# Where does this equation originate from? (found in the Big Bang Theory)

Recently, I've been watching "The Big Bang Theory" again and as some of you might know, it's a series with a lot of scientific jokes in it - mostly about Physics or Mathematics. I understand most of the things mentioned in the series and whenever I don't understand a joke, I just look up the knowledge I'm missing on wikipedia - e.g. I learnt about Schrödinger's cat in this way.

However, at one point, I did not know how to proceed, which is why I'm asking this question. At a Physics quiz, the participants are asked to "solve" the following equation:

The solution turns out to be $-8 \pi \alpha$. My questions are: What is the meaning of this equation? How does one obtain it? And of course, how does one solve it and is the solution given by the university janitor (in the series) correct?

Also, I am sorry if this is an equation found in Physics, I didn't know.

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## migrated from math.stackexchange.comJun 24 '11 at 22:05

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the feynman diagram kinda implies its physics, as does the fact that the characters are physics students (i think, never having seen the show). in other words, i dont know – yoyo Jun 24 '11 at 20:25
@yoyo: Two of the characters are physics professors/researchers; another is an astrophycist post-doc; the fourth works in the Engineering department. They all finished their degrees already. (-: – Arturo Magidin Jun 24 '11 at 20:27
There is no equals sign in this "equation". – Harry Stern Jun 24 '11 at 20:31
Googling gave the following: bigbangtheory.wikia.com/wiki/The_Bat_Jar_Conjecture they mention it a bit down, I hope that helps. – Dedalus Jun 24 '11 at 20:35
The physics consultant for the show has a blog at thebigblogtheory.wordpress.com which often discusses the details of the ideas talked about in the show -- though not for this particular episode. – Mark Betnel Jun 25 '11 at 3:20

This is a so called Feynman diagram you see on the board. It is a suggestive way to write the formula written below the diagram. Each aspect of the diagram directly translates to part of the formula via the so called "Feynman Rules"

With Feynman diagrams you can calculate the "amplitude" (that is related to the quantum mechanical probability for a process to happen). In this case the process it electron muon scattering. As you can see initially there is a electron and a muon (the lower part of the diagram where the time arrow starts). The electron is on the left and carries momentum $p_1$ and spin $s_1$. This translated to a so called spinor $u^{(s_1)}(p_1)$ in the formula.

The initial muon carries momentum $p_2$ and spin $s_2$ therefore its spinor is $u^{(s_2)}(p_2)$.

Now since electrons and muons are electrically charged (both are have negative charge), they will repel each other via the electromagnetic force. The "force carrier", also called gauge boson of the EM foce is the photon $\gamma$, the quantum of the EM vector field $A_\mu$. The electron and muon exchange a photon carrying momentum $q$ and represened by the term $\frac{-ig_{\mu\nu}}{q^2}$.

The outgoing electron with momentum $p_3$ and spin $s_3$ corresponds to the term $\bar{u}^{(s_3)}(p_3)$ and likewise the outgoing muon to $\bar{u}^{(s_4)}(p_4)$.

The strength with which the electrons and muons couple to the EM field is given by $-ie\gamma^\mu$, or as written here using the fine structure constant $\alpha$: $-i\sqrt{4\pi\alpha}\gamma^\mu$.

From these parts you construct an electron "current"

$\bar{u}^{(s_3)}(p_3) (-i\sqrt{4\pi\alpha}\gamma^\mu) u^{(s_1)}(p_1)$

and muon current

$\bar{u}^{(s_4)}(p_4) (-i\sqrt{4\pi\alpha}\gamma^\nu) u^{(s_2)}(p_2)$

which get couple by the photon. I.e.

$\bar{u}^{(s_3)}(p_3) i\sqrt{4\pi\alpha}\gamma^\mu u^{(s_1)}(p_1) \frac{-ig_{\mu\nu}}{q^2} \bar{u}^{(s_4)}(p_4) i\sqrt{4\pi\alpha}\gamma^\nu u^{(s_2)}(p_2)$

since the exchanged momentum can be anything you have to integrate over all possible momenta

$\int d^4q$ but make sure to conserve moentum via Dirac delta functions

$\delta^{(4)}(p_1-p_3-q)\delta^{(4)}(p_2+q-p_4)$

which is why you get

$\int {d^4q \bar{u}^{(s_3)}(p_3) i\sqrt{4\pi\alpha}\gamma^\mu u^{(s_1)}(p_1) \frac{-ig_{\mu\nu}}{q^2} \bar{u}^{(s_4)}(p_4) i\sqrt{4\pi\alpha}\gamma^\nu u^{(s_2)}(p_2) \delta^{(4)}(p_1-p_3-q)\delta^{(4)}(p_2+q-p_4)}$

The Lorentz indices and sign of the photon propagator in the screenshot are wrong btw.

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the integral is kind of redundant for this feynman diagram also.. – BjornW Jun 24 '11 at 23:25
yeah, true. The delta functions render it useless. – luksen Jun 24 '11 at 23:29

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