1
$\begingroup$

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 $\delta S=G(t_2)-G(t_1)$?

$\endgroup$

1 Answer 1

2
$\begingroup$

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

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 $$\int_{t_1}^{t_2} \text{d}x^0\,\partial_0 G(x^0)=G(t_2)-G(t_1)$$

$\endgroup$
1
  • $\begingroup$ Thanks a lot!! Was impossible for me to see it $\endgroup$
    – Rob Tan
    Oct 25, 2022 at 18:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.