- #1
Haorong Wu
- 415
- 90
- TL;DR Summary
- should ##\frac {\partial} {\partial x ^{\mu}}## transform to ##\frac {\partial} {\partial y ^{\mu}}## or remain unchanged?
Hello again. I am sorry I got another problem when learning QFT regarding the Lorentz transformation of derivatives.
In David Tong's notes, he says
I am not sure how to transform the partial derivatives. Explicitly, should ##\frac {\partial} {\partial x ^{\mu}}## transform to ##\frac {\partial} {\partial y ^{\mu}}## or remain unchanged? And why?
At first, I think since ##x## is transformed to ##y##, then ##\partial x^\mu## should transform to ##\partial y^\mu## as well. However, after I tried the calculation, I found that, the derivative does not transform and it is still ##\partial x^\mu##. I could not find a proper explanation to convince myself.
In David Tong's notes, he says
Consider a real scalar field transformed as ##\phi \left ( x \right ) \rightarrow \phi ^{'} \left ( x \right ) =\phi \left ( \Lambda ^{-1 }x \right )##.
Then the derivative of the scalar field transforms as a vector, meaning
##\left ( \partial _{\mu} \phi \right ) \left ( x \right ) \rightarrow \left ( \Lambda ^{-1 }\right ) ^{\nu} {}_{\mu} \left ( \partial _{\nu} \phi \right ) \left ( y \right )## where ##y=\Lambda ^{-1 }x ##
I am not sure how to transform the partial derivatives. Explicitly, should ##\frac {\partial} {\partial x ^{\mu}}## transform to ##\frac {\partial} {\partial y ^{\mu}}## or remain unchanged? And why?
At first, I think since ##x## is transformed to ##y##, then ##\partial x^\mu## should transform to ##\partial y^\mu## as well. However, after I tried the calculation, I found that, the derivative does not transform and it is still ##\partial x^\mu##. I could not find a proper explanation to convince myself.