Why can we approximate massive particles as massless or vice versa?

Our descriptions of massless and massive particles are very different. For example:

• Massless particles have only two polarizations, which we call helicities. Spin projection on axes different than that of momentum is not defined.
• Yang-Mills fields come equipped with a gauge symmetry group, which would be ruined if we added mass.

However, when considering processes at energy scales much larger than the rest masses of particles involved, we often treat particles as if they were massless, with good results. How is this possible if the massive and massless cases are so qualitatively different? Is there really a continuous transition between $m=0$ and $m= \epsilon$? If so, how is it reconciled with the facts above?

For example, consider the process $\pi^- \to e^- + \bar{\nu}_e$. Since the pion mass is much greater than the electron's or neutrino's, we can assume them to be massless. Due to properties of weak interaction, the electron will be left-handed and the neutrino will be right-handed. Since they are massless, chirality is the same as helicity. So, in the center of mass frame, where their momenta are antiparallel, their spins will be parallel. Therefore the total angular momentum is $1$, which is impossible because $\pi^-$ is a pseudoscalar. On the other hand, the $\mu ^-$ is much more massive (mass comparable to that of pion), so this argument doesn't work. Decay to muon is allowed and experimentally verified to be much more probable than decay to electron.

• " we often neglect those masses and treat particles as if they were masless, with good results" can you give an example? However, as long as observable quantities depend on $m$ as some analytical functions $f(m)$, you can always expand in power series and neglect higher order contributions in case $m$ is small. This said, most properties aren't so and peculiar differences do arise. – gented Jul 26 '16 at 20:38
• Gennaro Tedesco I added an example in the post. – Blazej Jul 26 '16 at 20:54
• In the example at hand the observables will depend on the ratio between the right hand sides masses and the left hand side one. As such, expanding in power series the higher order contributions vanish. – gented Jul 26 '16 at 21:24
• Do you know processess where such continuity is explicitly violated? Note that I am speaking here of directly measurable effects and not changes in mathemathical description (which I already listed). – Blazej Jul 26 '16 at 21:40
• If you mean "decay processes" then the sum of the masses (energies) in the LHS must always equal the sum of the masses (energies) on the RHS; therefore in any case additional sums and contributions appear and there's no violation (the total helicity, angular momentum etc will adjust accordingly and expand in power series accordingly). – gented Jul 26 '16 at 21:53

At the level of representation theory, massive and massless particles are fundamentally different. As you mentioned, massless particles have 2 d.o.f. (helicities $\pm h$) and massive particles have $2j+1$ d.o.f. This fundamental difference is quite important and leads to many structures - such as gauge invariance. It is also not possible for this reason to define an $S$-matrix (in the same way that we do now) for massless particles.

However, despite this, there is a reason that in the high energy limit, massive particles can be taken to be approximately massless. Take the example of a spin-1 particle with momentum $(E,0,0,p)$. This has 3 polarizations $$\epsilon^{\pm,\mu} = (0,1,\pm i,0)~, \qquad \epsilon^{L,\mu} = (p,0,0,E)~.$$ The spin summed amplitude squared is $$\big| \epsilon^{+,\mu} {\cal M}_\mu + \epsilon^{-,\mu} {\cal M}_\mu + \epsilon^{L,\mu} {\cal M}_\mu \big|^2$$ Now, in the high energy limit $p \to E$ in which case $\epsilon^{L,\mu} \to p^\mu$. Then, since $p^\mu {\cal M}_\mu = 0$ (coming from the fact that $\partial^\mu A_\mu = 0$ for massive fields), the last term simply drops out. This, in the high energy limit all contributions from the longitudinal polarization drops out and only the two photon polarizations contribute to the amplitude. Finally, the amplitude itself ${\cal M}_\mu$ reduces to the massless one since the massive propagator $\sim \frac{g_{\mu\nu} }{ p^2 + m^2 } \to \frac{g_{\mu\nu} }{p^2}$ reduces to the massless propagator in the high energy limit.

Thus, purely from the standpoint of amplitudes, at high energies all massive particles are effectively massless. Fundamentally though, the particles still have longitudinal polarizations. However, at high energies the particles are highly boosted and none of the longitudinal polarizations interact - only the transverse ones.

• I disagree with your statement about $p_\mu {\cal M}^\mu=0$ for massive gauge theories (see my answer). – JeffDror Jul 28 '16 at 15:48

IF the number of degrees of freedom in a theory are the same in the massive and massless limit then you can approximate particles as massless at high energies.

This is because position space $n$-point functions in quantum field theory do not have IR divergences (i.e., they don't blow up as you probe larger and larger distances). This means that physical quantities are not going to depend on long distance physics but only at physics at around the energy scale you are probing. This is a familiar property from other aspects of physics and transfers over the QFT. For example if you measure the spectra of hydrogen in a gas the details of the gas don't effect the spectra you find (to a good approximation).

For spin-$0$ and spin-$1/2$ particles the number of degrees of freedom are indeed the same in the massive and massless cases. Thus you can approximate the massive correlations functions with the massless ones at high energies. However, for spin-$1$ particles this is no longer true. This is because massless spin-$1$ particles have only $2$ degrees of freedom and must be protected by gauge symmetry. Massive spin-$1$ particles have $3$ degrees of freedom. In order to give a gauge boson a mass in a unitarity-preserving way it must arise from a spontaneously broken gauge symmetry. In such theories the longitudinal modes contributions not only don't go to zero, but actualy GROW with energy the fastest (see Goldstone boson equivalence theorem)! Thus they come to dominate the contribution amplitudes at high energies. Clearly in such theories it's not possible to use the massless limit to approximate the massive contributions at high energies.

Note: This last paragraph is in contradiction with the previous answer which claims that the longitudinal mode contributions vanish at high energies due to $\epsilon _{L, \mu} {\cal M} ^\mu \sim \frac{ p _\mu }{ m } {\cal M} ^{ \mu } = 0$. However, this last equality does not hold for spontaneously broken gauge symmetries since their charge is no longer conserved.