Coordinate transformation into a standard flat metric

[offscene]

Homework Statement

Starting from the metric $ds^2=-X^2dT^2+dX^2$, using a coordinate transformation of T and X to t and x, convert it to the form ds^2=-dt^2+dx^2 (standard flat 2D metric).

Relevant Equations

Not actually a homework problem ("confession" from example 7.3 in Hartle's book for Gravity and GR) and no other relevant equations I can think of besides the standard chain rule but thought that this was the most fitting place to ask.

I started by expanding $dx$ and $dt$ using chain rule:

$$dt = \frac{dt}{dX}dX+\frac{dt}{dT}dT$$
$$dx = \frac{dx}{dX}dX+\frac{dx}{dT}dT$$

and then expressing $ds^2$ as such:

$$ds^2 = \left(\left(\frac{dt}{dX}\right)^2+\left(\frac{dt}{dX}\right)^2\right)dX^2+\left(\left(\frac{dt}{dT}\right)^2+\left(\frac{dt}{dT}\right)^2\right)dT^2 + 2\left(\frac{dt}{dX}\frac{dt}{dT}+\frac{dx}{dT}\frac{dx}{dX}\right)$$

But after matching the coefficients to the original $ds^2$, I am unable to solve the equations to come up with the right transformation and was wondering if anyone could point me in the right direction/show me.

 

[offscene]

Nevermind, I found the transformation with $x = X\cosh(T)$ and $t = X\sinh(T)$ with some guess and check but is there a cleaner way to do this?

 

[martinbn]

The null geodesics in the usual Minkowski coordinates are given by $x\pm t = const$. You can find them in the given coordinates and use that to set the coordinate change that matches them. In your case the null curves are given by $dX^2=X^2dT^2$, which can be solved easily and gives $Xe^{\pm T} = const$ (you don't have to check that these are geodesics, if the change of variables works). So setting $x+t = Xe^T$ and $x-t = Xe^{-T}$ gives you the ones you found.

 

[pasmith]

Homework Statement:: Starting from the metric $ds^2=-X^2dT^2+dX^2$, using a coordinate transformation of T and X to t and x, convert it to the form ds^2=-dt^2+dx^2 (standard flat 2D metric).
Relevant Equations:: Not actually a homework problem ("confession" from example 7.3 in Hartle's book for Gravity and GR) and no other relevant equations I can think of besides the standard chain rule but thought that this was the most fitting place to ask.

I started by expanding $dx$ and $dt$ using chain rule:

But after matching the coefficients to the original $ds^2$, I am unable to solve the equations to come up with the right transformation and was wondering if anyone could point me in the right direction/show me.

Using the chain rule in ds^2 = -dt^2 + dx^2 = -X^2 dT^2 + dX^2 you should find <br /> \begin{split}<br /> \left(\frac{\partial x}{\partial X}\right)^2 - \left(\frac{\partial t}{\partial X}\right)^2 &amp;= 1 \\<br /> \left(\frac{\partial t}{\partial T}\right)^2 - \left(\frac{\partial x}{\partial T}\right)^2 &amp;= X^2 \\<br /> \frac{\partial t}{\partial X}\frac{\partial t}{\partial T} - \frac{\partial x}{\partial T}\frac{\partial x}{\partial X} &amp;= 0\end{split} Now the first two equations are satisfied by setting <br /> \begin{split}<br /> \frac{\partial x}{\partial X} = \cosh \zeta \quad \frac{\partial t}{\partial X} &amp;= \sinh \zeta \\<br /> \frac{\partial t}{\partial T} = X\cosh \eta \quad \frac{\partial x}{\partial T} &amp;= X\sinh \eta \end{split}<br /> due to the identify \cosh^2 u - \sinh^2 u = 1. At present \eta and \zeta are unknown functions of X and T, but the third equation gives <br /> X (\sinh \zeta \cosh \eta - \cosh \zeta \sinh \eta) = X \sinh(\zeta - \eta) = 0 so that \zeta = \eta. The problem is now reduced to finding \eta. We can either find by inspection that \eta = T will work, or we can use equality of mixed partials to find that <br /> \begin{split}<br /> \frac{\partial^2 t}{\partial X\,\partial T} - \frac{\partial^2 t}{\partial T\,\partial X} = <br /> \left(1 - \frac{\partial \eta}{\partial T}\right) \sinh \eta + X\frac{\partial \eta}{\partial X} \cosh \eta &amp;= 0 \\<br /> \frac{\partial^2 x}{\partial X\,\partial T} - \frac{\partial^2 x}{\partial T\,\partial X} = <br /> \left(1 - \frac{\partial \eta}{\partial T}\right) \cosh \eta + X\frac{\partial \eta}{\partial X} \sinh \eta &amp;= 0<br /> \end{split} and solving for the partial derivatives we find <br /> 1 - \frac{\partial \eta}{\partial T} = 0 = X\frac{\partial \eta}{\partial X}.