# Energy of a Gluon

1. Oct 4, 2009

### daviddanut

I'm currently doing a physics project on fusion for my A2, and need to show how to work out the range of the strong nuclear force.

So far I've got this

delta E*delta t > h/(4*Pi) (h = plancks constant, E = exchange particle energy and t = time.)

Multiply time by c to get the range.

However I've got no idea what the energy of a gluon is, and have been fruitlessly scouring wikipedia and google.

Anyone know?

2. Oct 4, 2009

### noblegas

Well, according to wikipedia, the mass of a gluon is 20 MeV/c^2. You can now probably figure out the energy of a gluon

3. Oct 4, 2009

### daviddanut

Damn, missed that bit, whoops...

Thanks, much appreciated.

4. Oct 4, 2009

### xepma

That mass is an experimental upper bound. If the gauge symmetry associated with the gluon is correct the mass of the gluon is zero.

Furthermore, the gluon doesn't carry a fixed energy. This is just like how photons behave. Rather, since they are massless particles, they obey the so-called on-shell condition which reads

$$E^2-p^2c^2 =0$$

This relation holds for all massless particles (gluons, photons..)

But the problem is that you cannot really say anything about the range of the strong force. One reason being is that this formula stems from special relativity, while the force itself is based on quantum theory.

http://en.wikipedia.org/wiki/Strong_interaction
http://en.wikipedia.org/wiki/Nuclear_force

It's probably the range of the nuclear force which you are after, since this (effective) force plays a major role in fusion processes.

5. Oct 4, 2009

### daviddanut

On the equation, the particles are massless, so surely p (Momentum I assume) must equal 0? Therefore E=c?

Force wise, I'm looking for the force that overcomes electromagnetism and pulls two nuclei together when they get to a certain range. I thought that this was the strong nuclear force.

I've never heard of the nuclear force before, I assume this is something they tell you at degree level? Unfortunately I don't yet understand most of the technical language on wikipedia. Is the nuclear force a side effect of the strong nuclear force or are there more than 4 fundamental forces?

6. Oct 4, 2009

### RedX

If gluons are massless, what is the mass gap talked about in the million dollar math problems, the one about the Yang-Mills Mass Gap?

As for the nuclear force versus the strong force, the nuclear force arises from an effective field theory built from the theory of quarks and gluons. In this scheme, instead of quarks and gluons, once has baryons and mesons. So the particle mass you need to look up is the mass of the pion which if I recall is around 0.1 GeV. This is the exchanged particle, called a pseudo-goldstone boson because the $$SU(2)_L \times SU(2)_R$$ chiral global symmetry is not exact. For more information on constructing the chiral Lagrangian try Wikipedia:

http://en.wikipedia.org/wiki/Chiral_perturbation_theory

http://en.wikipedia.org/wiki/Yukawa_potential

7. Oct 4, 2009

### clem

The "strong nuclear force" relevant to fusion is the force between nucleons and not the force between quarks. The range of the nuclear force is dominated by the pion of mass 140 MeV.

8. Oct 6, 2009

### humanino

The only issue is about what you call "mass". The current mass is the one appearing in the lagrangian and it is strictly zero for gluons, otherwise we are simply not talking about gluons (by definition). The so-called constituent mass is an effective mass, dressed after renormalisation, it is non-zero and can be calculated by various methods. This is shown in the flattening of the propagator in this lattice calculation for instance

blog page
The Infrared Behavior of QCD Green's Functions

Precisely because gluons are (nakedely) massless but might create (very) massive (pure) glueballs, and because this is relevant to more than 95% of our own body mass, it seems intuitive that the Clay prize is actually interesting.

9. Oct 6, 2009

### RedX

The mass that enters the Lagrangian is not necessarily the physical mass, right? If you are renormalizing at an energy that's far from the physical mass, then your bare mass will be a function of the cutoff.

So are you saying that the bare mass of the gluon is zero, but the physical mass is not?

From the graph it looks like the mass is around 4 GeV.

How can one say that gluons account for 95% of our body mass? Take the example of a pion. A pion is a condensate of a composite quark/antiquark field, and when you give mass to the quarks, the pion acquires a mass. If quarks have zero mass, pions have zero mass (the chiral symmetry would be exact). A proton is made of 3 quarks, but the formula for the proton mass is much more indirect than the formula for the pion mass: in order to calculate the proton mass, you have to calculate the amplitude for nucleon-nucleon scattering, and this can be theoretically related to the mass of the nucleon. Gluons do not seem to be appear anywhere however when it comes to hadrons - the baryons are made of quarks, and so are the exchanged particle, so how do gluons come into play?

10. Oct 6, 2009

### Parlyne

Gauge symmetry forbids the renormalization of the gluon (or, for that matter, photon) mass away from 0. What the plot is showing has to do with binding energy in physical states.

11. Oct 7, 2009

### RedX

You're right - I wasn't thinking clearly. So a term like $$m^2 A^\mu A_\mu$$ is not gauge symmetric.

SU(3) color and U(1) electromagnetism are unbroken also, so no mass is generated spontaneously.

The only time I've ever heard of a "dynamically generated mass" is when quantum corrections cause terms that generate symmetry breaking to appear in the effective action (without the quantum corrections from loop diagrams, there would not be symmetry breaking terms).

So is this true that a "dynamically generated mass" means a mass that is due to loop corrections acting to cause the system to spontaneously break a symmetry?

Zee gives as an example:$$V(\phi)=\frac{1}{2}m^2 \phi^2+\frac{\lambda}{4!} \phi^4$$. When m=0, the system still nevertheless exhibits symmetry breaking because of quantum corrections. So mass would still be generated since symmetry breaking occurs at the 1-loop level.

12. Oct 7, 2009

### humanino

What renormalization provides us with is the ability to make sense of effective field theories. It is true that at high energies we always keep exact gauge symmetry and zero mass gluon. But it is not the case for effective constituent gluons in various non-perturbative approaches. In fact, a dynamical gluon mass is not necessarily inconsistent with gauge symmetry in bound states, it is a matter of enforcing the massive Slavnov-Taylor identities. The high-energy perturbative vacuum is relevant to deep inside the hadrons. The vacuum, and especially the border between them where confinement and chiral symmetry breaking happens, may be different. It may be a dual color superconductor (abelian Higgs model). Already in electromagnetic superconductors the gauge symmetry is broken, true photon do not propagate, and effective photons (vectors of the interaction) become massive.

The dynamical generation of mass for the gluon is something investigated for years :

Dynamical mass generation in continuum quantum chromodynamics, Phys. Rev. D 26, 1453 - 1478 (1982)

13. Oct 7, 2009

### Parlyne

Since this is talking about gluons in bound states, I would assume that the interactions which are taking place can give gluons an effect mass in a manner similar to how photons can acquire an effective mass when passing through a medium. I was only responding to the idea that the gluon mass here could be purely the effect of the renormalization of QCD.

14. Oct 7, 2009

### blechman

gluons are massless. at least to the extent to which we believe that QCD is the correct description of the strong nuclear force (which the majority of us do).

HOWEVER, gluons are NOT asymptotic states, and therefore you cannot talk about them as "particles" in the usual sense. Every time we talk about "gluon scattering" we are pulling a fast one! As long as we are looking at "short-distance" effects, this is a valid approximation, but it is wrong to think of gluons as "physical particles" in the sense that there are no such thing as a free gluon. Contrast this to QED, where we can talk about free photons of arbitrarily low energy (radio waves).

The physical particles (and therefore the quantum states that we are allowed to use when writing down S-matrix elements, etc) are the hadrons (pions, protons, etc). In particular, even if we get rid of ALL the quarks, then the gluons would form hadrons (called "glueballs") and THESE would be the physical "asymptotic states" - that is, the particles that are free at infinite distance. Since these are bound states, they are expected to have mass.

For \$1M: prove it! Analytically (lattice numerics are not enough).

Hope that helps.