# Larmor Formula

### Short Topical Videos

### Reference Material

- Larmor Formula (Wikipedia)
- Larmor Radiation (NRAO)
- Magnetism, Radiation, and Relativity (Schroeder, Weber State), see S4.

### Need to Review?

- Maxwell Equations
- Electromagnetic Plane Waves for Poynting Flux.
- Energy Density

### Related Topics

- Synchrotron Radiation for the relativistic form.
- Introduction to Synchrotron Radiation and Relativistic Beaming for cyclotron polarization and relativistic beaming.
- Thermal Bremsstrahlung

<latex>
\documentclass[]{article}
\def\inv#1Template:1 \over
\def\ddtTemplate:D \over dt
\def\mean#1{\left\langle {#1}\right\rangle}
\def\sigot{\sigma_{12}}
\def\sigto{\sigma_{21}}
\def\eval#1{\big|_{#1}}
\def\tr{\nabla}
\def\dce{\vec\tr\times\vec E}
\def\dcb{\vec\tr\times\vec B}
\def\wz{\omega_0}
\def\ef{\vec E}
\def\ato{{A_{21}}}
\def\bto{{B_{21}}}
\def\bot{{B_{12}}}
\def\bfieldTemplate:\vec B
\def\apTemplate:A^\prime
\def\xp{{x^{\prime}}}
\def\yp{{y^{\prime}}}
\def\zp{{z^{\prime}}}
\def\tp{{t^{\prime}}}
\def\upxTemplate:U x^\prime
\def\upyTemplate:U y^\prime
\def\e#1{\cdot10^{#1}}
\def\hf{\frac12}
\def\^{\hat}
\usepackage[top=1in,bottom=1in,left=1in,right=1in]{geometry}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage{natbib}
\usepackage{eufrak}

\begin{document} \title{Larmor Formula} -- See also: Radiative Processes (Rybicki and Lightman) Chapter 3; Classical Electrodynamics (Jackson) Chapter 14\\ -- Section 2 and Subsection 1.3 are the main take-aways; Subsection 1.1 and 1.2 are just background information for knowing where subsection 1.3 comes from

\section{Electric Field of an Accelerated Charge} %% Edited by: Nick Kern, Fall 2015, nkern@berkeley.edu

\subsection{Potentials from a Non-Static Charge Distribution}

Recall that the electromagnetic fields can be expressed completely in terms of the scalar potential $\phi(r,t)$ and the vector potential $A(r,t)$: $$ \begin{aligned} \mathbf{B} &= \nabla\times\mathbf{A}\\ \mathbf{E} &= -\nabla\phi - \frac{1}{c}\frac{\partial\mathbf{A}}{\partial t}. \end{aligned} $$

Inserting the Maxwell Equations into the above expressions and solving for the potentials (Rybicki pg. 70) yields:
$$
\begin{aligned}
\phi(\mathbf{r},t) &= \int\frac{\rho_{\text{ret}}\ d^{3}\mathbf{r}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|}\\
\mathbf{A}(\mathbf{r},t) &= \frac{1}{c}\int\frac{\mathbf{j}_{\text{ret}}\ d^{3}\mathbf{r}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|},
\end{aligned}
$$
where $\rho_{\text{ret}}$ and $j_{\text{ret}}$ are the charge density and charge current at the \emph{retarded time}, $t-\frac{1}{c}|r-r^{\prime}|$. The ``retarded time* is another way of saying that because light travels at a finite speed, the potentials at an instantaneous time $t_{}$ and distance from charge source $|r-r^{\prime}|$ feel the charge distribution not as it is at $t$ but as it was at $t_{\text{ret}} = t-\frac{1}{c}|r-r^{\prime}|$ (see Figure 3.1 from Rybicki). Note that this is still generalized to a distribution of charges and currents. We will now specify the case for a moving point charge.*

\subsection{Li\'enard-Wiechert Potentials}

Let us now specify the charge distribution for a moving point charge. Consider a particle with charge $q$ moving along a trajectory $\mathbf{r} = \mathbf{r_{0}}(t)$. Its velocity is then $\mathbf{u}(t) = \dot{\mathbf{r}_{0}}(t)$. The charge and current densities are then, $$ \begin{aligned} \rho(\mathbf{r},t) &= q\cdot\delta(\mathbf{r} - \mathbf{r}_{0}(t))\\ \mathbf{j}(\mathbf{r},t) &= q\cdot\mathbf{u}(t)\cdot\delta(\mathbf{r}-\mathbf{r}_{0}(t)), \end{aligned} $$ where the Dirac Delta function has the effect of localizing the charge density and current to a specific point in space and time (i.e. a moving point charge). It is useful to recall that $\int f(x)\delta(x-x_{0})dx = f(x_{0})$.

\\

Let's now substitute this into our previous expressions for the EM scalar potential given the retarded charge density. Note that because we are using the retarded charge density, we will want to take the charge density over all times and isolate it to the retarded time. We can do this by integrating the charge density times a delta function over time: $$ \begin{aligned} \phi(\mathbf{r},t) &= \int dt^{\prime}\delta(t^{\prime}-t_{\text{ret}})\int d^{3}r^{\prime}\frac{\rho(\mathbf{r}^{\prime},t^{\prime})}{|\mathbf{r}-\mathbf{r}^{\prime}|}\\ &= \int dt^{\prime}\delta(t^{\prime}-t + |\mathbf{r}-\mathbf{r}^{\prime}|/c)\int d^{3}r^{\prime}\frac{\rho(\mathbf{r}^{\prime},t^{\prime})}{|\mathbf{r}-\mathbf{r}^{\prime}|}, \end{aligned} $$ where we substituted in for our definition of $t_{\text{ret}}$. If we now substitute our charge density for a point charge, $\rho$, and integrate over all space $d^{3}r^{\prime}$, we can see that because of the $\delta(\mathbf{r}-\mathbf{r_{0}(t)})$ in the charge density, we end up with $$ \begin{aligned} \phi(\mathbf{r},t) &= q\int\frac{dt^{\prime}}{|\mathbf{r}-\mathbf{r}_{0}(t^{\prime})|} \delta(t^{\prime}-t+|\mathbf{r}-\mathbf{r^{\prime}}|/c), \end{aligned} $$ which is an integral over only $dt^{\prime}$.

\\

To make this more compact, let's introduce the displacement vector $\mathbf{R}(t^{\prime}) = \mathbf{r}-\mathbf{r}_{0}(t^{\prime})$ and the displacement magnitude $R(t^{\prime}) = |\mathbf{R}(t^{\prime})|$. This gives us $$ \begin{aligned} \phi(\mathbf{r},t) &= q\int\frac{dt^{\prime}\ \delta(t^{\prime}-t+R(t^{\prime})/c)}{R(t^{\prime})}\\ \mathbf{A}(\mathbf{r},t) &= \frac{q}{c}\int \frac{dt^{\prime}\ \mathbf{u}(t^{\prime})\ \delta(t^{\prime}-t+R(t^{\prime})/c)}{R(t^{\prime})}, \end{aligned} $$ where we did identical calculations to derive the vector potential. These are known as the \emph{Li\'enard-Wiechert Potentials}, which describe the EM retarded potentials at a given point in space and time created by a moving point charge. It is instructive to simplify them further.

\\

Now let's make a variable substitution of $t^{\prime\prime} = t^{\prime} - t_{\text{ret}} = t^{\prime} - t + R(t^{\prime})/c$. This implies that $$ \begin{aligned} \frac{dt^{\prime\prime}}{dt^{\prime}} &= 1 + \frac{\dot{R}(t^{\prime})}{c}, \ \text{or:}\\ dt^{\prime\prime} &= dt^{\prime} + \frac{\dot{R}(t^{\prime})dt^{\prime}}{c}. \end{aligned} $$ Defining the unit vector $\mathbf{n} = \mathbf{R}/R$, we can see that $\dot{R} = \mathbf{n}\cdot\dot{\mathbf{R}} = -\mathbf{n}\cdot\mathbf{u}$, where we used the fact that the change in the displacement vector is equal to negative the particle velocity: $\dot{\mathbf{R}} = -\mathbf{u}$. This gives us the following relations for $t^{\prime}$ and $dt^{\prime}$: $$ \begin{aligned} t^{\prime} &= t^{\prime\prime} + t_{\text{ret}}\\ dt^{\prime} &= \frac{dt^{\prime\prime}}{1-\frac{1}{c}\mathbf{n}(t^{\prime\prime}+t_{\text{ret}})\mathbf{u}(t^{\prime\prime}+t_{\text{ret}})}. \end{aligned} $$

\\

Substituting this in to the above expression we can see that the Li\'enard-Wiechert scalar potential becomes $$ \begin{aligned} \phi(\mathbf{r},t) &= q\int\frac{dt^{\prime\prime}\ \delta(t^{\prime\prime})}{R(t^{\prime\prime}+t_{\text{ret}})\cdot(1-\frac{1}{c}\mathbf{n}(t^{\prime\prime}+t_{\text{ret}})\mathbf{u}(t^{\prime\prime}+t_{\text{ret}}))}. \end{aligned} $$ Recalling the properties of the Dirac Delta function, we can see that the argument of the integral is always zero except when $t^{\prime\prime} = 0$, which is equivalent of saying when $t^{\prime} = t_{\text{ret}}$. This makes the Li\'enard-Wiechert potentials simplify to $$ \begin{aligned} \phi(\mathbf{r},t) &= \frac{q}{\kappa(t^{\prime}) R(t^{\prime})}\Big|_{t^{\prime}=t_{\text{ret}}}\\ \mathbf{A}(\mathbf{r},t) &= \frac{q\mathbf{u}(t^{\prime})}{c\kappa(t^{\prime})R(t^{\prime})}\Big|_{t^{\prime}=t_{\text{ret}}}, \end{aligned} $$ where we did identical calculations to get the vector potential, and have defined $\kappa(t) = 1 - \frac{1}{c}\mathbf{n}(t)\cdot\mathbf{u}(t)$. Note that in the non-relativistic limit $\kappa \rightarrow 1$. In the relativistic limit, $\kappa$ concentrates the potentials into a cone about the particle velocity, which is related to the \emph{relativistic beaming effect} (see Introduction to Synchrotron Radiation and Relativistic Beaming).

\\

\subsection{EM Fields from Accelerating Charges}

To get the Electric and Magnetic fields, we need to know the particle position at the retarded time: $\mathbf{r}_{0}(t_{\text{ret}})$. Differentiation also gives us the particle velocity and acceleration at that time: $\mathbf{u}(t_{ret}) = \dot{\mathbf{r}}_{0}(t_{\text{ret}})$ and $\dot{\mathbf{u}}(t_{ret}) = \ddot{\mathbf{r}}_{0}(t_{\text{ret}})$ respectively. The steps are shown in Jackson (pg. 663), but here we we summarize the results, as done in Rybicki (pg. 80). We use the notation of $\vec{\beta} = \mathbf{u}/c$ and $\kappa =1-\mathbf{n}\cdot\vec{\beta}$ (Note that both $\vec{\beta}$ and $\mathbf{n}$ indicate vectors). Differentiating the Li\'enard-Wiechert potentials gives us the Electric and Magnetic fields for a moving point charge: $$ \begin{aligned} \mathbf{E}(\mathbf{r},t) &= q\left(\frac{(\mathbf{n}-\vec{\beta})(1-\beta^{2})}{\kappa^{3}R^{2}}\right)_{t_{\text{ret}}} + \frac{q}{c}\left(\frac{\mathbf{n}}{\kappa^{3}R}\times[(\mathbf{n}-\vec{\beta}]\times\dot{\vec{\beta}}]\right)_{t_{\text{ret}}}\\ \mathbf{B}(\mathbf{r},t) &= \left(\mathbf{n}\times\mathbf{E}(\mathbf{r},t)\right)_{t_{\text{ret}}}. \end{aligned} $$

\\

The electric field has two terms to it: one that falls off as $1/r^{2}$ called the velocity field because it is only a function of position and velocity, and one that falls off as $1/r$ called the acceleration field because it is a function of velocity and acceleration. Note that in the non-relativistic limit that $u<<c$, the velocity field simplifies to the standard E field from Coulomb's law. Also, when the particle is moving at a constant velocity, this is the only term that contributes to the E field. When the particle is accelerating, its E field has a contribution from both the velocity and acceleration fields. Use Figure 3.1 on page 80 of Rybicki and Lightman to see the direction that the E field points in after evaluating the cross products of the acceleration field.

\\

The electric and magnetic terms of the acceleration field are together referred to as the radiation field: $$ \begin{aligned} \mathbf{E}_{\text{rad}}(\mathbf{r},t) &= \frac{q}{c}\left(\frac{\mathbf{n}}{\kappa^{3}R}\times[(\mathbf{n}-\vec{\beta}]\times\dot{\vec{\beta}}]\right)_{t_{\text{ret}}}\\ \mathbf{B}_{\text{rad}}(\mathbf{r},t) &= \left(\mathbf{n}\times\mathbf{E}_{\text{rad}}(\mathbf{r},t)\right)_{t_{\text{ret}}}. \end{aligned} $$ Because the acceleration field falls of less steeply than the velocity field ($1/r$ versus $1/r^{2}$), it is this field that a distant observer would see from an accelerating charge. In fact, you can prove that the total spherically averaged power of the acceleration field through a spherical surface at radius $r$ is constant out to any $r$, meaning that \emph{radiation energy flows to infinite distances}. This is not true of the velocity field. This is related to the Larmor Formula up next.

\

\emph{Think about it:}

We typically define the near-field radiation zone of a dipole as where $r < \lambda$ and the far-field radiation zone as where $r > \lambda$, where $\lambda$ is the frequency of emitted radiation. Try to re-produce these results using the ratio of the radiation field to the velocity field given that the radiation field scales with velocity and acceleration and the velocity field scales with only velocity. In other words, $$ \begin{aligned} \frac{E_{\text{rad}}}{E_{\text{vel}}} \propto \dot{\beta}\cdot C \end{aligned} $$ where $C$ is a constant (solve for using dimensional analysis!). Using the approximation that $\dot{\beta} = \beta\cdot\nu$, what can we say about the E field contribution from the radiation and velocity fields in both the near and far-field zones?

\section{Derivation of Power Pattern (Cyclotron)}
See Figure 3.2 of Rybicki and Lightman for a visual guide of the problem.

Say that we have an $e^-$ moving in a straight line along the x axis, and
it gets accelerated, starting at $x=0$, for a duration $\Delta t$. Let
$\theta$ measure the angle of a field line emitted by the $e^-$, from the
acceleration axis. The
acceleration of $e^-$ causes a ``jog* in the electric field compared to where*
it would have been (this jog propagates outward at $c$). Say that $E_{tr}$
is the electric field carried in the transverse component
($\perp$ to $\theta$), and
$E_r$ measures the component in the radial component of the jog. Then:
$$\begin{aligned}{E_{tr}\over E_r}&={(a\Delta t)t\sin\theta\over c\Delta t}\\
&={at\sin\theta\over c}\\ \end{aligned}$$
Just to get that factor of $t$ out of there, say that $r=ct$:
$${E_{tr}\over E_r}={ar\sin\theta\over c^2}$$
We know that $E_r={e\over r^2}$, so:
$$\boxed{E_{tr}={ea\sin\theta\over c^2r}}$$
Note how the transverse field goes off as $r^{-1}$, so when
you go far enough away, it always wins out over the plain electric field.\par
What's the transverse magnetic field? The Maxwell Equations tell us:
$$\begin{aligned}\dce=-{1\over c}{\partial B\over\partial t}\\
&=\begin{vmatrix}\^x&\^y&\^r\\
{\partial\over\partial x}&{\partial\over\partial y}&{\partial\over\partial r}\\
E_{tr}&0&E_r\end{vmatrix}\\
&=\^x({\partial E_r\over\partial y})+\^y({\partial E_{tr}\over\partial r}
-{\partial E_r\over\partial y})+\^r({\partial E_{tr}\over\partial y})\\
&\approx\^y{\partial E_{tr}\over\partial r}\\ \end{aligned}$$
Say we define $f$ such that:
$$E_{tr}=f(r-ct)$$
then $\vec B=f(r-ct)\^y$, so:
$$\begin{aligned}-{1\over c}{\partial f(r-ct)\^y\over \partial t}
&={-\^y\over c}{\partial f(r-ct)\over \partial(r-ct)}{\partial(r-ct)\over
\partial t}\\
&=\^y{\partial f(r-ct)\over\partial(r-ct)}\\ \end{aligned}$$
Also:
$$\begin{aligned}\dce&={\partial E_{tr}\over\partial r}\\
&=\^y{\partial f(r-ct)\over\partial(r-ct)}{\partial(r-ct)\over\partial r}\\
&=\^y{\partial f(r-ct)\over\partial(r-ct)}\\ \end{aligned}$$
Thus, $E_{tr}$ and $B_{tr}$ have the same magnitude.\par
To cut to the chase, the power radiated from the entire sphere is:
$$\begin{aligned}P_{sphere}&={e^2a^2\over4\pi r^2c^3}\int_0^\pi{\sin^2\theta
2\pi\sin\theta r^2d\theta}\\
&={2\over3}{e^2a^2\over c^3}\\ \end{aligned}$$
This is the Larmor power formula. We can also calculate the Poynting flux:
$${\vec E\times\vec B\over4\pi}c=e^2a^2{\sin^2\theta\over4\pi r^2}$$
Incidentally, you get the same result if you accelerate the charge perpendicular
to the direction it was moving.

\section{Another derivation of Larmor}
Interestingly, one can derive the Larmor formula another way as well. Consider a charge $q$ oscillating within a one dimensional quantum harmonic oscillator (characterized by oscillator frequency $\omega$). Further suppose that the charge starts out in an energy eigenstate $|n\rangle$ of the oscillator, and decays by spontaneous emission to $|m\rangle$. The timescale for decay is set by the Einstein A coefficient:
\begin{equation}
A = \frac{\omega_0^3 |\alpha|^2}{3\pi\epsilon_0 \hbar c^3}.
\end{equation}
where $\alpha \equiv q\langle n | x | m \rangle$ (for a review of the relevant quantum mechanics, see Einstein Coefficients: a closer look ). This matrix element can be calculated trivially using the raising and lowering operators for the SHO,. Further note that that $m = n-1$ (we're talking about \textit{decay}), so we have:
\begin{equation}
\alpha = \sqrt{\frac{\hbar}{2m\omega}}\delta_{m, n-1} \hat{x}.
\end{equation}
The energy levels of the SHO are $E_n = (n+\frac{1}{2})\hbar\omega$. Therefore, the frequency of the emitted photon as the system decays is:
\begin{equation}
\omega_0 = \frac{E_n - E_m}{\hbar} = \omega.
\end{equation}
So, the emitted photon has a frequency which is equal to the oscillation frequency of the system! We can also calculate the lifetime of the state, $\tau \sim 1/A$:
\begin{equation}
\tau = \frac{6\pi\epsilon_0mc^3}{nq^2\omega^2}.
\end{equation}
The \textit{power} carried away is just the energy carried away divided by this timescale, i.e.:
\begin{equation}
P = \frac{q^2\omega^2}{6\pi\epsilon_0mc^3}n\hbar\omega = \frac{q^2\omega^2}{6\pi\epsilon_0mc^3}(E_n - \frac{1}{2}\hbar\omega)
\end{equation}
Dropping the $\frac{1}{2}\hbar\omega$ (this is the classical limit, $\hbar \rightarrow 0$), and using the fact that $E = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2x^2_0$, we have:
\begin{equation}
P = \frac{q^2\omega^4x_0^2}{12\pi\epsilon_0c^3}.
\end{equation}
Finally, we note that for a harmonic oscillator, differentiating $x(t)$ twice and time averaging gives us an acceleration $\langle a^2 \rangle = x_0\omega^2/2$. So, our expression above is just:
\begin{equation}
P = \frac{q^2a^2}{6\pi\epsilon_0c^3},
\end{equation}
which is exactly the classical Larmor formula!

\emph{For cyclotron radiation's polarization, see Introduction to Synchrotron Radiation and Relativistic Beaming}\\ \emph{For cyclotron radiation's relativistic counterpart, synchroton radiation, see Synchrotron Radiation}

\end{document} <\latex>