爱因斯坦场方程之Reissner-Nordstrom(电磁真空)解
爱因斯坦场方程之Reissner-Nordstrom解
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- 一点电动力学
- 静态球对称电磁场
- 电磁真空场方程
- 那个方程究竟是怎么解的
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- 第二种方法
一点电动力学
磁场是无源的，因此可以表示为旋度场，我们记
B ⃗ = ∇ ⃗ × A ⃗ \vec B = \vec \nabla \times \vec A B=∇×A
并称 A ⃗ \vec A A为磁矢势。再代入 ∇ ⃗ × E ⃗ = − ∂ B ⃗ ∂ t \vec\nabla\times\vec{E} = -{\partial\vec{B}\over\partial t} ∇×E=−∂t∂B，就有
∇ ⃗ × E ⃗ = − ∂ ∂ t ( ∇ ⃗ × A ⃗ ) = − ∇ ⃗ × ∂ A ⃗ ∂ t \vec\nabla\times\vec{E} = -{\partial\over\partial t}(\vec \nabla \times \vec A) = -\vec \nabla \times{\partial\vec A\over\partial t} ∇×E=−∂t∂(∇×A)=−∇×∂t∂A
也即
∇ ⃗ × ( E ⃗ + ∂ A ⃗ ∂ t ) = 0 \vec \nabla \times(\vec E + {\partial\vec A\over\partial t}) = 0 ∇×(E+∂t∂A)=0
这个无旋场又可以表示为梯度，记
E ⃗ + ∂ A ⃗ ∂ t = − ∇ ⃗ ϕ \vec E + {\partial\vec A\over\partial t} = -\vec\nabla\phi E+∂t∂A=−∇ϕ
并称 ϕ \phi ϕ为电磁场的标势。电磁4势定义为
A a = ( − ϕ c , A ⃗ ) A_a = (-\frac{\phi}{c}, \vec A) Aa=(−cϕ,A)
电磁张量定义为
F a b = ( d A ) a b F_{ab} = (dA)_{ab} Fab=(dA)ab
容易看出，给 A a A_a Aa加上任一函数的梯度不改变 F a b F_{ab} Fab的数值，也就是说若令 A ~ a = A a + ∇ a χ \tilde A_a = A_a + \nabla_a\chi A~a=Aa+∇aχ，则 d A ~ a = d A a + d ( d χ ) = d A a d\tilde A_a = dA_a + d(d\chi) = dA_a dA~a=dAa+d(dχ)=dAa。这被称为电磁4势的规范自由性。
麦克斯韦方程组可表示为
d F = 0 ∗ d ∗ F = μ 0 J \begin{array}{ll} dF &= 0\\ ^*d^*F &= \mu_0J \end{array} dF∗d∗F=0=μ0J
电磁能动张量的表达式(使用几何高斯单位制 c = G = ε 0 = 1 c = G = \varepsilon_0 = 1 c=G=ε0=1)为
T a b = 1 4 π ( F a c F b c − 1 4 g a b F c d F c d ) T_{ab} = \frac{1}{4\pi}(F_{ac}F_b{}^c - \frac{1}{4}g_{ab}F_{cd}F^{cd}) Tab=4π1(FacFbc−41gabFcdFcd)
静态球对称电磁场
回忆一下，静态球对称的Schwarzschild度规的表达式为
d s 2 = − e 2 α ( r ) d t 2 + e 2 β ( r ) d r 2 + r 2 d θ 2 + r 2 sin 2 θ d φ 2 ds^2 = -e^{2\alpha(r)}dt^2 + e^{2\beta(r)}dr^2 + r^2d\theta^2 + r^2\sin^2\theta d\varphi^2 ds2=−e2α(r)dt2+e2β(r)dr2+r2dθ2+r2sin2θdφ2
考虑静态球对称电磁4势，不难确信其在球坐标中分量不显含 t , θ , φ t, \theta, \varphi t,θ,φ而仅与 r r r坐标有关。而且由于球对称性， A a A_a Aa的第 2 , 3 2, 3 2,3分量 A 2 , A 3 A_2, A_3 A2,A3应为常数。注意到
A ~ 1 = ( ∂ ∂ r ) a ( A a + ∇ a χ ) = A 1 + ∂ χ ∂ r \tilde A_1 = (\frac{\partial}{\partial r})^a(A_a + \nabla_a\chi) = A_1 + \frac{\partial\chi}{\partial r} A~1=(∂r∂)a(Aa+∇aχ)=A1+∂r∂χ
因此我们可以通过选择适当的 χ \chi χ使得 A 1 A_1 A1为零。综上，非零的 F μ ν F_{\mu\nu} Fμν就只剩下
− F 01 = F 10 = ∂ 1 A 0 = d A 0 d r -F_{01} = F_{10} = \partial_1A_0 = \frac{dA_0}{dr} −F01=F10=∂1A0=drdA0
由此可得非零的电磁能动张量为
T 00 = 1 4 π [ F 01 F 0 1 − 1 4 g 00 ( F 01 F 01 + F 10 F 10 ) ] = 1 4 π ( F 01 F 0 1 − 1 2 g 00 F 10 F 10 ) = 1 4 π [ g 11 ( F 01 ) 2 − 1 2 g 00 g 00 g 11 ( F 10 ) 2 ] = 1 8 π g 11 ( F 10 ) 2 = 1 8 π e − 2 β ( F 10 ) 2 T 11 = 1 4 π [ F 10 F 1 0 − 1 4 g 11 ( F 01 F 01 + F 10 F 10 ) ] = 1 4 π ( F 10 F 1 0 − 1 2 g 11 F 10 F 10 ) = 1 4 π [ g 00 ( F 10 ) 2 − 1 2 g 11 g 00 g 11 ( F 10 ) 2 ] = 1 8 π g 00 ( F 10 ) 2 = − 1 8 π e − 2 α ( F 10 ) 2 T 22 = 1 4 π ( − 1 2 g 22 F 10 F 10 ) = − 1 8 π g 22 F 10 F 10 = − 1 8 π g 00 g 11 g 22 ( F 10 ) 2 = 1 8 π e − 2 ( α + β ) r 2 ( F 10 ) 2 T 33 = 1 4 π ( − 1 2 g 33 F 10 F 10 ) = − 1 8 π g 33 F 10 F 10 = − 1 8 π g 00 g 11 g 33 ( F 10 ) 2 = 1 8 π e − 2 ( α + β ) r 2 sin 2 θ ( F 10 ) 2 \begin{aligned} T_{00} &= \frac{1}{4\pi}[F_{01}F_0{}^1 - \frac{1}{4}g_{00}(F_{01}F^{01} + F_{10}F^{10})] = \frac{1}{4\pi}(F_{01}F_0{}^1 - \frac{1}{2}g_{00}F_{10}F^{10}) = \frac{1}{4\pi}[g^{11}(F_{01})^2 - \frac{1}{2}g_{00}g^{00}g^{11}(F_{10})^2] = \frac{1}{8\pi}g^{11}(F_{10})^2 = \frac{1}{8\pi}e^{-2\beta}(F_{10})^2 \\ T_{11} &= \frac{1}{4\pi}[F_{10}F_1{}^0 - \frac{1}{4}g_{11}(F_{01}F^{01} + F_{10}F^{10})] = \frac{1}{4\pi}(F_{10}F_1{}^0 - \frac{1}{2}g_{11}F_{10}F^{10}) = \frac{1}{4\pi}[g^{00}(F_{10})^2 - \frac{1}{2}g_{11}g^{00}g^{11}(F_{10})^2] = \frac{1}{8\pi}g^{00}(F_{10})^2 = -\frac{1}{8\pi}e^{-2\alpha}(F_{10})^2 \\ T_{22} &= \frac{1}{4\pi}(-\frac{1}{2}g_{22}F_{10}F^{10}) = -\frac{1}{8\pi}g_{22}F_{10}F^{10} = -\frac{1}{8\pi}g^{00}g^{11}g_{22}(F_{10})^2 = \frac{1}{8\pi}e^{-2(\alpha + \beta)}r^2(F_{10})^2 \\ T_{33} &= \frac{1}{4\pi}(-\frac{1}{2}g_{33}F_{10}F^{10}) = -\frac{1}{8\pi}g_{33}F_{10}F^{10} = -\frac{1}{8\pi}g^{00}g^{11}g_{33}(F_{10})^2 = \frac{1}{8\pi}e^{-2(\alpha + \beta)}r^2\sin^2\theta(F_{10})^2 \end{aligned} T00T11T22T33=4π1[F01F01−41g00(F01F01+F10F10)]=4π1(F01F01−21g00F10F10)=4π1[g11(F01)2−21g00g00g11(F10)2]=8π1g11(F10)2=8π1e−2β(F10)2=4π1[F10F10−41g11(F01F01+F10F10)]=4π1(F10F10−21g11F10F10)=4π1[g00(F10)2−21g11g00g11(F10)2]=8π1g00(F10)2=−8π1e−2α(F10)2=4π1(−21g22F10F10)=−8π1g22F10F10=−8π1g00g11g22(F10)2=8π1e−2(α+β)r2(F10)2=4π1(−21g33F10F10)=−8π1g33F10F10=−8π1g00g11g33(F10)2=8π1e−2(α+β)r2sin2θ(F10)2
我们还可以通过求解(无源)麦克斯韦方程组把 F 10 F_{10} F10求出来。第一个方程由于 F = d A F = dA F=dA自然满足，对于第二个方程，因为 ∗ F = 1 2 F a b ε a b c d , d ∗ F = 3 2 ∇ [ e F a b ε c d ] a b ^*F = \frac{1}{2}F^{ab}\varepsilon_{abcd}, ~ d^*F = \frac{3}{2}\nabla_{[e}F^{ab}\varepsilon_{cd]ab} ∗F=21Fabεabcd, d∗F=23∇[eFabεcd]ab，所以
∗ d ∗ F = 1 3 ! 3 2 ∇ [ e F a b ε c d ] a b ε c d e f = 1 4 ∇ [ e F a b δ a e δ b ] f 2 ! 2 ! = ∇ e F e f ^*d^*F = \frac{1}{3!}\frac{3}{2}\nabla^{[e}F_{ab}\varepsilon^{cd]ab}\varepsilon_{cdef} = \frac{1}{4}\nabla^{[e}F_{ab}\delta^a{}_e\delta^{b]}{}_f2!2! = \nabla^eF_{ef} ∗d∗F=3!123∇[eFabεcd]abεcdef=41∇[eFabδaeδb]f2!2!=∇eFef
我们取其分量形式为
F μ ν ; μ = 0 F^{\mu\nu}{}_{;\mu} = 0 Fμν;μ=0
又
F μ ν ; μ = F μ ν , μ + Γ μ μ σ F σ ν + Γ ν μ ρ F μ ρ F^{\mu\nu}{}_{;\mu} = F^{\mu\nu}{}_{,\mu} + \Gamma^\mu{}_{\mu\sigma}F^{\sigma\nu} + \Gamma^\nu{}_{\mu\rho}F^{\mu\rho} Fμν;μ=Fμν,μ+ΓμμσFσν+ΓνμρFμρ
注意到最后一项为零(因子一为对称，一为反称)，而
Γ μ μ σ = 1 ∣ g ∣ ∂ ∣ g ∣ ∂ x σ \Gamma^\mu{}_{\mu\sigma} = \frac{1}{\sqrt{|g|}}\frac{\partial\sqrt{|g|}}{\partial x^\sigma} Γμμσ=∣g∣1∂xσ∂∣g∣
其中 g g g是分量 g μ ν g_{\mu\nu} gμν对应的矩阵的行列式，所以就有
0 = F μ ν ; μ = 1 − g ∂ ( − g F μ ν ) ∂ x μ 0 = F^{\mu\nu}{}_{;\mu} = \frac{1}{\sqrt{-g}}\frac{\partial(\sqrt{-g}F^{\mu\nu})}{\partial x^\mu} 0=Fμν;μ=−g1∂xμ∂(−gFμν)
其中， − g = ∣ g ∣ = e α + β r 2 sin θ \sqrt{-g} = \sqrt{|g|} = e^{\alpha + \beta}r^2\sin\theta −g=∣g∣=eα+βr2sinθ，而要求的 F 10 = g 00 g 11 F 10 F_{10} = g_{00}g_{11}F^{10} F10=g00g11F10. 这个方程给出
d d r [ e − ( α + β ) r 2 F 10 ] = 0 \frac{d}{dr}[e^{-(\alpha + \beta)}r^2F_{10}] = 0 drd[e−(α+β)r2F10]=0
解得
F 10 = Q r 2 e α + β F_{10} = \frac{Q}{r^2}e^{\alpha + \beta} F10=r2Qeα+β
电磁真空场方程
容易验证 T μ ν T_{\mu\nu} Tμν是无迹的，即 g μ ν T μ ν = 0 g^{\mu\nu}T_{\mu\nu} = 0 gμνTμν=0. 于是 R = 0 R = 0 R=0，爱因斯坦场方程就可表示为
R μ ν = 8 π T μ ν R_{\mu\nu} = 8\pi T_{\mu\nu} Rμν=8πTμν
结合我们求得的 T μ ν T_{\mu\nu} Tμν和 R μ ν R_{\mu\nu} Rμν
R 00 = e 2 ( α − β ) ( α ′ ′ − α ′ β ′ + α ′ 2 + 2 r − 1 α ′ ) R 11 = − α ′ ′ + α ′ β ′ − α ′ 2 + 2 r − 1 β ′ R 22 = − e − 2 β [ 1 + r ( α ′ − β ′ ) ] + 1 R 33 = − { e − 2 β [ 1 + r ( α ′ − β ′ ) ] − 1 } sin 2 θ R_{00} = e^{2(\alpha - \beta)}(\alpha'' - \alpha'\beta' + \alpha'^2 + 2r^{-1}\alpha') \\ R_{11} = -\alpha'' + \alpha'\beta' - \alpha'^2 + 2r^{-1}\beta' \\ R_{22} = -e^{-2\beta}[1 + r(\alpha' - \beta')] + 1 \\ R_{33} = -\{e^{-2\beta}[1 + r(\alpha' - \beta')] - 1\}\sin^2\theta R00=e2(α−β)(α′′−α′β′+α′2+2r−1α′)R11=−α′′+α′β′−α′2+2r−1β′R22=−e−2β[1+r(α′−β′)]+1R33=−{e−2β[1+r(α′−β′)]−1}sin2θ
就给出如下方程
e 2 ( α − β ) ( α ′ ′ − α ′ β ′ + α ′ 2 − 2 r − 1 α ′ ) = e − 2 β ( F 10 ) 2 − α ′ ′ + α ′ β ′ − α ′ 2 + 2 r − 1 β ′ = − e − 2 α ( F 10 ) 2 − e − 2 β [ 1 + r ( α ′ − β ′ ) ] + 1 = e − 2 ( α + β ) r 2 ( F 10 ) 2 e^{2(\alpha - \beta)}(\alpha'' - \alpha'\beta' + \alpha'^2 - 2r^{-1}\alpha') = e^{-2\beta}(F_{10})^2 \\ -\alpha'' + \alpha'\beta' - \alpha'^2 + 2r^{-1}\beta' = -e^{-2\alpha}(F_{10})^2 \\ -e^{-2\beta}[1 + r(\alpha' - \beta')] + 1 = e^{-2(\alpha + \beta)}r^2(F_{10})^2 e2(α−β)(α′′−α′β′+α′2−2r−1α′)=e−2β(F10)2−α′′+α′β′−α′2+2r−1β′=−e−2α(F10)2−e−2β[1+r(α′−β′)]+1=e−2(α+β)r2(F10)2
前两个方程给出 α ′ = − β ′ \alpha' = -\beta' α′=−β′，即 α = − β + γ \alpha = -\beta + \gamma α=−β+γ，代入第三个方程得
1 − Q 2 r 2 = e − 2 β ( 1 − 2 β ′ r ) 1 - \frac{Q^2}{r^2} = e^{-2\beta}(1 - 2\beta'r) 1−r2Q2=e−2β(1−2β′r)
解得
e − 2 β = 1 + Q 2 r 2 + C r e^{-2\beta} = 1 + \frac{Q^2}{r^2} + \frac{C}{r} e−2β=1+r2Q2+rC
于是 e 2 α = e − 2 β e 2 γ e^{2\alpha} = e^{-2\beta}e^{2\gamma} e2α=e−2βe2γ，代入原度规表达式就有
d s 2 = − ( 1 + Q 2 r 2 + C r ) e 2 γ d t 2 + ( 1 + Q 2 r 2 + C r ) − 1 d r 2 + r 2 d θ 2 + r 2 sin 2 θ d φ 2 ds^2 = -(1 + \frac{Q^2}{r^2} + \frac{C}{r})e^{2\gamma}dt^2 + (1 + \frac{Q^2}{r^2} + \frac{C}{r})^{-1}dr^2 + r^2d\theta^2 + r^2\sin^2\theta d\varphi^2 ds2=−(1+r2Q2+rC)e2γdt2+(1+r2Q2+rC)−1dr2+r2dθ2+r2sin2θdφ2
考虑新坐标 t ^ = e γ t \hat t = e^\gamma t t^=eγt，并仍将其记作 t t t，上式就简化为
d s 2 = − ( 1 + Q 2 r 2 + C r ) d t 2 + ( 1 + Q 2 r 2 + C r ) − 1 d r 2 + r 2 d θ 2 + r 2 sin 2 θ d φ 2 ds^2 = -(1 + \frac{Q^2}{r^2} + \frac{C}{r})dt^2 + (1 + \frac{Q^2}{r^2} + \frac{C}{r})^{-1}dr^2 + r^2d\theta^2 + r^2\sin^2\theta d\varphi^2 ds2=−(1+r2Q2+rC)dt2+(1+r2Q2+rC)−1dr2+r2dθ2+r2sin2θdφ2
这就相当于取 α = − β \alpha = -\beta α=−β，在此条件下就有
F 10 = Q r 2 e α + β = Q r 2 F_{10} = \frac{Q}{r^2}e^{\alpha + \beta} = \frac{Q}{r^2} F10=r2Qeα+β=r2Q
注意到 F 10 F_{10} F10实际上就对应于径向场强 q r 2 \frac{q}{r^2} r2q，于是 Q Q Q的含义就是星体所带的电荷。另外，当 Q = 0 Q = 0 Q=0时，我们的度规应回到Schwarzschild度规，因此 C = − 2 M C = -2M C=−2M。最终Reissner-Nordstrom度规可以表示为
d s 2 = − ( 1 − 2 M r + Q 2 r 2 ) d t 2 + ( 1 − 2 M r + Q 2 r 2 ) − 1 d r 2 + r 2 d θ 2 + r 2 sin 2 θ d φ 2 ds^2 = -(1 - \frac{2M}{r} + \frac{Q^2}{r^2})dt^2 + (1 - \frac{2M}{r} + \frac{Q^2}{r^2})^{-1}dr^2 + r^2d\theta^2 + r^2\sin^2\theta d\varphi^2 ds2=−(1−r2M+r2Q2)dt2+(1−r2M+r2Q2)−1dr2+r2dθ2+r2sin2θdφ2
相应的电磁场和电磁4势为
F a b = − Q r 2 ( d t ) a ∧ ( d r ) b , A a = Q r ( d t ) a F_{ab} = -\frac{Q}{r^2}(dt)_a\wedge(dr)_b, ~ A_{a} = \frac{Q}{r}(dt)_a Fab=−r2Q(dt)a∧(dr)b, Aa=rQ(dt)a
那个方程究竟是怎么解的
我是说这个方程
1 − Q 2 r 2 = e − 2 β ( 1 − 2 β ′ r ) 1 - \frac{Q^2}{r^2} = e^{-2\beta}(1 - 2\beta'r) 1−r2Q2=e−2β(1−2β′r)
首先把方程形式改写一下
( 1 − e − 2 β − Q 2 r 2 ) d r + 2 e − 2 β r d β = 0 (1 - e^{-2\beta} - \frac{Q^2}{r^2})dr + 2e^{-2\beta}rd\beta = 0 (1−e−2β−r2Q2)dr+2e−2βrdβ=0
所谓恰当方程，指方程可以表达为某个函数的全微分的形式
d u = ∂ u ∂ r d r + ∂ u ∂ β d β = 0 du = \frac{\partial u}{\partial r}dr + \frac{\partial u}{\partial \beta}d\beta = 0 du=∂r∂udr+∂β∂udβ=0
此时应有
∂ u ∂ r ∂ β = ∂ u ∂ β ∂ r \frac{\partial u}{\partial r\partial \beta} = \frac{\partial u}{\partial \beta \partial r} ∂r∂β∂u=∂β∂r∂u
令 M = 1 − e − 2 β − Q 2 r 2 , N = 2 e − 2 β r M = 1 - e^{-2\beta} - \frac{Q^2}{r^2}, ~ N = 2e^{-2\beta}r M=1−e−2β−r2Q2, N=2e−2βr，容易看出 ∂ M ∂ β = ∂ N ∂ r \frac{\partial M}{\partial \beta} = \frac{\partial N}{\partial r} ∂β∂M=∂r∂N，因此该方程确实是一个恰当方程。我们有
u = ∫ N d β = − r e − 2 β + ϕ ( r ) u = \int Nd\beta = -re^{-2\beta} + \phi(r) u=∫Ndβ=−re−2β+ϕ(r)
其中 ϕ ( r ) \phi(r) ϕ(r)是积分常数，又
M = ∂ u ∂ r = − e − 2 β + d ϕ ( r ) d r M = \frac{\partial u}{\partial r} = -e^{-2\beta} + \frac{d\phi(r)}{dr} M=∂r∂u=−e−2β+drdϕ(r)
所以 d ϕ ( r ) d r = 1 − Q 2 r 2 , ϕ ( r ) = r + Q 2 r + C 1 \frac{d\phi(r)}{dr} = 1 - \frac{Q^2}{r^2}, ~ \phi(r) = r + \frac{Q^2}{r} + C_1 drdϕ(r)=1−r2Q2, ϕ(r)=r+rQ2+C1，代回 u u u的表达式就得到
u = − r e − 2 β + r + Q 2 r + C 1 u= -re^{-2\beta} + r + \frac{Q^2}{r} + C_1 u=−re−2β+r+rQ2+C1
原方程 d u = 0 du = 0 du=0的解就是 u = C 2 u = C_2 u=C2，所以
e − 2 β = 1 + Q 2 r 2 + C r e^{-2\beta} = 1 + \frac{Q^2}{r^2} + \frac{C}{r} e−2β=1+r2Q2+rC
第二种方法
若能看出
1 − Q 2 r 2 = ( r e − 2 β ) ′ 1 - \frac{Q^2}{r^2} = (re^{-2\beta})' 1−r2Q2=(re−2β)′
则
( 1 − Q 2 r 2 ) d r = d ( r e − 2 β ) (1 - \frac{Q^2}{r^2})dr = d(re^{-2\beta}) (1−r2Q2)dr=d(re−2β)
两边积分就有
r + Q 2 r + C = r e − 2 β r + \frac{Q^2}{r} + C = re^{-2\beta} r+rQ2+C=re−2β