# Finding out “the” coordinate transformation to Minkowski metric

This is a homework problem. I am given the following metric:

$$ds^2 = Fc^2dt^2 - \frac{1}{F}dr^2 - r^2d\phi^2 - dz^2~;~F>1, F = F(r)$$

They ask me to find "the" coordinate transformation that will make this metric Minkowski, in cylindrical coordinates. I am confused about two things:

1. Is there a single, unique coordinate transformation given some metric (which is Minkowski in disguise) such that it can be made Minkowski? If so, how to recognize such metrics (without calculating curvature)?

2. How to proceed in this problem? Finding out a $$4\times 4$$ general coordinate transformation that does this job seems like heavy lifting. There must be a clever way to do this!

Thanks.

I understand the quotation like this:

we want to transformed the metric to this shape (Minkowski metric) :

$$ds'^2=d\tau^2-dx^2-dy^2-dz^2$$

$$\left[ \begin {array}{cccc} F{c}^{2}&0&0&0\\0&-{F} ^{-1}&0&0\\ 0&0&-{r}^{2}&0\\0&0&0 &-1\end {array} \right]$$

so we are looking for a transformation matrix $$T$$ that fulfill the equations:

$$T^T\,G\,T=\eta\tag 1$$

with $$\eta$$ the signature matrix:

$$\eta=\left[ \begin {array}{cccc} 1&0&0&0\\ 0&-1&0&0 \\ 0&0&-1&0\\ 0&0&0&-1\end {array} \right]$$

it is easy to see that:

$$T=\left[ \begin {array}{cccc} {\frac {1}{\sqrt {F}c}}&0&0&0 \\ 0&\sqrt {F}&0&0\\ 0&0&{r}^{-1}&0 \\ 0&0&0&1\end {array} \right]$$

fulfill equation (1)

general solution ?

• But is this transformation unique? – Razor Feb 14 '19 at 18:06
• I am not sure. The solution must be real , but this is mathematical problem – Eli Feb 14 '19 at 18:32
• @Razor I took the general solution out – Eli Feb 14 '19 at 19:07