In a scientific paper, I need for my Bachelor thesis, the differential cross section for photo pair production

$\gamma\gamma\to e^{+}e^{-}$

is given as follow:

$\mathrm{d}\sigma = \frac{r_{0}^{2}\mathrm{d}\vec{p}_{+}\mathrm{d}\vec{p}_{-}}{2\epsilon_{+}\epsilon_{-}\omega_{1}\omega_{2}(1-\cos(\theta))}(B+4A-4A^{2})\delta(\vec{k}_{1}+\vec{k}_{2}-\vec{p}_{+}-\vec{p}_{-})\delta(\omega_{1}+\omega_{1}-\omega_{2}-\epsilon_{+}-\epsilon_{-})$

where $r_{0}$ is the classic electron radius,

$A=\big(\frac{1}{x_{1}}+\frac{1}{x_{2}}\big )$ and $B=\big(\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{1}}\big )$.

$x_{1}$ and $x_{2}$ are relativistic invariants:

$x_{1}=-2p^{(4)}_{+}k^{(4)}_{1}$ and $x_{1}=-2p^{(4)}_{+}k^{(4)}_{2}$

with the four momenta:

positron: $p_{+}^{(4)}=(\vec{p}_{+},i\epsilon_{+})$, electron: $p_{-}^{(4)}=(\vec{p}_{-},i\epsilon_{-})$,

photon 1: $k_{1}^{(4)}=(\vec{k}_{1},i\omega_{1})$, photon 2: $k_{2}^{(4)}=(\vec{k}_{2},i\omega_{2})$

The problem now is that these quantities are written in natural units $m=c=\hbar=1$, but I need this in SI-Units. But how can I rewrite this formula in SI units? A first step is to write for example

$p_{+}^{(4)}=(\vec{p}_{+},i\epsilon_{+}/c)$ and

$\hbar k_{1}^{(4)}=(\hbar\vec{k}_{1},i\omega_{1}\frac{\hbar}{c})$.

But this is not enough since B is dimensionless, but A is not. How is A+B defined as an example?

The paper cites the book "Quantum electrodynamics" from Akhiezer (1965), but I don't find the book on the internet or in a library.

Can anybody help me?


'A' has units of $1/(momentum)^2$. So in order to make this compatible with the dimensions of B. Transform A as $ A \rightarrow \hbar^2 A $

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  • $\begingroup$ But there have to be more constants......$\mathrm{d}\sigma$ schould have the dimension of an length^2.......When the term $(B+4A-4A^{2})$ is dimensionless and the dimension of the deltafunktion $\delta(x)$ is 1/(dimension(x)) i doesn't work... $\endgroup$ – Udalricus.S. Apr 14 '19 at 12:00
  • $\begingroup$ You can appropriately multiply the whole expression with $ \hbar^a c^b $ and work it out. For adding A's and B's, both should have the same dimensions. Hence, in order to make do for any other shortcomings to the units, need to multiply $ \hbar^a c^b $ to the whole expression $\endgroup$ – Rachana Apr 14 '19 at 12:14
  • $\begingroup$ Okay thank you......Also the electron mass is set to 1.....so I have to multiply with $m^{a}\hbar^{b}c^{d}$... $\endgroup$ – Udalricus.S. Apr 14 '19 at 12:28

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