0
$\begingroup$

If suppose we are given with two metrics of spacetime.

$$ds^2=(1+h)^2d\tau^2 -dh^2$$ and $$ds^2=\rho^2dt^2 -d\rho^2$$

Then is it possible to find the relationship between these two coordinate systems without dealing with metric transformation equation?

I was using the following equation to find the relationship.

$$ g^{'}_{\mu \nu} = g_{ij} \dfrac{\partial x^i}{\partial x^\mu } \dfrac{\partial x^j}{\partial x^\nu } $$

But it leads to quite complicated PDEs. Is there any other way?

$\endgroup$

1 Answer 1

1
$\begingroup$

In general you could propose a transformation

$$ t=f(\tau,h)\quad\rho=g(\tau,h)\Rightarrow dt=\partial_\tau f d\tau+\partial_hf dh\quad d\rho=\partial_\tau g d\tau+\partial_hg dh $$

and introduce this change in $ds^2$, which gives you a series of constraints in the functional form of the change of coordinates:

$$ ds^2=\rho^2dt^2+d\rho^2=g^2(\partial_\tau f d\tau+\partial_hf dh)^2+(\partial_\tau g d\tau+\partial_hg dh)^2=Ad\tau^2+Bdh^2+Cd\tau dh $$

with $A=g^2(\partial_\tau f)^2 +(\partial_\tau g)^2$, $B=g^2(\partial_h f)^2 +(\partial_h g)^2$ and $C=2(g^2\partial_\tau f\partial_h f+\partial_\tau g\partial_h f)$. By direct comparation you obtain a horrible system of partial equations, that in principle can be solved. In particular for this problem $A=(1+h)^2$, $B=1$, and $C=0$. Good luck.

I do not recommend this way, since even in very easy metrics, can end in awfully long calculations. However you do not need in general the most general change of coordinates that relates this two metric, so it is easy, in general, to impose, ad hoc, some extra conditions. For example, the metric is static ($t$ and $\tau$ do not appear explicitly in the metric components, so we can try to avoid including the time coordinate in our transformations

$$ t=\tau\quad\rho=g(h)\Rightarrow dt=d\tau\quad d\rho=\partial_hg dh $$

In this way the system looks tractable:

$$ A= g^2 = (1+h)^2, \\ B=(\partial_h g)^2 =1, \\ C=0=0 (\text{Trivially satisfied}) $$

With this simplification, the first equation gives the change of variables $\rho=1+h$.

It can be seen as if this trick of looking at shortcuts is it possible for specific situations, and for a general metric it will not be useful. And this is exactly the case for GR. A theory made to make the physics coordinate free will produce, complex problems if you do not prepare the examples correctly. Conclusion: If you can, try to look for simplifications in the change of variables or in the system of differential equations. If not, try a equation solver.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.