Explanation of Yukawa potential

707
0
I would appreciate an explanation of Yukawa potential
 

Tom Mattson

Staff Emeritus
Science Advisor
Gold Member
5,453
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The Yukawa potential is the potential that arises from a massive scalar field. It is:

[tex]V(r)=-k\frac{exp(-\mu r)}{r}[/tex]

where [itex]k>0[/itex] and [itex]\mu[/itex] is the mass of the mediating field.

Note that:

1.) It is attractive (that is, [itex]F=-\frac{\partial V}{\partial r}[/itex] is negative).

2.) It reduces to the Coulomb potential for [itex]\mu =0[/itex].
 
The Yukawa Potential can be roughly thought of as a generalization of an inverse-square force potential that takes into account a massive mediator or force. This would mean that instead of massless photons exchanging the force, as is the case with electromagnetism, some other particle with mass exchanges the force between two particles.
 
969
3
In the Wikipedia link:

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

it says that the Fourier transformation of the Yukawa potential is the amplitude for two fermions to scatter. But the Fourier transform ignores 4-momentum and only has 3-momentum. The amplitude to scatter should depend on a 4-momentum squared, and not 3-momentum. So how is this reconciled?
 
242
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There is a problem in Jackson's E&M book which asks you to derive the charge distribution corresponding to this potential. I could never quite get it right.
 
18
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may i know the page no and the problem to be solved in Jackson Book of E and M
 
In the Wikipedia link:

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

it says that the Fourier transformation of the Yukawa potential is the amplitude for two fermions to scatter. But the Fourier transform ignores 4-momentum and only has 3-momentum. The amplitude to scatter should depend on a 4-momentum squared, and not 3-momentum. So how is this reconciled?
I'm not as familiar with relativistic QM as I would like to be, but it seems like the given formula actually does depend on the four momentum. The k in transform corresponds to the three spatial components of the four-momentum while the m corresponds to mass, which in turn depends on the time component of the four-momentum. Why the fourth component of the four-momentum is left out of the Fourier Transform, on the other hand, is unfamiliar to me.

There is a problem in Jackson's E&M book which asks you to derive the charge distribution corresponding to this potential. I could never quite get it right.
It is known that, from the definition of V (in a static field) and the first of Maxwell's equations that:

[itex]\nabla\cdot\vec{E}=-\nabla ^2V=\frac{\rho}{\epsilon _0}[/itex]

So that to find the charge density, one must simply take the negative Laplacian of the potential. This will work every where except for the origin, where you have to apply gauss's law and gauss's vector calculus equation to find the charge. I found:

[itex]\rho (r)=4\pi g^2\epsilon _0 \delta (r)-g^2m^2\frac{e^{-mr}}{r}[/itex]
 

clem

Science Advisor
1,292
14
it says that the Fourier transformation of the Yukawa potential is the amplitude for two fermions to scatter. But the Fourier transform ignores 4-momentum and only has 3-momentum. The amplitude to scatter should depend on a 4-momentum squared, and not 3-momentum. So how is this reconciled?
The simplified version of the Yukawa derivation takes place in the barycentric system where the energy component of the 4-momentum transfer vanishes. Then the 3D Fourier T can be made.
 
969
3
The simplified version of the Yukawa derivation takes place in the barycentric system where the energy component of the 4-momentum transfer vanishes. Then the 3D Fourier T can be made.
Does barycentric system mean center of mass frame between the two fermions? Does this mean that the Yukawa potential is only valid in a center of mass frame, since you break Lorentz invariance by choosing a specific frame?
 
18
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The simplified version of the Yukawa derivation takes place in the barycentric system where the energy component of the 4-momentum transfer vanishes. Then the 3D Fourier T can be made.
May I come to know about the barycentric systems
I have listened the word for the first time and I am curious to know about it because you have mentioned that it reduces the 4 momentum to 3- momentum
 

clem

Science Advisor
1,292
14
The "barycentric system" is the Lorentz system in which the total momentum equals zero. It is usually loosely called (even by me) the center of mass (cm) system, even though the term "center of mass" has no clear meaning in relativity. The Yukawa potential is usually derived for two nucleons. Then the energy component of the 4-momentum transfer vanishes in the cm system because the two individual energies are equal. It is valid only in the cm system, where most calculations are made anyway. The simple Yukawa potential is used mainly in nonrelativitic calculations , because other effects become important at higher energies. The Yukawa potential by itself is now useful only for simple order of magnitude estimates because the full N-N interaction is more complicated. It does describe the long range part of the N-N potential.
 
18
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thanks a lot for such an eloborative reply
butit has increased my curicity
will you help me to tell what are the other factors responsible for N-N interactions
 

clem

Science Advisor
1,292
14
You really need to go to a book on strong interactions now.
 
18
0
I know it very well
that I am not known to all these things thats why I sk you for recommending me a book
(a specific book)
because I can have book from a library or may have it on rent before purchasing it
 

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