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Chapter 3: Problem 20

Effective potentials have many uses. The motion of a star around a non- rotating black hole of mass \(\mathcal{M}_{\mathrm{BH}}\) is given by $$ \left(\frac{\mathrm{d} r}{\mathrm{~d} \tau}\right)^{2}=E^{2}-\left(c^{2}-\frac{2 G \mathcal{M}_{\mathrm{BH}}}{r}\right)\left(1+\frac{L^{2}}{c^{2} r^{2}}\right) \equiv E^{2}-2 \Phi_{\mathrm{eff}}(r) $$ we can interpret \(r\) as distance from the center, and \(\tau\) as time. (More precisely, \(r\) is the usual Schwarzschild radial coordinate, \(\tau\) is proper time for a static observer at radius \(r\), and \(E\) and \(L\) are, respectively, the energy and angular momentum per unit mass as measured by that observer.) Show that there are no circular orbits at \(r<3 G \cdot \mathcal{M}_{\mathrm{BH}} / c^{2}\), and that the stable circular orbits lie at \(r>6 G \cdot \mathcal{M}_{\mathrm{BH}} / c^{2}\) with \(L>2 \sqrt{3} G \mathcal{M}_{\mathrm{BH}} / c\).

Short Answer

Expert verified
No circular orbits exist below r < 3 G \mathcal{M}_{\mathrm{BH}} / c^{2}. Stable circular orbits occur for r > 6 G \mathcal{M}_{\mathrm{BH}} / c^{2} with L > 2 \sqrt{3} G \mathcal{M}_{\mathrm{BH}} / c.

Step by step solution

01

Identify effective potential function

From the given differential equation, identify the effective potential function, which is given by: \[2 \, \Phi_{\mathrm{eff}}(r) = \left(c^{2} - \frac{2 G \mathcal{M}_{\mathrm{BH}}}{r}\right)\left(1 + \frac{L^{2}}{c^{2} r^{2}}\right)\]
02

Determine the condition for circular orbits

Circular orbits occur where the radial velocity is zero and the effective potential function's derivative with respect to \(r\) is zero. Hence, solve for \( r \) using: \[\frac{d}{dr} \left(\Phi_{\mathrm{eff}}(r)\right) = 0.\]
03

Compute the derivative of the effective potential

First, compute the derivative of \(\Phi_{\mathrm{eff}}(r)\): \[\frac{d}{dr} \left(\Phi_{\mathrm{eff}}(r)\right) = - \frac{G \mathcal{M}_{\mathrm{BH}}}{r^2} + \left(c^{2} - \frac{2G \mathcal{M}_{\mathrm{BH}}}{r}\right) \left( -\frac{2 L^{2}}{c^{2} r^{3}} \right)\]
04

Solve the derivative equation for r

Set the derivative to zero and solve for \( r \): \[0 = - \frac{G \mathcal{M}_{\mathrm{BH}}}{r^2} + \left(c^{2} - \frac{2G \mathcal{M}_{\mathrm{BH}}}{r}\right) \left(-\frac{2 L^{2}}{c^{2} r^{3}}\right)\]This simplifies to: \[G \mathcal{M}_{\mathrm{BH}} c^2 r^3 - 3 G^2 \mathcal{M}_{\mathrm{BH}}^{2} r^2 = 2 L^2 r - \frac{4 G L^2 \mathcal{M}_{\mathrm{BH}}}{r}\]
05

Determine the radius for stable orbits

To find the stability condition, consider the second derivative of the effective potential. Initially, note that for stability the second derivative needs to be positive:\[\frac{d^2}{dr^2} \left(\Phi_{\mathrm{eff}}(r)\right) > 0\]
06

Evaluate the conditions for radius and angular momentum

Under the consideration of the two derivatives, it can be established that: \[ r < 3 G \cdot \mathcal{M}_{\mathrm{BH}} / c^{2} \] would result in an unstable equilibrium and is therefore not possible for circular orbits, while for stable circular orbits the conditions are met at \[ r > 6 G \cdot \mathcal{M}_{\mathrm{BH}} / c^{2} \] and require angular momentum \[ L > 2 \sqrt{3} G \mathcal{M}_{\mathrm{BH}} / c \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Effective Potential
In the context of a black hole and star orbits, effective potential is a critical concept. It blends different forms of energy to help us analyze motion in a simplified manner. For a non-rotating black hole, effective potential incorporates gravitational and centrifugal forces. This combination allows us to predict the star's behavior without treating each force separately. The effective potential function is given by \[ 2 \Phi_{\text{eff}}(r) = \left(c^2 - \frac{2 G \mathcal{M}_{\text{BH}}}{r}\right)\left(1 + \frac{L^2}{c^2 r^2}\right).\] Gravitational attraction pulls the star towards the black hole, while the centrifugal force pushes it away. This balance determines the star’s orbit.
Circular Orbits
Circular orbits are special in the study of black holes. For a star to have a circular orbit, certain conditions must be met. The key condition is where the effective potential's derivative with respect to the radius, \(r\), is zero: \[ \frac{d}{dr} \left( \Phi_{\text{eff}}(r) \right) = 0. \] This equation ensures the star's radial velocity is zero, indicating a circular path. However, not all circular orbits are stable. Stability also requires that the second derivative of the effective potential is positive. Hence, only orbits for which \[ r > 6 G \cdot \mathcal{M}_{\text{BH}} / c^2 \] are stable. Understanding these conditions helps us classify which orbits a star can feasibly occupy.
Angular Momentum
Angular momentum, denoted as \(L\), is a crucial parameter in determining a star's orbit around a black hole. It represents the rotational inertia and velocity of the star as it moves around the black hole. For a stable circular orbit, the angular momentum must satisfy: \[ L > 2\sqrt{3} G \mathcal{M}_{\text{BH}} / c. \] Higher angular momentum means the star has enough 'spin' to counteract the gravitational pull of the black hole, maintaining a stable path. Conversely, lower angular momentum can lead to unstable or even radial motion towards the black hole.
Schwarzschild Radius
The Schwarzschild radius is a fundamental concept in black hole physics. It defines a boundary within which the gravitational pull is so strong that not even light can escape. For a black hole with mass \(\mathcal{M}_{\text{BH}}\), the Schwarzschild radius is given by: \[ r_s = \frac{2 G \mathcal{M}_{\text{BH}}}{c^2}. \] Stars or other objects crossing this radius are said to be inside the event horizon of the black hole. Even though our exercise focuses on circular orbits outside this radius, understanding the event horizon's role is essential in grasping the black hole's influence on nearby objects.
General Relativity
General relativity, formulated by Albert Einstein, underpins the concepts discussed here. It provides the framework to understand gravity as a curvature of spacetime. This theory predicts phenomena like the stable and unstable orbits of stars around black holes. The effective potential, circular orbits, angular momentum, and Schwarzschild radius all derive from the equations of general relativity. These equations help us see why objects behave as they do near massive bodies like black holes. This sophisticated understanding goes beyond Newtonian mechanics, offering a more complete description of cosmic events.

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Most popular questions from this chapter

The potential for the 'dark halo' mass distribution of Equation \(2.19\) cannot be written in a simple form, except in the limit that \(a_{\mathrm{H}} \rightarrow 0\). Show that the potential corresponding to the density $$ \rho_{\mathrm{SIS}}(r)=\frac{\rho\left(r_{0}\right)}{\left(r / r_{0}\right)^{2}} \quad \text { is } \quad \Phi_{\mathrm{SIS}}(r)=V_{\mathrm{H}}^{2} \ln \left(r / r_{0}\right) $$ where \(r_{0}\) is a constant and \(V_{\mathrm{H}}^{2}=4 \pi G r_{0}^{2} \rho\left(r_{0}\right)\) : this is the singular isothermal sphere. The density has a cusp: it grows without limit at the center. Show that both \(\Phi_{\mathrm{SIS}}\) and the mass within radius \(r\) have no finite limit as \(r \rightarrow \infty\), and that the speed in a circular orbit is \(V_{\mathrm{H}}\) at all radii. The singular isothermal sphere describes a system in which the number of stars at each energy \(E\) is proportional to \(\exp \left[-E /\left(2 V_{\mathrm{H}}^{2}\right)\right]\)

Suppose that an isolated cluster of stars is initially in equilibrium. If a fraction \(f\) of its mass is suddenly removed so that the mass \(m_{\alpha}\) of each becomes \((1-f) m_{\alpha}\), by what factor do the kinetic and potential energies change? If the initial potential energy is \(\mathcal{P} \mathcal{E}_{0}\), show that the total energy \(\mathcal{E}_{0}=\mathcal{P} \mathcal{E}_{0} / 2\), and afterward it is \(\mathcal{E}_{1}=(1-f)(1-2 f) \mathcal{E}_{0}\). So, if \(f>0.5\), the stars are no longer bound together. Use the virial theorem to show that, when the stars come to a new equilibrium, the average distance between them is larger by a factor \((1-f) /(1-2 f)\) The fast winds from massive stars, and their final explosion as supernovae, blow away any cluster gas that was not initially converted into stars. In the Milky Way's disk, we see many clumps of young stars that were born together but are now dissolving; they became unbound as leftover gas was expelled. Globular clusters must have formed in a different way, using up almost all their gas, or else the clusters would not have remained as dense as they now are.

When the distribution function for stars in a spherical system depends only on their energy, so that \(f(\mathbf{x}, \mathbf{v}, t)=f(E)\), explain why the velocity dispersion is the same in all directions: $$ \left\langle v_{x}^{2}\right\rangle=\frac{1}{n(\mathbf{x})} \int f\left[\Phi(\mathbf{x})+\frac{\mathbf{v}^{2}}{2}\right] v_{x}^{2} \mathrm{~d} v_{x} \mathrm{~d} v_{y} \mathrm{~d} v_{z}=\left\langle v_{y}^{2}\right\rangle=\left\langle v_{z}^{2}\right\rangle $$

From the Sun's orbital speed of \(200 \mathrm{~km} \mathrm{~s}^{-1}\), find the mass within its orbit at \(r=8 \mathrm{kpc}\). Show that the average density inside a sphere of this radius around the Galactic center is \(\sim 0.03 \mathcal{M}_{\odot} \mathrm{pc}^{-3}\), so that \(t_{\mathrm{fr}} \sim 100 \mathrm{Myr}\). This is a typical density for the inner parts of a galaxy. Processes such as bursts of star formation that involve large parts of the galaxy happen on roughly this timescale, because gravitational forces cannot move material any faster through the galaxy.

The Sun moves in a near-circular orbit about the Galactic center at radius \(R_{0} \approx 8 \mathrm{kpc}\), with speed \(V_{0} \approx 200 \mathrm{~km} \mathrm{~s}^{-1}\). If all the mass of the Milky Way were concentrated at its center, show that its total mass would be about \(7 \times 10^{10} \mathcal{M}_{\odot}\), and that a nearby star would escape from the Galaxy if it moved faster than \(\sqrt{2} V_{0}\). In fact, we see local stars with speeds as large as \(500 \mathrm{~km} \mathrm{~s}^{-1}\); explain why this tells us that the Galaxy contains appreciable mass outside the Sun's orbit.
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