
#1
Mar912, 04:37 AM

P: 56

Yes, I want to make sure that geodesics of a particle moving in curved space time is the same thing of projectile trajectories.
I start from assuming that [itex]1\frac{2GM}{r}\approx12gr[/itex] and then calculate the schwarzschild metric in this form [itex]\Sigma_{\mu\nu}=\begin{bmatrix}\sigma & 0\\ 0 & \sigma^{1}\end{bmatrix}[/itex] where [itex]\sigma = 12gr[/itex] and I calculated for the Christoffel symbols for this metric: [itex]\Gamma^0_{\mu\nu}=\sigma g\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}[/itex] [itex]\Gamma^1_{\mu\nu}=\frac{g}{\sigma^2}\Sigma_{\mu\nu}[/itex] I plugged them to a geodesics equation [itex]\partial^2_\tau x^\mu = \Gamma^\mu_{\alpha\beta}\partial_\tau x^\alpha\partial_\tau x^\beta[/itex] where [itex]d\tau^2 = dx^\mu dx^\nu\Sigma_{\mu\nu}[/itex] and I got these ugly conditions: [itex]\partial^2_\tau t = \sigma\partial_\tau t\partial_\tau \sigma[/itex] [itex]\partial^2_\tau \sigma = \frac{2g^2}{\sigma^2}[/itex] what I expect is just something like [itex]x=\frac{g}{2}t^2[/itex] I havn't finished these differential equations yet. But I want to know that I'm going through the right track, right? Any suggestion? 



#2
Mar912, 10:20 AM

Emeritus
Sci Advisor
PF Gold
P: 5,500

In the level of approximation you're using, you might as well make your life easier and approximate [itex]\sigma^{1}[/itex] as [itex]1+2gr[/itex]. I would also define a new coordinate [itex]\rho=gr[/itex] to avoid having to write all the factors of g.




#3
Mar1012, 01:50 AM

P: 56

I can't see why we can assume that [itex]\sigma^{1} = 1+2gr[/itex], they're not quitely equal.(at least at the earth's surface)
But i think i can apporximate σ to be 2gr because 1 is very small comparing with 2gr. But I still can't find a way to prove this. Please, any can help me? 



#4
Mar1012, 05:14 AM

Emeritus
Sci Advisor
P: 7,433

Geodesics VS Projectile 



#5
Mar1012, 05:43 AM

PF Gold
P: 4,081

I don't think this approach can give the Newtonian result because you are using kinematic equations. The full equation for r is
[tex] \ddot{r}=\frac{\left( m\,{r}^{2}4\,{m}^{2}\,r+4\,{m}^{3}\right) \,{\dot{t}}^{2}m\,{r}^{2}\,{\dot{r}}^{2}}{{r}^{4}2\,m\,{r}^{3}} [/tex] setting [itex]\dot{t}=1[/itex] and doing a MaclaurinTaylor expansion of the RHS in m [tex] \ddot{r}=\frac{\left( 1{\dot{r}}^{2}\right) \,m}{{r}^{2}}+\frac{2\,\left(1+{\dot{r}}^{2}\right) \,{m}^{2}}{{r}^{3}}+\frac{4\,{\dot{r}}^{2}\,{m}^{3}}{{r}^{4}}+ ... [/tex] assuming m << r we get [tex] \ddot{r}=\frac{\left( 1{\dot{r}}^{2}\right) \,m}{{r}^{2}} [/tex] which does not have a closed form solution. In this m=GM/c^{2}. However it is possible to deduce Newton's law of gravitation from GR by another approach. 



#6
Mar1012, 08:05 AM

P: 56





#7
Mar1012, 08:23 AM

PF Gold
P: 4,081

The geodesic equation is found by extremizing the action for a free particle which is
[tex] \int_{\lambda_1}^{\lambda_2}\frac{ds}{d\lambda}d \lambda = \int_{\lambda_1}^{\lambda_2}\sqrt{g_{\mu\nu}\frac{dx^\mu}{d\lambda}\fra c{dx^\nu}{d\lambda}}d\lambda [/tex] where s is the proper length. Look up 'weak field theory' in the context of GR to see how Newton's law can be inferred from GR. It's too involved for me to reproduce here. 



#8
Mar1112, 03:14 PM

PF Gold
P: 4,081

After some reading I found the correct procedure. From my post #5
[tex] \ddot{r}=\frac{\left( 1{\dot{r}}^{2}\right) \,m}{{r}^{2}}+\frac{2\,\left(1+{\dot{r}}^{2}\right ) \,{m}^{2}}{{r}^{3}}+\frac{4\,{\dot{r}}^{2}\,{m}^{3 }}{{r}^{4}}+ ... [/tex] Now for static rest particle [itex]\dot{r}=0[/itex] so the leading term gives the Newtonian value. [tex] \ddot{r}=\frac{m}{{r}^{2}}=\frac{GM}{{r}^{2}} [/tex] A longer way is to start with [tex] g_{\mu\nu}=\eta_{\mu\nu}+f_{\mu\nu} [/tex] and [tex] \frac{d^2 x^a}{d\tau^2}= \Gamma^a_{bc}\frac{dx^b}{d\tau}\frac{dx^c}{d\tau} [/tex] throwing away lots of stuff and setting the 4velocities to (1,0,0,0) getting [tex] \frac{d^2 x^a}{dt^2}= \Gamma^a_{00}= \frac{1}{2}\eta^{ab}g_{00,b}= \frac{1}{2}\eta^{ab}f_{00,b} [/tex] which works for [itex]f_{00}=2m/r[/itex] 



#9
Mar1212, 01:49 AM

P: 56

thx, that's clear :)



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