Monopole and dipole dynamics on Riemann surfaces

Last updated August 17, 2022.

We'll look at the dynamics of monopoles and dipoles on a Riemann surface. The presentation will build upon the papers Vortex pairs and dipoles on closed surfaces and Vortex motion and geometric function theory: the role of connections by Björn Gustafsson.

Notation: A prime denotes differentiation with respect to $z$. If the function that is differentiated is holomorphic, this is the holomorphic derivative: $f^{\prime }=df/dz$. If the function is not holomorphic, this is the Wirtinger derivative: $f^{\prime }=\partial f/\partial z=(1/2)(\partial f/\partial x-i\partial f/\partial y)$.

Setup

On a closed Riemann surface $M$ with genus $g\ge 0$ and conformal metric $ds=\lambda (z)|dz|$ we place $n$ monopoles at points $z_1,\dots ,z_n$ and $m$ dipoles at points $w_1,\dots ,w_m$. To each monopole we associate a complex strength ${\mathrm{\Gamma }}_k$. For real positive strength, the monopole is a source, for real negative strength, it is a sink and for imaginary strength it is a vortex. We require that the strengths sum to zero, $\sum _k{\mathrm{\Gamma }}_k=0$. To each dipole we associate a complex moment $\delta w_k$ that gives the orientation and strength of the dipole. The dipole moment value is coordinate dependent. If $\tilde{z}=\phi (z)$ is a coordinate transformation, the dipole moment transforms as $\delta \widetilde{w_k}={\phi }^{\prime }(w_k)\delta w_k$. The coordinate independent strength of a dipole is $\lambda (w_k)|\delta w_k|$.

The set of monopoles as two-point dipoles

Since the strengths of the monopoles sum to zero, we represent them with two-point dipoles instead. To the set of $n$ monopoles placed at points $\{z_k{\}}_{k=1}^n$ we have $n(n-1)/2$ two-point dipoles placed at point pairs $\{z_k,z_l{\}}_{k<l}$ where $z_k$ is the position of a unit source and $z_l$ is the position of a unit sink. We let the strengths of the two-point dipoles be $$\begin{array}{}\text{(1)}& {\mathrm{\Gamma }}_{kl}=\frac{{\mathrm{\Gamma }}_k({\mathrm{\Gamma }}_k-{\mathrm{\Gamma }}_l)}{\sum _{j=k+1}^n({\mathrm{\Gamma }}_k-{\mathrm{\Gamma }}_j)}\end{array}$$ This is the unique assignment such that ${\mathrm{\Gamma }}_{jk}+{\mathrm{\Gamma }}_{kl}-{\mathrm{\Gamma }}_{lj}=0$ for all triples $\{j,k,l{\}}_{j<k<l}$. A unit two-point dipole with source at $a_{+}$ and sink at $a_{-}$ has a unique potential $G(z,a_{+},a_{-})$ that is circulation free and holomorphic away from $a_{+}$ and $a_{-}$ and is the solution to $$\begin{array}{}\text{(2)}& \mathrm{\Delta }G(z,a_{+},a_{-})={\delta }_{a_{+}}-{\delta }_{a_{-}},\end{array}$$ where $\mathrm{\Delta }=-d\ast d$ is the Laplace operator taking zero-forms to two-forms and ${\delta }_a$ is the Dirac two-form defined as ${\int }_M\varphi {\delta }_a=\varphi (a)$ for any function $\varphi (z)$.

One-point dipoles

The function $$\begin{array}{}\text{(3)}& D(z,a)\delta a=\underset{h\to 0}{lim}\frac{G(z,a+h,a-h)}{2h}\delta a\end{array}$$ is the potential of a one-point dipole at $a$ with moment $\delta a$ since $$\begin{array}{}\text{(4)}& \mathrm{\Delta }D(z,a)& =\mathrm{\Delta }\underset{h\to 0}{lim}\frac{G(z,a+h,a-h)}{2h}=\underset{h\to 0}{lim}\frac{\mathrm{\Delta }G(z,a+h,a-h)}{2h}=\text{(5)}& & =\underset{h\to 0}{lim}\frac{{\delta }_{a+h}-{\delta }_{a-h}}{2h}=\frac{\partial {\delta }_a}{\partial a}.\end{array}$$

Renormalisation of the two-point dipole potential

Let $p_{\text{re}}(z)+\log |dz|=\log (\lambda (z)|dz|)$ be the logarithmic metric. The term $p_{\text{re}}(z)$ is a real 0-connection. At a point $a$ we can approximate the logarithmic metric with a complex logarithmic form $$\begin{array}{}\text{(6)}& p(z)+\log dz=\sum _{k=0}^{\mathrm{\infty }}\frac{p_{\text{re}}^{(k)}(a)}{k!}(z-a{)}^k+\log dz,\end{array}$$ where $p_{\text{re}}^{(k)}={\partial }^k{p}_{\text{re}}/\partial z^k$ is the iterated Wirtinger derivative and $p(z)$ is then a holomorphic 0-connection. The logarithmic form can be exponentiated to give a holomorphic one-form ${\lambda }_{\text{hol}}(z)dz=e^{p(z)}dz$. The one-form is the differential of the function $$\begin{array}{}\text{(7)}& \mathrm{\Lambda }(z,a)={\int }_a^z{\lambda }_{\text{hol}}(w)dw\end{array}$$ The function $F(z,a)=(-1/2\pi )\log \mathrm{\Lambda }(z,a)$ is the local potential of a monopole since $$\begin{array}{}\text{(8)}& \mathrm{\Delta }F(z,a)=-\mathrm{\Delta }\frac{1}{2\pi }\log \mathrm{\Lambda }(z,a)=-\mathrm{\Delta }\frac{1}{2\pi }(\log (z-a)+O(1))=-\frac{1}{2\pi }\mathrm{\Delta }\log (z-a)={\delta }_a\end{array}$$ The monopole potential has expansion $$\begin{array}{rl}\text{(9)}& F(z,a)& =-\frac{1}{2\pi }\log \mathrm{\Lambda }(z,a)=\text{(10)}& & =-\frac{1}{2\pi }\log (\lambda (a)(z-a)+\frac{{\lambda }^{\prime }(a)}{2}(z-a{)}^2+\frac{{\lambda }^{″}(a)}{6}(z-a{)}^3+O((z-a{)}^4))=& =-\frac{1}{2\pi }(\log (z-a)+\log \lambda (a)+\\ \text{(11)}& & +\log (1+\frac{1}{2}\frac{{\lambda }^{\prime }(a)}{\lambda (a)}(z-a)+\frac{1}{6}\frac{{\lambda }^{″}(a)}{\lambda (a)}(z-a{)}^2+O((z-a{)}^3)))=& =-\frac{1}{2\pi }(\log (z-a)+\log \lambda (a)+\frac{1}{2}\frac{{\lambda }^{\prime }(a)}{\lambda (a)}(z-a)+\\ \text{(12)}& & +\frac{1}{6}(\frac{{\lambda }^{″}(a)}{\lambda (a)}-\frac{3}{2}{\left(\frac{{\lambda }^{\prime }(a)}{\lambda (a)}\right)}^2)(z-a{)}^2+O((z-a{)}^3))\end{array}$$ The monopole potential has two differentials, the ordinary with respect to $z$, but also a differential with respect to $a$. We have the following relation between the two differentials $$\begin{array}{}\text{(13)}& F^{\prime }(z,a)dz=\frac{{\mathrm{\Lambda }}_a^{\prime }(z)dz}{2\pi \mathrm{\Lambda }(z,a)}=\frac{{\lambda }_{\text{hol}}(z)dz}{2\pi \mathrm{\Lambda }(z,a)}=\frac{{\lambda }_{\text{hol}}(a)da}{2\pi \mathrm{\Lambda }(z,a)}=\frac{-(\partial \mathrm{\Lambda }(z,a)/\partial a)da}{2\pi \mathrm{\Lambda }(z,a)}=-\frac{\partial F(z,a)}{\partial a}da.\end{array}$$ The ordinary differential of the monopole potential has expansion $$\begin{array}{}\text{(14)}& dF(z,a)& =d\frac{1}{2\pi }\log \mathrm{\Lambda }(z,a)=\text{(15)}& & =\frac{1}{2\pi }(\frac{1}{z-a}+\frac{1}{2}\frac{{\lambda }^{\prime }(a)}{\lambda (a)}+\frac{1}{3}(\frac{{\lambda }^{″}(a)}{\lambda (a)}-\frac{3}{2}{\left(\frac{{\lambda }^{\prime }(a)}{\lambda (a)}\right)}^2)(z-a)+O((z-a{)}^2))dz\end{array}$$ At the monopole points $a_{+}$ and $a_{-}$ the two-point dipole potential is singular and needs to be renormalised. We decompose it at a neighbourhood of the source $a_{+}$ as $$\begin{array}{}\text{(16)}& G(z,a_{+},a_{-})=\frac{1}{2\pi }\log {\mathrm{\Lambda }}_{a_{+}}(z)+H(z,a_{+},a_{-})\end{array}$$ where $H(z,a_{+},a_{-})$ is a holomorphic function. The differential of the potential is $$\begin{array}{}\text{(17)}& dG(z,a_{+},a_{-})=\frac{1}{2\pi }(\frac{1}{z-a_{+}}+\frac{1}{2}\frac{{\lambda }^{\prime }(a_{+})}{\lambda (a_{+})}+O(z-a_{+}))dz+H^{\prime }(z,a_{+},a_{-})dz\end{array}$$ The first term make up the singularity and we renormalise the differential by only keeping the regular part $$\begin{array}{}\text{(18)}& d{G}_{\text{renorm}}(a_{+},a_{+},a_{-})=H^{\prime }(a_{+},a_{+},a_{-})dz\end{array}$$ Since the two-point dipole potential is antisymmetrical in $a_{+}$ and $a_{-}$, $G(z,a_{+},a_{-})=-G(z,a_{-},a_{+})$, the renormalisation at the sink $a_{-}$ is $$\begin{array}{}\text{(19)}& d{G}_{\text{renorm}}(a_{-},a_{+},a_{-})=-H^{\prime }(a_{-},a_{-},a_{+})dz\end{array}$$

Renormalisation of the one-point dipole potential

The associated one-form in $a$ of a one-point dipole, $D(z,a)da$, has the decomposition at a neigbourhood of $a$ $$\begin{array}{}\text{(20)}& D(z,a)da& =\underset{h\to 0}{lim}\frac{G(z,a+h,a-h)}{2h}da=\text{(21)}& & =\underset{h\to 0}{lim}\frac{G(z,a+h,a-h)+G(z,a+h,a-h)}{4h}da=\text{(22)}& & =\underset{h\to 0}{lim}\frac{G(z,a+h,a-h)-G(z,a-h,a+h)}{4h}da=\text{(23)}& & =\underset{h\to 0}{lim}\frac{F(z,a+h)+H(z,a+h,a-h)-F(z,a-h)-H(z,a-h,a+h)}{4h}da=\text{(24)}& & =\underset{h\to 0}{lim}\frac{F(z,a+h)-F(z,a-h)+H(z,a+h,a-h)-H(z,a-h,a+h)}{4h}da=\text{(25)}& & =\frac{1}{2}\frac{\partial F(z,a)}{\partial a}da+\underset{h\to 0}{lim}\frac{H(z,a+h,a-h)}{4h}da-\underset{h\to 0}{lim}\frac{H(z,a-h,a+h)}{4h}da=\text{(26)}& & =-\frac{1}{2}{F}^{\prime }(z,a)dz+H_1(z,a)da,\end{array}$$ where $H_1(z,a)=\underset{h\to 0}{lim}H(z,a+h,a-h)/2h$. By evaluating $dz$ and $da$ on the vector fields $\delta a\partial /\partial z$ respectively $\delta a\partial /\partial a$ we get that the one-point dipole potential has the decomposition $$\begin{array}{}\text{(27)}& D(z,a)\delta a=-\frac{1}{2}{F}^{\prime }(z,a)\delta a+H_1(z,a)\delta a.\end{array}$$ The differential of the one-point dipole potential has expansion $$\begin{array}{}\text{(28)}& dD(z,a)\delta a& =d(-\frac{1}{2}{F}^{\prime }(z,a)\delta a+H_1(z,a)\delta a)=\text{(29)}& & =-\frac{1}{2}{F}^{″}(z,a)\delta adz+H_1^{\prime }(z,a)\delta adz=\text{(30)}& & =-\frac{1}{4\pi }(-\frac{1}{(z-a{)}^2}+\frac{1}{3}(\frac{{\lambda }^{″}(a)}{\lambda (a)}-\frac{3}{2}{\left(\frac{{\lambda }^{\prime }(a)}{\lambda (a)}\right)}^2)+O(z-a))\delta adz+H_1^{\prime }(z,a)\delta adz\end{array}$$ The first term make up the singularity and we renormalise the differential by only keeping the regular part $$\begin{array}{}\text{(31)}& d{D}_{\text{renorm}}(a,a)\delta a=H_1^{\prime }(a,a)\delta adz\end{array}$$

The flow

The flow is divergence free and irrotational outside the singularities and is described by a complex potential $\mathrm{\Phi }$ with differential $d\mathrm{\Phi }=d\phi +id\psi$, where $d\phi$ is the flow one-form and $d\psi$ is the conjugate flow one-form. Since the Riemann surface is not required to be simple connected, $\mathrm{\Phi }$ may be multivalued. The complex potential then fullfills the continuity equation

$$\begin{array}{}\text{(32)}& \mathrm{\Delta }\mathrm{\Phi }=\sum _{k=1}^n{\mathrm{\Gamma }}_k{\delta }_{z_k}+\sum _{k=1}^m\delta w_k\frac{\partial {\delta }_{w_k}}{\partial w_k}\end{array}$$

We can decompose the flow potential $\mathrm{\Phi }$ into

$$\begin{array}{}\text{(33)}& \mathrm{\Phi }(z)=\sum _{1\le k<l\le n}{\mathrm{\Gamma }}_{kl}G(z,z_k,z_l)+\sum _{k=1}^m\delta w_{k}D(z,w_k)+U(z)\end{array}$$ where $U(z)$ is the holomorphic part. The holomorphic potentials on $M$ have a basis of $g$ holomorphic multivalued functions $V_1,\dots ,V_g$ and $U$ can then be expressed as $U=\sum _{k=1}^g{c}_k{V}_k$.

The complex flow vector field $v(z)$ is given by

$$\begin{array}{}\text{(34)}& v(z)& =\frac{1}{\lambda (z{)}^2}\overline{{\mathrm{\Phi }}^{\prime }}=\text{(35)}& & =\frac{1}{\lambda (z{)}^2}(\sum _{1\le k<l\le n}{\bar{\mathrm{\Gamma }}}_{kl}\overline{G^{\prime }}(z,z_k,z_l)+\sum _{k=1}^m{\overline{\delta w}}_k\overline{D^{\prime }}(z,w_k)+\sum _{k=1}^g{\bar{c}}_k\overline{V_k^{\prime }}(z))\end{array}$$

Monopole dynamics

The monopoles are assumed to move with the flow, ${\dot{z}}_k=v(z_k)$, but the flow vector field is singular at $z_k$ and we replace $G(z_k,z_k,z_l)$ and $G(z_k,z_l,z_k)$ by their renormalisations $H(z_k,z_k,z_l)$ and $-H(z_k,z_k,z_l)$ which gives $$\begin{array}{rl}{\dot{z}}_k& =\frac{1}{\lambda (z_k{)}^2}(\sum _{j=k+1}^n{\bar{\mathrm{\Gamma }}}_{kj}\overline{H^{\prime }}(z_k,z_k,z_j)-\sum _{j=1}^{k-1}{\bar{\mathrm{\Gamma }}}_{jk}\overline{H^{\prime }}(z_k,z_k,z_j)+\\ \text{(36)}& & +\sum _{k+1\le j<l\le n}{\bar{\mathrm{\Gamma }}}_{jl}\overline{G^{\prime }}(z_k,z_j,z_l)+\sum _{1\le j<l\le k-1}{\bar{\mathrm{\Gamma }}}_{jl}\overline{G^{\prime }}(z_k,z_j,z_l)+\sum _{j=1}^m{\overline{\delta w}}_j\overline{D^{\prime }}(z_k,w_j)+\sum _{j=1}^g{\bar{c}}_j\overline{V_j^{\prime }}(z_k))\end{array}$$

Dipole dynamics

Also the one-point dipoles are assumed to move with the flow, ${\dot{w}}_k=v(w_k)$, and since the flow vector field is singular at $w_k$ we replace $D(w_k,w_k)$ by its renormalisation $H_1(w_k,w_k)$ which gives $$\begin{array}{}\text{(37)}& {\dot{w}}_k=\frac{1}{\lambda (w_k{)}^2}(\sum _{1\le j<l\le n}{\bar{\mathrm{\Gamma }}}_{jl}\overline{G^{\prime }}(w_k,z_j,z_l)+{\overline{\delta w}}_k\overline{H_1^{\prime }}(w_k,w_k)+\sum _{\begin{array}{c}j=1\\ j\ne k\end{array}}^m{\overline{\delta w}}_j\overline{D^{\prime }}(w_k,w_j)+\sum _{j=1}^g{\bar{c}}_j\overline{V_j^{\prime }}(w_k)).\end{array}$$ Since the flow is irrotational, the dipole moment $\delta w_k$ should be paralleltransported with the flow. We require that the covariant derivative $\mathrm{\nabla }$ of $\delta w_k$ in the direction of ${\dot{w}}_k$ vanishes, $$\begin{array}{}\text{(38)}& \frac{\mathrm{\nabla }\delta w_k}{dt}={\mathrm{\nabla }}_{{\dot{w}}_k}\delta w_k={\dot{\delta w}}_k+r(w_k){\dot{w}}_k\delta w_k=0,\end{array}$$ where $r(z)=2{\lambda }^{\prime }(z)/\lambda (z)$ is the affine connection of the surface and we have that $$\begin{array}{}\text{(39)}& {\dot{\delta w}}_k=-r(w_k){\dot{w}}_k\delta w_k.\end{array}$$

Motion of a single one-point dipole

To investigate the motion of a one-point dipole, we specialise to a setting with one one-point dipole at $a$ with moment $\delta a$, no monopoles and no background flow. We then have the system equations $$\begin{array}{}\text{(40)}& \{\begin{array}{rl}\dot{a}& =\frac{1}{\lambda (a{)}^2}\overline{\delta a}\overline{H_1^{\prime }}(a,a)\\ \dot{\delta a}& =-r(a)\dot{a}\delta a.\end{array}\end{array}$$ Solving for $\delta a$ in the first equation of $\text{(40)}$ gives $$\begin{array}{}\text{(41)}& \delta a=\frac{\lambda (a{)}^2}{H_1^{\prime }(a,a)}\dot{\overline{a}},\end{array}$$ which is substituted in the second equation of $\text{(40)}$ $$\begin{array}{}\text{(42)}& \frac{d}{dt}\left(\frac{\lambda (a{)}^2}{H_1^{\prime }(a,a)}\dot{\overline{a}}\right)=-r(a)\dot{a}\frac{\lambda (a{)}^2}{H_1^{\prime }(a,a)}\dot{\overline{a}}\end{array}$$ We calculate the derivative in the left hand side and solve for $\ddot{a}$ $$\begin{array}{}\text{(43)}& \frac{d}{dt}\left(\frac{\lambda (a{)}^2}{H_1^{\prime }(a,a)}\right)\dot{\overline{a}}+\frac{\lambda (a{)}^2}{H_1^{\prime }(a,a)}\ddot{\overline{a}}=-\frac{\lambda (a{)}^{2}r(a)}{H_1^{\prime }(a,a)}\dot{a}\dot{\overline{a}}\text{(44)}& \frac{\partial }{\partial a}\left(\frac{\lambda (a{)}^2}{H_1^{\prime }(a,a)}\right)\dot{a}\dot{\overline{a}}+\frac{\partial }{\partial \overline{a}}\left(\frac{\lambda (a{)}^2}{H_1^{\prime }(a,a)}\right){\dot{\overline{a}}}^2+\frac{\lambda (a{)}^2}{H_1^{\prime }(a,a)}\ddot{\overline{a}}=-\frac{\lambda (a{)}^{2}r(a)}{H_1^{\prime }(a,a)}\dot{a}\dot{\overline{a}}\text{(45)}& \frac{2\lambda (a){\lambda }^{\prime }(a)H_1^{\prime }(a,a)-\lambda (a{)}^2\partial H_1^{\prime }(a,a)/\partial a}{H_1^{\prime }(a,a{)}^2}\dot{a}\dot{\overline{a}}+\frac{2\overline{{\lambda }^{\prime }}(a)\lambda (a)}{H_1^{\prime }(a,a)}{\dot{\overline{a}}}^2+\frac{\lambda (a{)}^2}{H_1^{\prime }(a,a)}\ddot{\overline{a}}=-\frac{\lambda (a{)}^{2}r(a)}{H_1^{\prime }(a,a)}\dot{a}\dot{\overline{a}}\text{(46)}& (2\frac{{\lambda }^{\prime }(a)}{\lambda (a)}-\frac{\partial H_1^{\prime }(a,a)/\partial a}{H_1^{\prime }(a,a)})\dot{a}\dot{\overline{a}}+2\frac{\overline{{\lambda }^{\prime }}(a)}{\lambda (a)}{\dot{\overline{a}}}^2+\ddot{\overline{a}}=-r(a)\dot{a}\dot{\overline{a}}\text{(47)}& (r(a)-\frac{\partial H_1^{\prime }(a,a)/\partial a}{H_1^{\prime }(a,a)})\dot{a}\dot{\overline{a}}+\overline{r}(z){\dot{\overline{a}}}^2+\ddot{\overline{a}}=-r(a)\dot{a}\dot{\overline{a}}\text{(48)}& \ddot{\overline{a}}=-\overline{r}(z){\dot{\overline{a}}}^2+(\frac{\partial H_1^{\prime }(a,a)/\partial a}{H_1^{\prime }(a,a)}-2r(a))\dot{a}\dot{\overline{a}}\text{(49)}& \ddot{a}=-r(z){\dot{a}}^2+(\frac{\overline{\partial H_1^{\prime }(a,a)/\partial a}}{\overline{H_1^{\prime }}(a,a)}-2\overline{r}(a))\dot{a}\dot{\overline{a}}\end{array}$$ This can be interpreted as a geodesic equation of a connection coming from the dipole potential.