Question

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 ${\mathit{r}}_{\mathbf{c}\mathbf{r}\mathbf{i}\mathbf{t}\mathbf{i}\mathbf{c}\mathbf{a}\mathbf{l}}$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

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$r_{critical}$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}\left[L\right]={\left[M\right]}^a\frac{{\left[L\right]}^{3b}}{{\left[M\right]}^b{\left[T\right]}^{2b}}\frac{{\left[L\right]}^c}{{\left[T\right]}^c}\\ \left[L\right]={\left[M\right]}^{a-b}{\left[L\right]}^{3b+c}{\left[T\right]}^{-2b-c}\end{array}$$\begin{array}{c}\because G\to N{m}^{2}k{g}^{-2}\to m^{3}k{g}^{-1}{s}^{-2}\\ \because c\to m{s}^{-1}\end{array}$

Comparing coefficients of both sides,

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

$r_{critical}=\frac{MG}{c^2}$

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,

$r_s=\frac{2MG}{c^2}$