Harald Kirsch

genug Unfug.

2014-09-19

Wavelength and Schwarzschild Radius of a Photon

Inspired by the questions is there a smallest length, I wondered what it takes, at least formally, to have a photon that might contain itself in its own black hole.

Mass is able to deflect the path of light or photons. The more mass there is, the stronger the deflection. If a given mass $m$ is compressed into a sphere smaller than its Schwarzschild radius, it is no longer only a deflection but, the light cannot escape anymore from that sphere. The formula for the Schwarzschild radius $r_s$ of $m$ is $$r_s(m) = \frac{2Gm}{c^2}$$ where $G\approx 6.6\times 10^{-11}\frac{m^3}{kg\cdot s^2}$ is the gravitational constant and $c=299\,792\,458\frac{m}{s}$ is the speed of light. For a photon with frequency $\nu$, its energy is $h\nu$, where $h\approx 6.6\times 10^{-34} Js$ is Planck's constant. This energy can be related to a mass using Einsteins famous formula $E=mc^2$ to get $$m = h\nu/c^2 .$$ Due to the fixed relation $c=\lambda\nu$ between the frequency $\nu$ and the wave length $\lambda$ of a photon, we can express the mass also as $$ m = \frac{h}{\lambda c} .$$ We can insert this relation into the formula for $r_s(m)$ and get $$ r_s(m) = \frac{2Gh}{\lambda c^3} .$$

The interesting bit is that $r_s$ as well as $\lambda$ have the unit of length, so we are relating the wave length of a photon to its Schwarzschild radius. Further, as we decrease the wave length $\lambda$ of a photon, its frequency and thereby its energy increases — as does it Schwarzschild radius. Consequently we can ask when $\lambda$ and $r_s$ are equal. Or, rather, we can ask when a photon of wave length $\lambda$ "fits" into a sphere of radius $r_s$, i.e. $\lambda = 2r_s$ or $\lambda/2=r_s$.

This is the case when $r_s = a = \lambda/2$, where $a$ is an arbitrary new variable which we now enter into the last equation for $\lambda/2$ and $r_s(m)$ to get $$ a = \frac{Gh}{ac^3} .$$ We solve this for $a$ and get $$ a = \sqrt{\frac{Gh}{c^3}}. $$ So the wave length $\lambda$ of a photon "fits" into a sphere the size of its Schwarzschild radius $r_s(m)$ when both are equal to $a$, which is $$ r_s = \sqrt{\frac{Gh}{c^3}} = \lambda/2 .$$ This may not look very interesting, but physicists will recognize this square root as something they know, but not quite. Looking up the Planck length $$ l_p = \sqrt{\frac{G \hbar}{c^3}} $$ and knowing that $h = \hbar\cdot 2\pi$, we see that $$ r_s = \lambda/2 = \sqrt{\frac{G\cdot2\pi\hbar}{c^3}} = \sqrt{2\pi}\, l_p .$$

Does this mean that when we confine a photon of wave length $\lambda=2\sqrt{2\pi}\,l_p$ into a sphere with that radius $\lambda/2$, that it can not escape and in particular not spread out of this sphere?

I wonder whether this quite simple result is trival, given the definitions of Planck's units, or whether it is a more deeper consequence of the theory behind the Schwarzschild radius.

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