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PU-204 / L21 · lecture

The Larmor Formula

~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable with the Poynting vector, the concept of field energy density, and the basic structure of the Liénard–Wiechert potentials for a moving point charge.

Why this matters

Accelerating charges lose energy to radiation, a fundamental process governing everything from synchrotron beams to atomic transitions.

In the previous lecture, we examined the electric and magnetic fields generated by a point charge in arbitrary motion. We identified two distinct components: the "velocity" fields, which fall off as \( 1/R^2 \) and represent the static Coulomb field dragged along with the charge, and the "acceleration" fields, which fall off as \( 1/R \). It is this latter component that constitutes radiation. The Larmor formula quantifies the total power carried away by these radiation fields, providing the bridge between classical mechanics and the loss of mechanical energy due to electromagnetic emission.

The Physical Picture

Imagine a charge oscillating back and forth. When it accelerates, it disturbs the electromagnetic field around it. This disturbance propagates outward at the speed of light. Unlike the static field, which is tied to the charge's current position, the radiation field is a free wave that travels to infinity. The key insight is that the energy carried by this wave does not depend on how far away you measure it; the energy flux decreases as \( 1/R^2 \), but the area of the sphere increases as \( R^2 \), so the total power crossing any spherical surface surrounding the charge is constant. This conservation of radiated power is a direct consequence of the \( 1/R \) dependence of the acceleration fields.

From Fields to Power

To derive the total radiated power, we calculate the Poynting vector \( \vec{S} = \frac{1}{\mu_0} \vec{E} \times \vec{B} \) in the far zone. In this region, the magnetic field is related to the electric field by \( \vec{B} = \frac{1}{c} \hat{R} \times \vec{E} \). Substituting this into the Poynting vector expression, we find that the magnitude of the energy flux is proportional to the square of the electric field magnitude: \( S = \frac{E^2}{\mu_0 c} \). The electric field in the radiation zone is given by the acceleration term of the Liénard–Wiechert field. For non-relativistic speeds (\( v \ll c \)), the field simplifies to a form proportional to \( \frac{q}{4\pi\varepsilon_0 c^2 R} (\hat{R} \times (\hat{R} \times \vec{a})) \). The angular dependence of this field is crucial; it is zero along the direction of acceleration and maximum perpendicular to it.

\[ \vec{E}_{\text{rad}} = \frac{q}{4\pi\varepsilon_0 c^2 R} \left[ \hat{R} \times (\hat{R} \times \vec{a}) \right] \]

The power radiated is the integral of the Poynting vector over a sphere of radius \( R \): \( P = \oint \vec{S} \cdot d\vec{A} \). Because the field depends on the angle between the observation direction \( \hat{R} \) and the acceleration \( \vec{a} \), we must integrate over the solid angle. The term \( (\hat{R} \times (\hat{R} \times \vec{a}))^2 \) simplifies to \( a^2 \sin^2 \theta \), where \( \theta \) is the angle between \( \hat{R} \) and \( \vec{a} \). The integral of \( \sin^2 \theta \) over the sphere yields \( 8\pi/3 \). Combining these factors leads directly to the Larmor formula.

The Larmor Formula

The result of the integration is the Larmor formula, which states that the total instantaneous power radiated by a non-relativistic point charge is proportional to the square of its charge and the square of its acceleration. The formula is independent of the charge's velocity, provided that velocity is much less than the speed of light. This independence is a striking feature: a charge accelerating at \( 10 \ \mathrm{m/s^2} \) radiates the same power whether it is at rest or moving at \( 0.1c \), as long as the acceleration is the same. The constant of proportionality involves fundamental constants \( \varepsilon_0 \) and \( c \), reflecting the electromagnetic nature of the radiation.

\[ P = \frac{q^2 a^2}{6\pi\varepsilon_0 c^3} \]

Using the relation \( c^2 = 1/(\mu_0 \varepsilon_0) \), this can also be written in terms of the permeability of free space. This form is often more convenient when dealing with magnetic quantities or when comparing with other electromagnetic formulas. The Larmor formula is a cornerstone of classical electrodynamics, and the starting point for estimating radiation reaction.

Relativistic Corrections

The Larmor formula is strictly valid only in the non-relativistic limit. When the charge moves at speeds comparable to \( c \), relativistic effects become significant. The radiation pattern becomes beamed in the direction of motion, and the total power is enhanced by a factor involving the Lorentz factor \( \gamma \). The general relativistic Larmor formula is more complex and depends on the angle between the velocity and acceleration. However, for most introductory applications, such as electron orbits in atoms or radio antennas, the non-relativistic form is sufficient. It is important to remember that the Larmor formula gives the power radiated at a specific instant, determined by the acceleration at that instant. The energy radiated has to come from the charge's kinetic energy or from whatever is driving it; describing that recoil on the charge itself needs the radiation reaction (Abraham–Lorentz) force.

The derivation

◆ DERIVATION EMBED · D-170 · not re-derived here

Larmor Formula for Radiated Power

For a point charge \( q \) undergoing acceleration \( \vec{a} \) with speed \( v \ll c \), the radiation (far-zone) part of the Liénard–Wiechert fields falls off as \( 1/R \); integrating the Poynting flux of this field over a large sphere yields the total instantaneous power radiated, \[ P = \frac{q^2 a^2}{6\pi\varepsilon_0 c^3}, \] the Larmor formula: radiated power is proportional to the square of the charge and the square of its acceleration, and independent of the sphere radius.

\[ P = \frac{q^2 a^2}{6\pi\varepsilon_0 c^3} = \frac{\mu_0 q^2 a^2}{6\pi c} \]
Open full derivation →
Tier 3 ยท watch out

A common misconception is that the Larmor formula is a fair estimate for any accelerating charge. It is not, once \(v\) approaches \(c\). The relativistic (Liénard) result multiplies the power by \(\gamma^6\) for acceleration parallel to the velocity and by \(\gamma^4\) for acceleration perpendicular to it (circular motion), so for a synchrotron electron with \(\gamma \sim 10^4\) the non-relativistic formula is wrong by a factor of order \(10^{16}\). Check \(v/c\) before using \(P = q^2a^2/6\pi\varepsilon_0c^3\). C

Worked example

Problem. An electron with charge \( q = -1.60 \times 10^{-19} \ \mathrm{C} \) is accelerated by a constant electric field, resulting in an acceleration of \( a = 1.0 \times 10^{15} \ \mathrm{m/s^2} \). Calculate the total power radiated by the electron using the Larmor formula. Use \( \varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{F/m} \) and \( c = 3.00 \times 10^8 \ \mathrm{m/s} \).

  1. Identify the given values: \( q = 1.60 \times 10^{-19} \ \mathrm{C} \) (magnitude), \( a = 1.0 \times 10^{15} \ \mathrm{m/s^2} \), \( \varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{F/m} \), \( c = 3.00 \times 10^8 \ \mathrm{m/s} \).
  2. Substitute these values into the Larmor formula: \( P = \frac{q^2 a^2}{6\pi\varepsilon_0 c^3} \).
  3. Calculate the numerator: \( q^2 a^2 = (1.60 \times 10^{-19})^2 \times (1.0 \times 10^{15})^2 = 2.56 \times 10^{-38} \times 1.0 \times 10^{30} = 2.56 \times 10^{-8} \ \mathrm{C^2 \cdot m^2/s^4} \).
  4. Calculate the denominator: \( 6\pi\varepsilon_0 c^3 = 6 \times 3.14159 \times 8.85 \times 10^{-12} \times (3.00 \times 10^8)^3 \). First, \( c^3 = 27.0 \times 10^{24} \ \mathrm{m^3/s^3} \). Then, \( 6\pi \approx 18.85 \). So, \( 18.85 \times 8.85 \times 10^{-12} \times 27.0 \times 10^{24} \approx 4510 \times 10^{12} = 4.51 \times 10^{15} \ \mathrm{F \cdot m^3/s^3} \).
  5. Compute the final value: \( P = \frac{2.56 \times 10^{-8}}{4.51 \times 10^{15}} \approx 0.568 \times 10^{-23} \ \mathrm{W} = 5.68 \times 10^{-24} \ \mathrm{W} \).
\[ P \approx 5.7 \times 10^{-24} \ \mathrm{W} \]

Units check. The numerator has units \( \mathrm{C^2 \cdot m^2/s^4} \). The denominator has units \( \mathrm{F/m \cdot m^3/s^3} = \mathrm{C^2/(N \cdot m^2) \cdot m^3/s^3} = \mathrm{C^2 \cdot m / (N \cdot s^3)} \). Since \( \mathrm{N} = \mathrm{kg \cdot m / s^2} \), the denominator is \( \mathrm{C^2 / (kg \cdot s)} \). Dividing numerator by denominator gives \( \mathrm{kg \cdot m^3 / s^3} = \mathrm{J/s} = \mathrm{W} \). The units are consistent.

For scale: an electron would need this acceleration for a whole year (\(3.2 \times 10^{7}\ \mathrm{s}\)) to radiate about \(1.8 \times 10^{-16}\ \mathrm{J}\), roughly a kilo-electronvolt. Radiation losses only matter for enormous accelerations, as in synchrotrons and atomic-scale motion.

CHECK YOURSELF

Three quick tests

  1. If the acceleration of a charge is doubled, by what factor does the radiated power increase?
    AnswerIt increases by a factor of 4, since \( P \propto a^2 \).
  2. Does the Larmor formula depend on the velocity of the charge?
    AnswerNo, in the non-relativistic limit, the radiated power depends only on the acceleration, not the velocity.
  3. What is the angular dependence of the radiation field?
    AnswerThe radiation is zero along the direction of acceleration and maximum perpendicular to it, following a \( \sin^2 \theta \) pattern.
PROBLEMS

Problem set for PU-204 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-170 · Threads: force, energy, light