# Larmor formula

In electrodynamics, the Larmor formula 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 any charged particle (such as an electron, a proton, or an ion) accelerates, energy is radiated in the form of electromagnetic waves. For a particle whose velocity is small relative to the speed of light (i.e., nonrelativistic), the total power that the particle radiates (when considered as a point charge) can be calculated by the Larmor formula:

${\displaystyle P={\frac {2}{3}}{\frac {q^{2}}{4\pi \varepsilon _{0}c}}\left({\frac {\dot {v}}{c}}\right)^{2}={2 \over 3}{\frac {q^{2}a^{2}}{4\pi \varepsilon _{0}c^{3}}}={\frac {q^{2}a^{2}}{6\pi \varepsilon _{0}c^{3}}}=\mu _{0}{\frac {q^{2}a^{2}}{6\pi c}}{\text{ (SI units)}}}$
${\displaystyle P={\frac {2}{3}}{\frac {q^{2}a^{2}}{c^{3}}}{\text{ (cgs units)}}}$
where ${\displaystyle {\dot {v}}}$ or ${\displaystyle a}$ is the proper acceleration, ${\displaystyle q}$ is the charge, and ${\displaystyle c}$ 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:

${\displaystyle P={\frac {2}{3}}{\frac {m_{e}r_{e}a^{2}}{c}}}$

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)

${\displaystyle \mathbf {E} (\mathbf {r} ,t)=q\left({\frac {\mathbf {n} -{\boldsymbol {\beta }}}{\gamma ^{2}(1-{\boldsymbol {\beta }}\cdot \mathbf {n} )^{3}R^{2}}}\right)_{\text{ret}}+{\frac {q}{c}}\left({\frac {\mathbf {n} \times [(\mathbf {n} -{\boldsymbol {\beta }})\times {\dot {\boldsymbol {\beta }}}]}{(1-{\boldsymbol {\beta }}\cdot \mathbf {n} )^{3}R}}\right)_{\text{ret}}}$
and
${\displaystyle \mathbf {B} =\mathbf {n} \times \mathbf {E} ,}$
where ${\displaystyle {\boldsymbol {\beta }}}$ is the charge's velocity divided by ${\displaystyle c}$, ${\displaystyle {\dot {\boldsymbol {\beta }}}}$ is the charge's acceleration divided by c, ${\displaystyle \mathbf {n} }$ is a unit vector in the ${\displaystyle \mathbf {r} -\mathbf {r} _{0}}$ direction, ${\displaystyle R}$ is the magnitude of ${\displaystyle \mathbf {r} -\mathbf {r} _{0}}$, ${\displaystyle \mathbf {r} _{0}}$ is the charge's location, and ${\displaystyle \gamma =(1-\beta ^{2})^{-1/2}}$. The terms on the right are evaluated at the retarded time ${\displaystyle t_{\text{r}}=t-R/c}$.

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 ${\displaystyle {\boldsymbol {\beta }}}$ while the acceleration field depends on both ${\displaystyle {\boldsymbol {\beta }}}$ and ${\displaystyle {\dot {\boldsymbol {\beta }}}}$ and the angular relationship between the two. Since the velocity field is proportional to ${\displaystyle 1/R^{2}}$, it falls off very quickly with distance. On the other hand, the acceleration field is proportional to ${\displaystyle 1/R}$, which means that it falls off 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:

${\displaystyle \mathbf {S} ={\frac {c}{4\pi }}\mathbf {E} _{\text{a}}\times \mathbf {B} _{\text{a}},}$
where the 'a' subscripts emphasize that we are taking only the acceleration field. Substituting in the relation between the magnetic and electric fields and simplifying gives, in the non-relativistic case,
${\displaystyle \mathbf {S} ={\frac {q^{2}}{4\pi c}}\left|{\frac {\mathbf {n} \times (\mathbf {n} \times {\dot {\boldsymbol {\beta }}})}{R}}\right|^{2}\mathbf {n} .}$

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

${\displaystyle {\frac {\mathrm {d} P}{\mathrm {d} \Omega }}={\frac {q^{2}}{4\pi c}}{\frac {\sin ^{2}(\theta )\,a^{2}}{c^{2}}}.}$

The total power radiated is found by integrating this quantity over all solid angles (that is, over ${\displaystyle \theta }$ and ${\displaystyle \phi }$). This gives

${\displaystyle P={\frac {2}{3}}{\frac {q^{2}a^{2}}{c^{3}}},}$
which is the Larmor result for a nonrelativistic 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.

## Relativistic generalization

### Covariant form

Written in terms of momentum, p, the nonrelativistic Larmor formula is (in CGS units)

${\displaystyle P={\frac {2}{3}}{\frac {q^{2}}{m^{2}c^{3}}}|{\dot {\mathbf {p} }}|^{2}.}$

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 ${\displaystyle |{\dot {\mathbf {p} }}|^{2}}$ appearing in the nonrelativistic formula suggests that the relativistically correct formula should include the Lorentz scalar found by taking the inner product of the four-acceleration aμ = dpμ/ with itself [here pμ = (γmc, γmv) is the four-momentum]. The correct relativistic generalization of the Larmor formula is (in CGS units)

${\displaystyle P=-{\frac {2}{3}}{\frac {q^{2}}{m^{2}c^{3}}}{\frac {dp_{\mu }}{d\tau }}{\frac {dp^{\mu }}{d\tau }}.}$

It can be shown that this inner product is given by

${\displaystyle {\frac {dp_{\mu }}{d\tau }}{\frac {dp^{\mu }}{d\tau }}=\beta ^{2}\left({\frac {dp}{d\tau }}\right)^{2}-\left({\frac {d{\mathbf {p} }}{d\tau }}\right)^{2},}$

and so in the limit β ≪ 1, it reduces to ${\displaystyle -|{\dot {\mathbf {p} }}|^{2}}$, thus reproducing the nonrelativistic case. Expressed in terms of the Lorentz invariant proper acceleration, the relativistic Larmor power is (in CGS still)

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

### 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)

${\displaystyle P={\frac {2q^{2}\gamma ^{6}}{3c}}\left[({\dot {\boldsymbol {\beta }}})^{2}-({\boldsymbol {\beta }}\times {\dot {\boldsymbol {\beta }}})^{2}\right].}$

This is the Liénard result, which was first obtained in 1898. As ${\displaystyle \beta \rightarrow 1}$ the radiation grows like ${\displaystyle \gamma ^{6}}$, and the particle loses its energy in the form of EM waves. When the acceleration and velocity are orthogonal the power is reduced by a factor of ${\displaystyle 1-\beta ^{2}=1/\gamma ^{2}}$.

However, writing the Liénard formula in terms of the velocity gives a misleading implication. In terms of momentum instead of velocity, the Liénard formula for acceleration parallel to the velocity becomes

${\displaystyle P_{\parallel }={\frac {2q^{2}}{3cm^{2}}}\left({\frac {d\mathbf {p} }{dt}}\right)^{2}.}$

For acceleration perpendicular to the velocity, the radiated power is

${\displaystyle P_{\perp }={\frac {2q^{2}\gamma ^{2}}{3cm^{2}}}\left({\frac {d\mathbf {p} }{dt}}\right)^{2}.}$

This shows that the power emitted for acceleration perpendicular to the velocity is larger by a factor of ${\displaystyle \gamma ^{2}}$ than the power for acceleration parallel to the velocity.

### 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

${\displaystyle {\frac {\mathrm {d} P}{\mathrm {d} \Omega }}={\frac {q^{2}}{4\pi c}}{\frac {\left|\mathbf {\hat {n}} \times \left[(\mathbf {\hat {n}} -{\boldsymbol {\beta }})\times {\dot {\boldsymbol {\beta }}}\right]\right|^{2}}{(1-\mathbf {\hat {n}} \cdot {\boldsymbol {\beta }})^{5}}},}$
where ${\displaystyle \mathbf {\hat {n}} }$ is a unit vector pointing from the particle towards the observer. In the case of linear motion (velocity parallel to acceleration), this simplifies to
${\displaystyle {\frac {\mathrm {d} P}{\mathrm {d} \Omega }}={\frac {q^{2}a^{2}}{4\pi c^{3}}}{\frac {\sin ^{2}\theta }{(1-\beta \cos \theta )^{5}}},}$
where ${\displaystyle \theta }$ is the angle between the observer and the particle's motion.

## Radiation at the present time

In the formulations of Larmor's formula given above, the acceleration is given at the retarded time. This means that any acceleration in the earlier motion of the charged particle could be used in the formula, which makes it essentially undetermined. This difficulty has been resolved by a recent derivation that gives the acceleration in all of the formulas above at the present time.

## 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 Abraham-Lorentz force while its non-relativistic limit is known as the Lorentz self-force and relativistic forms are known as Lorentz-Dirac force or Abraham-Lorentz-Dirac force. The radiation reaction phenomenon is one of the key problems and consequences of the Larmor formula. According to classical electrodynamics, a charged particle produces electromagnetic radiation as it accelerates. The particle loses momentum and energy as a result of the radiation, which is carrying it away from it. The radiation response force, on the other hand, also acts on the charged particle as a result of the radiation.

The dynamics of charged particles are significantly impacted by the existence of this force. In particular, it causes a change in their motion that may be accounted for by the Larmor formula, a factor in the Lorentz-Dirac equation.

According to the Lorentz-Dirac equation, a charged particle's velocity will be influenced by a "self-force" resulting from its own radiation. Such non-physical behavior as runaway solutions, when the particle's velocity or energy become infinite in a finite amount of time, might result from this self-force.

The Lorentz-Dirac equation's self-force problem has generated a great deal of discussion and study in theoretical physics. Even though the equation has occasionally proved successful in describing the motion of charged particles, it is still a subject of current research.

### Atomic physics

The invention of quantum physics, notably the Bohr model of the atom, was able to explain this gap between the classical prediction and the actual reality. The Bohr model proposed that transitions between distinct energy levels, which electrons could only inhabit, might account for the observed spectral lines of atoms. The wave-like properties of electrons and the idea of energy quantization were used to explain the stability of these electron orbits.

The Larmor formula can only be used for non-relativistic particles, which limits its usefulness. The Liénard-Wiechert potential is a more comprehensive formula that must be employed for particles travelling at relativistic speeds. Additionally, the Larmor formula makes the unavoidable assumption that the charged particle is orbiting in a circle. In certain situations, more intricate calculations including numerical techniques or perturbation theory could be necessary to precisely compute the radiation the charged particle emits.

This page was last updated at 2024-04-16 09:32 UTC. . View original page.

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