1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

GR - longitude/latitude geodesics problem

  1. Feb 14, 2007 #1
    1. The problem statement, all variables and given/known data

    Consider a 2-sphere of radius R parametrized by the 2 spherical polar coordinates θ and φ. Write down the standard metric in these coordinates.

    1. Show that lines of constant longitude are geodesics, and that the only line of constant latitude is the equator.
    2. How does a tangent vector transform if it is parallel-transported around a line of latitude?

    2. Relevant equations

    i don't know how to make equations in here. this is going to be difficult. i used the equation for the christoffel connection, to find the christoffel components in these coordinates. now i am trying to prove that the longitude and latitude lines satisfy the geodesic equation, which are basically (i'll try):

    ((d^2)x^mu)/(dlambda)^2 + (GAMMA^rho)_munu((dx^mu)/dlambda)((dx^nu)/dlambda) = 0.

    that is probably incomprehensible.

    3. The attempt at a solution

    ok so i wrote down the std metric which is

    ds^2 = (R^2)(dθ)^2 + (R^2)(sin^2(θ))(dφ)^2

    then i calculated the christoffel components, using the equation, and they came out to be

    (GAMMA^θ)_φφ = sinθcosθ
    (GAMMA^φ)_θφ = (GAMMA^φ)_φθ = cot(θ)

    and all the others are zero.

    now i don't know much about lines of longitude and latitude except that for longitude are constant φ and latitude are constant θ. so i was working with that to try to show that the geodesic equations are satisfied by longitudes, but not latitudes unless θ = pi/2. now i got the geodesic equations to be:

    (d^2)θ/(dlambda)^2 + sinθcosθ(dφ/dlambda)^2 = 0

    and

    (d^2)φ/(dlambda)^2 + 2cotθ(dθ/dlambda)(dφ/dlambda) = 0

    so for longitude with phi constant, all the dφs are zero, so the second equation is automatically satisfied and the first we are left with

    (d^2)θ/(dlambda)^2 = 0.

    i said that this was okay because on longitudinal lines θ is changing at a constant rate so (d^2)(θ)/(dlambda)^2 is 0??? knowing the equations for latitude and longitude lines in terms of θ and φ would really help.

    so then it was on to latitude which is where i ran in to the problem. the dθs vanish, so we're left with

    sin(θ)cos(θ)(dφ/dlambda)^2 = 0

    and

    (d^2)φ/(dlambda)^2 = 0

    well if i use the same argument as i did for longitude, φ changes at constant rate so the second derivative should vanish and i get geodesics for all latitude lines, which is just wrong. alternatively, i can't figure out why it's only θ = pi/2 that satisfies the geodesic - i mean i know that it's true but i can't figure it out mathematically. when first envisioning the solution it was great because i thought i'd have to make both sinθcosθ and cotθ vanish, which would mean θ would have to be pi/2, but cotθ vanishes already because of the dθ vanishing. so i'm wrong somewhere but i don't know where please helppppppppppp

    i haven't even begun to think about the second part i just want to understand the first part first. please help please help.

    oh it would also be nice to understand how to write equations in here. thanks!

    wait do i really have to do it by attachment???
     
    Last edited: Feb 14, 2007
  2. jcsd
  3. Feb 14, 2007 #2

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

  4. Feb 14, 2007 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    sin(θ)cos(θ)(dφ/dlambda)^2 = 0

    What about this equation that you derived? Is that compatible with phi moving at a constant rate at constant theta?
     
  5. Feb 14, 2007 #4
    well i'm not sure that phi does change at a constant rate. i thought it did but it may not. does anyone know whether it does, in lines of constant latitude?
     
  6. Feb 15, 2007 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Whether phi changes at a constant rate depends on how you parametrize it and it doesn't matter. Let me rephrase the question. Is this equation compatible with phi changing AT ALL (at constant theta and geodesic)?
     
  7. Feb 15, 2007 #6
    well i am looking at it in the sense that we are keeping constant theta, and somehow mapping out a curve on the manifold, so one variable has to run from one value to another. with constant theta, to make a line of latitude (a circle, not necessarily great circle, on the 2-sphere), we need phi to run from 0 to 2pi. or from -pi to pi, i forget which variable runs on which interval.
     
  8. Feb 15, 2007 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Sure, sure. But according to your formula, can phi run at all at constant theta and still be a geodesic?
     
  9. Feb 15, 2007 #8
    oh do you mean can it satisfy the geodesic equation ? (sorry) if so then i believe it can only do so when theta equals a multiple of pi/2 (the equator). (or the poles, but those wouldn't count as lines of latitude).

    and i did mean to post that i realized last night that the way i did this problem actually comes out to the right answer (above, equator), but i wasn't sure it was a valid way to do the problem. basically i am picking a parametrization for phi (or theta, when phi is constant). is it valid for me to do that? can i just assume that on a line of constant latitude phi changes at a constant rate? i need to for the equations to be satisfied, because if not then the second derivative of phi with respect to lambda doesn't vanish as needed.
     
  10. Feb 15, 2007 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You can CHOOSE any parameterization for a path you want. But if you want the parameter to coincide with the affine parameter in the geodesic equations, then as you realize, you are constrained in the parametrization. As for the non-equatorial latitude lines, again, as you are seeing, no choice of parameter will make it a geodesic. Your approach is fine!
     
  11. Feb 15, 2007 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    BTW, if this helps, the affine parameter has a physical meaning. It's proportional to actual distance (not just coordinate distance) along the geodesic (or in GR, to proper time).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: GR - longitude/latitude geodesics problem
  1. Another GR problem (Replies: 17)

  2. Geodesic problem (Replies: 1)

Loading...