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Question about d'Alembert operator

  1. Nov 12, 2014 #1
    Hey guys,

    The expression [itex]\partial_{\mu}\partial^{\nu}\phi[/itex] is equal to [itex]\Box \phi[/itex] when [itex]\mu = \nu[/itex]. However when they are not equal, is this operator 0?

    I'm just curious cos this sort of thing has turned up in a calculation of mine...if its 0 I'd be a very happy boy
  2. jcsd
  3. Nov 12, 2014 #2


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    That thing is in fact a second rank tensor which in 1+1 dimensions becomes:

    \partial_\mu \partial^\nu \phi=\eta^{\nu \lambda}\partial_\mu\partial_\lambda \phi=\left( \begin{array}{cc} \eta^{00}\frac{\partial^2 \phi}{\partial x^0 \partial x^0}+\eta^{01}\frac{\partial^2 \phi}{\partial x^0 \partial x^1} \ \ \ \ \ \ \eta^{00}\frac{\partial^2 \phi}{\partial x^1 \partial x^0}+\eta^{01}\frac{\partial^2 \phi}{\partial x^1 \partial x^1} \\ \\ \eta^{10}\frac{\partial^2 \phi}{\partial x^0 \partial x^0}+\eta^{11}\frac{\partial^2 \phi}{\partial x^0 \partial x^1} \ \ \ \ \ \ \eta^{10}\frac{\partial^2 \phi}{\partial x^1 \partial x^0}+\eta^{11}\frac{\partial^2 \phi}{\partial x^1 \partial x^1} \end{array} \right)

    I think its obvious that there is no reason for it to be identically zero.
  4. Nov 13, 2014 #3


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    Usually ##\eta## is used for the Minkowski metric in Cartesian coordinates. Shyan wrote you a general expression for a general metric, in 2 dimensions. Usually we work in 4-dimensions, but by using the Minkowski metric, the expression is much simpler.

    It is perhaps easier to deal with ##\partial_\mu\partial_\nu\phi## in which case, this is just the Hessian matrix: http://en.wikipedia.org/wiki/Hessian_matrix (replace f with ##\phi##)
  5. Nov 13, 2014 #4
    Okay thank you. so would you say that [itex]\partial_{\mu}\partial^{\nu}\phi=\partial^{\nu}\partial_{\mu}\phi[/itex]?
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