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Scalar fields: why symmetric ener-mom. tensor?

  1. Dec 23, 2009 #1
    I'm studying the properties of the energy momentum tensor for a scalar field (linked to the electromagnetic field and corresponding energy-momentum tensor) and now I'm facing the statement:

    "for a theory involving only scalar fields, the energy-momentum tensor is always symmetric". But I've some doubts on how to demonstrate it. So I started with a particular example.

    Given a density of Lagrangian [tex]L[/tex] for a scalar field [tex]\phi[/tex] (our field is in Minkowsky space with metric tensor [tex] \eta^{\mu\nu}[/tex] +---) of the form:


    with real constant [tex]\lambda[/tex] and [tex]m[/tex], according to the definition of energy-momentum tensor [tex] T^{\mu\nu}[/tex], that is
    [tex]T^{\mu\nu}=\frac{\partial L}{\partial(\partial_{\mu}\phi_i)}\partial^{\nu}\phi_i-\eta^{\mu\nu}L} [/tex]

    (with latin index i running on space coordinates) for this [tex]L[/tex], substituing into the last formula, I should get (dropping the index i since [tex]\phi[/tex] is a scalar):


    Arrived at this point my first question is: is this [tex]T^{\mu\nu}[/tex] symmetric?

    Trying to answer by myself but I'm new in tensor computing and I know that it would be a joke with a few of QFT and GR knowledge...

    given that [tex]\partial^{\mu}\phi[/tex] is a vector, and then [tex]\partial^{\mu}\phi\partial^{\nu}\phi [/tex] is a dot product (but where are the contracted indices???), so a scalar, it should be


    so with all scalars my tensor should be symmetric, isn't?

    But, generally speaking, my second question is why a theory of only scalar field the energy-momentum tensor is always symmetric?

    Trying to answer by myself I think that being the Langrangian Lorentz invariant, in certain way this is reflected to the tensor... but I'm not sure...

    can someone be so kind to explain to help me in answering these two questions?
  2. jcsd
  3. Dec 23, 2009 #2
    well. i'm a little unsure but i'll give it a go

    [itex]\partial^\mu \phi[/itex] is a vector.
    [itex]\partial^\mu \phi \partial^\nu \phi[/itex] isn't a dot product. if there are contracted indices it reduces to a scalar.
    the indices are different so it corresponds to multiplying to matrices together.
    a 4x1 and a 1x4 giving an overall 4x4 matrix

    so the stress energy tensor is going to be a 4x4 matrix

    so now you have to ask : is [itex]T^{\mu \nu}=T^{\nu \mu}[/itex]

    but it appears not to be the case. hmmmmm, hopefully someone else can point out why!
  4. Dec 23, 2009 #3


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    How did you use the definition of the energy momentum tensor to get that minus sign between the two derivative terms? That's not right.
  5. Dec 23, 2009 #4
    Considering the terms between the brackets, the first one is obtained deriving the Lagrangian respect to [tex]\partial\phi[/tex]; the second one, with the minus sign, is obtained raising the index of the first 4-gradient... with + sign the terms in bracket with the 1/2 factor would be the common decomposition of a symmetric tensor. so everything would be right, right?
  6. Dec 23, 2009 #5


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    Think of it this way:
    [tex]L = \frac{1}{2}\eta^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi -\frac{1}{2}m^2\phi^2 - \frac{\lambda}{4!}\phi^4[/tex]
    In detail,
    [tex]L = \frac{1}{2}[-(\partial_t\phi)^2 + (\partial_x\phi)^2 + (\partial_y\phi)^2 + (\partial_z\phi)^2] -\frac{1}{2}m^2\phi^2 - \frac{\lambda}{4!}\phi^4[/tex]
    [tex]\frac{\partial L}{\partial(\partial_t \phi)} = -\partial_t\phi[/tex]
    [tex]\frac{\partial L}{\partial(\partial_x \phi)} = \partial_x\phi[/tex]
    [tex]\frac{\partial L}{\partial(\partial_y \phi)} = \partial_y\phi[/tex]
    [tex]\frac{\partial L}{\partial(\partial_z \phi)} = \partial_z\phi[/tex]
    which is equivalent to
    [tex]\frac{\partial L}{\partial(\partial_\mu \phi)} = \eta^{\mu\nu}\partial_\nu\phi = \partial^\mu\phi[/tex]
    Try working through the rest of it from there. Also remember that your result is a tensor, so if you get something like [itex]m^2\phi^2[/itex], with no indices, you've done something wrong.
  7. Dec 24, 2009 #6
    ok, I've understood my error. Thanks! So, finally I get

    T^{\mu\nu}=\partial^{\mu}\phi\partial^ {\nu}\phi-\eta^{\mu\nu}(\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-m^2\phi^2})-\frac{\lambda}{4!}\phi^4
    and I'm sure that until this point the tensor is correct. But how can I see that
    T^{\mu \nu}=T^{\nu \mu}
    Last edited: Dec 24, 2009
  8. Dec 24, 2009 #7
    [tex]-\eta^{\mu \nu}[/tex] should multiply the the last term as well. Also you could change the contracted index to something else so there is no confusion. Symmetricity follows from the fact that [tex]\eta^{\mu \nu}[/tex] is symmetric and that derivatives of the field commute.
    Last edited: Dec 24, 2009
  9. Dec 24, 2009 #8
    Writing it as:
    [tex]T^{\mu\nu}=\partial^{\mu}\phi\partial^ {\nu}\phi-\eta^{\mu\nu}L[/tex]
    should make you see it directly.

    Your metric is always symetric, and [tex]\partial^{\mu}\phi\partial^ {\nu}\phi=\partial^{\nu}\phi\partial^{\mu}\phi[/tex].
  10. Dec 24, 2009 #9
    your second term in post 6 is wrong-it violates the summation convetion. one of those indices should be a [itex]\nu[/itex] which means it will contract to an invariant.
  11. Dec 24, 2009 #10
    Ok thanks. I get everything, now. I was missing the point about the derivatives.

    Can you help me, please, to understand why I get always a symmetric energy-momentum for a theory of scalar fields?
  12. Dec 24, 2009 #11
    sorry, can you be more explicit? I've compared that formula with several books, and the error is the last term out of the brackets...
  13. Dec 24, 2009 #12


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    What, like this?
    [tex]\eta^{\mu\nu} \partial_\mu\phi \partial^\nu\phi[/tex]
    I think you meant to say something else...

    [tex]\eta^{\mu\nu} \partial_\alpha\phi \partial^\alpha\phi[/tex]
    is what we're after. It doesn't contract to a scalar, and can't, because the answer has to be a rank-2 contravariant tensor.
  14. Dec 24, 2009 #13
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