Question

The Heisenberg uncertainty principle applies to photons as well as to material particles. Thus, a photon confined to a very tiny box (size delta x) necessarily has a large uncertainty in momentum and energy (recall that for photons, E = pc) and hence a large average energy. Since energy is equivalent to mass, according to E = mc^2, a confined photon can create a large gravitational field. If delta x is sufficiently small, the energy density will be sufficiently large to create a black hole This size delta x is called the Plank length and defines the scale at which gravity and quantum mechanics are inextricably mixed. (The branch of physics that tries to combine quantum mechanics with gravity is called string theory.)
Show that the escape speed for a star of mass M and radius R is u_esc = Squareroot 2GM/R. By equating u_esc for a black hole to the speed of light, derive a formula for the Plank length in terms of c, G, and h.

Answer 1

Answer:

$v^2_{esc} = \sqrt{ \frac{2GM}{R}}$

$\Delta x =\sqrt{ \frac{Gh}{c^3}$

Explanation:

a) Let the average kinetic energy = gravitational potential energy

Then; by equation :

$\frac{1}{2}mv^2_{esc} = \frac{GmM}{R}$

here;

m = mass of the confined photons

$v_{esc}$ = escape velocity

M = mass of the star

R = radius

$\frac{1}{2}v^2_{esc} = \frac{GM}{R}$

$v^2_{esc} = \frac{2GM}{R}$

$v^2_{esc} = \sqrt{ \frac{2GM}{R}}$

b) Let $v_{esc} = c$ (where c = speed of light)

$c = \sqrt{ \frac{2GM}{R}}$

$c^2 ={ \frac{2GM}{R}}$

where

R =$\Delta x$  

$c^2 ={ \frac{2GM}{\Delta x}}$

By using the expression ;

$E = mc^2$ and $E = \rho \ c$

Then;

$mc^2 = \rho c$

$m = \frac{\rho}{c}$

$m = \frac{h}{2 \Delta xc}$     since ( $\rho = \frac{h}{2 \Delta x}$)

Replacing m for M in $c^2 ={ \frac{2GM}{\Delta x}}$; we have:

$c^2 = \frac{2 \ G}{\Delta x} \frac{h}{2 \Delta x}$

$c^2 = \frac{Gh}{\Delta x^2 c}$

$\Delta x^2 = \frac{Gh}{c^3}$

$\Delta x =\sqrt{ \frac{Gh}{c^3}$