Frank–Tamm formula

The energy ${\displaystyle dE}$ emitted per unit length travelled by the particle per unit of frequency ${\displaystyle d\omega }$ is:

${\displaystyle {\frac {d^{2}E}{dx\,d\omega }}={\frac {q^{2}}{4\pi }}\mu (\omega )\omega {\left(1-{\frac {c^{2}}{v^{2}n^{2}(\omega )}}\right)}}$

provided that ${\displaystyle \beta ={\frac {v}{c}}>{\frac {1}{n(\omega )}}}$. Here ${\displaystyle \mu (\omega )}$ and ${\displaystyle n(\omega )}$ are the frequency-dependent permeability and index of refraction of the medium respectively, ${\displaystyle q}$ is the electric charge of the particle, ${\displaystyle v}$ is the speed of the particle, and ${\displaystyle c}$ is the speed of light in vacuum.

Cherenkov radiation does not have characteristic spectral peaks, as typical for fluorescence or emission spectra. The relative intensity of one frequency is approximately proportional to the frequency. That is, higher frequencies (shorter wavelengths) are more intense in Cherenkov radiation. This is why visible Cherenkov radiation is observed to be brilliant blue. In fact, most Cherenkov radiation is in the ultraviolet spectrum; the sensitivity of the human eye peaks at green, and is very low in the violet portion of the spectrum.

The total amount of energy radiated per unit length is:

${\displaystyle {\frac {dE}{dx}}={\frac {q^{2}}{4\pi }}\int _{v>{\frac {c}{n(\omega )}}}\mu (\omega )\omega {\left(1-{\frac {c^{2}}{v^{2}n^{2}(\omega )}}\right)}d\omega }$

This integral is done over the frequencies ${\displaystyle \omega }$ for which the particle's speed ${\displaystyle v}$ is greater than speed of light of the media ${\displaystyle {\frac {c}{n(\omega )}}}$. The integral is convergent (finite) because at high frequencies the refractive index becomes less than unity and for extremely high frequencies it becomes unity.[1][2]

Consider a charged particle moving relativistically along ${\displaystyle x}$-axis in a medium with refraction index ${\displaystyle n(\omega )={\sqrt {\epsilon (\omega )}}}$ [3] with a constant velocity ${\displaystyle {\vec {v}}=(v,0,0)}$. Start with Maxwell's equations (in Gaussian units) in the wave forms (also known as the Lorenz gauge condition) and take the Fourier transformation:

${\displaystyle {\bigg (}k^{2}-{\frac {\omega ^{2}}{c^{2}}}\epsilon (\omega ){\bigg )}\Phi ({\vec {k}},\omega )={\frac {4\pi }{\epsilon (\omega )}}\rho ({\vec {k}},\omega )}$

${\displaystyle {\bigg (}k^{2}-{\frac {\omega ^{2}}{c^{2}}}\epsilon (\omega ){\bigg )}{\vec {A}}({\vec {k}},\omega )={\frac {4\pi }{c}}{\vec {J}}({\vec {k}},\omega )}$

For a charge of magnitude ${\displaystyle ze}$ (where ${\displaystyle e}$ is the elementary charge) moving with velocity ${\displaystyle v}$, the density and charge density can be expressed as ${\displaystyle \rho ({\vec {x}},t)=ze\delta ({\vec {x}}-{\vec {v}}t)}$ and ${\displaystyle {\vec {J}}({\vec {x}},t)={\vec {v}}\rho ({\vec {x}},t)}$, taking the Fourier transformation [4] gives:

${\displaystyle \rho ({\vec {k}},\omega )={\frac {ze}{2\pi }}\delta (\omega -{\vec {k}}\cdot {\vec {v}})}$

${\displaystyle {\vec {J}}({\vec {k}},\omega )={\vec {v}}\rho ({\vec {k}},\omega )}$

Substituting this density and charge current into the wave equation, we can solve for the Fourier-form potentials:

${\displaystyle \Phi ({\vec {k}},\omega )={\frac {2ze}{\epsilon (\omega )}}{\frac {\delta (\omega -{\vec {k}}\cdot {\vec {v}})}{k^{2}-{\frac {\omega ^{2}}{c^{2}}}\epsilon (\omega )}}}$ and ${\displaystyle {\vec {A}}({\vec {k}},\omega )=\epsilon (\omega ){\frac {\vec {v}}{c}}\Phi ({\vec {k}},\omega )}$

Using the definition of the electromagnetic fields in terms of potentials, we then have the Fourier-form of the electric and magnetic field:

${\displaystyle {\vec {E}}({\vec {k}},\omega )=i{\bigg (}{\frac {\omega \epsilon (\omega )}{c}}{\frac {\vec {v}}{c}}-{\vec {k}}{\bigg )}\Phi ({\vec {k}},\omega )}$ and ${\displaystyle {\vec {B}}({\vec {k}},\omega )=i\epsilon (\omega ){\vec {k}}\times {\frac {\vec {v}}{c}}\Phi ({\vec {k}},\omega )}$

To find the radiated energy, we consider electric field as a function of frequency at some perpendicular distance from the particle trajectory, say, at ${\displaystyle (0,b,0)}$, where ${\displaystyle b}$ is the impact parameter. It is given by the inverse Fourier transform:

${\displaystyle {\vec {E}}(\omega )={\frac {1}{(2\pi )^{3/2}}}\int d^{3}k{\vec {E}}({\vec {k}},\omega )e^{ibk_{2}}}$

First we compute ${\displaystyle x}$-component ${\displaystyle E_{1}}$ of the electric field (parallel to ${\displaystyle {\vec {v}}}$):

${\displaystyle E_{1}(\omega )={\frac {2ize}{\epsilon (\omega )(2\pi )^{3/2}}}\int d^{3}ke^{ibk_{2}}{\bigg (}{\frac {\omega \epsilon (\omega )v}{c^{2}}}-k_{1}{\bigg )}{\frac {\delta (\omega -vk_{1})}{k^{2}-{\frac {\omega ^{2}}{c^{2}}}\epsilon (\omega )}}}$

For brevity we define ${\displaystyle \lambda ^{2}={\frac {\omega ^{2}}{v^{2}}}-{\frac {\omega ^{2}}{c^{2}}}\epsilon (\omega )={\frac {\omega ^{2}}{v^{2}}}{\big (}1-\beta ^{2}\epsilon (\omega ){\big )}}$. Breaking the integral apart into ${\displaystyle k_{1},k_{2},k_{3}}$, the ${\displaystyle k_{1}}$ integral can immediately be integrated by the definition of the Dirac Delta:

${\displaystyle E_{1}(\omega )=-{\frac {2ize\omega }{v^{2}(2\pi )^{3/2}}}{\bigg (}{\frac {1}{\epsilon (\omega )}}-\beta ^{2}{\bigg )}\int _{-\infty }^{\infty }dk_{2}e^{ibk_{2}}\int _{-\infty }^{\infty }{\frac {dk_{3}}{k_{2}^{2}+k_{3}^{2}+\lambda ^{2}}}}$

The integral over ${\displaystyle k_{3}}$ has the value ${\displaystyle {\frac {\pi }{(\lambda ^{2}+k_{2}^{2})^{1/2}}}}$, giving:

${\displaystyle E_{1}(\omega )=-{\frac {ize\omega }{v^{2}{\sqrt {2\pi }}}}{\bigg (}{\frac {1}{\epsilon (\omega )}}-\beta ^{2}{\bigg )}\int _{-\infty }^{\infty }dk_{2}{\frac {e^{ibk_{2}}}{(\lambda ^{2}+k_{2}^{2})^{1/2}}}}$

The last integral over ${\displaystyle k_{2}}$ is in the form of a modified (Macdonald) Bessel function, giving the evaluated parallel component in the form:

${\displaystyle E_{1}(\omega )=-{\frac {ize\omega }{v^{2}}}\left({\frac {2}{\pi }}\right)^{1/2}{\bigg (}{\frac {1}{\epsilon (\omega )}}-\beta ^{2}{\bigg )}K_{0}(\lambda b)}$

One can follow a similar pattern of calculation for the other fields components arriving at:

${\displaystyle E_{2}(\omega )={\frac {ze}{v}}\left({\frac {2}{\pi }}\right)^{1/2}{\frac {\lambda }{\epsilon (\omega )}}K_{1}(\lambda b),\quad E_{3}=0\quad }$ and ${\displaystyle \quad B_{1}=B_{2}=0,\quad B_{3}(\omega )=\epsilon (\omega )\beta E_{2}(\omega )}$

We can now consider the radiated energy ${\displaystyle dE}$ per particle traversed distance ${\displaystyle dx_{\text{particle}}}$. It can be expressed through the electromagnetic energy flow ${\displaystyle P_{a}}$ through the surface of an infinite cylinder of radius ${\displaystyle a}$ around the path of the moving particle, which is given by the integral of the Poynting vector ${\displaystyle \mathbf {S} =c/(4\pi )[\mathbf {E} \times \mathbf {H} ]}$ over the cylinder surface:

${\displaystyle {\bigg (}{\frac {dE}{dx_{\text{particle}}}}{\bigg )}_{\text{rad}}={\frac {1}{v}}P_{a}=-{\frac {c}{4\pi v}}\int _{-\infty }^{\infty }2\pi aB_{3}E_{1}dx}$

The integral over ${\displaystyle dx}$ at one instant of time is equal to the integral at one point over all time. Using ${\displaystyle dx=vdt}$:

${\displaystyle {\bigg (}{\frac {dE}{dx_{\text{particle}}}}{\bigg )}_{\text{rad}}=-{\frac {ca}{2}}\int _{-\infty }^{\infty }B_{3}(t)E_{1}(t)dt}$

Converting this to the frequency domain:

${\displaystyle {\bigg (}{\frac {dE}{dx_{\text{particle}}}}{\bigg )}_{\text{rad}}=-ca\,{\text{Re}}{\bigg (}\int _{0}^{\infty }B_{3}^{*}(\omega )E_{1}(\omega )d\omega {\bigg )}}$

To go into the domain of Cherenkov radiation, we now consider perpendicular distance ${\displaystyle b}$ much greater than atomic distances in a medium, that is, ${\displaystyle |\lambda b|\gg 1}$. With this assumption we can expand the Bessel functions into their asymptotic form:

${\displaystyle E_{1}(\omega )\rightarrow {\frac {ize\omega }{c^{2}}}{\bigg (}1-{\frac {1}{\beta ^{2}\epsilon (\omega )}}{\bigg )}{\frac {e^{-\lambda b}}{\sqrt {\lambda b}}}}$

${\displaystyle E_{2}(\omega )\rightarrow {\frac {ze}{v\epsilon (\omega )}}{\sqrt {\frac {\lambda }{b}}}e^{-\lambda b}}$ and ${\displaystyle B_{3}(\omega )=\epsilon (\omega )\beta E_{2}(\omega )}$

Thus:

${\displaystyle {\bigg (}{\frac {dE}{dx_{\text{particle}}}}{\bigg )}_{\text{rad}}={\text{Re}}{\bigg (}\int _{0}^{\infty }{\frac {z^{2}e^{2}}{c^{2}}}{\bigg (}-i{\sqrt {\frac {\lambda ^{*}}{\lambda }}}{\bigg )}\omega {\bigg (}1-{\frac {1}{\beta ^{2}\epsilon (\omega )}}{\bigg )}e^{-(\lambda +\lambda ^{*})a}d\omega {\bigg )}}$

If ${\displaystyle \lambda }$ has a positive real part (usually true), the exponential will cause the expression to vanish rapidly at large distances, meaning all the energy is deposited near the path. However, this isn't true when ${\displaystyle \lambda }$ is purely imaginary - this instead causes the exponential to become 1 and then is independent of ${\displaystyle a}$, meaning some of the energy escapes to infinity as radiation - this is Cherenkov radiation.

${\displaystyle \lambda }$ is purely imaginary if ${\displaystyle \epsilon (\omega )}$ is real and ${\displaystyle \beta ^{2}\epsilon (\omega )>1}$. That is, when ${\displaystyle \epsilon (\omega )}$ is real, Cherenkov radiation has the condition that ${\displaystyle v>{\frac {c}{{\sqrt {\epsilon (\omega }})}}={\frac {c}{n}}}$. This is the statement that the speed of the particle must be larger than the phase velocity of electromagnetic fields in the medium at frequency ${\displaystyle \omega }$ in order to have Cherenkov radiation. With this purely imaginary ${\displaystyle \lambda }$ condition, ${\displaystyle {\sqrt {\frac {\lambda ^{*}}{\lambda }}}=i}$ and the integral can be simplified to:

${\displaystyle {\bigg (}{\frac {dE}{dx_{\text{particle}}}}{\bigg )}_{\text{rad}}={\frac {z^{2}e^{2}}{c^{2}}}\int _{\epsilon (\omega )>{\frac {1}{\beta ^{2}}}}\omega {\bigg (}1-{\frac {1}{\beta ^{2}\epsilon (\omega )}}{\bigg )}d\omega ={\frac {z^{2}e^{2}}{c^{2}}}\int _{v>{\frac {c}{n(\omega )}}}\omega {\bigg (}1-{\frac {c^{2}}{v^{2}n^{2}(\omega )}}{\bigg )}d\omega }$

This is the Frank–Tamm equation in Gaussian units. This derivation follows Jackson 3rd Edition[5]

• Cherenkov radiation (Tagged ‘Frank-Tamm formula’)