GR algebra pretty much (weak limit thm)

[binbagsss]

Homework Statement

Hi

I am stuck on a small algebra set in the weak limit theorem to recover Newtonian equations

The text I am looking at:

$\frac{d^2x^i}{ds^2}+\Gamma^i_{tt}\frac{dt}{ds}\frac{dt}{ds}=0$ (1)

$\Gamma^{i}_{tt}=-1/2 \eta^{ij}\partial_{j}h_{tt}$ (to first oder in the metric $h_{uv}$) (2)

$dt/ds \approx 1$
and so using (2), (1) becomes:

$\frac{d^2 x^i}{ds^2}=-1/2\partial_ih_{tt}$ (3)

MY QUESTION

$-1/2\eta^{ij}\partial_jh_{tt}$ in (2)
$= -1/2 \partial^{i} h_tt$

So for (3) I am getting

$\frac{d^2 x^i}{ds^2}=1/2\partial^ih_{tt}$

Im really confused how

$-1/2\eta^{ij}\partial_jh_{tt}=1/2\partial_{i}h_{tt}$ , or at least that is what it looks like has been done.

Many thanks

Homework Equations

see above[/B]

The Attempt at a Solution

see above

 

[stevendaryl]

There are two conventions for the flat space metric tensor (in Cartesian coordinates):

1. $\eta^{tt} = +1, \eta^{xx} = \eta^{yy} = \eta^{zz} =-1$ (all the other components zero)
2. $\eta^{tt} = -1, \eta^{xx} = \eta^{yy} = \eta^{zz} =+1$ (all the other components zero)

If they are using the first convention, then $\partial_i = - \partial^i$.

 

[binbagsss]

There are two conventions for the flat space metric tensor (in Cartesian coordinates):

1. $\eta^{tt} = +1, \eta^{xx} = \eta^{yy} = \eta^{zz} =-1$ (all the other components zero)
2. $\eta^{tt} = -1, \eta^{xx} = \eta^{yy} = \eta^{zz} =+1$ (all the other components zero)

If they are using the first convention, then $\partial_i = - \partial^i$.

I have $\eta^{ij}\partial_{j}=-\partial^i$ , don't know how to show $\partial_i = - \partial^i$. ( well since $\eta$ is diagonal I know I really have $i=j$ but to keep the index notation clear..)

 

[stevendaryl]

I have $\eta^{ij}\partial_{j}=-\partial^i$ , don't know how to show $\partial_i = - \partial^i$. ( well since $\eta$ is diagonal I know I really have $i=j$ but to keep the index notation clear..)

In the usual tensor notation, $\partial_\mu \equiv \sum_{\nu} \eta_{\mu \nu} \partial^\nu$ where $\eta_{\mu \nu}$ is the metric tensor. So if $\eta_{\mu \nu}$ is diagonal with diagonal entries $(+1, -1, -1, -1)$, then

$\partial_t = \partial^t$
$\partial_x = - \partial^x$
$\partial_y = - \partial^y$
$\partial_z = - \partial^z$