Recent content by miemie0205

So ##k\cdot k'=\sum_{over all \alpha} k_\alpha k'^\alpha## right? Within the same book when they mention the 4-vector, they wrote p (without the above arrow). Normal vector they wrote ##\vec p## or ##\bf p## and ##p^\mu## is the contravariant vector, which is ##p^\mu=(E, p_x, p_y, p_z)## Then I...

From the book that I am using: the metric tensor is: $$g_{\mu\nu}=g^{\mu\nu}= \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix}$$ and k, k' are the four-momenta of two photons in a final state of $$h(q) \leftarrow...

My bad. I need to correct some points. $$M=C/m(k\cdot k' g^{\mu\nu} - k^\nu k'^\mu)\epsilon^*_\mu(k,\lambda) \epsilon^*_\nu(k',\lambda') $$ with $$\epsilon^*_\mu(k,\lambda) \epsilon^*_\nu(k',\lambda')$$ are polarization vectors. I am fine with these polarization vectors conjugate. They are...

Homework Statement $$M=C/m(k.k'g^{\mu\nu} - k^{\nu}k'^{\mu})\epsilon ^*_{\mu}(k,\lambda)\epsilon _{\nu}(k',\lambda ')$$ Calculate $$\sum _{\lambda} |M|^2$$ Homework Equations $$\sum _{\lambda}\epsilon ^*_{\mu}\epsilon _{\nu}=-g_{\mu\nu}$$ The Attempt at a Solution Firstly, I find...