A Prove 2-D Lorentzian Metric is Locally Equivalent to Standard Form

[Pentaquark5]

Hi, how can I prove that any 2-dim Lorentzian metric can locally be brought to the form

$$g=2 g_{uv}(u,v) \mathrm{d}u \mathrm{d}v=2 g_{uv}(-\mathrm{d}t^2+dr^2)$$

in which the light-cones have slopes one?

Thanks!

 

[Orodruin]

By changing coordinates to a set of coordinates that diagonalise the metric locally. This should just be an exercise in diagonalising a 2x2 matrix.

 

[stevendaryl]

Your notation is a little mixed up. Typically, we would write:

ds^2 = \sum_{u v} g_{uv} dx^u dx^v

where ds is the line element, sort of an infinitesimal length, and u, v range over the indices for your coordinate system, and g_{uv} is a component of the metric tensor in that coordinate system. There is no factor of 2, and it doesn't make sense to write g_{uv} (dt^2 - dx^2). u and v are dummy indices, while t and x are specific coordinates. It would make sense to write:

ds^2 = g_{tt} dt^2 + g_{xx} dx^2

using t and x as both the coordinates and the indices, or you can write it as:

ds^2 = g_{00} (dx^0)^2 + g_{11} (dx^1)^2

where x^0 \equiv t and x^1 \equiv x

Let's not prejudice ourselves by writing x and t, but instead start with arbitrary coordinates u and v. Then we can write it as a matrix problem:

ds^2 = \left( \begin{array} \\ du & dv \end{array} \right) \left( \begin{array} \\ g_{uu} & g_{uv} \\ g_{vu} & g_{vv} \end{array} \right) \left( \begin{array} \\ du \\ dv \end{array} \right) = g_{uu} du^2 + g_{uv} du dv + g_{vu} dv du + g_{vv} dv^2

Then the issue is to change coordinates from u, v to x, t via a transformation matrix K:

\left( \begin{array} \\ du \\ dv \end{array} \right) = \left( \begin{array} \\ K_{ut} & K_{ux} \\ K_{vt} & K_{vx} \end{array} \right) \left( \begin{array} \\ dt \\ dx \end{array} \right)

Then you want to choose K such that \tilde{g} \equiv K^T g K is diagonal (as a matrix equation), where K^T means the transpose of K. Then in terms of \tilde{g}, you have:

ds^2 = \left( \begin{array} \\ dt & dx \end{array} \right) \left( \begin{array} \\ \tilde{g}_{tt} & 0 \\ 0 & \tilde{g}_{xx} \end{array} \right) \left( \begin{array} \\ dt \\ dx \end{array} \right) = \tilde{g}_{tt} dt^2 + \tilde{g}_{xx} dx^2

Once the metric is diagonal, you can get it into the form: ds^2 = g_{tt} (dt^2 - dx^2) by just scaling x: x \rightarrow \frac{\sqrt{g_{tt}}}{\sqrt{-g_{xx}}} x

Note, that all of this is taking place at a single point. That is, you can always pick coordinates so that g_{xt} = 0 at a single point. I'm actually not sure if it's always possible to make the metric diagonal at every point.

 

[Orodruin]

Your notation is a little mixed up. Typically, we would write:

ds^2 = \sum_{u v} g_{uv} dx^u dx^v

where ds is the line element, sort of an infinitesimal length, and u, v range over the indices for your coordinate system, and g_{uv} is a component of the metric tensor in that coordinate system.

The given metric is in light-cone coordinates where there is only one term (remembering that the metric tensor is symmetric) since both coordinate directions are null. The corresponding metric component is $g_{uv} = g_{vu}$. The problem is to rewrite this in another set of coordinates such that the metric components become $g_{tt} = -g_{xx} = 2g_{uv}$.

Edit: Note that physicists will often write $2 du\, dv$ when they really mean $du\otimes dv + dv\otimes du$.

I'm actually not sure if it's always possible to make the metric diagonal at every point.

The metric is symmetric. It is always possible to find an orthogonal basis to a real symmetric matrix.

 

[stevendaryl]

The metric is symmetric. It is always possible to find an orthogonal basis to a real symmetric matrix.

That wasn't the question.

It's certainly true that for any specific point \mathcal{P} there is a coordinate system x, t such that g_{x t} = 0 at point \mathcal{P}. But it doesn't obviously follow from that that g_{x t} = 0 everywhere.

 

[stevendaryl]

The given metric is in light-cone coordinates where there is only one term (remembering that the metric tensor is symmetric) since both coordinate directions are null.

Ah. I misunderstood. I thought that there was an implicit summation going on, so that g_{uv} du dv meant a sum over all possible u and v.

 

[martinbn]

More generally https://en.wikipedia.org/wiki/Isothermal_coordinates

 

[Orodruin]

That wasn't the question.

Sorry, I read your post a bit fast. I was already in "local mode" since it was mentioned in the OP.

 

[Pentaquark5]

Thanks everybody for your help!

I got thrown off by a Hint that said "use coordinates associated with right-going and left-going null geodesics." which made it sound much more complicated than it was.

 

[martinbn]

Thanks everybody for your help!

I got thrown off by a Hint that said "use coordinates associated with right-going and left-going null geodesics." which made it sound much more complicated than it was.

$u=r-t$ and $v=r+t$