Jul 25, 2014

Charge Symmetry

Classical Example

Under the charge conjugation (C for short), charge $q \to -q$, electric field $\mathbf E \to - \mathbf E$, but the solution of the dynamical equation, the position vector, remains invariant $\mathbf r(t) \to \mathbf r(t)$.

In the theories particle and antiparticle do not mix (e.g. Schroedinger's Equation, Pauli's Equation), charge symmetry is almost trivial.

Dirac Theory

Dirac equation, \[
(i\partial_\mu \gamma^\mu - eA_\mu\gamma^\mu + m)\psi(x) = 0
\] describes the relativistic motion of electrons as well as positrons. $e = 1.602176565(35)\times 10^{-19} \mathrm{C}$ is just a positive number, so-called elementary charge. In general, the solution $\psi(x)$ contains both the electron state and the positron state. Here we take $g_{\mu\nu} = \mathrm{diag}\{-,+,+,+\}$.

In free particle theory where $A = 0$, there are four independent plane wave solutions: two positive energy solutions describe electrons $u_s(p) e^{+ip\cdot x}, \; s=\pm\frac{1}{2}$ and two negative energy solutions describe the positrons $v_s(p) e^{-ip\cdot x},\; s=\pm\frac{1}{2}$, $E_{\mathbf p} = \sqrt{\mathbf p^2+m^2}$. A general solution (a wave packet) is a superposition of the two pieces: $\psi(x) \equiv \psi^+(x) + \psi^-(x)$ where \[
\psi^+(x) = \sum_{s=\pm}\int \frac{\mathrm{d}^3p}{(2\pi)^32E_{\mathbf p}} b_s(\mathbf p) u_s(p) e^{+ip\cdot x}, \\
\psi^-(x) = \sum_{s=\pm}\int \frac{\mathrm{d}^3p}{(2\pi)^32E_{\mathbf p}} d^*_s(\mathbf p) v_s(p) e^{-ip\cdot x}. \] $b, d^*$ are some c-number smooth functions.

As a relativistic wave function approach, the charge conjugation would be implemented as a "unitary" spinor matrix: $C \bar C \triangleq C (\beta C^\dagger \beta) \equiv 1$. This matrix should transform a plane wave electron to a plane wave positron or vice versa, i.e. $C u_s(p)e^{+ip\cdot x} \sim v_s(p) e^{-ip\cdot x}$. We can immediately see that this is not possible if the charge conjugation spares the exponential function. We conclude that in Dirac theory, the charge conjugation must be implemented as an anti-unitary spinor operator. Thus, we require \[
C \left( u_s(p) e^{+ip\cdot x}\right) =\eta_c v^*_s(p) e^{-ip\cdot x} \\
C \left( v_s(p) e^{-ip\cdot x} \right) = \xi_c u^*_s(p) e^{+ip\cdot x} \] where $|\eta_c| = |\xi_c| = 1$. For simplicity, we'll take these constant phases to unity. The charge conjugated field is denoted as, \[
C\psi(x) \equiv \psi^c(x) = \sum_{s=\pm}\int \frac{\mathrm{d}^3p}{(2\pi)^32E_{\mathbf p}} \Big[ b^*_s(\mathbf p) v^*_s(p) e^{-ip\cdot x} + d_s(\mathbf p) u^*_s(p) e^{+ip\cdot x} \Big]
It is conventional to define the unitary part of $C$ by a new spinor matrix $\mathcal C$ (curly C),
\[ C \psi(x) \triangleq \mathcal C \beta \psi^*(x) \equiv \mathcal C \bar\psi^T(x) \] (Here adding a $\beta$ is just a convention.). It is easy to see, $\mathcal C$ is unitary $\mathcal{C} \bar{\mathcal{C}}= 1$. An example of the choice of $\mathcal C$ is (Srednicki p. 242, 2007), \[
\mathcal C =
0 & -1 & 0 & 0 \\
+1 & 0 & 0 & 0 \\
0 & 0 & 0 & +1 \\
0 & 0 & -1 & 0
\] The theory must be invariant under the charge conjugation, or $\mathcal L \overset{C}{\to} \mathcal L$, which implies $\mathcal C \gamma_\mu = - \gamma_\mu^T \mathcal C$. Applying to the $u,v$ spinors, $\mathcal C \bar u^T_s(p) = v_s(p), \mathcal C \bar v^T_s(p) = u_s(p)$.

Heuristically, the transformation can be viewed as a two-step procedure:
  1. swap the particle species, by exchanging $u_s(p)$ and $v_s(p)$ (and of course also the sign of the charge);
  2. reverse the time, by conjugating the exponential factor. 

As stated before, the E&M field transforms under the charge conjugation as $A^\mu \overset{C}{\to} -A^\mu$. Then the charge conjugated of the Dirac wave equation is: \[
(i\partial_\mu \gamma^\mu {\color \red +} eA_\mu\gamma^\mu + m){\color \red {\psi^c(x)}} = 0
\] Note the sign change. It is easy to check that if $\psi(x)$ satisfies the original Dirac wave equation,
$\psi^c(x)$ satisfies this equation, which just confirms that the Dirac theory is invariant under the charge conjugation.

To compare with the classical charge conjugation, the dynamical equation (the equation of motion) is still invariant under the charge conjugation. But the solution of it would change under the charge conjugation, $\psi(x) \overset{C}{\to} \psi^c(x)$.

Quantum field theory

In quantum field theory, charge conjugation is implemented as a unitary operator, which we shall call $C$ (not be confused with the anti-unitary spinor operator introduced in the previous section). The definition is simple (We have also taken a particular choice of a possible phase factor. See Weinberg, 2005): \[
C^{-1} b_s(p) C = d_s(p), \quad C^{-1} d_s(p) C = b_s(p), \quad C^{-1} a_\lambda(p) C = - a_\lambda(p), \] where $b_s(p), d_s(p), a_\lambda(p)$ are the electron, positron and photon annihilation operators, respectively.

For the free Dirac field, \[

\psi(x) = \sum_{s=\pm}\int \frac{\mathrm{d}^3p}{(2\pi)^32E_{\mathbf p}} \Big[ b_s(\mathbf p) u_s(p) e^{+ip\cdot x} + d^\dagger_s(\mathbf p) v_s(p) e^{-ip\cdot x} \Big].
\] It can be shown, after some algebra (e.g., Srenicki p.225) that for the quantum fields, \[

C^{-1} \psi(x) C = \mathcal C \bar\psi^T(x) \equiv  \psi^c(x), \quad C^{-1} A^\mu(x) C = - A^\mu(x), \quad C^{-1}\varphi(x)C = \varphi^*(x). \] Then, it can be shown immediately that the Lagrangian is invariant under the charge conjugation.

Alot of notations are abused, although they are different quantities (quantum field theoretical vs. Dirac relativistic wavefunctional). Superficially, this resembles the Dirac equation result introduced in the previous section, although we should point out that here $\psi(x)$ is a field operator instead of a relativistic wave function. But this similarity means that we can do it one way or another. The same answer would be obtained. That's also the reason of the heavy abuse of notations.

To compare with the classical case and the Dirac case, the solution of the dynamical equation changes, $\psi(x) \overset{C}{\to} \psi^c(x)$. Notation-wise, this is similar to the Dirac case. See Table 1.
Table 1

Figure 1. $\langle p s \left| J^\mu(x) \right| p' s'\rangle \equiv e^{-i(p-p')\cdot x} \bar u_s(p)e \Gamma^\mu u_{s'}(p')$

A theory invariant under charge conjugation does not automatically implies a symmetric solution. The symmetry could be broken by either explicit theory (e.g. weak interaction, $\theta$-term etc.), or spontaneous symmetry breaking. In general, the charge distribution, the matrix element of the charge current operator between two physical states can be written as, \[
\langle p s \left| J^\mu(x) \right| p' s' \rangle \equiv e \bar u_s(p) \Gamma^\mu u_{s'}(p') \exp[-i(p-p')\cdot x].
\] Here $\Gamma^\mu = \Gamma^\mu(p-p')$ is some spinor operator. $\Gamma^\mu$ can be represented by the spinor basis, \[ \Gamma^\mu(q) = F_1(q^2) \gamma^\mu + F_2(q^2) \frac{i}{2m_e} \sigma^{\mu\nu} q_\nu + F_3(q^2) \frac{1}{2m_e}\gamma_5 \sigma^{\mu\nu} q_\nu,
\] here $q = p-p'$, $\sigma^{\mu\nu} = \frac{i}{2} [ \gamma^\mu, \gamma^\nu ]$. Under the charge conjugation,
$\gamma^\mu \overset{C}{\to} \mathcal C^{-1}(\gamma^\mu)^T \mathcal C = -\gamma^\mu$;
$\sigma^{\mu\nu} \overset{C}{\to} \mathcal C^{-1}(\sigma{^\mu\nu})^T \mathcal C = -\sigma^{\mu\nu}$;
$\gamma_5\sigma^{\mu\nu} \overset{C}{\to}\mathcal C^{-1}(\gamma_5\sigma{^\mu\nu})^T \mathcal C = -\gamma_5\sigma^{\mu\nu}$.
    The first term corresponds to charge with the charge normalization $F_1(0) = 1$. The second term represents the magnetic moment. $F_2(0) = \frac{\alpha}{2\pi} + \mathcal O(\alpha^2)$. The third term represents an electric dipole moment $d_e = \left|\frac{1}{2m_e}F_3(0)\right| < 10^{-27} e\cdot$ cm.

    Jul 9, 2014

    The Gravity Train Revisited


    The gravity train is a conceptual of design of transportation making use of the gravity of the planet, first proposed by 17th century scientist Robert Hooke in a letter to Issac Newton. It features a tunnel through the Earth.

    Suppose a straight tunnel was drill through the Earth. A train operating solely on gravity starts from one end of the tunnel. Assuming there is no friction inside the tunnel, and the density of the Earth is uniform. The problem asks how long it takes for the train to arrive at the other end.

    In the classic solution, the spin (rotation) of the Earth is not considered. (Strictly speaking, the rotation effect can be factorized in by using a locally measured gravitational acceleration $g$). The answer then is $t = \pi \sqrt{\frac{R_E}{g}} \simeq 42 \mathrm{min.}$, regardless of the exact location of the tunnel, where $R_E \simeq 6371 \mathrm{km}$ is the radius of the Earth, $g \simeq 9.8 \mathrm{m/s^2}$ is the gravitational acceleration. The rotation of the planet introduces a second time scale: $T_s \simeq 24 \mathrm{hr} \simeq 17 T_G$, where  $T_G = 2\pi \sqrt{\frac{R_E}{g}}$. The separation of the scales validates the neglecting of the Earth spin in the classical solution. Nevertheless, in this essay we want to look at the correction effect introduced by the spin of the Earth.

    Fig. 1

    General Settings

    Let the center of the Earth be the origin of our coordinate system. The tunnel is characterized by the position of the center of the tunnel $\vec d$, and the orientation of the tunnel $\hat \tau$, $\vec d \cdot \hat \tau = 0$. Let $\vec{r} = \vec{d} + x \hat{\tau}$ be the position of the train, where $x$ is the distance of the train from the center of the tunnel, $\vec{v} = \dot x\hat\tau$ the velocity of the train. The spin of the Earth is described by an angular velocity vector $\vec\Omega$. Suppose $\Omega \cos \theta \equiv \vec \Omega \cdot \hat \tau$, $\Omega d \cos\phi = \vec\Omega \cdot \vec d$.

    In the reference frame of the Earth (a non-inertial frame), there are four forces acting on the train: the gravity $\vec F_G = -\frac{GM_Em}{R^3_E} \vec r \equiv -mg \frac{\vec r}{R_E}$, the centrifugal force $\vec F_e = - m \vec\Omega \times (\vec\Omega \times \vec r)$, the Coriolis force $\vec F_c = 2 m \vec v \times \vec \Omega$, and the contact force. Since there is no friction, the contact force simply counter all forces in the direction that perpendicular to the tunnel. We only need to consider the force along the tunnel. $\hat \tau \cdot \vec F_G = -m g \frac{x}{R_E}$, $\hat \tau \cdot \vec F_e = m \Omega^2 x - m \Omega^2 (d \cos\phi + x \cos \theta)$, $\hat \tau \cdot \vec F_c = 0$. The total force along the tunnel is $F = -mg\frac{x}{R_E} + m \Omega^2 ( (1-\cos\theta)x - d\cos\phi)$.

    The equation of motion thus reads, \[
    \ddot x = \left( -\frac{g}{R_E} + \Omega^2 (1-\cos\theta) \right) x - \Omega^2 d \cos\phi
    \] The solution is a harmonic oscillator with angular frequency $\omega = \frac{g}{R_E} - \Omega^2 (1-\cos\theta)$ or period $T = 2\pi \sqrt{\frac{R_E}{g - \Omega^2R_E(1-\cos\theta)}}$. The center of the harmonic oscillator is not in the middle of the tunnel. In stead, it is shifted away from there by $\Delta = \frac{\Omega^2dR_E\cos\phi}{g - \Omega^2 R_E(1-\cos\theta)}$.

    To calculate the travel time, we need to consider two cases. If the train starts from the high altitude side, it arrives at the low altitude terminal in time $t = \pi \sqrt{\frac{R_E}{g - \Omega^2R_E(1-\cos\theta)}}\left(1 - \frac{1}{\pi}\arccos\frac{\sqrt{R_E^2-d^2}-\Delta}{\sqrt{R_E^2-d^2}+\Delta} \right)$. If the train starts from low altitude terminal (farther from the axis of the Earth) with zero velocity, it would never reach the high altitude side. Some minimum velocity is needed in order to arrive the other side, $v_{\min} = 2 \Omega \sqrt{\left( g - \Omega^2R_E (1-\cos\theta) \right) d\cos\phi} \sqrt[4]{1-\frac{d^2}{R^2_E}}$. by doing so, it also takes time $t$ to arrive at the destination.


    The spin of the Earth affects the problem in two ways. The correction in the period of the oscillator is about $\sim \frac{\Omega^2 R_E}{g}\sin^2\frac{\theta}{2} = 0.345\% \sin^2\frac{\theta}{2}$. The second correction is through the displacement of the center of the harmonic oscillator. We compare $\Delta$ and $l = \sqrt{R_E^2 - d^2}$ half the length of the tunnel. $\Delta \simeq \frac{\Omega^2 R_E}{g} \cos\phi d = 0.345371\% \cos\phi \sqrt{R_E^2 - l^2}$. Compare $\Delta_{\max}$ (with $\phi = 0$) and $l$ (see Fig. 2). The correction from $\Delta$ is only small for tunnels either of thousands of kilometers long or well oriented (i.e. choosing a good $\phi$). Thousands of kilometers long tunnel would be for transcontinental express.

    Fig. 2, the numbers for the tunnel length $l$ and maximum shift of the harmonic oscillator center $\Delta_{\max}$ are in kilometers.

    We have only considered the straight tunnels. It turns out, straight tunnels are not the fastest ones for gravity train. The fastest tunnels are the brachistochrone tunnels. The correction due to the Earth rotation can be considered accordingly.