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Question on Phenomenology of Einsteins Field Equations

  1. Oct 1, 2005 #1
    I am only just starting to realize that there is a correspondence between quantum field theory and Einsteins field equations.

    In QFT the approach is to write the Lagrangian and then to solve the Euler Lagrange equation to obtain the equations of motion of the field.

    In GR it seems that the starting point is not the Lagrangian but instead the metric - but that the end result, the stress energy tensor, is again, the equations of motion of the field.

    Now, in QFT the simplest lagrangian is for a scalar field.

    Does one take a similar approach in GR. When one solves the metric what do the eqtns of motion correspond to - ie are they for a field, or a particle or both?
     
  2. jcsd
  3. Oct 2, 2005 #2

    pervect

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    You can formulate relativity with a Lagrangian density using the Einstein-Hilbert action.

    according to

    http://www.lns.cornell.edu/spr/2000-05/msg0024719.html

    the answer to your question is the Ricci curvature, plus whatever you need for the Lagrangian density of your "source" fields.

    I'll quote the whole article,while it may be guilty of over-explaining, and it might come across as condescending, I think I've been way too terse above.

     
  4. Oct 2, 2005 #3
    This is fascinating, thank-you.

    So, let me check I have this correct - [tex]R_{\mu\nu}[/tex] is equivalent to [tex]L[/tex]

    That is the Ricci tensor is the lagrangian density.

    ...oh, and don't worry about 'over' explaining - thats exactly how I like it explained. :smile:
     
    Last edited: Oct 2, 2005
  5. Oct 2, 2005 #4

    pervect

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    Looking this up in my textbook, it looks like it can be even simpler. It's the Ricci scalar that's the Lagrangian density when you define things right.


    You get

    [itex]{\cal{L}} = \sqrt{ -g } R [/itex]
    You have to do some mucking around with volume elements and tensor densities to get this in this simple form - the sqrt term (which is the sqrt of the determinant of the metric) has to do with the issue of defining a volume element, and all tensor densities have to be proportional to [itex]\sqrt{-g}[/itex] to be tensor densities.

    There's another approach called the Palanti action which works the way Baez described, too, but it is probably harder to explain.

    Of course the vacuum Einstein equations aren't so interesting, unless you have fields in them to generate gravity. This however, requires more terms in the Lagrangian.

    Something like

    [itex]{\cal{L}} = -\frac{1}{4}\sqrt{-g} g^{ac} g^{bd} F_{ab}F_{cd}[/itex]
    for the electromagnetic field, for instance.
     
  6. Oct 2, 2005 #5
    Yes, I recognise the [tex]\sqrt g [/tex] term, I believe it corresponds to the Jocobian. Its like multiplying by [tex]r^2 sin\theta[/tex] when you work in spherical coordinates.

    So I just want to sumamrize:

    The Ricci scalar is equivalent to the lagrangian density

    the stress energy tensor is equivalent to the equations of motion.

    Now, what is the ricci tensor analagous to?

    Also, is the stress energy tensor equivalant to the EOM in field theory for a scalar field or some other field?

    Thanks!

    :smile:
     
  7. Oct 3, 2005 #6

    pervect

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    [itex]g^{ab}[/itex] are your "field" variables, and you get R by taking

    [itex]R_{ab} g^{ab} = R^a{}_a = R[/itex], so R_ab determines the mapping of the field variables to the action.
     
  8. Oct 5, 2005 #7
    when you say field variables - is this equivalent to talking about a scalar field in QFT?

    eg [tex]\phi \rightarrow \phi(x,t) [/tex] where x and t are you field variables.
     
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