u = 1 (B 2 + E2 E B (16.2) + N = j E (16.3) One might be tempted to put u and N into a 4-vector N and write the equation in the form

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1 Chater 6 Energy-momentum tensor (Version., 3 November 7 Earlier, we obtained for the energy density and flux u = (B + E µ c (6. We also had a continuity equation N = µ E B (6. u t + N = j E (6.3 One might be temted to ut u and N into a 4-vector N and write the equation in the form µ N ν =? (6.4 but j E isn t a roer 4-vector. If anything, it aears to be a time-like comonent of F µ J µ. This suggests that u and N are really art of another object. This is the stress-energy tensor, which describes momentum flows within a body. It was originally used to describe mechanical stresses and how forces change directions, but it alies to any system which can be described with a field. In the case of the electromagnetic field, it is often divorced from its mechanical origin and called the energy-momentum tensor. The elements of an arbitrary stress-energy tensor can be interreted as follows T T j T j T jj T ij, i j energy density energy flux momentum density ressure shear stress The sace elements of the tensor reresent the momentum flux. There s not a guarantee that this tensor is symmetric, but it s also ambiguous u to a 4-divergence, i.e., one can add a 4-divergence term which will not affect any hysics. 36

2 The units of the tensor elements are of force er unit area, or ressure. We would like to assign the energy density to T. And since the Poynting vector reresents energy flow, it should find its lace in T i. But to get a better icture of the tensor, let s look at some fluid examles first. (The symmetric rank- tensor field is a member of the (, reresentation sace. It shows u as a sin- graviton. 6. Fluid examles The simlest fluid we can envisage is something like a dust cloud: in its rest frame, there is no stress, and the only energy comes from the rest masses of the articles themselves. So in the cloud s rest frame, we have where ρ is the mass density. ρ c (6.5 Now let s give the cloud an overall motion 4-vector U. If we choose a direction ˆx, then the amount of dust crossing a lane of constant x is clearly roortional to u x. The amount of momentum crossing with each dust article is also roortional to u x. Similarly, the amount of y momentum crossing with each dust article is roortional to u y. This leads us to a simle form of the stress-energy tensor: ρ U µ U ν (6.6 which we also might have exected from the fact that we only had ρ and U out of which to fashion the tensor. For an ideal fluid, there is no heat conduction, and no viscosity. These conditions imly that T i = T i = and T ij = for i j. In its rest frame, there should be an energy density ρ c and ressure. ρ c (6.7 If we build the tensor out of 4-velocities as before, we exect it to have the form (ρ + c U µ U ν + g µν (6.8 (If we use a metric with a different signature, the sign of the second term can change. 37

3 6. *Energy-momentum tensor of the EM field Returning to the electromagnetic field, how can we fit the energy density and Poynting vector into the tensor and still have T µν constructed out of other tensors? Let s start with T i. Since it is a cross roduct between E and B, it suggests a contraction within F µν c (E Bi = g µν F µ F νi (6.9 We assign this to T i. But if we extend this to the time-time comonent, we get g µν F µ F ν = E /c (6. so we fix this u by adding an invariant term for diagonal terms. F µν F µν = (B E In the end, we work with the following: µ [ 4 (F αβf αβ g µν F µ γf γν ] c (6. (6. With any luck, we ll be able to look at some more methodical derivations in the last few lectures. When we look at the comonents, we find µ (B + E N c x /c N y /c N z /c N x /c P P P 3 N y /c P P P 3 N z /c P 3 P 3 P 33 (6.3 The satial elements P ij [ µ δ ij(b + E c (B ib j + E ] ie j c (6.4 then form the momentum flux tensor. Now the form-invariant generalization of Poynting s theorem is α T αβ = (6.5 It is also worth looking at the angular momentum of the electromagnetic field, L field = x (E Bd 3 x (6.6 The generalization is a rank-3 tensor M αβγ = T αβ x γ T αγ x β (6.7 38

4 Angular momentum conservation is α M αβγ = α T αβ x γ α T αγ x β (6.8 = ( α T αβ x γ + T γβ ( α T αγ x β T βγ (6.9 = T γβ T βγ (6. = (6. which is zero because of the symmetry of the stress-energy tensor. 6.3 *Alications with simle geometries 6.3. *Parallel-late caacitor A arallel-late caacitor with area A oriented with its ga along the x direction has a constant electric field Eˆx. The charge on each late is so the force ulling the lates together is Q = Aɛ E (6. f = QE = ɛ A E (6.3 The stress tensor is ɛ E (6.4 The energy density and ressure are as exected, the ressure being negative since the lates are being ulled together. There is also an outward ressure which reflects the tension among the field lines *Long straight solenoid This case is similar to the arallel-late caacitor, with a similar layout of field lines. The stress tensor is ɛ c B (6.5 39

5 6.3.3 *Plane waves Let s examine a lane wave travelling in the x direction, olarized along the y direction: E = (, E, cos(ωt kx (6.6 B = (,, B cos(ωt kx (6.7 The energy density is The Poynting vector is B = E c = ke ω B + E c (6.8 = E c (6.9 N = µ E B = µ ˆxEB cos (ωt kx (6.3 = ˆx E c cos (ωt kx (6.3 The lane wave therefore carries energy in the x direction (as exected, and we should see that it exerts a ressure (momentum flow in the x direction as well: P = E (B µ c x + E x µ c = E (6.3 µ c P = E (B µ c y + E y µ c = (6.33 P 33 = E (B µ c z + E z µ c = (6.34 The stress tensor is therefore E µ c cos (ωt kx (6.35 A more general way to write this tensor admits any direction: where E cos (K X µ c ω K µ K ν (6.36 K = (ω/c, k,, (6.37 is the 4-wavenumber vector. An interesting side-note on this way of writing the tensor is that the quotient rule, which states that an object which contracts with a tensor to roduce another tensor must itself be a tensor, imlies (E/ω is a (rank- tensor, i.e., a scalar. 4