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Home , Electromagnetic tensor

Fall 2019 Lecture 10: ; ; stress-energy

Yacine Ali-Ha¨ımoud October 1st 2019

CHARGE CONSERVATION

Last lecture we derived the 4-current J µ of an ensemble of particles, in a LICS. Its 0-th component is the (where “charge” can be , baryon number, etc...),

0 X (3) J (t, ~x) = qnδD (~x − ~xn(t)) n and its spatial component is the charge current,

X d~xn (3) J~(t, ~x) = q δ (~x − ~x (t)). n dt D n n We showed that the two expressions can be grouped in the 4-vector Z µ Z X dx (4) X (4) J µ(xν ) = q dt n δ (xν − xν ) = q dτ uµ δ (xν − xν ), n n dt D n n n n D n n n n µ where un is the 4-velocity of particle n. This is clearly a Lorentz 4-vector. Let us compute the following:

X d~xn (3) X d~xn (3) ∂ J µ = ∂ J 0 + ∇~ · J~ = − q · (∇~ δ )(~x − ~x (t)) + q · (∇~ δ )(~x − ~x (t)) = 0, µ t n dt D n n dt D n n n regardless of the trajectoris of the particles. Let us now rewrite these expressions in a way that is generally covariant, i.e. define generally tensorial expressions. First, we can rewrite Z X 1 (4) J µ = q dτ uµ √ δ (xν − xν ), (1) n n n −g D n n

(4) √ which is now a bona fide 4-vector, since we saw that δD / −g is invariant under arbitrary coordinate transforma- 0 ~ tions.√ You can work backwards and find that the expressions for J and J are the same as before, just with a factor 1/ −g. µ Second, the conservation of charge ∂µJ = 0 becomes

µ ∇µJ = 0 .

Both expressions match the ones we derived in a LICS, and are moreover generally covariant. Let us write the latter equation explicitly:

µ µ µ ν 0 = ∇µJ = ∂µJ + Γµν J . Now the relevant contraction of Christoffel symbols is 1 1 Γµ = gµλ (g + g − g ) = gµλg + gµλg . µν 2 µλ,ν νλ,µ µν,λ 2 µλ,ν ν[λ,µ] The last term vanishes: it is the contraction of a with an ! Now, for a M, Jacobi’s identity tells us 1 ∂  ∂  ∂ det(M) = tr M −1 M = (M −1)µν M . det(M) ∂x ∂x ∂x µν 2

Applying this to Mµν = gµν , we thus find √ ∂ −g Γµ = √ν . µν −g

Therefore the conservation of charge can be written as √ ∂ −g 1 √ 0 = ∂ J µ + √ν J ν = √ ∂ ( −gJ µ), µ −g −g µ which we could directly have derived from Eq. (1).

MAXWELL’S EQUATIONS

Maxwell’s equations for the electric field E~ and magnetic field B~ can be written in a simple, compact form, in terms µν 0i i ij ijk of the rank-2 antisymmetric electromagnetic tensor F , such that F = E and F =  Bk. Note that I no longer say rank-(0,2) or rank-(2,0) or rank-(1, 1), as such can be transformed into one another with the metric and inverse metric. The special-relativistic Maxwell’s equations (thus, in a LICS) are

νµ ν ∂µF = J , ∂[µFνλ] = 0, where J ν is the electric 4-current. Note that the second equation is equivalent to

∂µFνλ + ∂ν Fλµ + ∂λFµν = 0, because ∂(µFνλ) = 0, and the above combination is proprotional to ∂(µFνλ) + ∂[µFνλ]. Given an electromagnetic field, the motion of a particle of mass m and charge q, in a LCIS, is given by

duµ m = qF µν u . dτ ν The generally covariant equations are

αβ α β α αβ ∇βF = J , ∇[αFβγ] = 0, mu ∇βu = qF uβ .

STRESS-ENERGY TENSOR

α α Consider a set of massive particles (labeled by n) with 4-momenta pn = mnun. Remember that the 4-momentum is such that p0 is the energy and ~p is the 3-momentum. Just like we defined the density and flux of charge, we may define the density and flux of 4-momentum. In a LICS, they are given by

µ0 X µ (3) T (t, ~x) = mnunδD (~x − ~xn(t)), n and the flux of 4-momentum is

i X dx (3) T µi(t, ~x) = m uµ n δ (~x − ~x (t)). n n dt D n n

Just like before, these can be grouped into

ν Z X dx (3) X (4) T µν (xσ) = m uµ n δ (~x − ~x (t)) = m uµuν dτ δ (xσ − xσ). n n dt D n n n n n D n n n

µν This expression is manifestly a Lorentz tensor, and√ moreover shows that T is symmetric. This can be rewritten in a generally covariant way by multiplying by 1/ −g. 3

Let us now compute, in a LICS:

µ X du (3) ∂ T µν = ∂ T µ0 + ∂ T µi = m n δ (~x − ~x (t)), ν 0 i n dt D n n ~ (3) where the contributions proportional to ∇δD cancel out as before. Again, we may rewrite this in a manisfestly Lorentz-covariant form: Z µ Z X du (4) X (4) ∂ T µν = dτ m n δ (xσ − xσ) = dτ f µδ (xσ − xσ), ν n n dτ D n n n D n n n n

µ where fn is the 4-force acting on particle n. The covariant version of this equation is Z X 1 (4) ∇ T µν = dτ f µ √ δ (xσ − xσ) ≡ F µ, ν n n −g D n n where the right-hand-side is a 4-force density. More generally, we can define the stress-energy tensor of any substance, T µν , as the symmetric tensor such that:

T 00 = energy density T i0 = density of i-th component of momentum 0j T = energy flux along ∂(j) ij T = flux of i−th component of momentum along ∂(j)

The total stress energy tensor of all matter fields is conserved, i.e. there is no net creation or destruction of overal 4-momentum

µν ∇µT(total) = 0 .

However, as we saw in the case of a swarm of particles, the stress-energy tensor of any particular species s is not necessarily conserved:

µν X µ ∇ν T(s) = Fs0→s. s06=s

The conservation of the total stress-energy tensor is just a re-expression of the law of -reaction:

µν X X µ ∇ν T(total) = Fs0→s = 0. s s06=s

STRESS-ENERGY TENSOR OF THE

µ µν Let us apply the above results to charged particles in an electromagnetic field: the force is then fn = qnF uν , thus

X 1 (4) ∇ T µν = q F µν u √ δ (xσ − xσ) = F µν J , ν (particles) n ν −g D n ν n where Jν is the electric defined earlier. Let us define the following tensor:

1 T αβ ≡ F α F βδ − gαβ F ρδF . (em) δ 4 ρδ

We then find 1 ∇ T αβ = (∇ F α )F βδ + F α ∇ F βδ − gαβ(∇ F )F ρδ, β (em) β δ δ β 2 β ρδ 4 where we used metric compatibility of the , and the symmetry of the last term. Let’s now use Maxwell’s equations to simplify:

 1  ∇ T αβ = −F α J δ + gαγ F βδ∇ F + F δρ∇ F , β (em) δ β γδ 2 γ ρδ where I remnamed dummy indices and shuffled up-and-down indices with the metric. Doing some more index renam- ing, and using the antisymmetry of Fµν , we get

 1  ∇ T αβ = −F α J δ + gαγ F δβ ∇ F + ∇ F . β (em) δ β δγ 2 γ βδ

βδ Now because F is antisymmetric, we may replace ∇βFδγ by ∇[βFδ]γ , i.e. its antisymmetric part in β, δ. Using the antisymmetry of Fµν , the term in parenthesis then becomes 1/2(∇βFδγ + ∇δFγβ + ∇γ Fβδ) = 0, from the second of Maxwell’s equations. Thus we find

αβ αδ αβ ∇βT(em) = −F Jδ = −∇βT(particles) .

αβ We thus see that it makes sense to define T(em) as the stress-energy tensor of the electromagnetic field, since its satisfies

 αβ αβ  ∇β T(em) + T(particles) = 0 .

STRESS-ENERGY TENSOR OF AN IDEAL FLUID

An ideal fluid is isotropic in a preferred LICS, called the fluid’s rest frame. This means that, in this frame, 0i ij ij there is no preferred direction, thus T = 0 and T ∝ δ . We define ρrf to be the energy density in the fluid’s 00 ij ij rest-frame, i.e. such that T = ρrf , and denote by Prf the pressure in the fluid’s rest-frame, i.e. T = Prf δ in that frame. Let us now denote by uµ the 4-velocity of the fluid with respect to some arbitrary frame. In particular, in the fluid’s µ µν µ ν µν µ ν rest-frame, u = (1, 0, 0, 0) by definition, thus we can rewrite, in that frame, T = ρrf u u + Prf (η + u u ) since in that frame, ηµν + uµuν = diag(0, 1, 1, 1). This is a Lorentz-covariant expression. The generally covariant version is

αβ α β αβ α β T = ρrf u u + Prf (g + u u ) .

It is important to remember that ρrf and Prf are the fluid’s energy density and pressure in its restframe. In an arbitrary frame, the energy density is

00 0 2 00 0 2 ρ = T = ρrf (u ) + Prf g + (u ) .