Here's me following 1973 Misner, Thorne, Wheeler "Gravitation" through the Schwarzschild equations of structure:
Coordinate system is $t, r, \theta, \phi$
$\tau$ is the coordinate along the velocity curve.
Chapter 22
$\frac{d}{d\tau} (n V) = 0$ (eqn 22.1)
$V \frac{dn}{d\tau} + n \frac{dV}{d\tau} = 0$
$\frac{dV}{d\tau} = -\frac{V}{n} \frac{dn}{d\tau}$
$\frac{dV}{d\tau} = V {u^\mu}_{;\mu}$ (eqn 22.2 courtesy of exercise 22.1 - TODO show the details)
Using $\textbf{u} = \frac{d}{d\tau}$
$0 = V \frac{d}{d\tau} n + n \frac{d}{d\tau} V$
$0 = \frac{d}{d\tau} n + n \frac{1}{V} \frac{d}{d\tau} V$
substitute $\frac{1}{V} \frac{d}{d\tau} V = {u^\mu}_{;\mu}$ (courtesy of eqn 22.2)
$0 = n_{,\mu} u^\mu + n \frac{1}{V} \frac{d}{d\tau} V$
$0 = (n u^\mu)_{;\mu}$
Let $S^\mu = n u^\mu$
${S^\mu}_{;\mu} = 0$ (eqn 22.3)
starting with
$\frac{dV}{d\tau} = -\frac{V}{n} \frac{dn}{d\tau}$
$\frac{dn}{d\tau} = -\frac{n}{V} \frac{dV}{d\tau}$
substitute eqn 22.2 $\frac{dV}{d\tau} = V {u^\mu}_{;\mu}$
$\frac{dn}{d\tau} = -\frac{n}{V} V {u^\mu}_{;\mu}$
$\frac{dn}{d\tau} = -n {u^\mu}_{;\mu}$
eqn 22.9:
Stress-energy of matter:
$T^{\mu\nu} = (\rho + p) u^\mu u^\nu + p g^{\mu\nu}$
Conservation of stress-energy (eqn 22.10):
${T^{\mu\nu}}_{;\nu}$
$= ((\rho + p) u^\mu u^\nu + p g^{\mu\nu})_{;\nu}$
$= (\rho_{,\nu} + p_{,\nu}) u^\nu u^\mu
+ (\rho + p) ({u^\mu}_{;\nu} u^\nu + u^\mu {u^\nu}_{;\nu})
+ p_{,\nu} g^{\mu\nu}$
Unit vector derivative identity:
Now $u^\mu u_\mu = -1$
$(u^\mu u_\mu)_{;\nu} = 0$
${u^\mu}_{;\nu} u_\mu + u^\mu u_{\mu;\nu} = 0$
$u_{\mu;\nu} u^\mu + u^\mu u_{\mu;\nu} = 0$
$2 u^\mu u_{\mu;\nu} = 0$
$u^\mu u_{\mu;\nu} = 0$
Back to ${T^{\mu\nu}}_{;\nu} =
(\rho_{,\nu} + p_{,\nu}) u^\nu u^\mu
+ (\rho + p) ({u^\mu}_{;\nu} u^\nu + u^\mu {u^\nu}_{;\nu})
+ p_{,\nu} g^{\mu\nu}$
Project stress-energy divergence along velocity vector:
$u_\mu {T^{\mu\nu}}_{;\nu} =
u_\mu (\rho_{,\nu} + p_{,\nu}) u^\nu u^\mu
+ u_\mu (\rho + p) ({u^\mu}_{;\nu} u^\nu + u^\mu {u^\nu}_{;\nu})
+ u_\mu p_{,\nu} g^{\mu\nu}$
$ = - (\rho_{,\nu} + p_{,\nu}) u^\nu
+ (\rho + p) (u_\mu {u^\mu}_{;\nu} u^\nu - {u^\nu}_{;\nu})
+ u^\nu p_{,\nu}$
$ = - (\rho_{,\nu} + p_{,\nu}) u^\nu
- (\rho + p) {u^\nu}_{;\nu}
+ u^\nu p_{,\nu}$
$ = - \rho_{,\nu} u^\nu
- (\rho + p) {u^\nu}_{;\nu}$
Let $0 = u_\mu {T^{\mu\nu}}_{;\nu}$
$0 =- \rho_{,\nu} u^\nu
- (\rho + p) {u^\nu}_{;\nu} $
using $\textbf{u} \cdot \nabla = \frac{d}{d\tau}$
$0 =- \frac{d}{d\tau} \rho - (\rho + p) {u^\nu}_{;\nu} $
$\frac{d}{d\tau} \rho = -(\rho + p) {u^\nu}_{;\nu} $
using ${u^\mu}_{;\mu} = -\frac{1}{n} \frac{dn}{d\tau}$ above
$\frac{d\rho}{d\tau} = \frac{(\rho + p)}{n} \frac{dn}{d\tau}$ (eqn 22.11a)
Back to projection of stress-energy divergence - instead of along $u^\mu$, this time orthogonal to $u^\mu$:
Let ${P^\mu}_\nu = \delta^\mu_\nu + u^\mu u_\nu$ be the projection tensor.
${P^\alpha}_\mu {T^{\mu\nu}}_{;\nu}$
$ = (\delta^\alpha_\mu + u^\alpha u_\mu)
(
(\rho_{,\nu} + p_{,\nu}) u^\nu u^\mu
+ (\rho + p) ({u^\mu}_{;\nu} u^\nu + u^\mu {u^\nu}_{;\nu})
+ p_{,\nu} g^{\mu\nu}
)$
$ =
(\rho_{,\nu} + p_{,\nu}) u^\nu u^\alpha
+ (\rho + p) ({u^\alpha}_{;\nu} u^\nu + u^\alpha {u^\nu}_{;\nu})
+ p_{,\nu} g^{\alpha\nu}
+ u^\alpha u_\mu (\rho_{,\nu} + p_{,\nu}) u^\nu u^\mu
+ u^\alpha u_\mu (\rho + p) ({u^\mu}_{;\nu} u^\nu + u^\mu {u^\nu}_{;\nu})
+ u^\alpha u_\mu p_{,\nu} g^{\mu\nu}
$
$ =
(\rho_{,\nu} + p_{,\nu}) u^\nu u^\alpha
+ (\rho + p) ({u^\alpha}_{;\nu} u^\nu + u^\alpha {u^\nu}_{;\nu})
+ p_{,\nu} g^{\alpha\nu}
- u^\alpha (\rho_{,\nu} + p_{,\nu}) u^\nu
- u^\alpha (\rho + p) {u^\nu}_{;\nu}
+ u^\alpha u^\nu p_{,\nu}
$
$ = (\rho + p) {u^\alpha}_{;\nu} u^\nu
+ (g^{\alpha\nu} + u^\alpha u^\nu) p_{,\nu} $
$ = (\rho + p) {u^\alpha}_{;\nu} u^\nu
+ P^{\alpha\nu} p_{,\nu} $
Let ${P^\alpha}_\mu {T^{\mu\nu}}_{;\nu} = 0$
$0 = (\rho + p) {u^\alpha}_{;\nu} u^\nu
+ P^{\alpha\nu} p_{,\nu} $
$(\rho + p) {u^\alpha}_{;\nu} u^\nu = -P^{\alpha\nu} p_{,\nu}$ (eqn 22.13)
section 23.5 "Equations of Structure"
Spherically symmetric coordinate system:
$ds^2 = -dt^2 + dr^2 + r^2 d\Omega^2$ (eqn 23.1)
For $d\Omega = d\theta^2 + (sin \theta)^2 d\phi^2$ (eqn 23.2)
Next comes TOV modifications:
$ds^2 = -e^{2\Phi} dt^2 + e^{2\Lambda} dr^2 + R^2 d\Omega^2$ (eqn 23.3)
For $\Phi, \Lambda, R$ functions of $r$.
Static metric means $g_{\mu\nu,t} = 0$.
More generalized metric is equivalent. For example:
$ds^2 = -a^2 dt^2 - 2 ab dr dt + c^2 dr^2 + R^2 d\Omega^2$ (eqn 23.4)
Let $e^\Phi dt' = a dt + b dr$ (eqn 23.5)
Let $e^{2 \Lambda} = b^2 + c^2$
$ds^2 = -e^{2\Phi} dt^2 + e^{2\Lambda} dr^2 + R^2 d\Omega^2$
$= -(a dt + b dr)^2 + (b^2 + c^2) dr^2 + R^2 d\Omega^2$
$= -(a^2 dt^2 + 2 a b dr dt + b^2 dr^2) + (b^2 + c^2) dr^2 + R^2 d\Omega^2$
$= -a^2 dt^2 - 2 a b dr dt - b^2 dr^2 + b^2 dr^2 + c^2 dr^2 + R^2 d\Omega^2$
$= -a^2 dt^2 - 2 a b dr dt + c^2 dr^2 + R^2 d\Omega^2$
Tada, they are the same.
$g_{rt} = 0$ = coordinate condition.
Apply change of radial coordinate $r' = R(r)$ eqn 23.6
Finally we get: $ds^2 = -e^{2\Phi} dt^2 + e^{2\Lambda} dr^2 + r^2 d\Omega^2$ eqn 23.7
Therefore
$g_{tt} = -e^{2\Phi}$
$g_{rr} = e^{2\Lambda}$
$g_{\theta\theta} = r^2$
$g_{\phi\phi} = (r sin \theta)^2$
Diagonal inverse:
$g^{tt} = -e^{-2\Phi}$
$g^{rr} = e^{-2\Lambda}$
$g^{\theta\theta} = r^{-2}$
$g^{\phi\phi} = (r sin \theta)^{-2}$
Metric partials:
$g_{tt,r} = -2 \Phi_{,r} e^{2\Phi}$
$g_{rr,r} = 2 \Lambda_{,r} e^{2\Lambda}$
$g_{\theta\theta,r} = 2 r$
$g_{\phi\phi,r} = 2 r (sin \theta)^2$
$g_{\phi\phi,\theta} = 2 r^2 sin \theta cos \theta$
Connections:
$\Gamma_{\mu\mu\mu} = \frac{1}{2} g_{\mu\mu,\mu}$ (one unique index)
$\Gamma_{rrr} = \Lambda_{,r} e^{2\Lambda} $
$\Gamma_{\alpha\alpha\beta} = \Gamma_{\alpha\beta\alpha} = \frac{1}{2} (g_{\alpha\alpha,\beta} + g_{\alpha\beta,\alpha} - g_{\alpha\beta,\alpha})$
$ = \frac{1}{2} g_{\alpha\alpha,\beta}$ (two unique indexes)
$\Gamma_{ttr} = \Gamma_{trt} = -\Phi_{,r} e^{2\Phi}$
$\Gamma_{\theta\theta r} = \Gamma_{\theta r\theta} = r$
$\Gamma_{\phi\phi r} = \Gamma_{\phi r\phi} = r (sin \theta)^2$
$\Gamma_{\phi\phi\theta} = \Gamma_{\phi\theta\phi} = r^2 sin \theta cos \theta$
$\Gamma_{\beta\alpha\alpha} = \frac{1}{2} (g_{\beta\alpha,\alpha} + g_{\beta\alpha,\alpha} - g_{\alpha\alpha,\beta})$
$ = g_{\beta\alpha,\alpha} - \frac{1}{2} g_{\alpha\alpha,\beta}$
Since $g_{\alpha\beta} = 0$ for $\alpha \ne \beta$ and since we already considered connections for $\alpha = \beta$ above
$\Gamma_{\beta\alpha\alpha} = -\frac{1}{2} g_{\alpha\alpha,\beta}$ for $\alpha \ne \beta$
$\Gamma_{rtt} = -\Phi_{,r} e^{2\Phi}$
$\Gamma_{r\theta\theta} = -r$
$\Gamma_{r\phi\phi} = -r (sin \theta)^2$
Because $g_{\alpha\beta} = 0$ for $\alpha \ne \beta$, so does $g_{\alpha\beta,\gamma} = 0$ for $\alpha, \beta, \gamma$ all differing,
and likewise $\Gamma_{\alpha\beta\gamma} = 0$ for $\alpha, \beta, \gamma$ all differing.
2nd kind connections: (simple because $g_{\mu\nu}$ is diagonal)
${\Gamma^t}_{tr} = {\Gamma^t}_{rt} = g^{tt} \Gamma_{ttr} = -e^{-2\Phi} \cdot -\Phi_{,r} e^{2\Phi} = \Phi_{,r}$
${\Gamma^r}_{tt} = g^{rr} \Gamma_{rtt} = e^{-2\Lambda} \cdot -\Phi_{,r} e^{2\Phi} = -\Phi_{,r} e^{2(\Phi - \Lambda)}$
${\Gamma^r}_{rr} = g^{rr} \Gamma_{rrr} = e^{-2\Lambda} \Lambda_{,r} e^{2\Lambda} = \Lambda_{,r}$
${\Gamma^r}_{\theta\theta} = g^{rr} \Gamma_{r\theta\theta} = -r e^{-2\Lambda}$
${\Gamma^r}_{\phi\phi} = g^{rr} \Gamma_{r\phi\phi} = -r (sin \theta)^2 e^{-2 \Lambda}$
${\Gamma^\theta}_{\theta r} = {\Gamma^\theta}_{r\theta} = g^{\theta\theta} \Gamma_{\theta\theta r} = r / r^2 = \frac{1}{r}$
${\Gamma^\phi}_{\phi r} = {\Gamma^\phi}_{r\phi} = g^{\phi\phi} \Gamma_{\phi\phi r} = r (sin \theta)^2 (r sin \theta)^{-2} = \frac{1}{r}$
${\Gamma^\phi}_{\phi\theta} = {\Gamma^\phi}_{\theta\phi} = g^{\phi\phi} \Gamma_{\phi\phi\theta} = r^2 sin \theta cos \theta (r sin \theta)^{-2} = cot \theta$
exercise 23.1
Isotropic coordinate metric: $ds^2 = -e^{2 \Phi} dt^2 + e^{2 \mu} ( d \bar{r}^2 + \bar{r}^2 d\Omega^2 )$ (eqn 23.8)
for $\Phi = \Phi(\bar{r}), \mu = \mu(\bar{r})$
exercise 23.1a:
TOV metric: $ds^2 = -e^{2\Phi} dt^2 + e^{2\Lambda} dr^2 + r^2 d\Omega^2$ eqn 23.7
Find the change of coordinates from TOV to isotropic:
1) compare $d\Omega^2$ terms to find:
Let $r^2 = e^{2\mu} \bar{r}^2$
So $r = e^\mu \bar{r}$ and $\bar{r} = e^{-\mu} r$
So $\frac{r}{\bar{r}} = e^\mu$
So $\mu = log(\frac{r}{\bar{r}}) = log(r) - log(\bar{r})$
Likewise $d\mu = \frac{dr}{r} - \frac{d\bar{r}}{\bar{r}}$
Substituting this into the isotropic metric gives us:
$ds^2 = -e^{2 \Phi} dt^2 + e^{2 \mu} d \bar{r}^2 + r^2 d\Omega^2 $ (eqn 23.8)
Substituting this into the TOV metric gives us:
$ds^2 = -e^{2\Phi} dt^2 + e^{2\Lambda} dr^2 + e^{2\mu} \bar{r}^2 d\Omega^2$
Now we're left with equating $e^{2\mu} d\bar{r}^2 = e^{2\Lambda} dr^2$
$e^\Lambda dr = e^\mu d\bar{r}$
$e^\Lambda dr = \frac{r}{\bar{r}} d\bar{r}$
$\frac{1}{r} e^\Lambda dr = \frac{1}{\bar{r}} d\bar{r}$
if $\Lambda = \Lambda(r)$, or $\mu = \mu(\bar{r})$, then we can solve further ...
$\frac{1}{r} e^\Lambda \frac{dr}{d\Lambda} d\Lambda = \frac{1}{\bar{r}} d\bar{r}$
$\frac{1}{r \Lambda'(r)} e^\Lambda d\Lambda = \frac{1}{\bar{r}} d\bar{r}$
Next, substitute the Schwarzschild metric:
$ds^2 = -(1 - \frac{2 m}{r}) dt^2 + (1 - \frac{2 m}{r})^{-1} dr^2 + r^2 d\Omega^2$
$e^{\Lambda} = \frac{1}{\sqrt{1 - 2m/r}}$
$\frac{dr}{\sqrt{1 - 2m/r}} = \frac{r}{\bar{r}} d\bar{r}$
$\frac{dr}{\sqrt{r(r - 2m)}} = \frac{d\bar{r}}{\bar{r}}$
...
TODO solutions should be:
$r = \bar{r} (1 + \frac{m}{2\bar{r}})^2$
$r = \bar{r} (1 + \frac{m}{\bar{r}} + \frac{m^2}{4\bar{r}^2})$
$\bar{r}^2 + (m - r) \bar{r} + \frac{1}{4} m^2 = 0$
$\bar{r} = \frac{1}{2} (r - m \pm \sqrt{ (m - r)^2 - m^2 })$
$\bar{r} = \frac{1}{2} (r - m \pm \sqrt{ r^2 - 2 m r })$
$\bar{r} = \frac{1}{2} (r - m \pm \sqrt{ r (r - 2 m) } )$
TODO exercise 23.1b:
Show that, in the Newtonian limit, $\Phi$ becomes the Newtonian potential and $\mu$ becomes $-\Phi$.
Show that, in the Newtonian limit, $\Lambda = r \frac{d\Phi}{dr}$.
Surface area of a sphere:
$A = \int (r d\theta) (r sin \theta d\phi)$
$= \int_{\theta=0}^\pi r d\theta \cdot \int_{\phi=0}^{2\pi} r sin \theta d\phi$
$= \int_{\theta=0}^\pi r (r \phi sin \theta |_{\phi=0}^{2\pi} d\theta$
$= \int_{\theta=0}^\pi 2 \pi r^2 sin \theta d\theta$
$= (-2 \pi r^2 cos \theta|_{\theta=0}^\pi$
$= 4 \pi r^2$ eqn 23.9
Solving for $r$:
$r = \sqrt{\frac{A}{4 \pi}}$
Geometric properties:
$g_{\alpha\beta,t} = 0$
$g_{tr} = g_{t\theta} = g_{t\phi} = 0$
$\underset{r \rightarrow \infty}{\lim} \Phi(r) = 0$ eqn 23.10
eqn 23.11:
Five functions of radius:
$\rho(r)$ density of mass-energy in rest-frame of fluid.
$p(r)$ isotropic pressure in rest-frame of fluid.
$n(r)$ number density of baryons in rest-frame of fluid.
$u^\mu(r)$ = 4-velocity of fluid.
$T^{\mu\nu} = (\rho + p) u^\mu u^\nu + p g^{\mu\nu}$ = stress-energy tensor of fluid. eqn 23.12. Just like we saw before.
Static stars: eqn 23.13a
$u^r = \frac{dr}{d\tau} = 0$
$u^\theta = \frac{d\theta}{d\tau} = 0$
$u^\phi = \frac{d\phi}{d\tau} = 0$
Normalized 4-velocity: eqn 23.13b
$-1 = g_{\mu\nu} u^\mu u^\nu = g_{tt} u^t u^t = -e^{2\Phi} (u^t)^2$
$u^t = \frac{dt}{d\tau} = e^{-\Phi}$
$u_t = g_{tt} u^t = -e^{2\Phi} e^{-\Phi} = -e^\Phi$
Stress-energy: eqn 23.14
$T^{\mu\nu} = (\rho + p) u^\mu u^\nu + p g^{\mu\nu}$
$T^{tt} = (\rho + p) u^t u^t + p g^{tt}$
$= (\rho + p) e^{-2\Phi} + p (-e^{-2\Phi})$
$= \rho e^{-2\Phi}$
The book says $T^{00} = \rho e^{-2\Phi}$, but what is the difference between the $0$ and $t$ coordinate? The shift? Which is zero...
$T^{rr} = p g^{rr} = p e^{-2 \Lambda}$
$T^{\theta\theta} = p g^{\theta\theta} = p r^{-2}$
$T^{\phi\phi} = p g^{\phi\phi} = p (r sin \theta)^{-2}$
otherwise $T^{\alpha\beta} = 0$
Transform so the time axis is along the fluid rest velocity: eqns 23.15a-d
$e_\hat{t} = \frac{d}{d\tau} = e^{-\Phi} \frac{\partial}{\partial t}$, so ${e_{\hat{t}}}^t = e^{-\Phi}$
$e_\hat{r} = e^{-\Lambda} \frac{\partial}{\partial r}$, so ${e_{\hat{r}}}^r = e^{-\Lambda}$
$e_\hat{\theta} = r^{-1} \frac{\partial}{\partial \theta}$, so ${e_{\hat{\theta}}}^\theta = r^{-1}$
$e_\hat{\phi} = (r sin \theta)^{-1} \frac{\partial}{\partial \phi}$, so ${e_{\hat{\phi}}}^\phi = (r sin \theta)^{-1}$
$\omega^\hat{t} = e^\Phi dt$
$\omega^\hat{r} = e^\Lambda dr$
$\omega^\hat{\theta} = r d\theta$
$\omega^\hat{\phi} = r sin \theta d\phi$
${e_\hat{\mu}}^\nu = \left[\matrix{
e^{-\Phi} & 0 & 0 & 0 \\
0 & e^{-\Lambda} & 0 & 0 \\
0 & 0 & r^{-1} & 0 \\
0 & 0 & 0 & (r sin \theta)^{-1}
}\right]$
${\omega^{\hat{\mu}}}_\nu = \left[\matrix{
e^\Phi & 0 & 0 & 0 \\
0 & e^\Lambda & 0 & 0 \\
0 & 0 & r & 0 \\
0 & 0 & 0 & r sin \theta
}\right]$
In the new coordinate system: eqn 23.15 a-d
$T_{\hat{\mu}\hat{\nu}} = {e_{\hat{\mu}}}^\mu {e_{\hat{\nu}}}^\nu T_{\mu\nu}$
$= {e_{\hat{\mu}}}^\mu {e_{\hat{\nu}}}^\nu g_{\mu\alpha} g_{\nu\beta} T^{\alpha\beta}$
$T_{\hat{t}\hat{t}} = (e^{-\Phi})^2 \cdot (-e^{2\Phi})^2 \cdot \rho e^{-2\Phi} = \rho$
$T_{\hat{r}\hat{r}} = (e^{-\Lambda})^2 \cdot (e^{2\Lambda})^2 \cdot p e^{-2 \Lambda} = p$
$T_{\hat{\theta}\hat{\theta}} = (r^{-1})^2 \cdot (r^2)^2 \cdot p r^{-2} = p$
$T_{\hat{\phi}\hat{\phi}} = ((r sin \theta)^{-1})^2 \cdot ((r sin \theta)^2)^2 \cdot p (r sin \theta)^{-2} = p$
$T_{\hat{\alpha}\hat{\beta}} = 0$ otherwise
Determine functions $\Phi, \Lambda, \rho, p, n$ by their parameter $r$, by the EFE $G^{\mu\nu} = 8 \pi T^{\mu\nu}$, and by ${T^{\mu\nu}}_{;\nu} = 0$.
Also depend $p = p(n), \rho = \rho(n)$ on the number density of baryons, $n$ (eqn 23.16).
...as well either temperature $T$ or entropy $s$: $p = p(n, s), \rho = \rho(n, s)$.
From here it's handy to let $s = s(n)$, so $p = p(n, s(n)), \rho = \rho(n, s(n))$ (see box 22.1)
exercise 23.2a:
Verify the equations 23.15a,b form an orthonormal frame.
Argument: no terms are sums -- all are only scalars of the holonomic basis. Therefore $\omega^\hat\mu(e_\hat\nu) = 0$ for $\mu \ne \nu$.
This combined with the fact that the function coefficients of matching basis variables are inverses:
$\omega^\hat{t}(e_\hat{t}) = e^\Phi e^{-\Phi} \frac{\partial t}{\partial t} = 1$
$\omega^\hat{r}(e_\hat{r}) = e^\Lambda e^{-\Lambda} \frac{\partial r}{\partial r} = 1$
$\omega^\hat\theta(e_\hat\theta) = r \frac{1}{r} \frac{\partial \theta}{\partial \theta} = 1$
$\omega^\hat\phi(e_\hat\phi) = r sin(\theta) \frac{1}{r sin(\theta)} \frac{\partial \phi}{\partial \phi} = 1$
exercise 23.2b:
Verify that equation 23.15c gives the components of fluid relative to the tetrad.
$u^t = e^{-\Phi}$
Therefore $u = e^{-\Phi} \partial_t$
Therefore $u = e_\hat{t}$
exercise 23.2c:
Verify the equations 23.15d are the components of stress-energy tensor.
$T^{tt} = \rho e^{-2\Phi}$
$T^{rr} = p e^{-2\Lambda}$
$T^{\theta\theta} = \frac{p}{r^2}$
$T^{\phi\phi} = \frac{p}{r^2 (sin(\theta))^2}$
otherwise $T^{ab} = 0$
Therefore $T =
\rho e^{-2\Phi} \partial_t \otimes \partial_t
+ p e^{-2\Lambda} \partial_r \otimes \partial_r
+ \frac{p}{r^2} \partial_\theta \otimes \partial_\theta
+ \frac{p}{r^2 (sin(\theta))^2} \partial_\phi \otimes \partial_\phi
$
$T =
\rho \partial_\hat{t} \otimes \partial_\hat{t}
+ p \partial_\hat{r} \otimes \partial_\hat{r}
+ p \partial_\hat\theta \otimes \partial_\hat\theta
+ p \partial_\hat\phi \otimes \partial_\hat\phi
$
Therefore $T^{\hat{t}\hat{t}} = \rho, T^{\hat{r}\hat{r}} = p, T^{\hat\theta\hat\theta} = p, T^{\hat\phi\hat\phi} = p$
Since $g_{\hat\alpha\hat\beta} = \eta_{\alpha\beta}$, we find
$T_{\hat{t}\hat{t}} = \rho, T_{\hat{r}\hat{r}} = p, T_{\hat\theta\hat\theta} = p, T_{\hat\phi\hat\phi} = p$
Start with eqn 22.11a:
$\frac{d\rho}{d\tau} = \frac{(\rho + p)}{n} \frac{dn}{d\tau}$
Using $\frac{dn}{d\tau} = -n {u^\mu}_{;\mu}$
$\frac{d\rho}{d\tau} = -(\rho + p) {u^\mu}_{;\mu}$
For a static star, both sides are zero: $\frac{d\rho}{d\tau} = 0$, $(\rho + p) {u^\mu}_{;\mu} = 0$
Back to 22.13, the projection of the divergence of the stress-energy:
$(\rho + p) {u^\alpha}_{;\nu} u^\nu = -P^{\alpha\nu} p_{,\nu}$ eqn 22.13
$(\rho + p) {u^\alpha}_{;\nu} u^\nu = -(g^{\alpha\nu} + u^\alpha u^\nu) p_{,\nu}$ eqn 22.13
$(\rho + p) {u^\alpha}_{;\nu} u^\nu = -g^{\alpha\nu} p_{,\nu} - u^\alpha u^\nu p_{,\nu}$ eqn 22.13
$(\rho + p) u_{\alpha;\nu} u^\nu = -p_{,\alpha} - u_\alpha u^\nu p_{,\nu}$ eqn 22.13
$(\rho + p) (u_{\alpha,\nu} - {\Gamma^\beta}_{\alpha\nu} u_\beta) u^\nu = -p_{,\alpha} - u_\alpha u^\nu p_{,\nu}$ eqn 22.13
Look at the radial direction ($\alpha = r$):
$(\rho + p) (u_{r,\nu} - {\Gamma^\alpha}_{r\nu} u_\alpha) u^\nu = -p_{,r} - u_r u^\nu p_{,\nu}$ eqn 22.13
Substitute $u_i = u^i = 0$
$-(\rho + p) {\Gamma^t}_{rt} u_t u^t = -p_{,r}$
$-(\rho + p) \cdot \Phi_{,r} \cdot -e^{\Phi} \cdot e^{-\Phi} = -p_{,r}$
$(\rho + p) \Phi_{,r} = -p_{,r}$ (eqn 23.17)
Notice, for mass-energy density much greater than pressure, we get the limit $\rho \Phi_{,r} = -p_{,r}$ (eqn 23.17N)
Next comes the Einstein Field Equation contribution.
Look at $G_{\hat{t}\hat{t}}$ (book says $G_{\hat{0}\hat{0}}$, but how are the two different?)
TODO exercise 14.13: calculating $G_{\hat{\alpha}\hat{\beta}}$. Results are:
${G^{\hat{t}}}_{\hat{t}} = {G^{\hat{r}}}_{\hat{r}} = -(F + 2 \bar{F})$
${G^{\hat{\theta}}}_{\hat{\theta}} = {G^{\hat{\phi}}}_{\hat{\phi}} = -(E + \bar{E} + \bar{F})$
For $E = -e^{-2 \Lambda} (\Phi_{,rr} + (\Phi_{,r})^2 - \Phi_{,r} \Lambda_{,r})$
$\bar{E} = -\frac{1}{r} \Phi_{,r} e^{-2\Lambda}$
$F = \frac{1}{r^2} (1 - e^{-2\Lambda})$
$\bar{F} = \frac{1}{r} \Lambda_{,r} e^{-2\Lambda}$
...and the basis is now orthonormalized, so ${G^\hat{t}}_{\hat{t}} = \eta^{\hat{t}\hat{t}} G_{\hat{t}\hat{t}} = -G_{\hat{t}\hat{t}}$
So $G_{\hat{t}\hat{t}} = F + 2 \bar{F}$
$ =
\frac{1}{r^2} (1 - e^{-2\Lambda})
+ 2 \frac{1}{r} \Lambda_{,r} e^{-2\Lambda}$
$ =
r^{-2}
- r^{-2} e^{-2\Lambda}
+ r^{-1} 2 \Lambda_{,r} e^{-2\Lambda}$
$ =
r^{-2}
- r^{-2} e^{-2\Lambda}
- r^{-1} \frac{d}{dr} (e^{-2\Lambda})$
$ = r^{-2} (
1 - e^{-2\Lambda}
- r \frac{d}{dr} (e^{-2\Lambda})
)$
$ = r^{-2} \frac{d}{dr} (r (1 - e^{-2\Lambda}))$
Likewise $G_{\hat{t}\hat{t}} = 8 \pi T_{\hat{t}\hat{t}} = 8 \pi \rho$
So $8 \pi \rho = r^{-2} \frac{d}{dr} (r (1 - e^{-2\Lambda}))$
Let $m = m(r)$ be the mass contained within some radius.
Let $2 m = r (1 - e^{-2 \Lambda})$
so $e^{2\Lambda} = (1 - \frac{2 m}{r})^{-1}$ (eqn 23.18)
Now $G_{\hat{t}\hat{t}} = r^{-2} \frac{d}{dr} (2m) = \frac{2}{r^2} \frac{dm}{dr} = 8 \pi \rho$
$\frac{dm}{dr} = 4 \pi r^2 \rho$
Integrate:
$m(r) = \int_{r=0}^{r=r} 4 \pi r^2 \rho dr + m(0)$ (eqn 23.19)
$ = \frac{4}{3} \pi r^3 \rho + m(0)$
where $m(0) = 0$ for flat geometry, and $m(0) = $ otherwise for a singularity at the origin.
Applied to the spherical Schwarzschild metric:
$ds^2 = (1 - \frac{2 m}{r})^{-1} dr^2 + r^2 (d\theta^2 + (sin \theta)^2 d\phi^2)$
$= \frac{r}{r - 2m} dr^2 + r^2 (d\theta^2 + (sin \theta)^2 d\phi^2)$
For $r \approx 0$ and $m(0) \ne 0$:
$ds^2 \approx -\frac{r}{2m(0)} dr^2 + r^2 (d\theta^2 + (sin \theta)^2 d\phi^2)$
For $r \approx 0$ and $m(0) = 0$:
$= \frac{r}{r - 2 \frac{4}{3} \pi r^3 \rho} dr^2 + r^2 (d\theta^2 + (sin \theta)^2 d\phi^2)$
$= \frac{1}{1 - \frac{8}{3} \pi r^2 \rho} dr^2 + r^2 (d\theta^2 + (sin \theta)^2 d\phi^2)$
$\approx dr^2 + r^2 (d\theta^2 + (sin \theta)^2 d\phi^2)$ (eqn 23.20)
Let $m(0) = 0$, so $m(r) = \int_{r=0}^{r=r} 4 \pi r^2 \rho dr$ (box 23.1 eqn 1)
For a Newtonian star, $m(r)$ is just mass, however for a relativistic star we get...
$m(r) = m_0(r) + U(r) + \Omega(r)$ (box 23.1 eqn 3
For $m_0(r) =$ rest mass-energy,
$U(r) =$ internal energy,
$\Omega(r) =$ gravitational energy.
Next is something about splitting density $\rho$ into $\mu_0 n + (\rho - \mu_0 n)$
Where $\mu_0 =$ the average rest-mass of the baryonic species present,
And $n =$ the number of particles.
And $\rho - \mu_0 n$, the rest of the non-baryonic density, is due to internal thermal energy, compressional energy, etc.
Next is the proper volume of a shell thickness $dr$:
$dV = 4 \pi r^2 (e^\Lambda dr) = 4 \pi r^2 (1 - \frac{2 m}{r})^{-1/2} dr$
Total rest mass inside radius r:
$m_0(r) = \int_{r=0}^{r=r} \mu_0 n dV$
$m_0(r) = \int_{r=0}^{r=r} 4 \pi r^2 (1 - \frac{2 m}{r})^{-1/2} \mu_0 n dr$ (box 23.1 eqn 5)
Total energy inside radius r:
$U(r) = \int_{r=0}^{r=r} (\rho - \mu_0 n) dV$
$U(r) = \int_{r=0}^{r=r} 4 \pi r^2 (1 - \frac{2 m}{r})^{-1/2} (\rho - \mu_0 n) dr$ (box 23.1 eqn 6)
Gravitational energy inside radius r:
By box 23.1 eqn 3, $m = m_0 + U + \Omega$ means $\Omega = m - m_0 - U$, we get
$\Omega(r) = \int_{r=0}^{r=r} (4 \pi \rho r^2 - 4 \pi \rho r^2 (1 - \frac{2 m}{r})^{-1/2}) dr$
$\Omega(r) = -\int_{r=0}^{r=r} 4 \pi \rho r^2 ((1 - \frac{2 m}{r})^{-1/2} - 1) dr$ (box 23.1 eqn 7)
For the near-Newtonian limit where $\frac{m}{r} \approx 1$, therefore $(1 - \frac{2 m}{r})^{-1/2} \approx 1$, we find
$\Omega(r) \approx -\int_{r=0}^{r=r} \frac{\rho m}{r} 4 \pi r^2 dr$
$\Omega(r) \approx -\int_{r=0}^{r=r} 4 \pi r m \rho dr$
Next look at the $G_{\hat{r}\hat{r}}$ components of the EFE:
$G_{\hat{r}\hat{r}} = \eta_{\hat{r}\hat{r}} {G^\hat{r}}_\hat{r} = {G^\hat{r}}_\hat{r} = -(F + 2 \bar{F})$
$G_{\hat{r}\hat{r}} = -\frac{1}{r^2} (1 - e^{-2\Lambda}) - 2 \frac{1}{r} \Lambda_{,r} e^{-2 \Lambda}$
$G_{\hat{r}\hat{r}} = -r^{-2} + r^{-2} e^{-2\Lambda} + 2 r^{-1} e^{-2 \Lambda} (-\Lambda_{,r})$
The book now replaces $-\frac{d\Lambda}{dr}$ with $\frac{d\Phi}{dr}$ ... why is that? E's have $\Phi$'s, F's have $\Lambda$'s, and $G_{\hat{r}\hat{r}}$ has F's not E's ...
$G_{\hat{r}\hat{r}} = -r^{-2} + r^{-2} e^{-2\Lambda} + 2 r^{-1} e^{-2 \Lambda} \frac{d\Phi}{dr}$ (eqn above 23.21)
By the EFE, $G_{\hat{r}\hat{r}} = 8 \pi T_{\hat{r}\hat{r}} = 8 \pi p$ (eqn above 23.21)
Substitute:
$-r^{-2} + r^{-2} e^{-2\Lambda} + 2 r^{-1} e^{-2 \Lambda} \frac{d\Phi}{dr} = 8 \pi p$
Next we replace the Newtonian approximation again.
Substitute eqn 23.18, $e^{-2\Lambda} = 1 - \frac{2 m}{r}$
$-r^{-2} + r^{-2} (1 - \frac{2 m}{r}) + 2 r^{-1} (1 - \frac{2 m}{r}) \frac{d\Phi}{dr} = 8 \pi p$
$-\frac{2 m}{r} + (2 r - 4 m) \frac{d\Phi}{dr} = 8 \pi r^2 p$
$\frac{d\Phi}{dr} = \frac{ 4 \pi r^2 p + m/r }{ r - 2 m }$
$\frac{d\Phi}{dr} = \frac{ m + 4 \pi p r^3 }{ r (r - 2 m) }$ (eqn 23.21)
(This is similar to the Newtonian limit: $\frac{d\Phi}{dr} = m / r^2$ - eqn 23.21N)
Combine with eqn 23.17 $(\rho + p) \Phi_{,r} = -p_{,r}$ to get
(why do we go from partials to total derivatives?)
$\frac{dp}{dr} = -\frac{ (\rho + p)(m + 4 \pi r^3 p) }{ r (r - 2 m) }$ (eqn 23.22)
...which is the Oppenheimer-Volkoff equation of hydrostatic equilibrium.
(The Newtonian equivalent is $\frac{dp}{dr} = -\rho m / r^2$ - eqn 23.22N)
Next look at proper radial distance:
TODO better where proper radial distance comes from.
$\frac{dp}{d (proper radial distance)} = \frac{1}{e^\Lambda} \frac{dp}{dr}$
Substitute eqn 23.18 and 23.22 to get:
$\frac{dp}{d (proper radial distance)} = \frac{1}{e^\Lambda} \frac{dp}{dr}$
$\frac{dp}{d (proper radial distance)} = -\frac{ (\rho + p)(m + 4 \pi r^3 p) }{ r^2 \sqrt{1 - 2 m / r} }$ (eqn 23.23)
(The Newton equivalent is $\frac{dp}{d(proper radial distance)} = \frac{dp}{dr} = -\rho m / r^2$ - eqn 23.23N)
exercise 23.3:
${T^{\alpha\beta}}_{;\beta} = 0$
${T^{\alpha\beta}}_{,\beta} + {\Gamma^\alpha}_{\gamma\beta} T^{\gamma\beta} + {\Gamma^\beta}_{\gamma\beta} T^{\alpha\gamma} = 0$
Recap: stress energy components:
$T^{tt} = \rho e^{-2\Phi}$
$T^{rr} = p e^{-2\Lambda}$
$T^{\theta\theta} = \frac{p}{r^2}$
$T^{\phi\phi} = \frac{p}{r^2 (sin(\theta))^2}$
Connection coefficients:
${\Gamma^t}_{tr} = {\Gamma^t}_{rt} = \Phi_{,r}$
${\Gamma^r}_{tt} = -\Phi_{,r} e^{2(\Phi - \Lambda)}$
${\Gamma^r}_{rr} = \Lambda_{,r}$
${\Gamma^r}_{\theta\theta} = -r e^{-2\Lambda}$
${\Gamma^r}_{\phi\phi} = -r (sin \theta)^2 e^{-2 \Lambda}$
${\Gamma^\theta}_{\theta r} = {\Gamma^\theta}_{r\theta} = \frac{1}{r}$
${\Gamma^\phi}_{\phi r} = {\Gamma^\phi}_{r\phi} = \frac{1}{r}$
${\Gamma^\phi}_{\phi\theta} = {\Gamma^\phi}_{\theta\phi} = cot \theta$
Since $T^{\alpha\beta}$ is diagonal, for fixed $\alpha$:
${T^{\alpha\alpha}}_{,\alpha}
+ \underset{\beta}{\Sigma} ({\Gamma^\alpha}_{\beta\beta} T^{\beta\beta})
+ (\underset{\beta}{\Sigma} {\Gamma^\beta}_{\alpha\beta}) T^{\alpha\alpha}
= 0$
For $\alpha = t$:
${T^{tt}}_{,t} = 0$
$(\rho e^{-2\Phi})_{,t} = 0$
$(\rho_{,t} - 2 \rho \Phi_{,t}) e^{-2\Phi} = 0$
$\rho_{,t} = 2 \rho \Phi_{,t}$
For $\alpha = r$:
${T^{rr}}_{,r}
+ {\Gamma^r}_{tt} T^{tt}
+ {\Gamma^r}_{\theta\theta} T^{\theta\theta}
+ {\Gamma^r}_{\phi\phi} T^{\phi\phi}
+ {\Gamma^t}_{rt} T^{rr}
+ {\Gamma^\theta}_{r\theta} T^{rr}
+ {\Gamma^\phi}_{r\phi} T^{rr}
= 0$
$(p e^{-2\Lambda})_{,r}
- \Phi_{,r} e^{2(\Phi - \Lambda)} \rho e^{-2\Phi}
- r e^{-2\Lambda} \frac{p}{r^2}
- r (sin \theta)^2 e^{-2 \Lambda} \frac{p}{r^2 (sin(\theta))^2}
+ (\Phi_{,r}
+ \frac{1}{r}
+ \frac{1}{r}
) p e^{-2\Lambda}
= 0$
$p_{,r} e^{-2\Lambda}
- 2 p e^{-2\Lambda} \Lambda_{,r}
+ (p - \rho) \Phi_{,r} e^{-2\Lambda}
= 0$
$(p - \rho) \Phi_{,r} - 2 p \Lambda_{,r} = -p_{,r}$
For $\alpha = \theta$:
${T^{\theta\theta}}_{,\theta}
+ {\Gamma^\phi}_{\theta\phi} T^{\theta\theta}
= 0$
$( \frac{p}{r^2} )_{,\theta} + cot \theta \frac{p}{r^2} = 0$
$\frac{ p_{,\theta} r^2 - 2 p r r_{,\theta} }{ r^4 } + cot \theta \frac{p}{r^2} = 0$
$p_{,\theta} = -p cot \theta$
For $\alpha = \phi$:
${T^{\phi\phi}}_{,\phi} = 0$
$(\frac{p}{r^2 (sin(\theta))^2})_{,\phi} = 0$
$p_{,\phi} = 0$ for $\theta \ne 0 + k \pi$
TODO explain this in terms of the question.
exercise 23.4:
Based on the chapter 14 results:
$G_{\hat{t}\hat{t}} = F + 2 \bar{F}$
$G_{\hat{r}\hat{r}} = -(F + 2 \bar{F})$
$G_{\hat\theta\hat\theta} = -(E + \bar{E} + \bar{F})$
$G_{\hat\phi\hat\phi} = -(E + \bar{E} + \bar{F})$
For $E = -e^{-2 \Lambda} (\Phi_{,rr} + (\Phi_{,r})^2 - \Phi_{,r} \Lambda_{,r})$
$\bar{E} = -\frac{1}{r} \Phi_{,r} e^{-2\Lambda}$
$F = \frac{1}{r^2} (1 - e^{-2\Lambda})$
$\bar{F} = \frac{1}{r} \Lambda_{,r} e^{-2\Lambda}$
So $F + 2 \bar{F} $
$= r^{-2}
- r^{-2} e^{-2\Lambda}
+ r^{-1} 2 \Lambda_{,r} e^{-2\Lambda}
$
And $E + \bar{E} + \bar{F}$
$ =
- e^{-2 \Lambda} (\Phi_{,rr} + (\Phi_{,r})^2 - \Phi_{,r} \Lambda_{,r})
- \frac{1}{r} \Phi_{,r} e^{-2\Lambda}
+ \frac{1}{r} \Lambda_{,r} e^{-2\Lambda}
$
$ =
(
- \Phi_{,rr}
- (\Phi_{,r})^2
+ \Phi_{,r} \Lambda_{,r}
- \frac{1}{r} \Phi_{,r}
+ \frac{1}{r} \Lambda_{,r}
) e^{-2\Lambda}
$
So
$G_{\hat{t}\hat{t}} =
r^{-2}
- r^{-2} e^{-2\Lambda}
+ r^{-1} 2 \Lambda_{,r} e^{-2\Lambda}
$
$G_{\hat{r}\hat{r}} = -(
r^{-2}
- r^{-2} e^{-2\Lambda}
+ r^{-1} 2 \Lambda_{,r} e^{-2\Lambda}
)$
$G_{\hat\theta\hat\theta} = (
\Phi_{,rr}
+ (\Phi_{,r})^2
- \Phi_{,r} \Lambda_{,r}
+ \frac{1}{r} \Phi_{,r}
- \frac{1}{r} \Lambda_{,r}
) e^{-2\Lambda}$
$G_{\hat\phi\hat\phi} = (
\Phi_{,rr}
+ (\Phi_{,r})^2
- \Phi_{,r} \Lambda_{,r}
+ \frac{1}{r} \Phi_{,r}
- \frac{1}{r} \Lambda_{,r}
) e^{-2\Lambda}$
So $G_{tt} = G_{\hat{\mu}\hat{\nu}} {\omega^\hat{\mu}}_t {\omega^\hat{\nu}}_t$
$ = G_{\hat{t}\hat{t}} ({\omega^\hat{t}}_t)^2$
$ =
(
r^{-2}
- r^{-2} e^{-2\Lambda}
+ r^{-1} 2 \Lambda_{,r} e^{-2\Lambda}
) (e^\Phi)^2$
$ =
r^{-2} e^{2\Phi}
- r^{-2} e^{2\Phi-2\Lambda}
+ r^{-1} 2 \Lambda_{,r} e^{2\Phi-2\Lambda}
$
$G_{rr} = G_{\hat\mu\hat\nu} {\omega^\hat\mu}_r {\omega^\hat\nu}_r$
$ = G_{\hat{r}\hat{r}} ({\omega^\hat{r}}_r)^2$
$ = -(
r^{-2}
- r^{-2} e^{-2\Lambda}
+ r^{-1} 2 \Lambda_{,r} e^{-2\Lambda}
) (
e^\Lambda
)^2$
$ =
- r^{-2} e^{2\Lambda}
+ r^{-2}
- r^{-1} 2 \Lambda_{,r}
$
$G_{\theta\theta} = G_{\hat\mu\hat\nu} {\omega^\hat\mu}_\theta {\omega^\hat\nu}_\theta$
$= G_{\hat\theta\hat\theta} ({\omega^\hat\theta}_\theta)^2$
$= (
\Phi_{,rr}
+ (\Phi_{,r})^2
- \Phi_{,r} \Lambda_{,r}
+ \frac{1}{r} \Phi_{,r}
- \frac{1}{r} \Lambda_{,r}
) e^{-2\Lambda} (r)^2$
$= (
r \Phi_{,rr}
+ r (\Phi_{,r})^2
- r \Phi_{,r} \Lambda_{,r}
+ \Phi_{,r}
- \Lambda_{,r}
) e^{-2\Lambda} r$
$G_{\phi\phi} = G_{\hat\mu\hat\nu} {\omega^\hat\mu}_\phi {\omega^\hat\nu}_\phi$
$= G_{\hat\phi\hat\phi} ({\omega^\hat\phi}_\phi)^2$
$ = (
\Phi_{,rr}
+ (\Phi_{,r})^2
- \Phi_{,r} \Lambda_{,r}
+ \frac{1}{r} \Phi_{,r}
- \frac{1}{r} \Lambda_{,r}
) e^{-2\Lambda} (r sin\theta)^2 $
$ = (
r \Phi_{,rr}
+ r (\Phi_{,r})^2
- r \Phi_{,r} \Lambda_{,r}
+ \Phi_{,r}
- \Lambda_{,r}
) e^{-2\Lambda} r sin(\theta)^2 $
Doing the same thing by definition of $G_{\alpha\beta} = R_{\alpha\beta} - \frac{1}{2} g_{\alpha\beta} R$
and $R_{\alpha\beta} = {R^\mu}_{\alpha\mu\beta} =
{\Gamma^\mu}_{\alpha\beta,\mu}
- {\Gamma^\mu}_{\alpha\mu,\beta}
+ {\Gamma^\mu}_{\nu\mu} {\Gamma^\nu}_{\alpha\beta}
- {\Gamma^\mu}_{\nu\beta} {\Gamma^\nu}_{\alpha\mu}
$:
Connections:
${\Gamma^t}_{tr} = {\Gamma^t}_{rt} = \Phi_{,r}$
${\Gamma^r}_{tt} = -\Phi_{,r} e^{2(\Phi - \Lambda)}$
${\Gamma^r}_{rr} = \Lambda_{,r}$
${\Gamma^r}_{\theta\theta} = -r e^{-2\Lambda}$
${\Gamma^r}_{\phi\phi} = -r (sin \theta)^2 e^{-2 \Lambda}$
${\Gamma^\theta}_{\theta r} = {\Gamma^\theta}_{r\theta} = \frac{1}{r}$
${\Gamma^\phi}_{\phi r} = {\Gamma^\phi}_{r\phi} = \frac{1}{r}$
${\Gamma^\phi}_{\phi\theta} = {\Gamma^\phi}_{\theta\phi} = cot \theta$
Connection derivatives:
${\Gamma^t}_{tr,r} = {\Gamma^t}_{rt,r} = \Phi_{,rr}$
${\Gamma^r}_{tt,r} = (-\Phi_{,rr} - \Phi_{,r} (2 \Phi_{,r} - 2 \Lambda_{,r})) e^{2\Phi - 2\Lambda}$
${\Gamma^r}_{rr,r} = \Lambda_{,rr}$
${\Gamma^r}_{\theta\theta,r} = (2 r \Lambda_{,r} - 1) e^{-2\Lambda}$
${\Gamma^r}_{\phi\phi,\theta} = -2 r sin \theta cos \theta e^{-2 \Lambda}$
${\Gamma^r}_{\phi\phi,r} = (sin \theta)^2 (2 r \Lambda_{,r} - 1) e^{-2\Lambda}$
${\Gamma^\theta}_{\theta r,r} = {\Gamma^\theta}_{r\theta,r} = -\frac{1}{r^2}$
${\Gamma^\phi}_{\phi r,r} = {\Gamma^\phi}_{r\phi,r} = -\frac{1}{r^2}$
${\Gamma^\phi}_{\phi\theta,\theta} = {\Gamma^\phi}_{\theta\phi,\theta} = -\frac{1}{(sin\theta)^2}$
Ricci curvature:
$R_{tt} = {\Gamma^\mu}_{tt,\mu}
- {\Gamma^\mu}_{t\mu,t}
+ {\Gamma^\mu}_{\nu\mu} {\Gamma^\nu}_{tt}
- {\Gamma^\mu}_{\nu t} {\Gamma^\nu}_{t\mu}
$
$= {\Gamma^r}_{tt,r}
+ (
{\Gamma^t}_{rt}
+ {\Gamma^r}_{rr}
+ {\Gamma^\theta}_{r\theta}
+ {\Gamma^\phi}_{r\phi}
) {\Gamma^r}_{tt}
- {\Gamma^t}_{rt} {\Gamma^r}_{tt}
- {\Gamma^r}_{tt} {\Gamma^t}_{tr}
$
$= {\Gamma^r}_{tt,r}
+ (
-{\Gamma^t}_{rt}
+ {\Gamma^r}_{rr}
+ {\Gamma^\theta}_{r\theta}
+ {\Gamma^\phi}_{r\phi}
) {\Gamma^r}_{tt}
$
$= (-\Phi_{,rr} - \Phi_{,r} (2 \Phi_{,r} - 2 \Lambda_{,r})) e^{2\Phi - 2\Lambda}
- (
-\Phi_{,r}
+ \Lambda_{,r}
+ \frac{1}{r}
+ \frac{1}{r}
) \Phi_{,r} e^{2\Phi - 2\Lambda}
$
$= (
- \Phi_{,rr}
- (\Phi_{,r})^2
+ (\Lambda_{,r} - 2 \frac{1}{r}) \Phi_{,r}
) e^{2\Phi - 2\Lambda}
$
$R_{tr} =
{\Gamma^\mu}_{tr,\mu}
- {\Gamma^\mu}_{t\mu,r}
+ {\Gamma^\mu}_{\nu\mu} {\Gamma^\nu}_{tr}
- {\Gamma^\mu}_{\nu r} {\Gamma^\nu}_{t\mu}
$
$ = 0$
$R_{t\theta} =
{\Gamma^\mu}_{t\theta,\mu}
- {\Gamma^\mu}_{t\mu,\theta}
+ {\Gamma^\mu}_{\nu\mu} {\Gamma^\nu}_{t\theta}
- {\Gamma^\mu}_{\nu\theta} {\Gamma^\nu}_{t\mu}
$
$ = 0$
$R_{t\phi} =
{\Gamma^\mu}_{t\phi,\mu}
- {\Gamma^\mu}_{t\mu,\phi}
+ {\Gamma^\mu}_{\nu\mu} {\Gamma^\nu}_{t\phi}
- {\Gamma^\mu}_{\nu\phi} {\Gamma^\nu}_{t\mu}
$
$= 0$
$R_{rr} = {\Gamma^\mu}_{rr,\mu}
- {\Gamma^\mu}_{r\mu,r}
+ {\Gamma^\mu}_{\nu\mu} {\Gamma^\nu}_{rr}
- {\Gamma^\mu}_{\nu r} {\Gamma^\nu}_{r\mu}
$
$ =
{\Gamma^r}_{rr,r}
- {\Gamma^t}_{rt,r}
- {\Gamma^r}_{rr,r}
- {\Gamma^\theta}_{r\theta,r}
- {\Gamma^\phi}_{r\phi,r}
+ (
{\Gamma^t}_{rt}
+ {\Gamma^r}_{rr}
+ {\Gamma^\theta}_{r\theta}
+ {\Gamma^\phi}_{r\phi}
) {\Gamma^r}_{rr}
- {\Gamma^t}_{tr} {\Gamma^t}_{rt}
- {\Gamma^r}_{rr} {\Gamma^r}_{rr}
- {\Gamma^\theta}_{\theta r} {\Gamma^\theta}_{r\theta}
- {\Gamma^\phi}_{\phi r} {\Gamma^\phi}_{r\phi}
$
$ =
- {\Gamma^t}_{rt,r}
- {\Gamma^\theta}_{r\theta,r}
- {\Gamma^\phi}_{r\phi,r}
+ (
{\Gamma^t}_{tr}
+ {\Gamma^\theta}_{\theta r}
+ {\Gamma^\phi}_{\phi r}
) {\Gamma^r}_{rr}
- ({\Gamma^t}_{tr})^2
- ({\Gamma^\theta}_{\theta r})^2
- ({\Gamma^\phi}_{\phi r})^2
$
$ =
- \Phi_{,rr}
+ \frac{1}{r^2}
+ \frac{1}{r^2}
+ (
\Phi_{,r}
+ \frac{1}{r}
+ \frac{1}{r}
) \Lambda_{,r}
- (\Phi_{,r})^2
- (\frac{1}{r})^2
- (\frac{1}{r})^2
$
$ =
- \Phi_{,rr}
- (\Phi_{,r})^2
+ \Phi_{,r} \Lambda_{,r}
+ 2 \frac{1}{r} \Lambda_{,r}
$
$R_{r\theta} =
{\Gamma^\mu}_{r\theta,\mu}
- {\Gamma^\mu}_{r\mu,\theta}
+ {\Gamma^\mu}_{\nu\mu} {\Gamma^\nu}_{r\theta}
- {\Gamma^\mu}_{\nu\theta} {\Gamma^\nu}_{r\mu}
$
$ =
+ {\Gamma^\phi}_{\theta\phi} {\Gamma^\theta}_{r\theta}
- {\Gamma^\phi}_{\phi\theta} {\Gamma^\phi}_{\phi r}
$
$ =
cot\theta \frac{1}{r}
- cot\theta \frac{1}{r}
$
$ = 0$
$R_{r\phi} =
{\Gamma^\mu}_{r\phi,\mu}
- {\Gamma^\mu}_{r\mu,\phi}
+ {\Gamma^\mu}_{\nu\mu} {\Gamma^\nu}_{r\phi}
- {\Gamma^\mu}_{\nu\phi} {\Gamma^\nu}_{r\mu}
$
$ = 0$
$R_{\theta\theta} =
{\Gamma^\mu}_{\theta\theta,\mu}
- {\Gamma^\mu}_{\theta\mu,\theta}
+ {\Gamma^\mu}_{\nu\mu} {\Gamma^\nu}_{\theta\theta}
- {\Gamma^\mu}_{\nu\theta} {\Gamma^\nu}_{\theta\mu}
$
$ =
{\Gamma^r}_{\theta\theta,r}
- {\Gamma^\phi}_{\theta\phi,\theta}
+ (
{\Gamma^t}_{rt}
+ {\Gamma^r}_{rr}
+ {\Gamma^\phi}_{r\phi}
) {\Gamma^r}_{\theta\theta}
- ({\Gamma^\phi}_{\phi\theta})^2
$
$ =
(2 r \Lambda_{,r} - 1) e^{-2\Lambda}
+ \frac{1}{(sin\theta)^2}
+ (
\Phi_{,r}
+ \Lambda_{,r}
+ \frac{1}{r}
) (
-r e^{-2\Lambda}
)
- (cot \theta)^2
$
$ =
(
- 1
+ r \Lambda_{,r}
- r \Phi_{,r}
- e^{2\Lambda}
) e^{-2\Lambda}
$
$R_{\theta\phi} =
{\Gamma^\mu}_{\theta\phi,\mu}
- {\Gamma^\mu}_{\theta\mu,\phi}
+ {\Gamma^\mu}_{\nu\mu} {\Gamma^\nu}_{\theta\phi}
- {\Gamma^\mu}_{\nu\phi} {\Gamma^\nu}_{\theta\mu}
$
$ = 0$
$R_{\phi\phi} =
{\Gamma^\mu}_{\phi\phi,\mu}
- {\Gamma^\mu}_{\phi\mu,\phi}
+ {\Gamma^\mu}_{\nu\mu} {\Gamma^\nu}_{\phi\phi}
- {\Gamma^\mu}_{\nu\phi} {\Gamma^\nu}_{\phi\mu}
$
$ =
{\Gamma^r}_{\phi\phi,r}
+ (
{\Gamma^t}_{rt}
+ {\Gamma^r}_{rr}
+ {\Gamma^\theta}_{r\theta}
){\Gamma^r}_{\phi\phi}
$
$ =
(sin \theta)^2 (2 r \Lambda_{,r} - 1) e^{-2\Lambda}
+ (
\Phi_{,r}
+ \Lambda_{,r}
+ \frac{1}{r}
) (
-r (sin \theta)^2 e^{-2 \Lambda}
)
$ ... should be an extra term in here
$ =
(
- 1
+ r \Lambda_{,r}
- r \Phi_{,r}
- e^{2\Lambda}
) e^{-2 \Lambda} (sin \theta)^2
$
Next, Gaussian curvature:
$R = R_{\alpha\beta} g^{\alpha\beta}$
...
$ = e^{-2\Lambda} (
- \frac{1}{r^2} 2
+ 2 \frac{1}{r^2} e^{2\Lambda}
- 4 \frac{1}{r} \Phi_{,r}
+ 4 \frac{1}{r} \Lambda_{,r}
- 2 \Phi_{,rr}
+ 2 \Phi_{,r} \Lambda_{,r}
- 2 (\Phi_{,r})^2
)$
Next, the Einstein curvature tensor:
(TODO work this out -- I'm taking this from my TOV symmath worksheet)
$G_{tt} = {\frac{-{({{{2 - {{\Lambda_{,{{r}}}} {\Phi_{,{{r}}}} {{r}^{2}}}} - {4 {\Lambda_{,{{r}}}} r}} + {{\Phi_{,{{r}}{{r}}}} {{r}^{2}}} + {2 {\Phi_{,{{r}}}} r} + {{{{{({\Phi_{,{{r}}}})}}^{2}} {{r}^{2}}} - {2 {{e}^{{({2 \Lambda})}}}}}})}}{{{e}^{{({2 {({\Lambda - \Phi})}})}}} {{r}^{2}}}} $
$G_{rr} = {\frac{{2 - {2 {\Lambda_{,{{r}}}} r}} + {{{\Phi_{,{{r}}{{r}}}} {{r}^{2}}} - {{\Phi_{,{{r}}}} {\Lambda_{,{{r}}}} {{r}^{2}}}} + {{{{({\Phi_{,{{r}}}})}}^{2}} {{r}^{2}}} + {{4 {\Phi_{,{{r}}}} r} - {2 {{e}^{{({2 \Lambda})}}}}}}{{r}^{2}}} $
$G_{\theta\theta} = {\frac{{1 - {3 {\Lambda_{,{{r}}}} r}} + {{2 {\Phi_{,{{r}}{{r}}}} {{r}^{2}}} - {2 {\Phi_{,{{r}}}} {\Lambda_{,{{r}}}} {{r}^{2}}}} + {2 {{{({\Phi_{,{{r}}}})}}^{2}} {{r}^{2}}} + {{3 {\Phi_{,{{r}}}} r} - {{e}^{{({2 \Lambda})}}}}}{{e}^{{({2 \Lambda})}}}} $
$G_{\phi\phi} = {\frac{{{{sin\left(\theta\right)}}^{2}} {({{1 - {3 {\Lambda_{,{{r}}}} r}} + {{2 {\Phi_{,{{r}}{{r}}}} {{r}^{2}}} - {2 {\Phi_{,{{r}}}} {\Lambda_{,{{r}}}} {{r}^{2}}}} + {2 {{{({\Phi_{,{{r}}}})}}^{2}} {{r}^{2}}} + {{3 {\Phi_{,{{r}}}} r} - {{e}^{{({2 \Lambda})}}}}})}}{{e}^{{({2 \Lambda})}}}} $
Convert to tetrad basis:
$G_{\hat{t}\hat{t}} = e^{-2\Phi} \cdot {\frac{-{({{{2 - {{\Lambda_{,{{r}}}} {\Phi_{,{{r}}}} {{r}^{2}}}} - {4 {\Lambda_{,{{r}}}} r}} + {{\Phi_{,{{r}}{{r}}}} {{r}^{2}}} + {2 {\Phi_{,{{r}}}} r} + {{{{{({\Phi_{,{{r}}}})}}^{2}} {{r}^{2}}} - {2 {{e}^{{({2 \Lambda})}}}}}})}}{{{e}^{{({2 {({\Lambda - \Phi})}})}}} {{r}^{2}}}} $
$ = {\frac{-{({{{2 - {{\Lambda_{,{{r}}}} {\Phi_{,{{r}}}} {{r}^{2}}}} - {4 {\Lambda_{,{{r}}}} r}} + {{\Phi_{,{{r}}{{r}}}} {{r}^{2}}} + {2 {\Phi_{,{{r}}}} r} + {{{{{({\Phi_{,{{r}}}})}}^{2}} {{r}^{2}}} - {2 {{e}^{{({2 \Lambda})}}}}}})}}{{{e}^{{({2 {{\Lambda}}})}}} {{r}^{2}}}} $
$G_{\hat{r}\hat{r}} = e^{-2\Lambda} \cdot {\frac{{2 - {2 {\Lambda_{,{{r}}}} r}} + {{{\Phi_{,{{r}}{{r}}}} {{r}^{2}}} - {{\Phi_{,{{r}}}} {\Lambda_{,{{r}}}} {{r}^{2}}}} + {{{{({\Phi_{,{{r}}}})}}^{2}} {{r}^{2}}} + {{4 {\Phi_{,{{r}}}} r} - {2 {{e}^{{({2 \Lambda})}}}}}}{{r}^{2}}} $
$= {\frac{{2 - {2 {\Lambda_{,{{r}}}} r}} + {{{\Phi_{,{{r}}{{r}}}} {{r}^{2}}} - {{\Phi_{,{{r}}}} {\Lambda_{,{{r}}}} {{r}^{2}}}} + {{{{({\Phi_{,{{r}}}})}}^{2}} {{r}^{2}}} + {{4 {\Phi_{,{{r}}}} r} - {2 {{e}^{{({2 \Lambda})}}}}}}{ r^2 e^{2\Lambda} }} $
$G_{\hat\theta\hat\theta} = (r)^{-2} \cdot {\frac{{1 - {3 {\Lambda_{,{{r}}}} r}} + {{2 {\Phi_{,{{r}}{{r}}}} {{r}^{2}}} - {2 {\Phi_{,{{r}}}} {\Lambda_{,{{r}}}} {{r}^{2}}}} + {2 {{{({\Phi_{,{{r}}}})}}^{2}} {{r}^{2}}} + {{3 {\Phi_{,{{r}}}} r} - {{e}^{{({2 \Lambda})}}}}}{{e}^{{({2 \Lambda})}}}} $
$ = {\frac{{1 - {3 {\Lambda_{,{{r}}}} r}} + {{2 {\Phi_{,{{r}}{{r}}}} {{r}^{2}}} - {2 {\Phi_{,{{r}}}} {\Lambda_{,{{r}}}} {{r}^{2}}}} + {2 {{{({\Phi_{,{{r}}}})}}^{2}} {{r}^{2}}} + {{3 {\Phi_{,{{r}}}} r} - {{e}^{{({2 \Lambda})}}}}}{ r^2 {e}^{{({2 \Lambda})}}}} $
$G_{\hat\phi\hat\phi} = (r sin\theta)^{-2} \cdot {\frac{{{{sin\left(\theta\right)}}^{2}} {({{1 - {3 {\Lambda_{,{{r}}}} r}} + {{2 {\Phi_{,{{r}}{{r}}}} {{r}^{2}}} - {2 {\Phi_{,{{r}}}} {\Lambda_{,{{r}}}} {{r}^{2}}}} + {2 {{{({\Phi_{,{{r}}}})}}^{2}} {{r}^{2}}} + {{3 {\Phi_{,{{r}}}} r} - {{e}^{{({2 \Lambda})}}}}})}}{{e}^{{({2 \Lambda})}}}} $
$ = {\frac{ {{1 - {3 {\Lambda_{,{{r}}}} r}} + {{2 {\Phi_{,{{r}}{{r}}}} {{r}^{2}}} - {2 {\Phi_{,{{r}}}} {\Lambda_{,{{r}}}} {{r}^{2}}}} + {2 {{{({\Phi_{,{{r}}}})}}^{2}} {{r}^{2}}} + {{3 {\Phi_{,{{r}}}} r} - {{e}^{{({2 \Lambda})}}}}} }{ r^2 {e}^{{({2 \Lambda})}}}} $
exercise 23.5:
Show the total number of baryons in a star is equal to
should be the integral from 0 to R of the volume element times the density ...
The lapse is $e^\Phi$, the spatial volume is $e^\Lambda$.
The surface of a sphere at radius $r$ is $4 \pi r^2$.
Therefore the number of baryons within a volume of space on the surface of the sphere is $4 \pi r^2 \cdot n \cdot e^\Lambda$
Therefore the total within the volume of the sphere is $A = \int_0^R 4 \pi r^2 n e^\Lambda dr$
exercise 23.6:
Buoyant forces = gravitational forces.
Gravitational acceleration = $f^\mu = \rho a^\mu = -\rho {\Gamma^\mu}_{\alpha\beta} u^\alpha u^\beta$
An object at rest: $u = e_\hat{t} = e^{-\Phi} e_t$, so $u^t = e^{-\Phi}$
Substitute to find: $f^\mu = -\rho {\Gamma^\mu}_{tt} e^{-2\Phi}$
$f^\mu = -\rho {\Gamma^\mu}_{tt} (e^{-\Phi})^2$
$f^\mu = -\rho \cdot -\Phi_{,r} e^{2(\Phi - \Lambda)} \cdot e^{-2\Phi}$
$f^\mu = \rho \Phi_{,r} e^{-2\Lambda}$
...which looks like the Newtonian gravitational force, with the spatial volume squared.
...also, where do the $p$ and $m$ come into play?
...also, should I replace $e^{-\Lambda}$ with $V$?
TODO redo this one.
exercise 23.7:
TODO
section 23.6 "External Gravitational Field"
Outside a star:
$p = 0, \rho = 0$, only $\Lambda$ and $\Phi$ influence spacetime.
$m(r) = M$ for $r > R$
Start with $\frac{d\Phi}{dr} = \frac{ m + 4 \pi p r^3 }{ r (r - 2 m) }$ (eqn 23.21)
Substitute $p = 0$ and $m = M$
$\frac{d\Phi}{dr} = \frac{M}{ r (r - 2 M) }$
$\int d\Phi = \int \frac{M}{ r (r - 2 M) } dr$
$\Phi = C + \frac{1}{2} log(r - 2 M) - \frac{1}{2} log(r)$
$\Phi = C + \frac{1}{2} log(1 - 2 M / r)$
Let $\underset{r \rightarrow \infty}{\lim} \Phi(r) = 0$ eqn 23.10
$0 = \underset{r \rightarrow \infty}{\lim} \Phi(r)
= \underset{r \rightarrow \infty}{\lim} (C + \frac{1}{2} log(1 - 2 M / r))
= C + \frac{1}{2} log(1 - \underset{r \rightarrow \infty}{\lim} 2 M / r)
= C + \frac{1}{2} log(1)
= C
$
Therefore $C = 0$ and
$\Phi = \frac{1}{2} log(1 - 2 M / r)$ (eqn 23.26)
Determine $\Lambda$ outside of the star:
Start with equation 23.18:
$e^{2\Lambda} = (1 - \frac{2 m}{r})^{-1}$
Substitute our values:
$e^{2\Lambda} = (1 - \frac{2 M}{r})^{-1}$
Substitute $\Phi$ and $\Lambda$ into the TOV metric eqn 23.3 to find:
$ds^2 = -e^{2 \frac{1}{2} log(1 - 2 M / r) } dt^2 + e^{2\Lambda} dr^2 + R^2 d\Omega^2$
$ds^2 = -e^{2 \frac{1}{2} log(1 - 2 M / r) } dt^2 + (1 - \frac{2 M}{r})^{-1} dr^2 + R^2 d\Omega^2$
$ds^2 = -(1 - 2 M / r) dt^2 + \frac{1}{1 - \frac{2 M}{r}} dr^2 + R^2 d\Omega^2$ (eqn 23.27)
...Tada, Schwarzschild geometry.
TODO show how the Newtonian limit outside of stars is $\Phi = -M /r$ for $r > R, r >> 2 M$ (23.26N)
section 23.7 "How To Construct a Stellar Model"
Line element: $ds^2 = -(1 - 2 M / r) dt^2 + \frac{1}{1 - \frac{2 M}{r}} dr^2 + r^2 d\Omega^2$ (eqn 23.27 again, as eqn 23.27')
Mass equation: $m(r) = \int_{r=0}^{r=r} 4 \pi r^2 \rho dr$ (this is eqn 23.19 with $m(0) = 0$ -- eqn 23.28a)
OV Equation of Hydrostatic Equilibrium: $\frac{dp}{dr} = -\frac{ (\rho + p)(m + 4 \pi r^3 p) }{ r (r - 2 m) }$ (eqn 23.22 again, as eqn 23.28b)
with $p(r=0) = p_c = $ central pressure.
Equations of state: $p = p(n)$ (eqn 23.28c) and $\rho = \rho(n)$ (eqn 23.28d)
Source equation for $\Phi$: $\frac{d\Phi}{dr} = \frac{ m + 4 \pi p r^3 }{ r (r - 2 m) }$ (eqn 23.21 again, as eqn 23.28e)
with the boundary condition $\Phi = \frac{1}{2} log(1 - 2 M / r)$ from eqn 23.26
Next add more boundary conditions:
$\Phi(r = 0) = \Phi_0$
$p(r = 0) = p_c$
$m(r = 0) = 0$
Exercise 23.8 comes before box 23.2, but depends on content in the box, so I'm putting it after the box.
Box 23.2 comes before section 23.8, but depends on content in section 23.8, so I'm putting it after section 23.8.
section 23.8
Start with eqn 23.27':
$ds^2 = -(1 - 2 M / r) dt^2 + \frac{1}{1 - \frac{2 M}{r}} dr^2 + R^2 d\Omega^2$
Fix $d\theta = 0, d\phi = 0, dt = 0$ and let $l = s$ be the length:
$dl^2 = \frac{1}{1 - \frac{2 M}{r}} dr^2$
$dr = \pm \sqrt{1 - \frac{2 M}{r} } dl$ (eqn 23.29)
Same deal, but fix $\theta = \frac{\pi}{2}$, so $d\theta = 0$, and keep $\phi$:
$ds^2 = \frac{1}{1 - 2 m / r} dr^2 + r^2 d\phi^2$ (eqn 23.30)
Now we consider the embedding as a surface, $z$:
$ds^2 = dz^2 + dr^2 + r^2 d\phi^2$ (eqn 23.31)
Let $z = z(r)$, so $dz = \frac{dz}{dr} dr$:
$ds^2 = (\frac{dz}{dr} dr)^2 + dr^2 + r^2 d\phi^2$ (eqn 23.31)
$ds^2 = (1 + (\frac{dz}{dr})^2) dr^2 + r^2 d\phi^2$ (eqn 23.32)
Match this with the Schwarzschild line element $ds^2 = \frac{1}{1 - \frac{2 m(r)}{r}} dr^2 + r^2 d\phi^2$
to find $1 + (\frac{dz}{dr})^2 = \frac{1}{1 - \frac{2 m(r)}{r}}$ (eqn 23.33)
$dz = \sqrt{\frac{1}{1 - \frac{2 m(r)}{r}} - 1} dr$
$dz = \frac{1}{\sqrt{\frac{r}{2 m(r)} - 1}} dr$
$z(r) = \int_0^r \frac{1}{\sqrt{\frac{r}{2 m(r)} - 1}} dr$ inside a star (eqn 23.34a)
Outside a star, replace $m(r) = M$:
$z(r) = \int_0^r \frac{1}{\sqrt{\frac{r}{2 M} - 1}} dr$
$z(r) = \sqrt{ 8 M (r - 2 M) } + C$ (eqn 23.34b)
TODO explain: recall $m(r)$ varies by $\frac{4}{3} \pi \rho_c r^3$ near $r = 0$
TODO explain: therefore let $a = (3 / 8 \pi \rho_c)^{1/2}$ ... is that $\frac{3}{8} \pi \rho_c$ or is it $\frac{3}{8 \pi \rho_c}$ ?
TODO explain: $(a - z(r))^2 + r^2 = a^2$ for $r << a = (3 / 8 \pi \rho_c)^{1/2}$ (eqn 23.34c)
box 23.2
Assume the star has uniform density: $\rho = \rho_0$
mass equation:
$m = \left\{ \matrix{
r < R : & \frac{4}{3} \pi \rho_0 r^3 \\
r > R : & M = \frac{4}{3} \pi \rho_0 R^3
}\right\}$ box 23.2 eqn 2
$\frac{d (proper distance)}{dr} = e^\Lambda = (1 - 2 m(r) / r)^{-1/2}$ box 23.2 eqn 3
Lift equation:
TODO ... from section 23.8 ... should go before this.
$z(r) = \left\{\matrix{
r \le R : & \sqrt{\frac{ R^3 }{ 2 M }} (1 - \sqrt{1 - \frac{2 M r^2}{R^3}}) \\
r \ge R : & \sqrt{ \frac{ R^3 }{ 2 M } } (1 - \sqrt{ 1 - \frac{2 M}{R} } )
+ \sqrt{ 8 M (r - 2 M ) }
- \sqrt{ 8 M (R - 2 M ) }
}\right\}$ box 23.2 eqn 4
TODO this is supposed to come from box 23.2 eqn 2 and $\frac{dp}{dr}$ in eqn 23.28b:
Start with: $\frac{dp}{dr} = -\frac{ (\rho + p)(m + 4 \pi r^3 p) }{ r (r - 2 m) }$ from eqn 23.28b
Substitute box 23.2 eqn 2: $m(r) = \frac{4}{3} \pi \rho_0 r^3$ to find:
$\frac{dp}{dr} = -\frac{ (\rho + p)(\frac{4}{3} \pi \rho_0 r^3 + 4 \pi r^3 p) }{ r (r - 2 \frac{4}{3} \pi \rho_0 r^3) }$
TODO...
$p = \rho_0 \left(
\frac{
\sqrt{ 1 - \frac{2 M r^2 }{ R^3 } }
- \sqrt{ 1 - \frac{2 M}{ R } }
}{
3 \sqrt{ 1 - \frac{2 M }{ R } }
- \sqrt{ 1 - \frac{2 M r^2 }{ R^3 } }
}
\right)$ for $r < R$ -- box 23.2 eqn 5
TODO: show why...
$\frac{d(proper time)}{dt} = e^\Phi = \left\{\matrix{
r < R : & \frac{3}{2} (1 - \frac{2 M}{R})^{1/2} - \frac{1}{2} (1 - \frac{2 M r^2}{R^3})^{1/2} \\
r > R : & (1 - 2 M / r)^{1/2}
}\right\}$ -- box 23.2 eqn 6
TODO:
$p_c = \rho_0 \left(
\frac{
1 - (1 - 2 M / R)^{1/2}
}{
3 (1 - 2 M / R)^{1/2} - 1
}
\right)$ -- box 23.2 eqn 7
TODO:
$2 M / R = (8 \pi / 3) \rho_0 R^2$ -- box 23.2 eqn 8
TODO exercise 23.8: