0
$\begingroup$

I am trying to understand Gibbons and Manton's article. I am trying to understand the Lagrangian that they wrote down, from a physical point of view. Let me paraphrase from that article.

"Consider $n$ dyons, all with the same magnetic charge $g$. Let the $i$th dyon have electric charge $q_i$ and scalar charge $(g^2 + q_i^2)^{1/2}$, which we shall later approximate by $g + \frac{q_i^2}{2g}$; its rest mass is $m_i = v(g^2 + q_i^2)^{1/2}$. Suppose the dyons have positions $\mathbf{x}_i$ and velocities $\mathbf{v}_i$. Denote their separations $\mathbf{x}_j - \mathbf{x}_i$ by $\mathbf{r}_{ji}$, and set $r_{ji} = |\mathbf{r}_{ji}|$. The Lagrangian for the motion of the $n$th dyon in the background of dyons $1,\ldots,n-1$ is $$L_n = (-m_n + (g^2 + q_i^2)^{1/2}\phi)(1-\mathbf{v}_n^2)^{1/2} +q_n \mathbf{v}_n.\mathbf{A} - q_n A_0 +g \mathbf{v}_n.\tilde{\mathbf{A}} - g\tilde{A}_0.$$ Here $\phi$ is the scalar field due to dyons $1,\ldots,n-1$; its effect is to modify the rest mass of dyon $n$, with the coefficient of $(g^2 + q_n^2)^{1/2}$ being the scalar charge of dyon $n$. $\mathbf{A}$, $A_0$ are the vector and scalar Maxwell potentials due to all but the $n$th dyon, and they couple to the electric charge of the $n$th dyon, $q_n$. The magnetic charge of the $n$th dyon, $g$, couples to the dual Maxwell vector and scalar potentials $\tilde{\mathbf{A}}$, $\tilde{A}_0$ produced by dyons $1,\ldots,n-1$."

I would like to understand a little how they were able to write down this Lagrangian. My background is in Mathematics, but I know some Physics to some level. I kind of understand, roughly speaking, that the first term is the relativistic kinetic energy term, with the mass modified by the scalar field of the other dyons, by some kind of Higgs mechanism. The other two sets of terms are terms related to Maxwell fields and their duals. I am guessing they would induce perhaps Maxwell's equations probably. If someone could perhaps provide more explanation, so that "I could have written down this Lagrangian", or at least so that "ah yes, I understand this Lagrangian", then that would be much appreciated.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.