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Derivatives in a non-trivial metric

  1. Mar 31, 2014 #1
    I'm trying to work out:
    (∇f)^2 (f is just some function, its not really important)

    While working in curved space with a metric:
    ds^2 = α dt^2 + dr^2 + 2c√(α+1) dtdr

    I'm not really sure how to calculate a derivative in curved space, any help would be appreciated

    thanks
     
  2. jcsd
  3. Mar 31, 2014 #2
    Well, hmm...Your question is above my pay-grade. But it seems to me that you have a 2 dimensional space here.
    you have t and r. ∇f is DEFINED as ∑ ∂f/∂qi over all i (where qi are the coordinates).
    I am assuming you do really mean (∇f)² and not ∇²f which is completely different.
    In general, you are not going to get an invariant (coordinate independent) quantity by simply differentiating a variable. You need to use covariant differentiation (which includes the Christofel symbol) and results in a tensor (invariant). This may or may not be important in your problem, IDK.
    Oh my! In looking at the wikipedia article on covariant derivatives, I see that ∇ is also used to symbolize them! (I was familiar with using Df to symbolize them). So, now I'm not sure what your question is? Did you mean the square of the gradient or did you mean the square of the covariant derivative? Sorry, I almost never revisit a thread I've answered, but hopefully someone else will. Again, your question is above my competence level. Take it for what its worth.
     
  4. Mar 31, 2014 #3

    Bill_K

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    The gradient ∇f is ∂f/∂xi, just the partial derivatives wrt the coordinates.
    ∇f·∇f is where the metric comes in, ∇f·∇f = gij ∂f/∂xi ∂f/∂xj
     
  5. Mar 31, 2014 #4
    Are you trying to express that as a tensor? partial derivatives are not tensors in curved space time.
     
  6. Mar 31, 2014 #5

    Bill_K

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    Partial derivatives of a scalar are. The distinction between partial derivative and covariant derivative arises only when you're differentiating a vector or tensor quantity. I assume by "some function" the OP means a scalar.
     
  7. Apr 1, 2014 #6
    But is the square of the partials a tensor in general? To simplify, in Rn is Tμ=(∂f)^2 a true vector in general? I'd say it isn't in general, but I'm not sure exactly in wich cases it is.
     
  8. Apr 1, 2014 #7
    ah yeah thanks for the replies, think I've got it sorted now!
     
  9. Apr 1, 2014 #8
    I guess as long as Ti=δij ∂f/Xj
     
    Last edited: Apr 1, 2014
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