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Functional or regular (partial) taylor series in Field theory

  1. Jul 13, 2013 #1
    When expanding a function (for example the determinant of the space-time metric g) as a functional of a perturbation from the flat metric ##h_{\mu \nu}##, i.e. ##g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu} ## i would think that the thing to do is to recognize that ##g_{\mu \nu}## and thus also ##h_{\mu \nu}## is itself a function of x and thus ##g(g_{\mu \nu})## is a functional of ##g_{\mu \nu}## and therefore we can expand the functional Taylor series

    $$ g(g_{\mu \nu}) = g(\eta_{\mu \nu}) + \int d^4 x \frac{\delta g}{\delta g_{\mu \nu}(x)}|_{h_{\mu \nu} =0} \delta g_{\mu \nu} + \ldots $$

    However I've regularly encountered people seeming to treat ##g_{\mu \nu}## as a regular variable and thus expanding

    $$ g(g_{\mu \nu})= g(\eta_{\mu \nu}) + \frac{\partial g}{\partial g_{\mu \nu}}|_{h_{\mu \nu} =0} \delta g_{\mu \nu} + \ldots $$

    which also yields the correct answer. So I wondered is there something fundamentally wrong with the latter approach? I.e. what is the reason (if there is one) that one must thing of g as a functional and expand accordingly instead of thinking of it as a regular function?

    (I realize that this is perhaps a question more suited to the mathematics forum, but I suspect those who knows a little about the mathematics of field theory can answer perhaps even better from a physicists point of view than those coming from a formal mathematics background.)
     
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  3. Jul 13, 2013 #2

    Bill_K

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    They're the same because the thing you're expanding (the determinant of the metric) is a function of gμν alone, not its derivative. For objects that depend on gμν,σ you'll get an additional term in the functional derivative coming from ∂/∂gμν,σ.
     
  4. Jul 13, 2013 #3
    I do not really see how this is related to wheather the determinant is a functional of the derivative or not. I'm tinking of the functional taylor series as the continous version of the discrete multivariable expansion

    $$ g(\vec x) = g(\vec x_0) + \sum_i \frac{\partial g}{\partial x_i} (x - x_0)_i + \ldots $$

    In this sense, treating the 'continous vector' ##h_{\mu \nu}(x)## like a single variable would be like treating ##\vec x = (x_1, x_2, \ldots ,x_n)## like a single variable. Is what you are describing a condition where this is allowed?
     
  5. Jul 13, 2013 #4

    Bill_K

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    Perhaps you should look it up in Wikipedia in that case.

    The functional derivative is most familiar from deriving the Euler-Lagrange equations by variation of L. In taking the functional derivative δL/δf, one considers L to be a function of f and all its derivatives: f', f'', etc. Dependence on the coordinates xi is irrelevant.

    Then δL/δf = ∂L/∂f - d/dx(∂L/∂f') + d2/dx2(∂L/∂f'') - ... For the simple case you quoted, the determinant of gμν is just an algebraic function and qw have just the first term, δg/δgμν = ∂g/∂gμν.
     
  6. Jul 14, 2013 #5
    I think I've had a revelation regarding this issue. The determinant g is not a functional of ##g_{\mu \nu}## since in order to be a functional the map, from the function to a scalar, has to use a domain of the function, not just the function evaluated at one point x. Since the evaluation of the determinant does not require ##g_{\mu \nu}## to be evaluated at more than one point x, one can regard ##g_{\mu \nu}## as a variable and thus taylor expand according to a regular taylor expansion. In contrast for the action S[f,f',f''] one has an integration over a domain of f, f', and f'' and it is not possible to regard f, f' and f'' as just single variables when expanding as a taylor series.
     
  7. Jul 14, 2013 #6

    Bill_K

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    That's just what I've been saying! :smile:

     
  8. Jul 14, 2013 #7
    Ah, sorry I missed that point ;) Thank you anyway Bill.
     
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