
#1
Feb2705, 05:44 AM

P: 603

In Yukawas model the force between nucleons comes from the exchange of a pion. How precisely does this lower the energy? Has it something to do with confinement energy?
I know an exchange force can be understood in terms of a momentum exchange by the force carrying particle. But there must be a reason this particle is emitted. How does this lower the energy? 



#2
Feb2705, 06:09 AM

P: 4,008

[tex] E = \frac{ e^{mr}}{4 \pi r}[/tex] This is the expression for the energy that QFT gives us when we consider a massive free field theory (starting from the ordinary Klein Gordon equation) which we will submit to a perturbation expressed by two delta dirac functions. You can think of this as two massive lumps sitting on a matress (the field F to which those massive objects couple). This coupling means that the two massive objects will make the matress vibrate and this vibration (also called fluctuation or perturbation) exactly represents a certain particle of mass m and with integer spin (this will be the pimeson). Them massive lumps are the nucleons. This energy is negative, which means that the presence of the two delta dirac functions has lowered the energy of the free field theory. This really means that the two objects attract. Also dE/dr > 0, two massive lumps sitting on the mattress can lower the energy by getting closer to each other. The potential drops off exponentially over a distance 1/m : the range of the attractive force generated by the field F, is determined inversely by the mass m of the particle described by that field (ie the pi meson). These are the reasons for attraction : the lowering of energy. The pi meson (also called Ykawa meson) is only emitted (NOT TO LOWER THE ENERGY) but because of the influence of the two massive objects (ie two nuclei for example) on the field F (of which the fluctuation describes a particle of mass m : the pimeson.) regards marlon 



#3
Feb2705, 06:23 AM

P: 603

[tex]\psi = C \frac{e^{\frac{mc}{\hbar} r}}{r}[/tex] [Only considering the negative signed exponent as the physical one] What is the interpretation of psi? I know it is not susceptible to a statistical interpretation, but why can you interpret it as a (Yukawa) potential? And the sign of C, isn't that something you put in? So why attraction? 



#4
Feb2705, 06:41 AM

P: 4,008

Particle exchange forces
Hold on, you are forgetting a very important aspect here : namely the fact that you need to incorporate the nucleons. You just solved the KG equation.
I am gonna assume you know the Path integral formalism because this is the easiest way to prove the attraction mediated by a Yukawa meson. [tex]W(J) = \frac{1}{2} \int \int d^4x d^4y J(x) D(xy) J(y)[/tex] The J's express the two massive lumps i talked about. The D is the propagator of the free field (coming from the KG equation.) It's equation is [tex]D(xy)= \int \frac{d^4x}{(2 \pi)^4} \frac {e^{ik(xy)}}{k^2m^2 + i \epsilon}[/tex] Now plug this into the expression for W and make a Fouriertransform. Then write each J(x) = J1 + J2 and only look at the contribution of J1 and J2. So we neglect selfcouplings. You will get : [tex]W(J) = \frac{1}{2} \int \frac{d^4k}{(2 \pi)^4} J_2^*(k) \frac {1}{k^2m^2 + i \epsilon} J_1(k)[/tex] In order to compute the energy just realize that iW=iET (this comes from the path integral formalism), you will get : [tex] \int \frac{d^3 k}{(2 \pi)^3} \frac{e^{ik.(x_1 x_2)}}{k^2 +m^2}[/tex] In the above formula, the x and k are VECTORS where as they are 4vectors in the first formula's Now combine this with the insights of my first post and you will have all that you need marlon 



#5
Feb2705, 06:55 AM

P: 603

Thanks for the effort, but I am not too familiar with the path integral formalism. I would like to know 2 things:
1) Why can the solution to the KG equation be interpreted as a potential? 2) In the case of the hydrogen molecular ion eg I can understand where the exchange energy comes from. The electron is no longer confined to surrounding one proton but can smear out thus lowering the total energy, constiting the binding force between the two protons. But in the case of a proton and a neutron where does the exchange energy you mention come from? 



#6
Feb2705, 07:22 AM

P: 4,008

I don't get the example you gave on the hydrogen molecular ion, though ??? Two protons do not interact with each other by the exchange of electrons !!! Their EMinteractions are mediated by virtual photons. You can prove this by using the eact same way that i described above. However you WILL need to know your introductory QFT/QED in order to both prove and understand what is going on. marlon 



#7
Feb2705, 12:47 PM

P: 603

In Yukawas theory a proton can resonate between being a proton or a neutron and a positive pion. Does the energy lowering have a similar origin to the case of the hydrogen molecular ion? 



#8
Feb2705, 05:42 PM

P: 4,008

marlon 



#9
Feb2805, 01:35 AM

P: 603

But can you give me an intuitive physical reason (pls without advanced QFT...?!) why:
1) a particle would continuously emit gauge bosons? ["because of the influence of the two massive objects (ie two nuclei for example) on the field F" doesn't say much to me] 2) why this sometimes gives an attractive and sometimes repulsive force? [repulsive I can understand by momentumexchange, attractive I can understand because by the uncertainty principle you cant really tell from wich direction a gauge boson comes. But why in certain cases atrractive and in certain cases repulsive. If this lowers the energy, why?] 



#10
Feb2805, 02:02 PM

PF Gold
P: 2,884

Of course identical particles have equal charge. ElectronPositron, on the other hand, hmm, let me to think, could be seen as two electrons, one coming from the future, other from the past, and then its repulsive force should be seen as attractive. Uff, I can not picture this. Surely it is easier to get a minus sign multiplying the potential got in the previous step. 



#11
Feb2805, 02:07 PM

PF Gold
P: 2,884

Between electron and quark, I have never seen a calculation of the potential. But I believe I have seen it done between electron and pion, considered both as elementary particles.




#13
Feb2805, 03:37 PM

P: 4,008

A more thorough study can be made BUT NOT without QFT, so this is as intuitive as you can get regards marlon 



#14
Feb2805, 05:21 PM

P: 4,008

http://www.physicsforums.com/journal...90&action=view
Here is another vision on attracion and repulsion. I wrote it my journal regards marlon 



#15
Mar105, 03:37 AM

P: 603





#16
Mar105, 05:59 AM

PF Gold
P: 2,884





#17
Mar105, 06:03 AM

PF Gold
P: 2,884

The proof should be done by reversing calculation of this kind:
http://hrst.mit.edu/hrs/renormalizat...small/034.html so that instead of getting the scattering from the potential, you work yout the scattering in Feynmann style and then you reverse the steps to get the potential, so you will see if it is attractive or repulsive. 



#18
Mar105, 06:10 AM

P: 4,008

regards marlon 


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