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Parrallel transport on the 2-sphere

  1. Sep 6, 2014 #1
    I recently derived the Riemann tensor (Rabmv) for the 2 sphere.

    I then did RabmvUbVmWv to calculate dva (the change in the vector va as you parallel transport it around the loop of the sphere).

    The result I got for dv1 was 0. I got 0 for dv2 as well.

    I am just making this thread to verify if I am correct in getting 0. Should I get a change of <0,0> if I parallel transport a vector around a loop on the 2D surface of a sphere?

    Also, for the curvature scalar R, I got 2/r2. What info exactly does the curvature scalar give you besides telling you whether or not a space is curved?

    Here was the metric tensor I used for all of this:

    g11= r2
    g12 and g21= 0
    g22= r2sin2(θ)

    where r is actually a constant, x1 is θ and x2 is ø
  2. jcsd
  3. Sep 6, 2014 #2


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    You certainly should not get 0 since the Riemann tensor is not 0 on the 2-sphere.

    If you got a non-zero Riemann, I'm finding it hard to understand why you would obtain a 0 in your second calculation.
  4. Sep 6, 2014 #3


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  5. Sep 6, 2014 #4
    I did get a non-zero Riemann tensor. Here were the non-zero elements:

    R1212 = sin2(θ)

    R1221 = -sin2(θ)

    R2112 = -1

    R2121 = 1

    Every other element was 0.

    dv just ended up being <0,0> because addends canceled out when I did the summation. It wasn't because of anything involving the tensor itself.

    I have one idea on why <0,0> may be wrong:

    In the calculation of RabmvUbVmWv ,

    should this summation have 8 addends?

    Ra111U1V1W1 + Ra112U1V1W2 + Ra121U1V2W1 + Ra122U1V2W2 + Ra211U2V1W1 + Ra212U2V1W2 + Ra221U2V2W1 + Ra222U2V2W2
  6. Sep 6, 2014 #5


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    What "loop" did you calculate the above summation (which looks right) over?

    In terms of the Earth:

    Did you, for instance, start at some point on the equator, and go "east" all the way around in a great circle?

    Or did you start at some point on the equator, and make a loop by going east, north, west, south?

    Or was it some other loop?
  7. Sep 6, 2014 #6
    I didn't use any specific loop. I just set U1 or V1 or W1 equal to θ, while I set the vectors that have the 2 index equal to ø. I thought this was what you were supposed to do. Is it not? Don't you use x1 and x2 as the vectors for your loop?
  8. Sep 6, 2014 #7


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    The physical interpretation of what you are doing is computing the parallel transport of a vector around a small closed loop.

    See for instance http://www.physics.ucc.ie/apeer/PY4112/Curvature.pdf pg 5. There should be similar diagrams in your textbook, if you have one.

    So , using the web-page notation

    ##A^{\nu}## is a vector that forms one leg of the loop. By context it should be normalized to unit length.
    ##\delta a## is the length of vector A

    ##B^{\mu}## is a vector that forms another leg of the loop
    ##\delta b## is the length of vector B

    ##V^{\sigma}## is the vector that we transport around the loop formed by A and B

    Their #5 (similar to your equation, but I'm unclear of the significance of U, V, and W) gives a linear transform between the vector V and it's change ##(\delta V)## as

    ##(\delta V)^{\rho} = (\delta a) (\delta b) A^{\nu} B^{\mu} R^{\rho}{}_{\sigma \mu \nu} V^{\sigma} ##

    Ignoring the lengths of the vector now, let us say that A points in the ##\theta## direction, so it has components (1,0), since you've defined ##\theta## as the first coordinate, and that B points in the ##\phi## direction, so it has components (0,1),

    Note that because A has components (1,0) only components of the form ##R^x{}_{yz1}## will contribute to the sum. (x,y,z here are dummy variables). Components of the form ##R^x{}_{yz2}## will not contribute, because the second component of A is zero.

    Similar remarks apply to B, with the specified value of V only components of the form ##R^x{}_{y1z}## will contribute.

    Furthermore, to avoid the coordinate singularity (which shows up as some of the above vectors having zero length) , we want ##\sin \theta## to be nonzero, the best choice is ##\theta = \pi / 2##, the equatorial plane.

    Now let V point in the ##\theta## direction, so it has components (1,0)

    Then in a coordinate basis, ##R^2{}_{121}## is nonzero, demonstrating a nonzero component for the change in V.

    If V pointed in the ##\phi## direction, ##R^1{}_{221}## would be nonzero, demonstrating a nonzero component for the change in V.
  9. Sep 7, 2014 #8
    Okay, after reading this post and that link you gave me a few times, I think I get it now. I only have one remaining question.

    In the expression ([itex]\delta[/itex]V)a =([itex]\delta[/itex]c)([itex]\delta[/itex]d)A[itex]\mu[/itex]B[itex]\nu[/itex]Rab[itex]\mu[/itex][itex]\nu[/itex]Vb, do you absolutely have to have the infinitesimal lengths in the equation, or are the lengths similar to the cosmological constant in the Einstein field equations in that whether or not you put them into the equation depends on the situation?

    I only ask this because in videos and on web pages about tensor analysis and general relativity, I have seen people write this equation without putting the infinitesimal lengths into the equation. They wrote it like this:

    ([itex]\delta[/itex]V)a = A[itex]\mu[/itex]B[itex]\nu[/itex]Rab[itex]\mu[/itex][itex]\nu[/itex]Vb
  10. Sep 7, 2014 #9


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    The infinitesimal lengths are a matter of notation, really. By letting ##|A^\mu A_\mu| = (\delta a)^2## you can avoid the need for including ##\delta a## explicitly in your formula.

    The authors did not/ do not use my notation of a "hat" to indicate a unit length vector. But I think the notation makes things clearer in this instance, especially since all our vectors are spacelike so we don't need to worry about the sign issues involved with timelike vectors, and can assume ##A^\mu A_\mu## is always positive and equal to the length squared.

    Using the "hat" notation, we can write a general vector V either as just V or as ##(\delta L) \hat{V}##, where ##\delta L## is the length of the vector.

    Similarly we could write, with the hats
    ##(\delta V)^{\rho} = \left( (\delta a) \hat{A^{\nu}} \right) \left( (\delta b) \hat{B^{\mu}} \right) R^{\rho}{}_{\sigma \mu \nu} V^{\sigma} ##

    or without the hats as:

    ##A^\nu B^\mu R^{\rho}{}_{\sigma \mu \nu} V^{\sigma} ##
  11. Sep 8, 2014 #10
    I just thought about something today:

    In the equation:([itex]\delta[/itex]V)a= AmBvRabmvVb

    Setting Am = <θ , 0> , Bv= <0, ø> and Vb= <θ, ø> works for the 2 sphere or other sets 2 dimensional coordinates.

    However, this may be problematic in 3D or higher. This is because there are only two legs of the loop and 3 or more dimensions. Therefore, if your coordinates are r, θ and ø, then one leg could be

    Am = <r , 0, 0> . Another could be Bv= <0, θ, 0>. What would I do with the ø component though? What vector would account for it? There are only two legs of the loop and that link gave me the impression that the legs of the loop must be unit vectors multiplied by infinitesimal lengths.

    Could I have more than one type of coordinate in a leg of a loop (like <r, θ, 0> )?
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