Skip to main content
4 events
when toggle format what by license comment
Aug 31, 2018 at 14:04 comment added yalda @TaylorTiger $p$ is not naturlly a covector. However, the component of the corresponding covector is what is needed when computing the inner product $x\cdot p = x^\mu p_\mu \equiv x^0 p_0 + \vec{x} \cdot \vec{p}$. The thing is, that when you are in euclidean space the components of the covectors and vectors are the same, so index position does not matter. However, after the Wick rotation you are in Minkowski space and there the position of the index plays an important role.
Aug 30, 2018 at 18:28 comment added user148792 Interesting... I'm wondering why is p a covariant vector? Before the transformation p is just defined (in the middle of page 14 of the document in the link) to be the Fourier Transformation pair of x, but Cardy never mentioned that it is a covariant vector.
Aug 30, 2018 at 14:40 review First posts
Aug 30, 2018 at 17:38
Aug 30, 2018 at 14:36 history answered yalda CC BY-SA 4.0