Skip to content

Radiation from Accelerating Charges

A non-relativistic charge qq undergoing acceleration a\mathbf{a} radiates power:

P=q2a26πε0c3P = \frac{q^2 a^2}{6\pi\varepsilon_0 c^3}

For an oscillating dipole p=qdcosωt\mathbf{p} = q\mathbf{d}\cos\omega t with acceleration a=ω2da = \omega^2 d:

P=q2ω4d212πε0c3=ω4p0212πε0c3P = \frac{q^2 \omega^4 d^2}{12\pi\varepsilon_0 c^3} = \frac{\omega^4 p_0^2}{12\pi\varepsilon_0 c^3}

Where p0=qdp_0 = qd is the dipole moment amplitude.

Radiation resistance: Equating P=12I02RradP = \frac{1}{2}I_0^2 R_{\text{rad}} for an antenna of length \ell carrying current I0I_0 at frequency ω\omega:

Rrad=μ0c6π(ωc)2=π6Z0(λ)2197(λ)2 ΩR_{\text{rad} = \frac{\mu_0 c}{6\pi}\left(\frac{\omega \ell}{c}\right)^2 = \frac{\pi}{6}Z_0\left(\frac{\ell}{\lambda}\right)^2 \approx 197\left(\frac{\ell}{\lambda}\right)^2\ \Omega}

The radiation fields from an oscillating electric dipole at distance rλr \gg \lambda:

E=μ0ω2p04πrsinθei(krωt)θ^\mathbf{E} = -\frac{\mu_0 \omega^2 p_0}{4\pi r}\sin\theta\, e^{i(kr - \omega t)}\,\hat{\boldsymbol{\theta}}

B=μ0ω2p04πcrsinθei(krωt)ϕ^\mathbf{B} = -\frac{\mu_0 \omega^2 p_0}{4\pi c\, r}\sin\theta\, e^{i(kr - \omega t)}\,\hat{\boldsymbol{\phi}}

The angular distribution of radiated power:

dPdΩ=μ0p02ω432π2csin2θ\frac{dP}{d\Omega} = \frac{\mu_0 p_0^2 \omega^4}{32\pi^2 c}\sin^2\theta

The total power (integrating over solid angle):

P=μ0p02ω412πcP = \frac{\mu_0 p_0^2 \omega^4}{12\pi c}

The radiation pattern is toroidal (doughnut-shaped), with zero radiation along the dipole axis (θ=0,π\theta = 0, \pi) and maximum in the equatorial plane (θ=π/2\theta = \pi/2).

10.3 Relativistic Radiation: Liénard—Wiechert Potentials

Section titled “10.3 Relativistic Radiation: Liénard—Wiechert Potentials”

For a relativistic charge with velocity β=v/c\boldsymbol{\beta} = \mathbf{v}/c and acceleration β˙\dot{\boldsymbol{\beta}}:

P=q26πε0cγ6[(β˙)2(β×β˙)2]P = \frac{q^2}{6\pi\varepsilon_0 c}\gamma^6\left[(\dot{\boldsymbol{\beta}})^2 - (\boldsymbol{\beta} \times \dot{\boldsymbol{\beta}})^2\right]

For linear acceleration (ββ˙\boldsymbol{\beta} \parallel \dot{\boldsymbol{\beta}}):

P=q26πε0cγ6β˙2P = \frac{q^2}{6\pi\varepsilon_0 c}\gamma^6\dot{\beta}^2

For circular acceleration (ββ˙\boldsymbol{\beta} \perp \dot{\boldsymbol{\beta}}E.g., synchrotron):

P=q26πε0cγ4β˙2=q2c6πε0γ4R2P = \frac{q^2}{6\pi\varepsilon_0 c}\gamma^4\dot{\beta}^2 = \frac{q^2 c}{6\pi\varepsilon_0}\frac{\gamma^4}{R^2}

Where RR is the radius of curvature. The γ4\gamma^4 factor (vs. γ6\gamma^6 for linear) explains why synchrotron radiation is significant for relativistic electrons but negligible for protons at the same energy (γ\gamma is mp/me1836m_p/m_e \approx 1836 times smaller).

Synchrotron radiation spectrum: The critical frequency is ωc=32γ3cR\omega_c = \frac{3}{2}\gamma^3\frac{c}{R}. The spectrum peaks near ωc\omega_c and extends to high harmonics, making synchrotron radiation a powerful broadband source from infrared to X-rays.

Worked Example 10.1: Synchrotron Radiation from a Storage Ring

The Diamond Light Source operates at E=3E = 3 GeV electron energy with a ring circumference of 561.6 m.

(a) Lorentz factor: γ=E/(mec2)=3×109/(0.511×106)=5871\gamma = E/(m_e c^2) = 3 \times 10^9/(0.511 \times 10^6) = 5871.

(b) For a bending magnet with radius R=7.1R = 7.1 m:

P=e2c6πε0γ4R2=(1.6×1019)2×3×1086π×8.85×1012(5871)4(7.1)2P = \frac{e^2 c}{6\pi\varepsilon_0}\frac{\gamma^4}{R^2} = \frac{(1.6 \times 10^{-19})^2 \times 3 \times 10^8}{6\pi \times 8.85 \times 10^{-12}}\frac{(5871)^4}{(7.1)^2}

=2.56×1038×3×1081.669×10101.187×101550.4= \frac{2.56 \times 10^{-38} \times 3 \times 10^8}{1.669 \times 10^{-10}}\frac{1.187 \times 10^{15}}{50.4}

=4.60×1020×2.355×1013=1.08×106 Wperelectron= 4.60 \times 10^{-20} \times 2.355 \times 10^{13} = 1.08 \times 10^{-6}\ \text{W} per electron

With a beam current of 300 mA (I=0.3I = 0.3 A, N=I/e=1.875×1018N = I/e = 1.875 \times 10^{18} electrons/s):

Total power =1.08×106×1.875×1018×561.6(2π×7.1)= 1.08 \times 10^{-6} \times 1.875 \times 10^{18} \times \frac{561.6}{(2\pi \times 7.1)}

Wait: the power per electron is already the total radiated power. The total synchrotron radiation power from the ring is:

Ptotal=Nstored×Pperelectron×bendinglengthcircumferenceP_{\text{total} = N_{\text{stored} \times P_{\text{per} electron} \times \frac{\text{bending} length}{\text{circumference}}}}

For a rough estimate: Ptotal0.3×3×109×1.08×1061.6×1019×2π×7.1561.6500P_{\text{total} \approx 0.3 \times 3 \times 10^9 \times \frac{1.08 \times 10^{-6}}{1.6 \times 10^{-19}} \times \frac{2\pi \times 7.1}{561.6} \approx 500} kW.

The actual Diamond power is about 400 kW, consistent with this estimate.

(c) Critical frequency:

ωc=32γ3cR=32(5871)33×1087.1=1.5×2.02×1011×4.23×107=1.28×1019 rad/s\omega_c = \frac{3}{2}\gamma^3\frac{c}{R} = \frac{3}{2}(5871)^3\frac{3 \times 10^8}{7.1} = 1.5 \times 2.02 \times 10^{11} \times 4.23 \times 10^7 = 1.28 \times 10^{19}\ \text{rad}/s

ωc=1.055×1034×1.28×1019=1.35×1015 J=8.4 keV\hbar\omega_c = 1.055 \times 10^{-34} \times 1.28 \times 10^{19} = 1.35 \times 10^{-15}\ \text{J} = 8.4\ \text{keV}

This is in the hard X-ray range, suitable for protein crystallography and materials science.

  • The Larmor formula is the non-relativistic limit of the full relativistic expression.
  • The angular distribution sin2θ\sin^2\theta implies zero radiation along the acceleration axis.
  • Synchrotron radiation power scales as γ4\gamma^4 for circular motion but γ6\gamma^6 for linear motion.
  • The critical frequency ωcγ3\omega_c \propto \gamma^3 means higher energy electrons produce higher frequency radiation.
  • Confusing the radiated power with the energy density of the field. The power is the flux of the Poynting vector over a sphere, not the field energy.
  • Forgetting the γ4\gamma^4 vs γ6\gamma^6 distinction when comparing synchrotron and linear acceleration radiation.
  • Using the non-relativistic Larmor formula at relativistic speeds without applying the appropriate Lorentz transformation.
  • Neglecting that the radiation reaction force is small compared to the Lorentz force for most practical accelerator configurations.
  • Synchrotron light sources: Produce intense broadband radiation from infrared to X-rays for materials science, biology, and chemistry experiments.
  • Astrophysics: Explains radiation from pulsars, active galactic nuclei, and cosmic microwave background fluctuations.
  • Bremsstrahlung: X-ray production in medical imaging and industrial inspection uses radiation from decelerating electrons.
  • Antenna theory: The radiation resistance formula guides the design of dipole and monopole antennas for communication systems.

10.7 Worked Example: Bremsstrahlung Radiation

Section titled “10.7 Worked Example: Bremsstrahlung Radiation”

Problem. An electron decelerates from v=0.1cv = 0.1c to rest in a distance d=1d = 1 mm inside a metal target. Estimate the fraction of kinetic energy radiated as bremsstrahlung.

Solution

The kinetic energy is K=12mev2=12(9.109×1031)(3×107)2=4.10×1016K = \frac{1}{2}m_e v^2 = \frac{1}{2}(9.109 \times 10^{-31})(3 \times 10^7)^2 = 4.10 \times 10^{-16} J =2.56= 2.56 keV.

The stopping time is Δt=2d/v=2(103)/(3×107)=6.67×1011\Delta t = 2d/v = 2(10^{-3})/(3 \times 10^7) = 6.67 \times 10^{-11} s, so the average deceleration is a=v/Δt=4.50×1017a = v/\Delta t = 4.50 \times 10^{17} m/s2^2.

The radiated power is P=e2a2/(6πε0c3)=(1.6×1019)2(4.50×1017)2/(6π×8.85×1012×(3×108)3)2.31×1019P = e^2 a^2/(6\pi\varepsilon_0 c^3) = (1.6 \times 10^{-19})^2 (4.50 \times 10^{17})^2/(6\pi \times 8.85 \times 10^{-12} \times (3 \times 10^8)^3) \approx 2.31 \times 10^{-19} W.

The radiated energy is Erad=PΔt=2.31×1019×6.67×1011=1.54×1029E_{\text{rad}} = P \Delta t = 2.31 \times 10^{-19} \times 6.67 \times 10^{-11} = 1.54 \times 10^{-29} J.

The fraction is Erad/K3.76×1014E_{\text{rad}}/K \approx 3.76 \times 10^{-14}, which is negligible. \blacksquare

flowchart TD
A[10_Radiation From Accelerating Charges] --> B[Key Concepts]
A --> C[Core Principles]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
C --> F[Design patterns]
D --> G[Real-world usage]

Radiation is how accelerating charges shed energy. The Larmor formula shows that power radiated is proportional to the square of acceleration, so rapidly changing charges radiate intensely. The toroidal radiation pattern means antennas radiate maximally perpendicular to their axis, like a doughnut of energy. Synchrotron radiation becomes directional at relativistic speeds because the radiation cone narrows, like a lighthouse beam. Bremsstrahlung produces a continuous X-ray spectrum because the deceleration is随机. The radiation reaction force is the charge feeling its own field, a subtle effect that becomes important in extreme environments.

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

Ensure you have mastered the prerequisite material before attempting this advanced content.

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

Ensure you have mastered the prerequisite material before attempting this advanced content.