Diagonalizing a metric by a coordinate transformation

[Lilian Sa]

Homework Statement

gravity and mertices

Relevant Equations

coordinate transformation

I posted a thread yesterday and I think that I did not formulated it properly.
So I have a metric ${ds}^{2}=-{dt}^{2}+{dx}^{2}+2{a}^2(t)dxdy+{dz}^{2}$
I was asked to find the the coordinate transformation so that I can get a diagonalized metric.
so what I've done is I assumed a coordinate transformation $x=\tilde{x}+F(t,y)$ replaced it in the metric and equated to zero for the proper elements.
but what does that says that F have to be dependent on the other coordinates?
I've got complicated with it.
thanks for any help :)

 

[stevendaryl]

If you make the substitution $x = \tilde{x} + F(t,y)$, then $dx = d \tilde{x} + (\partial_t F) dt + (\partial_y F) dy$. So in your original expression for $ds^2$, try replacing $dx$ by the more complicated expression and see if you can find a choice of $F(x,y)$ that will make the cross-terms such as $dx dy$ vanish.

 

[Lilian Sa]

If you make the substitution $x = \tilde{x} + F(t,y)$, then $dx = d \tilde{x} + (\partial_t F) dt + (\partial_y F) dy$. So in your original expression for $ds^2$, try replacing $dx$ by the more complicated expression and see if you can find a choice of $F(x,y)$ that will make the cross-terms such as $dx dy$ vanish.

This is what I got at the end:
$ds^2=-dt^2(1-\partial_tF)+d\tilde{x}^2+(\partial_yF+2a^2\partial_yF)dy^2+dz^2+d\tilde{x}dy(sa^2+2\partial_yF)+dtdy(2a^2\partial_tF+2\partial_tF\partial_yF)+d\tilde{x}dt(2\partial_tF)$
that means that I got $F$ does not depend on any coordinate!
but this is not true, there's something wrong with the answer right?

 

[stevendaryl]

You didn't necessarily make a mistake. It might be that you just made an incorrect guess about the form of the transformation.

In general, you would have $t, x, y, z$ as functions of $\tilde{t}, \tilde{x}, \tilde{y}, \tilde{z}$, so there would be 4 different functions and 16 partial derivatives to worry about. Hopefully, you don't need that level of complication, but your first guess was too simple.

The next thing to try is to transform both $x$ and $y$ (and leave $t$ and $z$ alone).

Let $x = \tilde{x} + F(\tilde{y}, \tilde{t})$ and $y = G(\tilde{y}, \tilde{t})$.

Then $dx = d \tilde{x} + \dfrac{\partial F}{\partial \tilde{y}} d \tilde{y} + \dfrac{\partial F}{\partial t} dt$.
$dy = \dfrac{\partial G}{\partial \tilde{y}} d \tilde{y} + \dfrac{\partial G}{\partial t} dt$.

 

[stevendaryl]

I spent more time than I should have trying to solve this problem, and I don't actually know how to do it. If $a^2$ were constant, then you could let $x = \tilde{x} - a^2 y$, and that would diagonalize it. However, if $a^2$ is a function of $t$, then $dx = d \tilde{x} - \dfrac{d(a^2)}{dt} y dt - a^2 dy$. Then when you square it, you get additional cross-terms $-2 \dfrac{d(a^2)}{dt} y dt d \tilde{x}$.

 

[Lilian Sa]

I spent more time than I should have trying to solve this problem, and I don't actually know how to do it. If $a^2$ were constant, then you could let $x = \tilde{x} - a^2 y$, and that would diagonalize it. However, if $a^2$ is a function of $t$, then $dx = d \tilde{x} - \dfrac{d(a^2)}{dt} y dt - a^2 dy$. Then when you square it, you get additional cross-terms $-2 \dfrac{d(a^2)}{dt} y dt d \tilde{x}$.

but this additional cross term we have to equate it to zero and then we get the expression equal to zero, but that doesn't help :(
or I am wrong?

 

[Lilian Sa]

You didn't necessarily make a mistake. It might be that you just made an incorrect guess about the form of the transformation.

In general, you would have $t, x, y, z$ as functions of $\tilde{t}, \tilde{x}, \tilde{y}, \tilde{z}$, so there would be 4 different functions and 16 partial derivatives to worry about. Hopefully, you don't need that level of complication, but your first guess was too simple.

The next thing to try is to transform both $x$ and $y$ (and leave $t$ and $z$ alone).

Let $x = \tilde{x} + F(\tilde{y}, \tilde{t})$ and $y = G(\tilde{y}, \tilde{t})$.

Then $dx = d \tilde{x} + \dfrac{\partial F}{\partial \tilde{y}} d \tilde{y} + \dfrac{\partial F}{\partial t} dt$.
$dy = \dfrac{\partial G}{\partial \tilde{y}} d \tilde{y} + \dfrac{\partial G}{\partial t} dt$.

I'll try rhis today, hope it will work.
thanks very much!

 

[stevendaryl]

I'll try rhis today, hope it will work.
thanks very much!

I'm not convinced that there is any simple solution. The most general transformation involves three functions

$t = T(\tilde{t}, \tilde{x}, \tilde{y})$
$x = X(\tilde{t}, \tilde{x}, \tilde{y})$
$y = Y(\tilde{t}, \tilde{x}, \tilde{y})$

Then there are 9 partial derivatives:
$\partial_{\tilde t} T, \partial_{\tilde x} T, \partial_{\tilde y} T$
$\partial_{\tilde t} X, \partial_{\tilde x} X, \partial_{\tilde y} X$
$\partial_{\tilde t} Y, \partial_{\tilde x} Y, \partial_{\tilde y} T$

The constraint that $ds^2$ is diagonal in the $\tilde{t}, \tilde{x}, \tilde{y}$ coordinate system gives 3 equations. So we have more unknowns than equations, so we should have the flexibility to choose simplifying assumptions. For example, assume that $T$ does not depend on $\tilde{y}$ or that $Y$ does not depend on $\tilde{t}$. But every such choice that I have thought of has led to an inconsistent set of equations.

An approach that might give you an answer "in the limit" is this:

As I mentioned earlier, the case where $a^2$ is constant is diagonalizable. So my idea is to assume that

$a^2 = q(\lambda t)$

and expand $a^2$ in a power series in $\lambda$:

$a^2 = q(0) + \lambda q'(0) + \lambda^2/2 q''(0) + ...$

Now, you perform a sequence of transformations.

The first transformation attempts to make the non-diagonal terms of order $\lambda$.
The second transformation attempts to make the non-diagonal terms of order $\lambda^2$.
Etc.

 

[vela]

So I have a metric ${ds}^{2}=-{dt}^{2}+{dx}^{2}+2{a}^2(t)dxdy+{dz}^{2}$

Can you verify that there's no $dy^2$ term in the original metric? In your other thread, the matrix you started with included one.

 

[Lilian Sa]

Can you verify that there's no $dy^2$ term in the original metric? In your other thread, the matrix you started with included one.

I don't think so, because we have to get a diagonalized metric with the four coordinates, which include Y or the transformed Y.

 

[stevendaryl]

Wow! I found a mind-blowing solution to a related problem. Suppose instead of $a^2$ being a function of $t$, it was a function of $\tilde{t}$?

So we make the following simplifying assumptions for this case:

$t = T(\tilde{x}, \tilde{y})$ (note: $t$ does NOT depend on $\tilde{t}$)
$y = Y(\tilde{x}, \tilde{y})$ (note: $y$ also does not depend on $\tilde{t}$)
$x = X(\tilde{x}, \tilde{t})$ (note: $x$ does not depend on $\tilde{y}$

Then it is possible to choose a transformation that makes $ds^2$ diagonal, as long as $a^2$ is a function of only $\tilde{t}$ (rather than $t$).

 

[Lilian Sa]

Wow! I found a mind-blowing solution to a related problem. Suppose instead of $a^2$ being a function of $t$, it was a function of $\tilde{t}$?

So we make the following simplifying assumptions for this case:

$t = T(\tilde{x}, \tilde{y})$ (note: $t$ does NOT depend on $\tilde{t}$)
$y = Y(\tilde{x}, \tilde{y})$ (note: $y$ also does not depend on $\tilde{t}$)
$x = X(\tilde{x}, \tilde{t})$ (note: $x$ does not depend on $\tilde{y}$

Then it is possible to choose a transformation that makes $ds^2$ diagonal, as long as $a^2$ is a function of only $\tilde{t}$ (rather than $t$).

But it was given in the question that $a$ depends on $t$ , I don't know if its legal to suppose that it does not depend on $t$. :(

 

[stevendaryl]

But it was given in the question that $a$ depends on $t$ , I don't know if its legal to suppose that it does not depend on $t$. :(

No, it's not. I don't think the original problem has any good solutions, though.