Cherenkov radiation summary

28 Sep 2020 - tsp
Last update 21 Jun 2021
Reading time 11 mins

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:

Cherenkov radiation in the Triga reactor of Atominstitute of TU Vienna, Austria (at 250 kW thermical output)

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.

Constructive polarization shock wave

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.

Geometric relations

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) …

[ \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) ]

… 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 it’s 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. What’s 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 $k_1 = \frac{\omega}{v}$. It’s 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 it’s 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:

This article is tagged: Physics, Tutorial, Electrodynamics, Particle detectors

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Dipl.-Ing. Thomas Spielauer, Wien (webcomplains389t48957@tspi.at)

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