Functional or regular (partial) taylor series in Field theory

[center o bass]

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.)

 

[Bill_K]

They're the same because the thing you're expanding (the determinant of the metric) is a function of g_μν alone, not its derivative. For objects that depend on g_μν,σ you'll get an additional term in the functional derivative coming from ∂/∂g_μν,σ.

 

[center o bass]

They're the same because the thing you're expanding (the determinant of the metric) is a function of g_μν alone, not its derivative. For objects that depend on g_μν,σ you'll get an additional term in the functional derivative coming from ∂/∂g_μν,σ.

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 continuous 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?

 

[Bill_K]

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 x_i is irrelevant.

Then δL/δf = ∂L/∂f - d/dx(∂L/∂f') + d²/dx²(∂L/∂f'') - ... For the simple case you quoted, the determinant of g_μν is just an algebraic function and qw have just the first term, δg/δg_μν = ∂g/∂g_μν.

 

[center o bass]

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 x_i is irrelevant.

Then δL/δf = ∂L/∂f - d/dx(∂L/∂f') + d²/dx²(∂L/∂f'') - ... For the simple case you quoted, the determinant of g_μν is just an algebraic function and qw have just the first term, δg/δg_μν = ∂g/∂g_μν.

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.

 

[Bill_K]

I think I've had a revelation regarding this issue. The determinant g is not a functional of g_μν 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 gμν to be evaluated at more than one point x, one can regard gμν as a variable and thus taylor expand according to a regular taylor expansion.

That's just what I've been saying!

the thing you're expanding (the determinant of the metric) is a function of g_μν alone, not its derivative.

the determinant of g_μν is just an algebraic function

 

[center o bass]

That's just what I've been saying!

Ah, sorry I missed that point ;) Thank you anyway Bill.