Explanation of Yukawa potential

In summary, the Yukawa potential is the potential that arises from a massive scalar field. It is attractive (that is, F=-\frac{\partial V}{\partial r} is negative). It reduces to the Coulomb potential for \mu =0.
  • #1
wofsy
726
0
I would appreciate an explanation of Yukawa potential
 
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  • #2
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].
 
  • #4
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.
 
  • #5
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?
 
  • #6
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.
 
  • #7
may i know the page no and the problem to be solved in Jackson Book of E and M
 
  • #8
RedX said:
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.

Brian_C said:
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]
 
  • #9
RedX said:
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.
 
  • #10
clem said:
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?
 
  • #11
clem said:
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
 
  • #12
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.
 
  • #13
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
 
  • #14
You really need to go to a book on strong interactions now.
 
  • #15
I know it very well
that I am not known to all these things that's 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
 

1. What is the Yukawa potential?

The Yukawa potential is a mathematical model used in physics to describe the interaction between two particles through the exchange of a virtual particle. It is named after the Japanese physicist Hideki Yukawa who first proposed its use in the 1930s.

2. How is the Yukawa potential calculated?

The Yukawa potential is calculated using the formula V(r) = k * (q1 * q2 / r) * e^(-kr)/r, where k is the coupling constant, q1 and q2 are the charges of the two particles, r is the distance between them, and e is the base of natural logarithms. This formula takes into account the attractive and repulsive forces between the particles.

3. What is the significance of the Yukawa potential?

The Yukawa potential is significant because it is used to explain the behavior of the strong nuclear force, which is responsible for holding the nucleus of an atom together. It has also been used to model other fundamental forces, such as the weak nuclear force and the gravitational force.

4. What are the limitations of the Yukawa potential?

One limitation of the Yukawa potential is that it assumes the particles are point-like and do not have any internal structure. It also does not take into account the effects of quantum mechanics, which become important at very small distances.

5. How is the Yukawa potential related to the Higgs field?

The Yukawa potential is related to the Higgs field through the Higgs mechanism, which explains how particles acquire mass. In this mechanism, particles interact with the Higgs field and gain mass through their interactions with the Higgs boson, which is the particle associated with the Higgs field. The strength of this interaction is described by the Yukawa potential.

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