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The non-relativistic Breit-Wigner peak is given by: $$\sigma_{fi}=\frac{\pi\hbar^2}{q^2} \frac{2j+1}{(2S_1+1)(2S_2+1)} \frac{\Gamma_i \Gamma_f}{(E-Mc^2)^2+\Gamma^2/4}$$ where $q$ is the center of mass momentum of one of the particles and $E$ is the center of mass energy. From what I am aware $q$ is a function of $E$ (changing the center of mass energy will change the center of mass momentum) typically $q\propto E$ so we would get: $$\sigma_{fi}\propto \frac{1}{E^2} \frac{1}{(E-Mc^2)^2+\Gamma^2/4}$$ however in every source I can find (e.g. Martin, 2012 pg 27) it is stated that $$\sigma_{fi}\propto \frac{1}{(E-Mc^2)^2+\Gamma^2/4}$$ and is usually compared to been the same as a Lorentzian. My question is therefore what happens to the $E$ dependence of $q$ and why is it not considered?

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You are right that the relativistic Breit-Wigner distribution isn't functionally quite the same as the Lorenzian distribution. However, the approximation that you described and which compares the relativistic distribution to the non-relativistic, Lorentzian one may be extremely accurate.

Well, it is extremely accurate whenever $Mc^2\gg \Gamma$. Note that the distribution is almost entirely concentrated in the interval $$E=Mc^2 \pm {\rm few}\cdot \Gamma$$ because of the denominator that you kept in your last formula. But when $E$ belongs to this interval, $q$ may be well approximated by substituting $E=Mc^2$.

In other words, the relativistic Breit-Wigner distribution only "changes abruptly" in the same interval where most of the distribution is concentrated and when one accounts for all the abruptly changing factors in this region, the distribution becomes indistinguishable from the non-relativistic, Lorentzian one.

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@LubosMotl answered the question already, I just wanna chip in some physical intuition for why $\Gamma$ is usually small compared to $Mc^2$. The Breit-Wigner-Formula applies to resonances, in fact only isolated resonances. In the case of overlapping peaks the more general Fano or shape resonances become relevant.

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