Geodesics vs Projectile: Exploring the Equivalence in Curved Space-Time

In summary, geodesics of a particle moving in curved space time is the same thing of projectile trajectories. Schwarzschild metric is found using taylor series expansion of 1/(1-x), and Newton's law of gravitation can be inferred from the general relativity.
  • #1
Black Integra
56
0
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}\approx1-2gr[/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 = 1-2gr[/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?
 
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  • #2
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
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
Black Integra said:
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)

It IS an approximation. The taylor series expansion of 1/(1-x) is 1+x+x^2 + o(x^3), basically. So if x is small, it's a good approximation.
 
  • #5
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 Maclaurin-Taylor 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/c2.

However it is possible to deduce Newton's law of gravitation from GR by another approach.
 
Last edited:
  • #6
pervect said:
It IS an approximation. The taylor series expansion of 1/(1-x) is 1+x+x^2 + o(x^3), basically. So if x is small, it's a good approximation.

That's the point. I use 1-2GM/r = 1-2gr because I calculate in case where g=9.8 and r is around the Earth's radius (not small, is it?)


Mentz114 said:
I don't think this approach can give the Newtonian result because you are using kinematic equations.
Oh. I have never heard something like this before, it's new for me. What's the name of the other way, other than kinematic equation?

Mentz114 said:
However it is possible to deduce Newton's law of gravitation from GR by another approach.
What is the other approach to deduce the classical mechanics? Mainly, I just want to find out that Euler-Lagrange Equation and Geodesics Equation are the same concept.
 
  • #7
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}\frac{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
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 4-velocities 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]
 
Last edited:
  • #9
thx, that's clear :)
 

1. What is the difference between geodesics and projectiles in curved space-time?

Geodesics and projectiles are both paths that an object can take in a curved space-time. However, geodesics are the paths that objects naturally follow due to the curvature of space-time, while projectiles are paths that are affected by external forces such as gravity or other forces.

2. How are geodesics and projectiles related to the concept of equivalence in curved space-time?

The equivalence principle states that the effects of gravity and acceleration are indistinguishable in a small region of space-time. Geodesics and projectiles both follow the same laws of motion in this region, leading to the equivalence between the two in terms of their trajectories in curved space-time.

3. Can geodesics and projectiles have the same path in curved space-time?

Yes, in certain cases, the path of a geodesic and a projectile can be the same in curved space-time. This occurs when there are no external forces acting on the projectile, and it follows the natural path of a geodesic in the curved space-time.

4. How does the curvature of space-time affect the paths of geodesics and projectiles?

The curvature of space-time is responsible for the paths of geodesics and projectiles in curved space-time. The curvature affects the trajectory of objects by changing the spacetime geometry, causing objects to follow curved paths instead of straight lines.

5. Are geodesics and projectiles only applicable in the theory of general relativity?

No, geodesics and projectiles are concepts that can be applied in any curved space-time, not just in the theory of general relativity. They are used in various fields such as astronomy, cosmology, and space navigation, to understand the motion of objects in curved space-time.

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