Radiation

5.10.1 Radiation: E/M Dipoles

Option A. If your system can be reduced into a combination of elec-tric/magnetic dipoles, draw the picture of a dipole antenna/current loop and use the algorithm for radiation of a monochromatic source in the radiation zone:

Step 1. Write your system’s current density in the form J (~~ r⁰, t) = ~J_r(~r⁰)e^−iω0t

Start with ~J = ρ~v.

• Electric dipole: Dipole model is two tiny metal spheres separated by a distance d and connected by a fine wire. Driving a charge back and forth with angular frequency ω0, we have q(t) = q0cos (ω0t). ρ(~r⁰, t⁰) = qδ(~r⁰− ~w(t⁰)) where w(t⁰) is the position of the oscillating charge at time t⁰. We want to write ~J in terms of the dipole moment, which is p(t) = dq(t)ˆz = p₀cos ω₀tˆz where p₀= dq₀.

J (~~ r⁰, t) = ~Jr(~r⁰)e^−iω0t= −ip0ω0δ(~r⁰)e^−iω0t

• Magnetic dipole: We have a wire loop of radius b around which we drive an alternating current I(t) = I0cos (ω0t). This is a model for an oscillating magnetic dipole ~m(t) = πb²I(t)ˆz = m0cos (ω0t)ˆz where m0 = I0πb². After derivation, we find in Fourier space ~Jk( ~k0) = −i^ω0_c^m0sin θ ˆφ.

Step 2. Find the spatial Fourier transform of ~Jr(~r⁰) J~k( ~k0) =

Z

J~r(~r⁰)e^{−i ~k0· ~r0}d³r⁰

where we have defined a vector ~k0with its magnitude being the wavevector k0 = ω0/c. The direction ˆk0 ≡ ˆr that points to the observer at ~r; it is the direction the wave is propagating in.

Step 2.5 (if needed) The vector potential is A(~~ r, t) = µ0

4π e^ik0r

r

J~k(~k0)e^−iω0t Step 3. The magnetic field observed is

B(~~ r, t) = µ₀ 4π

ik₀e^ik0r r

ˆk₀× ~J_k(~k₀)e^−iω0t

When simplifying, write the exponent in terms of the retarded time tr = t − |~r − ~r⁰|/c, the moment at which the radiation left the source point. At tr, the

radiation was at a distance |~r − ~r⁰| away from the field point. Since radiation travels at the speed of light, altogether the radiation took a time |~r − ~r⁰|/c to get to the field point.

Step 4. The electric field is

E(~~ r, t) = c ~B(~r, t) × ˆk0

Step 5. The (time-averaged) Poynting vector is h~Si ≡ 1

2µ₀Re[ ~E^∗× ~B] = c

2µ₀[ ~B^∗· ~B]ˆr = c 2µ₀| ~B|²rˆ Step 6. The radiated power pattern

d hP i

dΩ = r²h~Si · ˆr

From here we can integrate over the sphere to get the radiated power hP i.

5.10.2 Radiation: Point Charges

Step 1. Write down hP i and hSi for electric/magnetic dipoles, and hP i for an arbitrary source (get the Larmor formula from this). Write down radiation reaction force.

h~Sie=µ0p²₀ω⁴₀ 32π²c

sin²θ r² rˆ hP ie= µ₀p²₀ω₀⁴

12πc h~Sim=m₀

p0c

² h~Sie

hP i =m₀ p0c

² hP ie

Prad(t0) ≈ µ₀

6πc[¨p(t0)]² F~reaction= µ₀q²

6πc~˙a

Note that you can get hSi and then ^{dhP i}_dΩ for a point charge by reverse engi-neering, using the Larmor formula and ^{dhP i}_dΩ = r²h~Si · ˆr.

Step 2. Get parameters that involve hP i or hSi:

• The time constant τ due to loss of KE due to EM radiation.

hP i = −^dU_dt. Find v(r) (Newton’s 2nd law) then U (r) to plug in, separate variables, integrate over t up to τ , integrate over r up to a0.

• Total cross section σtot for Thomson scattering (light scatters off free electron, scattered radiation is treated as a plane wave). Since ^{dhP i}_dΩ = r²h~Si · ˆr, we know that hP i ∝ h ~Si A, so σtot= ^{hP i}

|h ~Si|. Plug in the Larmor formula for hP i and recall that | h ~Si | for a plane wave is ¹₂c0E₀² where F = ma = qE0. We could also be asked to determine the power emitted per unit solid angle. For this we need to find ~S which we can do by plugging in plane wave solutions for ~E and ~B.

5.10.3 Radiation: Time Domain

If the time dependent current is too complicated to Fourier trans-form easily, stay in the time domain (e.g. radiation from a wire).

Step 0. Write down the definition of retarded time t_r= t − |~r − ~r⁰|/c, the moment at which the radiation left the source point. At t_r, the radiation was at a distance |~r − ~r⁰| away from the field point. Since radiation travels at the speed of light, altogether the radiation took a time |~r − ~r⁰|/c to get to the field point.

Step 1. Get current density ~J (~r⁰, tr). E.g. for an infinite wire along the z axis, ~J (~r⁰, t_r) = I(t_r)δ(x)δ(y)ˆz.

Step 2. Get the vector potential by plugging into the expression below and integrating.

A(~~ r, t) = µ0

4π

Z J (~~ r⁰, tr)

|~r − ~r⁰|dr⁰

Note that for t < r/c, ~A = 0 because no radiation has traveled far enough to reach the field point yet. For t ≥ r/c only radiation from a certain portion of the wire has traveled far enough to reach the field point: that where |z⁰| ≤ p(ct)²− r². So our integral simplifies: R dz → 2 R

√

(ct)²−r²

0 dz.

Step 3. Get ~E = − ~∇φ +^{∂ ~}_∂t^A

. Since φ = 0 at all times (the wire is neu-tral), ~E = −^{∂ ~}_∂t^A.

Step 4. Get ~B = ~∇ × ~A.

6 Thermodynamics

6.1 How to Solve a Thermo Problem

Step 1. Choose which thermodynamic potential you are going to use:

U (S, V ) = T S − pV

H(S, p) = U + pV F (T, V ) = U − T S

G(T, p) = F + pV

Note that H and G are only relevant to systems for which dW_rev = −pdV . U and F are more general: they involve dW_revitself, and dW_revdoes not always equal −pdV .

• When to use U ? When dS = 0/dV = 0.

• When to use H? When dS = 0/dp = 0. At constant pressure (e.g.

a chemical reaction in a test tube), dp = 0 and so dH = T dS = dQrev. Enthalpy is a easy quantity to measure at constant pressure- it is just the heat (e.g. the heat associated with a chemical reaction).

• When to use F ? When dT = 0/dV = 0.

• When to use G? When dT = 0/dp = 0 These are the conditions under which phase transitions (melting, boiling) take place, and are also relevant to chemical equilibrium. T and p are the easiest variables to measure and control.

Step 2. Write down all your relations:

• The laws of thermodynamics/the fundamental thermodynamic relation

• The chain rule and differential form for the potential(s) you are using

• The differential form of the equation of state for your gas

Step 3. Solve for thermodynamic quantities using those relations.

• Equate entire expressions if matching terms is too confusing.

• If you are given that a term is zero (e.g. T dS(x, t) = 0), expand that term according to the chain rule to see if you can learn anything new:

(^∂S_∂x)Tdx + (^∂S_∂T)xdT = 0.

• Extracting min/max amount of work from a system:

dQ = (dU − dW ) ≤ T dS

(−dW ), the work the system does on the surroundings, is minimum when dS = 0. It is maximum when dQ = T dS (the process is reversible).

• You can always use the first law/chain rule to derive relations for CV and Cp. If your system is an IMG, you can use the equipartition theorem result for U to derive CV and Cp, and you can use those in turn to derive the adiabatic constraint for an IMG.

• How to derive the Maxwell relations: e.g., dG = −SdT + V dp By the chain rule,

dG =dG dT



p

dT +dG dp



T

dp So

S = −∂G

∂T



p

V =∂G

∂p



T

You can take partial derivatives in any order:

 ∂²G

∂T ∂p

= ∂²G

∂p∂T



Therefore, the Maxwell relation for G is

−∂S

∂p



T =∂V

∂T



p

Similarly, we can derive the Maxwell relations for the other three potentials U , H and F . For non-hydrodynamic systems, we can obtain analogous relations involving, say, total magnetic moment m and magnetic field H instead of p and V : e.g. if dU = T dS−mdH, we find that (_∂H^∂T)S = (^∂m_∂S)H.