How can I calculate changes in amplitude using gauge transformations in GR?

[Logarythmic]

I have been told that using a metric

$$g_{00} = -a^2(\eta)(1+2\psi)$$
$$g_{oi} = g_{i0} = a^2(\eta)\omega_i$$
$$g_{ij} = a^2(\eta) \left[(1+2\phi)\gamma_{ij} + 2\chi_{ij} \right]$$

and a gauge transformation

$$x^{\bar{\mu}} = x^{\mu} + \xi^{\mu}$$

with

$$\xi^0 = \alpha$$
$$\xi^i = \beta^j$$

gives the changes in the amplitude as

$$\delta \psi = \alpha' + \frac{a'}{a} \alpha$$

and so on.

But how do I calculate these changes? How do I start?

 

[Dick]

The 'gauge' transformation is a coordinate transformation. You know how the metric changes in a coordinate transformation. That should certainly be enough info to start.

 

[Logarythmic]

Yes like

$$g_{\bar{\mu} \bar{\nu}} = \frac{\partial x^{\mu}}{\partial x^{\bar{\mu}}} \frac{\partial x^{\nu}}{\partial x^{\bar{\nu}}} g_{\mu \nu}$$

but I can't get it right. I get that the perturbations change like

$$\delta g = -\partial_{\bar{\nu}} \xi^{\nu} -\partial_{\bar{\mu}} \xi^{\mu}$$

but then what?

 

[Logarythmic]

If

$$g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$$

and

$$g_{00} = -(1+2\psi)$$

then

$$h_{\bar{\mu} \bar{\nu}} = h_{\mu \nu} -\partial_{\mu} \xi_{\nu} -\partial_{\nu} \xi_{\mu}$$

and

$$h_{00} = -2\psi$$.

But how do I get that

$$\psi \rightarrow \psi + \alpha' + \frac{a'}{a}\alpha$$??

If I use the above equation for $h_{\bar{\mu} \bar{\nu}}$ I just get that

$$\psi \rightarrow \psi + \alpha'$$.

Please help someone!

 

[Dick]

As usual, I'm pounding my head over the tex. The extra term comes from the effect of the transformation on the overall scale factor a. a(eta) -> a(eta+alpha) -> a(eta)+alpha*a'(eta) -> a(eta)(1+(a'/a)*alpha). (In case I never get the tex straightened out).

 

[Logarythmic]

I don't get it. Where does this come from?

 

[Dick]

I don't get it. Where does this come from?

The scale factor 'a' changes under the transformation.

 

[Logarythmic]

That I got, but how do I get your transformation?

 

[Dick]

a(eta) goes to a(eta+alpha). I just took the first term of the taylor series expansion of a(eta).

 

[Logarythmic]

I don't follow. How does this couple with my metric transformation?

 

[Dick]

I have to confess, I've only had to deal with this metric perturbation formalism once. And I found it pretty confusing myself. So I'm not sure I can clearly answer your question. But I do know that that is where your extra term comes from. It seems to me there is a review paper around by Brandenberger and Muhkanov that was pretty handy. But I don't have access to it right now.

 

[Logarythmic]

Ok. I think have all papers ever written about this here, but all they say is that "one can easily see that..."

 

[Dick]

Ok. I think have all papers ever written about this here, but all they say is that "one can easily see that..."

Annoying, isn't it?

 

[Logarythmic]

Yes. Very.

 

[nrqed]

Yes. Very.

I guess the point is what is $\eta$ ??

 

[Dick]

In problems like this eta is usually the conformal time. Just a specific parametrization of the time coordinate.

 

[Logarythmic]

Correct...

 

[nrqed]

In problems like this eta is usually the conformal time. Just a specific parametrization of the time coordinate.

Ok. Thanks. What is the definition? What is the relation with $x_0$?

 

[Logarythmic]

Conformal time is defined as

$$\eta = \int_0^{x_0} \frac{dx_0'}{a(x_0')}$$

 

[nrqed]

Conformal time is defined as

$$\eta = \int_0^{x_0} \frac{dx_0'}{a(x_0')}$$

Ok. Then why not simply do the change of coordinates in that expression?

 

[Logarythmic]

?

I have an expression for a gauge transformation for scalars:

\bar{Q}(x^{\mu}) = q(x^{\mu}) - \xi^{\nu}\partial_{\nu}Q[/tex].

This gives the transformation for the scale factor as above. Then I have the transformation for the matric which gives the above expression for $h_{\mu\nu}$, but how can I get these together to yield

$$\psi \rightarrow \psi + \alpha' + \frac{a'}{a}\alpha$$??

The next problem is to find a transformation

$$\omega_i \rightarrow \omega_i - \partial_i\alpha + \beta_i'$$

but that was easy.

The complete line element is

$$ds^2 = a^2(\eta) \left( -(1+2\psi) d\eta^2 + 2\omega_idx^id\eta + \left[ (1+2\psi)\gamma_{ij} + 2\chi_{ij} \right] dx^idx^j \right)$$