# 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 electrodynamics, the Larmor formula is used to calculate the total power radiated by a non relativistic 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 any charged particle (such as an electron, a proton, or an ion) accelerates, it 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:

$P={2 \over 3}{\frac {q^{2}a^{2}}{4\pi \varepsilon _{0}c^{3}}}={\frac {q^{2}a^{2}}{6\pi \varepsilon _{0}c^{3}}}{\mbox{ (SI units)}}$ $P={2 \over 3}{\frac {q^{2}a^{2}}{c^{3}}}{\mbox{ (cgs units)}}$ where $a$ is the proper acceleration, $q$ is the charge, and $c$ is the speed of light. A relativistic generalization is given by the Liénard–Wiechert potentials.

In either unit system, the power radiated by a single electron can be expressed in terms of the classical electron radius and electron mass as:

$P={2 \over 3}{\frac {m_{e}r_{e}a^{2}}{c}}$ One implication is that an electron orbiting around a nucleus, as in the Bohr model, should lose energy, fall to the nucleus and the atom should collapse. This puzzle was not solved until quantum theory was introduced.

## Derivation

### Derivation 1: Mathematical approach (using CGS units)

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)

$\mathbf {E} (\mathbf {r} ,t)=q\left({\frac {\mathbf {n} -{\boldsymbol {\beta }}}{\gamma ^{2}(1-{\boldsymbol {\beta }}\cdot \mathbf {n} )^{3}R^{2}}}\right)_{\rm {ret}}+{\frac {q}{c}}\left({\frac {\mathbf {n} \times [(\mathbf {n} -{\boldsymbol {\beta }})\times {\dot {\boldsymbol {\beta }}}]}{(1-{\boldsymbol {\beta }}\cdot \mathbf {n} )^{3}R}}\right)_{\rm {ret}}$ 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 $\mathbf {r} -\mathbf {r} _{0}$ , $\mathbf {r} _{0}$ is the charge's location, and $\gamma =(1-\beta ^{2})^{-1/2}$ . The terms on the right are evaluated at the retarded time $t_{\text{r}}=t-R/c$ .

The right-hand side is the sum of the electric fields associated with the velocity and the acceleration of the charged particle. 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]

$\mathbf {S} ={\frac {q^{2}}{4\pi c}}\left|{\frac {\mathbf {n} \times (\mathbf {n} \times {\dot {\boldsymbol {\beta }}})}{R}}\right|^{2}\mathbf {n} .$ 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 the power radiated per unit solid angle is

${\frac {dP}{d\Omega }}={\frac {q^{2}}{4\pi c}}{\frac {\sin ^{2}(\theta )\,a^{2}}{c^{2}}}.$ 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: Edward M. Purcell approach

The full derivation can be found here.

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):

$E_{t}={{ea\sin(\theta )} \over {4\pi \varepsilon _{0}c^{2}R}}.$ 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:

$P={{\mu _{0}e^{2}a^{2}} \over {6\pi c}}.$ Since $c^{2}=1/\mu _{0}\epsilon _{0}$ , we recover the result quoted at the top of the article, namely

$P={2 \over 3}{\frac {q^{2}a^{2}}{4\pi \varepsilon _{0}c^{3}}}={\frac {q^{2}a^{2}}{6\pi \varepsilon _{0}c^{3}}}.$ ## Relativistic generalization

### Covariant form

Written in terms of momentum, p, the non-relativistic Larmor formula is (in CGS units)

$P={\frac {2}{3}}{\frac {q^{2}}{m^{2}c^{3}}}|{\dot {\mathbf {p} }}|^{2}.$ The power P can be shown to be Lorentz invariant. Any relativistic generalization of the Larmor formula must therefore relate P to some other Lorentz invariant quantity. The quantity $|{\dot {\mathbf {p} }}|^{2}$ appearing in the non-relativistic formula suggests that the relativistically correct formula should include the Lorentz scalar found by taking the inner product of the four-acceleration aμ = dpμ/dτ with itself [here pμ = (γmc, γmv) is the four-momentum]. The correct relativistic generalization of the Larmor formula is (in CGS units)

$P=-{\frac {2}{3}}{\frac {q^{2}}{m^{2}c^{3}}}{\frac {dp_{\mu }}{d\tau }}{\frac {dp^{\mu }}{d\tau }}.$ It can be shown that this inner product is given by

${\frac {dp_{\mu }}{d\tau }}{\frac {dp^{\mu }}{d\tau }}=\beta ^{2}\left({\frac {dp}{d\tau }}\right)^{2}-\left({\frac {d{\mathbf {p} }}{d\tau }}\right)^{2},$ and so in the limit β ≪ 1, it reduces to $-|{\dot {\mathbf {p} }}|^{2}$ , thus reproducing the nonrelativistic case.

### Non-covariant form

The above inner product can also be written in terms of β and its time derivative. Then the relativistic generalization of the Larmor formula is (in CGS units)

$P={\frac {2q^{2}\gamma ^{6}}{3c}}\left[({\dot {\boldsymbol {\beta }}})^{2}-({\boldsymbol {\beta }}\times {\dot {\boldsymbol {\beta }}})^{2}\right].$ This is the Liénard result, which was first obtained in 1898. The $\gamma ^{6}$ means that when the Lorentz factor $\gamma =1/{\sqrt {1-\beta ^{2}}}$ is very close to one (i.e. $\beta \ll 1$ ) the radiation emitted by the particle is likely to be negligible. However, as $\beta \rightarrow 1$ the radiation grows like $\gamma ^{6}$ as the particle tries to lose its energy in the form of EM waves. Also, when the acceleration and velocity are orthogonal the power is reduced by a factor of $1-\beta ^{2}=1/\gamma ^{2}$ , i.e. the factor $\gamma ^{6}$ becomes $\gamma ^{4}$ . The faster the motion becomes the greater this reduction gets.

We can use Liénard'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. In CGS units, this formula is

${\frac {dP}{d\Omega }}={\frac {q^{2}}{4\pi c}}{\frac {|\mathbf {\hat {n}} \times [(\mathbf {\hat {n}} -{\boldsymbol {\beta }})\times {\dot {\boldsymbol {\beta }}}]|^{2}}{(1-\mathbf {\hat {n}} \cdot {\boldsymbol {\beta }})^{5}}},$ where $\mathbf {\hat {n}}$ is a unit vector pointing from the particle towards the observer. In the case of linear motion (velocity parallel to acceleration), this simplifies to

${\frac {dP}{d\Omega }}={\frac {q^{2}a^{2}}{4\pi c^{3}}}{\frac {\sin ^{2}\theta }{(1-\beta \cos \theta )^{5}}},$ where $\theta$ is the angle between the observer and the particle's motion.