Massachusetts Institute of TechnologyDepartment of PhysicsPhysics 8.962 Spring 2002Number-Flux Vector and Stress-EnergyTensorc°2000, 2002 Edmund Bertschinger. All rights reserved.1 IntroductionThese notes supplement Section 3 of the 8.962 notes “Introduction to Tensor Calculus forGeneral Relativity.” Having worked through the formal treatment of vectors, one-formsand tensors, we are ready to evaluate two particularly useful and important examples,the number-flux four-vector and the stress-energy (or energy-momentum) tensor for agas of particles. A good elementary discussion of these objects is given in chapter 4 ofSchutz, A First Course in General Relativity; more advanced treatments are in chapters5 and 22 of MTW. Some of the mathematical material presented here is formalized inSection 4 of the 8.962 notes; to avoid repetition we will present the computations here ina locally flat frame (orthonormal basis with locally vanishing connection) frame ratherthan in a general basis. However, the final results are tensor equations valid in any basis.2 Number-Flux Four-Vector for a Gas of ParticlesWe wish to describe the fluid properties of a gas of noninteracting particles of rest massm starting from a microscopic description. In classical mechanics, we would describethe system by giving the spatial trajectories xa(t) where a labels the particle and t isabsolute time. (An underscore is used for 3-vectors; arrows are reserved for 4-vectors.While the position xaisn’t a true tangent vector, we retain the common notation here.)The number density and number flux density aren =Xaδ3(x − xa(t)) , J=Xaδ3(x− xa(t))dxadt(1)1where the Dirac delta function has its usual meaning as a distribution:Zd3x f(x) δ3(x − y) = f(y) . (2)In order to get well-defined quantities when relativistic motions are allowed, we at-tempt to combine the number and flux densities into a four-vector~N. The obviousgeneralization of equation (1) is~N =Xaδ3(x− xa(t))d~xadt. (3)However, this is not suitable because time and space are explicitly distinguished: (t, x).A first step is to insert one more delta function, with an integral (over time) added tocancel it:~N =XaZdt0δ4(x − xa(t0))d~xadt0. (4)The four-dimensional Dirac delta function is to be understood as the product of thethree-dimensional delta function with δ(t − ta(t0)) = δ(x0− t0):δ4(x − y) ≡ δ(x0− y0)δ(x1− y1)δ(x2− y2)δ(x3− y3) . (5)Equation (4) looks promising except for the fact that our time coordinate t0is frame-dependent. The solution is to use a Lorentz-invariant time for each particle — theproper time along the particle’s worldline. We already know that particle trajectories inspacetime can be written xa(τ). We can change the parametrization in equation (4) soas to obtain a Lorentz-invariant object, a four-vector:~N =XaZdτ δ4(x − xa(τ))d~xadτ. (6)2.1 Lorentz Invariance of the Dirac Delta FunctionBefore accepting equation (6) as a four-vector, we should be careful to check that thedelta function is really Lorentz-invariant. We can do this without requiring the existenceof a globally inertial frame (something that doesn’t exist in the presence of gravity!)because the delta function picks out a single spacetime point and so we may regardspacetime integrals as being confined to a small neighborhood over which locally flatcoordinates may be chosen with metric ηµν(the Minkowski metric).To prove that δ4(x − y) is Lorentz invariant, we note first that it is nonzero only ifxµ= yµ. Now suppose we that perform a local Lorentz transformation, which maps dxµto dx¯µ= Λ¯µνdxνand d4x to d4¯x = | det Λ| d4x. Clearly, δ4(¯x − ¯y) is nonzero only if2x¯µ= y¯µand hence only if xµ= yµ. From this it follows that δ4(¯x − ¯y) = Sδ4(x − y) forsome constant S. We will show that S = 1.To do this, we write the Lorentz transformation in matrix notation as ¯x = Λx andwe make use the definition of the Dirac delta function:f(¯y) =Zd4¯x δ4(¯x − ¯y)f(¯x) =Zd4x | det Λ| Sδ4(x − y)f(Λx) = S | det Λ| f(¯y) . (7)Lorentz transformations are the group of coordinate transformations which leave theMinkowski metric invariant, η = ΛTηΛ. Now, det η = −1, from which it follows that| det Λ| = 1. From equation (7), S = 1 and the four-dimensional Dirac delta function isLorentz-invariant (a Lorentz scalar).As an aside, δ4(x) is not invariant under arbitrary coordinate transformations, be-cause d4x isn’t invariant in general. (It is invariant only for those transformations with| det Λ| = 1). In part 2 of the notes on tensor calculus we show that | det g|1/2d4x is fullyinvariant, so we should multiply the Dirac delta function by | det g|−1/2to make it in-variant under general coordinate transformations. In the special case of an orthonormalbasis, g = η so that | det g| = 1.3 Stress-Energy Tensor for a Gas of ParticlesThe energy and momentum of one particle is characterized by a four-vector. For a gasof particles, or for fields (e.g. electromagnetism), we need a rank (2, 0) tensor whichcombines the energy density, momentum density (or energy flux — they’re the same)and momentum flux or stress. The stress-energy tensor is symmetric and is defined sothatT(˜eµ, ˜eν) = Tµνis the flux of momentum pµacross a surface of constant xν. (8)It follows (Schutz chapter 4) that in an orthonormal basis T00is the energy density,T0iis the energy flux (energy crossing a unit area per unit time), and Tijis the stress(i-component momentum flux per unit area per unit time crossing the surface xj=constant. The stress-energy tensor is especially important in general relativity becauseit is the source of gravity. It is important to become familiar with it.The components of the number-flux four-vector Nν=~N(˜eν) give the flux of particlenumber crossing a surface of constant xν(with normal one-form ˜eν). From this, we canobtain the stress-energy tensor following equation (6). Going from number (a scalar) tomomentum (a four-vector) flux is simple: multiply by ~p = m~V = md~x/dτ. Thus,T =XaZdτ δ4(x − xa(τ))m~Va⊗~Va. (9)34 Uniform Gas of Non-Interacting ParticlesThe results of equations (6) and (9) include a discrete sum over particles. To go to thecontinuum, or fluid, limit, we suppose that the particles are so numerous that the sumof delta functions may be replaced by its average over a small spatial volume. To getthe number density measured in a locally flat (orthonormal) frame we must