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Chapter 18: Problem 26

In the Schwarzschild solution show the only possible closed photon path is a circular orbit at \(r=3 m\), and show that it is unstable.

Short Answer

Expert verified
The only closed photon path in the Schwarzschild solution is a circular orbit at \(r=3m\), and the orbit is unstable.

Step by step solution

01

Setting up the equation

In the Schwarzschild metric, the motion of a photon is determined by the equation \( \frac{1}{b^{2}} = \frac{1}{r^{2}} - \frac{1}{r^{3}} \). The letter 'b' is the impact parameter related to the angular momentum per unit energy, 'r' is the radial distance from the mass 'm', and 'm' is the mass of the object causing the gravitational field. We are looking for solutions where the photon moves in a circle of fixed radius. For this to happen, the photon needs to be at a local extremum of the effective potential. To derive this potential with respect to the radius 'r', set the derivative of the expression on the right-hand side of the equation to zero.
02

Deriving the closed path

The derivative of \( \frac{1}{r^{2}} - \frac{1}{r^{3}} \) with respect to 'r' is \(- \frac{2}{r^{3}} + \frac{3}{r^{4}} \). Setting this equal to zero gives the equation: \(- \frac{2}{r^{3}} + \frac{3}{r^{4}} = 0 \), which simplifies to \(- 2r + 3 = 0\). Solving for 'r' gives \(r=3m\). So, the only circular path is for \(r=3m\).
03

Proving instability

To show that the orbit at \(r=3m\) is unstable, we have to show that a small deviation from \(r=3m\) leads to progressively larger deviations, i.e., the potential must have a local maximum at \(r=3m\). Taking the second derivative of \( \frac{1}{r^{2}} - \frac{1}{r^{3}} \) gives \(\frac{6}{r^{4}} - \frac{12}{r^{5}}\). Replacing \(r\) with \(3m\) gives a positive value, indicating the potential has a local maximum and the orbit is indeed unstable.

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Key Concepts

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

Photon Circular Orbit
In the realm of general relativity, the Schwarzschild solution offers a fascinating insight into the paths taken by light or photons in the presence of a massive object. A photon circular orbit is an intriguing aspect of this solution. Here, the light travels in a perfect circle around the object. However, there's a catch. The only possible circular path for a photon in a Schwarzschild geometry occurs at a specific radial distance.For the Schwarzschild black hole, this magic number turns out to be at three times the Schwarzschild radius, represented as \( r = 3m \). This is derived by analyzing the Schwarzschild metric and determining where a photon can stably orbit given that the forces involved allow for such a circular path.When you think about a photon travelling in space, try to imagine how it navigates around a gravitational field without ever spiraling inward or flinging outward. At \( r = 3m \), this perfect balance is achieved. Yet, as mesmerizing as this precise pathway may seem, it is bound by certain rules of stability.
Effective Potential
To understand the behavior of photons moving under the influence of gravity, the concept of effective potential is crucial. Effective potential helps in analyzing whether paths are circular and if so, how stable they are. It's essentially a way to examine forces and energies without needing to dive into each individual force component.In our scenario, effective potential is defined for a photon moving in the Schwarzschild metric as\[ V(r) = \frac{1}{r^2} - \frac{1}{r^3} \]This equation emerges from translating the gravitational influences into a form that mimics conservative forces.For a circular orbit, the photon must be at a local extremum of this effective potential. By setting its derivative to zero, we establish the condition for the photon to "rest" at this orbit comfortably. Interestingly, at \( r = 3m \), the potential dictates that the forces acting on the photon are perfectly balanced, allowing it to maintain the circular orbit. However, just sitting at an extremum doesn't guarantee stability. That's where the next concept comes in.
Instability of Orbit
An important characteristic of the photon circular orbit at \( r = 3m \) is its instability. A simple nudge or deviation from this path can dramatically alter the orbit, causing the photon to depart from its perfect circle.The instability is confirmed through a second derivative test of the effective potential. Calculating the second derivative\[ V''(r) = \frac{6}{r^4} - \frac{12}{r^5} \]for \( r = 3m \), reveals a positive value indicating a local maximum. What this means is, unlike a stable orbit that would have a potential well (like a bowl), this orbit resides atop a "hill." Hence any slight disturbance sends the photon sliding off the edge.This instability is akin to a marble sitting on top of a hill; while stationary at the peak, any gentle breeze would roll it downhill. In the cosmos, this means that only when conditions are perfect does the photon maintain its path. Otherwise, tiny imperfections lead to divergent paths, showcasing the delicate equilibrium of gravitational forces.

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Most popular questions from this chapter

Show that a space is locally flat if and only if there exists a local basis of vector ficlds \(\left\\{e_{t} \mid\right.\) that are absolutely parallel, \(D e_{1}=0\).

Let \((M, \varphi)\) be a surface of revolution defined as a submantfold of \(\mathbb{E}^{3}\) of the form $$ r=g(u) \cos \theta, \quad y=g(u) \sin \theta, \quad z=h(u) $$ Show that the induced metric (see Example 18.1) is $$ \mathrm{d} s^{2}=\left(g^{\prime}(u)^{2}+h^{\prime}(u)^{2}\right) \mathrm{d} u^{2}+g^{2}(u) \mathrm{d} \theta^{2} $$ Picking the parameter \(u\) such that \(g^{\prime}(u)^{2}+h^{\prime}(u)^{2}=1\) (interpret this choice!), and setting the basis 1-forms to be \(\varepsilon^{\prime}=\mathrm{d} u, \varepsilon^{2}=g d \theta\), calculate the connection I-forms \(\omega^{\prime}\), the curvature 1 -forms \(\rho_{\prime}^{\prime}\), and the curvature tensor component \(R_{1212}\).

(a) Compute the components of the Ricer tensor } R_{\mu v} \text { for a space-tume that has a }\end{array}\( metric of the form $$ \mathrm{d} s^{2}=\mathrm{dx}^{2}+\mathrm{d} v^{2}-2 \mathrm{~d} u \mathrm{~d} v+2 H \mathrm{~d} v^{2} \quad(H=H(\mathrm{x}, y, u, v)) $$(b) Show that the space-time is a vacuum if and only if \)H=\alpha(x, y, v)+f(v) u\( where \)f(v)\( is an arbitrary function and \)\alpha\( sat?sfies the two-dimensional Laplace equation $$ \frac{\partial^{2} \alpha}{\partial x^{2}}+\frac{\partial^{2} \alpha}{\partial y^{2}}=0 $$ and show that it is possible to set \)f(v)=0\( by a coordunate transformation \)u^{\prime}=u g(v), v^{\prime}=h(v)\(. (c) Show that \)R_{\text {taj } 4}=-H_{v}\( for \)i, j=1,2$.

(a) A particle falls radially inwards from rest at in finity in a Schwarzschild solution. Show that it will arrive at \(r=2 m\) m a finite proper time after crossing some fixed reference position \(r_{0}\), but that coordinate time \(t \rightarrow \infty\) as \(r \rightarrow 2 m\). (b) On an infalling extended body compute the tidal force in a radual direction, by parallel propagating a tetrad (only the radial spacelike unit vector need he considered) and calculating \(R_{1414}\). (c) Estimate the total tidal force on a person of height \(1.8 \mathrm{~m}\), weighing \(70 \mathrm{~kg}\), fallang head-first into a solar mass black hole \(\left(M_{3}=2 \times 10^{10} \mathrm{~kg}\right)\), as he crosses \(r=2 \mathrm{~m}\).

(a) For a perfect flud in general relat?uty, $$ T_{\mu v}=\left(\rho c^{2}+P\right) U_{\mu} U_{v}+P g_{\beta v} \quad\left(U^{\mu} U_{y}=-1\right) $$ show that the conservation identities \(T^{\mu v}, v=0\) imply \(\rho_{v} U^{v}+\left(\rho c^{2}+P\right) U_{v^{2}}^{*}\) \(\left(\rho c^{2}+P\right) U_{,}^{\mu} U^{v}+P_{v}\left(g^{\mu \prime}+U^{\mu} U^{\nu}\right)\) (c) In the Newtonian approximatuon where $$ U_{\mu}=\left(\frac{1}{c},-1\right)+O\left(\beta^{2}\right), \quad P=O\left(\beta^{2}\right) \rho c^{2}, \quad\left(\beta=\frac{v}{c}\right) $$ where \(|\beta| \ll 1\) and \(g_{\mu \nu}=\eta_{h^{x}}+\epsilon h_{\mu \nu}\) with \(\epsilon \ll 1\), show that $$ h_{+\mu} \approx-\frac{2 \phi}{c^{2}}, \quad h_{i j} \approx-\frac{2 \phi}{c^{2}} \delta_{j} \quad \text { where } \quad \nabla^{2} \phi=4 \pi G \rho $$ and \(h_{t 4}=O(\beta) h_{44}\) Show in this approxumation that the equations \(T^{\mu t},=0\) approvimate to $$ \frac{\partial \rho}{\partial t}+\nabla \cdot(\rho v)=0, \quad \rho \frac{d v}{d t}=-\nabla P-\rho \nabla \phi $$
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