# Change of action after a transformation with space-time dependent parameters

I've been following David Tong's lecture on introduction to quantum field theory. In his lecture notes page number 19 (and his video class on Youtube), he talks about global transformation that rotates the phase of a complex scalar field $$\phi$$ but with spacetime dependent variable $$\alpha = \alpha(x)$$.

If the parameter alpha isn't spacetime dependent, the lagrangian $$\mathcal{L}$$ is invariant $$(\delta \mathcal{L}=0)$$, which makes the transformation a symmetry. But if it is spacetime dependent, then the resulting changes would have the form of: $$\delta\mathcal{L}=\partial_\mu \alpha(x) h^\mu(\phi) \tag{1.65}$$

He mentions that we can still make the transformation a symmetry as long as the action is invariant $$(\delta S =0)$$. Based on (1.65) the change in action is therefore:

$$$$\delta S=\int d^4x \enspace \delta \mathcal{L} = - \int d^4x \enspace \alpha(x) \partial_\mu h^\mu\tag{1.66}$$$$

and this is where I'm confused. My attempt to understanding it is here:

\begin{align} \delta S&=\int d^4x \enspace \delta \mathcal{L}\\ &= \int d^4x \enspace \partial_\mu \alpha(x) h^\mu(\phi)\\ &= \int d^4x \enspace \big[ \partial_\mu(\alpha(x)h^\mu)-\alpha(x)\partial_\mu h^\mu \big] \end{align}

Can anyone please correct me or tell me why I'm not arriving with the same formulation?

## 1 Answer

Yes, OP is right. Tong is suppressing a total derivative term in eq. (1.66). See also this related Phys.SE post.