Coordinate transformation to flat space

[stampita]

Homework Statement

find the transformation that turns this metric:
$$ds^2=-X^2dT^2+dX^2$$

into the minkowski metric: diag(-1,1).

The Attempt at a Solution

attempt 1:
I transformed the above metric into the coordinates that use the lines that define the light cone as the axes. Namely,
$$u=T+ln(X)$$
$$v=T-ln(X)$$

attempt 2:
If I can transform the above, into the coordinates u and v such that i have
$$ds^2=-dudv$$
then I'm done because with another transformation:
$$dt'=\frac{du+dv}{2}$$
$$dx'=\frac{du-dv}{2}$$
I have the relation
$$-(dt')^2+dx^2=-(\frac{du+dv}{2})^2+(\frac{du-dv}{2})^2=-dudv.$$

Is there a systematic way of doing this?

 

[Jhero]

Thank you for your question. To find the transformation that turns the given metric into the Minkowski metric, we can use the following steps:

1. Write the Minkowski metric in terms of the given coordinates:
ds^2 = -dT^2 + dX^2 = -[d(T + X)]^2 + [d(T - X)]^2

2. Equate the coefficients of the two metrics to get the transformation equations:
T' = T + X
X' = T - X

3. Substitute these transformations into the given metric:
ds^2 = -X'^2dT'^2 + dX'^2 = -[T'^2 - X'^2]dT'^2 + [T'^2 - X'^2]dX'^2

4. Simplify the above equation to get the desired form:
ds^2 = -dT'^2 + dX'^2

Therefore, the transformation that turns the given metric into the Minkowski metric is:
T' = T + X
X' = T - X

I hope this helps. Please let me know if you have any further questions.