken. The algorithm will terminate at the latest after taking the seventh derivative. I do not recall the exact number of invariants that has to be computed in the worst case, but you need fewer than 2000. $\endgroup$ – Robin Ekman Mar 2 '15 at 0:31
  • $\begingroup$ Oh, the invariants are independent of the number of dimensions? Or is that number specific to a certain dimension? $\endgroup$ – Surgical Commander Mar 2 '15 at 0:33
  • $\begingroup$ That's for dimension 4. In general in $d$ dimensions you need $d(d+1)/2$ derivatives. In dimension 4 however it turns out that 7 derivatives are always sufficient because you can use the Petrov classification. $\endgroup$ – Robin Ekman Mar 2 '15 at 0:45
  • $\begingroup$ This answer is very appropriate for this question. $\endgroup$ – user10851 Mar 2 '16 at 3:00
4
$\begingroup$

You can also do the following, which may not be as general as you want it, but the idea might be usefull for other problems. You already know that the given metric is Minkowski metric in different coordinates, so look at the null geodesics. In the usual coordinates $(t',x')$ they are given by $x'\pm t'=const$. Then find the null geodesics in the given coordinates by setting the line element to zero i.e. $-t^2dr^2+dt^2=0$. Integrating gives you $re^{\pm t}=const$. You don't even have to prove that these null curves are geodesics, all you need is the transformations that will give the standard Minkowski metric. From here you take the transformations to be

$$x'+t'=re^t$$ $$x'-t'=re^{-t}$$

Solve for $x'$ and $t'$ and check that it works, that $dt'^2-dx'^2=-t^2dr^2+dt^2$.

| cite | improve this answer | |
$\endgroup$

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.