Los factores objetivos de la violencia institucional

d⁴x F_ikF^ik−1 2



d⁴x m²A_iAⁱ, (2.137) with F_ik= ∂_iA_k− ∂kA_i.

(a) Show that when m = 0 the theory is not gauge invariant. Vary the action to obtain the equations of motion.

(b) Show that the equation of motion now implies the condition ∂iAⁱ = 0. Compare this with the case of electromagnetism. Can you think of a physical origin for this condition when m= 0?

(c) Using the result of (b), show that the equation of motion becomes ( − m²)Aⁱ = 0.

Solve this equation in the Fourier space and explain why m can be thought of as the mass of the field excitations.

(d) Add a term to the action coupling Aⁱto a current Jⁱ. Is it necessary that this current is conserved? Solve the field equations in the presence of Jⁱin the Fourier space and show that the matrix corresponding to the generalization ofEq. (2.122)now has a well-defined inverse. Can you take the limit m→ 0 smoothly?

Exercise 2.17

What isc if there are no massless particles? Show that, if the photon has a tiny mass m, then we will have an extra term m²A_jA^jin the electromagnetic Lagrangian and the veloc-ity of the electromagnetic waves will not be universal but will depend on the frequency.

Suppose there are no massless particles at all in nature (and the photon, for example, has a very tiny mass). How will one then formulate special relativity and interpret the universal speed c ?

2.9 Energy and momentum of the electromagnetic field

By including the electromagnetic fields in the action we are treating them as dynamical entities with degrees of freedom of their own. When a charge moves under the influence of an electromagnetic field, its momentum and energy will change. Since the total momentum (or energy) of a closed system is a constant, it follows that the change of energy and momentum of the charged particle must be compensated for by a change of the energy and momentum of the field. To show this explicitly, we need to obtain the expressions for the energy and momentum of the electromagnetic field.

Let us consider the energy-momentum tensor corresponding to the last term in Eq. (2.111)describing the action for the electromagnetic field. We have already seen that the canonical momentum corresponding to the vector field Al is given

2.9 Energy and momentum of the electromagnetic field 91 by P^(l)k =−(1/4π)F^kl. One should interpret this expression as giving the four-momentum (indexed by k) for each component of the field (indexed by l). The energy-momentum tensor inEq. (2.17)is now defined through

T_i^k=−[∂iAlP^(l)k− δ^kiL]. (2.138) Note that in the first term we are summing over l, which indicates the differ-ent compondiffer-ents of the vector field. This is an illustration of the commdiffer-ent we made earlier (at the end ofSection 2.3.1) that, in the case of a multi-component field, we merely sum up the expression for the energy-momentum tensor for each component. Substituting P^(l)k =−(1/4π)F^klinto this expression we get

T^ik = 1

4πF^k_l∂ⁱA^l− 1

16πη^ikFlmF^lm. (2.139) This is an example of an energy-momentum tensor which is not symmetric.

However, we can make it symmetric by adding the quantity

−F^k_l

4π ∂^lAⁱ =− 1

4π∂_l(AⁱF^kl), (2.140) which is of the form given inEq. (2.20). (We have used the fact that, in the absence of charges, ∂_lF^kl = 0.) This leads to the symmetrized energy-momentum tensor of the form:

T^ik = 1

4π(F^ilF^k_l−1

4η^ikF_lmF^lm). (2.141) While we have obtained a useful expression, the procedure that is adopted is inherently ambiguous and unsatisfactory. For example, there is no assurance that some other form of the energy-momentum tensor – obtained by adding some other tensor – will not be more appropriate physically. The fundamental reason for this ambiguity is that the energy-momentum tensor is not directly observable in the absence of gravitational interactions. What is usually relevant is the integral of this expression over a volume, which is quite unambiguous. However, since energy density is equivalent to mass density, we would expect different energy-momentum tensors to lead to different gravitational fields. In any theory which includes grav-ity, we need an unambiguous procedure to obtain a symmetric energy-momentum tensor. We will see that general relativity indeed provides such a prescription and leads (unambiguously) to the same expression as that in Eq. (2.141). This is the real justification for our choice.

Let us now consider the properties of T^ik. This tensor is obviously symmetric and is traceless: i.e. T_iⁱ = 0. Further, it has the components

T⁰⁰≡ W = E²+ B²

8π , T^0α≡ S^α = 1

4π(E× B)^α, (2.142)

and

It is possible to relate the divergence of the energy-momentum tensor to the source Jⁱ when the latter is nonzero. To do this, we differentiate the expression inEq. (2.141), getting

On usingEq. (2.115)as well asEq. (2.51)we can rewrite this in the form

∂_kT_i^k= 1 By permuting indices, we can easily show that the first three terms on the right cancel one another, thereby leading to

∂_kT_i^k=−FikJ^k. (2.146)

This equation relates the change in electromagnetic energy-momentum to the work done by the field on the charged particles. In particular, the i = 0 component of this equation gives: Integrating this expression over a three-dimensional volume and applying the Gauss theorem to the second term on the left, we get

∂ The term on the right hand side can be written as!

qava·E, where the sum is over all charges. Since this represents the amount of work done on the charged particles by the electromagnetic field, it is equal to the rate of change of the kinetic energy E of the charges so thatEq. (2.148)becomes

∂

The left hand side can be interpreted as the rate of change of the total energy of the system (made of the electromagnetic field and the charged particles) contained in a volume and the right hand side gives the flux of this energy through the surface bounding the given volume, demonstrating the conservation of energy. (Also see Exercise 2.18.)

2.9 Energy and momentum of the electromagnetic field 93

Exercise 2.18

Conserving the total energy (a) Consider a system of particles with charges qA and masses m_A(with A = 1, 2, ... labelling the particle) which are interacting through elec-tromagnetic forces. Argue that the energy-momentum tensor for the particles can be taken to be

T_part^ab =

A

mA



dτAu^a_A(τA)u^b_A(τA) δD[x− zA(τA)]. (2.150) (b) Compute ∂_aT_part^ab . Hence show that

∂a[T_part^ab + T_EM^ab] = 0, (2.151) where T_EM^ab is the energy-momentum tensor of the electromagnetic field.

Exercise 2.19

Stresses and strains Consider an electric field E which is constant along the x-axis.

Show that the energy-momentum tensor has the components T⁰⁰ = (E²/8π), T^xx =

−(E²/8π), T^yy = T^zz = (E²/8π). Consider now another frame moving with speed v = βc along the y-axis. Show that the components in this frame are given by

T^00= (E²/8π)γ²(1 + β²), T^0y=−(E²/4π)γ²β (2.152) T^xx=−(E²/8π) T^yy = (E²/8π)γ²(1 + β²), T^zz= (E²/8π). (2.153) This shows that, in the ultra-relativistic limit, the pressure in the y-direction dominates over the tension in the x- or z-directions. Explain why.

Exercise 2.20

Everything obeys Einstein Consider a parallel plate capacitor with plates of area A located normal to the x-direction with a small separation d. Assume that the capacitor is charged to give a uniform electric field E between the plates and ignore all the edge effects. The electromagnetic mass of the system is E²Ad/8π. If this capacitor moves along the x-direction, the electric field does not change but the separation undergoes Lorentz con-traction so that the electrostatic energy decreases to E²Ad/8πγ. Consider next the force required to hold the plates apart. One possibility will be to fill the region between the plates by an ideal gas of proper mass density ρ0such that its pressure provides the neces-sary force. The total rest mass of the system is now M = E²Ad/8π + ρ₀Ad. Show that under the Lorentz transformation this quantity goes over to γM as it should.

Exercise 2.21

Practice with the energy-momentum tensor (a) Prove that W²− |S|²is Lorentz invariant.

Note that W and S are not components of a four-vector.

(b) You saw inExercise 2.7that, except when (E·B)²+ (E²−B²)²= 0, it is possible to make a Lorentz transformation to a frame which will make E and B parallel. (That is, if you did that exercise.) Show that, in this frame, the energy-momentum tensor has the form T_b^a = W dia (−1, −1, +1, +1).

(c) When E· B = E²− B² = 0 show that one can always find a Lorentz frame in which the field has the form E = (0, f, 0) and B = (0, 0, f ). (Such a field is called null.) What is the form of energy-momentum tensor in this case?

(d) Show that the square of the electromagnetic energy-momentum tensor T_b^a, treated as a matrix, is proportional to the unit matrix. More explicitly, show that:

T_a^mT_n^a = δ^m_n

64π²[(E²− B²)²+ 4(E· B)²]. (2.154) (e) A symmetric 4× 4 tensor T_b^a, when represented as a matrix, has two properties: (i) T = 0 and (ii) T_b^aT_c^b = (M⁴/64π²)δ^a_c, where M is a constant. Show that by a suitable choice of an inertial frame, such a tensor can be reduced to the form in which the matrix elements of T_b^aare as given in parts (c) or (b) above.

(f) Show that the determinant of F^a_btreated as a matrix is det F^a_b= E· B. Use this as well as the results of (b) above to determine the eigenvalues of F^a_b. [Hint. Because of the antisymmetry of F_ab, we know that if λ is an eigenvalue so is−λ. So the equations deter-mining the eigenvalues must have the form λ⁴+ aλ²+ b = 0. Show that a = (1/2)FabF^ab and b =−(1/16)(abcdF^abF^cd)².]

(g) If an electromagnetic wave is propagating along a definite direction with unit vector n (so that k = ωn), it is called a monochromatic plane wave. (i) Show that the energy-momentum tensor for the plane wave which is given by T^ab= (W/ω²)k^ak^b, where W = (E²+ B²)/8π = E²/4π. Hence show that the combination (E/ω) is Lorentz invariant.

(This is a nontrivial result since it is valid for any Lorentz transformation – not merely for the boosts along the direction of propagation.) (ii) Find the momentum flux S of a plane wave and show that momentum density and energy density are related by S = W . Can you interpret this result?

(h) The standard duality operation, indicated by a∗, is defined as (∗F )ab = _abcdF^cd. The operation – called duality rotation – is defined through exp(∗α) = cos α + ∗(sin α).

How do the electric and magnetic fields change under a duality rotation? How does the electromagnetic energy-momentum tensor change under duality rotation?

(i) Define a complex antisymmetric tensor by Wab = Fab+ i(∗F )ab. How does Wab

transform under duality rotation? Show that 8πTab = −WacW_b^c where the overbar denotes complex conjugation.

(j) Express the energy-momentum tensor in terms of the electric and magnetic fields defined inExercise 2.3. How do you interpret this result?

Exercise 2.22

Poynting–Robertson effect A spherical particle of mass m and effective cross-sectional area A scatters all the electromagnetic radiation incident on it isotropically in its rest frame.

Determine the equation of motion for this particle when it is hit by a constant radiation field of intensity S (erg s⁻¹cm⁻²) incident from a given direction. What is the solution to this equation if the particle was initially at rest?

[Hint. Argue that if the four-velocity of the particle is uⁱand the four-vector along the direction of propagation of radiation is kⁱ, then the particle will absorb a four-momentum flux−AT^abua = −ASk^b(k^aua). The energy absorbed by the particle in the rest frame will be the zeroth component of this expression: SA(k^au_a)². This energy will be re-radiated away isotropically in the rest frame of the particle. Therefore, the equation of motion for the particle will be

dpⁱ

ds =−SA[(k^au_a)kⁱ+ (k^au_a)²uⁱ]. (2.155)