Isotropic neutrino-lepton scattering I'm a physics student and I'm attending an introductory course of particle physics. My professor stated that, in center of mass frame, the $\nu_\mu e^- \to \nu_\mu e^-$ elastic scattering has an isotropic angular distribution, while the $\bar{\nu}_\mu e^- \to \bar{\nu}_\mu e^-$ scattering has not.
I can't figure why this should be true. Any help would be appreciated.
 A: The link posted by akhmetelli suggests it's because of parity nonconservation: in the center-of-mass frame for the scattering, at high energy, only left-handed particles and right-handed antiparticles participate in the weak interaction.
The two particles may always scatter through an angle of 0º; the technical term for that is "they missed." So the interesting case is when they scatter through 180º. Using $\longrightarrow$ to show momentum direction and $\Rightarrow$ to show show spin direction, 
the matter-matter initial state is
$$
\underset{\nu} {\overset{\Leftarrow}{ \longrightarrow } } 
\quad
\underset{e^-}{\overset{\Rightarrow}{\longleftarrow}}
$$
and the final state for backscattering would be
$$
\underset{\nu} {\overset{\Rightarrow}{ \longleftarrow } } 
\quad
\underset{e^-}{\overset{\Leftarrow}{\longrightarrow}}
$$
Both of the matter particles must change both momentum and spin. On the other hand, for matter-antimatter backscattering, we go from
$$
\underset{\bar\nu} {\overset{\Rightarrow}{ \longrightarrow } } 
\quad
\underset{e^-}{\overset{\Rightarrow}{\longleftarrow}}
$$
to
$$
\underset{\bar\nu} {\overset{\Rightarrow}{ \longleftarrow } } 
\quad
\underset{e^-}{\overset{\Rightarrow}{\longrightarrow}}
$$
and we can see that an exchange of angular momentum is not required. This should be enough of a handwavy argument to see why 180º scattering is more likely in matter-antimatter weak interactions than in matter-matter interactions. If you really want to show that the $\bar\nu e^-$ amplitude is isotropic, you'll have to be more careful than this.
A: As weird as it may sound, I think that the statement is probably correct (I am not sure since I haven't check it directly, see below).  The kinematical dependence may be different in the two cases. For example,  imagine that the scattering amplitude for the first process depends only on the Mandelstam variable $s$, then the scattering amplitude for the second process is related to the first by crossing symmetry which, simplifying a little bit, exchanges $s$ and $u$ variables.  Now, $u$ depends on the scattering angle, whereas $s$ does not. You can thus see how the different behavior in the scattering angle may arise. I must say, however, that I haven't calculated directly either one of the two scattering amplitudes, so while I find the statement highly plausible, I can't be completely certain about its specific validity. If for example instead the amplitude for the first process was depending only on $t$, than there should be no difference in the two processes. 
