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Some intuitional questions about GR

  1. Jun 20, 2012 #1
    Hello fellow physicsists!
    I have some (and will probarbly get more) questions about differential geometry and general relativity which I would like to get answered.
    Most of my wonderings are usually intuitional errors so pure calculations are rarely needed.

    Here comes the first one:

    When generalizing to the covariant differentation a connection term is introduced. This term I think of as needed due to the global structure, such as curvature, of the manifold which makes the tangentspaces of the nearby points twist and turn. This explantion, however, doesnt seem consistent with the fact that a vector parallell transported along a closed cruve may differ from the original one. After all the initial and final set of basis vectors are the same.
    Is my way of thinking about the reason for the connection wrong or very simplified? How can I change my way of thinking so it is mathematically consistent?

    Thanks!
    //
    Kontilera
     
  2. jcsd
  3. Jun 20, 2012 #2

    Dale

    Staff: Mentor

    Good question. The basic idea is that the tangent space at each point in the manifold is a completely different vector space. So, in order to compare vectors at two different points in the manifold you need some "connection" between different tangent spaces. Once you have this connection then you can map vectors in one tangent space to vectors in another.
     
  4. Jun 20, 2012 #3
    This viewpoint about the manifold being curved is the prevailing framework for GR, but I think it can be instructive to consider a slightly different framework: one in which the manifold is explicitly flat but one imposes a generalization of rotational invariance: namely, that all objects in spacetime are subject to a position-dependent boost or rotation.

    Let [itex]\underline L(a)[/itex] denote this arbitrary, position-dependent linear operator that represents boosts and/or rotations (I'll just call say rotations from now on) which acts, for example, on a vector [itex]a[/itex]. Let [itex]A(x)[/itex] represent some vector field and [itex]A'(x) = \underline L(A)[/itex] represents its transformation.

    When we use just the ordinary partial derivative operator [itex]\nabla[/itex], we get an extra term:

    [tex]a \cdot \nabla \underline L(A) = \underline L(a \cdot \nabla A) + a \cdot \dot \nabla \dot{\underline L}(A)[/tex]

    Where the overdots denote what is being differentiated. As we can see by the product rule, there is an extra term introduced that starts mangling our derivative operator. The covariant derivative is defined to remedy this. It obeys the property

    [tex]a \cdot D' A' = \underline L(a \cdot D A)[/tex]

    under any position-dependent rotation [itex]\underline L[/itex]. The connection absorbs the [itex]a \cdot \dot \nabla \underline L(A)[/itex] term, but it's not in general exactly equal to this.

    In short, the connection is needed to define a derivative operator with logical transformation properties under arbitrary position-dependent rotations throughout spacetime. The covariant derivative is defined as a matter of convenience because it has nice transformation properties that make it easy to work with when transforming between frames.
     
  5. Jun 21, 2012 #4
    Ah thanks for the answers! I believe I follow your motivation for the covariant derivative Muphrid but does it really solve the problem? After all is your transformation again spacetime dependent.. doesnt this imply that we return to the same basis as we started in when considering parallell transportation on a closed curve? :/
     
  6. Jun 22, 2012 #5
    Another question concerns the explanation of the riemann tensor as a machine that calculates the difference of a vector when parallell transported around an infinitesimal loop from, and ending in, that point. Lets not focus on which the meaning we should put to the input vectors. Our spacetime manifold is in every small enough region approximatly flat so the curvature is a phenomena that can only be detected in non-infinitesimal regions... why is there a difference when transporting a vector in the loop?
     
  7. Jun 22, 2012 #6

    Ben Niehoff

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    Science Advisor
    Gold Member

    The Riemann tensor computes the second-order corrections to the vector as it travels around infinitesimal loops. That is, the corrections are proportional to the area of the loop. If the loop shrinks to truly zero size, then there is no correction.

    On a 2-dimensional manifold, one can explicitly calculate holonomies (i.e., differences between initial and final state of a vector as it travels around macroscopic loops) by integrating the curvature 2-form (a version of the Riemann tensor) over the area enclosed by the loop.

    A great example of this is Foucault's pendulum. If you have a pendulum at a certain latitude, its plane of motion will precess as the Earth rotates. The amount by which it precesses in one day is proportional to the area enclosed by the line of latitude (which is a circle).
     
  8. Jun 24, 2012 #7
    A common misconception is that covariant differentiation applies only to curved manifolds. This is not the case. It also applies to flat manifolds in which curvilinear coordinates are being employed. Of course, in the case of curved manifolds, use of curvilinear coordinates is essential.

    Let's see how this all plays out. See attachment.
     

    Attached Files:

  9. Jun 24, 2012 #8
    I think that's a very misleading statement. If there's a transformation from one coordinate system to another encoded in [itex]x' = f(x)[/itex], then only the Jacobian operator [itex]\underline f(a)[/itex] (or more precisely, its transpose [itex]\overline f(a)[/itex] is needed to have a coodinate system independent notion of differentiation.

    Specifically, one can show that [itex]\overline f(\nabla') = \nabla[/itex]. While the explicit use of Christoffel symbols isn't incorrect, I think it leads to a very misleading picture of what's actually going on. There's a reason Christoffel symbols aren't true tensors--they mix up terms based on coordinate system changes with terms actually related to curvature.
     
  10. Jun 25, 2012 #9
    I'm disappointed that the analysis I presented did not work for you. I can tell you that I actually performed the exercise that I alluded to in my Word Document for the case of a spherical surface. First I imagined myself as a 2D being embedded within the surface of a sphere, unaware that the radial dimension existed. I calculated the cross partial derivatives of the latitudinal and longitudinal coordinate basis vectors, and found that the order of differentiation mattered. I then approached the same calculation as a 3D being, recognizing that the derivatives of the coordinate basis vectors with respect to position within the surface produced components in the radial direction. By doing this, I was able to identify what was omitted from the analysis of the 2D beings, and to precisely establish the source of the components of the Riemann tensor in 2D (of course, in the 3D analysis, the components of the Riemann tensor were all zero, since the 3D space in this analysis was flat). If you would like to see the detailed 2D and 3D analysis, I would be glad to provide them.

    Chet
     
  11. Jun 25, 2012 #10
    I'm not saying the math is incorrect. I'm saying that the use of Christoffel symbols to help describe derivatives even in flat space curvillinear coordinates, while necessary for the traditional approach, contributes to making the geometric picture of what's going on muddied. I dislike it from a pedagogical standpoint.
     
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