Simplifying Equations for Lorentz Boosts with Small Beta

[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}

 

[George Jones]

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

g_{\alpha&#039; \beta&#039;} = (\delta^{\mu}_{\alpha} - \xi^{\mu}_{,\alpha})(\delta^{\nu}_{\beta} - \xi^{\nu}_{,\beta})(\eta_{\mu \nu} + h_{\mu \nu})<br />

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

<br /> \xi^{\nu}_{,\beta}h_{\mu \nu}.<br /> [/itex]<br /> <br /> Regards,<br /> George

 

[pervect]

A specific example I recently worked might be of some interest
Starting with the metric

<br /> \left[ \begin {array}{cccc} 1-2\,{\it \Phi0} \left( {\it x1},{\it y1},{\it z1} \right) &amp;0&amp;0&amp;0\\\noalign{\medskip}0&amp;-1-2\,{\it \Phi0} \left( {<br /> \it x1},{\it y1},{\it z1} \right) &amp;0&amp;0\\\noalign{\medskip}0&amp;0&amp;-1-2\,{<br /> \it \Phi0} \left( {\it x1},{\it y1},{\it z1} \right) &amp;0<br /> \\\noalign{\medskip}0&amp;0&amp;0&amp;-1-2\,{\it \Phi0} \left( {\it x1},{\it y1},{<br /> \it z1} \right) \end {array} \right] <br />

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

<br /> \left[ \begin {array}{cccc} 1+{\frac { \left( 2+2\,{\beta}^{2} \right) {\it \Phi0} \left( {\it x1},{\it y1},{\it z1} \right) }{-1+{<br /> \beta}^{2}}}&amp;-4\,{\frac {\beta\,{\it \Phi0} \left( {\it x1},{\it y1},{<br /> \it z1} \right) }{-1+{\beta}^{2}}}&amp;0&amp;0\\\noalign{\medskip}-4\,{\frac {<br /> \beta\,{\it \Phi0} \left( {\it x1},{\it y1},{\it z1} \right) }{-1+{<br /> \beta}^{2}}}&amp;{-1+\frac { \left( 2+2\,{\beta}^{2} \right) {\it &#039;\Phi0}<br /> \left( {\it x1},{\it y1},{\it z1} \right) }{-1+{\beta}^{2}}}}&amp;0&amp;0\\\noalign{\medskip}0&amp;0&amp;-1-2\,{\it <br /> \Phi0} \left( {\it x1},{\it y1},{\it z1} \right) &amp;0\\\noalign{\medskip}0<br /> &amp;0&amp;0&amp;-1-2\,{\it \Phi0} \left( {\it x1},{\it y1},{\it z1} \right) <br /> \end {array} \right]<br />

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

 

[pervect]

Here is eqn 8.21 with the non-linear terms deleted

\Lambda^{\alpha}_{\beta&#039;} = \delta^{\alpha}_{\beta} - \xi^{\alpha}_{,\beta}

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.

 

[Jimmy Snyder]

What sort of transformation is \Lambda supposed to be?

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

 

[Jimmy Snyder]

I think you need to transform \eta too

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.

 

[pervect]

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.

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.

While your relation 8.21 requires


\Lambda^{\alpha}{}_{\beta&#039;} = \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)?

 

[Jimmy Snyder]

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)?

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.