genug Unfug.

2014-09-21

# Photon contained in its own Schwarzschild radius

As a followup to my previous post, where I showed that the Schwarzschild radius $r_s$ of a photon with wave length

$l_p$: Planck length
$G$: gravitational constant
$c$: speed of light
$h$: Planck's constant

$$\lambda_o =2\sqrt{2\pi}\, l_p = 2\sqrt{\frac{Gh}{c^3}}$$ is $\lambda_o/2$, I want to add a few simple fun calculations.

The energy of a photon of frequency $\nu$ is $E_p = h\nu$, where $\nu=c/\lambda$ for a given wave length $\lambda$. With the specific $\lambda_o$ we get \begin{align*} E_o &= h c/\lambda_o \\ &= \frac{1}{2} hc \sqrt{\frac{c^3}{Gh}} \\ &= \frac{1}{2} \sqrt{\frac{h^2 c^5}{Gh}} = \frac{1}{2} \sqrt{\frac{h c^5}{G}} \end{align*}

Using Einstein's formula $E=mc^2$ relating energy $E$ and mass $m$, the mass of this photon is \begin{align*} m_o &= E_o/c^2 \\ &= \frac{1}{2} \sqrt{\frac{h c^5}{c^4 G}} = \frac{1}{2} \sqrt{\frac{h c}{G}} \end{align*} Replacing $h$ by $2\pi\hbar$ we arrive at $$m_o = \frac{1}{2}\sqrt{\frac{2\pi\hbar c}{G}} = \frac{1}{2} \sqrt{2\pi}\, m_p$$ where $m_p$ is the Planck mass.

Putting everything together we get \begin{align*} \lambda_o &= 2\sqrt{2\pi}\, l_p\\ m_o &= \frac{1}{2} \sqrt{2\pi}\, m_p \end{align*}

I wanted the wave length of a photon to be identical to its Schwarzschild diameter, that is two times its Schwarzschild radius and end up with the factors $1/2$ and $2$. If I had set $r_s=\lambda_o$ these two factors would also disappear, but anyway.

I also wanted to know the numerical value of the frequency of such a photon to see where it is located in the electromagnetic spectrum. The frequency is $$\nu_o = c/\lambda_o = \frac{1}{2} \sqrt{\frac{c^5}{G h}}$$ which Google happily computes for us without the need to type in all those digits for the constants to be $$\nu_o = 3.70003533\cdot 10^{42}\, \text{Hz} .$$ It is fun to note that this contains the Answer to the Ultimate Question of Life, the Universe, and Everything.

A more serious note is that this frequency is 23 orders of magnitude beyond gamma rays, where Wikipedia's description of the electromagnetic spectrum ends. My hunch is that we are not going to generate such a photon anytime soon.