Coordinate transformation into a standard flat metric

In summary, 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.
  • #1
offscene
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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.
 
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  • #2
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?
 
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  • #3
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.
 
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  • #4
offscene said:
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 [tex]ds^2 = -dt^2 + dx^2 = -X^2 dT^2 + dX^2[/tex] you should find [tex]
\begin{split}
\left(\frac{\partial x}{\partial X}\right)^2 - \left(\frac{\partial t}{\partial X}\right)^2 &= 1 \\
\left(\frac{\partial t}{\partial T}\right)^2 - \left(\frac{\partial x}{\partial T}\right)^2 &= X^2 \\
\frac{\partial t}{\partial X}\frac{\partial t}{\partial T} - \frac{\partial x}{\partial T}\frac{\partial x}{\partial X} &= 0\end{split}[/tex] Now the first two equations are satisfied by setting [tex]
\begin{split}
\frac{\partial x}{\partial X} = \cosh \zeta \quad \frac{\partial t}{\partial X} &= \sinh \zeta \\
\frac{\partial t}{\partial T} = X\cosh \eta \quad \frac{\partial x}{\partial T} &= X\sinh \eta \end{split}
[/tex] due to the identify [itex]\cosh^2 u - \sinh^2 u = 1[/itex]. At present [itex]\eta[/itex] and [itex]\zeta[/itex] are unknown functions of [itex]X[/itex] and [itex]T[/itex], but the third equation gives [tex]
X (\sinh \zeta \cosh \eta - \cosh \zeta \sinh \eta) = X \sinh(\zeta - \eta) = 0[/tex] so that [itex]\zeta = \eta[/itex]. The problem is now reduced to finding [itex]\eta[/itex]. We can either find by inspection that [itex]\eta = T[/itex] will work, or we can use equality of mixed partials to find that [tex]
\begin{split}
\frac{\partial^2 t}{\partial X\,\partial T} - \frac{\partial^2 t}{\partial T\,\partial X} =
\left(1 - \frac{\partial \eta}{\partial T}\right) \sinh \eta + X\frac{\partial \eta}{\partial X} \cosh \eta &= 0 \\
\frac{\partial^2 x}{\partial X\,\partial T} - \frac{\partial^2 x}{\partial T\,\partial X} =
\left(1 - \frac{\partial \eta}{\partial T}\right) \cosh \eta + X\frac{\partial \eta}{\partial X} \sinh \eta &= 0
\end{split}[/tex] and solving for the partial derivatives we find [tex]
1 - \frac{\partial \eta}{\partial T} = 0 = X\frac{\partial \eta}{\partial X}.[/tex]
 
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