# Infrared-free QED and Higgsless standard model phenomenology

+ 7 like - 0 dislike
122 views

This is one of those "what if" fantasy world type questions. I like hard sci-fi so please no "well, you changed one thing about the world so now anything goes." :)

What if the Higgs had no vev?

That is essentially this question and I was charmed by Ron Maimon's excellent answer to it, but I was under the impression from this statement of his:

The massless electron will lead the electromagnetic coupling (the unHiggsed U(1) left over below the QCD scale) to logarithmically go to zero at large distances, from the log-running of QED screening. So electromagnetism, although it will be the same subgroup of SU(2) and U(1) as in the Higgsed standard model, will be much weaker at macroscopic distances than it is in our universe.

...

The combination of a long-ranged attractive nuclear force and a log-running screened electromagnetic force might give you nuclear bound galaxies, held at fixed densities by the residual slowly screened electrostatic repulsion. These galaxies will be penetrated by a cloud of massless electrons and positrons constantly pair-producing from the vacuum.

and recently reading about the log running in Shifman, Advanced Topics in Quantum Field Theory, that massless QED would be very weak at macroscopic distances. Of course, log running is slow so this seemed a little odd as well, so I decided to calculate.

The (one loop) running coupling of QED with one massless charged fermion is

$$e^2(p) = \frac{e^2(\mu)}{1-\frac{e^2(\mu)}{6\pi^2}\ln\frac{p}{\mu}},$$

where $\mu$ is the arbitrary renormalization point and $p\sim 1/\ell$ is the scale of the probe. I took $\mu=1\ \mathrm{MeV}$ and set $e^2(\mu)=4\pi\alpha$ and got this:

Note the scale. So the log running really is slow. Even at the Hubble scale the effective charge is only reduced by about 6%! Including $N$ massless charged flavours multiplies the beta function (yeah?):

$$\frac{\mathrm{d}e}{\mathrm{d \ln\mu}} = N \frac{e^3}{12\pi^2},$$

so the coupling becomes

$$e^2(p) = \frac{e^2(\mu)}{1-\frac{N e^2(\mu)}{6\pi^2}\ln\frac{p}{\mu}}.$$

For the SM $N=3$ (assuming all the charged mesons are above 1 MeV - is this true?) this increases the running, but not much:

So EM is still sizable at cosmological distances if:

1. I didn't make any mistakes,

2. And the number of massless charged particles isn't too great.

This was somewhat surprising to me, though perhaps only because I misunderstood the assertion. It is certainly the case that atoms blow up (Bohr radius $a\sim1/\alpha m_e\to\infty$) so life is definitely not as we know it, but EM is still an important interaction at large scales - large enough to expect matter to form neutral conglomerates much smaller than galaxies. It is not at all clear to me that the dominant effect is an electrostatic repulsion that would stabilize any kind of nuclear bound galaxy. So:

What is the real phenomenology of the Higgs-less SM? Can anyone elaborate on Ron Maimon's vision?

This post imported from StackExchange Physics at 2014-03-07 14:33 (UCT), posted by SE-user Michael Brown
retagged Apr 19, 2014

+ 0 like - 0 dislike

It is small wonder there is no definitive answer to this question here. But I can pass along the best answer I got.

I asked the same question, in a slightly different format, to a blog "Of Particular Significance" that is popular with Physics Stack Exchange. My question was whether the famous Goldstone Mexican Hat potential had a value of 245 GeV at the rim (vacuum expectation value) of the hat?

The answer given there (by Matt himself) was in effect: "no". The energy associated with the rim of the hat is still the same as the energy Goldstone himself posited long before the Higgs was discovered. Its value is zero.

Astrophysicists really need this value in order to make sense of the 10^116 discrepancy in the estimation of the dark energy. The bottom line seems to be that science doesn't know whether the energy of the vacuum is the equivalent of a proton mass per cubic meter, or something much larger, as suggested by the Higgs formulation of the vacuum expectation value.

To Matt's answer I would only add that 245 GeV would make perfect sense, relatively speaking, to a free Higgs energy of 126 Gev. This is analogous to the idea that while a single photon, all by itself has indeterminate energy, the energy of two photons traveling in opposite directions will always have a defined energy that can actually be observed. Think Doppler shift. How would an observer know if a single photon was red or blue shifted (different energies), or even there at all, depending on inertial reference frame. The situation is much different with two photons, and I suspect, also with two Higgs bosons, and for the same reason.

This post imported from StackExchange Physics at 2014-03-07 14:33 (UCT), posted by SE-user Daniel Shawen
answered Mar 6, 2014 by (0 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsO$\varnothing$erflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.