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Energy-momentum tensor for a scalar field (sign problem!)

  1. Jun 23, 2008 #1
    Hi

    I have a small subtle problem with the sign of the energy-momentum tensor for a scalar field as derived by varying the metric (s.b.). I would appreciate very much if somebody could help me on my specific issue. Let me describe the problem in more detail:

    I conform to the sign convention [itex]g_{\mu \nu} = (+,-,-,-)[/itex]. The Lagranagian for a real scalar field is

    [tex] \mathcal{L} = \frac{1}{2} \dot{\Phi}^2- (\nabla \Phi)^2 - V(\Phi ) = \frac{1}{2} g^{\mu \nu} \partial_\mu\Phi \;\partial_\nu\Phi- V(\Phi ).[/tex]

    From Noether Theorem we find the energy-momentum tensor


    [tex]T^{\mu \nu} = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \Phi)} \: \partial^\nu \Phi - \mathcal{L} g^{\mu \nu} = \partial^\mu \Phi \partial^\nu \Phi - \mathcal{L} g^{\mu \nu}.[/tex]

    Now I want to derive this via varying the action

    [tex]S = \int \mathcal{L} \sqrt{-g}\; dx^4[/tex]

    in respect to [itex]g_{\mu \nu}[/itex]. In particular it holds

    [tex]\delta S = \delta\int \mathcal{L} \sqrt{-g}\; dx^4 = -\frac{1}{2}\int T_{\mu \nu} \delta g^{\mu\nu} \sqrt{-g}\; dx^4.[/tex]

    [itex]T_{\mu \nu}[/itex] is defined so that varying the action derived from the total Lagrangian

    [tex] \mathcal{L_{\rm tot}} = \frac{1}{16\pi G} R + \mathcal{L}[/tex]

    yields the Einstein field equations

    [tex]G_{\mu \nu} = 8\pi G T_{\mu \nu}.[/tex]

    (Note that

    [tex]\delta\int\frac{1}{16\pi G} R \sqrt{-g}\; dx^4 = \int G_{\mu \nu} \delta g^{\mu \nu}\sqrt{-g}\; dx^4, [/tex]

    therefore the - sign in the definition of [itex]T_{\mu \nu}[/itex].)

    Now let's vary the lagrangian of the scalar field:

    [tex]\delta \int \mathcal{L} \sqrt{-g}\; dx^4[/tex]
    [tex] = \int \delta(\mathcal{L}) \sqrt{-g} + \mathcal{L} \delta(\sqrt{-g})\; dx^4[/tex]
    [tex] = \int \delta \left( \frac{1}{2} g^{\mu \nu} \partial_\mu\Phi \;\partial_\nu\Phi- V(\Phi ) \right) \sqrt{-g} + \mathcal{L} \left(-\frac{1}{2} g_{\mu \nu} \delta g^{\mu \nu}\right) \sqrt{-g}\; dx^4[/tex]
    [tex] = \frac{1}{2}\int \left( \delta g^{\mu \nu} \partial_\mu\Phi \;\partial_\nu\Phi - \mathcal{L} g_{\mu \nu} \delta g^{\mu \nu} \right) \sqrt{-g}\; dx^4[/tex]
    [tex] = \frac{1}{2}\int \left(\partial_\mu\Phi \;\partial_\nu\Phi - \mathcal{L} g_{\mu \nu} \right) \delta g^{\mu \nu} \sqrt{-g}\; dx^4.[/tex]

    Comparing this with the definition of the [itex]T_{\mu \nu}[/itex] yields

    [tex]T_{\mu \nu} = -\partial_\mu \Phi \partial_\nu \Phi + \mathcal{L} g_{\mu \nu}[/tex]

    leading to the opposite sign as derived by the Noether Theorem.

    I would appreciate very much if somebody could explain why I get the sign wrong. I know this is a subtle (and possibly unimportant) issue but getting the wrong sign without understanding why gives a bad feeling. Thank you for any help!
     
  2. jcsd
  3. Jun 23, 2008 #2

    George Jones

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    According to Wald the Klein-Gordon energy-momentum tensor from Noether's theorem agrees with the Klein-Gordon energy-momentum tensor from varying the metric "up to a numerical factor." I do not know if the numerical factor is -1.

    Wald says that in others cases, there is less agreement, and it is the energy-momentum arrived at by varying g that appears on the right of Einstein's equation.

    If you have Wald, look near the bottom of page 457.

    I first ran into differences between the canonical and symmetric energy-momentum tensors in section 12.10 of Jackson.
     
    Last edited: Jun 23, 2008
  4. Jun 23, 2008 #3
    I think, I got the reason for the wrong sign. Since I used the signature [itex]g_{\mu \nu} = (+,-,-,-)[/itex] my definitions of [itex]T^{\mu \nu}[/itex] and [itex]\mathcal{L_{\rm tot}}[/itex] are not correct. With my signature the correct expressions read as

    [tex]\delta S = \delta\int \mathcal{L} \sqrt{-g}\; dx^4 = +\frac{1}{2}\int T_{\mu \nu} \delta g^{\mu\nu} \sqrt{-g}\; dx^4.[/tex]

    and

    [tex] \mathcal{L_{\rm tot}} = -\frac{1}{16\pi G} R + \mathcal{L}.[/tex]

    With this I get everything right. :-)
     
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