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In the discussion of external bremsstrahlung, the following amplitude is used:

$$M=i\bar{u}_e(k')e \left(\gamma^\nu\epsilon_\nu \left[ \frac{i \gamma^\nu(k'_\nu+\omega_\nu) + m}{(k'+\omega)^2-m^2}\right]\right)\gamma^\nu u_e(k) \frac{e^2}{q^2}\bar{u}_p(p')\Gamma_\mu(q)u_p(p)$$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$enter image description here

If one assumes the emitted photon's energy is small and the electron mass negligible, we can write the term in brackets as: $$ i\frac{\gamma^\nu\epsilon_\nu \gamma^\nu k'_\nu}{2k'\cdot\omega}$$ So far so good. In the literature however, this is somehow reduced to: $$ i\frac{\epsilon_\nu \cdot k'_\nu}{k'\cdot\omega}$$ How is this done? the rules for gamma matrices would suggest that the multiplication of two Feynman slashes should be more involved.

What am I missing? How did the gamma matrices disappear?

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    $\begingroup$ Good question. (I mean, well written question, though I think the answer is standard, I just don't have time to look up the details.) One suggestion: don't reuse the index $\nu$. Maybe change it to $\mu$ in $\gamma^\nu \epsilon_\nu$? (Unless you're quoting that from a textbook or something, but if so, that's really sloppy notation) $\endgroup$ – David Z Nov 21 '15 at 16:58
  • $\begingroup$ @DavidZ - Thanks. I'm quoting from a specific paper, but the result is standard. I think the issue could be something specific to the polarization vector gauge, but I just can't see it. $\endgroup$ – nbubis Nov 21 '15 at 17:00
  • $\begingroup$ Which specific paper? $\endgroup$ – Qmechanic Nov 21 '15 at 17:02
  • $\begingroup$ @Qmechanic - There are a couple with the result, as well as textbooks, but the notation I used is taken from here (eq. 12): authors.library.caltech.edu/5588/1/ENTprc01.pdf $\endgroup$ – nbubis Nov 21 '15 at 17:08
  • $\begingroup$ @qftishard - See below :) $\endgroup$ – nbubis Nov 21 '15 at 17:25
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Think I got it:

One uses the anti-commutation relationship: $$\gamma^\nu \epsilon_\nu \gamma^\mu k'_\mu = 2\epsilon \cdot p - \gamma^\mu k'_\mu \gamma^\nu \epsilon_\nu$$ And then uses the fact that the spinors satisfy the Dirac equation, i.e: $$\bar{u} \gamma^\mu k'_\mu \approx 0$$ So that we are only left with $2\epsilon \cdot k'$ in the numerator.

$\blacksquare$

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