Gravitation and Cosmology Lecture 6: The energy-momentum tensor The energy-momentum tensor Reading: Ohanian, §2.4, 2.5 In this Chapter and subsequently, we shall follow the convention c=1. Currents The equation of current conservation (electrical, particle number, probability or whatever) is ® ¶r + Ñ × j = 0 . (6.1) Written in 4-dimensional notation, this is m ¶m J = 0 , (6.2) ¶ where ¶m º . Because it is the contraction of 2 tensor indices, and because we already know that ¶xm ¶m transforms as a covariant 4-vector), we see that Eq. 6.1 would be manifestly Lorentz invariant if Jm were a contravariant 4-vector (more precisely, a vector density; but we shall leave this detail for a future lecture). Of course, something as basic as a conservation law must be Lorentz invariant----that is, it cannot depend on the frame within which we make our observations. To prove it is so we must determine that Jm is indeed a 4-vector. ® Consider the density for a point particle located at x (t): ® r(®r , t) = d(®r - x (t)) (6.3) Its time derivative is ® ® dx ® ® édx ® ® ù ¶t r = - × Ñd(r - x (t)) = - Ñ × ê d(r - x (t))ú (6.4) dt ë dt û or ® ® ® ® ® dx ® ® ® j (r , t) = d(r - x (t)) º d(r - x (t)) u(t) . (6.5) dt Now, as we have seen, it is possible to define the proper time t as a function of t----and vice versa----by integrating the equation dt = dt Ö````1 - `®u2 , (6.6) ® so, defining x0(t) = t(t), we can rewrite r and j in the combined form m ® ® ® dx (t) Jm (r , t) = dt d(3)(r - x(t)) d(t - x0(t)) . (6.7) ò dt 23 Gravitation and Cosmology Energy-momentum tensor dxm (t) Since dt is manifestly a scalar under Lorentz transformation, and since is manifestly a dt contravariant 4-vector, it remains only to show that ® d(4)æ m - xm(t)ö º d(3) ® - x( ) d - x0(t) èx ø (r t ) (t ) is a Lorentz invariant density. To do this we note that 4 d(4)æ m - xm(t)ö = ò d x èx ø 1 , (which is the same in any coordinate system!) so all we have left to show is that the 4-dimensional volume element d4x is Lorentz invariant. But this is easy: in our special case, d4x = dy dz dx dt d4x’ = dy’ dz’ dx’ dt’ = dy dz dx’ dt’ . Hence we must show dx dt = dx’ dt’. Now, when changing variables of integration, we may write ï¶(x’, t’)ï dx’ dt’ = ï ï dx dt . (6.8) ï ¶(x, t) ï The Jacobian, of the transformation é¶x’/¶x ¶x’/¶tù é g -gvù ï¶(x’, t’)ï ê ú ê ú ï ï = det = det ¶(x, t) ê ú ê ú ï ï ê¶ ¶ ¶ ¶ ú ê-g g ú ë t’/ x t’/ tû ë v û (6.9) = g2 (1 - v2) = 1 . Hence, dx’ dt’ = dx dt , and so Jm is indeed a 4-vector with respect to Lorentz transformations. Energy-momentum tensor We now consider the object (for a point particle of mass m) m n ® dx (t) dx (t) Tmn (r , t) = m dt d(4)æxm - xm(t)ö . (6.10) ò è ø dt dt Manifestly, again, this is a second-rank contravariant tensor (density) with respect to Lorentz transformation. We now want to consider its physical meaning. First of all, we see that 24 Gravitation and Cosmology Lecture 6: The energy-momentum tensor m n ® dx (t) dx (t) ¶ Tmn (r , t) = m dt ¶ d(4)æxm - xm(t)ö m ò dt dt m è ø dxn (t) d = -m dt d(4)æxm - xm(t)ö (6.11) ò dt dt è ø d2xm(t) or, since for a free particle, = 0 , we find upon integrating by parts and discarding the dt2 end-point contribution, 2 n mn (4)æ m m ö d x (t) ¶m T = m ò dt d x - x (t) = 0 . (6.12) è ø dt2 That is, Tmn is conserved. Clearly, Tm0 is a conserved 4-vector density. If we integrate it over volume (over all space) we obtain dxm d3x Tm0 = m = pm . (6.13) ò dt Hence we can interpret Tm0 as the 4-momentum density of a point particle. The tensor Tmn is called the energy-momentum tensor. It is symmetric in mn. 4-momentum density of a gas The energy-momentum tensor of a collection of non-interacting point particles is N m n mn ® (3) ® ® pk pk T (r , t) = å d (r - xk(t)) (6.14) Ek k=0 On the other hand, the energy momentum tensor of a perfect fluid has the form† Tmn = Um Un (r + p) - p hmn (6.15) where Um is the 4-velocity of the rest-frame of the fluid with respect to the observer’s frame (the ‘‘Laboratory’’). The parameters r and p are the ‘‘proper’’ energy density and pressure, respectively. This means that for a relativistic perfect gas in its rest frame, Tij = p d ij (6.16) T00 = r and therefore, in this frame, † See Ohanian and Ruffini, 2nd. ed., prob. 2.28. 25 Gravitation and Cosmology Sound in an ultrarelativistic gas N ® ® 1 (3) ® ® pk × pk p = å d (r - xk(t)) , (6.17) 3 Ek k=0 N (3) ® ® r = å d (r - xk(t)) Ek . (6.18) k=0 For any gas, p £ 1 r ; for an ultra-relativistic gas (for example, a gas of photons) 3 1 p = r . (6.19) 3 Sound in an ultrarelativistic gas Eq. 6.19 has the following interesting consequence† for the propagation of sound: from the second law of thermodynamics, if n is the proper particle-number density and s is the proper entropy, then æ1ö ærö kTds = p dç ÷ + dç ÷ , (6.20) ènø ènø where k is Boltzmann’s constant. We consider small disturbances dr, dp, dn and d®v to the average values of r, n, p, and ®v (=0). The conservation of particle number gives ® ¶t dn + n Ñ × dv = 0 (6.21) A sound wave involves adiabatic compression, hence no change in entropy. Thus -pdn + ndr - rdn = 0 . (6.22) mn ® But we also have, from ¶m T = 0 (keeping only terms to first order in dv) that ® Ñp ¶ dv + = 0 . (6.23) t p + r Now, supposing that dr (the change in internal energy) » l dp, we have ® 1 Ñr ® 1 ædnö ¶t dv + = ¶t dv + Ñç ÷ = 0 (6.24) l p + r l è n ø which, together with Eq. 6.21, yields the wave equation ¶2 1 dn - Ñ2 dn = 0 . (6.25) ¶t2 l We can therefore interpret the square of the sound velocity as l-1 (in units of c). For a gas of ultrarelativistic particles, l = 3, hence 1 usound /c = Ö`` `¤3 . (6.26) † See, e.g., Weinberg, §2.10. 26.