Larmor Formula — Physica

Equivalent forms

P = \frac{\mu_0 q^2 a^2}{6 \pi c}

P = \frac{2}{3} \frac{q^2 a^2}{4\pi\varepsilon_0 c^3}

Power depends on a² — not v². Uniform motion radiates nothing (relativity demands it), but any wiggle bleeds energy at a rate set only by charge, acceleration, and the constants of free space.

Symbol	Name	SI unit	Dimension	Range
$P$	Radiated poweroutput Total electromagnetic power emitted in all directions	W	M·L²·T⁻³	0 – 1e-15
$q$	Charge Electric charge of the accelerating particle	C	T·I	1.602176634e-19 – 1.602176634e-18
$a$	Acceleration Magnitude of the particle's acceleration	m/s²	L·T⁻²	0 – 10000000000000000000

Unit systems

Symbol	SI	CGS	Imperial
$P$	W	erg/s	W
$q$	C	statC	C
$a$	m/s²	cm/s²	ft/s²

Where it holds

Valid for point charges at non-relativistic speeds (v ≪ c) observed in the far field. For relativistic motion use Liénard's generalization; at atomic scales quantum electrodynamics replaces it (discrete photon emission).

Dimensional analysis

$[q^{2}a^{2}/(\varepsilon _{0}c^{3})]$ = $(A^{2}\cdot s^{2})(m^{2}\cdot s^{-4})$ / $((A^{2}\cdot s^{4}\cdot kg^{-1}\cdot m^{-3})(m^{3}\cdot s^{-3})) = kg\cdot m^{2}\cdot s^{-3} = W \checkmark$

Discovery

Joseph Larmor · 1897

Larmor derived the radiated power of an accelerating charge in 1897, the same year Thomson identified the electron. The formula promptly broke classical physics: applied to Rutherford's orbiting electron it predicted atomic collapse in ~10⁻¹¹ s, a crisis only Bohr's quantum postulates (1913) resolved.

Try this

Classical physics predicted every atom should collapse in 16 picoseconds as its electron radiates away — why don't they?

An electron undergoes acceleration 10¹⁸ m/s². How much power does it radiate, and why did this result terrify physicists circa 1910?

Research status: stable

Real-world applications

Synchrotron light sources producing X-rays for protein crystallography
X-ray tubes (bremsstrahlung from electrons slamming into tungsten)
Radio antennas — oscillating charges radiating by design
Free-electron lasers and inverse-Compton gamma sources

Common misconceptions

Velocity doesn't matter — only acceleration radiates; a charge moving at constant 0.99c emits nothing
The radiation pattern is a donut around the acceleration vector (zero along it, maximum perpendicular), not a sphere
Classical atomic collapse was a real prediction, not a mistake in the formula — its failure for atoms is evidence for quantum mechanics, not against Larmor

Experimental verification

Synchrotron light sources measure radiated power matching the (relativistic) Larmor prediction to high precision. X-ray tube bremsstrahlung output and antenna radiation resistance

(73 \Omega for

a half-wave dipole) both follow from the same formula.

Derivation

An accelerated charge's far field has a transverse 'radiation' component

E_rad = qa \sin \theta /(4\pi \varepsilon _{0}c^{2}R)

.

The Poynting flux

S = E^{2}/(\mu _{0}c)

integrated over a sphere — with

\int \sin ^{2}\theta d\Omega = 8\pi /3

— gives

P = q^{2}a^{2}/(6\pi \varepsilon _{0}c^{3})

.

Limiting cases

a = 0

(uniform velocity)⟶

P = 0

A uniformly moving charge radiates nothing — it must, or relativity would let you detect absolute motion.

v \to c

⟶

P \to \gamma ^{4} or \gamma ^{6}

enhancementThe relativistic (Liénard) generalization multiplies

P by \gamma ^{6} for

linear

and \gamma ^{4} for

circular acceleration — why synchrotrons lose energy ferociously.

Oscillating charge,

a = \omega ^{2}x_{0}

⟶

P \propto \omega ^{4}

Radiated power scales as frequency to the fourth — the heart of Rayleigh scattering and the reason the sky is blue.

What if…

What if the charge oscillates twice as fast at the same amplitude?

Acceleration $a = \omega ^{2}x_{0}$ quadruples, so power jumps 16-fold — $the \omega ^{4}$ scaling that makes blue light scatter $\sim 10\times$ more than red and paints the sky.

What if you double the charge?

Power quadruples $(P \propto q^{2})$ . This is why heavy ions in synchrotrons radiate far more per particle than protons at the same acceleration.

What if the electron orbits at relativistic speed in a synchrotron?

The Liénard factor $\gamma ^{4}$ takes over: a 5 GeV electron $(\gamma \approx 10^{4})$ radiates $\sim 10^{16}$ times more than Larmor alone suggests — the dominant cost of circular electron colliders, and why the LHC uses protons.

1

Electron under strong acceleration

Given ·

q:: 1.602176634e-19
a:: 1000000000000000000

Find · P

Steps

$P = q^{2}a^{2}/(6\pi \varepsilon _{0}c^{3})$
Numerator: $(1.602\times 10^{-19})^{2} \times (10^{18})^{2} = 2.567\times 10^{-2}$
Denominator: $6\pi \times 8.854\times 10^{-12} \times (2.998\times 10^{8})^{3} = 4.497\times 10^{15}$
$P = 2.567\times 10^{-2}/4.497\times 10^{15} \approx 5.7\times 10^{-18} W$ — tiny, but ruinous over atomic timescales

Result ·

P \approx 5.7\times 10^{-18} W

2

Classical hydrogen collapse time

Given ·

r0:: 5.29e-11

Find · Time for the electron to spiral into the proton

Steps

Coulomb acceleration at the Bohr radius: $a = ke^{2}/(m_e r^{2}) \approx 9\times 10^{22} m/s^{2}$
Radiated power $P \approx 4.6\times 10^{-8} W$ drains the orbital energy
Integrating the energy loss gives $t = r_{0}^{3}/(4r_e^{2}c) \approx 1.6\times 10^{-11} s$
Atoms are stable — classical electromagnetism must yield to quantum mechanics

Result ·

t \approx 1.6\times 10^{-11} s

Poynting Vector→Cyclotron Frequency→Speed of Electromagnetic Waves→Coulomb's Law→