The leading order Weinberg soft factor is found to be $$\displaystyle\sum_{a=1}^{n}\frac{\epsilon^{\mu\nu}p_{\mu}p_{\nu}}{q.k_a},$$ where $a$ labels the external particles and $p$ denotes the external particle 4-momenta, $q$ is the soft momenta. I want to express this in spinor helicity formalism. I choose $x$ and $y$ as 2 arbitrary spinors and express the polarisation $\epsilon^{\mu\nu}$ as $\epsilon^{\mu}\epsilon^{\nu}$. For positive helicity $$\epsilon^{\mu}_{+}=\frac{\langle y|\gamma^{\mu}|x]}{\langle yx\rangle}.$$ Therefore $$\epsilon^{\nu}_{+}=\frac{\langle y|\gamma^{\nu}|x]}{\langle yx\rangle}.$$ Denominator of the soft factor becomes $[qp]\langle qp \rangle$.Consider the factor in the denominator $$\epsilon^{\mu}_{+}p_{\mu}=\frac{\langle y|\gamma^{\mu}|x]}{\langle yx\rangle}p_{\mu}=\frac{\langle yp\rangle [px]}{\langle yx\rangle}.$$ Substituting this in the soft factor I am finally getting $$\frac{\langle yp\rangle[px]\langle yp\rangle[px]}{\langle yx\rangle^2[qp]\langle qp\rangle}.$$ But the factor which I found in the paper was $$\frac{[qp]\langle xp\rangle\langle yp\rangle}{\langle qp\rangle\langle xq\rangle\langle yq\rangle}.$$ Are these both equivalent or am I wrong in the computation anywhere?

  • $\begingroup$ $\uparrow$ Which paper? $\endgroup$ – Qmechanic Jun 13 '17 at 4:25
  • $\begingroup$ arxiv.org/abs/1404.4091 this is link to the paper. Thanks for the help. I got the answer though. We should set one of the reference spinors as that of the soft graviton,ie set $x=q$ and you will get the answer. $\endgroup$ – Anupam Ah Jun 13 '17 at 14:00
  • $\begingroup$ If you have figured it out, you are encouraged to write up an answer. $\endgroup$ – Qmechanic Jun 14 '17 at 18:44

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