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where E is average kinetic energy of atoms. Eventually, fusion at the core must stop, after which the star cools and contracts. Consider the possible final state of a star at T = 0. The pressure P does not go to zero as T → 0 because of degeneracy pressure . Since me mp the electrons become degenerate first, at a number density of one electron in a cube of side ∼ Compton wavelength.
− ne 1/3 ∼...
For any vector field, k, local coordinates can be found such that k = ∂ /∂ ξ where ξ is one of the coordinates. In such a coordinate system £k gµν = ∂ gµν ∂ξ (2.29) (2.28)...
so hypersurfaces of constant Schwarzschild time t are straight lines through the origin in the Kruskal spacetime....
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Since l is normal to N , N is a Kil ling horizon of k. Since l · Dl = 0, the surface gravity is ∂ 1 4M V ∂ V ln |V | on U = 0 κ = k · ∂ ln |f | = (2.104) 1 ∂ − U ln |U | on V = 0 4M ∂ U 1 on {U = 0} 4M = (2.105) − 1 on {V = 0} 4M So κ2 = 1/(4M )2 is indeed a constant on N . Note that orbits of k lie either entirely in {U = 0} or in {V = 0} or are fixed points on B , which allows a difference of sign in κ on the two branches of N . c3 ] [N.B. Reinstating factors of c and G, |κ| = 4GM Normalization of κ If N is a Killing horizon of ξ with surface gravity κ, then it is also a Killing horizon of cξ with surface gravity c2κ [from formula (2.89) for κ] for any constant c. Thus surface gravity is not a property of N alone, it also depends on the normalization of ξ . There is no natural normalization of ξ on N since ξ 2 = 0 there, but in an asymptotically flat spacetime there is a natural normalization at spatial infinity, e.g. for the time-translation Killing vector field k we choose k2 → −1 as r → ∞ (2.106) This fixes k, and hence κ, up to a sign, and the sign of κ is fixed by requiring k to be future-directed. Degenerate Killing Horizon (κ = 0) In this case, the group parameter on the horizon is also an affine parameter, so there is no bifurcation 2-sphere. More on this case later....
so the Hawking temperature (i.e. temperature as measured at spatial ∞) is actually zero. This is expected because Rindler is just Minkowski in unusual coordinates, there is nothing inside which could radiate. But for a black hole Tlocal → TH at infinity (2.149)...
Since k = ∂ /∂ t in static coordinates we have k2 → −1 as r → ∞. So we identify κ± as the surface gravities of N± ....
Note that a static multi black hole solution is possible only when there is an exact balance between the gravitational attraction and the electrostatic repulsion. This occurs only for M = |Q|...
Causal structure Because we now have only axial symmetry we really need a 3-dim spacetime diagram to encode the causal structure, but the θ = 0, π /2 submanifolds are total ly-geodesic , i.e. a geodesic that is initially tangent to the submanifold remains tangent to it, so we can draw 2-dim CP diagrams for them...
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= Pµ λB(λρ) P ρν ≡ Pµλ D(ρlλ)P ρν ∂ ξ ρ = Pµ λ (ρ f −1 λ) Pν (since D(ρξλ) = 0) = 0 (since P ξ = ξ P = 0) In particular θ = 0, everywhere on N , so dθ/dλ = 0. Corollary For Killing horizon N of ξ Rµν ξ µ ξ ν |N = 0...
φα (Dµ ∂ µ φβ ) − (Dµ ∂ µ φα ) φβ φ − m2 m2 φα β = 0 , φβ = φα...
These conditions are not implied by (7.16); the additional information contained in them is the invertibility of the change of basis. In a general spacetime there is no ‘preferred’ choice of basis satisfying (7.9) and so no preferred choice of vacuum. In a stationary spacetime, however, we can choose the basis {ui } of positive frequency eigenfunctions of k, i.e. kµ ∂µ ui = −iωi ui , Notes (1) Since k is Killing it maps solutions of the Klein-Gordon equation to solutions (Proof: Exercise). (2) k is anti-hermitian, so it can be diagonalized with pure-imaginary eigenvalues. (3) Eigenfunctions with distinct eigenvalues are orthogonal so u
∗ i , uj...
H
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Assuming the bound P ∼ ( c)n4/3 , e where ne (r) is the electron number density, show that the total mass, M , of the star satisfies 2 h 3/2 µ c < e M∼ G mN where mN is the nucleon mass and µe is the number of electrons per nucleon. Why is it reasonable to bound the pressure as you have done? Compare your bound with Chandresekhar’s limit. 3. A particle orbits a Schwarzschild black hole with non-zero angular momentum per unit mass h. Given that σ = 0 for a massless particle and σ = 1 for a massive particle, show that the orbit satisfies d2 u Mσ + u = 2 + 3M u2 2 dφ h where u = 1/r and φ is the azimuthal angle. Verify that this equation is solved by 2ω 2 2ω 2 1 + − u= , 6M 3M M cosh2(ω φ) where ω is given by 4ω = ±
2 <...
if ξ = ξ µ ∂µ is a conformal Killing vector. Show that ξ is the operator corresponding to the conserved charge implied by Noether’s theorem. 3. Show that the extreme RN metric in isotropic coordinates is 1 −2 2 1 ( d2 M M 2 †) ρ + ρ2dΩ2 dt2 + + ds = − + ρ ρ...
for a point particle of mass m is equivalent, for q = 0, to the action of Q.4. Show that S is invariant to first order in α under the transformation δ xµ = αK µν pν 1 δ pµ = − α pρ pσ ∂µ K ρσ 2...
Asymptotically-flat solutions are stationary and axisymmetric if the metric ∂ admits Killing vectors k and m that can be taken to be k = ∂ t and m = ∂∂φ near infinity, and if (for some choice of electromagnetic gauge) Lk A = L m A = 0 , where the Lie derivative of A with respect to a vector ξ , Lξ A, is as defined in Q.4 of Example Sheet 2. The event horizon of such a solution is necessarily a Killing horizon of ξ = k + ΩH m, for some constant ΩH . What is the physical interpretation of ΩH ? What is its value for the Kerr-Newman solution? The co-rotating electric potential is defined by Φ = ξ µ Aµ . Use the fact that Rµν ξ µ ξ ν = 0 on a Killing horizon to show that Φ is constant on the horizon. In particular, show that for a choice of the electromagnetic gauge for which Φ = 0 at infinity, ΦH =
2 r+...
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