When talking about 3-point amplitudes (e.g. three gluons) I`ve come across the fact, that if we have three lightlike 4-momenta $p_1,p_2$ and $p_3$ they should satisfy

$$p_1^{\mu} + p_2^{\mu} + p_3^{\mu} = 0,\tag{1}$$

which looks like one of the momenta has a negative energy.

This makes me wonder, because I thought, that

$$p_1^{\mu} + p_2^{\mu} = p_3^{\mu} \tag{2}$$

makes more sence. I can´t figure it out on my own, so help would be appreciated.


1 Answer 1


Yes, in e.g. eq. (1) by convention all 4-momentum vectors are directed as incoming and the zero-component is positive (negative) if the particle is incoming (outgoing), respectively.$^1$


  1. M. Srednicki, QFT, 2007; Chapter 16. A prepublication draft PDF file is available here.


$^1$ Or vice-versa, depending on the author's conventions.

  • $\begingroup$ Would that mean, that e.g. in case of the s-channel diagram of Bhabha-scattering, that the physical amplitude transforms from $M(e^{+}(p_1)e^{-}(p_2)\rightarrow e^{+}(p_3)e^{-}(p_4))$ into $M(e^{+}(p_1)e^{-}(p_2) \rightarrow e^{+}(-p_3)e^{-}(-p_4))$? $\endgroup$
    – maxxam
    May 5, 2021 at 21:51

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