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Covariant derivative

  1. Jul 26, 2011 #1
    How can the derivative of a basis vector at a point be the linear combination of tangent vectors at that point?

    For example, if you take a sphere, then the derivative of the polar basis vector with respect to the polar coordinate is in the radial direction. How can something in the radial direction be written as a linear combination of vectors in a plane perpendicular to the radial direction?

    [tex]\partial_i e_j=\Gamma^{k}_{ij} e_k [/tex]

    Or is this a misunderstanding of the Christoffel symbols? The above equation seems to make sense in spherical coordinates in 3-dimensional space, but not in spherical coordinates on a 2-dimensional sphere.
     
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  3. Jul 26, 2011 #2

    pervect

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    If you have a spherical coordinate system, you'll have three basis vectors at every point that span the space.

    I.e. if your coordinates are [itex]r, \theta, \phi[/itex] you'll have the coordinate basis vectors of:

    [itex]
    \frac{\partial}{\partial r}, \frac{\partial}{\partial \theta}, \frac{\partial}{\partial \phi} [/itex]
     
  4. Jul 26, 2011 #3

    Mentz114

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    Why shouldn't it be ? The derivative of a vector does not have to point the same direction as the vector itself.

    If the derivative is a vector then it can be written in terms of the tangent vectors which are a basis for all vectors in the tangent space.

    This only true for parallel transported vectors, of course since
    [tex]
    \partial_i e_j=\Gamma^{k}_{ij} e_k\ \ \rightarrow\ \nabla_ie_j = 0
    [/tex]

    In the notation I'm used to, your [itex]e_\mu[/itex]'s would be covectors (1-forms) and would form the cobasis.
     
    Last edited: Jul 26, 2011
  5. Jul 27, 2011 #4
    The derivative straddles two different tangent spaces, since it is a difference of vector fields at two different points. For example in uniform circular motion, the derivative of the velocity points towards the center of the circle. Only a tangential acceleration can be written as a linear combination (trivial in this case, since a linear combo is a rescaling of the velocity vector) of vectors in the tangent space.

    Does this mean once you choose basis vectors at a point, you are no longer free to choose the basis vectors at other points? It seems you're saying that all basis vectors must be parallel transported, so if you choose the basis vector at one point, then at all other points the orientation of the basis should be determined?
     
  6. Jul 27, 2011 #5

    George Jones

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    Embedded submanifolds don't have the same covariant derivative operator as the space in which they are embedded.

    As an example, consider [itex]S^2[/itex] embedded in [itex]\mathbb{R}^3[/itex], i.e., take [itex]S^2[/itex] to be the surface in [itex]\mathbb{R}^3[/itex] given by the equation of constraint [itex]x^2 + y^2 + z^2 = 1[/itex]. Give [itex]\mathbb{R}^3[/itex] the standard Euclidean metric, which, in spherical coordinates (used by physicists), is
    [tex]g = dr^2 + r^2 \left( d\theta^2 + \sin^2 \theta d \phi^2 \right).[/tex]
    This metric for [itex]\mathbb{R}^3[/itex] induces a metric [itex]\tilde{g}[/itex] on [itex]S^2[/itex] given by
    [tex]\tilde{g} = d\theta^2 + \sin^2 \theta d \phi^2 .[/tex]
    Associated with each of these metrics there is metric-compatible, torsion-free connection, [itex]\nabla[/itex] for [itex]\mathbb{R}^3[/itex] and [itex]\tilde{\nabla}[/itex] for [itex]S^2[/itex]. These are different animals!

    Let a particle move on the path in [itex]\mathbb{R}^3[/itex] given by [itex]r = 1[/itex], [itex]\theta = \pi/2[/itex], [itex]\phi = \omega t[/itex], where [itex]\omega[/itex] is constant, so the particle also moves on the embedded [itex]S^2[/itex]. At any time, the tangent vector to the particle's path, [itex]v = d/dt[/itex], lives both in a tangent space for [itex]\mathbb{R}^3[/itex] and a tangent space for [itex]S^2[/itex]. Note that any point in [itex]S^2[/itex], the tangent space for [itex]S^2[/itex] is a 2-dimensional subspace of of a tangent space for [itex]\mathbb{R}^3[/itex].

    Now compute [itex]\tilde{\nabla}_v v[/itex] and [itex]\nabla_v v[/itex]. The former is a vector in a tangent space for [itex]S^2[/itex], while the latter is a vector in a tangent space for [itex]\mathbb{R}^3[/itex].

    In [itex]S^2[/itex], the particle moves on geodesic and the tangent vector to its path is parallel tansported, [itex]\tilde{\nabla}_v v = 0[/itex]. In [itex]\mathbb{R}^3[/itex], it is easy to computer that [itex]\nabla_v v = -\omega^2 \partial/\partial r[/itex]. There is no reason to expect that [itex]\nabla_v v[/itex] lies in a tangent space for [itex]S^2[/itex], as it is not computed using the connection for [itex]S^2[/itex].
     
  7. Jul 27, 2011 #6

    Ben Niehoff

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    To add to George Jones, there is a neat trick you can do to get the connection on a submanifold from the connection on the ambient manifold. This is called the "pullback". To compute it, you just take the change-of-basis formula for the Christoffel symbols:

    [tex]\tilde{\Gamma}^k_{ij} = \frac{\partial x^p}{\partial y^i} \frac{\partial x^q}{\partial y^j} \Gamma^r_{pq} \frac{\partial y^k}{\partial x^r} + \frac{\partial y^k}{\partial x^s} \frac{\partial^2 x^s}{\partial y^i \partial y^j}[/tex]

    and interpret it as follows: Let [itex]\Gamma^r_{pq}[/itex] be the Christoffel symbols on [itex]\mathbb{R}^3[/itex], and [itex]\tilde{\Gamma}^k_{ij}[/itex] be the Christoffel symbols on [itex]S^2[/itex]. The [itex]x^p[/itex] are the coordinates on [itex]\mathbb{R}^3[/itex], and the [itex]y^i[/itex] are the coordinates on [itex]S^2[/itex]. Hence (p,q,r,s) run from 1 to 3, whereas (i,j,k) run from 1 to 2.

    For fun, you should try to compute the Christoffel symbols on [itex]S^2[/itex] using this formula, and also compute them from the metric on [itex]S^2[/itex], and check that you get the same answer in each case.
     
  8. Jul 27, 2011 #7

    Mentz114

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    A linear combo can be more than rescaling. It can mix dimensions ( directions) as well.

    When you refer to points, I assume you mean points along a curve, yes ?

    I'm struggling to understand you, because I'm no expert in this. Thinking in terms of vector spaces carried along a curve - if you choose a basis at one point then the basis vectors will in general change along the curve. It depends on the path. The equation in your original post implies parallel transport, in which case the basis vectors are transported unchanged ( no rotation or rescaling).

    Ben and George, thanks for the informative posts. I'm not sure how embedded manifolds are relevant to the OP's question.
     
  9. Jul 27, 2011 #8

    Ben Niehoff

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    Because the OP was confused by the fact that the covariant derivative on R^3 wasn't giving him a covariant derivative on S^2.
     
  10. Jul 27, 2011 #9

    WannabeNewton

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    If we have some map [itex]\phi :\mathbb{R}^{3}\rightarrow S^{2} [/itex] then would the pullback of the connection from [itex]\mathbb{R}^{3}[/itex] to [itex]S^{2}[/itex] be analogous to the pullback of a function from [itex]\mathbb{R}^{3}[/itex] to [itex]S^{2}[/itex], so that [itex]\phi ^{*}\boldsymbol{\Gamma } = \boldsymbol{\widetilde{\Gamma }}[/itex]?
     
  11. Jul 27, 2011 #10

    Mentz114

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    Thanks. Maybe I should have read his last sentence.
     
  12. Jul 27, 2011 #11

    Ben Niehoff

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    The map should be the other way around: [itex]\varphi : S^2 \rightarrow \mathbb{R}^3[/itex]. This is just the embedding map. In this case, yes,

    [tex]\nabla_{S^2} = \varphi^* \nabla_{\mathbb{R}^3}[/tex]

    However, you must remember that the pullback of a connection has an extra piece involving second derivatives. The connection form goes as

    [tex]\omega_{S^2} = A \omega_{\mathbb{R}^3} A^{-1} + A d(A^{-1})[/tex]

    where A is the Jacobian matrix of [itex]\varphi[/itex] (or it might be its inverse, I forget). Also, since A is not generally square, the "inverse" here is either the left or right inverse, whichever one makes sense (again I forget, but it's obvious in context).

    The direction of the map [itex]\varphi[/itex] is important, because the manifolds on either side are of different dimensions. A map from a larger space to a smaller space is not invertible. Or stated another way, the Jacobian matrix A is not square, and can only have a one-sided inverse.
     
  13. Jul 27, 2011 #12

    WannabeNewton

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    Right sorry. I was reading the part of my text on diffeomorphisms and completely forgot about mappings not being necessarily invertible for different manifolds. Thanks by the way.
     
  14. Jul 27, 2011 #13
    Consider a vector field on a sphere given by [itex]\vec{V}= \hat{e}_\phi [/itex]. This vector field is not parallel transported on a circle at any latitude except the equator. Yet your equation above seems to say that it's parallel transported along any path. That is, [itex]\nabla_ie_j = 0 [/itex] implies [itex]u^i \nabla_ie_j = 0 [/itex] for any velocity ui.

    So can you really view basis vectors as covectors? I thought [itex]\partial_i e_j=\Gamma^{k}_{ij} e_k [/itex] was practically the definition of the Christoffel symbol so that it always holds no matter what, but if you view [itex]e_i[/itex] as a covector, then it suggests it's a parallel transport equation.
     
    Last edited: Jul 27, 2011
  15. Jul 27, 2011 #14
    The Christoffel symbols for your first metric in R3 are the same as the ones for your second metric in S2, provided you ignore all Christoffel symbols that have a radius index. Ben Niehoff provided a general formula, but is it just coincidence that the Christoffel symbols are the same for the two metrics above?

    I see what you mean. I just find it odd that in S2 a vector field might not be changing, but when you view it from R3 it's changing, albeit in the radial direction. But in either case, the change can always be expressed as a linear combination of the basis vector at a point. In your example at the equator, for R3 the change in the vector field is entirely in the radial direction of the tangent plane (or rather tangent solid), but in S2 the change in the vector field is also entirely in the tangent plane (namely the zero vector since there is no change).
     
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