Feeds:
Posts

On Tuesday of last week I blogged about the H.E.S.S. telescopes in Namibia which detect Cherenkov radiation produced by high energy rays (cosmic rays or gamma rays) entering the Earth’s atmosphere from outer space. These high energy rays can hit atoms in the Earth’s atmosphere; it is these collisions which can give rise to Cherenkov radiation. Today I am going to give more detail about Cherenkov radiation; this is based on my lecture notes from teaching High Energy Astrophysics at Cardiff University, a Masters level astrophysics course.

## Who was Cherenkov and why does he have a type of radiation named after him?

Cherenkov radiation is named in honour of Pavel Alekseyevich Cherenkov, a Soviet scientist who won the 1958 Nobel Prize for his experimental observation of this type of radiation.

Figure 1: Pavel Alekseyevich Cherenkov (1904-1990), who won the 1958 Nobel Prize in Physics for his theoretical prediction of what we now call Cherenkov radiation

Cherenkov worked at the Lebedev Physical Institute, which is based in Moscow. His observation of Cherenkov radiation was made in 1934, and 24 years later, in 1958, he won the Nobel Prize for this work. He shared the prize with fellow Russian-scientists Igor Tamm and Ilya Frank who worked out the theory of the radiation. Cherenkov was also awarded two Stalin prizes, the first in 1946 and the second in 1952.

Figure 2: The characteristic blue light produced by Cherenkov radiation in a nuclear reactor

What Cherenkov observed in 1934 was a blue light being give off by a bottle of water which was being subjected to bombardment by radioactivity (what we call irradiated). By the 1950s, with the advent of nuclear reactors, this blue glow had become a common sight – an example is shown in Figure 2. It shows some nuclear fuel rods submerged in water (the moderator), and  the interaction of the high-energy radiation from the fuel rods with electrons in the water causes the water to glow blue – this is Cherenkov radiation.

## The details of Cherenkov radiation

Cherenkov radiation is only one type of interaction between either high-energy particles, or high-energy radiation, and matter. Other such interactions include when a high-energy nucleon (proton or neutron) interacts with an electron, or a high-energy electron interacts with an electron, or a high-energy electron interacts with a gamma-ray. But, in this blogpost I am going to concentrate on Cherenkov radiation.

Cherenkov radiation is produced when a charged particle travels faster than the speed of light in that medium. As nothing can travel faster than the speed of light in a vacuum, Cherenkov radiation can only happen in a medium such as air, water or glass. Many of you are probably familiar with the idea of the refractive index $n$ of a medium. It crops up for example in Snell’s law, which allows us to calculate the angle through which radiation deviates when it travels from one medium to another. Snell’s law states that

$\frac{ n_{2} }{ n_{1} } = \frac{ sin \theta_{1} }{ sin \theta_{2} }$

where $n_{1} \text{ and } n_{2}$ are the refractive indices of the two media and $\theta_{1} \text{ and } \theta_{2}$ are the angles between the direction of the ray and the normal, as shown in Figure 3.

Figure 3: Snell’s law gives the relationship between the angles between the direction of the rays and the refractive indices $n$ of the two media

Snell’s law is usually taught in A-level physics, but what is not taught is the relationship between the refractive index and the speed of light. It is a very simple relationship; if $n_{1}$ is the refractive index in medium 1, we can simply write

$n_{1} = \frac{ c }{ v_{1} }$

where $v_{1}$ is the speed of light in that medium (and $c$ is, of course, the speed of light in a vacuum). We should note that, in general, the refractive index $n$ is a function of wavelength; this means that red light and blue light travel at different speeds in e.g. glass (or water). This is why these media disperse (split up) light and you see the colours of the rainbow when light passes through a prism, the blue and red lights are refracted different amounts in the glass because they travel at different speeds.

This page gives a list of refractive indices for different media. For pure water at 20 Celsius, $n=1.3325$, whereas for heavy water (deuterium oxide, each hydrogen atom is replaced with deuterium which has a neutron in the nucleus in addition to the proton present in hydrogen), the refractive index according to this page is $n=1.328$, so slightly less than for pure water. Heavy water is used as a moderator in nuclear reactors as the additional neutron helps slow down the fast neutrons from the nuclear reactions more effectively than pure water.

Assuming we are dealing with heavy water, this means that the speed of light in heavy water is $v = 3 \times 10^{8} / 1.328 = 2.259 \times 10^{8}$ metres per second. If a charged particle travels faster than this speed in heavy water, it will emit Cherenkov radiation. A cone of constructive interference will form at an angle $\theta$ to the direction of travel of the charged particle.

If $v$ is the speed of the particle in the medium and $v_{1}$ is the speed of light in that medium (where $v_{1} = c/n$), we can write that

$\boxed{ cos \theta = \frac{ v_{1} }{ v } = \frac{ c }{ nv } } \; \; \text{(1)}$

Clearly, as $cos \theta$ can never be greater than 1, $v$ (the speed of the particle) cannot be less than $c/n$. This is the threshold, Cherenkov radiation will only happen if $v > c/n$.

Figure 4: In Cherenkov radiation, a wavefront forms at an angle $\theta$ to the direction of travel of the charged particle. In this figure, the particle is travelling at a velocity $v$ horizontally; the radiation in the medium travels at a velocity $v_{1}$

This cone of light is very analogous to the cone of sound which forms when an object travels faster than the speed of sound in that medium. In Figure 5 we show an aeroplane travelling at less than the speed of sound (on the left), at the speed of sound (in the middle) and in excess of the speed of sound (on the right). The cone of sound produced at speeds greater than the speed of sound has the same relationship for the angle as with Cherenkov radiation, it is given by the ratio of the speed of the object to the speed of the waves in that medium.

Figure 5: The sound waves created as an aeroplane travels at different speeds. When it exceeds the speed of sound (figure on the right), a cone of sound is formed around the aeroplane, the angle depends on the ratio of the aeroplane to the speed of sound, just like the angle of the cone of light in Cherenkov radiation

Another way to illustrate the radiation given off in Cherenkov radiation is shown in figure 6. It is just a different way of showing the same thing as shown in Figure 4, but may be clearer to some of you. In Figure 6 the wavefronts are shown after a time $t$, where the particle will have travelled a distance $vt$.

$c/n$ in Figure 6 is the same as $v_{1}$ in Figure 4, the speed of the radiation in the medium. $\beta$ is defined as $v/c$ which means that $\beta c = v$, the speed of the particle in the medium. Figure 6 shows nicely why a cone of light develops which shines forwards at an angle of $\theta$ to the direction of travel of the charged particle.

Figure 6: This is a different way of illustrating the emission of Cherenkov radiation.

The energy emitted per unit length travelled by an electron per angular frequency interval $d\omega$ is given by the Frank-Tamm formula (see here)

$\boxed{ \frac{ dE }{ dx d\omega } = \frac{ e^{2}}{ 4 \pi } \mu(\omega) \omega \left( 1 - \frac{ c^{2} }{ v^{2} n^{2}(\omega) } \right) }$

where $v$ is the speed of the particle, $e$ is the charge on an electron, $\mu(\omega)$ is the magnetic permeability of the medium (which may be frequency dependent), $n(\omega)$ is the refractive index of the medium (which may be frequency dependent) and $c/n(\omega)$ is the speed of light in the medium. Frank and Tamm are the two other physicists who shared the 1958 Nobel with Cherenkov.

As we saw for Equation (1), which gave the angle of the wavefronts to the direction of the charged particle, the specific intensity will go to zero if the speed of the particle $v$ is equal to $c/n_{\nu}$, and becomes negative if $v$ becomes less than $c/n_{\nu}$.

Rather than show the energy, it is more usual to calculate the number of photons emitted per particle length per frequency interval or wavelength interval. We are going to show it per wavelength interval $d\lambda$. We start with Equation (2), and first we change from angular frequency $\omega$ to temporal frequency $\nu$. To do this we remember that $\omega = 2 \pi \nu$, and so $d \omega = 2 \pi d\nu$. This gives us

$\frac{ dE }{ dx d\nu } = \frac{ e^{2}}{ 4 \pi } \mu(\nu) 2 \pi \nu \left( 1 - \frac{ c^{2} }{ v^{2} n^{2}(\nu) } \right) \cdot 2 \pi$

which simplifies to

$\frac{ dE }{ dx d\nu } = \frac{ \pi e^{2}}{ 1 } \mu(\nu) \nu \left( 1 - \frac{ c^{2} }{ v^{2} n^{2}(\nu) } \right)$

We are now going to assume that the permeability $\mu(\nu)$ is unity. To go from the energy to the number of photons we remember that the energy of each photon is $h \nu$, where $h$ is Planck’s constant, so $dE = N h \nu$, where $N$ is the number of photons. So, we can write

$\frac{ d^{2}N }{ dx d\nu } = \frac{ \pi e^{2} }{ h \nu } \nu \left( 1 - \frac{ c^{2} }{ v^{2} n^{2}(\nu) } \right)$

We are now going to express this per wavelength interval. Remember $c = \lambda \nu$ so $\nu = c/\lambda$ and $d\nu = - (c/\lambda^{2})d\lambda$. We can ignore the minus sign, this is just telling us that wavelength decreases as frequency increases, so then we can write

$\frac{ d^{2}N }{ dx d\lambda } = \frac{ \pi e^{2} }{ h } \frac{ \lambda }{ c } \frac{ c }{ \lambda }\left( 1 - \frac{ c^{2} }{ v^{2} n^{2}(\lambda) } \right) \frac{ c }{ \lambda^{2} }$

which simplifies to

$\frac{ d^{2}N }{ dx d\lambda } = \frac{ \pi e^{2} }{ hc \lambda^{2} } \left( 1 - \frac{ c^{2} }{ v^{2} n^{2}(\lambda) } \right)$

Finally, we multiply by $4 \pi$, the solid angle in a sphere, to give us our final expression

$\boxed{ \frac{ d^{2}N }{dxd\lambda} = \frac{ 4 \pi^{2} e^{2} }{ h c \lambda^{2} } \left( 1 - \frac{ c^{2} }{ v^{2} n_{\lambda}^{2} } \right) } \; \; (3)$

In Figure 7 I use Equation (3) to show the numbers for two different particle speeds, (a) $v=2.8 \times 10^{8}$ m/s and (b) $2.3 \times 10^{8}$ m/s. I have illustrated the visual part of the spectrum by the shaded box.

Figure 7: the number of photons emitted in Cherenkov radiation per unit length per wavelength interval for two different particle velocities – $v=2.8\times 10^{8}$ m/s (green) and (b) $v=2.3 \times 10^{8}$ m/s (blue), both for heavy water with $n_{\lambda}=1.328$. The shaded area represents the visible part of the spectrum.

As can be seen from Figure 7, (a) the number of photons is more when the charged particles are moving quicker and (b) the number of photons increases with shorter wavelength, which is why Cherenkov radiation has its characteristic blue appearance. As the number goes as $1/\lambda^{2}$, the number of blue photons is roughly four times as many as the number of red photons. In fact, the intensity is even higher in the ultraviolet, but of course we cannot see this light with our eyes.

### 9 Responses

1. A little difficult for me as a layman but I think I have the gist of the theory.
Does this imply that the colours we see in stars are due to variations of the speed of light as it escapes from the star?
At the moment we have the great constellation of Orion due south with the red giant betelgeuse rcontrasted with the brilliant star of Rigel.

• No, those colours have nothing to do with Cherenkov radiation. I’ve blogged about why stars have different colours, I’ll find the post(s). It’s all to do with surface temperature

• Forgive my ignorance but if stars get their energy from nuclear action surely they must give off Cherenkov radiation. It still seems odd that only some stars appear to be blue.
Sirius at the moment is brilliant. I’m veiwing from southern England.
Perhaps the blue colour is masked by a temperature at the surface effect.

• No, the photons take many thousands of years to get from the core to the surface. There are no charged particles travelling faster than (c/n) in the Sun or inside any other star. There is no Cherenkov radiation involved in stellar emission.

• Thanks for that reply; when I first read your post I thought where neuclear reactions were taking place there would be Cherenkov radiation.
I overlooked what goes on inside reactors is man made but in stars it is a natural process.
Knowing that stars are relatively small I find it difficult to see how photons can take so long to surface unless they are held tight by gravity.
My science is quite rudimentary having had no higher education.

• Photons random walk inside stars as the interiors of stars are plasmas (ionised gases), so the photons cannot get anywhere as they scatter off of the free electrons. The photons eventually escape at the “surface”, which is where the hydrogen is cool enough to become neutral, allowing the photons to escape.

• I won’t badger you with more questions , interestingly science seems to uncover a multitude of questions but I suspect that is its nature to question.
Thanks

• Science is all about questioning!

• Only some stars appear blue because only some stars are hot enough to appear blue. Blue stars last for much less time than cooler, red stars. So, they are quite rare.