Examples of Tensors. February 3, 2013

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1 Examples of Tensors February 3, 2013 We will develop a number of tensors as we progress, but there are a few that we an desribe immediately. We look at two ases: (1) the spaetime tensor desription of eletromagnetism, and (2) energy-momentum tensors. 1 Eletromagnetism in speial relativity A omplete treatment of this topi is readily available on wikipedia. The link is given in the Notes. The 4-vetor potential is built from the magneti vetor potential A and the eletri potential ϕ as ( ϕ ) A α =, A where the eletri and magneti fields are given by E = ϕ A t B = A In order for the spatial omponents to arise from the 4-dimensional desription, we onsider the magneti field in omponents: B i = ε ij k x j Ak = ε ijk j A k There is no 4-dimensional equivalent of the ross-produt, beause the 4-dimensional Levi Civita tensor, ε αβµν, annot turn the derivatives of a vetor α A β, into another vetor. Nonetheless, we an still write α A β β A α. In three dimensions, we an also write this by inverting the ε ijk above: B i = ε ijk j A k ε imn B i = ε imn ε ijk j A k B i ε imn = ( δ j mδ k n δ k mδ j n) j A k = m A n n A m This is the lue we need. Define the Faraday tensor, Then the spatial omponents, F ij, beome F αβ α A β β A α F jk = j A k k A j = B i ε ijk 0 B 3 B 2 = B 3 0 B 1 B 2 B 1 0 1

2 For the remaining omponents, set α = 0. Then, notiing that we have A α = η αβ A β = ( ϕ ), A F 0β = 0 A β β A 0 F 00 = 0 A 0 0 A 0 = 0 F 0i = 0 A i i ( ϕ) = 1 t A i i ( ϕ ) = E i and, from the antisymmetry, F i0 = E i = δ ij E j, showing that the eletri and magneti fields are both parts of a single rank-2 tensor, 0 E1 E2 E3 E 1 F αβ = 0 B 3 B 2 E 2 B 3 0 B 1 E 3 B 2 B E 1 E 2 E 3 = E 1 0 B 3 B 2 E 2 B 3 0 B 1 E 3 B 2 B 1 0 where in the seond form we have set = 1. We may now write Maxwell s equations in terms of these. Four equations follow as identities. Sine we have F αβ α A β β A α it follows that taking a derivative and antisymmetrizing on all three indies gives zero: To antisymmetrize, notie that µ F αβ = µ α A β µ β A α [µ F αβ] 1 3! ( µf αβ + α F βµ + β F µα µ F βα α F µβ β F αµ ) = 1 3! (( µf αβ µ F βα ) + ( α F βµ α F µβ ) + ( β F µα β F αµ )) = 1 3 ( µf αβ + α F βµ + β F µα ) This simplifiation always ours: when antisymmetrizing an expression, the number of terms is redued if part of the expression (in this ase, F αβ = F βα ) is already antisymmetri. Writing this in terms of the potential, µ F αβ + α F βµ + β F µα = µ α A β µ β A α + α β A µ α µ A β + β µ A α β α A µ = ( µ α A β α µ A β ) + ( β µ A α µ β A α ) + ( α β A µ β α A µ ) = 0 2

3 beause partial derivatives ommute, 2 A β x µ x α = 2 A β x α x µ. Therefore, we have [µ F αβ] = 0 This gives two of the Maxwell equations. To see this, first notie that beause of the antisymmetry, µ, α, β must all be different. This gives two ases. First let α = 0, β = i, µ = j and write out the omponents 0 = j F 0i + 0 F ij + i F j0 Contrating with the 3-dimensional Levi-Civita tensor, The seond ase is when α, β, µ are all spatial, Contrat this with ε ijk, = j E i + 0 B m ε ijm + i E j 0 = ε ijk ( i E j j E i ) + 0 B m ε ijk ε ijm = 2ε ijk i E j + 0 B m ( 2δm k ) [ = 2 E + 1 ] k B t 0 = j F ki + k F ij + i F jk = j (B m ε mki ) + k (B m ε mij ) + i (B m ε mjk ) 0 = ε ijk ε mki j B m + ε ijk ε mij k B m + ε ijk ε mjk i B m 0 = 2δ j m j B m + 2δ k m k B m + 2δ i m i B m 0 = 2 ( m B m + m B m + m B m ) 0 = 6 ( B) We now have the two homogeneous Maxwell equations, B = 0 E + 1 B t = 0 written as a single 4-dimensional equation, [µ F αβ] = 0. For the remaining Maxwell equations, notie that we need the divergene of the eletri field, E. Sine F αβ = η αµ F µν η νβ = = E 1 E 2 E 3 E 1 0 B 3 B 2 E 2 B 3 0 B 1 E 3 B 2 B E 1 E 2 E 3 E 1 0 B 3 B 2 E 2 B 3 0 B 1 E 3 B 2 B we an write so we guess that the objet to onsider is E = i F 0i β F αβ 3

4 Then the time omponent, α = 0, gives the divergene of E, while the spatial omponents, α = i, beome β F iβ = 0 F i0 + j F ij = 1 t ( 1 Ei ) + x j εij k Bk = 1 E i 2 + ε ij k t = 1 E 2 t + B x j Bk These give the inhomogeneous Maxwell equations if we define the 4-urrent, J α = (ρ, J) so that β F αβ = 4π J α is equivalent to E = 4πρ B 1 E 2 t = 4π J For the energy ontent of the eletromagneti field, see below. 2 Energy-momentum tensor The energy-momentum tensor desribes the energy ontent of a region of spaetime. Its non-relativisti spatial part is the stress tensor from lassial mehanis. Read in Shutz for more detail on these omponents. The T 00 kg 2 ( omponent ) is the energy density m, while the T 0i omponents give momentum flux density 3 kg m se 1 m 1 2 se. 2.1 Eletrodynami energy-momentum For the energy-momentum ontent of the eletromagneti field, we first onsult Jakson, Setion 6.7. Here we find the energy density of the eletromagneti field in vauum eq , u = 1 2 ( E 2 + B 2) and the momentum density (proportional to the Poynting vetor) eq , g = 1 2 E B while the spatial omponents of the Maxwell stress tensor take the form E ( E) E ( E) These are the parts of the full energy-momentum tensor, T αβ = F αµ F β µ 1 4 ηαβ (F µν F µν ) 4

5 2.2 The energy-momentum tensor of a salar field For other field theories, there are similar expressions. For example, a Klein-Gordon (massive salar) field satisfies 1 2 ϕ 2 t ϕ = m2 2 2 ϕ The non-relativisti limit of this equation is the Shrödinger equation. The energy-momentum tensor for this field is given by T αβ = α ϕ β ϕ 1 4 ηαβ ( µ ϕ µ ϕ) where α = η αβ β 2.3 The energy-momentum tensor of a perfet fluid For osmologial models, we approximate the distribution of matter in universe as a fluid. A perfet fluid is one with no visosity or heat flow, and show in Chapter 4 or Shutz to have an energy-momentum tensor of form ρ T αβ = p p p where the pressure p is spatially isotropi and the density is ρ. We may write this tensor in terms of the loal 4-veloity u α = (, 0, 0, 0) of the matter and the metri. We an get the pressure piees using the metri, p pη αβ = 0 p p p but this gives the wrong expression for T 00. We fix this by adding (ρ + p ) 2 u α u β = ( ) ρ + p Cheking units, we see that ρ 2 and p both have units of energy density: Adding the two expressions, we have T αβ = [ p 2 ] = F A 2 = kg m s 2 m 2 s 2 m 2 = kg m 3 [ρ] = kg m 3 (ρ + p 2 ) u α u β + pη αβ = (ρ + p) u α u β + pη αβ 5

6 2.4 General properties of the energy-momentum tensor The energy momentum tensor for any matter will have two essential properties. The first is symmetry, T αβ = T βα This follows from onservation of angular momentum, guaranteeing that a small volume element of the material will not spontaneously start to rotate. The seond property is the onservation of energy and momentum, whih may always be expressed as the vanishing of the divergene of the tensor, α T αβ = 0 To see how this implies onservation of energy, we integrate the β = 0 omponent of the 3-divergene over an arbitrary spatial volume, V 3, and use the divergene theorem, i T i0 d 3 x = n i T i0 d 2 x V 3 S 2 =δv 3 where n i is the outward normal to the 2-dimensional spatial boundary of V 3. That integral gives the energy flowing out aross the boundary surfae. When the 4-divergene of T αβ vanishes, we have i T iβ = 0 T 0β and therefore, n i T i0 d 2 x = 0 T 00 d 3 x S 2 =δv 3 V 3 = d ρ 2 d 3 x = de where E is the total energy in the volume V 3. Therefore, the rate of hange of energy in V 3 equals the negative of the rate at whih energy flows aross the boundary of V 3, de = n i T i0 d 2 x S 2 =δv 3 Similarly, the rate of hange of momentum in a volume V 3 is given by the spatial omponenets, β = k, V 3 V 3 d T 0k d 3 x = n i T ik d 2 x 0 T 0k d 3 x = i T ik d 3 x V 3 S 2 Again, the hange in momentum of the volume is the integral of the momentum flux over the boundary. V 3 6