So since I have a strong interest in particle physics as well as nuclear physics - and
since someone raised some ideas about measurement applications recently - I thought
it was time to take a look back at the physics behind Cherenkov radiation and that
it might be a good idea to summarize the basics again as a blog entry.
Addition in June 2021: The following photograph shows Cherenkov radiation
originating from the core of an Triga Mark-II reactor running at
approximately $250 kW$ (thermal). This setting corresponds approximately to:
- $2.1 * 10^{17} \frac{n}{m^2 * s}$ overall
- $6.1 * 10^{16} \frac{n}{m^2 * s}$ for $E < 0.55 eV$
- $7.6 * 10^{16} \frac{n}{m^2 * s}$ for $E > 0.1 MeV$
- $4.0 * 10^{16} \frac{n}{m^2 * s}$ for $E > 1 MeV$

What is Cherenkov radiation anyways?
Basically the (historic) idea behind Cherenkov radiation is based on the fact
that a charged particle that passes through a polarizable medium - i.e. a
dielectric like water - does polarize the medium for a short time. The electromagnetic
field change is spreading out with the maximum speed of an electromagnetic wave
inside the material $c_m$. Usually - for particles slower than the propagating
wave - spherical waves emitted from different positions reached by the charged
particle destructively interfere periodically so no visible light is emitted.
For particles that move faster than the local speed of light - note that this is
always slower than the speed of light in vacuum - a shock wave forms at which all
spherical waves interfere constructively. Inside a cone formed shape they add
up to a electromagnetic wave that might have a wavelength inside visible range.

As one can see from the geometric construction the distance traveled by the
particle $s_p = v * t$ and the distance traveled by the spherical wave $s_w = c_m * t$
form a triangle. The tangent that moves along the wavefront is perpendicular to
the radius of the spheres.

As one can see the angle at which the radiation spreads can simply be calculated
via
[
\cos(\alpha) = \frac{c_m * t}{v * t} = \frac{c_m}{v}
]
The speed of light (phase velocity) inside the medium $c_m$ can also - and
usually is - be expressed by the refractive index $n = \frac{c}{c_m}$
which leads to the expression
[
n = \frac{c}{c_m} \to c_m = \frac{c}{n} \\
\cos(\alpha) = \frac{c_m}{v} = \frac{c}{n * v}
]
Note that this construction only describes emission of Cherenkov radiation for
speeds $\frac{c}{n} \leq v \leq c$. As has been shown Cherenkov radiation also
radiates in meta materials into arbitrary directions.
Spectrum and energy loss
To calculate the spectrum (Frank-Tamm formula) one usually starts with the
Maxwell equations and Lorenz gauge.
[
\Box \vec{A}(\vec{r},t) = \mu_0 j(\vec{r},t) \\
\Box \phi(\vec{r},t) = \frac{1}{\epsilon_0} \rho(\vec{r},t)
]
The derivation of these equations is given in the blog article linked above. Then
one usually applies the Fourier transformation:
[
(\nabla^2 - \frac{1}{c^2} \partial_t^2) \vec{A}(\vec{r},t) = \mu_0 j(\vec{r},t) \\
\to (k^2 - \frac{1}{c^2} \omega^2) \vec{A}(\vec{k}, \omega) = \mu_0 j(\vec{k}, \omega) \\
(\nabla^2 - \frac{1}{c^2} \partial_t^2) \phi(\vec{r},t) = \frac{1}{\epsilon_0} \rho(\vec{r},t) \\
\to (k^2 - \frac{1}{c^2} \omega^2) \phi(\vec{k}, \omega) = \frac{1}{\epsilon_0} \rho(\vec{k}, \omega)
]
Now one is capable of modeling a single particle with charge $z * e$ moving with
speed $\vec{v}$:
[
\rho(\vec{x}, t) = z * e * \delta(\vec{x} - \vec{v} * t) \\
\vec{j}(\vec{x}, t) = \vec{v} * \rho(\vec{x},t)
]
This time domain representation can be - again - Fourier transformed into frequency
space:
[
\rho(\vec{k}, \omega) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \rho(\vec{x},t) e^{-i \omega t} dt \\
\rho(\vec{k}, \omega) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} z * e * \delta(\vec{x} - \vec{v} t) e^{-i \omega t} dt \\
\rho(\vec{k}, \omega) = \frac{1}{\sqrt{2 \pi}} z * e * \delta(\omega - \vec{k}\vec{v}) \\
\vec{j}(\vec{k}, \omega) = \vec{v} \rho(\vec{k}, \omega)
]
Inserting into the Maxwell equation for the scalar potential above yields an expression
for the scalar potential:
[
\left(k^2 - \frac{\omega^2}{c^2} \right) \psi(\vec{k}, \omega) = \frac{1}{\sqrt{2 \pi} \epsilon_0} z e \delta(\omega - \vec{k} \vec{v}) \\
\to \psi(\vec{k},\omega) = \frac{z * e * \delta(\omega - \vec{k} \vec{v})}{\sqrt{2 \pi} \epsilon_0 (k^2 - \frac{\omega^2}{c^2})}
]
The same can be done using the current $j$ and the Maxwell equation for the vector potential:
[
(k^2 - \frac{\omega^2}{c^2}) \vec{A}(\vec{k}, \omega) = \mu_0 \vec{j}(\vec{k}, \omega) \\
(k^2 - \frac{\omega^2}{c^2}) \vec{A}(\vec{k}, \omega) = \mu_0 \vec{v} \rho{j}(\vec{k}, \omega) \\
(k^2 - \frac{\omega^2}{c^2}) \vec{A}(\vec{k}, \omega) = \mu_0 \vec{v} \frac{z e}{\sqrt{2 \pi}} \delta(\omega - \vec{k}\vec{v}) \\
\vec{A} = \frac{\mu_0 \vec{v} z e \delta(\omega - \vec{k} \vec{v})}{\sqrt{2 \pi} (k^2 - \frac{\omega^2}{c^2})} \\
\vec{A} = \mu_0 \vec{v} \epsilon_0 \underbrace{\frac{z e \delta(\omega - \vec{k} \vec{v})}{\epsilon_0 \sqrt{2 \pi} (k^2 - \frac{\omega^2}{c^2})}}_{\psi(\vec{k}, \omega)} \\
\vec{A} = \mu_0 \vec{v} \epsilon_0 \psi(\vec{k}, \omega)
]
Using the knowledge of the general transformation of time derivatives and spatial derivatives
which can easily be calculated by integrating by parts (just take the derivative of the exponential
function and the integral of the derivative part) โฆ
[
\int_{-\infty}^{\infty} \partial_x f(x,t) e^{i \omega t} dt = ik f(k, \omega) \\
\int_{-\infty}^{\infty} \partial_t f(x,t) e^{i \omega t} dt = i \omega f(k, \omega)
]
โฆ this can now be used to calculate and expression for the Fourier transform of
the electric field:
[
E(\vec{r}, t) = - \vec{\nabla} \psi(\vec{r}, t) - \partial_t \vec{A} \\
E(\vec{k}, \omega) = - \int_{-\infty}^{\infty} \vec{\nabla} \psi(\vec{r}, t) e^{i \omega t} dt - \int_{-\infty}^{\infty} \partial_t \vec{A} e^{i \omega t} dt \\
E(\vec{k}, \omega) = i( - k - \frac{\omega \vec{v} \epsilon_0}{c}) \psi(\vec{k}, \omega)
]
The same can be done with the magnetic field:
[
\vec{B}(\vec{r}, t) = \vec{\nabla} \times \vec{A}(\vec{r}, t) \\
\vec{B}(\vec{k}, \omega) = \int_{-\infty}^{\infty} \vec{\nabla} \times \vec{A}(\vec{r}, t) e^{i \omega \vec{k}} dt \\
\vec{B}(\vec{k}, \omega) = i \vec{k} \times \vec{A}(\vec{r}, t) \\
\vec{B}(\vec{k}, \omega) = i \vec{k} \times \frac{\vec{v} \epsilon_0 \psi(\vec{k},\omega)}{c}
]
With this approach one can (rather) easily calculate the electric field in spatial
space by inverse Fourier transformation.
Solving this integral is a rather long operation - I might add it in the near future,
basically itโs an evaluation of the function using the delta function for the first
dimension followed by a simple integral over the second dimension as well as an
Bessel integral. One should really not try to integrate all components in a combined
fashion but one after each other. Whatโs often done in literate too is to decide
to evaluate the field at a given position, usually chosen to be $(0,b,0)$ - in which
case $b$ is called the impact parameter.
The impact parameter is a distance between an impacting particle trajectory and the
center of a potential that the projectile is approaching - for example an ion or
atomic nucleus. The choice of this impact parameter allows one to model either
a central collision ($b=0$) - for which calculations often turn out not to work
as expected - or a scattering process ($b \neq 0$) Since one assumes that sources of
potentials are usually points the probability of a head on collision is assumed to be
zero.
This choice makes the inverse Fourier transform a little easier since the first and
last component (for the choice $(0,b,0)$) vanish for the spatial coordinate:
[
\vec{E}(\omega) = \frac{1}{\sqrt{2 pi}^3} \int \int \int \vec{E}(\vec{k}, \omega) * e^{i b k_2} d^3 k
]
[
E_1(\omega) = \frac{z e i}{\sqrt{2 \pi}^3 \sqrt{2 \pi}} \int \int \int e^{i b k_2} (- k_1 - \frac{\omega v_1 \epsilon_0}{c}) \frac{\delta(\omega - v k_1)}{k^2 - \frac{\omega^2}{c^2}} d^3 k
]
The first integral over $k_1$ is simple since the evaluation of the delta
function $\delta(\omega - v k_1)$ simply yields an evaluation at $k_1 = \frac{\omega}{v}$.
Itโs also assumed that the particle travels along the $x$ direction $\vec{v} = (v, 0, 0)$.
[
E_1(\omega) = \frac{z e i}{\sqrt{2 pi}^2 \sqrt{2 \pi}} (-\frac{\omega}{v} - \frac{v \epsilon_0}{c}) \int \int e^{i b k_2} \frac{1}{\frac{\omega^2}{v^2} - \frac{\omega^2}{c^2} + k_2^2 + k_3^2} dk_2 dk_3 \\
E_1(\omega) = \frac{z e i}{\sqrt{2 pi}^2 \sqrt{2 \pi}} (-\frac{\omega}{v} - \frac{v \epsilon_0}{c}) \int_{-\infty}^{\infty} e^{i b k_2} \int_{-\infty}^{\infty} \frac{1}{(\frac{\omega^2}{v^2} - \frac{\omega^2}{c^2}) + k_2^2 + k_3^2} dk_3 dk_2
]
Using the identity
[
\int \frac{u'(x)}{1 + u^2(x)} = \arctan(u) \\
\lim_{x \to \infty} \arctan(x) = \frac{\pi}{2} \\
\lim_{x \to -\infty} \arctan(x) = -\frac{\pi}{2}
]
one can arrive at the last integral for the first component:
[
E_1(\omega) = \frac{z e i \pi}{\sqrt{2 pi}^2 \sqrt{2 \pi}} (-\frac{\omega}{v} - \frac{v \epsilon_0}{c}) \int_{-\infty}^{\infty} e^{i b k_2} \frac{1}{\sqrt{\frac{\omega^2}{v^2} - \frac{\omega^2}{c^2} + k_2^2}} dk_2
]
This integral is not trivial to solve any more - one can see itโs result from
studying the Bessel differential equations as a modified Bessel function $K_0$:
[
\Gamma(x) = (x-1)! \\
I_{\alpha}(x) = i^{-\alpha} J_{\alpha}(ix) = \sum_{m = 0}^{\infty} \frac{1}{m! \Gamma(m+\alpha+1)} (\frac{x}{2})^{2m + \alpha} \\
K_{\alpha}(x) = \frac{pi}{2} \frac{I_{- \alpha}(x) - I_{\alpha}(x)}{\sin \alpha \pi}
]
[
E_1(\omega) = \frac{z e i \pi}{\sqrt{2 pi}^2 \sqrt{2 \pi}} (-\frac{\omega}{v} - \frac{v \epsilon_0}{c}) * 4 \pi K_0(b * \sqrt{\frac{\omega^2}{v^2} - \frac{\omega^2}{c^2}}) \\
E_2(\omega) = \frac{z e \sqrt{2 \pi}^3}{v} 4 \pi \frac{\sqrt{\frac{\omega^2}{v^2} - \frac{\omega^2}{c^2}}}{\epsilon_0} K_1(b * \sqrt{\frac{\omega^2}{v^2} - \frac{\omega^2}{c^2}}) \\
E_3 = 0 \\
B_1 = 0 \\
B_2 = 0 \\
B_3 = \epsilon_0 \underbrace{\frac{v^2}{c^2}}_{\beta} E_2(\omega)
]
Now that the fields are known in spatial coordinates one can determine the Poynting
vector:
[
\vec{S} = \vec{E} \times \vec{H}
]
The Poynting vector describes the directional energy flux of an electromagnetic field,
i.e. it contains information about the amount of radiated energy as well as the direction
of radiated energy. To calculate the electromagnetic energy flow $P$ over an surface
one can simply integrate over the Poynting vector over an infinitely long cylinder
with given radius $r_a$ around the particle path:
[
P_a = \int_{-\infty}^{\infty} 2 \pi r_a B_3 E_1 dx
]
Using this definition one can describe the radiated energy per traveled distance
of the particle $x_p$:
[
\frac{dE}{dx_p} = \frac{1}{v} P_a \\
\frac{dE}{dx_p} = -\frac{1}{v} 2 \pi r_a \int_{-\infty}^{\infty} B_3(t) E_1(t) dx \\
\frac{dE}{dx_p} = -\frac{1}{v} 2 \pi r_a \int_{-\infty}^{\infty} B_3(t) E_1(t) v dt \\
\frac{dE}{dx_p} = -2 \pi r_a \int_{-\infty}^{\infty} B_3(t) E_1(t) v dt \\
\frac{dE}{dx_p} = -4 \pi r_a \int_{0}^{\infty} B_3(\omega) E_1(\omega) d\omega \\
]
Using a series expansion of Bessel functions for $b$ much larger than the typical
radius of an atom one can see that for real values of $\sqrt{\frac{\omega^2}{v^2} - \frac{\omega^2}{c^2}}$,
i.e. speeds of particles slower than the local speed of light, one deposits all
electromagnetic energy in vicinity of the particle path. As the value gets imaginary
the radiated fields are independent of the radius of the integration cylinder, i.e.
gets radiated away. As one can see the energy deposited inside the material is then
given (after transformation to Gaussian units) by the expression
[
\frac{dE}{dx_{p}} = \frac{z^2 e^2}{c^2} \int_{v > \frac{c}{n}} \omega(1 - \frac{c^2}{v^2 n^2}) d\omega
]
This expression is also known as the Frank-Tamm formula.
Applications of Cherenkov detectors
Just a last word on why Cherenkov radiation is particular interesting for particle physics
as well as possibly other fields:
- One can - for example - use Cherenkov radiation to measure the speed of particles.
If one also knows the momentum (for example by using an varying magnetic field or
the bending angle of a particle under magnetic field) one can determine particle mass - which
is particular interesting for particle physics. Examples of such detectors
are the ring imaging cherenkov detector at AMS
thatโs capable of measuring charge and velocity of particles, the two
RICH detectors at LHCb and
the DIRC detector at BaBar.
- On the other hand Cherenkov detectors are pretty simple and compared to other
technologies rather cheap to construct - even on a large scale. They can be used to
detect the presence of primary or secondary charged particles in case the speed
is high enough and they can be used to determine speed and charge of a particle in
parallel. This is exploited on telescopes like the Cherenkov telescope array (CTA)
which is used to detect charged particle showers caused by particle decay inside
the atmosphere as well as the IceCube neutrino observatory
that captures Cherenkov light generated by particles that are created during
neutrino interaction in a really huge ice volume. Another famous detector
array is the ANTARES telescope located in the Mediterranean Sea
as well as many other - primarily Neutrino - telescopes currently developed.
- Theyโre used in medical equipment like positron emission tomography
- Fiber optic Cherenkov detectors are sometimes used in particle accelerators for
beam current monitoring.
This article is tagged: Physics, Tutorial, Electrodynamics, Particle detectors