Why is the effective action $\Gamma[\phi_c] \propto -(VT)$ (spacetime volume)?

Question

I'm asking about equation $(11.50)$ in Peskin and Schroeder where they state that the effective action evaluated at the classical field is given by

$$\tag{11.50}\Gamma[\phi_c] = -(VT)\cdot V_{eff}(\phi_c).$$

They claim that the $\Gamma$ is, in thermodynamic terms, an extensive quantity. Where do they get this information from?

Does is simply come from the fact that $$\tag{11.47}\Gamma[\phi_c] = -E[J]-\int d^4y\, J(y)\phi_c(y),$$ where $$E[J]\sim \log \int \mathcal{D}\phi \exp\bigg[\mathrm{i}\int\mathrm{d}^4x\,(\mathcal{L}+J\phi)\bigg].$$

Then I guess one could say that $\Gamma\propto\int d^4x=VT.$

Answer 1

Use the following expansion of effective action

$$\Gamma[\phi_c]=\int d^4x [-V_{eff}(\phi_c)+(\partial_\mu\phi_c)^2A(\phi_c)+...]$$

where $...$ represents higher order derivatives of $\phi_c$. When $J\rightarrow 0$, $\phi_c(x)=constant$. This constant is the VEV $\langle\phi_c(x)\rangle=\phi_0$ and we obtain from above $$\Gamma[\phi_0]=-VT. V_{eff}(\phi_0)$$ where all the derivatives of $\phi_c(x)$ vanish. I hope this is clear.