A SURVEY OF PHYSICS
Physics 401 - Spring 2015

HOME THE HIGGS PARTICLE   >   ELECTROMAGNETISM

George Siopsis - April 21, 2015

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I start by reviewing the familiar electromagnetic forces. Electromagnetism is governed by the four Maxwell equations

 $\vec\nabla\cdot\vec E = \frac{\rho}{\epsilon_0}$ (Gauss's Law) $\vec\nabla\times\vec E = - \frac{\partial\vec B}{\partial t}$ (Faraday's Law) $\vec\nabla\cdot\vec B = 0$ (No-name Law) $\vec\nabla\times\vec B = \mu_0 \vec J + \mu_0\epsilon_0 \frac{\partial\vec E}{\partial t}$ (Ampere's Law)

Maxwell's contribution was the last term on the right-hand side of the last equation, which he added to make the system consistent. In doing so, he unified electricity and magnetism. This was an achievement that had far-reaching consequences. It explained light as electromagnetic radiation. Let's see how: suppose there are no charges, so set $\rho = 0$, $\vec J = \vec 0$. We obtain $\mu_0\epsilon_0 \frac{\partial^2 \vec E}{\partial t^2} = \vec\nabla \times \frac{\partial\vec B}{\partial t} = - \vec\nabla\times (\vec\nabla\times\vec E) = \nabla^2 \vec E - \vec\nabla (\vec\nabla\cdot \vec E) = \nabla^2 \vec E$ This is the wave equation. To solve it, consider a plane wave $\vec E = \vec E_0 e^{i(\vec k\cdot \vec r - \omega t)}$ Substituting into the wave equation, we obtain $-\mu_0\epsilon_0 \omega^2 = - k^2$ where $k= |\vec k|$, showing that the speed of the wave is $c = \frac{\omega}{k} = \frac{1}{\sqrt{\mu_0\epsilon_0}}$ which coincides with the speed of light. Surprisingly, $c$ is a constant of Nature, a fact that violates Newton's Laws. As Einstein first realized, Maxwell is right and Newton is wrong - Nature is relativistic.

Nature is also quantum: the electric and magnetic fields are not continuous, but consist of point particles, the photons. Each photon has energy and momentum, respectively, $E = \hbar\omega \ \ , \ \ \ \ \vec p = \hbar \vec k$ therefore, $E = c |\vec p|$ showing that the photon is massless, has no rest mass, therefore, it always travels at the speed of light and never slows down. This fact can be seen directly from the wave equation, if we substitute $- i\hbar \vec \nabla \to \vec p \ \ , \ \ \ \ i\hbar \frac{\partial}{\partial t} \to E$ This kind of substitution can be applied in reverse to a classical equation in order to obtain the corresponding quantum mechanical description. E.g., for a free particle, $E = \frac{p^2}{2m}$ leads to the Schroedinger equation $- \frac{\hbar^2}{2m} \nabla^2 \Psi = i\hbar \frac{\partial\Psi}{\partial t}$ This is, of course, the wrong equation (a fact already known to Schroedinger himself), because it is non-relativistic, but for practical purposes, it is good enough as long as no high speeds are involved. The correct equation corresponding to Einstein's equation $E^2 = p^2 c^2 + m^2 c^4$ is $\frac{1}{c^2} \frac{\partial^2\Psi}{\partial t^2} = \nabla^2 \Psi - \frac{1}{\lambda^2} \Psi$ where $\lambda = \frac{\hbar}{mc}$ is the Compton wavelength (shortest wavelength the particle of mass $m$ can have) divided by $2\pi$. In fact, this is the first equation Schroedinger wrote (even though it is now known as the Klein-Gordon equation). He abandoned it, because he could not make sense out of it, so he settled for its non-relativistic version, which is only an approximation. It took years of effort to combine quantum mechanics and the theory of relativity, but physicists succeded, and the result was quantum field theory, a triumph of the human mind in the 20th century.

Back to the photon. Even though it is a point particle, it carries a vector with it, the polarization $\vec E_0$ (just like the electron, which is also a point particle, carries spin). Notice that from Gauss's Law, setting $\rho =0$, we obtain $\vec p \cdot \vec E_0 = 0$ therefore, the polarization is always transverse (perpendicular to the direction of motion, defined by the momentum $\vec p$). Therefore, even though a vector in general has 3 degrees of freedom (three axes to point into), the polarization, and therefore the photon only has 2 degrees of freedom. If the photon had mass, we would be able to stop it and then `transversality' would be meaningless, since $\vec p = \vec 0$. A massive photon, if it exists, will have 3 degrees of freedom.

It is often convenient to express the electric and magnetic fields in terms of potentials $V$ and $\vec A$, as $\vec E = - \vec\nabla V - \frac{\partial\vec A}{\partial t} \ \ , \ \ \ \ \vec B = \vec\nabla\times\vec A$ In the simplest case, we have a point charge $q$ which creates a Coulomb potential, $V = \frac{q}{4\pi\epsilon_0 r} \ \ , \ \ \ \ \vec A = \vec 0$ Under the gauge transformation $V \to V + \frac{\partial\chi}{\partial t} \ \ , \ \ \ \ \vec A \to \vec A - \vec\nabla \chi$ where $\chi (t, \vec r)$ is an arbitrary function of space and time, the fields $\vec E$ and $\vec B$ do not change. Two potentials related to each other by a gauge transformation represent the same physical system. We say that the equations of electromagnetism are gauge invariant.

This enormous amount of symmetry looks superficial, because the potentials were merely introduced for convenience and no compelling physical reason. It turns out that the potentials are not just convenient mathematical devices when quantum mechanics comes into play. Indeed, consider a point charge $q$. It feels the Lorentz force $\vec F = q (\vec E + \vec v\times \vec B)$ What Hamiltonian can this be derived from? This question is not so important in classical physics, but if one wants to define the quantum theory, the Hamiltonian becomes very important. It turns out that the Hamiltonian that gives the Lorentz force is $H = \frac{1}{2m} (\vec p - q\vec A)^2 + qV$ You can show this by applying Hamilton's equations, $\dot{x} = \frac{\partial{H}}{\partial p_x} \ \ , \ \ \ \ \dot{p}_x = - \frac{\partial H}{\partial x}$ twice to find $F_x = m\ddot{x}$, and similarly for the other two components. Thus, for a charged particle, in the quantum theory we ought to replace $-i\hbar \vec\nabla \to -i\hbar \vec\nabla - q\vec A \ \ , \ \ \ \ i\hbar \frac{\partial}{\partial t} \to i\hbar \frac{\partial}{\partial t} - qV$ to accommodate its interaction with the electric and magnetic fields. Thus, the potential is elevated to a prominent role in the quantum theory. Nevertheless, the system must remain gauge invariant, otherwise we cannot make sense of the theory.

As an example, consider the Klein-Gordon equation that describes a relativistic particle of mass $m$ and spin 0 (because there is no polarization, and only 1 degree of freedom). Endowing the particle with charge $q$, we obtain the modified wave equation $\frac{1}{c^2} \left( \frac{\partial}{\partial t} - \frac{q}{i\hbar} V \right)^2 \Psi = \left( \vec\nabla + \frac{q}{i\hbar} \vec A \right)^2 \Psi - \frac{1}{\lambda^2} \Psi$ Check that it is invariant under the gauge transformation $V \to V + \frac{\partial\chi}{\partial t} \ \ , \ \ \ \ \vec A \to \vec A - \vec\nabla \chi \ \ , \ \ \ \ \Psi \to e^{-iq\chi/\hbar} \Psi$

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