GR algebra pretty much (weak limit thm)

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SUMMARY

The discussion centers on the weak limit theorem in the context of recovering Newtonian equations, specifically examining the relationship between the metric tensor components and derivatives in the equations of motion. The user is confused about the equivalence of the expressions ##-1/2\eta^{ij}\partial_jh_{tt}## and ##1/2\partial^{i} h_{tt}##. It is clarified that the confusion arises from the conventions used for the flat space metric tensor, where the sign of the derivatives changes based on the chosen convention. The user is advised that if the first convention is used, then ##\partial_i = - \partial^i## holds true.

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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
 
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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.
 
stevendaryl said:
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..)
 
binbagsss said:
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
 

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