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Strings, Particles, and Surfaces

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  • Started 2 years ago by lee

  1. lee
    Key Master

    Conformal Surfaces in $\mathbb{R}^3$

    Let $D\subset \mathbb{R}^3$ be a domain (connected open set) and $\varphi: D(x,y)\longrightarrow\mathbb{R}^3$ a parametric surface with metric $ds^2=e^u(dx^2+dy^2)$. It is often convenient to work with complex variables $z,\bar z$, so let $z:=x+iy$. Then the metric can be written as $ds^2=e^udzd\bar z$.

    Let us introduce the following quantities that are typical in differential geometry of surfaces:
    \begin{align*}
    E&=\langle\varphi_x,\varphi_x\rangle,\ F=\langle\varphi_x,\varphi_y\rangle,\ G=\langle\varphi_y,\varphi\rangle\\
    l&=\langle\varphi_{xx},N\rangle,\ m=\langle\varphi_{xy},N\rangle,\ n=\langle\varphi_{yy},N\rangle
    \end{align*}
    Then as well-known in differential geometry the Gaussian curvature $K$ and the mean curvature $H$ are computed by the formulas:
    $$K=\frac{ln-m^2}{EG-F^2},\ H=\frac{Gl+En-2Fm}{2(EG-F^2)}\ \ \ \ \ (1)$$
    Definition. $\varphi: D\longrightarrow\mathbb{R}^3$ is called conformal if
    $$\langle\varphi_x,\varphi_x\rangle=\langle\varphi_y,\varphi_y\rangle=e^u,\ \langle\varphi_x,\varphi_y\rangle=0$$

    Let us introduce the differential operators (Wirtinger operators)
    $$\frac{\partial}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right),\ \frac{\partial}{\partial\bar z}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)$$
    Then $\varphi_z$, $\varphi_{\bar z}$, $N$ (unit normal) satisfy
    \begin{align*}
    \langle\varphi_z,\varphi_z\rangle&=\langle\varphi_{\bar z},\varphi_{\bar z}\rangle=0,\ \langle\varphi_z,\varphi_{\bar z}\rangle=\frac{1}{2}e^u,\\
    \langle\varphi_z,N\rangle&=\langle\varphi_{\bar z},N\rangle=0,\ \langle N,N\rangle=1
    \end{align*}

    The moving frame $(\varphi_z,\varphi_{\bar z},N)$ satisfies the Gauss-Weingarten equations
    \begin{align*}
    \sigma_z&=\mathcal{U}\sigma,\ \sigma_{\bar z}=\mathcal{V},\ \sigma=(\varphi_z,\varphi_{\bar z},N)^T\\
    \mathcal{U}&=\begin{pmatrix}
    u_z & 0 & Q\\
    0 & 0 & \frac{1}{2}He^u\\
    -H & -2e^{-u}Q & 0
    \end{pmatrix},\ \mathcal{V}=\begin{pmatrix}
    0 & 0 & \frac{1}{2}He^u\\
    0 & u_{\bar z} & \bar Q\\
    -2e^{-u}\bar Q & -H & 0
    \end{pmatrix}
    \end{align*}
    where $H$ and $Q$ are , respectively, the mean curvature and the coefficient of the Hopf differential $Qdz^2$. $Q$is defined to be
    $$Q=\langle F_{zz},N\rangle$$
    In terms of complex variables, the mean curvature $H$ can be written as
    $$\langle F_{z\bar z},N\rangle=\frac{1}{2}He^u$$
    by the formula in (1). The Gaussian curvature $K$ is calculate to be
    $$K=H^2-4Q\bar Qe^{-2u}\ \ \ \ \ (2)$$

    Quaternionic Description

    Let $\mathbb{H}$ be the algebra of quaternions with standard basis $\{\mathbf{1},\mathbf{i},\mathbf{j},\mathbf{k}\}$ where
    \begin{align*}
    \mathbf{i}^2&=\mathbf{j}^2=\mathbf{k}^2=-\mathbf{1},\\
    \mathbf{i}\mathbf{j}&=-\mathbf{j}\mathbf{i}=\mathbf{k},\ \mathbf{j}\mathbf{k}=-\mathbf{k}\mathbf{j}=\mathbf{i},\ \mathbf{k}\mathbf{i}=-\mathbf{i}\mathbf{k}=\mathbf{j}
    \end{align*}
    This basis can be represented by the pauli matrices $\sigma_\alpha$ as
    \begin{align*}
    \sigma_1&=\begin{pmatrix}
    0 & 1\\
    1 & 0
    \end{pmatrix}=i\mathbf{i},\ \sigma_2=\begin{pmatrix}
    0 & -i\\
    i & 0
    \end{pmatrix}=i\mathbf{j}\\
    \sigma_3&=\begin{pmatrix}
    1 & 0\\
    0 & -1
    \end{pmatrix}=i\mathbf{k},\ \mathbf{1}=\begin{pmatrix}
    1 & 0\\
    0 & 1
    \end{pmatrix}
    \end{align*}
    $\mathbb{H}$ is then identified with $\mathbb{R}^4$:
    $$q=q_0\mathbf{1}+q_1\mathbf{i}+q_2\mathbf{j}+q_3\mathbf{k}\longleftrightarrow(q_0,q_1,q_2,q_3)\in\mathbb{R}^4$$
    The sphere $S^3\subset\mathbb{R}^4$ is naturally identified with the group of unitary quaternions $\mathbb{H}_1=\mathrm{SU}(2)$. $\mathbb{R}^3$ is identified with the space of imaginary quaternions $\mathrm{Im}\mathbb{H}$:
    $$X=-i\sum_{\alpha=1}^3 X_\alpha\sigma_\alpha\in\mathrm{Im}\mathbb{H}\longleftrightarrow X=(X_1,X_2,X_3)\in\mathbb{R}^3$$
    The scalar product of vectors in terms of quaternions and matrices:
    $$\langle X,Y\rangle=-\frac{1}{2}(XY+YX)=-\frac{1}{2}\mathrm{tr}XY$$

    Conformal Framing

    The moving frame $(e^{-u/2}\varphi_x,e^{-u/2}\varphi_y,N): D\longrightarrow\mathrm{SO}(3)$ can be lifted to a map $F: D\longrightarrow\mathrm{SU}(2)$ using the fact that $\mathrm{SU}(2)$ is a double cover of $\mathrm{SO}(3)$
    $$\begin{array}{ccc}
    & & \mathrm{SU}(2)\\
    & \nearrow &\downarrow\\
    F:D & \longrightarrow & \mathrm{SO}(3)
    \end{array}$$
    Take the Weyl gauge fixing $\Phi=e^{u/4}F: D\longrightarrow\mathbb{H}_\ast$, where $\mathbb{H}_\ast=\mathbb{H}\setminus\{0\}$. Then
    \begin{align*}
    \varphi_x&=e^{u/2}\Phi^{-1}\mathbf{i}\Phi,\ \varphi_y=e^{u/2}\Phi^{-1}\mathbf{j}\Phi,\ N=\Phi^{-1}\mathbf{k}\Phi,\\
    \varphi_z&=-ie^{u/2}\Phi^{-1}\begin{pmatrix}
    0 & 0\\
    1 & 0
    \end{pmatrix}\Phi,\ \varphi_{\bar z}=-ie^{u/2}\Phi^{-1}\begin{pmatrix}
    0 & 1\\
    0 & 0\end{pmatrix}\Phi
    \end{align*}
    Thus we have the moving frame $(\varphi_z,\varphi_{\bar z},N)$ in terms of the quaternion $\Phi$.

    The quaternion $\Phi$ satisfy linear differential equations called Lax equations:
    $$U=\Phi_z\Phi^{-1},\ V=\Phi_{\bar z}\Phi^{-1}$$
    where
    $$\begin{pmatrix}
    u_z/2 & -Qe^{-u/2}\\
    \frac{1}{2}He^{u/2} & 0
    \end{pmatrix},\ V=\begin{pmatrix}
    0 & -\frac{1}{2}He^{u/2}\\
    \bar Qe^{-u/2} & u_{\bar z}/2
    \end{pmatrix}$$
    The matrices $U$ and $V$ are called Lax pair. The Lax pair $U$ and $V$ satisfy
    $$U_{\bar z}-V_z+[U,V]=0$$
    which is equivalent to the Gauss-Codazzi equations
    \begin{align*}
    &\mbox{Gauss equation}\ u_{z\bar z}+\frac{1}{2}H^2e^u-2|Q|^2e^{-u}=0,\ \ \ \ \ (3)\\
    &\mbox{Codazzi equation}\ Q_{\bar z}=\frac{1}{2}H_ze^u\ \ \ \ \ (4)
    \end{align*}
    The Gauss-Codazzi equations are necessary and sufficient conditions for the existence of the corresponding surface with metric $e^udzd\bar z$, Hopf differential $Qdz^2$, and mean curvature $H$.

    From Codazzi equation (4) it is clear that the mean curvature $H$ is constant if and only if $Q_{\bar z}=0$ i.e. $Q$ is holomorphic. Constant mean curvature surfacess will play an important role in our discussion.

    Using the Gauss equation (3) and (2), we obtain the Liouville's equation
    $$\triangle u=-Ke^{2u}\ \ \ \ \ (5)$$

    2-D Dirac Equation

    Definition. A first-order differential operator $\mathcal{D}$ on a Riemannian manifold is called a Dirac operator if $\mathcal{D}^2=-\triangle$ where $\triangle$ denotes the Laplacian operator.

    Example. [2-D Dirac Operator]
    \begin{align*}
    \mathcal{D}&=-i\sigma_1\partial_y+i\sigma_2\partial_x\\
    &=2\begin{pmatrix}
    0 & \partial_z\\
    -\partial_{\bar z} & 0
    \end{pmatrix}\ \ \ \ \ (6)\end{align*} is a 2-D Dirac operator. The 2-D Dirac equation
    $$\hbar c\mathcal{D}\psi=mc^2\psi$$ describes a fermion with spin-$\frac{1}{2}$ and mass $m$ confined in a plane. Here $\hbar$ is the reduced Planck constant and $c$ is the speed of light in vacuum. If $m\to 0$, we obtain 2-D Weyl equation
    $$\mathcal{D}\psi=0$$ which describes a neutrino with spin-$\frac{1}{2}$ confined in a plane. Hereafter (6) is the Dirac operator we will use.

    Theorem. The conformal frame $\Phi: D\longrightarrow\mathbb{H}_\ast$ satisfies the Dirac equation
    $$\mathcal{D}\Phi=He^{u/2}\Phi$$ where $\mathcal{D}=2\begin{pmatrix}
    0 & \partial_z\\
    -\partial_{\bar z} & 0
    \end{pmatrix}$.

    Note that the conformal framing $\Phi$ can be written as
    $$\Phi=\begin{pmatrix}
    s_1 & -s_2\\
    \bar s_2 & \bar s_1
    \end{pmatrix}$$ for some $s_1,s_2: D\longrightarrow\mathbb{C}$. Let $\psi:=\begin{pmatrix}
    s_1\\
    \bar s_2
    \end{pmatrix}: D\longrightarrow\mathbb{C}^2$. Then one can readily see that $\psi$ satisfies the Dirac equation
    $$\mathcal{D}\psi=He^{u/2}\psi$$
    We may regard $s_1$, $\bar s_2$ as spinors and the wave function $\psi$ as a Dirac spinor field (2-spinor) with $s_1$ and $\bar s_2$ indicating spin-up and spin-down states. The Dirac spinor $\psi$ describes a fermion confined in a plane with spin-$\frac{1}{2}$ and mass
    $$m=\frac{\hbar}{c}He^{u/2}$$
    or the energy
    $$E=\hbar cHe^{u/2}$$
    Mathematically spinors $s_1$ and $s_2$ are the smooth sections of a spin bundle $K^\frac{1}{2}$ over $D$, where $K$ is the canonical bundle. If $m=0$ or equivalently $H=0$ (i.e. $\varphi$ is a minimal surface), then $\psi$ is a Weyl spinor field in which case $s_1$ and $s_2$ are holomorphic sections. Thus for every surface with mean curvature of the form $H=\frac{c}{\hbar}me^{-u/2}$ for some constant $m>0$, there is a corresponding fermion with spin-$\frac{1}{2}$ and mass $m$. For every minimal surface, there is a corresponding neutrino with spin-$\frac{1}{2}$.

    The whole construction can be reversed, namely

    Theorem. Let $D\subset\mathbb{R}^2$ be a simply connected domain and the wave function (field) $\psi=(s_1,\bar s_2)^T: D\longrightarrow\mathbb{C}^2$ a solution of the Dirac equation with the mass $m\leq 0$
    $$\hbar c\mathcal{D}\psi=mc^2\psi$$
    Then
    $$\Phi:=\begin{pmatrix}
    s_1 & -s_2\\
    \bar s_2 & \bar s_1
    \end{pmatrix}: D\longrightarrow\mathbb{H}_\ast$$
    is a conformal frame of the surfaces
    \begin{align*}
    \varphi_1+i\varphi_2&=\int s_1^2dz-\int\bar s_2^2dz,\\
    \varphi_3&=\int s_1s_2dz+\bar s_1\bar s_2d\bar z
    \end{align*}
    The metric and the mean curvature are given by
    $$e^udzd\bar z=(|s_1|^2+|s_2|^2)^2dzd\bar z,\ H=\frac{c}{\hbar}me^{-u/2}$$

    What the theorem says is that given a fermion with mass $m>0$ (or a neutrino with $m=0$) and spin-$\frac{1}{2}$ there is a corresponding surface represented by the integral formula (Weierstrass representation) as stated in the Theorem.

    Now, we have seen a peculiar correspondence between certain particles (neutrinos or fermions with spin-$\frac{1}{2}$) and surfaces, and the question is why? The reason is that those surfaces are (Euclidean) string worldsheets. (I'll discuss more details in a separate thread.) Hence what we have seen so far is a correspondence between strings and particles. There can be infintely many surfaces with prescribed mean curvature, so does that mean there are infinitely many neutrinos and fermions with spin-$\frac{1}{2}$? The answer is no and here is why. The conformal factor $e^u$ is written in terms of spinors $s_1$, $s_2$ as
    $$e^{u/2}=|s_1|^2+|s_2|^2=\psi^\dagger\psi$$
    On the other hand, the wave function $\psi=(s_1,\bar s_2)^T$ describes a quantum mechanical system of two states (spin-up or spin-down). Since the particle with spin-$\frac{1}{2}$ must exist, $e^{u/2}=\psi^\dagger\psi=1$ i.e. $u=0$. This implies that $K=0$, i.e. the surface is flat and $H$ is constant which gives rise to the mass-mean curvature relation
    $$m=\frac{\hbar}{c}H\ \ \ \ \ (7)$$
    and the energy-mean curvature relation
    $$E=\hbar cH\ \ \ \ \ (8)$$
    The relation (8) is a reminiscence of the Planck relation or the Planck-Einstein equation $E=h\nu=h\frac{c}{\lambda}$ and it hints us that $H=\frac{2\pi}{\lambda}$ for non-zero constant mean curvature $H$. For minimal case ($H=0$), the only flat minimal surface is a plane, so that means the string that corresponds to a neutrino with spin-$\frac{1}{2}$ is a straight line. For non-zero constant mean curvature case, since $K=k_1k_2=0$ either one of principal curvatures $k_1$, $k_2$ is zero and the other is not. So, the constant mean curvature surface is a right circular cylinder, i.e. a tube like worldsheet swept by a closed string which corresponds to a fermion with spin-$\frac{1}{2}$. A right circular cylinder can be given by the parametric equation
    $$\varphi(\sigma,\tau)=(R\cos\sigma,R\sin\sigma,R^2\tau)$$
    where $0\leq\sigma\leq 2\pi$ and $-\infty<\tau<\infty$. The mean curvature is computed to be $H=\frac{1}{R}$. Thus (7) and (8) can be written as
    \begin{align*}
    m&=\frac{\hbar}{Rc}=\frac{h}{2\pi Rc},\ \ \ \ \ (9)\\
    E&=\frac{\hbar c}{R}=\frac{hc}{2\pi R}\ \ \ \ \ (10)
    \end{align*}
    Richard Feynman did not like string theory and he said to string theorists if you can calculate the mass of electron, I will believe you.'' Using (10) we sort of attempt to calculate the mass of electron. We actually do not know the radius of the closed circular string. Instead we can try the classical electron radius $r_e=2.8179\times 10^{-15}\mbox{m}$ for $R$.
    Now,
    \begin{align*}
    m&=\frac{\hbar}{r_ec}\\
    &=\frac{1.0546\times 10^{-34}\mbox{J}\cdot\mbox{s}}{(2.8179\times 10^{-15}\mbox{m})(3\times 10^8\mbox{m/s})}\\
    &=1.248\times 10^{-28}\mbox{kg}
    \end{align*}
    The known electron mass is $m_e=0.9109\times 10^{-30}\mbox{kg}$, so the mass I obtained is about 100 times heavier.

    Posted 2 years ago #

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