10.5: Is $c$ the Speed of Light?

We have seen that an electromagnetic wave in the $z$ direction satisfying Maxwell’s equations in free space has the dispersion relation (8.47), so that light, at least in vacuum, travels at the speed of light. But is the theory right? How do we test the dispersion relation? In fact, the most sensitive tests of Maxwell’s equations do not involve traveling waves. They come from observations of magnetic fields that extend over astrophysical distances (like the galaxy!). However, there is an interesting, if not very sensitive, way of looking for corrections to (8.47) that involves the speed of light directly. Before discussing this, let us digress briefly to talk in more detail about photons, the particles of light that we described briefly in chapter 8.

Light is a wave phenomenon, as we have seen. Indeed, the wave properties of light are obvious in our everyday experience. It is less obvious from our everyday experience, but equally true, that light also consists of photons. This becomes obvious when you work with light at very low intensities and/or very high energies. That both of these statements can be true simultaneously is one of the (many) miracles of quantum mechanics.

Quantum mechanics tells us that all particles have wave properties. A particle with momentum $p$ and energy $E$ has an associated angular frequency and angular wave number related by $$E=\hbar \omega, \quad p=\hbar k ,$$

where $\hbar$ is Planck’s constant divided by $2 \pi$. This combination appears so ubiquitously in quantum mechanics that it has its own symbol, and we physicists almost always use $\hbar$ rather than $h$. The reason is just that $h$ is related to the frequency $ν$ rather than the angular frequency, $\omega$, and we have seen that $\omega$ is the more convenient measure for most purposes. In addition, the energy and momentum of the particle are related as follows: $$E^{2}=p^{2} c^{2}+m^{2} c^{4}, \quad v=c \frac{p c}{E}$$

where $m$ is the rest mass and $v$ is the classical velocity.

If we put (10.105) into (10.106), we get a dispersion relation for the quantum mechanical wave associated with the particle $$\omega^{2}=c^{2} k^{2}+\omega_{0}^{2}, \quad \omega_{0}=\frac{m c^{2}}{\hbar} .$$

The classical velocity is the group velocity of the quantum mechanical wave! $$v=\frac{\partial \omega}{\partial k}=c^{2} \frac{k}{\omega}=c \frac{p c}{E}$$

In fact, particles, in a quantum mechanical picture, correspond to wave packets that move with the group velocity.

The quantum mechanical dispersion relation, (10.107), agrees with (8.47) only if $m = 0$. Thus we can restate the question of whether (8.47) is correct by asking “Is the photon mass really zero?”

It would seem that we ought to be able to test this idea by looking at two photons with different frequencies emitted at the same time from a far away object and checking whether they arrive at the same time. There is an obvious flaw in this scheme. If the object is so far away that we cannot get there, how do we know that the two photons were emitted at the same time? In fact, astrophysics has provided us with a way around this difficulty. We can look at pulsars. Pulsars are (presumably) rotating neutron star remnants of supernova explosions that emit light toward the earth at regular intervals. For example, pulsar 1937+21 is so regular that the departure time of photons can be determined to within a few microseconds ($\mu \mathrm{s}$).⁵ It is also about 16,000 light years away, so the photons with the higher frequency (the faster ones) have plenty of time to get ahead. When this experiment is done, one finds a nonzero $\omega_{0}$, of about $1.7 \times 10^{4} \mathrm{~s}^{-1}$, corresponding to a mass of about $1.26 \times 10^{-49} \mathrm{~g}$. That seems like a rather small mass, but in fact, it is ridiculously large for a photon. From studies of the galactic magnetic field, we suspect that it is less than $4 \times 10^{-65} \mathrm{~g}$!⁶ Thus something else is going on.

The problem with this measurement as a test of the dispersion relation is that there are electrons lying around out there — free electrons in interstellar space ($10^{-1}$ to $10^{-2} \mathrm{~cm}^{-3}$). These electrons in space will wiggle in the $E$ field — this will produce a current density that will affect Maxwell’s equations, and that, in turn, will affect the dispersion relation. Let us analyze the effect of this dilute plasma assuming that the electron density is constant. Then (at least for the long wavelength radio waves of interest in these experiments) we can still use translation invariance to understand what is happening. Consider a plane wave in the $z$ direction and suppose that the electric field of the plane wave is in the $x$ direction. Then it is still true that at a given $\omega$ $$E_{x}(\vec{r}, t)=E_{0} e^{i(k z-\omega t)}, \quad B_{y}(\vec{r}, t)=B_{0} e^{i(k z-\omega t)} ,$$

for some $k$. To find $k$, we must look at the effect of the electric fields on the electrons, and then go back to Maxwell’s equations. The fields are very small, and for small fields the induced electron velocities, $v$ are small. Thus we can neglect $B$. Then the force on an electron at the point$(\vec{r}, t)$ is $$F_{x}(\vec{r}, t)=e E_{x}(\vec{r}, t)=e E_{0} e^{i(k z-\omega t)}=m a_{x}(\vec{r}, t)$$

The displacement of the electron has the same form: $$d_{x}(\vec{r}, t)=d_{0} e^{i(k z-\omega t)}$$

which implies $$a_{x}(\vec{r}, t)=-\omega^{2} d_{0} e^{i(k z-\omega t)}$$

comparing (10.110) and (10.112) gives $$d_{0}=-\frac{e E_{0}}{m \omega^{2}} .$$

Thus the electrons are displaced $180^{\circ}$ out of phase with the electric field and in the same direction. Then the electron velocity is $$v_{x}=\frac{i e E_{0}}{m \omega} e^{i(k z-\omega t)} .$$

The movement of the electrons gives rise to a current density:⁷ $$\mathcal{J}_{x}=\frac{i e^{2} N E_{0}}{m \omega} e^{i(k z-\omega t)}$$

where $N$ is the electron number density.

Putting this into the relevant Maxwell’s equations, we find $$k E_{0}=\omega B_{0}, \quad-k B_{0}=-\omega \mu_{0} \epsilon_{0} E_{0}+\mu_{0} \frac{e^{2} N E_{0}}{m \omega} ,$$

or using $c=1 / \sqrt{\mu_{0} \epsilon_{0}}$, (8.47), $$B_{0}=\frac{k}{\omega} E_{0}, \quad-\frac{k^{2}}{\omega}=-\frac{\omega}{c^{2}}+\frac{e^{2} N}{c^{2} m \epsilon_{0} \omega} ,$$

or solving for $\omega^{2}$ $$\omega^{2}=c^{2} k^{2}+\omega_{0}^{2}, \quad \text { with } \quad \omega_{0}^{2}=\frac{e^{2} N}{\epsilon_{0} m} .$$

The constant $\omega_{0}$ in (10.118) is called the “plasma frequency.” The amazing thing is that it looks just like a photon mass. For $N \approx 10^{-2} \mathrm{~cm}^{-3}$, this is consistent with the observation from the pulsar.

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⁵See G. Barbiellini and G. Cocconi, Nature 329 (1987) 21.
⁶Chibisov, Soviet Physics - Uspekhi, 19 (1986) 624.
⁷Notice that the result is inversely proportional to the electron mass. This why we are concentrating on electrons rather than protons. The protons don’t move as fast!