# Symmetry and range of interaction

1. Jul 20, 2009

### Lepton_vn

I have just read a book. It said that: "if symmetry is exact, interaction is a long range". Could you explain me more detailed? Thanks!

2. Jul 20, 2009

### genneth

You're going to need to provide a longer context.

3. Jul 20, 2009

### Phiphy

Does it refer to gauge symmetry?
If gauge symmetry is exact, the gauge bosons are massless so interaction is long range; if gauge symmetry is broken, the gauge bosons get mass and interaction becomes short range.

4. Jul 21, 2009

### blechman

that's probably what the book means, although it's wrong! QCD has an "exact" gauge symmetry and yet the force is short-ranged due to confinement. And while we're at it, the phrase "exact symmetries" includes "spontaneously broken" symmetries!

I think your book lied to you!

5. Jul 21, 2009

### Parlyne

I think you need to be little more careful here. It's certainly true that the force between nucleons is short-ranged. However, nucleons have 0 charge under the strong force. Using nucleons to judge the range of a force is like saying the electromagnetic force is short-ranged because atoms have to be close to each other to feel such forces. What matters is the range of the force which acts on the quarks making up nucleons; and, this is a much subtler question.

It is, however, possible to create a quark-gluon plasma (if you're at high enough temperature and pressure), in which quarks no longer remain bound. Since there's no confinement here, it should be much clearer that the strong force is, in fact, long-ranged.

6. Jul 21, 2009

### humanino

Please note that the careful assessment released by experimental groups does not claim discovery of a QGP. This seems to contradict at least "naive" expectations dating back to the design of the machines. It therefore not so obvious that your claim holds even in principle.

7. Jul 21, 2009

### genneth

Besides, it's important to maintain distinction of phases. Gluons, photons, etc. are massless in the absence of any other matter, including any potential condensates.

However, I think Phiphy managed to guess the OP's intent --- good work sir :-) For a first level understanding of things, I think it's fairly to say that exact gauge symmetry => massless bosons. It's a useful rule of thumb until you have to face reality

8. Jul 21, 2009

### Phiphy

Yep, I was just trying to guess the meaning of the sentence. And I guess the OP's book may well be a popular science book. I agree that it is too sloppy to make a statement like this using the not well defined terms like "long range".

9. Jul 22, 2009

### Lepton_vn

That's right. But i don't understand why if gauge bosons are massless then interaction is long range and vice versa ? Could you explain more clearly? If it is possible, can you prove by using mathematics?

10. Jul 22, 2009

### xepma

The introduction of a mass causes a dropoff of the effective range of the mediated particle. Effectivel, the range drops as ~$$e^{-x m}$$ (give or take a few parameters). For instance, a massive photon would give rise to a modified Coulomb potential that looks like: $$\frac{1}{r}e^{-x m}$$. For m=0, the exponential dropoff vanishes and we have long-ranged behavior again.

11. Jul 30, 2009

### Abbas Sherif

I agree with you in full. I think exact symmetry means when you cannot distinguish the three types of intermediate vector bosons (W+,W-, Z0 particles). They are long range since existing as quarks, they are massless but acquire masses as the symmetry is broken. In this case, they become short range forces. The strong force is an exception as you rightly said since the property of confinement restricts the actions of the glouns making up the quarks. great insight

12. Jul 30, 2009

### Abbas Sherif

I think we could make this more logical for him than mathematical. We know that gravity is brought on by the presence of mass and that gravity attracts and thus drops distances a mass is to travel. If we have, let's say, zero mass, there is very little gravitational pullback on the particle and the particle is capable of travelling a relatively longer distance. think you got me.

13. Jul 30, 2009

### Abbas Sherif

Here we are not discussing the color force's broken symmetry. Infact there's no way to achieve such high temperatures. Put simply, because the gluons are contained in the neutron and proton. they cannot be used to predict the range. anything coming out of the neutron or proton will be a quark-antiquark pair and we could presume that since the pion is the lightest possible meson, it's possible to use them for range calculation. the range of the gluon is indeed short (it is considered the acrrier of the strong force).

Last edited: Jul 30, 2009
14. Jul 30, 2009

### Parlyne

The color force doesn't have a broken symmetry. That's kind of the point here. The finite range of internucleon interactions is essentially a result of the fact that nucleons are uncolored states. This is much like the fact that a neutral atom will neither be attracted nor repelled by a net charge.

As for quark-gluon plasmas, I was under the impression that the consensus view was that they have, in fact, been created in heave atom collisions at RHIC.

Finally, confinement, in and of itself, is not sufficient to say that the interaction is short-ranged. It is, in fact, quite possible to write down a model of an SU(3) gauge theory in which the quarks are able to be separated to arbitrary distance (given, of course, that there's a way to give them sufficient energy). What is necessary is that all quarks in the model have mass much greater than the strong coupling scale of the theory (which, in QCD, is approximately the pion mass). In this case, the fragmentation of a bound state (which is what really restricts the range of the fundamental interaction in normal QCD) is suppressed by a factor of $e^{-m_q^{\phantom{q}2}/\Lambda^2}$. Models like this are generally referred to as "quirk" models. (See, for instance, http://arxiv.org/abs/0805.4642.)