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How would torsion affect the particle equation of motion?

  1. Apr 18, 2013 #1
    Background for the question:
    In standard GR the equation of motion of a (classical) particle is a geodesic. A geodesic is a curve along which the tangent vector is parallel transported. If the tangent vector is denoted by V this condition is

    [tex]\nabla_V V=0[/tex]

    In components this is


    where s is the parameter of the curve. [itex]{\Gamma^\alpha}_{\mu\nu}[/itex] are the connection coefficients. The connection coefficients are the component representation of a given connection (covariant derivative). Only the symmetric part [itex]{\Gamma^\alpha}_{(\mu\nu)}[/itex] contributes to the equation above.

    For a metric connection, the connection coefficents can be expressed as

    [itex]{\Gamma^\alpha}_{(\mu\nu)}=\frac{1}{2}g^{\alpha\beta}(g_{\mu\beta,\nu}+g_{\nu\beta,\mu}-g_{\mu\nu,\beta})+\frac{1}{2}( {{T_\mu}^\alpha}_\nu + {{T_\nu}^\alpha}_\mu)[/itex]

    where T is the torsion tensor.

    If the length of a curve between two fixed points is


    then it can be shown that the equation of the curve which minimizes (or maximizes maybe?) this length is


    This equation matches the geodesic condition in the absence of torsion.

    so now my question: which one do we use if torsion is non-zero? I would imagine we would use the geodesic equation since, otherwise, the presence of torsion would have no measurable effect. But if so we then find that particle paths do NOT follow extremal curves. Is that acceptable?

    Or maybe we need to use another measure of length in the presence of torsion?
  2. jcsd
  3. Apr 18, 2013 #2
  4. Apr 18, 2013 #3


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    If you are looking to solve for the affine geodesics of a non-torsion free connection ##\nabla## then you use the first of what you wrote, ##\nabla_{U}U = 0##. The concept of a connection has nothing to do with a riemannian metric; I can have a non-torsion free connection on my smooth manifold with no need to ever endow a riemannian structure on the manifold. If you explicitly endow your smooth manifold with a riemannian structure and consider the geodesics which come out of first variations of arc-length with respect to that riemannian metric then for the case of the levi-civita connection it turns out that both definitions are equivalent but this need not hold for connections with non-vanishing torsion as you have noticed.

    It seems like your question is, if our manifold is given a metric ##g_{ab}## then how does parallel transport play out. Well it is very easy to show (Wald Problem 3.1 ) that there still exists a unique derivative operator with torsion, ##\nabla_{a}##, such that ##\nabla_{a}g_{bc} = 0##. This is where the formula you listed at the beginning comes from and this is what you would use in solving for the affine geodesics of the connection ##\nabla## associated with the derivative operator ##\nabla_{a}## if you wanted ##u^{a}\nabla_{a}u^{b} = 0## expressed in terms of the formula for the coefficients of ##\nabla## in terms of ##g_{ab}##.

    EDIT: I remembered this pdf that I read after first completing chapter 3 of Wald that neatly extends things to the case where the derivative operator ##\nabla_{a}## has non-vanishing torsion. Hope you find it helpful: http://www.slimy.com/~steuard/teaching/tutorials/GRtorsion.pdf See in particular section 3.3. We want the affine geodesics of ##\nabla_{a}## so we can study the physical effects of torsion.

    EDIT 2: Just as a little side note, one can also very easily show (Wald problem 3.1) that given a derivative operator ##\nabla_{a}## with non-vanishing torsion and any two vector fields ##X^{a},Y^{a}##, ##T^{c}_{}{}_{ab}X^{a}Y^{b} = X^{a}\nabla_{a}Y^{c} - Y^{a}\nabla_{a}X^{c} - \mathcal{L}_{X}Y^{c}## where ##T^{c}_{}{}_{ab}## is the torsion tensor. This isn't necessarily related to your primary question but it does give more physical insight into the effects of a connection with non-vanishing torsion on vector fields.
    Last edited: Apr 18, 2013
  5. Apr 19, 2013 #4


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    I think the first observation is that in the presence of torsion the identification of 'straightest' lines and 'shortest' lines is no longer valid.
  6. Apr 19, 2013 #5
    Excellent links, guys. Thanks for sharing.

    Right. So the question is then, what would we expect a particle follow? The straightest path or the shortest path?
  7. Apr 19, 2013 #6


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    In Einstein-Cartan-Theory which is an extension to Einstein-Hilbert allowing for torsion the two trajectories will coincide ;-)

    Torsion is non-dynamical i.e. non propagating, so torsion vanishes in vacuum and therefore the two e.o.m. should be identical (except for particles travelling not in vacuum ;-)
  8. Apr 19, 2013 #7


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    Again, if you are solving for the affine geodesics of the connection with non-vanishing torsion then you can say with certainty, by definition, that the particle will follow the "straightest possible path". What you cannot say with certainty in general is that the geodesics of this connection are also the "shortest possible paths". This depends on the physical situation / theory at hand.

    EDIT: Just to supplement what Tom said, read this: http://ebooks.cambridge.org/pdf_vie...bCode=CUP&urlPrefix=cambridge&productCode=cbo
    I found it an interesting read.
    Last edited: Apr 19, 2013
  9. Apr 20, 2013 #8
  10. Apr 21, 2013 #9


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    Imagine a 2d manifold.

    Putting things together, it seems to me that in the presence of torsion, "straight" lines as defined by the connection, projected onto the tangent space, must appear to be curved to order two in that tangent space by the "shortest distance" defintions.

    Otherwise, I can't see how parallelograms could fail to close to order 3 - parallelograms do close to order three for the Leva Civita connection, but only close to order two in the presence of torsion.
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