Christoffel symbol from Variational Principle

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  • #1
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Homework Statement



It's not exactly a homework question. I can find Christoffel Symbols using general definition of Christoffel symbol. But, when I try to find Christoffel Symbols using variational principle, I end up getting zero.

I have started with the space-time metric in a weak gravitational field (with the assumption of low velocity):
ds2=−(1+2ϕ)dt2+(1−2ϕ)(dx2+dy2+dz2)

Where
ϕ<<1 is the gravitational potential.

Homework Equations



Euler Lagrange(EL) Equation: [itex]\frac{d}{d\tau}[/itex]([itex]\frac{dL}{d\dot{x^{a}}}[/itex]) = [itex]\frac{dL}{dx}[/itex]


The Attempt at a Solution


Lagrangian:
L = −(1+2ϕ)[itex]\dot{t}[/itex]2+(1−2ϕ)([itex]\dot{x}[/itex]2+[itex]\dot{y}[/itex]2+[itex]\dot{z}[/itex]2)

using EL:[itex]\frac{d}{d\tau}[/itex]([itex]\frac{dL}{d\dot{t}}[/itex]) = [itex]\frac{dL}{dx}[/itex]


−(1+2ϕ)[itex]\ddot{t}[/itex]= 0

I repeated the similar process for x, y, and z and I got zero for all. Can someone please help me?

Answer should be:

Γtti=ϕ,i

Γi00=ϕ,iijk=δjkϕ,i−δijϕ,k−δikϕ,j
 
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Answers and Replies

  • #2
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You are taking phi to be constant in your calculation. The Christoffel symbols are indeed zero if phi is constant, but not if it's not.
 
  • #3
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I know that phi is function of r and I am having hard time to differentiate it. I just got stuck. Since r = sqrt (x^2 + y^2+ z^2) . How can it depend on t. The way it can depend on t by using the definition of proper time. Sorry, I am just lost.
 
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  • #4
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So you just use chain rule:

[itex] \frac{d}{d\tau} \frac{\partial L}{\partial \dot{x}_i} = 2\frac{d}{d\tau} [(1+2\phi) \dot{x}_i] = 2((1+2\phi)\ddot{x}_i + 2\dot{x}_i \phi_{,\mu} \dot{x}^{\mu}) [/itex]

and as usual,

[itex]\frac{\partial L}{\partial x_i}= 2 \phi_{,i} ( \dot{t}^2 +\dot{x}_j \dot{x}^j) [/itex]

Note that I can be careless with my index notation here, because the errors I'm making would be proportional to phi squared, which I can drop in first order perturbation theory. You should probably be more careful though, it's very easy to make stupid mistakes with this kind of calculations
 
  • #5
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I didn't get how you get [itex]{\mu}[/itex]...why are you using lower index [itex] \dot{x}_i [/itex]


[itex] \frac{d}{d\tau} \frac{\partial L}{\partial \dot{x}_i} = 2\frac{d}{d\tau} [(1+2\phi) \dot{x}_i] = 2((1+2\phi)\ddot{x}_i + 2\dot{x}_i \phi_{,\mu} \dot{x}^{\mu}) [/itex]


What is [itex]\dot{x}_j \dot{x}^j[/itex]?


[itex]\frac{\partial L}{\partial x_i}= 2 \phi_{,i} ( \dot{t}^2 +\dot{x}_j \dot{x}^j) [/itex]
 
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  • #6
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I didn't get how you get [itex]{\mu}[/itex]

What is [itex]\dot{x}_j \dot{x}^j[/itex]?

Suppose I have a function f(t,x) and t and x depend on the proper time, t=t(λ) and x=x(λ) along my path. Then

df/dλ = ∂f/∂x dx/dλ + ∂f/∂t dt/dλ.

The Einstein summing convention is just my shorthand way of writing this for all 4 indices. So I would write x,y,z,t as a 4-vector xμ and sum over repeating indices.

It's just a shorthand notation. So latin indices get values 1,2,3 (x,y,z) and greek indices get values 0,1,2,3 (t,x,y,z). I also use Einstein summation convention, so for example
[itex] \dot{x}_j \dot{x}^j = \sum_{j} g_{jj} \dot{x}^j \dot{x}^j = \dot{x}^2+\dot{y}^2+\dot{z}^2 + O(\phi) [/itex]


...why are you using lower index [itex] \dot{x}_i [/itex]

You're right, my indices are all over the place in that equation :-) You should really do the calculation more carefully.
 
  • #7
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For time, Christoffel Symbol should be:
[itex]{\Gamma^t}_{it}[/itex] = [itex]x_i\frac{\phi'(r)}{r}[/itex]

But by doing the way you suggest I didn't get the answer:

[itex] \frac{d}{d\tau} \frac{\partial L}{\partial \dot{t}} = -2\frac{d}{d\tau} [(1+2\phi) \dot{t}] = -2((1+2\phi)\ddot{t} + 2\dot{t} \phi_{,\mu} \dot{x}^{\mu}) [/itex]

[itex]\frac{\partial L}{\partial t}= 0 [/itex]

[itex] (1+2\phi)\ddot{t} + 2\dot{t} \phi_{,\mu} \dot{x}^{\mu}=0 [/itex]

The μ does not really fit, why do you used μ..can't we use {i,j,k}?
 
  • #8
938
9
For time, Christoffel Symbol should be:
[itex]{\Gamma^t}_{it}[/itex] = [itex]x_i\frac{\phi'(r)}{r}[/itex]

But by doing the way you suggest I didn't get the answer

The μ does not really fit, why do you used μ..can't we use {i,j,k}?

My formula gives the same result, so you must be doing something wrong.

You can use absolutely whatever notation you like, as long as you calculate the derivatives right.
 
  • #9
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I have my solution in the above post I am getting

[itex]\frac{1}{(1+2ϕ)}[/itex]ϕ,μ[itex]\dot{x}[/itex]μ=0

I think my derivatives are correct
 
  • #10
938
9
I have my solution in the above post I am getting

[itex]\frac{1}{(1+2ϕ)}[/itex]ϕ,μ[itex]\dot{x}[/itex]μ=0

I think my derivatives are correct

[itex]\frac{1}{1+2\phi} \phi_{,\mu} x^{\mu} = -\frac{\partial \phi}{\partial t} \dot{t} + \frac{\partial \phi}{\partial x} \dot{x} + \frac{\partial \phi}{\partial y} \dot{y} + \frac{\partial \phi}{\partial x} \dot{y} + O(\phi^2)[/itex]
 

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