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Christoffel symbol from Variational Principle

  1. Mar 16, 2012 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

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


    3. 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
     
    Last edited: Mar 16, 2012
  2. jcsd
  3. Mar 16, 2012 #2
    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.
     
  4. Mar 16, 2012 #3
    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.
     
    Last edited: Mar 16, 2012
  5. Mar 16, 2012 #4
    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
     
  6. Mar 16, 2012 #5
    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]
     
    Last edited: Mar 16, 2012
  7. Mar 16, 2012 #6
    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]


    You're right, my indices are all over the place in that equation :-) You should really do the calculation more carefully.
     
  8. Mar 16, 2012 #7
    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}?
     
  9. Mar 16, 2012 #8
    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.
     
  10. Mar 16, 2012 #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
     
  11. Mar 17, 2012 #10
    [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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