Chapter 2: Q100E (page 69)

If it is fundamental to nature that a given mass has a critical radius at which something extraordinary happens (i.e., a black hole forms), we might guess that this radius should depend only on mass and fundamental constants of nature. Assuming that $r_{c r i t i c a l}$ depends only on M, G, and c, show that dimensional analysis gives the equation for the Schwarzschild radius to within a multiplicative constant.

Short Answer

Expert verified

The Schwarzschild radius is the radius below which a body's gravitational attraction between its components must force it to collapse irreversibly.

Step by step solution

Perform a dimensional analysis of the critical radius.

The $r_{c r i t i c a l}$ is dependent on mass, gravitational constant, and speed of light.

$\begin{array}{l} r = M^{a} G^{b} c^{c} \end{array}$

In dimensional form, it can be written as,

$\begin{array}{l} [L] = {[M]}^{a} \frac{{[L]}^{3 b}}{{[M]}^{b} {[T]}^{2 b}} \frac{{[L]}^{c}}{{[T]}^{c}} \\ [L] = {[M]}^{a - b} {[L]}^{3 b + c} {[T]}^{- 2 b - c} \end{array}$ $\begin{matrix} ∵ G \to N m^{2} k g^{- 2} \to m^{3} k g^{- 1} s^{- 2} \\ ∵ c \to m s^{- 1} \end{matrix}$

Comparing coefficients of both sides,

$\begin{matrix} a - b = 0 \Rightarrow a = b \\ - 2 b - c = 0 \Rightarrow c = - 2 b \\ 3 b + c = 1 \Rightarrow b = 1 = a \Rightarrow c = - 2 \end{matrix}$

$r_{c r i t i c a l} = \frac{M G}{c^{2}}$

Define Schwarzschild radius.

The Schwarzschild radius is the radius below which a body's gravitational attraction between its components must force it to collapse irreversibly. This phenomenon is supposed to represent the most massive stars’ final fate.

The Schwarzschild radius ( $r_{s}$ ) of a mass M object is calculated using the formula below,

$r_{s} = \frac{2 M G}{c^{2}}$

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Short Answer

Step by step solution

Perform a dimensional analysis of the critical radius.

Define Schwarzschild radius.

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