Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Interpretation of the covariant derivative

  1. Mar 4, 2010 #1
    Is this the right way to think about the covariant derivative, and if not, what improvements would you suggest to visualize the meaning of the covariant derivative?

    [tex](1)\mbox{ }\vec{e}_i(x')=\vec{e}_i(x)+\frac{\partial \vec{e}_i(x)}{\partial x^j}dx^j
    =\vec{e}_i(x)+\Gamma^{k}_{ij}\vec{e}_k(x)dx^j[/tex]

    so that [tex]\vec{e}_i(x)=\vec{e}_i(x')-\Gamma^{k}_{ij}\vec{e}_k(x)dx^j [/tex].

    This means that the vector V at x, when written in the x' basis, is given by contracting both sides with Vi

    [tex](2) \mbox{ }\vec{V}(x)=V^i(x) [\vec{e}_i(x')-\Gamma^{k}_{ij}\vec{e}_k(x)dx^j]=
    V^i(x) [\vec{e}_i(x')-\Gamma^{k}_{ij}\vec{e}_k(x')dx^j]
    [/tex]

    (in the last step I replaced the basis in x with the basis in x' since the difference is a product of two differentials which is small)

    Now, suppose that V(x) changes to V(x') as you move from x to x'. V(x') can be written in the x' basis as:

    [tex]\vec{V}(x')=[V^i(x)+\frac{\partial V^i(x)}{\partial x^j}dx^j]\vec{e}_i(x') [/tex]

    Subtracting V(x') from V(x):

    [tex] \left[\frac{\partial V^i(x)}{\partial x^j}+V^k\Gamma^{i}_{kj}\right]\vec{e}_i(x')dx^j=\left[\frac{\partial V^i(x)}{\partial x^j}+V^k\Gamma^{i}_{kj}\right]\vec{e}_i(x)dx^j[/tex]

    I also have a question about equation (1). To relate the basis vectors at two different points, do you have to use an embedding space? How is it possible to write the basis at a point as a linear combination of the basis at different point without an embedding space?

    Also, on something that I think is related to my 1st question, for equation (2), shouldn't this equality be exact? But if (2) represents a parallel transport of V(x) to x' (as it should since the covariant derivative is the difference between the new vector and the parallel transported old vector, so (2) has to represent the parallel transported old vector), then the vectors aren't equal. A parallel transported vector is different from the initial vector in the embedding space.

    edit:

    I was doing some reading on Wikipedia, and I ran across the article on an "affine connection". It would make sense to me if the Christoffel symbols "lifted" the basis vectors from one tangent space to another. However, the Christoffel symbols do not behave this way, since they are just an array of ordinary numbers! In other words, a linear combination of the basis vectors in the tangent space still exists in the tangent space, so how can the Christoffel symbols form an affine connection to a different tangent space (I'm not sure if I'm using the terminology right)?
     
    Last edited: Mar 4, 2010
  2. jcsd
  3. Mar 5, 2010 #2
    Here's an example. Consider spherical coordinates in R3. Some of the basis are:

    [tex]\vec{e_r}=(sin \theta cos\phi,sin\theta sin\phi, cos \theta) [/tex]

    [tex]\vec{e_\theta}=(r cos \theta cos \phi, r cos \theta \sin \phi, -r sin\theta)[/tex]

    Now

    [tex]\partial_\theta\vec{e_\theta}=-r\vec{e_r}[/tex]

    follows immediately from the above.

    So if you restrict yourself to the surface of the sphere S2, the derivatives of the basis vectors cannot be expressed as a linear combination of [tex]\vec{e_\theta} [/tex] and [tex]\vec{e_\phi} [/tex]. You are required to have [tex]\vec{e_r} [/tex] of the embedding space. Therefore the Christoffel symbols can't be just pure numbers, because pure numbers don't lift the vector to the new tangent space: you need [tex]\vec{e_r} [/tex] to do that.
     
  4. Mar 5, 2010 #3

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    Christoffel originally thought of his connection this way, as simply a projection onto the surface from the embedding space. Christoffel studied the differential geometry of surfaces in R^3.

    But you don't NEED the embedding space. All you need is a linear, invertible map between tangent spaces at nearby points, [itex]\varphi_X: T_pM \rightarrow T_{p + \epsilon X}M[/itex]. Of course, this allows you to define more general kinds of connections than just the Christoffel connection. You can pretty much make up a set of correspondences at will, subject only to the requirement that they be continuous.

    Think of [itex]\varphi[/itex] as a "parallel transport operator". It tells you what vectors should be considered "parallel" when comparing nearby points. So if you feed it a vector Y at point P, then [itex]Y' = \varphi_X(Y)[/itex] is the vector that is parallel to Y, at the point [itex]P + \epsilon X[/itex]. The covariant derivative is just the difference between Y' and Y, so

    [tex]\phi_X(Y) = Y + \epsilon \nabla_X Y = (1 + \epsilon \nabla_X) Y[/tex]

    So, you should think of [itex]\nabla_X[/itex] as the "infinitesimal generator of parallel transport" in curved space in the same way that you think of [itex]\partial / \partial x[/itex] as the "infinitesimal generator of translations" in flat space. The reason it is called a "connection" is because it tells you how the tangent spaces of nearby points are connected to each other, by way of saying which vectors are parallel.

    But as mentioned, you can define any old connection you like on a manifold. If your manifold has a metric, you might ask, which connection can be considered solely due to the metric? After all, the metric tells you the distances between nearby points, and if the tangent vectors are supposed to be considered as lying "in" the manifold, then that distance information ought to be sufficient to completely determine parallel transport. It turns out there is a unique answer to this question, and that is the Christoffel connection (also called the "metric connection"). And just as suspected, the Christoffel connection can be computed entirely from the information in the metric itself. This is why you don't require an embedding space: precisely because the metric (which does not depend on the embedding space) contains all the information needed to completely determine the metric connection! Basically, once you fix all the distances between points, the space is sufficiently rigid that you can determine what happens to any vectors as you move them around, subject to the requirement that they stay "in" the space.


    However, if you do happen to have an isometric embedding into R^n, it can be useful, because it makes some things easier to compute. For example, one can obtain the metric AND the Christoffel connection coefficients by simply pulling back from R^n onto the manifold. And since R^n has a very simple metric and a trivial connection, this makes the calculation very easy.
     
  5. Mar 5, 2010 #4
    Thanks. That was very helpful. I have a few questions.
    How would you define continuity? Say you are mapping two vectors A and B in the tangent plane at point p to the tangent plane in point p'. Then would it be the typical way in analysis, where gij(A-B)i(A-B)j less than delta implies the same thing with the two mapped vectors at the metric p' less than epsilon?

    That seems to be the obvious way, but I just want to make sure.
    I think this is where I'm having problems. Are tangent spaces vector spaces? If so, does it make sense to take the difference of two vectors that belong to different vector spaces? For flat space, I suppose it kind of makes sense to take the difference of two vectors at two different points, since the vector spaces at each point are copies of each other.

    The Christoffel symbol seems to go one step further as it seems to say that not only is there a mapping from one tangent space to another, but that the vector in one tangent space can be written as a linear combination of the vectors in the other tangent space!

    [tex]\vec{e}(P + \epsilon X)=\vec{e}(P)+\epsilon\Gamma\vec{e}(P) [/tex]

    In the notation you have, it does not suggest this:

    [tex]\phi_X(Y) = Y + \epsilon \nabla_X Y = (1 + \epsilon \nabla_X) Y[/tex]

    as [tex]\nabla[/tex] can be an operator, rather than a real number like the Christoffel symbol. Real numbers have meaning in vector spaces as they keep the vector in the space, while operators can do anything.

    I tried pulling back from R^3 spherical coordinates to S^2 and it didn't work! You can't express changes in the tangent space of points on the sphere without using the radial vector, but the radial vector does not exist in the intrinsic space.
     
  6. Mar 5, 2010 #5

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    Note that the pullback map for a connection is not just multiplying by Jacobian matrices. Connections transform differently. Also, the connection in spherical coordinates is not trivial. I would recommend going from Cartesian coordinates to S^2.

    I have to go, but I'll address your other questions another time.
     
  7. Mar 7, 2010 #6
    I was reading Penrose's layperson book, "The Road to Reality", and he says that the covariant derivative is the difference between the vector at the new point, and the old vector (at the nearby point) parallel transported to the new point.

    Using that interpretation, I tried to derive the covariant derivative (my original post), but I got confused about parallel transport, which entails writing a vector in one vector space as a linear combination of vectors in a different space.

    I suspect that maybe the Christoffel symbols should be interpreted only with components, and not with basis vectors. But all the books I've checked define the Christoffel symbols as acting on basis vectors and not components of vectors.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Interpretation of the covariant derivative
  1. Covariant Derivatives (Replies: 1)

  2. Covariant derivative (Replies: 1)

  3. Covariant Derivative (Replies: 8)

  4. Covariant derivative (Replies: 8)

Loading...