Dyon solution in BPS limit (differential equations)

In summary, the differential equations arising from a dyon within a SU(2) Yang-Mills theory coupled to a Higgs field are:-r^2J'' = 2JK^2-r^2H'' = 2HK^2 - \mu^2r^2H(1 - \frac{\lambda}{e^2\mu^2r^2}H^2)-r^2K'' = K(K^2 - J^2 + H^2 - 1)subject to boundary conditions that J and H approach 0 and K approaches 1 as r approaches zero, and K approaches zero and H/r approaches ev as r approaches infinity.
  • #1
Ian Lovejoy
7
0
Investigating a dyon within a SU(2) Yang-Mills theory coupled to a Higgs field, the following differential equations arise (Julia and Zee, Phys. Review. D, 11:2227, 1975):

[tex]r^2J'' = 2JK^2[/tex]
[tex]r^2H'' = 2HK^2 - \mu^2r^2H(1 - \frac{\lambda}{e^2\mu^2r^2}H^2)[/tex]
[tex]r^2K'' = K(K^2 - J^2 + H^2 - 1)[/tex]

The BPS limit corresponds to taking [tex]\mu=0[/tex], leading to the simpler equations:

[tex]r^2J'' = 2JK^2[/tex]
[tex]r^2H'' = 2HK^2[/tex]
[tex]r^2K'' = K(K^2 - J^2 + H^2 - 1)[/tex]

Subject to the boundary conditions that J and H approach 0 and K approaches 1 as r approaches zero, and K approaches zero and H/r approaches ev as r approaches infinity.

One solution is J=0, H = evr(coth evr) - 1, K = evr/(sinh evr). This is the magnetic monopole. A solution with [tex]J\neq 0[/tex] would correspond to a monopole with an electric charge (a dyon).

However, I am completely stumped trying to find the explicit solution, which apparently exists according to A. Zee, Quantum Field Theory, page. 288. I have tried every straightforward generalization of the monopole I can think of, without success. Actually I'm unsure how to even derive the monopole solution, although it is easy to verify that the above solution works.

Can anyone give me a clue? Is there a general procedure for solving coupled differential equations of this form? Any hints or references would be much appreciated. Especially if there is a book outlining how to solve differential equations like this, I would very much like to purchase a copy for reference.

Thanks in advance for any help.
 
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  • #2
Moderator's Note: Thread moved to Differential Equations forum
 
  • #3
What about the following solution

{H(r) = -((C-r)*exp(2/C*r)-C-r)/(-1+exp(2/C*r))*ev,
J(r) = -(-1+ev^2*C^2)^(1/2)*((C-r)*exp(2/C*r)-C-r)/C/(-1+exp(2/C*r)),
K(r) = 2*r/C/(-1+exp(2/C*r))*exp(1/C*r)}

where C>0.
 
  • #4
I think that there is a misprint in first system as it is not converted to sys2 under mu=0.
By the way, there is a sense to attack sys1?
 
  • #5
Hi kosovtsov, many thanks for your reply!

First, thanks for pointing out my error, I should have said both [tex]\lambda[/tex] and [tex]\mu[/tex] approach zero - this gives the sys2 from my original post. My apologies for the confusion, I got mixed up because the notation is slightly different between the article and the textbook I referenced.

Regarding your question of solving sys1, I would be very interested in seeing how to do it. In the paper I referenced (Julia and Zee 1975) it is only solved numerically, so I just assumed that there was no exact solution. On the other hand maybe better techniques have been developed in the time since the paper was written.

Finally, thanks for the solution to sys2! To match the references more closely I wrote it as:

J = (1 - 1/(evC)^2)^1/2 * (-evC + evr * coth(r/C))
H = -evC + evr * coth(r/C)
K = r/C / sinh(r/C)

I checked it out and sure enough it works! May I ask how you derived it? I'm interested in learning to solve systems like this.

Thanks,
Ian
 
  • #6
Ian,

The total oder of your system is 6. But you give only 5 boundary conditions. How it is grounded by physics? What sense have the excess arbitrary constants?

I suspect that another boundary conditions may exist, for example K(0)=-1 (as "anti-Dyon") or K(0)=0 (as "massless Dyon"). Such solutions are in more general solution of the second sistem

{H(r) = -(exp(2/C*(r+_C2))*(C-r)-C-r)*ev/(-1+exp(2/C*(r+_C2))),
J(r) = -(exp(2/C*(r+_C2))*(C-r)-C-r)*(-1+ev^2*C^2)^(1/2)/C/(-1+exp(2/C*(r+_C2))),
K(r) = (+or-) 2*r/(-1+exp(2/C*(r+_C2)))/C*exp(1/C*(r+_C2))}

if _C2<>0 then K(0)=0, (if _C2=0 but sign is minus, then K(0)=-1) the rest of boundary conditions are as yours.
 

Related to Dyon solution in BPS limit (differential equations)

1. What is a dyon solution?

A dyon solution is a type of mathematical solution to a set of differential equations that describes the behavior of a system in which both electric and magnetic charges are present. In other words, it is a solution that takes into account both electric and magnetic fields and their interactions.

2. What is the BPS limit in relation to dyon solutions?

The BPS limit, also known as the Bogomol'nyi-Prasad-Sommerfield limit, is a special case in which a dyon solution becomes particularly simple and symmetric. In this limit, the energy of the system is minimized and the equations governing its behavior become simpler and easier to solve.

3. What are the applications of dyon solutions in physics?

Dyon solutions are used in various areas of physics, including particle physics, cosmology, and condensed matter physics. They have been studied in the context of string theory, supergravity, and gauge theories, and have applications in understanding the behavior of black holes and other astrophysical objects.

4. How are dyon solutions different from monopole or instanton solutions?

Dyon solutions differ from monopole and instanton solutions in that they take into account both electric and magnetic fields, whereas monopole solutions only consider magnetic fields and instanton solutions only consider electric fields. Dyon solutions also have different boundary conditions and symmetries compared to monopole and instanton solutions.

5. What are the challenges in finding dyon solutions in the BPS limit?

One of the main challenges in finding dyon solutions in the BPS limit is the complexity of the equations involved. These solutions often require advanced mathematical techniques and numerical methods to solve. Additionally, the presence of singularities and other complications can make it difficult to find exact solutions, and approximations may need to be used.

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