3+1 formalism and bases of numerical relativity - LUTh ...
3+1 formalism and bases of numerical relativity - LUTh ... 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)
5.3 Perfect fluid 81 Finally, by applying formula (4.7), we get the fluid stress tensor with respect to the Eulerian observer: or, taking into account Eq. (5.52), 5.3.4 Energy conservation law S = γ ∗ T = (ρ + P) γ ∗ u =ΓU ⊗ γ ∗ u =ΓU +P γ ∗ g =γ = P γ + Γ 2 (ρ + P)U ⊗ U, (5.62) S = P γ + (E + P)U ⊗ U . (5.63) By means of Eqs. (5.61) and (5.63), the energy conservation law (5.12) becomes ∂ − Lβ ∂t E +N D · [(E + P)U] − (E + P)(K + KijU i U j ) +2(E +P)U ·DN = 0 (5.64) To take the Newtonian limit, we may combine the Newtonian limit of the baryon number conservation law (5.50) with Eq. (5.18) to get ∂E ′ ∂t + D · [(E′ + P)U] = −U · (ρ0DΦ), (5.65) where E ′ := E − E0 = Ekin + Eint and we clearly recognize in the right-hand side the power provided to a unit volume fluid element by the gravitational force. 5.3.5 Relativistic Euler equation Injecting the expressions (5.61) and (5.63) into the momentum conservation law (5.23), we get ∂ − Lβ [(E + P)Ui] + NDj Pδ ∂t j i + (E + P)Uj Ui + [Pγij + (E + P)UiUj]D j N −NK(E + P)Ui + EDiN = 0. (5.66) Expanding and making use of Eq. (5.64) yields ∂ − Lβ Ui + NU ∂t j DjUi − U j DjN Ui + DiN + NKklU k U l Ui + 1 ∂ NDiP + Ui − Lβ P = 0. (5.67) E + P ∂t Now, from Eq. (5.41), NU j DjUi = V j DjUi + β j DjUi, so that −Lβ Ui + NU j DjUi = V j DjUi − UjDiβ j [cf. Eq. (A.7)]. Hence the above equation can be written ∂Ui ∂t + V j DjUi + NKklU k U l Ui − UjDiβ j = − 1 ∂P NDiP + Ui E + P ∂t −DiN + UiU j DjN. ∂P − βj ∂xj (5.68)
- Page 30 and 31: 30 Geometry of hypersurfaces Since
- Page 32 and 33: 32 Geometry of hypersurfaces 2.4.3
- Page 34 and 35: 34 Geometry of hypersurfaces 2.5 Ga
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- Page 40 and 41: 40 Geometry of foliations Figure 3.
- Page 42 and 43: 42 Geometry of foliations 3.3.2 Nor
- Page 44 and 45: 44 Geometry of foliations means Eq.
- Page 46 and 47: 46 Geometry of foliations Remark :
- Page 48 and 49: 48 Geometry of foliations Note that
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- Page 86 and 87: 86 Conformal decomposition As an ex
- Page 88 and 89: 88 Conformal decomposition 6.2.4 Co
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- Page 94 and 95: 94 Conformal decomposition to write
- Page 96 and 97: 96 Conformal decomposition hence Lm
- Page 98 and 99: 98 Conformal decomposition 6.5.2 Ha
- Page 100 and 101: 100 Conformal decomposition discuss
- Page 102 and 103: 102 Conformal decomposition Remark
- Page 104 and 105: 104 Asymptotic flatness and global
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5.3 Perfect fluid 81<br />
Finally, by applying formula (4.7), we get the fluid stress tensor with respect to the Eulerian<br />
observer:<br />
or, taking into account Eq. (5.52),<br />
5.3.4 Energy conservation law<br />
S = γ ∗ T = (ρ + P) γ ∗ u<br />
<br />
=ΓU<br />
⊗ γ ∗ u<br />
<br />
=ΓU<br />
+P γ ∗ g<br />
<br />
=γ<br />
= P γ + Γ 2 (ρ + P)U ⊗ U, (5.62)<br />
S = P γ + (E + P)U ⊗ U . (5.63)<br />
By means <strong>of</strong> Eqs. (5.61) <strong>and</strong> (5.63), the energy conservation law (5.12) becomes<br />
<br />
∂<br />
− Lβ<br />
∂t<br />
<br />
E +N D · [(E + P)U] − (E + P)(K + KijU i U j ) +2(E +P)U ·DN = 0 (5.64)<br />
To take the Newtonian limit, we may combine the Newtonian limit <strong>of</strong> the baryon number<br />
conservation law (5.50) with Eq. (5.18) to get<br />
∂E ′<br />
∂t + D · [(E′ + P)U] = −U · (ρ0DΦ), (5.65)<br />
where E ′ := E − E0 = Ekin + Eint <strong>and</strong> we clearly recognize in the right-h<strong>and</strong> side the power<br />
provided to a unit volume fluid element by the gravitational force.<br />
5.3.5 Relativistic Euler equation<br />
Injecting the expressions (5.61) <strong>and</strong> (5.63) into the momentum conservation law (5.23), we get<br />
<br />
∂<br />
<br />
− Lβ [(E + P)Ui] + NDj Pδ<br />
∂t j<br />
i + (E + P)Uj <br />
Ui + [Pγij + (E + P)UiUj]D j N<br />
−NK(E + P)Ui + EDiN = 0. (5.66)<br />
Exp<strong>and</strong>ing <strong>and</strong> making use <strong>of</strong> Eq. (5.64) yields<br />
<br />
∂<br />
− Lβ Ui + NU<br />
∂t j DjUi − U j DjN Ui + DiN + NKklU k U l Ui<br />
+ 1<br />
<br />
∂<br />
NDiP + Ui − Lβ P = 0. (5.67)<br />
E + P<br />
∂t<br />
Now, from Eq. (5.41), NU j DjUi = V j DjUi + β j DjUi, so that −Lβ Ui + NU j DjUi = V j DjUi −<br />
UjDiβ j [cf. Eq. (A.7)]. Hence the above equation can be written<br />
∂Ui<br />
∂t + V j DjUi + NKklU k U l Ui − UjDiβ j = − 1<br />
<br />
∂P<br />
NDiP + Ui<br />
E + P<br />
∂t<br />
−DiN + UiU j DjN.<br />
∂P<br />
− βj<br />
∂xj <br />
(5.68)