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Calculation of Proper Acceleration

  1. Oct 10, 2012 #1

    pervect

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    In another thread, Bill_K writes:

    Wald, is one of the sources that does it the "wrong" way. But it seems that while the wrong way is rather "sloppy", it seems to work in practice. I' would like to see if this "slopiness" could lead to incorrect answers.

    If we imagine that parallel transport was defined by a vector field, rather than a curve, would you agree that we could use aμ = UααUμ ?

    Wald's argument is that parallel transport turns out to depend only on the values of the vector field along the curve. If I'm reading this right and Wald is correct, then _any_ vector field will give the right answer, as long as its orbit generates the curve in question.

    This would imply that while the formulation is sloppy, it's not so bad as to give wrong answers.

    I'd like to be sure I'm not missing something though,.
     
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  3. Oct 10, 2012 #2

    George Jones

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    If a the 4-velocity defined along the worldline is extendable to a vector field defined on a neighbourhood of (the image of) the worldline, then aμ = UααUμ makes sense, but when the 4-velocity is not extendable to a neigbourhood, then this expression cannot be used, but the absolute derivative can be used. For (a non-physical) example, consider a closed timelike curve. At the event where the curve intersects itself there are, in general, two different 4-velocities.
     
  4. Oct 10, 2012 #3

    Bill_K

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    As I said in the earlier thread, the issue is a conceptual one.

    Expressions involving covariant derivatives are appropriate when dealing with a fluid, where the dynamical variables are in fact fields. But, in order to calculate the orbit of mercury, or Thomas precession say - anything involving the world line of a particle - am I really required to imagine that the velocity vector has been extended continuously off the world line? And then afterwards say the result is independent of how I extended it! Doing it in terms of a covariant derivative is just a habit, and perhaps a lack of familiarity with the other concepts, we wouldn't dream of doing it that way in Newtonian mechanics. If anything, doing it the "right" way is simpler.

    In these days we are so careful about saying exactly what we mean in other contexts - drawing the distinction between a vector and a 1-form, for example - I don't think it's being overly pedagogic to distinguish between a function of one variable and a function of four variables! :wink:
     
  5. Oct 10, 2012 #4

    pervect

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    No, it's a very good point, and the fact that a major textbook apparently got it wrong makes it an even better one.

    But from the point of damage control, it sounds like as long as I'm not dealing with timelike curves, I won't have a problem with the sloppy formula giving "wrong" answers, rather similar to the way that Feynman's trick of differentiating under the integral sign doesn't usually cause problems (given that most functions in physics are differentiable).
     
  6. Oct 10, 2012 #5

    George Jones

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    I could agree more! Students often get the multivariable chain rule wrong, because they don`t understand domains of functions.
     
  7. Oct 11, 2012 #6

    stevendaryl

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    I agree that it is kind of weird to do that, but it seems to happen in more than one place in the GR literature. For example, the definition of the Riemann curvature tensor is:

    [itex]R(u,v,w) = \nabla_u \nabla_v w - \nabla_v \nabla_u w - \nabla_[u,v] w[/itex]

    Conceptually, that doesn't make any obvious sense, because [itex]u[/itex], [itex]v[/itex] and [itex]w[/itex] are vectors, not vector fields, so it doesn't make sense to use covariant derivatives. However, one can imagine the process of:
    1. Extend [itex]u[/itex], [itex]v[/itex] and [itex]w[/itex] to vector fields.
    2. Compute the above expression for [itex]R(u,v,w)[/itex]
    3. Note that the value of [itex]R(u,v,w)[/itex] doesn't depend on the vector fields except their value at one point (so the value is actually a function of three vectors, rather than three vector fields).
     
  8. Oct 11, 2012 #7
    This is not correct. In that formula w IS a vector field, and that is all is needed to use the covariant derivative. While tangent vectors u and v are the linear arguments of the tensor.
     
  9. Oct 11, 2012 #8

    stevendaryl

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    No, it's not. In the expression [itex]R(u,v,w)[/itex] all three are vectors, not vector fields.

    --
    Daryl McCullough
    Ithaca, NY
     
  10. Oct 11, 2012 #9

    George Jones

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    In math books, u,v, and w are vector fields. In many references, the word "field" is implicit. Meaning is then given by context. For example, "consider vector v", could mean that v is an element of a tangent space (just a vector), or that v lives in the tangent bundle (a vector field), depending on context.
     
  11. Oct 11, 2012 #10

    stevendaryl

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    But the Riemann curvature tensor acts on vectors, not vector fields.
     
  12. Oct 11, 2012 #11

    stevendaryl

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    Actually, I suppose it's a little bit of a moot point. Any operator on vectors immediately can be considered an operator on vector fields. But [itex]R(u,v,w)[/itex] has a meaning for any three vectors, there is no need to think of them as vector fields.
     
  13. Oct 11, 2012 #12
    First of all, the expression is R(u,v)w, and for Riemann manifolds this equals the covariant derivative for the vector field w in the direction of the tangent vectors u and v minus the same thing in the direction of v and u, the failure of the commutativity of the covariant derivatives leads to nonvanishing curvature.
     
  14. Oct 11, 2012 #13
    Acts on both.
     
  15. Oct 11, 2012 #14

    George Jones

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    From "Riemannian Maniforlds: an Introduction to Curvature" by Lee:

    "If [itex]M[/itex] is any Reimannian manifold, the (Riemann) curvature endomorphism is the map [itex]R: T\left(M\right) \times T\left(M\right) \times T\left(M\right) \rightarrow T\left(M\right)[/itex] defined by ..."

    Here, [itex]T\left(M\right)[/itex] is the tangent bundle, so, vector fields.
     
  16. Oct 11, 2012 #15

    stevendaryl

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    Are you arguing about the parentheses? There is really no difference in differential geometry between the three types of tensors
    (1) A linear function of two vector arguments that returns a function from vectors to vectors, (2) A linear function of three vector arguments that returns a vector, and (3) a linear function of 3 vectors and 1 covector that returns a scalar. In all three cases, you write the components of the tensor [itex]R^i_{jkl}[/itex]

    I don't understand the point of that paragraph. It seems to be just expressing in words exactly what I wrote in symbols:

    [itex]R(u,v,w) = \nabla_u \nabla_v w - \nabla_v \nabla_u w - \nabla_{[u,v]} w[/itex]

    My point was the weirdness that the left-hand side is a function of three vectors, while the right-hand side depends on three vector fields.
     
  17. Oct 11, 2012 #16

    stevendaryl

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    This might be a quibble of terminology, but at every point on the manifold, there is a corresponding tangent space, and a corresponding Riemann tensor that is an operator on vectors of that tangent space. It's ALSO true that the Riemann tensor can be thought of as an operator on vector fields, as well.
     
  18. Oct 11, 2012 #17

    stevendaryl

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    My point is that if at some point on the manifold you have three tangent vectors u,v,w, you can compute R(u,v,w). You don't need a vector field.
     
  19. Oct 11, 2012 #18
    Why do you keep saying that?
     
  20. Oct 11, 2012 #19

    stevendaryl

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    Because it's true. Let me quote Misner, Thorne and Wheeler on this point:

    I'm really just repeating their point; although the mathematical definition of Riemann involves extending the vectors A, B and C to vector fields and using covariant derivatives, Riemann itself is a tensor, a function of vectors at a point.
     
  21. Oct 11, 2012 #20

    stevendaryl

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    Sigh. Every operator on vectors can be considered trivially an operator on vector fields. My point is that the other way around doesn't hold; not every operator on vector fields is a legitimate operator on vectors. An example is the covariant derivative. It doesn't make any sense to take a covariant derivative of a vector, only of a vector field. An operator on vector fields is only considered a tensor if it acts as an operator on vectors at each point. The Riemann curvature tensor is a tensor (hence the name "curvature tensor").
     
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