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I forgot what this is called and what is is written out.

  1. Aug 22, 2011 #1
    [tex]\partial_u \partial^u [/tex]

    Isn't this like a box symbol? How is it written out again? You can write it out in 2 dimensions x,t no need to include all 3 spacial dimensions.
     
  2. jcsd
  3. Aug 22, 2011 #2

    WannabeNewton

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    [itex]\partial _{\mu }\partial ^{\mu } = \square ^{2}[/itex] or [itex]\square [/itex]. Both notations are used (with and without the 2). As in, [itex]\square ^{2}A^{\mu } = \partial ^{2}_{t}A^{\mu } - \triangledown ^{2}A^{\mu }[/itex] in natural units.
     
  4. Aug 22, 2011 #3

    G01

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    It's the D'Alembert Operator, essentially the 3+1 generalization of the Laplacian. Using a space-negative metric:

    [tex]\partial_{\mu}\partial^{\mu}=\Box=\frac{\partial^2}{\partial t^2}-\nabla^2[/tex]

    http://en.wikipedia.org/wiki/D'Alembertian
     
  5. Aug 22, 2011 #4
    I am confused about why it is not written [tex]\partial_u \partial_u[/tex]

    I thought an upper index on a derivative implied that [tex]\partial x^u[/tex] is in the numerator and not the denominator.
     
  6. Aug 22, 2011 #5

    WannabeNewton

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    [itex]\partial _{\mu }\partial _{\mu }[/itex] is simply the second partial derivative with respect to a single component [itex]\mu [/itex] and won't give you the divergence which is [itex]\partial ^{\mu }\partial _{\mu }[/itex] (notice that here you are summing over all the possible values of [itex]\mu [/itex], not just with respect to a single component like the other form). Don't try to write [itex]\partial ^{\mu }[/itex] as you would the right hand side of [itex]\partial _{\mu } = \frac{\partial }{\partial x^{\mu }}[/itex], it won't help. Just think of it as [itex]\partial ^{\mu } = g^{\mu \nu }\partial _{\nu }[/itex].
     
  7. Aug 22, 2011 #6
    Oh, right. Thanks everyone.
     
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