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Chapter 3: Problem 3

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 \(\exp \left[-E /\left(2 V_{\mathrm{H}}^{2}\right)\right]\)

Short Answer

Expert verified
The potential is \( \Phi_{\mathrm{SIS}}(r)=V_{H}^{2} \ln \left(r / r_{0}\right) \), it diverges as \( r \to \infty \), and the circular speed is \( V_{H} \).

Step by step solution

01

Understand the Given Density Function

The density function is given as \( \rho_{\text{SIS}}(r) = \frac{\rho(r_0)}{\left( r / r_0 \right)^2} \)where \(r_0\) is a constant.
02

Derive the Potential from the Density Function

We need to find the potential \( \Phi_{\text{SIS}}(r) \) that corresponds to \( \rho_{\text{SIS}}(r) \). Using Poisson's equation in spherical coordinates: \[ abla^2 \Phi = 4 \pi G \rho \]For a spherically symmetric system, this becomes: \[ \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{d\Phi}{dr} \right) = 4 \pi G \rho \]Substitute \( \rho(r) = \frac{\rho(r_0)}{\left( r / r_0 \right)^2} \) into Poisson's equation.
03

Solve the Differential Equation

Simplify and solve the differential equation: \[ \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{d\Phi}{dr} \right) = 4 \pi G \frac{\rho(r_0)}{\left( r / r_0 \right)^2} \]This implies:\[ \frac{d}{dr} \left( r^2 \frac{d\Phi}{dr} \right) = 4 \pi G r_0^2 \rho(r_0) \]Integrate both sides with respect to \(r\):\[ r^2 \frac{d\Phi}{dr} = 4 \pi G r_0^2 \rho(r_0) r + C \]Divide both sides by \(r^2\) and integrate again:\[ \frac{d\Phi}{dr} = \frac{4 \pi G r_0^2 \rho(r_0) r}{r^2} + \frac{C}{r^2} \right] \ = 4 \pi G r_0^2 \rho(r_0) \ln(r) + C \right] \]Finally, integrate with respect to \(r\):\[ \Phi(r) = V_H^2 \ln\left( \frac{r}{r_0} \right) + \text{constant} \right] \]where \( V_H^2 = 4 \pi G r_0^2 \rho(r_0) \).
04

Show the Limits of the Potential and Mass

Show that \( \Phi_{\mathrm{SIS}} \) diverges as \( r \to \infty \):\[ \Phi_{\mathrm{SIS}}(r) = V_H^2 \ln\left( \frac{r}{r_0} \right) \]As \( r \to \infty \), the logarithm term \( \ln(\frac{r}{r_0}) \to \infty \), thus \( \Phi_{\mathrm{SIS}}(r) \to \infty \).Similarly, the mass within radius \( r \) also has no finite limit because the density function \( \rho_{\mathrm{SIS}} (r) \sim r^{-2} \).
05

Calculate the Circular Orbital Speed

The speed in a circular orbit is given by:\[ v_c^2 = r \frac{d\Phi}{dr} \]Using our derived potential:\[ \Phi(r) = V_H^2 \ln\left( \frac{r}{r_0} \right) \]\[ \frac{d\Phi}{dr} = \frac{V_H^2}{r} \]Then:\[ v_c^2 = r \frac{V_H^2}{r} = V_H^2 \]Thus, the circular speed \( v_c = V_H \) at all radii.

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Key Concepts

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

Dark Halo Mass Distribution
The 'dark halo' refers to a region around galaxies where dark matter dominates the mass distribution. This concept is crucial for understanding the overall gravitational influence of a galaxy. The mass within the dark halo does not emit, absorb, or reflect light, making it invisible in electromagnetic observations.
In this exercise, the mass distribution of the dark halo follows a specific density function, which must be analyzed to derive the potential function and other related properties. The density function describes how mass is distributed within the halo, directly influencing gravitational forces.
Density Function
The density function provided in this exercise is \$ \rho_{\text{SIS}}(r) = \frac{\rho(r_0)}{\left( r / r_0 \right)^2} \$ where \$$ r_0 \$$ is a constant. This specific form represents a singular isothermal sphere (SIS) - a simplified model to describe mass distribution in galaxies.
The term 'isothermal' suggests that the system is at a constant temperature. The singularity at the center means the density becomes infinitely large as you approach the center. This feature is called a 'cusp'. The SIS model is useful because it is simple and provides insights into the dynamics of dark matter-dominated regions.
  • \$ r_0 \$ - A characteristic radius
  • \$ \rho(r_0) \$ - Density at radius \$ r_0 \$
Potential Function
Using Poisson's equation, we derive the potential function corresponding to the given density function. Poisson's equation in spherical coordinates is given by:
\[ abla^2 \Phi = 4 \pi G \rho \]
For spherical symmetry, this simplifies to:
\[ \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{d\Phi}{dr} \right) = 4 \pi G \rho \]
Substituting the density function \$ \rho(r) = \frac{\rho(r_0)}{\left( r / r_0 \right)^2} \$ leads to a differential equation that we solve to find the potential:
\[ \Phi(r) = V_H^2 \ln\left( \frac{r}{r_0} \right) \]
where \$ V_H^2 = 4 \pi G r_0^2 \rho(r_0) \$. This potential function diverges as \$ r \to \infty \$, indicating that there is no finite limit to the potential at large radii.
Poisson's Equation
Poisson's equation relates the gravitational potential \$ \Phi \$ to the mass density \$ \rho \$:
\[ abla^2 \Phi = 4 \pi G \rho \]
In this particular problem, Poisson's equation helps us find the gravitational potential for the given density of a dark halo. The equation reflects how mass distribution affects gravity in a region.
For a spherically symmetric system, Poisson's equation can be simplified, making it easier to solve. Through integration, one can derive the potential function \$ \Phi(r) \$, which provides crucial information about the gravitational influence at any point in the halo.
Circular Orbital Speed
The speed of an object in a circular orbit within the dark halo can be determined using the derived potential function. The circular orbital speed \$ v_c \$ is given by:
\[ v_c^2 = r \frac{d\Phi}{dr} \]
Using our derived potential, we find:
\[ \Phi(r) = V_H^2 \ln\left( \frac{r}{r_0} \right) \]
which results in:
\[ \frac{d\Phi}{dr} = \frac{V_H^2}{r} \]
Thus,
\[ v_c^2 = r \frac{V_H^2}{r} = V_H^2 \]
Hence, the circular speed \$ v_c = V_H \$ is constant at all radii, implying that the gravitational influence remains consistent throughout the dark halo.

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Most popular questions from this chapter

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.

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 \(L>2 \sqrt{3} G \mathcal{M}_{\mathrm{BH}} / c\).

Show that, at radius \(r\) inside a uniform sphere of density \(\rho\), the radial force \(F_{\mathrm{r}}=-4 \pi G \rho r / 3\). If the density is zero for \(r>a\), show that $$ \Phi(r)=-2 \pi G \rho\left(a^{2}-\frac{r^{2}}{3}\right) \quad \text { for } r \leq a $$ so that the potential energy is related to the mass \(\mathcal{M}\) by $$ \mathcal{P E}=-\frac{16 \pi^{2}}{15} G \rho^{2} a^{5}=-\frac{3}{5} \frac{G \mathcal{M}^{2}}{a} . $$ Taking \(a=R_{\odot}\), the solar radius, and the mass \(\mathcal{M}=\mathcal{M}_{\odot}\), show that \(\mathcal{P E} \sim L_{\odot} \times 10^{7} \mathrm{yr}\); approximately this much energy was set free as the Sun contracted from a diffuse cloud of gas to its present size. Since the Earth is about 4.5 Gyr old, and the Sun has been shining for at least this long, it clearly has another energy source - nuclear fusion.

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: $$ \left\langle v_{x}^{2}\right\rangle=\frac{1}{n(\mathbf{x})} \int f\left[\Phi(\mathbf{x})+\frac{\mathbf{v}^{2}}{2}\right] v_{x}^{2} \mathrm{~d} v_{x} \mathrm{~d} v_{y} \mathrm{~d} v_{z}=\left\langle v_{y}^{2}\right\rangle=\left\langle v_{z}^{2}\right\rangle $$

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.
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