Register to reply 
Schwarzschild Christoffels 
Share this thread: 
#1
Mar912, 10:08 PM

P: 7

I needed some help with the Christoffel symbols for the Schwarzschild metric. I used the metric in wikipedia with signature (+). For some reason, I get different Christoffel symbols when I use Mathimatica so I'm not sure if it's my calculations that are wrong or not. This isn't homework or anything of the sorts. I'm just trying to find the Ricci tensor and scalar for this metric partly for fun, partly for getting used to the operations and partly to learn more about this metric. In the attached file you'll find the Christoffel symbols that were calculated by me.
Thanks in advance. 


#2
Mar912, 10:51 PM

Sci Advisor
P: 2,698

The Schwarzschild solution is a vacuum solution, what does that tell you about the Ricci tensor and scalar right off the bat?



#3
Mar912, 10:56 PM

P: 7

They should be zero. Sorry that part completely slipped my mind. But my overall goal remains the same nevertheless. I don't seem to be getting the correct Christoffels for some reason.
This is the third metric I'm solving and GR is very new to me so I apologize. 


#4
Mar912, 11:01 PM

Sci Advisor
P: 2,698

Schwarzschild Christoffels
Which method are you using to find the Christoffel symbols? Are you just brute forcing it from the equation where they are equal to derivatives of g?



#5
Mar912, 11:04 PM

P: 7

I don't know what you mean exactly but I am using the equation, the partials of g.
Is that a wrong approach to this? 


#6
Mar912, 11:50 PM

Sci Advisor
P: 2,698

No, that's a perfectly valid approach, but sometimes other approaches are quicker. For example, you can use the Euler Lagrange equations on the integral [itex]\frac{1}{2}\int g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu d\lambda[/itex] to get the geodesic equations. By comparing with the regular geodesic equations, one can often quickly read off all the Christoffel symbols.



#7
Mar912, 11:51 PM

P: 7

Yeah well the problem is that I have no knowledge of Lagrangian mechanics so my hands are quite tied.



#8
Mar1012, 12:09 AM

Sci Advisor
P: 2,698

Interestingly enough, I can't seem to find a nice concise tabulation of the results online...and since I'm way too lazy to try to do this myself, I hope someone else will come along and help you. XD



#9
Mar1012, 12:33 AM

P: 7

I have all the time in the world. I just wanna make sure I get the Christoffel symbols right before I do the Riemann tensor calculation since getting non zero Ricci tensor would mean a lot of wasted time and I hate it when I get stuff wrong after working so hard for it :(



#10
Mar1012, 04:34 AM

P: 11

There are maybe all wrong, for example [itex]\Gamma_{00}^1=\frac{1}{2}g^{11}\partial_r g_{00}=\frac{c^2 r_s (rr_s)}{2 r^3}[/itex]
So maybe something is wrong when you define the inverse, e.g. [itex]g^{11}=(1r_s/r)[/itex] 


#11
Mar1012, 04:48 AM

PF Gold
P: 4,087

[tex]
\begin{align} {mcs}_{1,1,2}&=\frac{m\,\left( r2\,m\right) }{{r}^{3}}\\ {mcs}_{1,2,1}&=\frac{m}{r\,\left( r2\,m\right) }\\ {mcs}_{2,2,2}&=\frac{m}{r\,\left( r2\,m\right) }\\ {mcs}_{2,3,3}&=\frac{1}{r}\\ {mcs}_{2,4,4}&=\frac{1}{r}\\ {mcs}_{3,3,2}&=2\,mr\\ {mcs}_{3,4,4}&=\frac{cos\left( \theta\right) }{sin\left( \theta\right) }\\ {mcs}_{4,4,2}&=\left( r2\,m\right) \,{sin\left( \theta\right) }^{2}\\ {mcs}_{4,4,3}&=cos\left( \theta\right) \,sin\left( \theta\right) \end{align} [/tex] where [tex] mcs_{abc}={\Gamma^c}_{ab} [/tex] and t=x^{1}, r=x^{2} etc. This is from Maxima, using the ctensor package. 


Register to reply 
Related Discussions  
About Schwarzschild geodesics  Special & General Relativity  6  
Schwarzschild curvature  Special & General Relativity  3  
Schwarzschild Four Velocity  Advanced Physics Homework  12  
Schwarzschild in 5D  Special & General Relativity  4  
Schwarzschild radius  Special & General Relativity  6 