# 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. 11 in Schutz
Why is there a gap between the disk and the neutron star?

## Methods in GR

• So we derived Einstein's field equations, when do we use them?
• Freely falling particles travel along geodesics
• If the metric tensor is independent of a degree of freedom, there is a conserved momentum associated with that degree of freedom
• Or simplified metrics

## Isometries and Killing Vectors

• In all metrics, one constant of the motion comes from the constant value of $\vec{U} \cdot \vec{U}$
• If the metric is independent of a coordinate (isometry) then there is a vector pointing along the direction called the "Killing vector".
• The existence of a Killing vector implies some quantity is conserved along a geodesic

## Dark Stars in Newtonian Gravity

• If escape velocity $$\frac{1}{2} v^2 = \frac{G M}{R}$$ exceeds the speed of light
• Set $v=c$, then $$R_S = \frac{2 G M}{c^2}$$ this is the Schwarzchild radius
• This classical result is identical to the radius of a black hole in GR
• For a 1 solar mass star, this radius is $\sim 2.96$ km
• Compute mean density for dark stars $$\bar{\rho} = \frac{3 M}{4 \pi R^3} = \frac{3 c^6}{32 \pi G^3 M^2}$$ and while for a $1~\mathrm{M}_{\odot}$ star this is $10^{18}~\mathrm{g}/\mathrm{cm}^3$, for $10^{9}~\mathrm{M}_{\odot}$, this is the density of water

## Orbits in GR

• Metric $$ds^2 = - \left( 1-\frac{2 M}{r} \right) dt^2 + \left( 1-\frac{2M}{r}\right)^{-1} dr^2 + r^2 d\Omega^2$$
• Treat particles (left) with and without mass (right) separately
• Metric is time independent, thus $-p_0$ is conserved. $$\tilde{E} \equiv -p_0/m \quad ; \quad E = -p_0$$
• Metric is $\phi$-independent (axial symmetry), therefore angular momentum is conserved $$\tilde{L} \equiv p_{\phi}/m \quad ; \quad L = p_{\phi}$$
• Construct the momentum, $p^{\mu}$ $$p^{0} = m \left( 1- \frac{2 M}{r} \right)^{-1} \tilde{E} \quad ; \quad p^{0} = \left( 1- \frac{2 M}{r} \right)^{-1} \tilde{E}$$ $$p^{r} = m \frac{dr}{d \tau} \quad ; \quad p^{r} = \frac{dr}{d \lambda}$$ $$p^{\phi} = m \frac{\tilde{L}}{r^2} \quad ; \quad p^{\phi} = \frac{d\phi}{d \lambda} = \frac{L}{r^2}$$

## Orbits in GR II

• Use $\vec{p} \cdot \vec{p} = -m^2$ and solve $$\left(\frac{dr}{d \tau}\right)^2 = \tilde{E}^2 - \left(1-\frac{2 M}{r} \right) \left( 1 + \frac{\tilde{L}^2}{r^2}\right) \quad ; \quad \left(\frac{dr}{d \lambda}\right)^2 = E^2 - \left(1-\frac{2 M}{r} \right) \frac{L^2}{r^2}$$
• Define RHS as $E^2-V^2$, where $V$ is an effective potential
• Where are stable/unstable orbits?

## Orbits in Schwarzchild Spacetime

• Scattering orbits are not parabolas as in Newtonian gravity
• "Plunging orbits" do not appear in Newtonian gravity
Guidry ch. 18

## Circular Orbits

• Differentiate the equation defining the orbit $$\frac{d^2 r}{d \tau^2} = - \frac{1}{2} \frac{d\tilde{V}^2(r)}{dr} \quad ; \quad \frac{dd^2 r}{d \lambda^2} = - \frac{1}{2} \frac{V^2(r)}{dr}$$
• Thus circular orbits happen only if $$\frac{d\tilde{V}^2(r)}{dr} =0 \quad ; \quad \frac{d V^2(r)}{dr} =0$$
• These are solved at $$r = \frac{\tilde{L}}{2 M} \left[ 1 \pm \left( 1 - \frac{12 M^2}{\tilde{L}^2} \right)^{1/2}\right] \quad ; \quad r = 3 M$$
• And there is a stable orbit for particles only if $\tilde{L}^2 = 12 M^2$, at which point $r_{\mathrm{MIN}} = 6 M$
• This is the innermost stable circular orbit

## Innermost stable circular orbits

• ISCOs are not limited to GR, but occur in Newtonian gravity for ellipsoidal stars (see Torok et al. (2014))
• Also modified by rotation, see e.g. gr-qc/0312070
• kHz QPOs in X-ray binaries: may be observing the frequency of particles traveling around the ISCO
• If $r_{\mathrm{ISCO}} < 9$ km, evidence for a compact object which is not a neutron star
• If $r_{\mathrm{ISCO}} = 12~\mathrm{km}$, then we know $r_{\mathrm{NS}} < 12$ km

## Stable Circular Orbits

• For a particle in a stable circular orbit, solve $$r = \frac{\tilde{L}}{2 M} \left[ 1 + \left( 1 - \frac{12 M^2}{\tilde{L}^2} \right)^{1/2}\right]$$ for $\tilde{L}^2$, giving $$\tilde{L}^2 = \frac{M r}{1-3 M/r}$$ and then using $\tilde{E}^2=\tilde{V}^2$ we know the energy $$\tilde{E}^2 = \left( 1-\frac{2 M}{r}\right)^2 \left( 1-\frac{3 M}{r}\right)^{-1}$$
• Compute angular velocity by first computing $$\frac{d \phi}{d \tau} = \frac{\tilde{L}}{r^2} \quad \mathrm{and} \quad \frac{dt}{d \tau} = \frac{\tilde{E}}{1-2 M/r}$$ and then taking the ratio $$\frac{dt}{d \phi} = \left(\frac{r^3}{M}\right)^{1/2}$$
• (Same as Newtonian expression)

## Approximate Newtonian Orbits

• Obtain $dr/d\phi$ $$\left(\frac{dr}{d \phi} \right)^2 = \left(\frac{r^4}{\tilde{L}^2}\right) \left[ \tilde{E}^2 - \left(1-\frac{2 M}{r}\right) \left(1 + \frac{\tilde{L}^2}{r^2}\right)\right]$$
• Define $u\equiv 1/r$ and then negelect terms of order $u^3$ or higher to obtain the Newtonian result $$\left(\frac{du}{d\phi}\right)^2 = \frac{\tilde{E}^2}{\tilde{L}^2}- \frac{\left(1-2 M u\right)}{\tilde{L}^2} -u^2$$
• And solve to obtain the equation for an ellipse $$\frac{1}{r} = \frac{M}{\tilde{L}^2} + A \cos \left( \phi + B \right)$$

• Go back to $$\left(\frac{dr}{d \phi} \right)^2 = \left(\frac{r^4}{\tilde{L}^2}\right) \left[ \tilde{E}^2 - \left(1-\frac{2 M}{r}\right) \left(1 + \frac{\tilde{L}^2}{r^2}\right)\right]$$ and use $y=u-M/\tilde{L}^2$, but instead of ignoring $u^3$ term, neglect the $y^3$ terms to get $$\left(\frac{dy}{d\phi}\right)^2 = \left(\frac{\tilde{E}^2+M^2/\tilde{L}^2-1} {\tilde{L}^2}\right) + \frac{2 M^4}{\tilde{L}^6} + \frac{6 M^3 y}{\tilde{L}^2} - y^2 \left( 1-\frac{6 M^2}{\tilde{L}^2}\right)$$
• This has a solution of the form $$y = y_0 + A \cos \left( k \phi + B \right)$$ where previously we had $k=1$

• The change in $\phi$ from one perihelion to the next is $$\Delta \phi = \frac{2 \pi}{k} \approx 2 \pi \left( 1+ \frac{3 M^2}{\tilde{L}^2}\right)$$ thus the perihelion advance is $$\Delta \phi = 6 \pi M^2/\tilde{L}^2$$
• Then using $\tilde{L} \approx M r$ we have $\Delta \phi \approx 6 \pi M/r$ or 43 arc seconds per century

## Mercury and Earth

Clemence (1947)
• Largest contribution from Venus and Jupiter
• Earth's perihelion precession is larger (but the GR part is smaller)

## Tests of GR

 Kramer et al. (2006) Larger periastron precession in double NS systems (16 degrees per year) This system, PSR J0737-3039A/B Double pulsar systems allow observation of GR corrections Shapiro delay: radio pulses delayed by curved space-time Tests of GR have always passed See also Ch. 19 in Guidry's book

## Group Work

• Examine Figs. 7 and 8 of arxiv.org/1205.1450. How are the constraints on $\alpha_0$ given in the abstract related to the limits given in the figures?
• What will happen to these figures as more observations are obtained?

## Gravitational Deflection of Light

• Remember that $d\phi/d\lambda = L/r^2$ and $$\left(\frac{dr}{d \lambda}\right)^2 = E^2 - \left(1-\frac{2 M}{r} \right) \frac{L^2}{r^2}$$ thus defining $b\equiv L/E$ we have $$\frac{d\phi}{dr} = \pm \frac{1}{r^2} \left[ \frac{1}{b^2} - \frac{1}{r^2} \left(1-\frac{2 M}{r}\right)\right]^{-1/2}$$
• The impact parameter $b$ is the minimum value of $r$ in the Newtonian theory
• Defining $u=1/r$ as before $$\frac{d \phi}{ d u} = \left( b^{-2} - u^2 + 2 M u^3\right)^{-1/2}$$
• Assume $M u \ll 1$ and define $y \equiv u(1-M u)$, then $$\frac{d \phi}{dy} \approx \frac{1 + 2 M y} {\left(b^{-2}-y^2\right)^{1/2}}$$

## Gravitational Deflection of Light II

• This gives $$\phi = \phi_0 + \frac{2 M}{b} + \mathrm{arcsin}(by) - 2 M \left(b^{-2}-y^2\right)^{1/2}$$
• The total deflection is $4 M/b$