Newtonian elliptical orbits

Newtonian elliptical orbits: sketch 1

Newtonian elliptical orbits: equation

• Work in polar coordinates $(r, \phi)$.
• Start with the orbit equation $$ \frac{\mathrm{d}^2 u}{\mathrm{d} \phi^2} + u = \frac{GM}{h^2}, $$ where:
  • $u(\phi)$ is $1/r(\phi)$ (makes equation cleaner).
  • $G$ is Newton's constant.
  • $M$ is the mass of the massive object.
  • $h = L/m$, angular momentum per unit mass.
• Angular momentum is conserved: motion in a plane.

Newtonian elliptical orbits: solution

• Equation $ \frac{\mathrm{d}^2 u}{\mathrm{d} \phi^2} + u = \frac{GM}{h^2} $ has solution $ u = \frac{GM}{h^2} ( 1 + e\cos \phi).$
• Eccentricity: for $0 \leq e < 1$, "how elliptical" the orbit is.
• Extrema:
  • Maximum $u$ (smallest $r$, closest approach): $$ u_\text{max} = \frac{GM}{h^2} ( 1+ e)$$
  • Minimum $u$ (largest $r$, greatest separation): $$u_\text{min} = \frac{GM}{h^2} (1 - e)$$

Newtonian elliptical orbits: sketch 2

What does it mean?

Precession of the perihelion

• Precession: a gradual change of orbital parameters
• Perihelion: closest approach of an object to the Sun

Previous observations

• In 1859, Urbain Le Verrier studied 14 transits of Mercury:
  • Reanalysed transits of the Sun between 1697 and 1848.
  • Mercury's perihelion advances by 565" per century.
  • But the motion of other planets only accounts for 527".
  
  
• Various explanations, none satisfactory:
  • Planet Vulcan (even closer to the Sun than Mercury).
  • Adjust Newton's inverse-square law to: $$ F = \frac{G M m}{r^{2.00000016}}.$$
  • Theories of gravity inspired by Maxwell's equations.

General relativity
to the rescue?

Q: Can we use the Schwarzschild metric for the Sun?

Which of these are conditions on the use of
the Schwarzschild metric? [True/False]

1. The massive object must have negligible angular momentum and electrical charge. True.
2. Schwarzschild metric is only valid for black holes. False.
3. The massive object must be spherical. True.

Does the Sun satisfy these conditions?

Massive particle, Schwarzschild metric

• Worldline $x^\mu(\tau)$ of particle, mass $m$ and energy $E$ in plane:
  $$ \left( \frac{\mathrm{d} r}{\mathrm{d} \tau} \right)^2 + \frac{h^2}{r^2} \left(1 - \frac{2GM}{c^2 r} \right) - \frac{2GM}{r} = c^2 \left( \frac{E^2}{m^2 c^4} -1 \right).$$
  $$ \left( \color{red}{\frac{h}{r^2} \frac{\mathrm{d} r}{\mathrm{d} \phi}} \right)^2 + \frac{h^2}{r^2} \left(1 - \frac{2GM}{c^2 r} \right) - \frac{2GM}{r} = c^2 \left( \frac{E^2}{m^2 c^4} -1 \right).$$
• Use the chain rule (remember $h = r^2 \frac{\mathrm{d} \phi}{\mathrm{d} \tau} $): $$\frac{\mathrm{d}r}{\mathrm{d}\tau} = \frac{\mathrm{d} r}{\mathrm{d} \phi} \frac{\mathrm{d} \phi}{\mathrm{d} \tau} = \frac{h}{r^2} \frac{\mathrm{d}r}{\mathrm{d} \phi} $$
• Let $u = 1/r$ as before, $\frac{\mathrm{d}u}{\mathrm{d}\phi} = \frac{1}{r^2} \frac{\mathrm{d}r}{\mathrm{d}\phi}$: $$ h^2 \left( \frac{\mathrm{d} u}{\mathrm{d} \phi} \right)^2 + h^2 u^2 \left ( 1 -\frac{2GM}{c^2} u \right) - 2GM u = c^2 \left(\frac{E^2}{m^2 c^4} - 1\right).$$

Tidying up the orbit equation

• Starting with the first-order nonlinear equation $$ h^2 \left( \frac{\mathrm{d} u}{\mathrm{d} \phi} \right)^2 + h^2 u^2 \left ( 1 -\frac{2GM}{c^2} u \right) - 2GM u = c^2 \left(\frac{E^2}{m^2 c^4} - 1\right).$$
• Differentiate both sides with respect to $\phi$: $$ 2 h^2 \frac{\mathrm{d}^2 u}{\mathrm{d}\phi^2} \frac{\mathrm{d} u}{\mathrm{d} \phi} + 2 h^2 u\frac{\mathrm{d}u}{\mathrm{d}\phi} - \frac{6 GM}{c^2} h^2 u^2 \frac{\mathrm{d}u}{\mathrm{d}\phi} - 2 GM \frac{\mathrm{d}u}{\mathrm{d}\phi} = 0 $$
• Factor out $2 h^2 \frac{\mathrm{d}u}{\mathrm{d}\phi}$ and rearrange:
  $$ \frac{\mathrm{d}^2 u}{\mathrm{d}\phi^2} + u - \frac{3 GM}{c^2} u^2 - \frac{GM}{h^2} = 0$$
  $$ \frac{\mathrm{d}^2 u}{\mathrm{d} \phi^2} + u = \frac{GM}{h^2} + \frac{3 GM}{c^2} u^2.$$

General-relativistic orbit equation

So GR adds an extra term to the equation of motion

For Mercury: $$\begin{align} \frac{GM}{h^2} & \approx 1.8 \times 10^{-11} \, \mathrm{m}^{-1} \\ \frac{3GM}{c^2} u^2 & \approx 2.1\times 10^{-18} \, \mathrm{m}^{-1} \quad \text{at perihelion; maximum $u$} \end{align}$$

Very small perturbation!

Consider a small perturbation $\Delta u$...

• Write the solution as $ u = u_0(\phi) + \color{red}{\Delta u(\phi)},$
  where $u_0(\phi)$ obeys the Newtonian equation of motion $$ \frac{\mathrm{d}^2 u_0}{\mathrm{d} \phi^2} + u_0 = \frac{GM}{h^2}.$$
• Substitute into the equation of motion:
  $$\frac{\mathrm{d}^2 u_0}{\mathrm{d} \phi^2} + \frac{\mathrm{d}^2 \Delta u}{\mathrm{d} \phi^2} + u_0 + \Delta u = \frac{GM}{h^2} \color{blue}{+ \frac{3 G M}{c^2}(u_0 + \Delta u)^2}$$
  $$\color{gray}{\frac{\mathrm{d}^2 u_0}{\mathrm{d} \phi^2}} + \frac{\mathrm{d}^2 \Delta u}{\mathrm{d} \phi^2} + \color{gray}{u_0} + \Delta u = \color{gray}{\frac{GM}{h^2}} \color{blue}{+ \frac{3 G M}{c^2}(u_0 + \Delta u)^2}$$
• Subtract Newtonian equation of motion for $u_0$: $$ \frac{\mathrm{d}^2 \Delta u}{\mathrm{d} \phi^2} + \Delta u = \color{blue}{\frac{3GM}{c^2} (u_0 + \Delta u)^2}. $$

... at first order in perturbation theory

• The evolution of a perturbation at first order is $$ \frac{\mathrm{d}^2 \Delta u}{\mathrm{d} \phi^2} + \Delta u = \color{red}{\frac{3GM}{c^2} u_0^2}, $$ (terms with $\Delta u$ on right hand side are at higher order); where $$ u_0(\phi) = \frac{GM}{h^2} (1 + e \cos \phi).$$
• Substituting in and defining $\alpha = 3 (GM)^2/(h^2 c^2) $: $$ \frac{\mathrm{d}^2 \Delta u(\phi)}{\mathrm{d} \phi^2} + \Delta u(\phi) = \overbrace{\frac{3 GM}{c^2} \frac{(GM)^2}{h^4}}^{GM \alpha /h^2} (1 + 2 e \cos \phi + e^2 \cos^2 \phi).$$

Particular solution

• The inhomogeneous ordinary differential equation $$ \frac{\mathrm{d}^2 \Delta u(\phi)}{\mathrm{d} \phi^2} + \Delta u(\phi) = \frac{GM}{h^2} \alpha (1 + 2 e \cos \phi + e^2 \cos^2 \phi).$$ has the particular solution
  $$ \Delta u(\phi) = \frac{GM}{h^2} \alpha \left[ 1 + e^2 \left(\frac{1}{2} - \frac{1}{6} \cos 2\phi \right) + e \phi \sin \phi \right] $$
  $$ \Delta u(\phi) = \frac{GM}{h^2}\alpha \left[ \color{gray}{1 + e^2 \left(\frac{1}{2} - \frac{1}{6} \cos 2\phi \right)} + e \phi \sin \phi \right] $$ of which only the last term grows over one period.
• Then $$ u(\phi) = u_0(\phi) + \Delta u(\phi) = \frac{GM}{h^2} \left[ 1+ e(\cos \phi + \alpha \phi \sin \phi) \right].$$

Perihelion shift

• Start from $ u(\phi) = \frac{GM}{h^2} \left[ 1+ e(\cos \phi + \alpha \phi \sin \phi) \right]$.
• With $\alpha$ small, we can use $ \cos \phi + \alpha \phi \sin \phi \approx \cos \left[ \phi (1 - \alpha) \right]$: $$ u \approx \frac{GM}{h^2} \left( 1 + e \cos \left[ \phi ( 1 -\alpha) \right] \right).$$
• The orbit returns to the same $u=1/r$ at an angle $$ \Delta \phi = \frac{2\pi}{1 - \alpha} - 2\pi \approx 2\pi \alpha = \frac{6 \pi (GM)^2 }{h^2 c^2}$$
• Mercury's orbit and mass give $\alpha = 3\frac{GM}{h^2 c^2} = 8.0\times 10^{-8}$.
• Mercury's orbital period is 88 days: 415 orbits/century.
• Cumulative $\Delta \phi$ is then: $415\times 2\pi \times 8.0\times 10^{-8} = 0.00021 \, \text{rad/century} = 43''/\text{century}$

Upshot: precession of the
perihelion of Mercury

• Observed value is $574 \pm 1 \text{"/century}$, of which
  • $532 \text{''/century}$ is due to other solar system objects.
  • $43 \text{"/century}$ is explained by the calculation here.
• GR explained the precession, it did not predict it.

Next time: predictions of GR

• Bending of light and the 1919 Eddington expedition.
• Gravitational waves.