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Inventing Lagrangians (D'Inverno)

  1. Nov 19, 2012 #1

    TerryW

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    Gold Member

    I am steadily working my way through D'Inverno and have reached Chapter 20. On page 272 there is a problem which goes along the lines of ... here is a Lagrangian ... show that the Einstein tensor can be derived from it. The Lagrangian in question is a 'quadratic' Lagrangian and has four terms along the lines of hab,bhcc,a, ie four products of h with a variety of subscripts and superscript.

    So my question is - where the hell does that come from? What is the physical significance of the terms? Is this a case of someone working backwards from the Einstein tensor to produce the equivalent Lagrangian?

    Earlier in the book there was a problem relating to the Eddington Lagrangian, again the question is - What motivates the construction of these objects?

    Can anyone shed some light?


    TerryW
     
  2. jcsd
  3. Nov 20, 2012 #2

    haushofer

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    It would help if you show us the explicit Lagrangian; not everyone has acces to this book. But I suspect that you're talking about the Fierz-Pauli Lagrangian, which is a spin-2 theory on a Minkowski background. Take a look at these notes,

    http://arxiv.org/abs/1105.3735

    read the introduction because it's so nice, and then look at page 13 and further, where this Lagrangian (and the corresponding Hamiltonian!) is analyzed thoroughly. In the massless case this Fierz-Pauli Lagrangian corresponds to the Einstein-Hilbert action, with

    [tex]
    g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}
    [/tex]

    and up to quadratic interactions of h.

    If it's not the Fierz-Pauli Lagrangian d'Inverno is talking about, you should give us the explicit Lagrangian here, but these notes give a pretty good overview Fierz-Pauli and the motivation behind these Lagrangians.

    So you can look upon it from two different ways:

    1) bottom-up: construct a spin-2 self-interacting theory on a Minkowksi background. This demands e.g. invariance the gauge symmetry

    [tex]
    \delta h_{\mu\nu} = 2 \partial_{(\mu}\xi_{\nu)}
    [/tex]
    The relative coefficients in the Lagrangian are such that you don't have any propagating ghosts.

    2) top-down: perturb the Einstein-Hilbert action.

    In both cases you arrive at (massless) Fierz-Pauli.

    You also might want to check out Zee's QFT in a Nutshell, one of the last chapters about gravity (8?).
     
    Last edited: Nov 20, 2012
  4. Nov 20, 2012 #3

    atyy

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    It's not clear what the "physical significance" of any particular Lagrangian is. In classical general relativity, it is the field equations that make testable predictions. There is more than one Lagrangian producing the field equations, so which one is "correct" at the classical level is unknown (and irrelevant). http://www.phy.olemiss.edu/~luca/Topics/gr/action_1st.html [Broken] lists many different actions. Some speculative theories of quantum gravity generalize the classical actions, in which case the quantum theories based on different actions may be different, even though the classical theories are the same, eg . http://arxiv.org/abs/1012.4280 (published in Physics Letters B).
     
    Last edited by a moderator: May 6, 2017
  5. Nov 22, 2012 #4

    TerryW

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    Thanks for the replies Haushofer and atyy,

    I've downloaded the documents suggested. It looks like I have a bit of reading to do!

    I did take a quick look at the Fierz-Pauli action and it isn't quite the same as the problem I was working on.

    The Lagrangian in d'Inverno is:
    1/2ε(hab,bhcc,a - hab,chcb,a +1/2hcd,ahcd,a - 1/2hcc,ahdd,a)

    So it is like the Fierz Pauli action in terms of the components involving the derivatives of h but with some bits missing!

    As I said, I was able to derive the Einstein tensor from the Lagrangian above. Maybe once I've read Hinterbichler's paper, I'll be able to relate D'Inverno's Lagrangian to the Fierz-Pauli action.

    Thanks again for your suggestions,


    TerryW
     
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