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Stress tensor from action in Landau-Ginzburg field theory

  1. Mar 23, 2015 #1
    I would appreciate any help with the following question:

    I know that for relativistic field theories, the stress tensor can be obtained from the classical action by differentiating with respect to the metric, as is explained on the wikipedia page

    http://en.wikipedia.org/wiki/Stress–energy_tensor
    If I understand this correctly, the energy-momentum tensor is defined as the functional derivative of the action w.r.t. the metric,

    ##T^{\mu \nu} = \frac{2}{\sqrt{-g}} \frac{\sqrt{-g}\delta \mathcal L}{\delta g_{\mu\nu}}.##

    Here ##g_{\mu \nu}## is the standard Minkowski metric in the case of special relativity. (I need to add here that my relativity background is not good, and that I don't understand this derivation very well).

    I was wondering whether one can do this for a Landau-Ginzburg theory for some (scalar) order parameter field. In this case the stress tensor is defined in terms of partial derivatives of the Hamiltonian density,

    ##T_{ij} = \delta_{ij} \mathcal H - \partial_i \frac{\partial \mathcal H}{\partial (\partial_j \phi)}##

    where the Hamiltonian is

    ##H = \int d\vec x \mathcal H.##

    E.g. consider a simple Gaussian Hamiltonian

    ##H = \int d\vec x \; [\nabla \phi(\vec x)]^2##

    whose corresponding stress tensor is

    ## T_{ij} = \frac 1 2 \delta_{ij} \nabla^2 \phi - \partial_i \phi \partial_j \phi. ##

    Could anyone perhaps help me to show in detail how (or whether) I can obtain this stress tensor from the Hamiltonian by differentiation in the case of the Gaussian Hamiltonian with respect to the Euclidean metric? In particular, how do I obtain the relative minus sign between the two terms in the stress tensor?

    Any help would be appreciated.
     
    Last edited: Mar 23, 2015
  2. jcsd
  3. Mar 24, 2015 #2

    vanhees71

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    It is important that the ##\sqrt{-g}## factor is under the functional derivative,
    $$T^{\mu \nu}=\frac{2}{\sqrt{-g}} \frac{\delta}{\delta g_{\mu \nu}} \left (\sqrt{-g} \mathcal{L} \right ).$$
    This implies that for the potential part of the Lagrangian density
    $$\mathcal{L}_I=-V(\phi)$$
    you get the contribution
    $$T_I^{\mu \nu} =-V(\phi) \frac{2}{\sqrt{-g}} \frac{\delta}{\delta g_{\mu \nu}} \left (\sqrt{-g} \right ) =+g^{\mu \nu} V(\phi),$$
    because
    $$\frac{\delta \sqrt{-g}}{\delta g_{\mu \nu}} =-\frac{1}{2 \sqrt{-g}} \frac{\delta g}{\delta g_{\mu \nu}}=-\frac{\sqrt{-g}}{2} g^{\mu \nu},$$
    because ##g^{\mu \nu}## is the inverse of ##g_{\mu \nu}##.

    Note that all this is given in terms of the east-coast metric with mostly positive signs when in diagonal form.
     
  4. Mar 24, 2015 #3
    Hi @vanhees71

    Thanks for your reply. Does your statement also hold for a Euclidean metric? Specifically, the problem I'm considering is on 3-D Euclidean space, where I guess all factors of ##\sqrt g ## are trivial, so my hamiltonian is

    ##H = \int d\vec x [\nabla \phi (\vec x) ] ^2 ##,

    with ##\vec x \in \mathbb R ^3##. So what I am trying to figure out is whether I can retrieve the stress tensor ##T_{ij}## that I wrote down in my first post from this hamiltonian by differentiating with respect to ##g_{ij} = \delta_{ij}##, with ##i,j \in \{1,2,3\}##.

    I suppose one could write for my hamiltonian that with ##g_{ij} = \delta_{ij}## that

    ##H = \int d\vec x \sqrt{g} [\nabla \phi (\vec x) ] ^2 = \int d\vec x \sqrt{g} g^{ij} (\nabla \phi)_i ( \nabla \phi)_j , ##

    i.e.,

    ## T_{ij} \propto \frac{\delta }{\delta g_{ij}} \big( \sqrt{g} g^{ij} (\nabla \phi)_i ( \nabla \phi)_j \big). ##

    If I apply the product rule to the derivative, I guess I am quite close to obtaining ##T_{ij}## from my first post, but formally I'm not quite sure how to fill in the steps... Any help with this last bit would be really appreciated.

    Thanks!
     
  5. Mar 24, 2015 #4

    vanhees71

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    Here the metric is varied. So it doesn't make a difference whether you have a Euclidean (Wick rotated) metric or the original space-time metric.
     
  6. Mar 24, 2015 #5
    Thanks @vanhees71

    So my problem reduces to evaluating two functional derivatives:

    1)
    ## \frac{\delta \sqrt{g} }{\delta g_{ij}} ##, which you evaluate above. Could you explain your result to me?

    2)
    ## \frac{\delta g_{ij} }{\delta g_{ij}} ## which I assume is equal to 1 or otherwise some functional delta-function?

    Sorry, I somehow have no intuition for differentiating a functional of the metric with respect to the metric.

    These results combined should give what I am looking for :)

    Thanks again.
     
  7. Mar 24, 2015 #6

    vanhees71

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    This is a bit tedious since you derive with respect to two-indexed quantities, and one has to be careful with the indices, and how to understand it. You have to keep in mind that you should consider, e.g., the two quantities ##g_{12}## and ##g_{21}## as independent although of course the general metric is symmetric. So the formula you want is
    $$\frac{\delta g_{\mu \nu}}{\delta g_{\rho \sigma}}=\delta_{\mu}^{\rho} \delta_{\nu}^{\sigma}.$$
     
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