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W and Z masses

by ChrisVer
Tags: masses
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ChrisVer
#1
Jun21-14, 10:45 AM
P: 899
Because I am a little tired to think...
For where does the mass difference between the W and Z bosons come from?
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Bill_K
#2
Jun21-14, 10:59 AM
Sci Advisor
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MW = v g
MZ = v (g2 + g'2)1/2

so the difference basically comes from the weak mixing angle,

MW/MZ = cos θW
ChrisVer
#3
Jun21-14, 11:08 AM
P: 899
would that mean that it's because the Z -as chargeless- gets contributions from more interactions? because it gets the g' of U1

ofirg
#4
Jun21-14, 06:12 PM
P: 93
W and Z masses

would that mean that it's because the Z -as chargeless- gets contributions from more interactions? because it gets the g' of U1
Unlike the charged components of the [itex]W[/itex] [itex]SU(2)[/itex] Triplet, The Neutral component mixes with the Hypercharge gauge boson ( [itex]B[/itex] ) to give two mass eigenstates, the [itex]Z[/itex] and the photon. This Hypercharge gauge boson component is the origin of the contribution to the [itex]Z[/itex] mass which is proportional to [itex]g^{'}[/itex]
nikkkom
#5
Jun24-14, 05:16 AM
P: 611
While we are on this topic, I have a question.

Layman explanation of W and Z masses are that they "ate" Goldstone bosons produced by electroweak symmetry breaking.
What is meant by "eating" here? Thats not a scientific term for sure :D
ofirg
#6
Jun24-14, 09:54 AM
P: 93
Layman explanation of W and Z masses are that they "ate" Goldstone bosons produced by electroweak symmetry breaking
A full understanding of this would require a thorough study of spontaneously broken gauge thoeries.

However, in short:

In General when a theory has a spontaneously broken continues symmetry (Are you familiar with the concept?) it has massless particle which is called a goldstone boson ( see http://en.wikipedia.org/wiki/Goldstone_boson )

Masses for the W and Z require the gauge symmetry of the standard model to be spontaneously broken. This breaking would naively introduce goldstone bosons to theory. However, It turns out the theory can be written without any reference to these goldstone bosons, i.e. , they are not physical. This happens because the symmetry is a gauge symmetry and not a global symmetry.

On the other handed When massless spin 1 particle becomes massive it has in addition a longitudinal polarization ( Where the spin's projection on its momentum is zero). Therefore there are 3 additional degrees of freedom after the symmetry is broken.

The phrase "The W and Z ate the goldstone bosons" refers to the 3 unphysical goldstone boson degrees of freedom appearing as the W and Z longitudinal polarization.
ChrisVer
#7
Jun24-14, 12:22 PM
P: 899
In other words, you have the Higgs field belonging to (2,1) representation of SU(2)xU(1) (is a doublet under SU(2) and singlet under U(1) ). That means that you have 2 scalar complex fields (upper lower component in SU(2) doublet) which give you four degrees of freedom. The physical degrees of freedom though are 1, because the other 3 can be gauged out by choosing gauge.
The procedure of SSB leads to 3 massive/1 massless spin-1 particles from 4 massless... the extra degree of freedoms (3) came from the gauged out fields of the Higgs doublet - the Goldstone bosons were those fields...They were "eaten"/disappeared by 3 gauge bosons and the 3 became massive (or acquired longitudial dof).

Wouldn't it be correct to see the Goldstone boson as a ghost field?
I mean the initial gauge bosons, massless, would have to be represented by a state:
[itex]|Ψ>= |T> + |S-L> [/itex]
(T: transverse, L: longitudia, S:scalar , and the + is not the "normal" sum sign).
The norm of the second space is 0, while the other's is positive definite.
In that case I could say that the Goldstone bosons should belong to some scalar
[itex] |Φ> = |S> [/itex]
in order to kill out the (negatively normed S states and leave L untouched).
I am not so keen in Ghosts that's why I am making this question. To see how well I understand it :)


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