GR algebra pretty much (weak limit thm)

  • Thread starter Thread starter binbagsss
  • Start date Start date
  • Tags Tags
    Algebra Gr Limit
binbagsss
Messages
1,291
Reaction score
12

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
 
Physics news on Phys.org
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
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top