**Sidebar: Why do photons take so long to escape the core of the Sun?**

The claim one may read is that a gamma ray bounces around in the core for about 130,000 or 170,000 years or so. If a photon moves at the speed of light, 300,000 km per second and the Sun’s radius is only 696,000 km, it could move the distance of the radius in a bit over 2 seconds. The photon has a very contorted path to take, as it constantly is bumping into protons and electrons. At each collision with a charged particle it has a chance of being diverted, or scattered, possibly also with a significant change in energy, often a loss. We may view the trajectory of the photon as a random walk. Over a number of steps, there is an essentially equal probability of it going forward or backward in any of the three orthogonal directions along axes we may call x, y ,and z. How is escape then effected if the expected value of progress in any given direction is zero? The mathematics may be explored with analytical formulas or with numerical simulations. I offer simple programs in Python on my website to run a random walk in one or two dimensions and to have a plot made of the position; another program simply calculates the average *square* of the displacement. In brief, we have to consider a group or ensemble of photons, all bouncing around. Their average displacement will be zero but some go to any distance one may consider – some in the positive direction, some in the negative direction. This is the process of diffusion. If we consider uniform steps in distance for each scattering, the problem becomes rather simple while giving the same style of result as for the case with scattering lengths having their own statistical distribution. Consider a step length that is a small increment ∆x and the distance to escape in x. The number of steps all in one direction needed to escape is n=x/∆x. With a random walk, the number of steps needed is the square of this, n_{r} = (x/∆x)^{2}.

Let’s apply this to the gamma ray in the core. One website proposed a step size, or mean free path, or 2 cm. If the dense core is the hard part to escape, let’s make the escape distance to be 50,000 km. That’s 2.5 billion steps. With a random walk the number of steps needed is the square of this, or 6.25×10^{18} steps, for a total path length of 2.5 billion times 50,000 km. That’s 1.25×10^{14} km. The photon, however, needs a very short time to take a step, 2×10^{-2}m/3×10^{8} m s^{-1}, about 6.7×10^{-11} s. Multiply that by 6.25×10^{18} steps to get 42×10^{7} s. A year is 3.15×10^{7}s, so that’s about 13 years. Something’s wrong. One author who found the same mismatch noted that the figure of 2 cm for the mean free path came from a textbook, and textbook authors copy each other. The error propagated. We need to up the calculated escape time by a factor of about 10,000, so we need a step size that’s smaller by a factor of =100. That’s 0.2 mm. Is that plausible, given the physics of scattering of gamma rays?

In the core out to 50,000 km the average density of matter has been inferred as 100,000 km per square meter. As hydrogen nuclei each weighing 1.67×10^{-27} kg, that’s about 6×10^{31} protons per cubic meter! Now, the probability of a gamma ray being scattered as it runs a distance *L* in this dense mass is the density of particles, ρ, (both protons and equally abundant electrons) multiplied by the scattering cross-section, σ (an area, as if the gamma ray scoops out this area) and multiplied by the distance *L*. The probability is then ρσL. We can write this as L/F, with F = mean free path = 1/(ρσ). For protons scattering gamma rays the cross-section is near 1 barn, 10^{-24} cm^{2} or 10^{-28} m^{2}. The mean free path is then approximately 1/(6×10^{31}m^{-3} x 10^{-28}m^{2}) = 0.17 mm. That looks right. By the way, that name “barn” has an amusing history. When physicists were exploring the capture cross-sections they expected small values, near the projected area of a nucleus, up to tens of square femtometers, about 10^{-31}m^{2}. The large values led one researcher to say that the cross-section was as big as the side of a barn, using the old metaphor.

The photon actually gives rise to other photons. The energy it loses in collisions with matter heats that matter, which then emits its own lower-energy photons; think black-body radiation. Eventually, all the energy in photons with some MeV of energy ends up creating nearly a million photons with only 2 to 3 eV of energy, those light particles that leave the Sun and speed toward Earth (and other destinations). So, the escape of a photon is really the escape of a cascade of many photons. None of the original gamma rays make it out. The Sun is not a gamma-ray source, except for cosmic rays from the Galaxy hitting its atmosphere and creating new gammas.