Coordinate transformations on the Minkowski metric

[amathie]

Homework Statement

Take the line element ds^2= (a^2t^2−c^2)dt^2+ 2atdtdx+ dx^2+ dy^2+ dz^2, where a and c are constants. Calculate the components of the inverse metric. By identifying a suitable coordinate transformation, show that the line element can be reduced to the Minkowski line element.

Relevant Equations

Minkowski line element: ds^2 = -dt^2 + dx^2 + dy^2 + dz^2
Matrix diagonalisation: S^-1AS=D

The line element given corresponds to the metric:

$$g = \begin{bmatrix}a^2t^2-c^2 & at & 0 & 0\\at & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{bmatrix}$$

Using the adjugate method: $g^{-1}=\frac{1}{|g|}\tilde{g}$ where $\tilde{g}$ is the adjugate of $g$. This gives me:

$$g^{-1}=\begin{bmatrix}-\frac{1}{c^2} & \frac{at}{c^2} & 0 & 0\\\frac{at}{c^2} & -\frac{a^2t^2}{c^2}+1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{bmatrix}$$

The question now asks me to identify a coordinate transformation to reduce the line element given to the Minkowski line element.

I think what I need to do is find a matrix that operates on $g$ to give $\begin{bmatrix}-1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{bmatrix}$.

I've tried to find the eigenvalues of $g$ by the standard method (i.e. finding the characteristic equation and attempting to solve it), but as far as I can see it just makes an unfactorisable mess.

I get $(a^2t^2-c^2-\lambda)(1-\lambda)(1-\lambda)^2=a^2t^2(1-\lambda)^2$, which gives one degenerate eigenvalue of $\lambda=1$ with multiplicity...3? But I can't find the fourth.

I'm self-teaching this stuff so have no idea what I'm doing really. Any help very appreciated!

 

[Antarres]

Well, finding the eigenvalues of g isn't really the best thing to do. This is because even though we represent $g$ by matrix, we have to keep in mind that $g$ is a (0,2) tensor. What does this mean?
It means that when we act with $g$ on some vector, we don't get another vector. Instead, we get a dual vector. Or in otherwords, we need to act with $g$ on two vectors to get a scalar(this action is called scalar product). So, since we don't get a vector when acting with $g$, we can't talk about eigenvectors, since those don't exist. This is a confusion that can arise when we're not taking into account that this matrix is just a convenient representation of $g$, but we must know how to act with it on vectors in order to make use of it.

Now on to your question about the transformation of $g$. Law of transformation of $g$ under coordinate transformations, reads:

$$g'_{\mu\nu} = \frac{\partial x^\rho}{\partial x'^\mu}\frac{\partial x^\sigma}{\partial x'^\nu}g_{\rho\sigma}$$

Here indices represent different components of $g$, so in your case you can take $g'$ to be standard Minkowski metric, and take $g$ to be your own metric. Summation convention is in use, which means we're summing over indices that are repeating. I will give you example of the first component, so you get the gist of how to use this, in case you haven't used this before.

$$g'_{00} = \frac{\partial x^\rho}{\partial x'^0}\frac{\partial x^\sigma}{\partial x'^0}g_{\rho\sigma} = \frac{\partial x^0}{\partial x'^0}\frac{\partial x^0}{\partial x'^0}g_{00} +2\frac{\partial x^0}{\partial x'^0}\frac{\partial x^1}{\partial x'^0}g_{01} + \frac{\partial x^1}{\partial x'^0}\frac{\partial x^1}{\partial x'^0}g_{11} + \frac{\partial x^2}{\partial x'^0}\frac{\partial x^2}{\partial x'^0}g_{22} + \frac{\partial x^3}{\partial x'^0}\frac{\partial x^3}{\partial x'^0}g_{33} $$

Now you proceed the same thing for the other components that should be transformed, and from the system of equations that you get, you find what the coordinate transformation is. This is like the most straightforward(brute-force) method that you get, so in case you have no other intuition on how to do it, it should work. Either way, even when guessing the transformation, this is how you check if it is right or wrong.

 

[amathie]

Thanks Antarres, that's helpful. And clears up some conceptual issues I was having about the metric.

I've followed what you've suggested, and have ended up with the following set of equations:

$$g'_{00}=-1=(a^2t^2-c^2)(\frac{\partial{t}}{\partial{t'}})^2+2at\frac{\partial{x}}{\partial{t'}}\frac{\partial{t}}{\partial{t'}}+(\frac{\partial{x}}{\partial{t'}})^2+(\frac{\partial{y}}{\partial{t'}})^2+(\frac{\partial{z}}{\partial{t'}})^2$$
$$g'_{11}=1=(a^2t^2-c^2)(\frac{\partial{t}}{\partial{x'}})^2+2at\frac{\partial{x}}{\partial{x'}}\frac{\partial{t}}{\partial{x'}}+(\frac{\partial{x}}{\partial{x'}})^2+(\frac{\partial{y}}{\partial{x'}})^2+(\frac{\partial{z}}{\partial{x'}})^2$$
$$g'_{22}=1=(a^2t^2-c^2)(\frac{\partial{t}}{\partial{y'}})^2+2at\frac{\partial{x}}{\partial{y'}}\frac{\partial{t}}{\partial{y'}}+(\frac{\partial{x}}{\partial{y'}})^2+(\frac{\partial{y}}{\partial{y'}})^2+(\frac{\partial{z}}{\partial{y'}})^2$$
$$g'_{33}=1=(a^2t^2-c^2)(\frac{\partial{t}}{\partial{z'}})^2+2at\frac{\partial{x}}{\partial{z'}}\frac{\partial{t}}{\partial{z'}}+(\frac{\partial{x}}{\partial{z'}})^2+(\frac{\partial{y}}{\partial{z'}})^2+(\frac{\partial{z}}{\partial{z'}})^2$$

Can anyone give me some tips on how I solve for $t', x', y', z'$? I haven't encountered sets of simultaneous PDEs like this before.

 

[PeroK]

Thanks Antarres, that's helpful. And clears up some conceptual issues I was having about the metric.

I've followed what you've suggested, and have ended up with the following set of equations:

$$g'_{00}=-1=(a^2t^2-c^2)(\frac{\partial{t}}{\partial{t'}})^2+2at\frac{\partial{x}}{\partial{t'}}\frac{\partial{t}}{\partial{t'}}+(\frac{\partial{x}}{\partial{t'}})^2+(\frac{\partial{y}}{\partial{t'}})^2+(\frac{\partial{z}}{\partial{t'}})^2$$
$$g'_{11}=1=(a^2t^2-c^2)(\frac{\partial{t}}{\partial{x'}})^2+2at\frac{\partial{x}}{\partial{x'}}\frac{\partial{t}}{\partial{x'}}+(\frac{\partial{x}}{\partial{x'}})^2+(\frac{\partial{y}}{\partial{x'}})^2+(\frac{\partial{z}}{\partial{x'}})^2$$
$$g'_{22}=1=(a^2t^2-c^2)(\frac{\partial{t}}{\partial{y'}})^2+2at\frac{\partial{x}}{\partial{y'}}\frac{\partial{t}}{\partial{y'}}+(\frac{\partial{x}}{\partial{y'}})^2+(\frac{\partial{y}}{\partial{y'}})^2+(\frac{\partial{z}}{\partial{y'}})^2$$
$$g'_{33}=1=(a^2t^2-c^2)(\frac{\partial{t}}{\partial{z'}})^2+2at\frac{\partial{x}}{\partial{z'}}\frac{\partial{t}}{\partial{z'}}+(\frac{\partial{x}}{\partial{z'}})^2+(\frac{\partial{y}}{\partial{z'}})^2+(\frac{\partial{z}}{\partial{z'}})^2$$

Can anyone give me some tips on how I solve for $t', x', y', z'$? I haven't encountered sets of simultaneous PDEs like this before.

The first thing to do is to let $y' =y, z' = z$. That should simplify things.

You need an equation for $g'_{01}$ as well.