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Connection of the unit sphere

  1. Sep 23, 2007 #1
    1. The problem statement, all variables and given/known data
    I am confused about this question and probably the connection in general. When we only have two independent coordinates here what do the three indices in [tex] \Gamma ^a_{bc} [/tex] stand for?

    I found the metric for this space, but every formula for Gamma has indices a,b,c and I do not know what to plug in for them?

    2. Relevant equations



    3. The attempt at a solution
     

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    Last edited: Sep 23, 2007
  2. jcsd
  3. Sep 23, 2007 #2

    Dick

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    a,b and c stand for either phi or theta. I can say this without even seeing your attachment.
     
  4. Sep 24, 2007 #3
    So, when it asks me to calculate the coefficients, i need to calculate for

    (a,b,c) = (theta, theta, theta)
    (a,b,c) = (theta, theta, phi)
    (a,b,c) = (theta, phi, theta)
    (a,b,c) = (theta, phi, phi)

    and then four more with a = phi ?
     
  5. Sep 24, 2007 #4

    Dick

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    Yes. but you know case (a,b,c)=(a,c,b) right? And didn't you just do an exercise about diagonal metrics that would come in handy here?
     
  6. Sep 24, 2007 #5
    So, to do the next part, should I just plug the connection into 3.41 and integrate from 0 to 2 pi?
     

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  7. Sep 25, 2007 #6

    Dick

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    Your attachments still haven't been approved, but you should plug the connection into the parallel transport equations, if that's what you mean. And then solve them.
     
    Last edited: Sep 25, 2007
  8. Sep 27, 2007 #7
    Which equations do you mean? I know the connection and I think that I know the initial vector, so which equation relates these and the rotation angle to the final vector?
     
  9. Sep 27, 2007 #8

    Dick

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    Look up the 'parallel transport' operation mentioned in the problem.
     
  10. Sep 28, 2007 #9
    What to you mean look up the parallel transport operation? I already looked it up and I even posted that entire section from my book. Did you mean look it up elsewhere? Because I have tried that also.

    I just want help figuring out which equations to plug the connection and the initial vector into and get the final vector.
     
  11. Sep 28, 2007 #10

    Dick

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    Equation (3.42). Start trying to use it and then people can start trying to help you. Did you read what you posted? It's a differential equation for the rate of change of the transported vector components in terms of the tangent vector to the curve and the connection coefficients. You have to solve this differential equation.
     
  12. Sep 28, 2007 #11
    Here is my attempt:

    [tex] \Gamma^{\phi}_{\phi \theta} = \Gamma^{\phi}_{\theta \phi} = \cot \theta[/tex]

    [tex] \Gamma^{\theta}_{\phi \phi} = - \sin \theta \cos \theta[/tex]

    The rest of the connection vanishes.

    I plug this into the equation along with the initial position and the theta equation vanishes and the phi equation is

    [tex] \frac{ dv^\phi}{du} = - \theta_0 \cot \theta \frac{dx^\phi}{du} [/tex]

    So, is the next step finding dx^phi/du with some parametrization?
     
  13. Sep 29, 2007 #12

    Dick

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    The theta equation does not vanish. And why theta_0? Shouldn't that be v^theta? And use the obvious parametrization. Use phi as the path parameter.
     
  14. Sep 29, 2007 #13
    You're right. I was using the initial conditions for v^theta and v^phi. That is why I thought the theta equation vanishes.

    So, the two equations are:

    [tex] \frac{ dv^\theta}{du} = \sin \theta \cos \phi v^{\theta} \frac{d\phi}{du} [/tex]


    [tex] \frac{ dv^\phi}{du} = - v^{\theta} \cot \theta \frac{d\phi}{du} [/tex]

    So when I replace u with theta as my parameter, why don't both of these equation vanish since they both would have a d\phi/d\theta and theta and phi are supposed to be independent?
     
  15. Sep 30, 2007 #14

    Dick

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    That's why theta would not be a good choice of parameter. Your path is along constant theta with varying phi. Use phi. You also have some phi's and theta's in the wrong place in the first equation.
     
  16. Oct 1, 2007 #15
    You're right, the equations should be:

    [tex] \frac{ dv^\theta}{du} = \sin \theta \cos \theta v^{\phi} \frac{d\phi}{du} [/tex]
    [tex] \frac{ dv^\phi}{du} = - v^{\theta} \cot \theta \frac{d\phi}{du} [/tex]

    And when I apply the phi parametrization (I don't know what I was thinking with the theta parametrization), I get


    [tex] \frac{ dv^\theta}{d\phi} = \sin \theta \cos \theta v^{\phi} [/tex]
    [tex] \frac{ dv^\phi}{d\phi} = - v^{\theta} \cot \theta [/tex]

    So if we integrate both of these equations w.r.t. phi, we get:

    [tex] \Delta v^\theta = \sin \theta_0 \cos \theta_0 \int_{0}^{2\pi}v^{\phi}d\phi [/tex]
    [tex] \Delta v^\phi = - \cot\theta_0 \int_{0}^{2\pi} v^{\theta} d\phi [/tex]

    How do I evaluate the integrals on the RHS?
     
  17. Oct 1, 2007 #16

    Dick

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    You can't just integrate them like that. You have two ODE's wrt to two different functions. You want to get a single ODE in terms of a single function to start out. Hint: differentiate the second one wrt phi so you get a second order ODE and then substitute the first one in to eliminate v_theta.
     
  18. Oct 1, 2007 #17
    I get

    [tex] \frac{d^2v^{\phi}}{d\theta} = - \cos \theta v^{\phi} [/tex]

    Since [tex] \theta = \theta_0 [/tex] is a constant all along this parametrization, this DE has solutions:

    [tex] v^{\phi} = A e^{i \cos \theta_0 \phi} + B e^{-i \cos \theta_0 \phi} [/tex]

    So, now I need to plug this into the other DE, find A and B, and plug in 2 pi for phi to get the final value of v^phi and v^theta, right?

    EDIT: the first equation should be [tex] \frac{d^2v^{\phi}}{d\phi^2} = - \cos^2 \theta v^{\phi} [/tex]
     
    Last edited: Oct 2, 2007
  19. Oct 1, 2007 #18

    Dick

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    Basically right. Except I get a cos^2(theta) in your ODE. And life might be a little simpler if you use the sin,cos form of the solutions instead of complex exponentials, but it's up to you.
     
  20. Oct 1, 2007 #19
    I am having trouble finding A and B.

    I get [tex] v^{\phi}_{initial} = A + B [/tex] from the starting point.

    The other equation should come from the fact that the vector is parallel to the circle phi = 0. Doesn't that condition just imply that the vector has only a phi component?
    Should I use the unit size of the vector to get the other equation instead?
     
  21. Oct 1, 2007 #20

    Dick

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    The other condition is the initial values of the derivatives of v_phi and v_theta. You aren't free to choose them arbitrarily. Your transport equations specify initial values of derivatives in terms of initial values of the v vectors. I don't know what makes you think v will have only a phi part.
     
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