Rutherford Scattering

Abstract: Server sky thinsats deflect particles in the van Allen belt, and over time cause them to precipitate into the upper atmosphere. 1000 tons of thinsatellites precipitates 63% of the protons at 1MeV in 20 years ( the electrons much more quickly). The time is inversely proportional to the mass, and proportional to proton energy to the 3/2 power. The process is significant, but far slower than first estimates suggested.

the details

Thinsats orbit in the van Allen Belt, and encounter very high fluxes of trapped protons and electrons. An array as a whole has a very high surface area to weight ratio. An array of thinsats made of 50 micron thick glass have an area to mass ratio of 7 m² per kilogram, or 7 km² per thousand metric tonnes (about 300 Arianne launches).

When space is populated by a huge area of thinsats, then electrons and protons bouncing back and forth through the L=2 region of the equatorial plane will have a significant chance of encountering the area of the thinsats. Some fraction of those particles will come close to the nuclei of the silicon and oxygen atoms that make up the glass substrate of the thinsats, and they will be scattered. Some particles will change pitch angle, causing the distribution of pitch angle to diffuse and spread out. Particles whose pitch angle decreases enough will move their mirror point (the altitude at which they are reflected by the lower, denser magnetic field) below the top of the atmosphere - those particles will be absorbed. Over time, this reduces the radiation flux in the van Allen belt.

Define:

• \Large \sigma is area in m²

• \Large \Omega is solid angle in radians²

• \Large e is the electronic charge = 1.60921E-19 Colombs

• \large E_{eV} is the particle energy in electron volts

• \large Z is the atomic number of the target atom

• \Large \epsilon_0 is the permittivity of vacuum = 8.85E-12 Colombs² / Joule-meter

• \Large \theta is the scattering angle

• \large D \LARGE = {{eZ}\over{4\pi\epsilon_0E_{eV}}} ~ ~ is the collision parameter the closest approach for a head-on collision between particle and nucleus.

• \large D_{1MeV} = 1.44E-15 × Z ~ ~ ... in meters

D_{1MeV} is much smaller than the atomic radius (111pm for silicon, 48pm for oxygen), the nuclear charge is not shielded by the electron charge of the atoms.

For singularly charged particles like protons and electrons, Rutherford scattering follows the formula:

\large s = (D/2) \cot ( \theta / 2 )

If s is zero, then \large \theta is \large \pi radians, or 180 degrees. If \theta is 5 degrees, then s is 11.45 D . If \theta is 0, then s is infinite - that is, particles passing by far away are not deflected.

We don't care so much about \theta - we care about the component of \theta in the perpendicular direction, or \Delta\alpha . As we will see, that is the perturbation of the pitch angle, which will cause particles to "fall out" of the van Allen belt. That component is \Delta\alpha = \theta \cos ( \phi ) where \phi is the angle between the plane of the deflected particle and the magnetic field line. Our equation becomes:

\large s = ( D / 2 ) \cot ( \Delta\alpha / 2 \cos ( \phi ) )

Let s_x ~ = ~ s ~ cos ( \phi ) and s_x ~ = ~ s ~ sin ( \phi ) . The s_x component is in the direction of the field line. This function can be evaluated between \pm \phi_{max} \equiv \arccos ( \Delta\alpha / \pi ) , and forms a monotonic curve between 0 and \phi_max .

If \Delta\alpha is less than 20° ( \phi_{max} \approx 83.6 \deg ), s_x and s_y approximately follow a circular locus with a radius s_0 ~ = ~ ( D / 4 ) \cot ( \Delta\alpha / 2 ) around a center point at ( s_0, 0 ) . This encloses a scattering cross section of:

\large \sigma = \pi s_0^2 = ( \pi D^2 / 16 ) \cot ^2 ( \Delta\alpha / 2 )

For larger \Delta\alpha , the area is reduced, only 95% of the above at 33°, 90% at 52°, and 50% at 142°, reaching 0% at 180°. However, these cross sections are already tiny, and the deflections do not contribute much to the precipitation of particles from the van Allen belt.

The derivative of the cross section with area versus deflection angle (in radians), is:

\LARGE { { \partial\sigma } \over { \partial\Delta\alpha } } = { { - \pi D^2 ~ \cos ( \Delta\alpha / 2 ) } \over { 16 ~ \sin^3 ( \Delta\alpha / 2 ) } }

Or for small angles (again in radians):

\LARGE { { \partial\sigma } \over { \partial\Delta\alpha } } \approx { { - \pi D^2 } \over { 2 ~ \Delta\alpha ^ 3 } } ~ ~ ~ ... For angles less than 30 degrees, this approximation is accurate to within 0.1%, the 5% error dominates.

Here is a table of scattering angle \large\theta and cross section versus the ratio of impact parameter to collision parameter, s/D :

For Carbon, Oxygen, Aluminum, Silicon, Gold, and SiO₂ glass at 1MeV:

For example, for oxygen atoms with D_{100KeV} = 1.15E-13 meters and an atomic diameter 1.32E-10 meters, then s reaches the atomic diameter when s/D = 1148 and \theta = 0.05 degrees . This is a low energy and a small angle, which justifies the "no electrons, nucleus only" assumption.

The Radiation Belts

The energetic particles trapped in the radiation belt spiral helically around magnetic field lines. Their velocity is divided into two components. \large v_{\|}\: is the parallel velocity, along the field line, and \large v_{\bot}\: is the tangential velocity, spiraling around the field line. The radius of the helix is on the order of kilometers. The pitch angle \large \alpha is defined by the ratio of \large v_{\|}\: and \large v_{\bot}\: :

\LARGE \alpha = \large \arctan \LARGE \left( {{v_{\bot}} \over {v_{\|}}} \right)

If the particle has no parallel velocity, the pitch angle is 90° (or \large\pi/2 radians ). If the particle has only parallel velocity, the pitch angle is zero.

The magnetic field lines are described by the McIlwain model. The field lines close through the inside of the earth, and one might expect particles to follow them to the surface. However, the field intensifies closer to the earth, and this causes the particles to exchange parallel velocity for tangential velocity as they approach the surface, increasing the pitch angle. If the angle reaches 90 degrees before reaching the upper atmosphere, the particle is "mirrored" as shown:

If the particle pitch angle at the equator (defined as \large \alpha_{eq} ) is small enough, it will not accumulate enough pitch angle to mirror, and it will collide with the upper atmosphere and leave the van Allen belt. The particles remaining in the belt have large enough pitch angles to always mirror, unless they are scattered. Define the "loss cone angle" when the particle crosses the magnetic equator (see Piscane equation 6.77 page 150 ):

\LARGE \alpha_{lc} = \large \arcsin \left( [ 4{L^6}\:-\:3{L^5} ]^{-0.25} \right) .

Any particles with a loss cone angle less than \large \alpha_{lc} or greater than 180\deg\large\:-\:\alpha_{lc} will fall into the atmosphere.

At m288, the L value varies between 1.93 and 2.14, a band in the equatorial plane between 12310 km and 13650 km from the center of the dipole. This is a disk with an area A_D of 1.1E7 km² . The associated \large \alpha_{lc} values are 17.4° for L=1.93, and 14.6° for L=2.14 . For simplicity, we will assume L=2 and \large \alpha_{lc} = 16°

A scattering event can change the pitch angle, sometimes increasing it, sometimes reducing it. This is why Rutherford scattering by server sky satellites can slowly deplete the van Allen belt, by "diffusing the pitch angle". Thus, a particle with a pitch angle of 20° may be scattered 5° to an angle of 15° and enter the atmosphere.

The bounce period of particles, between the north and the south, is a strong function of the energy and a weak function of the pitch angle (after Piscane equation 6.75 page 149 ):

\LARGE \tau_{bounce} \: \approx { { 4 L R } \over { \sqrt { q E_{eV}/m } } } ( 1.3 - 0.56 \sin \alpha_{eq} )

Here is a table of bounce periods (in seconds) for electrons and protons, at pitch angles of 16° and 89° (ignoring relativistic effects) :

This is the time to make a complete round trip between upper and lower mirrors. The particles will pass through the equator twice per bounce time. For the following calculations, we will use the larger bounce time per particle, at 16 degrees, and the following encounter period, collision cross section (1000 tons SiO₂ , and time to collision ( 100 bounces * encounter period * 1.1E7km² / ((sec/year) * cross section ).

The distribution of equatorial pitch angles for particles can probably be computed from measured intensity versus latitude, since the particles with pitch angles near 90° will not be detected farther north and south. It is likely that the pitch angled distribution follows a parabola, like most diffusion processes, with a peak near 90°, and zero at 16° and 174°. The vast majority of the particles will pass through the large equatorial disk twice per bounce time without encountering a thinsat, but some fraction will.

A crude model based on a random walk

Assume that a particle must make an average of 0.5*(90/5)² or about 200 passes at 5 degrees on average to be bounced out - a random walk. We will do a more precise calculation, involving multiple convolutions, after this table. Assume that the maximum s is about half the average SiO₂ atomic size, approximately 1.6E-10 meters.

5 degrees, 1000 tonnes SiO₂ , 200 collisions

The less energetic particles are reduced by 1/e within the lifetime of the server sky satellites, with 1000 metric tons of thinsats in orbit, a population of 300 million first generation thinsats. As we approach 100 billion 0.2 gram satellites, or 20 thousand metric tons, the population decay times go down by a factor of 200. Within a couple of decades, the van Allen belt at server sky altitudes essentially vanishes, with new particles formed from cosmic rays swept out in a few days.

A more accurate calculation

Instead of assuming 5 degrees and a random walk, let's compute the actual diffusion of the initial distribution by computing the convolution of our starting distribution with a probability distribution for scattering versus \Delta\alpha , twice per bounce period.

The probability computed above is for a single scattering center - with tons of thinsats in orbit, there will be many. For scattering angles \Delta\alpha less than 0.1°, temporarily assume no scattering at all. Multiply the rest of the distribution between 0.1° and 30° by the sum of D times N, the collision parameter for each type of atom at a given energy, times the number of those atoms (huge!). Integrate the result, then subtract from one. The remainder is multiplied by a delta impulse function and placed at zero - that is, everything that is not deflected between 0.1° and 30° is assumed to not be deflected. For the actual convolution, we can set the delta to the sum of all the other deflections - whatever is at the zero deflection point is reduced by what is deflected away from it.

Assume both the pitch angle distribution and the deflection probability curve are digitized to 0.1&deg pitch angle accuracy. The pitch angle distribution looks like so:

GRAPH HERE

And the initial deflection probability curve looks like so:

GRAPH HERE

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