3+1 formalism and bases of numerical relativity - LUTh ...

80 3+1 equations for matter and electromagnetic field
Remark : For pressureless matter (dust), the above formula reduces to E = Γ 2 ρ. The reader
familiar with the formula E = Γmc 2 may then be puzzled by the Γ 2 factor in (5.52).
However he should remind that E is not an energy, but an energy per unit volume: the
extra Γ factor arises from “length contraction” in the direction of motion.
Introducing the proper baryon density nB, one may decompose the proper energy density ρ
in terms of a proper rest-mass energy density ρ0 and an proper internal energy εint as
ρ = ρ0 + εint, with ρ0 := mBnB, (5.53)
mB being a constant, namely the mean baryon rest mass (mB ≃ 1.66 × 10 −27 kg). Inserting
the above relation into Eq. (5.52) and writting Γ 2 ρ = Γρ + (Γ − 1)Γρ leads to the following
decomposition of E:
E = E0 + Ekin + Eint, (5.54)
with the rest-mass energy density
the kinetic energy density
the internal energy density
E0 := mBNB, (5.55)
Ekin := (Γ − 1)E0 = (Γ − 1)mBNB, (5.56)
Eint := Γ 2 (εint + P) − P. (5.57)
The three quantities E0, Ekin and Eint are relative to the Eulerian observer.
At the Newtonian limit, we shall suppose that the fluid is not relativistic [cf. (5.25)]:
Then we get
P ≪ ρ0, |ǫint| ≪ ρ0, U 2 := U · U ≪ 1. (5.58)
Newtonian limit: Γ ≃ 1 + U2
2 , E ≃ E + P ≃ E0 ≃ ρ0, E − E0 ≃ 1
2 ρ0U 2 + εint. (5.59)
The fluid momentum density as measured by the Eulerian observer is obtained by applying
formula (4.4):
p = −T(n,γ(.)) = −(ρ + P) 〈u,n〉 〈u,γ(.)〉 −P g(n, γ(.))
=−Γ =ΓU =0
= Γ 2 (ρ + P)U, (5.60)
where Eqs. (5.31) and (5.34) have been used to get the second line. Taking into account
Eq. (5.52), the above relation becomes
p = (E + P)U . (5.61)