Erich Poppitz). Radiation reaction force for a dipole system.

PHY450H1S. Relativistic Electrodynamics Lecture 26 (Taught by Prof.

Erich Poppitz). Radiation reaction force for a dipole system.

Originally appeared at:

http://sites.google.com/site/peeterjoot/math2011/relativisticElectrodynamicsL26.pdf Peeter Joot — [email protected]

April 5, 2011 relativisticElectrodynamicsL26.tex

1. Reading.

Covering chapter 8 §65 material from the text [1].

Covering pp. 181-195: (182-189) [Tuesday, Mar. 29]; the EM potentials to order (v/c)² (190- 193); the “Darwin Lagrangian. and Hamiltonian for a system of nonrelativistic charged particles to order(v/c)²and its many uses in physics (194-195) [Wednesday, Mar. 30]

Next week (last topic): attempt to go to the next order(v/c)³- radiation damping, the limita- tions of classical electrodynamics, and the relevant time/length/energy scales.

2. Recap.

A system of N charged particles m_a, q_a; a∈ [1, N]closed system and nonrelativistic, v_a/c1.

In this case we can incorporate EM effects in a Largrangian ONLY involving particles (EM field not a dynamical DOF). In general case, this works to O((v/c)²), because at O((v/c))system radiation effects occur.

In a specific case, when

m₁ q₁ = ^m2

q₂ = ^m3

q₃ = · · · (1)

we can do that (meaning use a Lagrangian with particles only) to O((v/c)⁴)because of specific symmetries in such a system.

The Lagrangian for our particle after the gauge transformation is L_a = ¹

2m_av²_a+ ^ma 8

v⁴_a c² −

∑

b6=a

q_aq_b

|xa(t) −_xb(t)|+

∑

b

q_aq_bv_a·v_b+ (n·v_a)(n·v_b)

2c²|x−_xb| ^{. (2)} Next time we’ll probably get to the Lagrangian for the entire system. It was hinted that this is called the Darwin Lagrangian (after Charles Darwin’s grandson).

We find for whole system

L =

∑

a

L_a+¹ 2

∑

a

L_a(interaction) (3)

L = ¹ 2

∑

a

mav²_a+

∑

a

ma

8 v⁴_a

c² −

∑

a<b

qaq_b

|x_a(t) −x_b(t)|+

∑

b

qaq_bva·v_b+ (n·va)(n·v_b)

2c²|x−x_b| ^{. (4)} This is the Darwin Lagrangian (also Charles). The Darwin Hamiltonian is then

1

H=

∑

a

pa

2m_av²_a+

∑

a

p⁴_a

8m³_ac² +

∑

a<b

qaq_b

|x_a(t) −x_b(t)|−

∑

a<b

qaq_b 2c²m_am_b

pa·p_b+ (n_ab·pa)(n_ab·p_b)

|x−x_b| ^{. (5)}

3. Incorporating radiation effects as a friction term.

To O((v/c)³)obvious problem due to radiation (system not closed). We’ll incorporate radia- tion via a function term in the EOM

Again consider the dipole system

m¨z= −kz (6)

ω²= ^k

m (7)

or

m¨z= −ω²mz (8)

gives

d dt

 m

2 ˙z²+ ^mω

2

2 z²



=0 (9)

(because there’s no radiation).

The energy radiated per unit time averaged per period is P= ^2e

2

3c³ ¨z²

(10) We’ll modify the EOM

m¨z= −ω²mz+ f_radiation (11)

Employing an integration factor ˙z we have

m¨z ˙z = −ω²mz˙z+ f_radiation˙z (12)

or d

dt m˙z²+ω²mz²

= f_radiation˙z (13)

Observe that the last expression, force times velocity, has the form of power

md²z dt²

dz dt = ^d

dt m

2

 dz dt

2!

(14) So we can make an identification with the time rate of energy lost by the system due to radia- tion

d

dt m˙z²+ω²mz²

≡ ^dE

dt. (15)

Average over period both sides

2

 dE dt

=hf_radiation˙zi = −^2e

2

3c³ ¨z²

(16) We demand this last equality, by requiring the energy change rate to equal that of the dipole power (but negative since it is a loss) that we previously calculated.

Claim:

f_radiation = ^2e

2

3c³

...z (17)

Proof: We need to show

hf_radiationi = −^2e

2

3c³ ¨z²

(18) We have

2e²

3c³h^...z ˙zi = ^2e

2

3c³ 1 T

Z _T

0 dt...

z ˙z

= ^2e

2

3c³ 1 T

Z _T

0 dt





 d

dt(¨z ˙z) − ^2e

2

3c³ 1 T

Z _T

0 dt(¨z)²

We first used (¨z ˙z)⁰ = ^...z ˙z+ (¨z)². The first integral above is zero since the derivative of ¨z ˙z = (−ω²z₀sin ωt)(ωz₀cos ωt) = −ω³z²₀sin(2ωt)/2 is also periodic, and vanishes when integrated over the interval.

2e²

3c³ h^...z ˙zi = −^2e

2

3c³

(¨z)² (19)

We can therefore write

m¨z= −mω²z+ ^2e

2

3c³

...z (20)

Our “frictional” correction is the radiation reaction force proportional to the third derivative of the position.

Rearranging slightly, this is

¨z= −ω²z+ ² 3c

 e² mc²

 ...

z = −ω²z+ ² 3c

r_e c

...z , (21)

where r_e ∼ 10⁻¹³cm is the “classical radius” of the electron. In our frictional term we have re/c, the time for light to cross the classical radius of the electron.

There are lots of problems with this. One of the easiest is with ω =0. Then we have

¨z= ² 3

r_e c

...z (22)

with solution

z∼e^αt, (23)

3

where

α∼ ^c r_e ∼ ¹

τ_e. (24)

This is a self accelerating system! Note that we can also get into this trouble with ω 6= 0, but those examples are harder to find (see: [2]).

FIXME: borrow this text again to give that section a read.

The sensible point of view is that this third term ( f_rad) should be taken seriously only if it is small compared to the first two terms.

References

[1] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980. 1 [2] D.J. Griffith. Introduction to Electrodynamics. Prentice-Hall, 1981.3

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