I'm reading "Quantum Field Theory in Curved Spacetime" by Parker, Toms and I'm stuck in the very last part of the demonstration of the Schwinger action principle. I arrived at eq. 1.34

$$ \delta S = \int \text{d}^n{x} \, \partial_\alpha \left( \delta \phi^\mu \frac{\partial\mathscr{L}}{\partial(\partial_\alpha \phi^\mu)} - \delta x^\beta {T^\alpha}_\beta \right) \tag{1.34} $$

and there I completely miss why I should obtain $$\delta S=G(t_2)-G(t_1),$$ where $t_1,t_2$ are the time boundaries for the integral that defines the action. The "generator" $G(t)$ is defined in eq. 1.32 through a spatial integration with volume element $\text{d}V$ $$ G(t) \doteq \int \text{d}{V} \left( \pi_\mu \delta \phi^\mu - {T^0}_\beta \delta x^\beta \right). \tag{1.32}$$ What brings to the conclusion that $\delta S=G(t_2)-G(t_1)$?


1 Answer 1


$$ \delta S = \int \text{d}^n{x} \, \partial_\alpha \left( \delta \phi^\mu \frac{\partial\mathscr{L}}{\partial(\partial_\alpha \phi^\mu)} - \delta x^\beta {T^\alpha}_\beta \right)=\\=\int \text{d}^n{x} \,\Bigg[ \partial_0 \left( \delta \phi^\mu \frac{\partial\mathscr{L}}{\partial(\partial_0\phi^\mu)} - \delta x^\beta {T^0}_\beta \right)+\partial_i \left( \delta \phi^\mu \frac{\partial\mathscr{L}}{\partial(\partial_i\phi^\mu)} - \delta x^\beta {T^i}_\beta \right)\Bigg] $$

Second term, using Gauss theorem, is

$$\int_{t_1}^{t_2} \text{d}x^0\,\int \text{d}V\,\partial_i F^i=\int_{t_1}^{t_2} \text{d}x^0 \int_{\partial V}\text{d}\sigma_i~F^i |_{\partial V}=0,$$ $|_{\partial V}$ means "evaluated at the border" (spatial infinity), where fields are supposed to vanish.

Using $$\frac{\partial\mathscr{L}}{\partial(\partial_0\phi^\mu)}=\pi_\mu,$$ the first term is $$\int_{t_1}^{t_2} \text{d}x^0\,\partial_0 G(x^0)=G(t_2)-G(t_1)$$

  • $\begingroup$ Thanks a lot!! Was impossible for me to see it $\endgroup$
    – Rob Tan
    Oct 25, 2022 at 18:09

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