The distribution function and its moments

Exercise 1.12

Photon–electron scattering (a) Use four-vector techniques to show that when a photon of wavelength λ scatters off a stationary electron of mass me, its wavelength will change to λ^such that λ^− λ = (h/mec)(1− cos θ) where θ is the scattering angle.

(b) A related process, called inverse Compton scattering, occurs when a charged particle of mass m and energy E (in the lab frame) collides head-on with a photon of frequency ν.

Show that when E mc², the maximum energy that is transfered to the photon is given by E[1 + (m²c⁴/4hνE)]⁻¹.

Exercise 1.13

More practice with collisions Prove the following results.

(a) The threshold of energy for the production of an e⁺e⁻ pair in a collision between a photon and an electron at rest is 4mec².

(b) A high energy electron strikes an electron at rest in an elastic encounter and the two electrons share the energy equally. Then the angle between their directions of travel will be π/2 in non-relativistic scattering but will be less than π/2 in relativistic mechanics.

(c) If a particle of mass M hits a stationary target of mass m, the γ factor of the inci-dent particle after the collision cannot exceed (m²+ M²)/2mM . Compare this with the corresponding situation in the non-relativistic situation.

Exercise 1.14

Relativistic rocket A relativistic rocket has a variable rest mass m(τ ) and obeys the equa-tion of moequa-tion d(muⁱ)/dτ = Jⁱ where Jⁱ is the rate of emission of four-momentum through the burning of the fuel.

(a) Show that this requires the condition

mfinal< minitialexp



g(τ ) dτ



, (1.87)

where g is the magnitude of the acceleration.

(b) Consider a motion in (1+1) dimension with g(τ ) = dχ/dτ where χ is the rapidity.

If the rocket starts from rest and reaches a final velocity vfinalshow that

m_final< m_initial

1− vfinal

1 + vfinal

. (1.88)

1.9 The distribution function and its moments

So far, we have discussed the dynamics of a single, free particle. Often in physics, one has to deal with a large collection of particles undergoing nearly identical physical processes. In non-relativistic mechanics, we deal with this situation using a distribution function. It is necessary to generalize this concept in a Lorentz invariant manner to take into account a system of relativistic particles.

In order to do that, we shall first obtain several Lorentz invariant quantities which will serve as basic building blocks. Let us consider a set of N particles, each of mass m, described by a distribution function f (pⁱ) at any given location in space.

The total number of particles can be written in terms of the distribution function as N = particles have mass m and the theta function θ(p⁰) (which is unity for p⁰ > 0 and vanishes for p⁰ < 0) ensures that p⁰ > 0 so that the energy is positive. The quanti-ties N, d⁴p, θ and δ_D(p^ap_a+m²c²) are all individually Lorentz invariant, implying f is Lorentz invariant. (It is obvious from their definitions that N, d⁴p, θ(p⁰) are Lorentz invariant. To prove that the Dirac delta function is invariant we only need to use the fact that Lorentz transformation has unit Jacobian.) Introducing the energy E_p≡ (m²c⁴+ p²c²)^1/2corresponding to momentum p, we write the Dirac delta Noting that integration over dp⁰ in Eq. (1.89) will merely replace p⁰ by (Ep/c) due to the condition p⁰ > 0, we get

Since N and f are invariant, the combination (d³p/Ep) must be invariant under Lorentz transformations.

We noted earlier (see page26) that u⁰d³x = d⁴x/dτ is Lorentz invariant. Since E = mcu⁰, it follows that the combination Epd³x is also an invariant. Combined with the result that d³p/Ep is Lorentz invariant, we conclude that the product (Epd³x)(d³p/Ep) = d³xd³p is Lorentz invariant. In other words, an element of phase volume is Lorentz invariant even though neither the spatial volume nor the volume in momentum space is individually invariant.

This result allows us to introduce distribution functions in relativistic theory in exact analogy with non-relativistic mechanics. We define the distribution function f such that

dN = f (xⁱ, p)d³xd³p (1.92)

1.9 The distribution function and its moments 37 represents the number of particles in a small phase volume d³xd³p. The xⁱ here has the components (ct, x) while p is the three-momentum vector; the fourth com-ponent of the momentum vector (Ep/c) does not appear since it is completely determined by p and mass m of the particle. Each of the quantities dN , f and d³xd³p are individually Lorentz invariant.

Given the Lorentz invariant distribution function f , one can construct sev-eral other invariant quantities by taking moments of this function. Of particular importance are the moments constructed by integrating the distribution function over various powers of the four-momentum. We shall now construct a few such examples.

The simplest Lorentz invariant quantity which can be obtained from the distri-bution function by integrating out the momentum, is the harmonic mean ¯E_har of the energy of the particles at an event xⁱ. This is defined by the relation

1 E¯har(xⁱ) ≡

 d³p

E_pf (xⁱ, p), (1.93) which is clearly Lorentz invariant because of our earlier results. Unfortunately, this quantity does not seem to play any important role in physics.

Taking the first power of the four-momentum, we can define the four-vector S^a(xⁱ)≡ c

 d³p

E_pp^af (xⁱ, p). (1.94) The components of this vector are (S⁰, S) where

S⁰(xⁱ) =



d³pf (xⁱ, p)≡ n(xⁱ);

S(xⁱ) = 1 c



d³pf (xⁱ, p)v≡ c⁻¹n(xⁱ)v , (1.95) where we have used the relation (p^α/E) = (v^α/c²). The time component of this vector, S⁰, gives the particle number density n in a given frame; the spatial com-ponents give the flux of the particles in each direction. The factor c was introduced in the definitionEq. (1.94)to facilitate such an interpretation.

Taking quadratic moments allows us to define the quantity T^ab(xⁱ)≡ c²

 d³p

E_pp^ap^bf (xⁱ, p), (1.96) called the energy-momentum tensor of the system. This tensor is clearly symmetric.

When one of the indices is zero, we get, T^b0(xⁱ) = T^0b(xⁱ) = c

 d³p Ep

(Epp^b)f (xⁱ, p) = c



d³pp^bf (xⁱ, p), (1.97)

which is (c times) the sum of the four-momentum of all the particles per unit volume. The time–time component, T⁰⁰(xⁱ), gives the energy density and the time–space component, T^0α(xⁱ), gives the density of the α-component of the three-momentum. The total four-momentum of the system is defined as the integral over all space:

Pⁱ=



d³x T⁰ⁱ. (1.98)

The space–space components of the energy-momentum tensor represent the stresses within the medium. The component T^αβis

T^αβ(xⁱ)≡ c²

 d³p Ep

p^αp^βf (xⁱ, p) =



d³pv^αp^βf (xⁱ, p) =



d³pv^βp^αf (xⁱ, p).

(1.99) Since f denotes the phase space density of particles, p^αf represents the density of the α-component of the momentum and v^βp^αf denotes the flux of this momen-tum.Equation (1.99)gives the α-component of the momentum that crosses a unit area orthogonal to the β direction per unit time. Therefore, T^αβ represents the α-component of the net force acting across a unit area of a surface, the normal to which is in the direction denoted by β. The symmetry of T^αβ implies that this is also equal to the β-component of the net force acting across a unit area of a surface the normal to which is in the direction denoted by α.

The symmetry of the energy-momentum tensor is necessary – in general – for the angular momentum of the system to be conserved. In three dimensions, angular momentum is usually defined through the cross product (x× p). But as we saw in Section 1.5 the cross product of two vectors is a special construction which works only in three dimensions. It is therefore better to think of the components of the angular momentum J^μin three dimensions as the dual (seeEq. (1.51)) of the tensor product J_αβ ≡ (xαp_β− xβp_α) defined by:

J^μ= 1

2^μαβ(x_αp_β− xβp_α) = 1

2^μαβJ_αβ = (x× p)^μ. (1.100) In four dimensions, the tensor product generalizes to an antisymmetric tensor J^ik = xⁱp^k− x^kpⁱ. (But, of course, we cannot take its dual to get another vector which only works in three dimensions.) When we proceed from a single particle to a continuous medium, we need to work with an integral over dp^a = d³x T^0aetc.

So the angular momentum tensor is now defined as:

J^ik≡



d³σ_l(xⁱT^kl−x^kT^il) =



d³x (xⁱT^k0−x^kTⁱ⁰)≡



dσ_lM^ikl. (1.101) The second equality shows that J^ik is indeed the moment of the momentum den-sity integrated over all space and hence represents the total angular momentum.

1.9 The distribution function and its moments 39 The conservation of this quantity requires ∂_lM^ikl= 0. A simple computation now shows that this requires T^ab= T^baand – in particular – we need T^αβ = T^βα. This symmetry ensures that the angular momentum of an isolated system is conserved and the internal stresses cannot spontaneously rotate a body.

The angular momentum tensor J^ik is clearly antisymmetric and hence has six independent components. Its spatial components have clear meaning as the angular momentum of the system since they essentially generalize the expression x× p.

The other three components

J^0α= tP^α−



d³x x^αT⁰⁰, (1.102)

where P^α is the total three-momentum of the system, however, do not play an important role. They give the location of the centre of mass at t = 0. It is possible to choose the coordinate system such that at t = 0 the integral in the above expression vanishes.

While the angular momentum tensor is Lorentz covariant, it changes under the translation of coordinates xⁱ → x^i= xⁱ+ ⁱ. It is easy to see that

J^ik → J^ik = J^ik+ ⁱP^k− ^kPⁱ. (1.103) This result arises because J^ikincludes the orbital angular momentum of the system as well as any intrinsic angular momentum and the former depends on the choice of origin of coordinates. It is, however, straightforward to obtain the intrinsic angular momentum of the system by defining a spin four-vector as

Σa≡ 1

2abcdJ^bc

 P^d

(−PjP^j)^1/2



≡ 1

2abcdJ^bcU^d. (1.104) This quantity is expressed in terms of the (dimensionless) four-velocity Uⁱ of the system which, in turn, is defined in terms of the total four-momentum. Under the translation of the coordinates, when J^bc changes as inEq. (1.103), Σk does not change because of the antisymmetry of the -tensor. In the centre of mass frame of the system in which Uⁱ = (1, 0), each spatial component of the spin vec-tor Σ_α are related to the spatial components of the angular momentum tensor by Σ_α = (1/2)_αβγJ^βγ; the time component vanishes, Σ₀ = 0. In any frame, the definition inEq. (1.104) ensures that UⁱΣ_i = 0 so that the spin vector has only three independent components.

Given a distribution function, we can construct the current four-vector S^a(xⁱ) at any given event, throughEq. (1.94). It is also always possible to choose a Lorentz frame such that the spatial components of this vector vanish at that event (i.e.

v = 0) so that an observer at rest in that Lorentz frame does not see any mean flux of particles around a given event. If the gradient of the mean velocityv is

sufficiently small, then such a Lorentz frame can be defined even globally for the whole system. (Such a definition is approximate; it is valid and useful when physi-cal processes which depend on the gradients of mean velocity, mean kinetic energy, etc., are ignored; also seeProject 1.1.) Let us suppose that we are working in such a Lorentz frame and also that the distribution function is isotropic in momentum in this frame; that is, it depends only on the magnitude, p, of the momentum p. In such a frame, As regards the space–space part of the energy-momentum tensor, it has to be an isotropic, symmetric, three-dimensional tensor. Hence, T_β^α must have the form T_β^α = P (xⁱ)δ^α_β, since δ^α_β is the only tensor available satisfying these conditions.

(The symbol P should not be confused with the total four-momentum Pⁱ used earlier.) To find an expression for P (xⁱ), note that

T_α^α = P (xⁱ)δ_α^α= 3P (xⁱ) = c² This quantity represents the pressure of the fluid and has simple limits in two extreme cases. In the non-relativistic limit, the energy of the particle is E(p) ∼= mc² + (p²/2m). Substituting in the expression for T⁰⁰, we find that the energy density can be written T⁰⁰ ≡ mc²n + _nr where the non-relativistic contribution

_nrto the kinetic energy is

nr ≡ 4π In the same limit, the expressionEq. (1.108)for pressure reduces to

Pnr∼= 4πc² Comparing the two expressions,Eq. (1.109) andEq. (1.110), we see that Pnr = (2/3)nr which is the relation between energy density and pressure in non-relativistic theory. (Note that pressure has nothing to do, a priori, with inter-particle

1.9 The distribution function and its moments 41 collisions but is defined in terms of the momentum transfer across a surface.) In the other extreme limit of highly relativistic particles we have

E(p) ∼= pc. (1.111)

Then

ρ≡ Trel⁰⁰= 4πc

 _∞

0

p³f (xⁱ, p)dp; P = 4πc 3

 _∞

0

p³f (xⁱ, p)dp, (1.112) which shows that, for extreme relativistic particles, the pressure and energy density are related by

P = 1

3ρ. (1.113)

In particular, this equation is exact for particles with zero mass (e.g. a gas of photons) for which E(p) = pc is an exact relation.

Given the components of the energy-momentum tensor in the special frame in which bulk flow vanishes, it is easy to obtain the results in any other frame in which the observer has a four-velocity u^a. The result, obtained by a Lorentz transformation (with c = 1 for simplicity), is

T_b^a= (P + ρ)u^au_b+ P δ^a_b; S^a= n_propu^a. (1.114) Here n_prop is the proper number density – i.e. the number density in the frame comoving with the particles – and is a scalar; it is related to n in Eq. (1.95) by n = γnprop. This momentum tensor is usually called the energy-momentum tensor of an ideal fluid. The trace of this energy-energy-momentum tensor T ≡ Ta^a= 3P−ρ and vanishes for a fluid of ultra-relativistic particles or radiation with the equation of state P = (1/3)ρ.

This energy-momentum tensor in Eq. (1.114) can be expressed in a different form which brings out its physical meaning more clearly. We can write

T_b^a= ρu^au_b+ P (δ_b^a+ u^au_b) = ρu^au_b+ PP_b^a, (1.115) where the symmetric tensorP_b^a= δ^a_b + u^au_bis called the projection tensor. When any other other vector v^ais contracted on one of the indices of this tensor, the resul-tant vectorP_j^av^j will be the part of v^awhich is orthogonal to uⁱ. Mathematically, for any four-vector v^j, we have

v_⊥^a ≡ Pj^av^j = v^a+ u^a(v^ju_j). (1.116) Since v^ju_j is the component of vector v^a along the vector u^a (note that the lat-ter has a norm uⁱui = −1), this expression is clearly the part of the vector v^a which is orthogonal to u^a and we do get v_⊥^aua = 0 from the above equation as expected. The projection tensorPj^a itself is orthogonal to the four-velocity uⁱ in

the sense thatPj^au^j = 0. Therefore, in the instantaneous rest frame of the particle in which uⁱ= (1, 0), the tensorPj^ahas only (nonzero) spatial components. In this frameEq. (1.115)shows a clear separation of the two contributions to the energy-momentum tensor: the time–time component arises from the first term and is equal to ρ. The second term involving the projection tensor has only spatial contribution and along each of the three axes it contributes a pressure P .

In the absence of collisions or external forces, the distribution function f (x^a, p) satisfies the equation (df /dτ ) = 0 (called the Vlasov equation) which can be written in four-dimensional notation as

df dτ = dxⁱ

dτ ∂_if = uⁱ∂_if =−E m

∂f

∂t − p E · ∇f



=−E m

∂f

∂t − v · ∇f 

= 0, (1.117)

where we have used v = (p/E). Since the proper time derivative along a stream-line of a fluid is (d/dτ ) = (uⁱ∂_i), this shows that f is conserved along the streamlines.

It is also easy to show that the current vector S^aas well as the energy-momentum tensor T^ab are conserved; that is, ∂_aS^a = 0, ∂_aT^ab = 0. More generally, these equations will lead to the standard equations governing the dynamics of the fluid.

To see this, we substitute the explicit form of T^ab inEq. (1.115)into ∂_aT^ab = 0 and simplify the terms to obtain

u^muⁿ∂_m(ρ + P ) + (ρ + P ) [uⁿ(∂_mu^m) + u^m(∂_muⁿ)] =−η^mn∂_mP. (1.118) On the other hand, differentiating the relation u^juj = −1 we get un∂muⁿ = 0.

(This condition is equivalent to a^juj = 0.) This suggests projectingEq. (1.118) along unand perpendicular to it. Taking the dot product ofEq. (1.118)with unand collecting terms, we get

∂_m(ρu^m) + P ∂_mu^m = 0. (1.119) This is the relativistic generalization of the continuity equation in fluid mechanics.

Using this inEq. (1.118)we get

(ρ + P )u^m∂muⁿ= (η^mn+ u^muⁿ)∂mP =P^mn∂mP. (1.120) This is the relativistic Euler equation giving the acceleration of the fluid element in terms of the pressure gradient along the spatial directions. The occurrence of the projection tensor makes this clear. In normal units, ρ has the same dimensions as P/c²and the combination (ρ + P/c²) becomes just ρ in the c→ ∞ limit. In this case, the equations reduce to ∂m(ρu^m) ≈ 0 and ρu^m∂muⁿ ≈ P^mn∂mP , which

1.9 The distribution function and its moments 43 can be easily shown to be equivalent to the standard continuity equation and Euler equation of non-relativistic fluid mechanics.

In the study of radiative processes, one often has to deal with a photon gas using our formalism. Considering its practical utility, we shall briefly describe this special case. If the number of photons in a phase space volume d³xd³p is dN , then we have

dN = f (xⁱ, p) d³xd³p = f [xⁱ, (hν/c)ˆk] d³xd³p

= n

xⁱ, p d³xd³p

(2π)³ = n[xⁱ, (hν/c)ˆk]d³xd³p

(2π)³ , (1.121) where n is the number of photons in a particular quantum state labelled by the wave vector k and momentum (hν/c)ˆk, where ˆk is the unit vector in the direction of propagation. In conformity with the usual practice, we are now using the frequency ν = (ω/2π) instead of energy. The energy-momentum tensor corresponding to this distribution function is

T^ab(xⁱ) =

 d³p

E(p)c²p^ap^bf (xⁱ, p). (1.122) The integration over p in d³p = p²dpdΩ can be converted into an integration over the frequency ν by using p = (hν/c). Defining the symbol ˆk^a = k^a/k⁰, where k^a is the wave vector of the photons, T^abbecomes

T^ab(xⁱ) =

 h⁴ν³

c³ kˆ^aˆk^bf (xⁱ, ν, ˆk) dνdΩ. (1.123) This expression suggests defining a quantity (called the specific intensity of radiation) by

I_ν(xⁱ, ˆk) = (h⁴ν³/c²)f = (hν³/c²)n, (1.124) so that the energy-momentum tensor becomes

T^ab(xⁱ) = 1 c



dνdΩ ˆk^aˆk^bI_ν(xⁱ, ˆk). (1.125) Note that ˆk^a (which is not a four-vector) has the four components (1, ˆk).

Since T⁰⁰ = (dE/dV ) is the energy per unit volume, it is clear that I_ν = (cdE/dV dνdΩ) = (dE/dtdAdνdΩ) is the energy flowing per unit area per sec-ond per unit frequency range into a solid angle dΩ. The units for I_ν will be erg cm⁻²s⁻¹Hz⁻¹ steradian⁻¹, and is extensively used in astrophysics when dealing with radiative processes. From the definition of intensity Iν in terms of the photon occupation number, we also find that Iν ∝ ν³n. Since n is Lorentz invariant it follows that

Iν/ν³

is invariant.

Exercise 1.15

Practice with equilibrium distribution functions Consider a distribution function, describ-ing particles in thermal equilibrium, given by

f (xⁱ, p) = dN

d³x d³p =2j + 1 h³

exp(−θ − βpⁱui)− ₋₁

, (1.126)

where h is the Planck constant, j is the spin of the particle, uⁱis the mean four-velocity of the gas,  = 1, 0,−1 for the Bose–Einstein, Maxwell–Boltzmann or Fermi–Dirac statistics, β = (1/k_BT ) and θ is a parameter independent of pⁱ.

(a) Obtain integral expressions for S^a and T^ab. Using these express n, ρ and P as one-dimensional integrals.

(b) Manipulate the expressions to show that dP = [(ρ + P )/T ]dT + nkBT dθ.

(c) Show that θk_BT is actually the chemical potential μ = (ρ + P )/n− T s, where s is the entropy density.

(d) For an MB gas, show that P = nkBT . Also find an exact expression for ρ/n.

[Hint. The required expressions can be obtained by using appropriate dot products like n = −uiSⁱ, P = (1/3)PabT^ab and ρ − 3P = −ηabT^ab. Using the variable χ = sinh⁻¹(p/m), one gets the integral expressions

n = 4πgm³ Part (b) can be proved directly from these expressions. For part (c) evaluate dμ from the definition of μ and use the result of part (b). Part (d) can be obtained directly by putting

 = 0. The exact expression for ρ/n when  = 0 is given by

where Kn(z) is the modified Bessel function.]

Exercise 1.16

Projection effects LetS be a surface with normal ni. Show thatP_b^a = δ_b^a+ n^an_b is the projection tensor whenS is a spacelike surface, while P_b^a = δ^a_b − n^anbis the projection tensor whenS is a timelike surface. Is there a unique projection tensor associated with a null surface?

Exercise 1.17

Relativistic virial theorem Using the conservation law ∂iT^ij = 0, show that for any system which exists in a finite region of space (i.e. T^ij = 0 outside a compact region in space) we have:

In document Gravitation by Paddy (Page 65-75)