# Why is the electroweak cross-section zero at Z boson mass?

1. Mar 29, 2015

### sk1105

The Standard Model formula for the cross-section of $e^+e^-\rightarrow\mu^+\mu^-$ is made up of 3 components: QED interactions, weak interactions, and electroweak interference. I understand the behaviour of the QED part, and the resonance that occurs in the weak part at $\sqrt{s}=M_Z$, but can anyone explain why the electroweak part goes to zero at this energy?

2. Mar 29, 2015

### Staff: Mentor

What do you mean with "electroweak part goes to zero"? If you look at the cross-section, it has a maximum at that energy value, and this maximum comes from the electroweak interaction.
Splitting it in three components seems to be a bit arbitrary.

3. Mar 29, 2015

### sk1105

I suppose it is a bit arbitrary, but it's just what we were taught. If I plot the differential cross-sections with respect to polar scattering angle, the electroweak component is zero for all angles. After integrating to get the full cross-section and plotting as a function of C.O.M. energy, the electroweak part goes from negative to positive (I understand this comes from forward-backward asymmetry) and passes through zero at $M_Z$. I've been searching for a reason for these two observations and nothing I've found so far has completely answered my question.

4. Mar 29, 2015

### Orodruin

Staff Emeritus
I believe what he calls "electroweak part" refers to the interference term between the diagram with the photon propagator and the diagram with the Z propagator.

Now this is my guess on what is going on, I did not really have time to check it in detail: The Z is an unstable particle and at the Z mass its propagator is regulated by the decay width of the Z, which adds an imaginary part to the denominator of the propagator. Far away from the resonance, this addition does not do much and the phase of the propagator would be essentially the same without it. However, at resonance, it rotates the phase of the propagator by 90 degrees, meaning it no longer interferes with the photon term.

5. Mar 30, 2015

### RGevo

The integrated cross section for the interference term is negative below resonance and positive above.

Nothing to do with an asymmetry. Simply that the z propagator is 1/(shat-mz^2) and this changes sign as you go through the propagator.

If you want to understand the form of the propagators on resonance, take 1/(shat-mz^2- i gam_z mz) and find the real part of this. For the z-z interference compute the real part of this propagator time it's complex conjugate.