A charged particle A, moving with v << c, decelerates
uniformly. A second particle B has one-half the mass, twice the charge, three
times the velocity, and four times the acceleration of particle A. Find the
ratio P_{B}/P_{A} of the powers radiated.

Solution:

- Concepts:

The Larmor formula - Reasoning:

The Larmor formula is an expression for the total power radiated by a a point charge moving non-relativistically. - Details of the calculation:

Larmor formula: : P = -dE/dt = [q^{2}/(6πε_{0}c^{3})]a^{2}.

P_{B}/P_{A}= q_{B}^{2}a_{B}^{2}/(q_{A}^{2}a_{A}^{2}) = 4*16 = 64.

Electrons in a computer monitor CRT are accelerated to a final
kinetic energy of 30 keV over a distance of 1 cm, then are rapidly
decelerated to zero speed in collisions with the screen phosphor.
Assume both acceleration and deceleration are constant. Consider
the energy radiated by accelerated electrons (which has nothing directly
to do with the light emitted by the phosphor).

(a) Can this problem be treated non-relativistically? Explain
why or why not.

(b) Develop an expression for the ratio r of the energy
radiated during the acceleration phase, E_{rad}, to the
final kinetic energy E_{kin}, assuming constant acceleration a.
Also calculate a numeric value for r under the conditions
pertaining to the acceleration of electrons in the monitor CRT described
above.

(c) Again assuming constant acceleration, estimate the maximum
total fraction of kinetic energy that is radiated during the stopping of
the electrons in the phosphor, and from that, the average power radiated
per stopped electron in watts. Assume all the kinetic
energy is consumed in single collisions in a distance of 0.05 nm within
single atoms of the phosphor.

Solution:

- Concepts:

Radiation produced by accelerating charges, the Larmor formula - Reasoning:

We are asked to determine the energy radiated by an accelerating electron. - Details of the calculation:

(a) The rest energy of the electron is 510 keV, its maximum kinetic energy is 30 keV.

γ_{max}510 = 540, γ_{max}= 1.06 ≈ 1. The problem can be treated nonrelativistically.

(b) An accelerating charged particle radiates away energy. If for a time T the magnitude of the acceleration is a, then the energy radiated E_{rad}= 2e^{2}a^{2}T/(3c^{3}) is given by the Larmor formula. If the particle is initially at rest, and we neglect radiation, then the kinetic energy of the particle after time T will be E_{kin}= ½ma^{2}T^{2}.

E_{rad}<< E_{kin}, if 2e^{2}a^{2}T/(3c^{3}) << ½ma^{2}T^{2}, T >> (4/3)e^{2}/(mc^{3}) = (4/3)r_{0}/c.

For an electron r_{0}≈ 3*10^{-15}m.

If T is much greater than the time it takes light to travel 4*10^{-15}m ( or if T >> 10^{-23}s) then we can neglect radiation losses when considering the short term motion.

E_{rad}/E_{kin}= 4e^{2}/(3c^{3}mT).

½ma^{2}T^{2}= 30000*1.6*10^{-19}J. a = (30000V/0.01m)(1.6*10^{-19}C/9.1*10^{-31}kg).

T = 1.94*10^{-10}s. E_{rad}/E_{kin}= 6.4*10^{-14}.

(c) E_{kin}= ½ma^{2}t^{2}= 30 keV. From kinematics: d = ½at^{2}. Therefore a = E_{kin}/md.

t^{2}= 2d^{2}m/E_{kin}= 2(5*10^{-11}m)^{2}(9.1*10^{-31}kg)/(3*10^{4}*1.6*10^{-19}J). t = 9.7*10^{-19}s.

E_{rad}/E_{kin}= 1.3*10^{-5}. <P> = E_{rad}/t = 0.06W.

An electron is released from rest and falls under the influence of gravity. In the first centimeter, what fraction of the potential energy lost is radiated away?

Solution:>

- Concepts:

The radiation field of a point charge moving non-relativistically, the Larmor formula - Reasoning:

We are asked to determine the energy radiated by an accelerating electron. -
Details of the calculation:

The rest energy of the electron is 510 keV, the change in gravitational potential energy is

mgΔh = 9.1*10^{-31}*9.8*10^{-2}J = 5.6*10^{-13}eV.

This problem can be treated non-relativistically.

An accelerating charged particle radiates away energy.

If for a time T the magnitude of the acceleration is a, then the energy radiated

E_{rad}= 2e^{2}a^{2}T/(3c^{3}) is given by the Larmor formula.

If the particle is initially at rest, and we neglect radiation,

then the kinetic energy of the particle after time T will be E_{kin}= ½ma^{2}T^{2}.

E_{rad}/E_{kin}= 4e^{2}/(3c^{3}mT).

We find T from kinematics: d = ½gt^{2}. Here t = 4.52*10^{-2}s.

E_{rad}/E_{kin}= 2.7 10^{-22}.

Since the fraction of the energy radiated away is so small, E_{kin}is equal to the potential energy lost.

In a large region of space the electric field is constant
and homogeneous, **E** = E**i**, and gravity can be neglected. A point
mass m with charge q moves through the origin at

t_{i} = 0 with velocity **v** = (v_{0}cosα, 0, v_{0}sinα),
with v_{0} << c and cosα > 0.

At some later time t_{f} the x coordinate of the particle is L and the
particle is still moving with v_{f }<< c.

(a) Find the z-coordinate of the particle at t = t_{f}.

(b Find the total energy radiate between t_{i} and t_{f} as a
function of the variables given.

Solution:

- Concepts:

Kinematics, the Larmor formula - Reasoning:

We have motion with constant acceleration acceleration**a**= (qE/m)**i**. - Details of the calculation:

(a) Neglecting radiation, the equation of motion is**F**= q**E**.

a_{x}= qE/m, a_{y}= a_{z}= 0. L = v_{0}cosα t_{f}+ ½(qE/m)t_{f}^{2}, z(t_{f}) = v_{0}sinα t_{f}.

We can solve for t_{f}in terms of L.

t_{f}= [-mv_{0}cosα + (m^{2}v_{0}^{2}cos^{2}α + 2qEmL)^{½}]/(qE)

z(t_{f}) = v_{0}sinα [-mv_{0}cosα + (m^{2}v_{0}^{2}cos^{2}α + 2qEmL)^{½}]/(qE).

(b) The power is given by the Larmor formula. P = q^{2}a^{2}/(6πε_{0}c^{3})

The energy radiated is

E = Pt_{f}= μ_{0}q^{3}Ev_{0}[-cosα + (cos^{2}α + 2qEL/(mv_{0}^{2}))^{½}]/(6πmc).

A ball with a total charge of 1 Coulomb and a mass 1 kg is dropped from the top of a tall building of height 100 m. What is the total power radiated as a function of height? (Ignore air friction).

Solution:

- Concepts:

The radiation field of a point charge moving non-relativistically, the Larmor formula - Reasoning:

We are asked to determine the energy radiated**per unit time**by an accelerating charge. - Details of the calculation:

This problem can be treated non-relativistically.

An accelerating charged particle radiates away energy.

Power radiated = q^{2}a^{2}/(6πε_{0}c^{3}). Here a^{2}= g^{2}.

P = (1 C)^{2}(9.8 m/s^{2})^{2}/(6πε_{0}c^{3}) = 2.13*10^{-14}J/s.

The power radiated is independent of the height.

The total energy radiated as a function of distance s fallen is E_{rad}= Pt = P(2s/g)^{½}.

For s = 100 m we have E_{rad}= 9.6*10^{-14}J, completely negligible compared to the potential energy lost or kinetic energy gained.

A linear accelerator of length 10 m uniformly accelerates
protons to kineticenergy 100 MeV. Ignore relativistic effects.

(a) What is the power radiated by each proton (Watts)?

(b) What fraction of the energy imparted to the protons is lost to radiation?

(c) Sketch the normalized (1 = max) power pattern of the radiation [use a polar
plot, indicate the direction of motion of the protons].

Solution:

- Concepts:

The radiation field of a point charge moving non-relativistically, the Larmor formula - Reasoning:

We are asked to determine the power radiated by an accelerating proton. - Details of the calculation:

(a) An accelerating charged particle radiates away energy.

Power radiated = q^{2}a^{2}/(6πε_{0}c^{3}).

Here a = v^{2}/(2Δx) = E/(mΔx) (kinematics).

a = (1.6*10^{-11}J)/(1.672*10^{-27}kg * 10 m) = 9.57*10^{14}m/s^{2}.

P = (1.6*10^{-19}C)^{2}( 9.57*10^{14}m/s^{2})^{2}/(6πε_{0}c^{3}) = 5.2*10^{-24}J/s.

(b) If for a time T the magnitude of the acceleration is a, then the energy radiated is

E_{rad}= q^{2}a^{2}T/(6πε_{0}c^{3}).

If the particle is initially at rest, and we neglect radiation,

then the kinetic energy of the particle after time T will be E_{kin}= ½ma^{2}T^{2}.

E_{rad}/E_{kin}= q^{2}/(3πε_{0}c^{3}mT).

We find T from kinematics: d = ½aT^{2}. Here T = 1.45*10^{-7}s.

E_{rad}/E_{kin}= 4.7*10^{-20}.

(c) If the direction of motion is the z-direction, the direction of the acceleration is the z-direction.

P(θ,φ)dΩ is proportional to (**a**_{⊥})^{2}, therefore P(θ,φ) is proportional to sin^{2}θ, where θ is the usual polar angle. P(θ,φ) has no φ dependence.