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Let's take a functional $F[\phi]$ as defined in this answer

$$ F[\phi] = \int d^4x \, \phi\, \partial^2 \phi $$

whose dimensions are, if the coordinates have dimensions of a length, as it's customary, are $[F]\, = [\phi^{2}]\,[L^{-2}] [L^{4}]$

It's functional derivative is $$ \frac{\delta F[\phi]}{\delta \phi} = 2\, \partial^2 \phi $$

This means that the dimensions of the $\displaystyle{\frac{\delta F[\phi]}{\delta \phi}}$ are $[\phi]\, [L^{-2}]$

which is equivalent to say that, assuming $[\delta\phi] = [\phi]$, $$ [\delta F] = [\phi^2]\,[L^{-2}] = [F]\, [L^{-4}] $$

Is this correct that $F$ and $\delta F$ have different dimensions? I find it a bit weird and I don't actually don't get why this happens.

RenatoRenatoRenato
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1 Answers1

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No, $F$ and the infinitesimal variation $\delta F$ have the same dimension. And $\phi$ and the infinitesimal variation $\delta \phi$ have the same dimension. But by definition of the functional/variational derivative $\frac{\delta F}{\delta\phi}$, the infinitesimal variation is $$ \delta F ~=~\color{red}{\int\!d^dx}~ \frac{\delta F}{\delta\phi}~\delta\phi. \tag{1} $$ So the functional derivative $\frac{\delta F}{\delta\phi}$ does not have the dimension $\frac{[F]}{[\phi]}$, despite the notation. There is also the dimension of spacetime. Therefore $$ \left[\frac{\delta F}{\delta\phi}\right]~=~\frac{[F]}{[\phi]\color{red}{[x]^d}}.\tag{2}$$

See also e.g. this related Phys.SE post.

Qmechanic
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