# Physics 616

• Prof. Andrew W. Steiner
(or Andrew or "Dr. Steiner")
• Office hour: 103 South College, Thursday 11am
• Email: awsteiner@utk.edu
• Homework: Electronically as .pdf
• You may work with each other on the homework, but you must write the solution in your own words

Use down and up arrows to proceed to the next or previous slide.

## Outline

• Ch. 10 in Schutz

## Spherically symmetric spacetimes

• Flat space in spherical coordinates $$ds^2 = -dt^2 + dr^2 + r^2\left( d\theta^2 + \sin^2 \theta~d \phi^2\right) \equiv -dt^2 + dr^2 + r^2 d\Omega^2$$
• This metric has no intrinsic curvature (how would one verify this?)
• A spacetime is spherically symmetric if every point of spacetime is on a two-sphere
• Schutz shows that a spherically symmetric metric must be of the form $$ds^2 = g_{00} dt^2 + 2 g_{0r} dr dt + g_{rr} dr^2 + r^2 d\Omega^2$$ where the metric functions depend only on $r$ and $t$.

## Static spherically symmetric spacetimes

• All metric functions independent of $t$ and time-reversal symmetry
• The latter condition rules out $g_{0r} \neq 0$ (and also rules out rotation)
• Thus the metric is $$ds^2 = g_{00}(r) dt^2 + g_{rr}(r) dr^2 + r^2 d\Omega^2$$ but it is useful to replace the metric functions with new definitions $$ds^2 = -e^{2 \Phi(r)} dt^2 + e^{2 \Lambda(r)} dr^2 + r^2 d\Omega^2$$
• Again, the notation for this metric is not standard, and different authors choose different notations
• Ensure system is isolated, thus $$\lim_{r\rightarrow \infty} \Phi(r) = \lim_{r\rightarrow \infty} \Lambda(r) = 0$$

## Using our metric

• The metric function $\Lambda(r)$ tells us how to measure radial distances $$\ell_{12} = \int_{r_1}^{r_2}~e^{\Lambda(r)}~dr$$
• Since the metric is independent of $t$, we know that geodesics have constant $p_0 \equiv -E$
• An inertial observer at rest has a four velocity with $U^{i}=0$, but also $\vec{U} \cdot \vec{U} = -1$, thus $U^{0}=e^{-\Phi(r)}$.
• Our observer measures an energy of $$E^{*} = - \vec{U} \cdot \vec{p} = e^{-\Phi}E$$
• (We will find $e^{-\Phi} > 1$.) This extra energy is the kinetic energy from falling into the gravitational field (or the extra energy required to escape it)

## Gravitational Potential and Photons

• A photon is emitted at radius $r_1$ and its energy is $$E=h \nu_{\mathrm{emit}} \exp \left[ \Phi( r_1) \right]$$
• The observer measures a photon with the same energy, but a different frequency $$\nu_{\mathrm{obs}} = \nu_{\mathrm{emit}} \exp \left[ \Phi( r_1) \right]$$
• Since $\exp \left[ \Phi( r_1) \right] < 1$, $\nu_{\mathrm{obs}} < \nu_{\mathrm{emit}}$ and the photon is redshifted
• We define the redshift $z$ as $$z \equiv e^{-\Phi(r)} - 1$$

## Stress Energy Tensor and EOS

• The normalization $$\vec{U} \cdot \vec{U} = -1$$ implies $$U^{0} = e^{-\Phi} \quad \mathrm{and} \quad U_{0} = -e^{\Phi}$$ Thus, the stress energy tensor is $$\begin{eqnarray} T_{00} &=& \rho e^{2 \Phi} \\ T_{rr} &=& p e^{2 \Lambda} \\ T_{\theta \theta} &=& r^2 p \\ T_{\phi \phi} &=& r^2 p \sin^2 \theta \end{eqnarray}$$
• We showed earlier that one can write $$p=p(\rho,S)$$

## Energy-momentum Conservation and Einstein's Field Equations I

• Energy-momentum conservation $${T^{\alpha \beta}}_{;\beta} = 0$$ implies $$\left(\rho + P\right) \frac{d\Phi}{dr} = - \frac{dp}{dr}$$
• Using the $(0,0)$ component of EFEs gives $$\frac{dm}{dr} = 4 \pi r^2 \rho$$ where $$m \equiv \frac{r}{2} \left( 1 - e^{-2\Lambda} \right)$$ The quantity $m$ is just the enclosed (gravitational) mass, as we will verify in a moment

## Einstein's Field Equations II and the TOV equation

• The $(r,r)$ component of the EFEs gives $$\frac{d \Phi}{dr} = \frac{m+4 \pi r^3 p} {r \left(r-2 m\right)}$$
• This immediately gives the Tolman-Oppenheimer-Volkov equation $$\frac{dp}{dr} = -\left(\rho + P\right) \left[\frac{m+4 \pi r^3 p} {r \left(r-2 m\right)}\right]$$ or (putting in $G$ but leaving $c=1$) $$\frac{dp}{dr} = -\frac{G \rho m}{r^2} \left(1 + \frac{P}{\rho}\right) \left(1+\frac{4 \pi r^3 p}{m}\right) \left(1-\frac{2 G m}{r}\right)^{-1}$$ (the latter form makes the connection to the Newtonian limit clear)

## Boundary Conditions Outside

• Outside, we have $\rho=p=0$, thus $$\frac{dm}{dr}=0$$ and $$\frac{d\Phi}{dr}=\frac{m}{r(r-2m)}$$ and these can be solved by $$m(r) = M$$ and $$e^{2 \Phi} = 1-\frac{2 M}{r}$$
• Outside the object, the metric is the "Schwarzchild metric" $$ds^2 = - \left( 1-\frac{2 M}{r} \right) dt^2 + \left(1 - \frac{2 M}{r} \right)^{-1} dr^2 + r^2 d\Omega^2$$
• This is the only spherically symmetric and asymptotically flat metric

## Birkhoff's theorem

• Any spherically symmetric solution of the vacuum field equations must be static and asymptotically flat

## More Boundary Conditions and Mass

• Inside the star, $m(r=0)=0$, outside $m(r=R)=M$ and $p(r=R)=0$ (but not necessarily $\rho(r=R)=0$).
• Given any specified $p_c$, one obtains a solution of the TOV equations
• This one-parameter family gives a relation between mass and radius
• The total gravitational mass is $$M = \int_0^{R} 4 \pi r^2 \rho~dr$$
• This result is deceptive because the proper volume element in GR in spherical coordinates is $$|g|^{1/2}~d^3x = e^{\Lambda} r^2 \sin \theta~dr~d\theta~d\phi$$ but the factors of $e^{\Lambda}$ and $e^{\Phi}$ have canceled
• As a demonstration of this, the total number of baryons in a star is $$N_B = \int_0^{R} 4 \pi r^2 n_B(r) \left(1-\frac{2 m}{r}\right)^{-1/2} ~dr$$

## Group Work

• Complete 10.7, 10.9a and 10.14a in Schutz.