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 Joseph 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:
P = 2 3 q 2 4 π ε 0 c ( v ˙ c ) 2 = 2 3 q 2 a 2 4 π ε 0 c 3 = q 2 a 2 6 π ε 0 c 3 = μ 0 q 2 a 2 6 π c (SI units) P = 2 3 q 2 a 2 c 3 (cgs units) {\displaystyle {\begin{aligned}P&={\frac {2}{3}}{\frac {q^{2}}{4\pi \varepsilon _{0}c}}\left({\frac {\dot {v}}{c}}\right)^{2}={\frac {2}{3}}{\frac {q^{2}a^{2}}{4\pi \varepsilon _{0}c^{3}}}\\[0.6ex]&={\frac {q^{2}a^{2}}{6\pi \varepsilon _{0}c^{3}}}=\mu _{0}{\frac {q^{2}a^{2}}{6\pi c}}&{\text{ (SI units)}}\\[1.5ex]P&={\frac {2}{3}}{\frac {q^{2}a^{2}}{c^{3}}}&{\text{ (cgs units)}}\end{aligned}}}
where v ˙ {\displaystyle {\dot {v}}} or a {\displaystyle a} is the proper acceleration, q {\displaystyle q} is the charge, c {\displaystyle c} is the speed of light, and μ 0 = 1 / ε 0 c 2 {\displaystyle \mu _{0}=1/\varepsilon _{0}c^{2}} is the vacuum permeability. 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 3 m e r e a 2 c {\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 To calculate the power radiated by a point charge q {\displaystyle q} at a position r ( t ) {\displaystyle \mathbf {r} (t)} , with a velocity, v ( t ) , {\displaystyle \mathbf {v} (t),} we integrate the Poynting vector over the surface of a sphere of radius R, to get:
P = c R 2 4 π ∮ r ^ ⋅ [ E ( r , t ) × B ( r , t ) ] d Ω . {\displaystyle P={\frac {cR^{2}}{4\pi }}\oint \mathbf {\hat {r}} \cdot [\mathbf {E} (\mathbf {r} ,t)\times \mathbf {B} (\mathbf {r} ,t)]d\Omega .}
The electric and magnetic fields are given by the Liénard–Wiechert field equations,
E ( r , t ) = q ( r ^ r − β r ) r r 2 γ r 2 ( 1 − r ^ r ⋅ β r ) 3 + q c r ^ r × [ ( r ^ r − β r ) × β ˙ r ] r r ( 1 − r ^ r ⋅ β r ) 3 B ( r , t ) = r ^ × E ( r , t ) . {\displaystyle {\begin{aligned}\mathbf {E} (\mathbf {r} ,t)&={\frac {q\left(\mathbf {\hat {r}} _{r}-{\boldsymbol {\beta }}_{r}\right)}{r_{r}^{2}\gamma _{r}^{2}{\left(1-\mathbf {{\hat {r}}_{r}} \cdot {\boldsymbol {\beta }}_{r}\right)}^{3}}}+{\frac {q}{c}}\;{\frac {\mathbf {\hat {r}} _{r}\times \left[\left(\mathbf {\hat {r}} _{r}-{\boldsymbol {\beta }}_{r}\right)\times {\dot {\boldsymbol {\beta }}}_{r}\right]}{r_{r}{\left(1-\mathbf {\hat {r}} _{r}\cdot {\boldsymbol {\beta }}_{r}\right)}^{3}}}\\[1ex]\mathbf {B} (\mathbf {r} ,t)&={\hat {\mathbf {r} }}\times \mathbf {E} (\mathbf {r} ,t).\end{aligned}}}
The radius vector, r r {\displaystyle {\mathbf {r} }_{r}} , is the distance from the charged particle's position at the retarded time to the point of observation of the electromagnetic fields at the present time, β {\displaystyle {\boldsymbol {\beta }}} is the charge's velocity divided by c {\displaystyle c} , β ˙ {\displaystyle {\dot {\boldsymbol {\beta }}}} is the charge's acceleration divided by c {\displaystyle c} , and γ = ( 1 − β 2 ) − 1 / 2 {\displaystyle \gamma =(1-\beta ^{2})^{-1/2}} . The variables, r r {\displaystyle {\mathbf {r} }_{r}} , β r {\displaystyle {\boldsymbol {\beta }}_{r}} , γ r {\displaystyle \gamma _{r}} , and a r {\displaystyle {\mathbf {a} }_{r}} are all evaluated at the retarded time, t r = t − r r / c {\displaystyle t_{r}=t-r_{r}/c} . We make a Lorentz transformation to the rest frame of the point charge where v ′ = 0 {\displaystyle \mathbf {v} '=0} , and
a ′ ∥ = a ∥ γ 3 , a ⊥ ′ = a ⊥ γ 2 . {\displaystyle {\begin{aligned}\mathbf {a'} _{\parallel }&=\mathbf {a} _{\parallel }\gamma ^{3},&\mathbf {a} '_{\perp }&=\mathbf {a} _{\perp }\gamma ^{2}.\end{aligned}}}
Here, a ′ ∥ {\displaystyle {\mathbf {a} '}_{\parallel }} is the rest frame acceleration parallel to v {\displaystyle \mathbf {v} } , and a ⊥ ′ {\displaystyle \mathbf {a} '_{\perp }} is the rest frame acceleration perpendicular to v {\displaystyle \mathbf {v} } . We integrate the rest frame Poynting vector over the surface of a sphere of radius R', to get.
P ′ = c R ′ 2 4 π ∮ r ^ ′ ⋅ [ E ′ ( r ′ , t ′ ) × B ′ ( r ′ , t ′ ) ] d Ω ′ , {\displaystyle P'={\frac {cR'^{2}}{4\pi }}\oint \mathbf {\hat {r}} '\cdot \left[\mathbf {E} '(\mathbf {r} ',t')\times \mathbf {B} '(\mathbf {r} ',t')\right]d\Omega ',}
We take the limit R ′ → 0. {\displaystyle R'\to 0.} In this limit, t r ′ = t ′ {\displaystyle t'_{r}=t'} , and a ′ r = a ′ , {\displaystyle {\mathbf {a} '}_{r}={\mathbf {a} '},} so the electric field is given by
E ′ ( r ′ , t ′ ) = q r ^ ′ r ′ 2 + q [ r ^ ′ ( r ^ ′ ⋅ a ′ ) − a ′ ] c 2 r ′ , {\displaystyle \mathbf {E} '(\mathbf {r} ',t')={\frac {q\mathbf {\hat {r}} '}{r'^{2}}}+{\frac {q\left[\mathbf {\hat {r}} '\left(\mathbf {\hat {r}} '\cdot \mathbf {a} '\right)-\mathbf {a} '\right]}{c^{2}r'}},}
with all variables evaluated at the present time. Then, the surface integral for the radiated power reduces to
P ′ = q 2 4 π c 3 ∮ r ^ ′ ⋅ [ a ′ × ( r ^ ′ × a ′ ) ] d Ω ′ = 2 q 2 a ′ 2 3 c 3 . {\displaystyle P'={\frac {q^{2}}{4\pi c^{3}}}\oint {\mathbf {\hat {r}} '}\cdot \left[\mathbf {a} '\times \left({\hat {\mathbf {r} }}'\times \mathbf {a} '\right)\right]d\Omega '={\frac {2q^{2}a'^{2}}{3c^{3}}}.}
The radiated power can be put back in terms of the original acceleration in the moving frame, to give
P ′ = 2 q 2 3 c 3 ( a ∥ 2 γ 6 + a ⊥ 2 γ 4 ) = 2 q 2 γ 6 3 c 3 [ a 2 − ( v × a / c ) 2 ] . {\displaystyle P'={\frac {2q^{2}}{3c^{3}}}\left(a_{\parallel }^{2}\gamma ^{6}+a_{\perp }^{2}\gamma ^{4}\right)={\frac {2q^{2}\gamma ^{6}}{3c^{3}}}\left[a^{2}-{\left(\mathbf {v} \times \mathbf {a} /c\right)}^{2}\right].}
The variables in this equation are in the original moving frame, but the rate of energy emission on the left hand side of the equation is still given in terms of the rest frame variables. However, the right-hand side will be shown below to be a Lorentz invariant, so radiated power can be Lorentz transformed to the moving frame, finally giving
P = 2 q 2 γ 4 3 c 3 ( a ∥ 2 γ 2 + a ⊥ 2 ) = 2 q 2 γ 6 3 c 3 [ a 2 − ( v × a / c ) 2 ] . {\displaystyle P={\frac {2q^{2}\gamma ^{4}}{3c^{3}}}\left(a_{\parallel }^{2}\gamma ^{2}+a_{\perp }^{2}\right)={\frac {2q^{2}\gamma ^{6}}{3c^{3}}}\left[a^{2}-{\left(\mathbf {v} \times \mathbf {a} /c\right)}^{2}\right].}
This result (in two forms) is the same as Liénard's relativistic extension of Larmor's formula, and is given here with all variables at the present time. Its nonrelativistic reduction reduces to Larmor's original formula. For high-energies, it appears that the power radiated for acceleration parallel to the velocity is a factor γ 2 {\displaystyle \gamma ^{2}} larger than that for perpendicular acceleration. However, writing the Liénard formula in terms of the velocity gives a misleading implication. In terms of momentum instead of velocity, the Liénard formula becomes
P = 2 q 2 3 c 3 m 2 [ ( d p d t ) ∥ 2 + γ 2 ( d p d t ) ⊥ 2 ] . {\displaystyle P={\frac {2q^{2}}{3c^{3}m^{2}}}\left[\left({\frac {d\mathbf {p} }{dt}}\right)_{\parallel }^{2}+\gamma ^{2}\left({\frac {d\mathbf {p} }{dt}}\right)_{\perp }^{2}\right].}
This shows that the power emitted for d p / d t {\displaystyle d\mathbf {p} /dt} perpendicular to the velocity is larger by a factor of γ 2 {\displaystyle \gamma ^{2}} than the power for d p / d t {\displaystyle d\mathbf {p} /dt} parallel to the velocity. This results in radiation damping being negligible for linear accelerators, but a limiting factor for circular accelerators.
Covariant form The radiated power is actually a Lorentz scalar, given in covariant form as
P = − 2 3 q 2 m 2 c 3 d p μ d τ d p μ d τ . {\displaystyle P=-{\frac {2}{3}}{\frac {q^{2}}{m^{2}c^{3}}}{\frac {dp_{\mu }}{d\tau }}{\frac {dp^{\mu }}{d\tau }}.}
To show this, we reduce the four-vector scalar product to vector notation. We start with
d p μ d τ d p μ d τ = γ 2 [ ( d p 0 d t ) 2 − ( d p d t ) 2 ] . {\displaystyle {\frac {dp_{\mu }}{d\tau }}{\frac {dp^{\mu }}{d\tau }}=\gamma ^{2}\left[\left({\frac {dp_{0}}{dt}}\right)^{2}-\left({\frac {d\mathbf {p} }{dt}}\right)^{2}\right].}
The time derivatives are
d p 0 d t = d d t ( m γ ) = m γ 3 v a ∥