# Another question from Schutz

1. Oct 4, 2005

### Jimmy Snyder

I got so much excellent help from my last question that I have decided to take advantage as much as I can. Here is another question.

On page 202, I combine equations 8.18 and 8.19 and change bars for primes (to match eqn. 8.22)

$$g_{\alpha' \beta'} = \eta_{\alpha' \beta'} + \Lambda^{\mu}_{\alpha'}\Lambda^{\nu}_{\beta'}h_{\mu \nu}$$

where $\Lambda$ is a boost. Here is eqn 8.21 with the non-linear terms deleted

$$\Lambda^{\alpha}_{\beta'} = \delta^{\alpha}_{\beta} - \xi^{\alpha}_{,\beta}$$

Combining these I get:

$$g_{\alpha' \beta'} = \eta_{\alpha' \beta'} + (\delta^{\mu}_{\alpha} - \xi^{\mu}_{,\alpha})(\delta^{\nu}_{\beta} - \xi^{\nu}_{,\beta})h_{\mu \nu}$$

using $\eta_{\alpha' \beta'} = \eta_{\alpha \beta}$, expanding the factors, and dropping the term quadratic in $\xi$ I get:

$$g_{\alpha' \beta'} = \eta_{\alpha \beta} + \delta^{\mu}_{\alpha}\delta^{\nu}_{\beta}h_{\mu \nu} - \xi^{\mu}_{,\alpha}\delta^{\nu}_{\beta}h_{\mu \nu} - \delta^{\mu}_{\alpha}\xi^{\nu}_{,\beta}h_{\mu \nu}$$

$$g_{\alpha' \beta'} = \eta_{\alpha \beta} + h_{\alpha \beta} - \xi^{\mu}_{,\alpha}h_{\mu \beta} - \xi^{\nu}_{,\beta}h_{\alpha \nu}$$

Now, finally, comes my question:

How can I use eqn 8.23 to simplify this to 8.22? It looks like Schutz is using h as if it were $\eta$

Here is 8.23

$$\xi_{\alpha} = \eta_{\alpha \beta}\xi^{\beta}$$

Here is 8.22

$$g_{\alpha' \beta'} = \eta_{\alpha \beta} + h_{\alpha \beta} - \xi_{\alpha,\beta} - \xi_{\beta,\alpha}$$

Last edited: Oct 5, 2005
2. Oct 4, 2005

### George Jones

Staff Emeritus
I think you need to transform $\eta$ too, i.e.,

$$g_{\alpha' \beta'} = (\delta^{\mu}_{\alpha} - \xi^{\mu}_{,\alpha})(\delta^{\nu}_{\beta} - \xi^{\nu}_{,\beta})(\eta_{\mu \nu} + h_{\mu \nu})$$

After multiplying everything out and dropping terms that are "second order small", I get (8.22).

Because of (8.13),what I call "second order small" includes terms like

\xi^{\nu}_{,\beta}h_{\mu \nu}. [/itex] Regards, George 3. Oct 4, 2005 ### pervect Staff Emeritus A specific example I recently worked might be of some interest Starting with the metric [tex] \left[ \begin {array}{cccc} 1-2\,{\it \Phi0} \left( {\it x1},{\it y1},{\it z1} \right) &0&0&0\\\noalign{\medskip}0&-1-2\,{\it \Phi0} \left( { \it x1},{\it y1},{\it z1} \right) &0&0\\\noalign{\medskip}0&0&-1-2\,{ \it \Phi0} \left( {\it x1},{\it y1},{\it z1} \right) &0 \\\noalign{\medskip}0&0&0&-1-2\,{\it \Phi0} \left( {\it x1},{\it y1},{ \it z1} \right) \end {array} \right]

when boosted in the 'x' direction by the substitutions for a Lorentz boost
x = (x1+beta*t1)/sqrt(1-beta^2), t=(t1+beta*x1)/sqrt(1-beta^2)

gives

\left[ \begin {array}{cccc} 1+{\frac { \left( 2+2\,{\beta}^{2} \right) {\it \Phi0} \left( {\it x1},{\it y1},{\it z1} \right) }{-1+{ \beta}^{2}}}&-4\,{\frac {\beta\,{\it \Phi0} \left( {\it x1},{\it y1},{ \it z1} \right) }{-1+{\beta}^{2}}}&0&0\\\noalign{\medskip}-4\,{\frac { \beta\,{\it \Phi0} \left( {\it x1},{\it y1},{\it z1} \right) }{-1+{ \beta}^{2}}}&{-1+\frac { \left( 2+2\,{\beta}^{2} \right) {\it '\Phi0} \left( {\it x1},{\it y1},{\it z1} \right) }{-1+{\beta}^{2}}}}&0&0\\\noalign{\medskip}0&0&-1-2\,{\it \Phi0} \left( {\it x1},{\it y1},{\it z1} \right) &0\\\noalign{\medskip}0 &0&0&-1-2\,{\it \Phi0} \left( {\it x1},{\it y1},{\it z1} \right) \end {array} \right]

of course, for a general transformation $\Lambda$, $\eta_{uv}$ might not turn out to be the same after the transform as it did for this example, you have to assume that the transform preserves the lengths of space-time intervals to make this happen.

ps a minor Latex point - note that Lambda^a{}_b gives $$\Lambda^a{}_b$$, properly aligned, as opposed to Lambda^a_b which gives $$\Lambda^a_b$$

4. Oct 5, 2005

### pervect

Staff Emeritus
What sort of transformation is $\Lambda$ supposed to be? It looks like my boost example, for instance, does not have $\Lambda$ in this form for large beta, because the diagonal coefficients are 1/sqrt(1-beta^2), not unity.

5. Oct 5, 2005

### Jimmy Snyder

A boost. I will edit my original post to indicate this.

6. Oct 5, 2005

### Jimmy Snyder

Thanks. As long as you guys are so willing to help, I intend to continue to take advantage. I have no access to a teacher or tutor.

7. Oct 5, 2005

### pervect

Staff Emeritus
OK, that's exactly what I did in my specific example. But note that for large values of beta, the term that maps x into x' is

x' = x/sqrt(1-beta^2)

which has a magnitude of 1/sqrt(1-beta^2), which can be very large, approaching infinity as beta->1.

$$\Lambda^{\alpha}{}_{\beta'} = \delta^{\alpha}{}_{\beta} - \xi^{\alpha}{}_{,\beta}$$

which means that you are assuming that the mapping from x to x' is unity, or close to it.

Perhaps this is a Lorentz boost restricted to small beta, and that only the linear terms in beta are being kept (because beta is assumed to be <<1)?

8. Oct 6, 2005

### Jimmy Snyder

This is my fault. I left out that information (I assumed the reader had a copy of Schutz even though I knew that you don't). There is an extra condition:

$$|\xi^{\alpha}{}_{,\beta}| \ll 1$$

That is why George called the product of this term with h, second order small.