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This is clearly symmetric!

  1. Nov 10, 2008 #1
    I was reading about the momentum-energy tensor (or stress-energy tensor), at one point the author says,
    [tex]\theta^{\mu\nu} = (\partial^\mu\phi)(\partial^\nu\phi) - g^{\mu\nu}L[/tex]

    This is clearly symmetric in [tex]\mu[/tex] and [tex]\nu[/tex]."

    [tex]\theta^{\mu\nu}[/tex]: is the stress-energy tensor
    [tex]\phi[/tex] is a scalar field
    [tex]g^{\mu\nu}[/tex] is the metric (+---)
    L is the lagrangian density

    My question (I'm not an expert in tensors) is how do you see that it's "clearly" symmetric? another silly question: when do we need to symmetrize and antisymmetrize tensors?

    Please, tell me guys if this isn't the right place for my question.
  2. jcsd
  3. Nov 10, 2008 #2


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    It's symmetric because:

    - g is symmetric since it's the metric, thus the second term is symmetric.
    - The first term is the product of two derivatives, so is symmetric (i.e. [itex]\partial^0\phi \partial^1\phi \equiv \partial^1\phi\partial^0\phi[/itex], and similarly for other values of mu and nu.)
  4. Nov 10, 2008 #3
    Sometimes one has a product of a symmetric tensor with another tensor which is not symmetric nor antisymmetric, then one can show that the antisymmetric part of the second is killed by the first, the same thing occurs for the antisymmetric case, this is why we need to antisymmetrise and symmetrise tensors: to see which part remains and which part is killed...
  5. Oct 25, 2011 #4
    Suppose to have a generic tensor T, which isn't symmetric or antisymmetric.
    You can ALWAYS write it as sum of its symmetric part with its antisymmetric part.

    e.g. for a rank-2 tensor

    [tex] T^{\mu\nu}=\frac{1}{2}(T^{\mu\nu}+T^{\nu\mu}) + \frac{1}{2}(T^{\mu\nu}-T^{\nu\mu})[/tex]

    On the right-hand side, the first term [itex]\left(\bullet+\bullet\right)[/itex] is a symmetric tensor, while the second [itex]\left(\bullet-\bullet\right)[/itex] is antisymmetric!

    Hence, in general: [itex]T=T_S+T_A[/itex].

    Now...if you have a product between tensors, and you know that one of them is explicitly symmetric [itex]S[/itex] or antisymmetric [itex]A[/itex]...you can write:



    Of course the product between a symmetric and an antisymmetric tensor is zero, i.e. [itex]SA=AS=0[/itex]!

    And thus


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