# A photon travels in space for 10 billion years. What are the odds it will arrive here without interacting with a atom on the way?

Space isn't a perfect vacuum and I wonder how an image of a galaxy can travel billions of years without becoming diffused by photon collisions with space matter?

• The probability is $\exp\left(-\alpha d\right)$ where $\alpha$ is the absorption coefficient along the photons path and $d$ is the distance travelled. – boyfarrell May 30 '14 at 1:09
• Did the photon start out in a direction that would intersect with "here"? Because that is truly a tiny fraction of all photons that travel for 10 billion years... You might want to be a bit more precise in stating the question. – Floris May 30 '14 at 1:16
• Important to note that the universe was considerably denser 10 billion years ago. See also What does ionization of neutral Hydrogen have to do with “transparency”? – user10851 May 30 '14 at 16:52

If you say density of outer space is 50 atoms per cubic meter (assumed to be hydrogen), then the mass of the "pillar of universe" that light traverses in 10 billion years is 9.4E25 meters times 50 atoms times atomic mass unit = $7.8 kg/m^2$
The photon can travel a long way before scattering. The Rayleigh cross section of hydrogen interacting with a photon in the visible region is about $\sigma=1.9 \times 10^{-32} m^2$. The distance a photon travels before hitting the hydrogen is then $l=\frac{1}{n_e \sigma}$ where $n_e$ is the number of molecules per cubic metre. If $n_e=10^6 m^{-3}$ then $l=5.2 \times 10^{25}$ or about 5 billion light years. There are of course variations in the density of space, dust clouds etc and the cross-sectional area is highly dependent on wavelength.
The likelihood a photon after traveling a distance $x$ arrives without been absorbed is $T=\exp (-x/l)$. If the photon has travelled 10 billion years, $x=9.46 \times 10^{25}m$ so $T=0.16$. So the answer is about $16 \%$ (with some large error bars!)