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Using the 4-gradient

  1. Feb 6, 2013 #1
    lets say we have a covariant 4-gradient ∂[itex]_{μ}[/itex] = (1/c ∂/∂t, ∇). How do I actually apply this to a function, say x^2 where x is an arbitrary 4-vector (ct,-x,-y,-z)?
     
  2. jcsd
  3. Feb 6, 2013 #2
    What is x^2? Is it the scalar product of x*x or the contravariant vector ((ct)^2, -x^2, -y^2, -z^2)? If it is the latter, just contract:

    [itex]\partial_\mu x^2^\mu[/itex]

    If it is the former, I suppose you would apply it just like an ordinary 3-gradient on a scalar function.
     
  4. Feb 6, 2013 #3
    that is what is bothering me as well. I assumed it was the scalar product, since it seems to me that you need a scalar function to apply a gradient operator, but in an example problem, the 4-gradient was applied to a 4-vector.

    In an example I'm looking at, they tell me ∂[itex]_{μ}[/itex] x[itex]^{μ}[/itex] = 1 where x[itex]^{μ}[/itex] = (ct,-x,-y,-z). How did they get this?
     
  5. Feb 6, 2013 #4
    Assuming your indices are in the right place, that inner product should give -2, not 1.

    You can apply a gradient to a vector, all the time in fluid dynamics. It produces a tensor.

    If it is instead ∂[itex]_{μ}[/itex] x[itex]_{μ}[/itex], using the minkowski metric with signature +,-,-,-, you could get something that looks like (1,1,1,1). If somehow that "1" you have in your example is supposed to be a vector with just 1's, this is what they did, otherwise I have no clue.
     
    Last edited: Feb 6, 2013
  6. Feb 6, 2013 #5
    thank you greatly. i believe i figured it out.
     
  7. Feb 6, 2013 #6
    What was it, out of curiosity?
     
  8. Feb 6, 2013 #7

    WannabeNewton

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    That is the divergence. The gradient is not being involved at all. [itex]\partial _{\mu }x^{\mu } = \partial _{t}x^{t} + \triangledown \cdot \vec{x}[/itex].
     
  9. Feb 6, 2013 #8

    dextercioby

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    No, [itex] \partial_{\mu}x^{\mu} = 4 [/itex].
     
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