advanced32 min studyLesson 256 in Physics

Match radiation power to the kinematic regime

Connect Larmor power to its relativistic extension and preserve the distinction between emission time and reception time.

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What you’ll understand

  • Apply Larmor power only in its nonrelativistic regime.
  • Resolve parallel and perpendicular acceleration in the Liénard formula.
  • Explain the reception-to-emission Jacobian in angular power.

Acceleration radiates, but at relativistic speed its direction relative to velocity matters sharply. The Larmor formula is the slow-motion limit of the Liénard expression, not a universal laboratory power rule. There is another subtlety: field intensity is measured per reception time, whereas emitted power is counted per source time. Keeping these distinctions visible prevents both missing Lorentz factors and an incorrect angular denominator. Throughout this lesson acceleration means a laboratory coordinate-time derivative of velocity.

Recover the slow-motion result

For a slowly moving point charge, the transverse radiation field has magnitude proportional to charge times transverse acceleration divided by speed squared and distance. Squaring it, multiplying by the vacuum wave factor, and integrating the sine-squared pattern yields Larmor power. This is an instantaneous result for the source's retarded acceleration, not necessarily a cycle average. A sinusoid needs its own averaging; a constant acceleration magnitude does not automatically introduce a factor one half.

At prescribed acceleration, power is quadratic in charge magnitude and acceleration and independent of charge sign. This does not imply that changing mass has no effect in a specified force field. The acceleration itself depends on mass and the chosen force law. Likewise, the formula computes radiated power from a given trajectory. If radiation loss significantly changes the trajectory, one must determine that motion consistently rather than use an indefinitely maintained orbit without an energy source.

PL=Cqa2,Cq=q26πϵ0c3,β≪1P_{\mathrm L}=C_q a^2,\qquad C_q=\frac{q^2}{6\pi\epsilon_0c^3},\qquad \beta\ll1
q is charge in coulombs, a is acceleration magnitude in m/s², and P is in watts. Cq has SI units kg·s. The expression describes classical vacuum radiation from a prescribed slow point charge.↔ Scroll wide equations horizontally.

Relativity distinguishes acceleration directions

The relativistic extension contains a Lorentz factor to the sixth power multiplying acceleration squared minus the squared cross product of beta and acceleration. Split acceleration into components parallel and perpendicular to the velocity. The perpendicular coefficient loses two Lorentz powers because one minus beta squared equals the reciprocal of gamma squared. Thus equal laboratory acceleration magnitudes radiate different powers according to their orientation at relativistic speed, even for the same charge and speed.

The distinction between equal acceleration and equal force is consequential. A parallel force produces coordinate acceleration inversely proportional to gamma cubed times mass, whereas a perpendicular force gives a different response. Comparing particles at fixed field, fixed radius, fixed energy, or fixed acceleration therefore leads to different scalings. State what is held fixed before interpreting a gamma enhancement. In the limit of small beta, both direction-dependent coefficients approach one and the Larmor formula is recovered.

P=Cqγ6 ⁣(a2−∣β×a∣2)=Cq(γ6a∥2+γ4a⊥2),γ=(1−β2)−1/2P=C_q\gamma^6\!\left(a^2-|\boldsymbol\beta\times\mathbf a|^2\right)=C_q\left(\gamma^6a_{\parallel}^2+\gamma^4a_{\perp}^2\right),\qquad \gamma=(1-\beta^2)^{-1/2}
a=dv/dtᵣ is laboratory coordinate acceleration in m/s². The components are relative to laboratory velocity. This Liénard result is classical and does not include quantum recoil or solve self-consistent motion.↔ Scroll wide equations horizontally.

Specify whose clock defines angular power

Squaring the exact radiation field gives a sixth power of kappa in the denominator of radius squared times radial Poynting flux. That is received energy per observer time. The corresponding emitted energy per source time requires multiplication by dt/dtᵣ, equal to kappa. The angular emission formula consequently has a fifth power. Integrating the wrong sixth-power expression over directions does not give the Liénard emitted power for a moving charge.

This conversion assumes a fixed remote observer direction for each ray during the infinitesimal comparison. It expresses the compression or stretching of arrival intervals, not energy creation in transit. A practical audit begins by identifying the source time, the detection time, and the acceleration convention. The quadratic lab can test acceleration-amplitude scaling in a normalized slow-motion example, while the gamma and kappa factors must be evaluated analytically; its controls contain no relativistic trajectory or retarded-time calculation.

dPemitdΩ=q216π2ϵ0c3∣n×[(n−β)×a]∣2(1−n⋅β)5,dtdtr=1−n⋅β\frac{dP_{\mathrm{emit}}}{d\Omega}=\frac{q^2}{16\pi^2\epsilon_0c^3}\frac{|\mathbf n\times[(\mathbf n-\boldsymbol\beta)\times\mathbf a]|^2}{(1-\mathbf n\cdot\boldsymbol\beta)^5},\qquad \frac{dt}{dt_r}=1-\mathbf n\cdot\boldsymbol\beta
The numerator uses a in m/s², not dβ/dtᵣ. This is emitted energy per source laboratory time per steradian, for a prescribed subluminal trajectory and its outgoing far field.↔ Scroll wide equations horizontally.

Compare equal acceleration at relativistic speed

For a fixed charge and acceleration magnitude let Cq a²=2 W. At β=0.6, compute emitted power for acceleration parallel and perpendicular to velocity. Compare both with the naive Larmor value.

  1. Calculate gamma from one minus beta squared: 1−0.36=0.64, so gamma=1.25. The slow-motion reference is the supplied 2 W, independent of acceleration direction.

  2. For parallel acceleration multiply by gamma to the sixth: 2×1.25⁶≈7.629 W. The perpendicular cross product vanishes for the parallel case.

  3. For perpendicular acceleration use gamma to the fourth: 2×1.25⁴≈4.883 W. Their ratio is gamma squared, 1.5625. These values compare equal coordinate accelerations, not equal applied forces.

The result

Parallel and perpendicular accelerations emit about 7.63 W and 4.88 W. Both exceed 2 W, with a ratio 1.5625 under the stated fixed-acceleration comparison.

Explore the model

Interactive lab

Interactive lab · function

Check the acceleration square

For each fresh run Reset and set BOTH coefficient a=0.5, position x=1; then Reset and set coefficient a=0.5, position x=2. Interpret x only as normalized acceleration. Predict readout y values 0.5 and 2, a fourfold increase. This y=ax² instrument has no relativistic factors.

Check the acceleration squareThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forQuadratic acceleration scaling is separate from the direction-dependent relativistic correction.

Check your understanding

At fixed relativistic speed and equal laboratory acceleration magnitude, what is Pparallel/Pperpendicular?

Try it yourself

A slow charge radiates 5 mW at acceleration a. Its charge magnitude is doubled and acceleration becomes 3a. Find the new Larmor power. Would this scaling alone justify using it at β=0.95?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Apply both squares independently, then examine the speed assumption.

Compare with the solution

Doubling charge multiplies power by four and tripling acceleration multiplies it by nine. The product is 36, so the result is 180 mW. This scaling uses a slow prescribed trajectory. At β=0.95 the nonrelativistic approximation is unjustified; the relativistic expression also needs acceleration direction and gamma, plus classical-validity assumptions.

Take it with you

  • Larmor power is the nonrelativistic limit of Liénard power.
  • Equal laboratory acceleration gives γ⁶ and γ⁴ enhancements for parallel and perpendicular motion.
  • Angular emitted power uses κ⁻⁵ after converting reception time to source time.

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