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Riemann curvature of a unit sphere

  1. Aug 11, 2014 #1
    The Riemann curvature of a unit sphere is sine-squared theta, where theta is the usual azimuthal angle in spherical co-ordinates, and this is shown in many textbooks. But since a sphere is completely specified by its radius, then as far as I can see its curvature should be a function of its radius only and I cannot see how theta is involved or what value it would be given.
     
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  3. Aug 11, 2014 #2

    robphy

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    Don't confuse the scalar curvature with one component of the Riemann tensor in some set of coordinates.
     
  4. Aug 11, 2014 #3
    This doesn't answer my question. How would you assign a value to theta in this case?
     
  5. Aug 11, 2014 #4

    stevendaryl

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    Could you clarify what you mean by "the Riemann curvature"? There is a tensor called the Riemann curvature tensor, which has 4 indices. There is another measure of curvature, Gaussian curvature, which is just a single real number. What is it that you are saying is equal to [itex]sin^2(\theta)[/itex]?
     
  6. Aug 11, 2014 #5
    I mean the Riemann curvature tensor which as you say has 4 indices. The calculation for the unit sphere is standard and is given in many textbooks, presumably because it is the simplest possible case.
     
  7. Aug 11, 2014 #6

    stevendaryl

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    Okay, well for a two-dimensional surface, the Riemann curvature tensor

    [itex]R_{abcd} = K (g_{ac} g_{db} - g_{ad}g_{bc})[/itex]

    where [itex]K[/itex] is the Gaussian curvature, which for a sphere is indeed a function of the sphere's radius:

    [itex]K = 1/r^2[/itex]

    But if we calculate [itex]R_{\theta \phi \theta \phi}[/itex], we find:

    [itex]R_{\theta \phi \theta \phi} = K (g_{\theta \theta} g_{\phi \phi} - g_{\theta \phi}g_{\phi \theta})[/itex]

    The metric tensor's components tell us that

    [itex]g_{\theta \theta} = r^2[/itex]
    [itex]g_{\phi \phi} = r^2 sin^2(\theta)[/itex]
    and all the other components are zero. So that gives:

    [itex]R_{\theta \phi \theta \phi} = K (r^4 sin^2(\theta)) = r^2 sin^2(\theta)[/itex]

    To raise the first index, you multiply by [itex]g^{\theta \theta} = 1/g_{\theta \theta} = 1/r^2[/itex]. So

    [itex]R^\theta_{\phi \theta \phi} = sin^2(\theta)[/itex]

    So the answer works. It's just that [itex]R^\theta_{\phi \theta \phi}[/itex] doesn't really give you an intuitive feel for the strength of the curvature. For 2D surfaces, the Gaussian curvature is a better measure.
     
  8. Aug 11, 2014 #7

    robphy

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    Possibly useful:
    http://www.physics.usu.edu/wheeler/genrel/lectures/2sphere.pdf
    http://www.physics.ohio-state.edu/~mathur/grsolprob3.pdf
    https://www.pa.msu.edu/courses/2012spring/AST860/03-27.pdf
    http://www.physics.ucc.ie/apeer/PY4112/Curvature.pdf (p.14)

    When you say "curvature should be a function of its radius [for a sphere]",
    you are (or should be) talking about the "Ricci scalar" (a.k.a. "scalar curvature").
    The [itex]sin^2\theta[/itex] quantity you quote is one component of the Riemann curvature tensor in some coordinate system... that is, one piece of a complicated object [the curvature tensor]... and there is no guarantee that this piece carries the symmetry of the entire sphere.

    Because of symmetries of the curvature tensor _and_ symmetries of the sphere _and_ choice of coordinates, the other components may be multiples of [itex]sin^2\theta[/itex]...so there is essentially one component to work out for the curvature tensor. However, you must construct from Riemann, Ricci (by tracing) and Ricci-scalar by contracting with the metric... only then, you have eliminated all of the coordinate-dependence. So, that final Ricci-scalar should reflect the symmetry of the entire sphere.
     
  9. Aug 12, 2014 #8
    Yes I know all that - you have just repeated what it says in my textbook, but it still doesn't answer my question. Since the radius of the sphere is 1, that makes this a numerical example, so theta should have a value, but what value is it, and why?
     
  10. Aug 12, 2014 #9

    pervect

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    I think you may be referring to the fact that if you use the line element

    ##ds^2 = r^2 d\theta^2 + r^2 \sin^2 \theta d \phi^2##

    for a metric, the nonzero components for the Riemann curvature tensor in a coordinate basis are:

    ##R_{\theta\phi\theta\phi} = R_{\phi\theta\phi\theta} = -R_{\theta\phi\phi\theta} = -R_{\phi\theta\theta\phi} = r^2 \sin^2 \theta ##

    The other 12 components are zero, only these four are nonzero.

    If this isn't what you mean, you'll need to be more precise about your claim and copy it more directly from your textbook with all the necessary details. Unfortunately you've omitted enough details where it's not quite clear to me what you're trying to figure out.


    Assuming this is your claim, if you calculate the Riemann curvature tensor in an orthonormal basis instead of a coordinate basis you get the following

    ##R_{\hat{\theta}\hat{\phi}\hat{\theta}\hat{\phi}} = R_{\hat{\phi}\hat{\theta}\hat{\phi}\hat{\theta}} = -R_{\hat{\theta}\hat{\phi}\hat{\phi}\hat{\theta}} = -R_{\hat{\phi}\hat{\theta}\hat{\theta}\hat{\phi}} = \frac{1}{r^2}##

    which is independent of ##\theta## and ##\phi##.

    Here ##\hat{\phi}## and ##\hat{\theta}## are unit length vectors (actually, because they are in the lower index, they are more like dual vectors).

    Note that the coordinate dual vectors ##d\phi## and ##d\theta## are NOT unit length.
     
  11. Aug 12, 2014 #10

    Matterwave

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    The Riemann tensor's component that Steve calculated is obviously of function of θ, and so what whatever value theta has at the point you are evaluating this component of the Riemann, is the value of θ that you plug into the equation. This component depends on where on the sphere you are, which really just says it depends on the coordinate system you imposed on your sphere.
     
  12. Aug 12, 2014 #11
    Good- we are getting somewhere at last. I didn't realise that the evaluation is done at a particular point and I have never seen that made clear in the textbooks. It still seems strange that it should be different at different places on the sphere. So - question answered, and my thanks to all!
     
  13. Aug 12, 2014 #12

    Matterwave

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    The fact that this component of the Riemann is different on different places on the sphere is really an artifact of the coordinate system and the coordinate basis vectors that you are using. For spherical coordinates, the length of the basis vectors are not unit-length (see post by pervect), and their lengths depend on where on the sphere you are (again, this just means their lengths depend on the way you imposed your coordinate system). It's really just this change in lengths of the basis vectors that give you this theta dependence. This is why previous members pointed you to the Gaussian curvature which is much more intuitive for this problem. And this is also why if you calculate the Riemann curvature tensor in an orthonormal basis, then it is no longer dependent on theta.
     
  14. Aug 12, 2014 #13
    Yes, I had worked some of that out for myself, starting with the fact that the sphere itself does not have any preferred directions. My goal is to arrive at the Einstein field equations, and I can see a much bigger headache coming when I tackle the stress energy tensor.
     
  15. Aug 12, 2014 #14

    pervect

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    It may be too late now, but let me expand on this a bit.

    The components of the Riemann, in an orthonormal basis, don't depend on ##\theta##. The unit length vector ##\hat{\theta}##, equal to ##r d\theta## (covariant) or ##\frac{1}{r} \frac{\partial}{\partial \theta}## (contravariant), doesn't depend on ##\theta## either. However, the unilt length vector ##\hat{\phi}##, equal to ##r \sin \theta d\phi## (covariant) or ##\frac{1}{r \sin \theta} \frac{\partial}{\partial \phi}## (contravariant) DOES depend on ##\theta##.

    Because it is a rank 4 tensor, the Riemann is a map from 4 vectors to a scalar, written with lower indices it's a map from four contravariant vectors to a scalar.

    Written in an orthonormal basis where the length of the basis vectors is not dependent on ##\theta##, the components of the Riemann do not depend on ##\theta##. Written in a basis where the lengths of some of the basis vectors DO depend on ##\theta##, the components of the Riemann also depend on ##\theta##.

    In particular, the length of the basis vector ##\frac {\partial}{\partial \phi}##, which is a coordinate basis vector, does depend on ##\theta##

    Abstractly, I would say it's not the Riemann that depends on ##\theta##, it's the choice of basis vectors when one uses a coordinate basis.

    My notation, taken mostly from MTW, may or may not match the OP's notation. I'm not sure what textbook he took it from. I have a feeling that the OP"s failure to specify what basis vectors he (or his textbook) used to calculate the values of the components of the Riemann indicates a lack of appreciation of their significance :(, in which case a lot of my post may not be comprehensible - especially if my notation is different from his text. Unfortunately, I don't see how to avoid that. Since he's apparently left anyway, it's all sort of moot.
     
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