Problems Chapter 1

Quantum Mechanics
 K. Konishi, G. Paffuti

Problem

Form all possible quantities having the dimension of a length, by using (Subscript[m, e]), ℏ, (e^2).  Try the same by allowing also the velocity of light, c.

Solution

Having at our disposal (Subscript[m, e]), ℏ and (e^2) only, the unique combination possible is the Bohr radius

a = ((ℏ^2)/(m (e^2))) ≃ 5.29 (10^(-9)) cm.

(e^2) has dimension Energy × Length = m (v^2)L;   ℏ has dimensions p L= m v L, then

((ℏ^2)/(m (e^2)))=(( (m^2)(v^2)(L^2))/(m m (v^2) L))= L .

When c is also allowed,  other quantities can be formed.  One is the "classical eletron radius":

(Subscript[r, e])= ((e^2)/(m (c^2)))≃ 2.8 (10^(-13)) cm ;

another is the "Compton length" of the electron

(Subscript[r, c]) = (ℏ/(m c))≃ 3.8 (10^(-11))cm    or  (Subscript[λ, c])=(h/(m c))≃ 2.43 (10^(-10))cm

Note that (Subscript[r, c]) is tipically 100 times the standard nuclear radius, (10^(-13))cm. This was one of the reasons why the electron was considered to be absent in the nucleus.

As the combination

α= ((ℏ^2)/(e c))≃ (1/137)

("fine structure constant") is dimensionless, actually we can contruct an infinite number of quantities by using (Subscript[m, e]), ℏ, (e^2) and c.

Problem

Consider the scattering of an X-ray on an electron at rest. The X ray with wavelength λ is regarded as a flux of photons, each with energy hν and momentum k = h v/c, where ν = c/λ . Let p and E the momentum and the energy of the electron in the final state and let ϕ and θ the angle between the directions of the final photon momentum and electron momentum, respectively, with respect to the incident photon direction. The final photon has energy and momentum (hν)'and  h ν'/c. Use energy and momentum conservation to obtain the Compton formula

λ'- λ = ((2 h)/(m c))(Sin^2)[(ϕ/2)]= (h/(m c))(1-Cos[ϕ]).

Solution

The proof of equation () requires only energy and momentum conservation. We use momentum conservation to eliminate the electron scattering angle and energy conservation to eliminate its velocity.

Energy conservation implies

m (c^2)+ h ν =  h ν' + (Subscript[E, e]) = h ν' + ((m (c^2))/(Sqrt[1- ((v^2)/(c^2))])).

Let the x axis be in the direction of the incident X ray and y perpendicular to it, such that the momenta of the final particles are contained in the x-y plane. Using the notation of the text,  momentum conservation reads

((h ν)/c) = (Subscript[p, x])+ (Subscript[k, x]^′)= p Cos[θ] + ((h ν')/c)Cos[ϕ] = ((m v)/(Sqrt[1- ((v^2)/(c^2))]))Cos[θ]+((h ν')/c)Cos[ϕ]
0 = (Subscript[p, y]) + (Subscript[k, y]^′) = ((m v)/(Sqrt[1- ((v^2)/(c^2))]))Sin[θ] + ((h ν')/c)Sin[ϕ]

The electron angle θ can be eliminated from the two equations, giving

(((m^2) (v^2))/(1- ((v^2)/(c^2))))= (((((h ν)/c) - ((h ν')/c)Cos[ϕ]))^2) + (((((h ν')/c)Sin[ϕ]))^2)= (((((h ν)/c)))^2)+ (((((h ν')/c)))^2) - 2 ((h ν)/c)((h ν')/c) Cos[ϕ].

The left hand side can be written as

(((m^2) (v^2))/(1- ((v^2)/(c^2))))= (((m^2) (c^2)((v^2)/(c^2)))/(1- ((v^2)/(c^2))))= - (m^2)(c^2)+ (((m^2)(c^2))/(1- ((v^2)/(c^2)))) ,

while energy conservation () yields

((m c)/(Sqrt[1- ((v^2)/(c^2))]))=m c+ ((h ν )/c)-  ((h ν')/c) ⇒ (((m^2)(c^2))/(1- ((v^2)/(c^2)))) = (((m c+ ((h ν )/c)-  ((h ν')/c)))^2).

The left hand side of eq.() is then

(((m c+ ((h ν )/c)-  ((h ν')/c)))^2) - (m^2)(c^2) = (((((h ν )/c)))^2)+ (((((h ν')/c)))^2) + 2m (h ν -  h ν') - 2 ((h ν )/c)((h ν')/c) .

Substitution in equation () gives

2m (h ν -  h ν') - 2 ((h ν )/c)((h ν')/c) = - 2 ((h ν)/c)((h ν')/c) Cos[ϕ].

Multiplication by  c/h ν c /(hν)' finally gives eq.():

((m (c^2))/(h ν')) - ((m (c^2))/(h ν)) = 1 - Cos[ϕ]    ⇒     λ'- λ = (h/(m c))(1- Cos[ϕ]).

The use of invariants

This derivation is not the easiest one;  it is actually better to work directly with four - momentum. We have

k = (((h ν)/c), k);  p = ((E/c), p);    (k^2)= 0;   (p^2)= (m^2)(c^2).

The four - momentum conservation reads

k + p = k'+ p' ⇒ (((k - k'))^2)= (((p-p'))^2) .

Expanding this relation and using the invariants () one has immediately

((h ν)/c) ((h ν')/c)(1-Cos[ϕ]) = (E/c) m c - (m^2)(c^2).

Using the energy conservation

(E/c) = ((h ν)/c)+m c - ((h ν')/c)

in () reproduces ().

Problem

Consider a hydrogen atom  (made of a proton and an electron; (Subscript[m, p])∼ 1836 (Subscript[m, e])).  Calculate the reduced mass μ and compare it with (Subscript[m, e]). Solve the classical equation of motion assuming a circular orbit of radius r and write the solution for the angle θ in the form θ[t] = 2 π ν t + φ. Compute ν for r= 5 (10^(-9))cm. Compute hν and compare it with the thermal energy k T for T = 290 K, k is the Boltzman's constant, k = 1.38 (10^(-16))erg/K.

Solution

The potential energy between the electron and proton is

U[r] =- (( (e^2))/r)

the corresponding radial force is

(Subscript[F, r]) = - (( (e^2))/(r^2)).

It is known from classical mechanics that a two-body problem can be reduced to a one-body problem by using the reduced mass μ in an external potential () where

μ = (((Subscript[m, e]) (Subscript[m, p]))/((Subscript[m, e]) +(Subscript[m, p])))= (Subscript[m, e])  (1/(1+((Subscript[m, e])/(Subscript[m, p]))))≃ (Subscript[m, e])(1-((Subscript[m, e])/(Subscript[m, p]))) ∼ (Subscript[m, e])  ; as    ((Subscript[m, e])/(Subscript[m, p]))∼ 2 (10^(-3)) .

The radial equation of motion is

μ r(((((dθ)/dt)))^2)= (( (e^2))/(r^2))

For circular motion (r = const.), this equation suffices to determine the time evolution:

((d θ)/(d t))= (Sqrt[(( (e^2))/(μ (r^3)))])  ⇒   θ[t] = ω t + φ ≡ 2π ν t + φ

with

ν =(1/(2π)) (Sqrt[(( (e^2))/(μ (r^3)))]).

For the data of the problem, with  e ≃ 4.8 (10^(-10))e.s.u. we find

ν ≃ 7.17 (10^15) (sec^(-1))  ;   h ν ≃ 4.74 (10^(-11))erg.

The thermal energy k T at T = 290 K is

k T ≃ 4 (10^(-14))erg

which is three orders of magnitude smaller than  h ν .   Thus in the consideration of the specific heat for the hydrogen gas at romm temperatures the neglect of the electron degrees of freedom was justified.

An energy unit often used is the electron volt,  eV , defined as the energy acquired by an electron in a potential difference of 1 Volt (300 StatVolt). In CGS units

1 eV = 1.60218 (10^(-12)) erg.

A useful mnemonics is to note that 1 eV correspond to a wavelength

(Subscript[λ, 1]) = c (h/eV)∼ 12400 A   (1 Å =1 Angstrom = (10^(-8)) cm)

while for room temperatures T∼ 300 K

k T ∼ (1/38.7)∼ (1/40)eV.

Problem

Verify that the Hamiltonian for a charged particle moving in an external electromagnetic field

H = (1/(2m))(((p - (e/c)A))^2) + e ϕ ,

follows from the Lagrangian

L = (m/2)(v^2) + (e/c) v·A - e ϕ.

Compare the Euler-Lagrange and Hamilton's equations.

Solution

The Hamiltonian can be constructed from the Lagrangian by a Legendre transformation:

H[p,q] =(∑_i) (Subscript[p, i]) (Subscript[(OverDot[q]), i])- L[q,(OverDot[q])] ;           (Subscript[p, i])= ((∂L)/(∂(Subscript[(OverDot[q]), i]))).

On the right hand side it is understood that the velocities   (Subscript[(OverDot[q]), i]) are expressed in terms of p and q by inverting the last equation.

In our case

p = ((∂L)/(∂ v))= m v + (e/c)A  ⇒  v = (p/m) - (e/(m c))A .

The Hamiltonian is then

H =p·((p/m) - (e/(m c))A)- { (m/2)((((p/m) - (e/(m c))A))^2) + (e/c)((p/m) - (e/(m c))A)·A - e ϕ }  = (1/(2m))(((p - (e/c)A))^2) + e ϕ

Euler - Lagrange equations

The Lagrangian equations of motions are

(d/dt)((∂L)/(∂(Subscript[(OverDot[q]), i]))) = ((∂L)/(∂(Subscript[q, i])))

i.e. in our case, in components (sum over repeated index is understood)

(d/dt)(m (Subscript[v, i]) + (e/c)(Subscript[A, i]))= (e/c) (Subscript[v, k]) (∂/(∂(x^i)))(Subscript[A, k])- e (∂/(∂(x^i)))ϕ.

Expanding the time derivative

(d/dt)(Subscript[A, i]) = (∂/(∂t))(Subscript[A, i]) + ((dx^k)/dt) (∂/(∂(x^k)))(Subscript[A, i]) ,

we have

m ((d (Subscript[v, i]))/dt)=(e/c) (Subscript[v, k]) (((∂(Subscript[A, k]))/(∂(x^i)))- ((∂(Subscript[A, i]))/(∂(x^k)))) + e(-(1/c)(∂/(∂t))(Subscript[A, i]) - ((∂ϕ)/(∂(x^i)))).

Introducing the electric field and the magnetic field

(Subscript[E, i])= (-(1/c)(∂/(∂t))(Subscript[A, i]) - ((∂ϕ)/(∂(x^i))));     (Subscript[B, i])= (Subscript[ϵ, ijk]) (∂_j)(Subscript[A, k])

and noticing that

(Subscript[v, k]) (((∂(Subscript[A, k]))/(∂(x^i)))- ((∂(Subscript[A, i]))/(∂(x^k)))) =(Subscript[( v⋀ B), i])

the equations of motion take the usual form

m ((dv)/dt) = (e/c) v ⋀ B + e E .

To proof the relation () consider for example the first component, the others follow by cyclic permutations:

(Subscript[v, 1])0 + (Subscript[v, y])(((∂(Subscript[A, y]))/(∂x))- ((∂(Subscript[A, x]))/(∂y))) + (Subscript[v, z])(((∂(Subscript[A, z]))/(∂x))- ((∂(Subscript[A, x]))/(∂z))) = (Subscript[v, y])(Subscript[B, z])- (Subscript[v, z])(Subscript[B, x])= (Subscript[( v⋀ B), x]) .

Hamilton's equations

Hamilton' s equations are

((Subscript[dq, i])/dt)= ((∂H)/(∂(Subscript[p, i]))) ;   ((Subscript[dp, i])/dt) =- ((∂H)/(∂(Subscript[q, i]))).

In our case

((dx)/dt)= (1/m)(p - (e/c)A);

((Subscript[dp, i])/dt) = (1/m)((Subscript[p, k])- (e/c)(Subscript[A, k]))(e/c)((∂(Subscript[A, k]))/(∂(x^i))) - e ((∂ϕ)/(∂(x^i))).

Problem

Show that in classical mechanics the volume in the phase space occupied by the states between (q,p) and (q + δq, p + δp) remains invariant during the time evolution of the system  (Liouville's theorem).

Solution

Consider a small volume Ω in phase space. During a time evolution for a small time dt each point (q,p) evolves into (Q,P). The correspondence

(q,p) → (Q,P)

is one-to-one due to the uniqueness theorem for the solution of the equations of motion. The volume in the phase space at time t+dt is obtained by integrating the measure dQ dP, expressing these variables through (q,p) using () we can write

dQ dP = ((∂(Q,P))/(∂(q,p)))dq dp.

The Jacobian of the transformation is easily computed as, for small Δt

Q = q + (dq/dt)Δt = q + ((∂H)/(∂p))Δt;   P = p + (dp/dt)Δt = p - ((∂H)/(∂q))Δt.

We have

((∂(Q,P))/(∂(q,p)))= det(({{((∂Q)/(∂q)), ((∂Q)/(∂p))}, {((∂P)/(∂q)), ((∂P)/(∂p))}}))= (1+(((∂^2)H)/(∂p∂q))Δt)(1- (((∂^2)H)/(∂p∂q))Δt) - ((((∂^2)H)/(∂(p^2)))Δt)(- (((∂^2)H)/(∂(q^2)))Δt)≃ 1 + ((O[Δt])^2)

Thus to first order in Δt the volume in phase space is

Ω[t+Δt] = ∫dQ dP = ∫((∂(Q,P))/(∂(q,p)))dq dp ≃ ∫dq dp +  ((O[Δt])^2)=  Ω[t]+ ((O[Δt])^2) ;

i.e.,

((dΩ)/dt) = 0 .

Problem

Compute the Planck's distribution function for an observer moving with velocity V.

Solution

Remarks

Let (Subscript[ν, 0]),(Subscript[θ, 0]),  f[(Subscript[ν, 0]),(Subscript[θ, 0])] be the frequency, azimuthal angle and energy density distribution for a bunch of unpolarized photons, in a 
frequency interval between ν and (Subscript[ν, 0]) + (Subscript[dν, 0]), in a given Lorentz frame. The energy distribution is given by

(Subscript[dE, 0]) = (Subscript[f, 0])[(Subscript[ν, 0]),(Subscript[Ω, 0])] (Subscript[dν, 0]) ((Subscript[dΩ, 0])/(4π))

An observer moving with velocity V will find for the same quantity:

dE = f[ν,Ω] dν ((dΩ)/(4π)).

We must find the relation between the functions f and (Subscript[f, 0]) , in particular for the case of Planck's distribution.

Solution: first method

The first method uses the full power of Lorentz invariance (see the text of Weinberg [1] for a similar approach). The energy density is a component of the energy momentum tensor:  dE = (Subscript[T, 00]).  (Subscript[T, μν])  denotes the energy momentum tensor for the bunch of photons we are considering. In general, for particles with 4-momentum (Subscript[p, μ])   the  energy momentum tensor is proportional to (Subscript[p, μ])(Subscript[p, ν]). In our case, in the original Lorentz frame:

(Subscript[(T^((0))), μν]) = (((Subscript[(p^((0))), μ])(Subscript[(p^((0))), ν]))/(((h (Subscript[ν, 0])))^2)) (Subscript[f, 0])[(Subscript[ν, 0]),(Subscript[Ω, 0])] (Subscript[dν, 0]) ((Subscript[dΩ, 0])/(4π)).

The coefficient is fixed by requiring dE = (Subscript[T, 00]) and by noting that (Subscript[(p^((0))), 0])= h (Subscript[ν, 0]) .
Under a Lorentz transformation ((Subscript[Λ, μ])^α) we have:

(Subscript[T, μν]) = ((Subscript[Λ, μ])^α) ((Subscript[Λ, ν])^β)(Subscript[(T^((0))), αβ]) = (((Subscript[p, μ])(Subscript[p, ν]))/(((h (Subscript[ν, 0])))^2)) (Subscript[f, 0])[(Subscript[ν, 0]),(Subscript[Ω, 0])] (Subscript[dν, 0]) ((Subscript[dΩ, 0])/(4π)).

This formula is general and the variables in the RHS must be expressed through the variables in the new frame: ν, Ω. 
Let us now specialize to the case of an isotropic distribution in the original frame, i.e. with (Subscript[f, 0]) independent of (Subscript[Ω, 0]). The components of 4-momentum p are given by

(Subscript[p, 0])= h ν;  p = ((h ν)/c)(Sin[θ]Cos[φ],Sin[θ]Sin[φ],Cos[θ]);
				(Subscript[(p^((0))), 0])= h ν;  (p^((0))) = ((h (Subscript[ν, 0]))/c)(Sin[(Subscript[θ, 0])]Cos[(Subscript[φ, 0])],Sin[(Subscript[θ, 0])]Sin[(Subscript[φ, 0])],Cos[(Subscript[θ, 0])]).

V is along the z axis. The usual Lorentz transformations give (Doppler effect and aberration of light):

(Subscript[ν, 0]) = ((1+ β  Cos[θ])/(Sqrt[1- (β^2)])) ν;       Cos[(Subscript[θ, 0])] = (((β + Cos[θ]))/(1+ β  Cos[θ]));   (Subscript[φ, 0]) = φ;       β = V/c ;

From the previous equation:

(((Subscript[dΩ, 0])(Subscript[dν, 0]))/(dΩ dν))= ((Sqrt[1- (β^2)])/(1+ β  Cos[θ]))= (ν/(Subscript[ν, 0]))  ⟹  (Subscript[ν, 0]) (Subscript[dΩ, 0])(Subscript[dν, 0]) = ν dΩ dν

For the Planck's distribution:

(Subscript[f, 0])[(Subscript[ν, 0]),(Subscript[Ω, 0])]= ((8π h ((Subscript[ν, 0])^3))/(c^3))(1/(Exp[h (Subscript[ν, 0])/(Subscript[kT, 0])]-1)); ⟹ (Subscript[(T^((0))), μν]) = (Subscript[(p^((0))), μ])(Subscript[(p^((0))), ν]) ((8π )/(h (c^3)))(1/(Exp[h (Subscript[ν, 0])/(Subscript[kT, 0])]-1))(Subscript[ν, 0])(Subscript[dν, 0]) ((Subscript[dΩ, 0])/(4π)).

From eq.() and ():

(Subscript[T, μν]) = (Subscript[p, μ])(Subscript[p, ν]) ((8π )/(h (c^3)))(1/(Exp[h (Subscript[ν, 0])/k (Subscript[T, 0])]-1))ν dν ((dΩ)/(4π)) .

Using the expression of  (Subscript[ν, 0]) (eq.()) :

(Subscript[T, μν]) = (Subscript[p, μ])(Subscript[p, ν]) ((8π )/(h (c^3)))(1/(Exp[h ν/kT]-1))ν dν ((dΩ)/(4π)) ;    T[θ] = ((Sqrt[1- (β^2)])/(1+ β  Cos[θ]))(Subscript[T, 0]).

In the moving frame we have a Planck-like distribution with an angle - dependent temperature.

N.B.: In  eq.() θ is the angle between the photon momentum and the observer's velocity. The incidence angle of the photon, as seen by the observer, is  (Subscript[θ, in]) = π - θ.

For small velocities

T[θ]= (Subscript[T, 0]) (1 - β Cos[θ]+ ((β^2)/2)Cos[2θ] + O((β^3))) = (Subscript[T, 0]) (1 + β Cos[(Subscript[θ, in])]+ ((β^2)/2)Cos[2(Subscript[θ, in])] + O((β^3))).

Solution: second method

The second method of solution is more elementary. The energy density for the radiation is proportional to (E^2) + (B^2)  where E and B are the electric and magnetic field respectively. The radiation is unpolarized so, using rotation invariance around z - axis, we can restrict ourselves to the following two situations:

In the two cases, for a given amplitude A, the nonzero components are:

(1) (Subscript[E, z])=-A Sin[θ], (Subscript[E, x])= A Cos[θ];  (Subscript[B, y])= A;   (2) (1) (Subscript[B, z])=-A Sin[θ], (Subscript[B, x])= A Cos[θ];  (Subscript[E, y])= -A;

The Lorentz transformation for E and B are given by

(Subscript[(E^0), z]) = (Subscript[E, z]),  (Subscript[(E^0), x]) = γ ( (Subscript[E, x])+ β (Subscript[B, y])),   (Subscript[(E^0), y]) = γ ( (Subscript[E, y])- β (Subscript[B, x]));
(Subscript[(B^0), z]) = (Subscript[B, z]),  (Subscript[(B^0), x]) = γ ( (Subscript[B, x])- β (Subscript[E, y])),   (Subscript[(B^0), y]) = γ ( (Subscript[B, y])+ β (Subscript[E, x]));

An easy algebraic computation shows that:

(1/2)(((E^0)^2) + ((B^0)^2)) = ((Subscript[A, 0])^2)= (A^2)(γ^2)(((1+β Cos[θ]))^2)= (A^2) (((((Subscript[ν, 0])/ν)))^2);

The same result is obtained for the polarization (2). Putting together the two results we find that if (A^2)is the energy density, the quantity (A^2)/(ν^2) is invariant under the Lorentz transformation. This means, for the Planck distribution:

(u/(ν^2))dν dΩ = ((Subscript[u, 0])/((Subscript[ν, 0])^2))(Subscript[dν, 0]) (Subscript[dΩ, 0]) ⟹ u = ((ν^2)/((Subscript[ν, 0])^2))((8π h ((Subscript[ν, 0])^3))/(c^3))(1/(Exp[h (Subscript[ν, 0])/(Subscript[kT, 0])]-1)) (((Subscript[dΩ, 0])(Subscript[dν, 0]))/(dΩ dν));

using the equation ()

u = ((8π h (ν^3))/(c^3))(1/(Exp[h (Subscript[ν, 0])/(Subscript[kT, 0])]-1))

(Subscript[ν, 0]) is given as a function of ν and θ by eq.() and we obtain again the result ().

References

1) S. Weinberg, Gravitation and Cosmology, John Wiley & Sons (1972).

Problem

Solution

This problem was proposed and solved by Einstein in [2.1]  and has been used to study the fluctuations of the radiation in quantum theory.

Origin of the force

The mirror reflects the radiation with the usual optics reflection law in its rest frame. The radiation in this frame is Doppler shifted, photons coming from the front of the mirror have higher frequency than photons coming behind, there is a net transfer in momentum. The Figure shows a reflection of an incoming photon with an angle 0 < θ < π/2.

Let u be the radiation enrgy density. The Poynting vector (momentum density) is u/c. The momentum of the radiation hitting an area δA in a time interval  δτ and in a frequency interval δν is

(u/c) δA Cos[θ]  c δτ δν((dΩ)/(4π)).

In the reflection the radiation momentum changes and that of the mirror momentum are respectively:

(Subscript[ΔP, γ])= - 2 u δA (Cos^2)[θ] δν δτ ((dΩ)/(4π));  ΔP= + 2 u δA (Cos^2)[θ] δν δτ ((dΩ)/(4π)).

On the right side of the mirror the sign is changed.

An equivalent statement is that there is a change of momentum 2 hν/c Cos[θ] for each scattered photon, and the number of photons for unit interval of frequency, solid angle and unit volume is  u/ h ν.

Calculation of the force

The force on a unit surface of the mirror (pressure) for unit interval of frequency on the left and right sides of the mirror are:

((Subscript[δf, L])/(δν)) = (∫_0^(π/2))2 u (Cos^2)[θ]((dΩ)/(4π));    ((Subscript[δf, R])/(δν)) = -(∫_(π/2)^π)2 u (Cos^2)[θ]((dΩ)/(4π)).

In the rest frame of the mirror u depends on θ to order V/c. To the leading order u is isotropic and the integral is trivial:

((Subscript[((δf)^0), L])/(δν)) = ((Subscript[u, 0])/3) ;     ((Subscript[((δf)^0), R])/(δν)) = - ((Subscript[u, 0])/3).

The total force is zero. If the mirror is the boundary of the cavity the integral over ν gives the well known equation of state for the radiation

p = (1/3)(U/V)

From the results of the previous exercise we know that the only dependence on θ is in the effective temperature T[θ]. At first order in β:

T -> T (1- β Cos[θ]);  δT = - T β Cos[θ].

For the energy density it follows that:

δu = ((∂u)/(∂T))δT = - β Cos[θ]T ((∂u)/(∂T))= - β Cos[θ](ν^(3 ))T ((∂(u/(ν^3)))/(∂T))= 
+ β Cos[θ](ν^(3 ))ν ((∂(u/(ν^3)))/(∂ν))= -β Cos[θ](3 u - ν ((∂u)/(∂ν))).

We used the fact that  u/(ν^3) is a homogenous function of degree 0 in ν/T. In this formula u is the usual Planck distribution. We can now compute, to first order in β:

((Subscript[δf, L])/(δν))= ((Subscript[((δf)^0), L])/(δν)) - β (3 u - ν ((∂u)/(∂ν)))(∫_0^(π/2))2 u (Cos^3)[θ]((dΩ)/(4π))=
((Subscript[((δf)^0), L])/(δν))- (1/4)β (3 u - ν ((∂u)/(∂ν)));
((Subscript[δf, R])/(δν)) = -(∫_(π/2)^π)2 u (Cos^2)[θ]((dΩ)/(4π)) = ((Subscript[((δf)^0), R])/(δν))+ β (3 u - ν ((∂u)/(∂ν)))(∫_(π/2)^π)2 u (Cos^3)[θ]((dΩ)/(4π))=
((Subscript[((δf)^0), R])/(δν))- (1/4)β (3 u - ν ((∂u)/(∂ν)));

and summing the two,  the net force is given by:

((δf)/(δν))=- (1/2)β (3 u - ν ((∂u)/(∂ν)))= - (1/2)β (1/kT)(hν u + (((c^3)(u^2))/(8π (ν^2)))).

N.B. The force is an invariant under the Galileian transformation (i.e. to order V/c) so the result of the calculation is the same in the mirror's frame and in the radiation frame.

References

1) A. Einstein, Physikalische Zeitschrift 10, 185 (1909).