# Relativistic correction for strong nuclear force?

1. Jul 5, 2008

### emb6150

I'm a physics undergraduate student who has recently finished a special relativity class, but haven't done a nuclear physics class yet, so if I'm mistaken anything about nuclear physics, please object.

I was pondering about a situation while doing my special relativity class which I haven't gotten a solid answer yet. The problem is this:

Two particles who are not electrically attracted but may be subjected to a nuclear force attraction are seperated, in their frame of reference where they are both immobile, by a distance longer than the minimal active distance of the strong force (thereby there is no attraction).

Now a minisculle traveller (who is on a linear path) is travelling at a speed close to the speed of light, close enough so that in his frame of reference, he mesures (using the usual method of length mesurement in special relativity) the distance between both particles to be shorter than the minimal strong force distance. Paradoxally, he 'should' see a force between them, and they should be attracted and collide, but in their frame of reference they are immobile and there is no force.

So my question is this: Is there a relativistic correction for the strong nuclear force, or is one(or many) of my initial assumptions false?

References(books, articles, etc.) would be good, but an expert's word is just as appreciated!

2. Jul 5, 2008

### nrqed

the problem is that it is not clear cut like this. It's not true that the force acts up to a certain distance and then does not act anymore. People say that sometimes to model the situation (i.e. to simplify it enough to be able to have simple discussions) but this is a very crude picture. In actuality, the potential tampers off and the moving observer will simply see a different force that will correspond to the potential Lorentz transformed in his frame (so the actual value of the potential will be different and the distance will be different and in the end, everything works out between the frames (any physical measurement made in the two frames will agree with one another if one includes how they transform under Lorentz transformations). This has to be true for any fundamental theory. If you use models to represent potentials an dyour model is not Lorentz invariant then all bets are off. But you are not talking about a fundamental theory anymore.

(I am setting aside the issue of confinement here just to make the point clear).

3. Jul 5, 2008

### emb6150

Through physical intuition I know this, but is there some equations, calculations, or anything written about this. I know that physically, in both frame of reference, the same 'event' will happen, but what exactly makes that correction. I say this because all I know is that there is a limited distance for the strong for to act, but haven't ever seen an equation on it to do a Lorentz transformation or any other derivation to verify it. I'd like better proof than intuition.

4. Jul 5, 2008

### humanino

Hi emb6150, and welcome to PF,

this is an excellent question

There are relativistic corrections to the strong nuclear force indeed.

To warm up the discussion, let me describe what is fairly well known and measured. I am simplifying a bit, but keeping the essential aspects relevant here. We usually say that a proton is made up of three quarks, two up quarks and one down quark. They are tied together by exchanging gluons (gluons are very similar to quark-antiquark pairs). There are actually distributions of gluons, quarks, and antiquarks, which depend on (say) the fraction of momentum carried by the active quark. So, you probe your proton with (say) a virtual photon, you have mathematical theorems ensuring you that the virtual photon interacts with only one quark, this active quark has a fraction of momentum of the proton, and you measure the distribution of them. This fraction of momentum we call $$x_{B}$$ (x-Bjorken). Now, the virtuality ($$Q^{2}$$, just the invariant mass square) of the photon defines the (inverse) scale (typical size) at which your process occurs. You can separate up and down by measuring on protons and neutrons. You get distributions as a function of $$x_{B}$$ at fixed $$Q^{2}$$. Look up fig 16.4 here :
Structure functions (PDG)
This is a log plot, so don't get too worried if you see a lot of gluons. Comparing left and right, you contemplate two scales, two values of the photon virtuality $$Q^{2}$$. The closer you look at the nucleon, the more you will find gluons and quarks anti-quarks pairs. What we mean when we say that the proton is made up of three quarks, is that if you count the number of quarks and antiquarks, make a grand total adding quarks and subtracting antiquarks, you will always end up with a net sum of three.

Look now at fig 16.7 here :
Please realize, this is log-log scale measured with incredible precision over many orders of magnitudes.

The basics is this : the strong nuclear force vanishes at very high energy, or very small scale. So the plot 16.7 should display flat sets of points. The deviations from flatness are the relativistic corrections, and are remarkably well described by the theoretical calculations.

With there in mind, now you probably know that there is no such thing as a free quark. So you need to be more specific in your question : are you asking about two free quarks sitting around separated by more than the confinement scale, or are you talking about (say) two neutrons ?

If you are wondering about two free quarks, first it is impossible, and second it would be awfully complicated if you insisted to "make a thought experiment where you lay them, in this initial position, and see what happens" (which is doable, say, on a computer).

So I will assume you are talking about two free neutrons, sitting here on the table separated by more than the typical nuclear interaction. What if now I happen to be flying by at a speed enough with respect to the table, so that the two neutrons seem much closer to each other ? Although I did not do any calculation, I would strongly suspect that time dilatation will freeze their interaction The reason I claim that is, if you remember I mentioned theorem ensuring you that the photon interacts with only one quark. Those "factorization theorems" are of utmost importance in this business. You must always make sure they hold before making any claim, and any measurement beyond their validity usually reveals a great deal. One way to think about those theorems is that time dilation between the referential of the proton and the one of the virtual photon freezes quark interaction. You work in light cone coordinates, so it is not exactly the same kind of time we usually talk about, yet this "hand waving" argument can be made quite rigorous. Just a guess.

edit
must have taken me quite some time, I did not see nrqed before I started this post !

5. Jul 6, 2008

### clem

The first postulate of SR says that if there is no force in the rest system, there is no force if you run past the two particles. There is no need to try to calculate a simple result in a complicated system.

6. Jul 6, 2008

Staff Emeritus
While that's true, one has to be careful with that. For example, suppose you had a system of charges arranged such that the combined electrostatic force on one charge is zero. Now set the system in motion. While it's true that the force on that charge is still zero, it's not because the combined electrostatic force is still zero - it's because the electric and magnetic forces on it are equal and opposite.

To the OP, you are being presented with a cartoon version of the theory, one with numerous approximations and oversimplifications. The fact that the cartoon version is not Lorentz invariant is one of them.

7. Jul 6, 2008

### humanino

Can you provide a reference for that ? I have never seen, be it in the old books or the most recent ones, such a postulate.

8. Jul 6, 2008

Staff Emeritus
Method 1: Imagine two objects that cause a bell to ring when they touch. In frame K, they are a fixed distance apart, at rest with respect to each other, and feel no force. They will never touch. If in frame K' they feel an attraction towards each other, eventually they will touch and the bell will ring. Since you can't have a bell ring in one frame and remain unrung in another, in frame K', they must not feel a force.

Method 2: Look at the Lorentz transformation of zero acceleration into any arbitrary inertial frame. You will see that the acceleration is still zero.

9. Jul 6, 2008

Staff Emeritus
Method 2.5 Note that a fixed velocity remains fixed after a Lorentz transformation.

10. Jul 6, 2008

### clem

How would you state the first postulate?
Better than "force", I should have said acceleration.
If there is no acceleration in the rest system, a LT cannot induce acceleration in any Lorentz system.

11. Jul 6, 2008

### humanino

I bet you would not expect particles to be created by a LT either...

The fact that a LT will not produce a force is not listed in SR postulates.

The second postulate concerning the speed of light is irrelevant here. The first postulate states that the mathematical formulation of the laws must have the same form in one inertial reference frame and in another. It is by the way clear where the word "generalized" comes from in this context. So one can derive things from this postulate, among which the fact that a LT will not create a force, but this is not in the postulates as far as I am concerned.

12. Jul 7, 2008

### emb6150

Thanks humanino! I can't honestly say you answered my question, because I don't completely understand it yet. But you gave me good guidance on how to do so!

Merci!