4
$\begingroup$

In the electroweak theory there is a mixing between forces which results in the photon being part weak and mostly electromagnetic and the Z0 being mostly weak and partly electromagnetic. Why doesn't the neutral gluon and photon (and for that matter the Z0) have a mixing angle? If the forces are all united in a GUT, how does the GUT not combine the stong force with the other forces with a mixing angle for the neutral particles?

$\endgroup$
1
  • 2
    $\begingroup$ "the photon being part weak and mostly electromagnetic" certainly sounds bizarre. Are you sure you are not misnaming the hypercharge U(1) as "electromagnetic? (If you were discussing currents, you could call the neutral current "part weak isospin and part electromagnetic", using a practical elimination of the hypercharge, but the phonon has never been anything but electromagnetic, for a century.) $\endgroup$ Sep 23, 2017 at 19:44

4 Answers 4

8
$\begingroup$

In the standard model the Higgs field only carries hypercharge and isospin, but no color. As a result the gluon does not mix with the photon or $Z$.

Remember that the original gauge fields, $U(1)$ hypercharge $B_\mu$ and $SU(2)$ isospin $W_\mu^a$, both couple to the Higgs. The photon $A_\mu$ and the $Z_\mu$ emerge as diagonal eigenstates.

The requirement on a GUT is simply to reproduce the hypercharge and color assignments in the SM.

If there was a colored Higgs, or a composite scalar field that carries color, then mixing would occur. It has been suggested that this does happen in QCD at very high baryon density. Quark form Cooper pairs, and the pair condensate is a composite colored Higgs field, see Goldstone bosons, quark and glon masses counting in color-flavor locking QCD

$\endgroup$
5
3
$\begingroup$

"...Why doesn't the neutral gluon and photon (and for that matter the Z0) have a mixing angle?.."

The gluon belongs to the representation of the QCD color group $SU_{c}(3)$ while the photon belongs to the representation of the electroweak $SU_{W}(2)\times U_{Y}(1)$. Moreover, the Higgs doublet it uncharged under the QCD color group and there is no possibility to obtain the mixing term through the Higgs mechanism.

"...If the forces are all united in a GUT, how does the GUT not combine the stong force with the other forces with a mixing angle for the neutral particles?.."

The answer is very simple. Absense of the mixing between the photon and the gluon is the fact which must be incorporated by any of the realistic GUTs.

$\endgroup$
2
$\begingroup$

That has to do with gauge invariance, and the fact that the electroweak and strong groups are distinct. Let's recall the two types of interactions that occur in gauge theories. The most important one is between matter and gauge bosons: say if $\psi$ is a fermion of charge $e$ with respect to a gauge group that has gauge boson $A_\mu$, then there is an interaction $$e \times A_\mu \bar{\psi} \gamma^\mu \psi$$ (suppressing indices etc.). Clearly, this type of interaction doesn't mix two different gauge bosons. Second, there can be self-interactions, in the case of non-abelian gauge groups (cubic and quartic interactions between gluons in Yang-Mills). Again, different gauge bosons don't mix there.

$\endgroup$
1
$\begingroup$

The photon is a superposition of B and W3. Both belong to the electroweak symmetry group and for this reason can mix. In contrast gluons belong to the SU(3) symmetry group that is defined as independent from SU(2). Therefore photons cannot interact or "mix" with gluons directly. Higher order processes may be possible, but have not been observed due to their extremely low probability and high energies required.

$\endgroup$
1
  • 1
    $\begingroup$ @CosmasZachos: Yes, you are correct, of course. I have edited the answer to say "electroweak" instead of SU(2). $\endgroup$
    – safesphere
    Sep 23, 2017 at 19:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.