- #1
craigthone
- 59
- 1
I am reading Polchinski's review on AdS/CFT https://arxiv.org/abs/1010.6134.
I have a very simple question, and please help me out. Thanks in advanced.
The question abou formula (3.19)
The scalar effective bulk action is given by
$$ S_0=\frac{\eta}{2}\epsilon^{1-D}\int d^Dx \phi_{\rm cl} \partial_\epsilon \phi_{\rm cl}$$
The variation of ##S_0## is given by
$$ \delta S_0={\eta}\epsilon^{1-D}\int d^Dx \delta \phi_{\rm cl} \partial_\epsilon \phi_{\rm cl}$$
My question is why the two terms from Leibnitz are equal?
The variation of ##S_0## is given by
$$\int d^Dx \delta \phi_{\rm cl} \partial_\epsilon \phi_{\rm cl}=\int d^Dx \phi_{\rm cl} \partial_\epsilon \delta \phi_{\rm cl}$$
I have a very simple question, and please help me out. Thanks in advanced.
The question abou formula (3.19)
The scalar effective bulk action is given by
$$ S_0=\frac{\eta}{2}\epsilon^{1-D}\int d^Dx \phi_{\rm cl} \partial_\epsilon \phi_{\rm cl}$$
The variation of ##S_0## is given by
$$ \delta S_0={\eta}\epsilon^{1-D}\int d^Dx \delta \phi_{\rm cl} \partial_\epsilon \phi_{\rm cl}$$
My question is why the two terms from Leibnitz are equal?
The variation of ##S_0## is given by
$$\int d^Dx \delta \phi_{\rm cl} \partial_\epsilon \phi_{\rm cl}=\int d^Dx \phi_{\rm cl} \partial_\epsilon \delta \phi_{\rm cl}$$