Larmor formula

A Yagi-Uda antenna. Radio waves can be radiated from an antenna by accelerating electrons in the antenna. This is a coherent process, so the total power radiated is proportional to the square of the number of electrons accelerating.

In physics, in the area of electrodynamics, the Larmor formula (not to be confused with the Larmor precession from classical nuclear magnetic resonance) is used to calculate the total power radiated by a nonrelativistic point charge as it accelerates. It was first derived by J. J. Larmor in 1897, in the context of the wave theory of light.

When accelerating or decelerating, any charged particle (such as an electron) radiates away energy in the form of electromagnetic waves. For velocities that are small relative to the speed of light, the total power radiated is given by the Larmor formula:

$P = \frac{e^2 a^2}{6 \pi \varepsilon_0 c^3} \mbox{ (SI units)}$
$P = {2 \over 3} \frac{e^2 a^2}{ c^3} \mbox{ (cgs units)}$

where $a$ is the acceleration, $e$ is the charge, and $c$ is the speed of light. A relativistic generalization is given by the Liénard–Wiechert potentials.

Derivation

Derivation 1: Mathematical Approach

We first need to find the form of the electric and magnetic fields. The fields can be written (for a fuller derivation see Liénard–Wiechert potential)

$\mathbf{E}(\mathbf{r},t) = q\left(\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1-\boldsymbol{\beta}\cdot\mathbf{n})^3 R^2}\right)_{\rm{ret}} + \frac{q}{c}\left(\frac{\mathbf{n}\times[(\mathbf{n}-\boldsymbol{\beta})\times\dot{\boldsymbol{\beta}}]}{(1-\boldsymbol{\beta}\cdot\mathbf{n})^3R}\right)_{\rm{ret}}$

and

$\mathbf{B} = \mathbf{n}\times\mathbf{E},$

where $\boldsymbol{\beta}$ is the charge's velocity divided by $c$, $\dot{\boldsymbol{\beta}}$ is the charge's acceleration divided by c, $\mathbf{n}$ is a unit vector in the $\mathbf{r} - \mathbf{r}_0$ direction, $R$ is the magnitude of $\mathbf{r} - \mathbf{r}_0$, and $\mathbf{r}_0$ is the charge's location. The terms on the right are evaluated at the retarded time $t_\text{r} = t - R/c$.

These field equations divide themselves up into velocity and acceleration fields. The velocity field depends only upon $\boldsymbol{\beta}$ while the acceleration field depends on both $\boldsymbol{\beta}$ and $\dot{\boldsymbol{\beta}}$ and the angular relationship between the two. Since the velocity field is proportional to $1/R^2$, it falls off very quickly with distance. On the other hand, the acceleration field is proportional to $1/R$, which means that it falls much more slowly with distance. Because of this, the acceleration field is representative of the radiation field and is responsible for carrying most of the energy away from the charge.

We can find the energy flux density of the radiation field by computing its Poynting vector:

$\mathbf{S} = \frac{c}{4\pi}\mathbf{E}_\text{a}\times\mathbf{B}_\text{a},$

where the 'a' subscripts emphasize that we are taking only the acceleration field. Substituting in the relation between the magnetic and electric fields while assuming that the particle instantaneously at rest at time $t_\text{r}$ and simplifying gives[note 1]

$\mathbf{S} = \frac{q^2}{4\pi c}\left|\frac{\mathbf{n}\times(\mathbf{n}\times\dot{\boldsymbol{\beta}})}{R}\right|^2.$

If we let the angle between the acceleration and the observation vector be equal to $\theta$, and we introduce the acceleration $\mathbf{a} = \dot{\boldsymbol{\beta}} c$, then

$\mathbf{S} = \frac{q^2}{4\pi c}\frac{\sin^2(\theta)\, a^2\, \hat{\mathbf{n}}}{c^2 R^2}.$

This quantity the power radiated per unit solid angle by the charge. The total power radiated is found by integrating this quantity over all solid angles (that is, over $\theta$ and $\phi$). This gives

$P = \frac{2}{3}\frac{q^2 a^2}{c^3},$

which is the Larmor result for a non-relativistic accelerated charge. It relates the power radiated by the particle to its acceleration. It clearly shows that the faster the charge accelerates the greater the radiation will be. We would expect this since the radiation field is dependent upon acceleration.

Derivation 2: Using Edward M. Purcell approach

The full derivation can be found here.[1]

Here is an explanation which can help understanding the above page.

This approach is based on the finite speed of light. A charge moving with constant velocity has a radial electric field $E_r$ (at distance $R$ from the charge), always emerging from the future position of the charge, and there is no tangential component of the electric field $(E_t=0)$. This future position is completely deterministic as long as the velocity is constant. When the velocity of the charge changes, (say it bounces back during a short time) the future position "jumps", so from this moment and on, the radial electric field $E_r$ emerges from a new position. Given the fact that the electric field must be continuous, a non-zero tangential component of the electric field $E_t$ appears, which decreases like $1/R$ (unlike the radial component which decreases like $1/R^2$).

Hence, at large distances from the charge, the radial component is negligible relative to the tangential component, and in addition to that, fields which behave like $1/R^2$ cannot radiate, because the Poynting vector associated with them will behave like $1/R^4$.

The tangential component comes out (SI units):

$E_t = {{e a \sin(\theta)} \over {4 \pi \varepsilon_0 c^2 R}}.$

And to obtain the Larmour formula, one has to integrate over all angles, at large distance $R$ from the charge, the Poynting vector associated with $E_t$, which is:

$\mathbf{S} = {{E_t^2 \over \mu_0 c}}\mathbf{\hat{r}} = {{e^2 a^2 \sin^2(\theta)} \over {16 \pi^2 \varepsilon_0 c^3 R^2}} \mathbf{\hat{r}}$

giving (SI units)

$P = {{e^2 a^2} \over {6 \pi \varepsilon_0 c^3}}.$

This is mathematically equivalent to:

$P = {{\mu_0 e^2 a^2} \over {6 \pi c}}.$

Relativistic Generalisation

Covariant Form

We can do this by rewriting the Larmor formula in terms of momentum and then using the four vector generalisation of momentum (see four momentum), $P^{\mu}$. We know that the power is a Lorentz invariant, so all we have to show is that our generalisation is also invariant and that it reduces to the Larmor formula in the low velocity limit. So;

$P = \frac{2}{3}\frac{q^2}{c^3m^2}\left(\frac{d\vec{p}}{dt}\cdot\frac{d\vec{p}}{dt}\right).$

Assume the generalisation;

$P = -\frac{2}{3}\frac{q^2}{m^2c^3}\frac{dP^{\mu}}{d\tau}\frac{dP_{\mu}}{d\tau}.$

When we expand and rearrange the energy-momentum four vector product we get;

$\frac{dP^{\mu}}{d\tau}\frac{dP_{\mu}}{d\tau} = \frac{v^2}{c^2}\left(\frac{dp}{d\tau}\right)^2 - \left(\frac{d\vec{p}}{d\tau}\right)^2$

where I have used the fact that $\frac{dE}{d\tau} = \frac{pc^2}{E}\frac{dp}{d\tau} = v\frac{dp}{d\tau}$. When you let $\beta$ tend to zero, $\gamma$ tends to one, so that $d\tau$ tends to dt. Thus we recover the non relativistic case.

This is an interesting equation. It says that the power radiated by the particle into space depends upon its rate of change of momentum with respect to its time. It also says that the power radiated is proportional to the charge squared and inversely proportional to the mass squared. Thus for a highly charged, extremely small particle the radiation will be much greater than that for a large particle with a small charge.

Non-Covariant Form

To obtain the non-covariant form of the generalisation we first substitute $p^{\mu} = (\gamma mc^2, \gamma m \vec{v})$ in to the above and then performing the differentiation as follows (for brevity I have omitted the constants from the calculation below);

$\frac{dp^{\mu}}{d\tau}\frac{dp_{\mu}}{d\tau} = -\left(\frac{d\vec{p}}{d\tau}\right)^2 + \frac{1}{c^2}\left(\frac{dE}{d\tau} \right)^2$
$\frac{dp^{\mu}}{d\tau}\frac{dp_{\mu}}{d\tau} = -\gamma^2\left(\frac{d\gamma m\vec{v}}{dt}\right)^2 + \frac{\gamma^2}{c^2}\left(\frac{d\gamma mc^2}{dt}\right)^2$
$\frac{dp^{\mu}}{d\tau}\frac{dp_{\mu}}{d\tau} = -\gamma^2[-(\gamma m\vec{\dot{v}} + \gamma^3m\vec{v}(\vec{\beta}\cdot\vec{\dot{\beta}}))^2 + \frac{1}{c^2}(\gamma^3\vec{\beta}\cdot\vec{\dot{\beta}}mc^2)^2]$
$\frac{dp^{\mu}}{d\tau}\frac{dp_{\mu}}{d\tau} = \gamma^8m^2c^2[(\vec{\beta}\cdot\vec{\dot{\beta}})^2 - (\vec{\beta}(\vec{\beta}\cdot\vec{\dot{\beta}}) + \frac{\vec{\dot{\beta}}}{\gamma^2})^2]$
$\Rightarrow\frac{dp^{\mu}}{d\tau}\frac{dp_{\mu}}{d\tau} = \gamma^8m^2c^2\left(-\frac{1}{\gamma^2}(\vec{\beta}\cdot\vec{\dot{\beta}})^2 - \frac{\vec{\dot{\beta}}^2}{\gamma^4}\right).$

Although the above is correct as it stands, it is not immediately obvious what sort of relationship the radiated power has to the velocity and the acceleration of the particle. If we make this relationship more explicit then it will be clear how the radiation depends on the particle's motion, and what happens in different cases. We can obtain this relation by adding and subtracting $\frac{\vec{\beta}^2\cdot\vec{\dot{\beta}}^2}{\gamma^2}$ to the above giving;

$\gamma^6m^2c^2[(\vec{\beta}^2\vec{\dot{\beta}}^2 - (\vec{\beta}\cdot\vec{\dot{\beta}})^2) - \vec{\dot{\beta}}^2].$

If we apply the vector identity;

$(\vec{\beta}\times\vec{\dot{\beta}})\cdot(\vec{\beta}\times\vec{\dot{\beta}}) = (\vec{\beta}^2\vec{\dot{\beta}}^2 - (\vec{\beta}\cdot\vec{\dot{\beta}})^2).$

Then we obtain;

$P = \frac{2q^2\gamma^6}{3c}\left((\vec{\dot{\beta}})^2 - (\vec{\beta}\times\vec{\dot{\beta}})^2\right)$

where I have replaced all the constants and the negative sign dropped earlier.

This is the Lienard result, which was first obtained in 1898. The $\gamma^6$ means that when $\gamma$ is very close to one (i.e. $\beta << 1$) the radiation emitted by the particle is likely to be negligible. However when $\gamma$ is greater than one (i.e. $\beta \rightarrow 1$) the radiation explodes as the particle tries to lose its energy in the form of EM waves. It's also interesting that when the acceleration and velocity are orthogonal the power is reduced by a factor of $\vec{\beta}\cdot\vec{\dot{\beta}}$. The faster the motion becomes the greater this reduction gets. In fact, it seems to imply that as $\beta$ tends to one the power radiated tends to zero (for orthogonal motion). This would suggest that a charge moving at the speed of light, in instantaneously circular motion, emits no radiation. However, it would be impossible to accelerate a charge to this speed because the $\gamma^6$ would explode to $\infty^6$, meaning that the particle would radiate a gigantic amount of energy which would require you to put more and more energy in to keep accelerating it. This would imply that there is a cosmic speed limit, namely c. Such a connection was not made until 1905 when Einstein published his paper on Special Relativity.

We can use Lienard's result to predict what sort of radiation losses to expect in different kinds of motion.

Angular distribution

The angular distribution of radiated power is given by a general formula (applicable whether or not the particle is relativistic):[2]

$\frac{d P}{d\Omega} = \frac{q^2}{4\pi c} \frac{|\hat{n} \times ((\hat{n} - \vec{\beta})\times \dot{\vec{\beta}})|^2}{(1-\hat{n}\cdot\vec{\beta})^5}$

where $\hat{n}$ is a unit vector pointing from the particle towards the observer. In the case of linear motion (velocity parallel to acceleration), this simplifies to[3]

$P = \frac{q^2a^2}{4\pi c^3}\frac{\sin^2 \theta}{(1-\beta \cos\theta)^5}$

where $\theta$ is the angle between the observer and the particle's motion.

Issues and implications

The radiation from a charged particle carries energy and momentum. In order to satisfy energy and momentum conservation, the charged particle must experience a recoil at the time of emission. The radiation must exert an additional force on the charged particle. This force is known as the Abraham-Lorentz force in the nonrelativistic limit and the Abraham-Lorentz-Dirac force in the relativistic limit.

Atomic physics

A classical electron orbiting a nucleus experiences acceleration and should radiate. Consequently the electron loses energy and the electron should eventually spiral into the nucleus. Atoms, according to classical mechanics, are consequently unstable. This classical prediction is violated by the observation of stable electron orbits. The problem is resolved with a quantum mechanical or stochastic electrodynamic description of atomic physics.

Notes

1. ^ The case where $\beta\left(t_\text{r}\right) \neq 0$ is more complicated and is treated, for example, in Griffiths's Introduction to Electrodynamics.

References

1. ^ Purcell Simplified
2. ^ Jackson eq (14.38)
3. ^ Jackson eq (14.39)
• J. Larmor, "On a dynamical theory of the electric and luminiferous medium", Philosophical Transactions of the Royal Society 190, (1897) pp. 205–300 (Third and last in a series of papers with the same name).
• Jackson, John D. (1998). Classical Electrodynamics (3rd ed.). Wiley. ISBN 0-471-30932-X. (Section 14.2ff)
• Misner, Charles; Thorne, Kip S. & Wheeler, John Archibald (1973). Gravitation. San Francisco: W. H. Freeman. ISBN 0-7167-0344-0.
• R. P. Feynman, F. B. Moringo, and W. G. Wagner (1995). Feynman Lectures on Gravitation. Addison-Wesley. ISBN 0-201-62734-5.