# The Rest of the Yukawa's Theory

1. Dec 22, 2007

### ophase

I had a conversation with my nuclear physics professor and she got mad bcs of me. I claimed there's no pi meson in the nucleus and i know a calculation (which is completed by Yukawa) that predicts an exchange particle mass about 100 MeV but a pion has a mass about 140 MeV. So I claimed that yukawa's theory is not completely correct.

Where did i go wrong?? She talks something about shell ?? what did she mean? what's the complete story of Yukawa's theory???

2. Dec 22, 2007

### blechman

Yukawa's particle should have a mass ROUGHLY of order the size of the proton (in natural units). This is about 1 fm. Plugging in factors of hbar and c, this corresponds to a mass of about 130 MeV. And the pion is 135 and 139 MeV. Sounds really good for a one-significant-figure, back-of-the-envelope calculation to me!

I don't know what your prof got "mad"! But Yukawa's theory does a remarkably good job. More careful calculations get even better results. Of course, at the end of the day, the "REAL" answer is non-perturbative QCD, but we don't know how to calculate with that.

3. Dec 23, 2007

### ophase

Yukawa theory

Here's the original proof by Yukawa

Yukawa potential $$U(r)=\frac{-g^{2}_{s}}{4\pi}\frac{e^{-r/a}}{r}$$
gs: Yukawa constant

Yukawa proposed that nuclear force has to be like elektromagnetic force. So the potential above need to satisfy second green equation with a source term:

$$(\nabla^{2}-\frac{1}{a^{2}})U(r)=g^{2}_{s}\delta(r)$$

Yukawa generalized the equation for non-static states.

$$(\nabla^{2}-\frac{d^{2}}{c^{2}dt^{2}}-\frac{1}{a^{2}})U(r,t)=0$$ (*)

This equation is also relativistical invariant. Then Yukawa quantized the potential:

$$U(r)=\frac{-g^{2}_{s}}{4\pi}\frac{e^{ipr/\hbar-iEt/\hbar}}{r}$$

Now we put that potential expression in the second green equation (*) and we get:

$$\frac{E^{2}}{c^{2}\hbar^{2}}=\frac{p^{2}}{\hbar^{2}}+\frac{1}{a^{2}}$$

$$E^2 =p^2c^2+\frac{c^{2}\hbar^{2}}{a^{2}}$$

Here the last term should be the mass term:
$$m^{2}_{u}c^{4}=\frac{c^{2}\hbar^{2}}{a^{2}}$$
If we assume a=2 fm, then the exchange particle mass is mu= 100 MeV.

In 1947 Pion discovered at 140 MeV and it's proved that there is no meson in the nucleus according to Yukawa theory.

I don't know the rest of that story. Probably someone made a correction about the calculation above. Any ideas??

4. Dec 23, 2007

### blechman

Where did your assumption for "a" come from?????? Why is it 2.0000000 fm???

You only have 1 sig fig, ophase, so "computing" $m_\pi=100$ GeV is what we physicists call WONDERFUL agreement!! Being off by less than a factor of 2 - that's a GREAT bit of predicting!

Last edited: Dec 23, 2007
5. Dec 23, 2007

### ophase

Yukawa theory

a is the average nuclear radius.

So you'r saying that a wonderful agreement. But physicists made predictions for W,Z bosons as exchange particle for weak interaction. And yukawa expanded his theory for W,Z bosons.
Whati want to know is Why did the physicists make predictions about W,Z bosons??
We have a great theory above... So why bother for W,Z bosons which are quite massive (approx90GeV)???

Last edited: Dec 24, 2007
6. Dec 24, 2007

### malawi_glenn

If we assume a=2 fm, then the exchange particle mass is mu= 100 MeV.

That is wrong duder, a is not the mean nuclear radius nor is the mean nuclear radius 2fm. The thing one did is as blechman told you, that you get that the force carrier has mass of approx 132MeV/c^2

"In 1947 Pion discovered at 140 MeV and it's proved that there is no meson in the nucleus according to Yukawa theory." Is just bull-talking.

The meson exchange potentials are good models, the nucleons are colorless so the force carriers sholud also be colourless (gluons have colour). But this is an approximation due to the fact that we dont know how to calculate these things yet (see blechmans posts).

W and Z bosons are for the WEAK interaction. So I dont understand your:
"Why did the physicists make predictions about W,Z bosons then??
We have a great theory above... So why bother for W,Z bosons??? "

This just follows from quantization of fields etc..

Also the pions in the pion exhange potential are virtual.

You should trust and learn from your teacher, if you dont understand these things now, come back to them later when you have more knowledge.

7. Dec 24, 2007

### blechman

The W and Z bosons were proposed 30 years later in an attempt to explain the weak nuclear force, as malawi_glenn says. Pions were proposed (long before we knew about gluons) to explain the strong nuclear force. These are quite different beasts. SNF keeps the protons bound in a nucleus, while the WNF mediates beta-decay (for example).

The thing you have to remember, ophase, is that if you only have one sig fig in your calculation, then you are only allowed one sig fig result!! Electroweak has MANY sig figs (we've done VERY careful measurements of beta decay, for example) so by the time W and Z were proposed in the 60's, we could pin them down theoretically much better than Yukawa could his pions back in the 30's.

8. Dec 24, 2007

### blechman

This is true, but it's not relevant. The Yukawa potential is still $e^{-m_\pi r}/r$ with $m_\pi=$140 MeV (in natural units), irregardless of the offshellness of the pion. That's why I didn't bring it up.

9. Dec 24, 2007

### ophase

Yukawa theory

That's what i'm trying to say from the beginning. We know the feymann diagram for d quark --> u quark + W(-) boson (Beta decay) and u quark --> d quark + W(+) boson (positron emisson)

And that means there is virtual pion exchange between nucleons. But in nuclear physical aspect, that also means "there is no meson" in a nucleus as long as the pion is virtual.

On the other hand, in field theory aspect, i don't know how to explain after calculating a pion has approx 140 MeV mass (above) while a W boson has 80 GeV mass !!

Do you know any theoretical predictions for W boson mass ???

10. Dec 24, 2007

### blechman

I don't understand any of that, ophase. Saying "there's no mesons int he nucleus as long as the pion is virtual" is like saying "there are no photons in a beam of light!" They're there!

What's REALLY happening is - forget pions! There are quarks and gluons and they are exchanging NON-perturbatively. We have no idea how to calculate with such a theory, so the pion language makes more sense (this concept goes under the name "Effective Field Theory"). But this explanation is very nice, and quantitatively accurate. So, it's "right".

Calculating the mass of the mesons comes from experiment. We know the mass of the W-boson is so large because we measure beta-decay. This measurement gives us the Fermi coupling constant, which is related to 1/m_W^2. This is opposed to the DIFFERENT phenomena of nuclear bounding energy, which proceeds (effectively) through a Yukawa interaction with a range 1/m_pi. We measure these quantities, and extract the masses from these measurements. And they do a great job!

11. Dec 24, 2007

### ophase

Yukawa theory

I cannot forget mesons bcs i wrote it on the exam paper... I'd rather talked about in quark scale but that's just where i'm in conflict with nuclear physicists. They don't talk in quark scale.

Saying "there's no mesons in the nucleus as long as the pion is virtual" I meant in quark scale you can explain virtual pion exchange between nucleons with W,Zbosons instead of creating a Pion. And you'r right there are quarks and gluons. So we don't need a pion ! And it means "No mesons in a nucleus".
But it cost me 5 points in the exam. So am i wrong now??

Last edited: Dec 24, 2007
12. Dec 24, 2007

### ophase

And i said yukawa theory is not complete bcs Yukawa predicted wrong that all 4 fields interacts via a common mediating quatum field. Because in 1935, Yukawa didn't have any ideas about gauge theories that will unify the three field.

All the weakness of Standard Model comes from Yukawa theory: S.Symmetry breaking, Higgs mechanism, Massive bosons, hierarchy problem...etc.

13. Dec 25, 2007

### malawi_glenn

Iam not an expert on this yet, but my teacher in nuclear physics (professor in Hadron physics) have told us what a wonderful job the pion exchange potentials does, but that we should not expect this as the truth, that we "must" get down to quark scale in order to figure out how colourless objects interact. This is the future for us who are students now :) So there is a reason why nuclear physicsts dont talk on quark scale (yet), does bio-chemistrits talk about the DNA molecule on electron scale and use maxwells equations? ;)

Do you have references to "you can explain virtual pion exchange between nucleons with W,Zbosons instead of creating a Pion" ? W,Z are much heavier, so it should be easier to produce mesons, also mesons have longer range than W and Z.

14. Dec 25, 2007

### blechman

Nor should they. You can't resolve quarks and gluons in nuclear physics, so you should not talk about them (see malawi_glenn's post).

Hang on a second. Forget W and Z - they have NOTHING to do with this! You can think of "pion exchange" as being some sort of non-perturbative thing where an infinite number of GLUONS are being exchanged. On the scales of which we are speaking, these "infinite number of gluons" form these sort-of mesonic packets that we acknowledge as the pion (and also the kaon, rho, eta, omega, phi, a1, f1, f2...). The details of this kind of analysis goes under the name of "Regge Trajectories" and is a very advanced subject in nuclear physics (and, in case you're interested, the historical beginnings of string theory!).

Anyway, these pion (and other meson) exchanges generate a Yukawa potential between the protons and neutrons in the nucleus. And the resulting force dominates the electromagnetic repulsion between the protons. In order for this force to match physical observations, we find (see previous posts) that the LIGHEST (and therefore (naively) the most important) meson must have a mass OF ***ORDER*** 100 MeV. Later, they discovered two particles that fit the description: one at 100 MeV (muon) and one at 140 MeV (pion). Later it was realized that the muon was NOT a Yuakwa meson, although some older textbooks (and nuclear physicists!) still call it the "mu meson". The pion, however, was exactly where it belonged, and Yukawa won (and deserved!) his Nobel Prize.

Yukawa's model of the SNF has nothing to do with unification of forces! I don't know if historically that was Yukawa's goal (I didn't think so!). But you are right in that the pion has NOTHING to do with the W and Z bosons, the photon or gravity! But it *IS* an effective description of the gluon force at low energies. And it does remarkably well for itself, especially when you add the other mesons (see above). Let me say it again: NONE OF THIS HAS **ANYTHING** TO DO WITH W/Z BOSONS!!! That's the WRONG FORCE!

That is just wrong!! The Yukawa interactions in the SM have absolutely NOTHING to do with the gauge hierarchy problem, or any of the problems you mentioned, for that matter. To use the correct terminology: Yukawa interactions are "technically natural".

Last edited: Dec 25, 2007
15. Dec 25, 2007

### blechman

This just can't be. The weak nuclear force at such low energies can NEVER compete with the HUGE electromagnetic repulsion between the protons! Only the gluon force (effectively described by the pions and other mesons, as I said above) can overpower this repulsion.

16. Dec 25, 2007

### ophase

Yukawa Theory

Ok, blechman... Regge Theory and corresponding connection between Fermi Theory and Yukawa theory helped me so much to understand exchange mechanism. These were all unknown to me. But i'll still critise nuclear physicist not to explain which force uses which mechanism. They keep confusing my mind when they say smthg anyhow.

In conclusion, Yukawa Theory is quite essential and applicable to Weak interactions as well as Strong interactions. Yukawa built a potential and calculated the approx exchange particle mass for strong interaction. Then he expanded his theory to weak interactions.

Moreover he predicted graviton as exchange particle for gravitational interaction in non-relativistic limit. But we know one cannot explain GF using Yukawa theory. So doesn't it make Yukawa theory uncomplete??

Last edited: Dec 25, 2007
17. Dec 26, 2007

### blechman

The "Yukawa Theory" is the following potential:

$$V(r)=\frac{g}{4\pi}\frac{e^{-mr}}{r}$$

That's all there is to it. It's a very generic situation (to use a technical term, "universality class"), that has relevance in all aspects of physics, including nuclear, particle, solid-state, atomic, etc. Which "Yukawa interaction" you are talking about is just a statement of what values of g and m to use. But the Yukawa interactions are just this exponential damped + inverse linear potential. That's all it means.

What more do you want?!

I didn't know that Yukawa tried to describe gravity this way. If he did, then he was wrong. But I suspect there's more to it. Surely he knew that the nonrelativistic potential for gravity was an inverse linear (m=0) potential, aka Newton! He might have been suggesting that the massless graviton generated the Newtonian potential in analogy to how the photon generates the Coulomb potential (which it does). But I think that idea goes back before Yukawa. Although maybe it was his idea. I don't know.

It is correct that the graviton, a massless excitation of the "quantum gravity field," does describe gravity. Non-relativistic expansions of the graviton exchange force do generate the Newtonian potential, so that's that.

18. Dec 26, 2007

### malawi_glenn

"But i'll still critise nuclear physicist not to explain which force uses which mechanism. They keep confusing my mind when they say smthg anyhow. "

Calm down :)

19. Dec 26, 2007

### blechman

I should mention, BTW, for all those listening, that when a nuclear physicist says "Yuakwa forces", he or she is referring to the PION exchange and the *STRONG* nuclear force. I have never heard of nuclear physicists referring to the weak nuclear force as a Yukawa interaction:

1) *Technically* the WNF is not a Yukawa interaction, since the W and Z bosons are spin 1, and Yukawa implicitly assumes that the mediating particle is spin 0. This is a nit-picky technicality, I'll grant you, but...

2) In order to resolve the Yukawa-nature (exponential potential) coming from a W or Z boson exchange, you have to be able to probe a distances smaller than roughly 1 am (attometer) which is $10^{-3}$ fm. These scales are beyond "nuclear physics" scales, and have no business in "nuclear physics;" this is where us "particle physicists" can shine!

Anyway, when nuclear physicists want to talk about the WNF, they use the Fermi Theory to do it (a contact interaction between 4 fermions, for those that are not familiar). This is a *very* different universality class from the Yukawa theories. Roughly speaking: nuclear physicists cannot resolve the "exchanged boson" that mediates the WNF, although they can do it for the SNF (the pion).

BTW: I made some rather grandiose claims about what nuc physicists will and will not say. Not being a true nuclear physicist myself, I hope that no one thinks that presumptuous!

Last edited: Dec 26, 2007
20. Dec 27, 2007

### malawi_glenn

blechman: I think it depends on who you ask (age) and in what field they are in. Nuclear physics is a big field today. Span from mesons to nuclear power.