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Derivation of Christoffel symbol 
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#1
May705, 08:17 AM

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Has anyone a derivation at hand of the Christoffel symbol by permuting of indices in a free fall system?
Roland 


#2
May705, 11:47 AM

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If you have the equations for geodesic motion in a coordinate basis, you can "read off" the Christoffel symbols from the equation using the geodesic deviation equation, i.e if [tex]x^a(\tau)[/tex] is a geodesic, and we represent the differentaition with respect to [tex]\tau[/tex] by a dot, we can write
[tex]\ddot{x^a} + \Gamma^a{}_{bc} \dot{x^b}\dot{x^c} = 0[/tex] MTW gives the example on pg 345 in "Gravitation" If [tex]\ddot{\theta}  sin(\theta) cos(\theta) (\dot{\phi})^2 = 0 [/tex] then [tex]\Gamma^{\theta}{}_{\phi \phi} = sin(\theta}) cos(\theta) [/tex] and the other Christoffel symbols are zero. I'm not sure if this is what you're looking for, though. 


#3
May705, 11:55 AM

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Section 6.4 of [1] gives exacly what u want.Isn't that
[tex] \Gamma^{\sigma}{}_{\mu\rho}=:e_{\mu}{}^{m}D_{\rho}e_{m}{}^{\sigma} [/tex] ,where the covariant derivative uses the ordinary [itex] \partial_{\rho} [/itex]and the spinconnection...? Daniel.  [1]Pierre Ramond "Field Theory:A Modern Primer",AddisonWesley,2nd ed.,1989 


#4
May705, 12:20 PM

P: 2,954

Derivation of Christoffel symbol
This is strictly a mathematical derivation so there is no mention of freefall. However the condition which is equivalent to it is in Eq. (3). Please note that one does not "derive" the Christoffel symbols (of the second kind). They are "defined." Once they are defined then one demonstrates relationships between them and other mathematical objects such as the metric tensor coefficients etc. In the link above the Christoffel symbols are defined in the same way Kaplan defines them in his advanced calculus text. In my page that is given in Eq. (8). I know of at least 3 different ways to define them though. Pete 


#5
May705, 12:39 PM

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Well,Pete,either u or Dirac[1] have it all mixed up.I'd go for you,as Dirac got a Nobel prize and i've been taught GR from his book[1].
Your formula #2 is valid for contravariant vectors ([1],eq.3.3,page 6) (a.k.a.vector,which is defined on the tangent bundle to a flat/curved [itex] \mathbb{M}_{4} [/itex])...So how about getting it all done correctly or,don't give that link anymore and exlude it from your post. Daniel.  [1]P.A.M.Dirac,"General Relativity",1975. 


#6
May705, 01:23 PM

P: 2,954

As I said, their are several definitions of the Christoffel symbols. Dirac uses one definition, Kaplan another, Ohanian yet another, Lovelock and Rund yet another. However, I'm looking at the text you refer to and I don't see what you're talking about. That equations is not directly related to this topic. Also what you refer to as "valid for contravariant vectors" (really the components of such a vector in a particular coordinate system) is identical in meaning to what I refered to above as the Christoffel symbols of the first kind. The other ones are the Christoffel symbols of the first kind. So how about first learning the subject completely before you make another attempt at correcting me in such a rude manner? 


#7
May705, 01:28 PM

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Alright,u didn't get it...Let's take it slowly:What's formula #2 about...?
Daniel. 


#8
May705, 01:35 PM

P: 2,954

However I see your point in Dirac. It was confusing since he never named them. Dirac and Ramond, two authors you quote, use different definitions of the Christoffel symbols of the second kind. However it appears that you're having a bad day since I find your comments to be irritating. Later 


#9
May705, 01:37 PM

P: 2,954

So why bother with me? Seems that you're unwilling to entertain the possibility that you made an error. In any case Eq. #2 in my page is the transformation properties of the components of a contravariant vector.
However I see your point in Dirac. It was confusing since he never named them. Dirac and Ramond, two authors you quote, use different definitions of the Christoffel symbols of the second kind. However it appears that you're having a bad day since I find your comments to be irritating. A long time ago I found that discussing anything with someone posting in such a grating manner not worth posting to. Li'fe's too short and I have many other irritating things which deserve more attention. Later 


#10
May705, 01:38 PM

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You specifically use "covariant vector",and 2ice...And now u turn it and use "contravariant" vector...What should i understand,that I'm having a bad day...?
Daniel. 


#11
May705, 04:07 PM

P: 2,954

See http://www.geocities.com/physics_wor...c_geometry.htm I'm not 100% sure which symbols are which since I'm not sure everyone uses the same definition of each. In any case they are the same when the manifold has both a connection and metric defined on it and the metric geodesics and the affine geodesics are the same. Pete 


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