FeaturedI Imagining a Higgsless Universe

1. Aug 15, 2016

nikkkom

I would like to imagine how would physics look in a Universe governed by Standard model, sans Higgs. IOW: how would the unbroken SU(2)*U(1) be different from our usual broken one?

Matt Strassler has a similar article here:

https://profmattstrassler.com/artic...known-particles-if-the-higgs-field-were-zero/

(It's a bit different - he retains Higgs field, but posits that it's VEV is zero.)

I'll describe how I understand things would be in such a case. Please correct me whereever I'm wrong.

There is no "usual" electromagnetism in such a Universe. Instead, there are two new forces, SU(2) weak isospin and U(1) weak hypercharge. (I'll be skipping word "weak" from now on).

All elementary particles are massless.
Left- and right- fermions are different particles.
The antiparticle of a left-fermion is a right-antifermion.

Unbroken SU(2) is in some ways analogous to the familiar color SU(3) in our Universe:
* there are three W massless bosons corresponding to generators of SU(2)
* left-fermions are affected by isospin force - they form isospin doublets (similar to how quarks form color triplets). Namely, (e-left,ve-left) is one such doublet. (u-left,d-left) is another.
* there are two "isospin charges" (similar to three colors of quarks). I'll call them "up-charge" and "down-charge". They are NOT similar to +- electric charges! e-left has -1 up-charge, but positron-right, the antiparticle of e-left, has -1 anti-up-charge, NOT +1 up-charge.
* right-fermions have isospin charge 0.

In essense, in this Universe leptons and W bosons would act as "two-color quarks and three-color gluons". Will this SU(2) interaction exhibit asymptotical freedom? Will there be confinement? Will leptons form two-particle "leptohadrons"?

Also, in this Universe, left quarks are also isospin-charged, and will feel the effect of isospin force. This is going to make hadrons much more complicated.

Right-fermions have zero isospin-charge and are not affected by isospin force. For instance, u-right and d-right quarks would probably still form "protons" and "neutrons", but they will have the same mass.

Unbroken U(1) works a lot like electromagnetism, but with a stronger coupling constant. (By how much stronger?) Force carrier is the B boson, similar to photons we know.

Most fermions are affected by this force. They have a hypercharge, similar to electric charge:
Left-leptons (yes, including left-neutrinos!) have -1 hypercharge.
Left-quarks are +1/3.
Right-up-leptons (electrons et al) are −2.
Right-up-quarks are +4/3.
Right-down-quarks are −2/3.
(If right-down-leptons (IOW, right-neutrinos) exist, they are the only particles with hypercharge 0).

Note that for left-fermions, hypercharge does not depend on up/down-ness. This is another manifestation of the fact that, say, (e-left,ve-left) is more like one particle, not two - similar to how we usually talk about "u-quark" as one particle, not three differently colored quarks.

u-right and d-right quarks would probably still form "protons" and "neutrons", but they will have the same mass. This mass will be lower than our usual protons and neutrons, since quarks are massless.

But u-left and d-left quarks should be unable to form a nucleon-like composite particle - while they can be color-neutral, three isospin-charged particles can't be isospin-neutral, and "isospin-confinement" prohibits this! So, either one of the quarks needs to be a right-quark; or this "left-nucleon" needs to pair with another "left-nucleon" to achieve isospin-neutral state.

Mesons, on the other hand, can be formed out of left (anti)quarks. There will be more varieties of both hadrons and mesons, because (u-left,anti-d-right) "pion" is different from (u-right,anti-d-left) one.

Binding energies will be different because left-quarks are also affected by isospin interaction, which should be attractive (about 1/2 strength of color force?).

Thus, it's likely during baryogenesis only the least-energetic hadron states will survive. What those will be? left+left+right?

left-leptons would probably form two-particle "leptohadrons", bound by isospin interaction. For example, (e-left,ve-left) pair is an analog of three-quark hadron. (e-left,anti-e-right) is an analog of pion0 (and it's also the usual positronium).

Hadrons and leptohadrons will have mass due to gluons and W bosons bouncing inside, just like hadrons get most of their mass from gluons in our Universe.

(u-left,d-left,d-right) hadron would have hypercharge +1/3 +1/3 -2/3 = 0
(u-left,d-left,u-right) hadron would have hypercharge +1/3 +1/3 +4/3 = +2
(e-left,ve-left) leptohadron would have hypercharge -1 -1 = -2

These two last composite particles can form a hypercharge-neutral "hydrogen atom".

It seems that 2nd and 3rd generation of leptons and quarks will be indistinguishable from the 1st - they are all massless, and have the same charges. IOW: generations would be unobservable, there will be just one generation?

Last edited: Aug 15, 2016
2. Aug 15, 2016

nikkkom

Probably someone is going to come and show me that my ideas about isospin charges in unbroken SU(2) are all wrong...
But if I'm not terribly wrong. I looked at it and it seems in my post I was wrong, in left-lepton SU(2) pairs neutrinos are "up" particles, and electrons are "downs".

If we assign isospin charges I so that ve-left gets +1up isospin charge, and we define I3 (the usual isospin in our Universe) as (up charge - down charge)/2, everything lines up:
Code (Text):
Y    I      I3=(Iu-Id)/2            Y    I I3  Q=I3+Y/2
ve-left  -1    +1up   +1/2          ve-right  0    0 0   0
e-left   -1    +1down -1/2          e-right  -2    0 0  -1
u-left   +1/3  +1up   +1/2          u-right  +4/3  0 0  +2/3
d-left   +1/3  +1down -1/2          d-right  -2/3  0 0  -1/3
Y - hypercharge
I - isospin (two-component charge of unbroken SU(2))
I3 - third projection of isospin as defined in the usual SM with broken SU(2)*U(1)
Q - electric charge (in unbroken Universe, it's not a very useful concept, included for cross-checking)

I3 and Q are correct for all particles.
The table includes right neutrino too, to show that its charges are also ok (all zero), if it exists.

Last edited: Aug 15, 2016
3. Aug 15, 2016

nikkkom

It's interesting that this "Higgsless SM" has only three (!) parameters: gauge coupling strengths g1, g2 and g3.

All fermion masses (or alternatively, Higgs Yukawa couplings) are gone since there is no Higgs.
Higgs VEV and Higgs self-coupling are gone since there is no Higgs.
CKM matrix is gone since generations are unobservable.

4. Aug 15, 2016

Staff Emeritus
Chris Quigg and I spent an hour on an airplane discussing this. He wrote a paper or two on this.

This is not so simple to answer, as any answer has a set of assumptions built in. Here's my view with my own assumptions. It's easiest to see if I turn the Higgs off one piece at a time.

Step 1: set the Higgs coupling to the W and Z to zero. The W and Z masses don't go to zero: they go to about 30 MeV because they still get a QCD mass from the quark condensate.

Step 2: Let's set the Higgs couplings to the quarks to zero. This should set all the 0- masses to a 36-fold degenerate zero, because they are all Goldstones. Other mesons will still be massive, as will the baryons, because their mass is governed by LambdaQCD and not the quark current masses. Surely that will set the W and Z masses to zero. And...now they weigh about 100 keV.

What went wrong?

You now have a hadron condensate, dominated by N-Nbar. This gives a tiny piece of the QCD mass, and it's not zeroed out when chiral symmetry is restored.

Step 3: Look at the fermion masses. Suddenly they are non-zero as well, but still very small (probably eV ballpark). This is because the (massive) W can turn a f_left into a f_right through a box diagram. That couples them together just as the Higgs mechanism does today.

Interestingly, there is now no physics to break the degeneracy of the mesons, so you find that you have a pi+, pi- and pi0, but the production cross-section e+ e- --> pi + pi- is 9x higher than it is today.

Step 4: Consider the implications for atoms. The strong force is much longer range than it is today - so instead of nuclei having tens of nucleons, they will have millions. Atoms will be huge - the Bohr radius will be macroscopic - and there will be so many electrons that each atom will act as a metal. You will have energy bands instead of energy levels and metallic bonds instead of ionic or covalent.

5. Aug 16, 2016

nikkkom

If quark condensate gives some mass to W bosons (because quarks have SU(2) charges), why doesn't it give any mass to photons? Quarks are electrically charged too, right?

6. Aug 16, 2016

vanhees71

The photon mass is protected by electromagnetic gauge symmetry, which is presumably unbroken in your hypothetical deformation of the Standard Model.

7. Aug 16, 2016

nikkkom

Then quark condensate should also give masses to gluons, right? But everyone says that they are massless.
I'm confused. Are gluons massless or not? And if massless, why QCD condensates don't give them masses?

8. Aug 16, 2016

Staff Emeritus
I think this only happens if you have a condensate that carries net color. You don't get one, and I think ultimately it goes back to gauge invariance.

9. Aug 16, 2016

nikkkom

Are you saying that quark condensate is color-neutral (and therefore doesn't make gluons massive), but not weak isospin-neutral (and therefore does give mass to Ws)?
If so, why does it have weak isospin charge?

10. Aug 16, 2016

vanhees71

In the vacuum of interacting QCD the "quark condensate" means that $\langle \overline{q} q \rangle \neq 0$. This leads to the spontaneous breaking of global (approximate) chiral symmetry in the light-quark sector but has no influence on the gluon mass.

11. Aug 16, 2016

nikkkom

Quarks have both SU(3) charges and SU(2) charges.
QCD vacuum at low temperatures contains quark condensate.
Quark condensate makes SU(2) bosons massive.

Shouldn't situation be symmetric?
Unbroken QFD (quantum flavordynamic, iow: SU(2)) vacuum should also have quark condensate.
Therefore, quark condensate should make SU(3) bosons massive too?

Where analogy fails?

12. Aug 16, 2016

vanhees71

The quark condensate is meant as the sum over all flavors, i.e.,
$$\overline{q} q = \overline{u} u + \overline{d} d + \overline{s} s+\cdots,$$
which is a WISO singlet.

13. Aug 16, 2016

MathematicalPhysicist

What is WISO?

14. Aug 16, 2016

vanhees71

WISO = "weak isospin"

15. Aug 16, 2016

ChrisVer

is the SU(2) that quark condensates break the SU(2) of the SM?

16. Aug 16, 2016

vanhees71

In principle, it contributes to the "Higgsing" of the electroweak gauge symmetry, but the contribution to the W- and Z-boson masses is way too small (at a scale of a few MeV) to explain their larger masses of around 80-90 GeV. There are models, called technicolor models, that assume the Higgs to be composite of socalled techni-quarks. So far there's no hint of their existence, and as far as I know most TC models are ruled out. For a review, see

https://arxiv.org/abs/hep-ph/0203079

17. Aug 17, 2016

nikkkom

Can you explain this a bit further? W does not interact with right-fermions, it shouldn't be able to produce them.

18. Aug 17, 2016

Haelfix

I love these scenarios! One interesting possible problem with the 2nd step is that you have an inconsistent limit. In the extreme case where you turn off bare fermion masses you now have a theory with unconfined, massless charged particles. So via a Schwinger effect, something like positron electron pair production will destabilize the vacuum.

Even accounting for dynamical masses, one would have to carefully run a stability analysis of the resulting theory in different thermodynamic regimes.

19. Aug 17, 2016

Staff Emeritus
That's true for transverse W's. Longitudinal W's are different. This is the same reason why W's produced in top decays are not transversely polarized. Essentially, things work backwards from the SM: in the SM the Higgs mechanism causes the W's longitudinal component, here the W's longitudinal component gives you a Higgs-like mechanism. Of course this is just words - what really matters is the mathematics and the equations are the same. They don't know about "cause" and "effect".

20. Aug 26, 2016

nikkkom

What does it tell us that "Higgsless SM" has only 3 parameters?

IOW: adding Higgs field brought in 16 (!!) additional parameters to the SM (23 parameters if neutrinos are massive).

There is something we miss about Higgs, there may be a way to add it with fewer parameters. Since most of them are masses ("couplings to Higgs field"), "adding Higgs with fewer parameters" means "explaining fermion rest masses systematically".

21. Aug 26, 2016

Staff: Mentor

It tells us that (a) nearly all nonzero particle masses are free parameters and (b) some parameters (like the CKM matrix) are meaningless if particles are massless.

There are some curious approximate (within uncertainties) relationships between the masses, like the Koide formula, but no deeper physical motivation for them.

22. Aug 26, 2016

houlahound

My spare time for the next 6 months, decoding this insight article.

Honestly thanks for the article, nice idea just pretty full on for the average enthusiast.

23. Aug 27, 2016

nikkkom

It can well be "we don't yet have a theory for these relationships". Just like mesons were showing curious relationships, and quark model was eventually developed to explain them.

24. Sep 6, 2016

john baez

The questions you raise are a huge amount of fun and a great way to learn physics. I don't know the answer to your question about confinement, but you can answer the question about asymptotic freedom by going here. This gives an approximate formula for the beta function. If we believe this approximation is okay, and most people do, we see that we get asymptotic freedom in SU(N) Yang-Mills theory with m different fermions transforming in the fundamental representation if and only if

11 N / 3 > 2 m / 3

I've resisted simplifying the fractions so you can more easily see how I'm getting this inequality. Please check my work!

In your scenario, N = 2 and m is the number of left-handed fermions, since those transform in the fundamental representation of SU(2). Right-handed fermions transform trivially under SU(2) so they don't show up in this calculation. I'm getting

22 < 24

so we don't get asymptotic freedom (and perhaps not confinement either).

25. Sep 6, 2016