# The Electron at the End of the Universe

In A Passion for Science, Michael Berry's essay "The Electron at the End of the Universe" poses two scenarios.

1. Assume that a box of gas particles obeys Newtonian mechanics and that we precisely know the particles' initial positions and velocities. Suppose we want to predict the future state of these particles, but we leave out the gravitational pull of a single electron at the edge of the universe. How much time must pass before this damages our predictions? Berry claims gravity from a single electron ten thousand million light years away will change the predicted angle that a gas particle leaves its fiftieth collision by ninety degrees. Fifty collisions happen "in a tiny, tiny fraction of a microsecond."
2. Assume a billiard player is calculating a shot on an idealized pool table (perfectly flat and smooth). After how many billiard ball collisions will the player need to factor in the gravity of the people standing around the table? Berry claims the answer is six or seven.

Are these claims true? If so, could you please provide calculations to support them?

• Are the collisions themselves supposed to be relevant, or are they just being used as a means of picturing a time interval or a distance travelled? Nov 6 '21 at 22:01
• I updated the question with a link to the original claim. I believe they are supposed to be relevant; it's a system of colliding billiard balls / gas particles that we're trying to model. Nov 6 '21 at 22:12
• If Berry makes a claim it is surely for Berry to demonstrate it. Why should someone on this site do it ? Would it not be better to email Berry and ask for proof ? Nov 6 '21 at 23:21
• For the billiards case he is probably talking about colliding with the walls of the table rather than other balls. There are many idealized dynamical systems of billiards that demonstrate sensitive dependence on initial conditions and I think that's what he's getting at. Nov 6 '21 at 23:35
• @octonion Quite the opposite. It is the collisions between two curved objects that lead to exponential increase of the deviation. Nov 7 '21 at 14:51

The calculations are done in

Here's a summary of the key equations, using the notation and equation numbers from those lecture notes.

Consider a classical system of spherical objects, each of radius $$R$$. Let $$\ell$$ be the average distance an object travels between collisions (also called the mean free path), and let $$v$$ be the typical relative velocity between colliding objects. Suppose that we make an error $$\Delta\theta_1$$ in our estimate of the direction of one object's initial velocity. After $$N$$ collisions, the error $$\Delta\theta_N$$ in our estimate of the direction of its final velocity is $$\Delta\theta_N\approx\left(\frac{\ell}{R}\right)^N\Delta\theta_1. \tag{4}$$ Now consider the gravitational effect of a distant mass $$m$$, which could be a distant electron or a person walking near the pool table. The difference between the gravity-induced accelerations of two objects at different distances from the mass $$m$$ has magnitude $$\Delta a \approx 2G\frac{m}{r^3}\Delta r, \tag{5}$$ where $$G$$ is Newton's constant, $$r$$ is the overall distance to $$m$$, and $$\Delta r$$ is the difference between the distances of the two objects to $$m$$. Over the time $$\ell/v$$ between collisions, this gives an error $$\Delta\theta_1\sim (\Delta a)(\ell/v)^2/R \tag{text below 6}$$ in the estimated angle after the first collision, which we can use as the initial error in equation (4).

## Application to scenario 1

Consider a gas of typical atoms at $$20$$ $${}^\circ$$C and a density comparable to atmospheric density. Then:

• $$R\sim 10^{-10}$$ meter

• $$\ell\sim 10^{-7}$$ meter (https://en.wikipedia.org/wiki/Mean_free_path)

• $$v\sim 10^3$$ m/s (https://en.wikipedia.org/wiki/Thermal_velocity)

• $$m\sim 10^{-30}$$ kg

• $$\Delta r\sim 1$$ meter (the size of the box of gas)

• $$r\sim 10^{26}$$ meters

• $$G\sim 10^{-10}$$ m$$^3$$ kg$$^{-1}$$ s$$^{-2}$$

Using these inputs in the preceding equations gives $$\Delta\theta_N\sim (10^3)^N \times 10^{-128}$$ so we get $$\Delta\theta_N\sim 1$$ ($$\sim 90^\circ$$) after $$N\sim 43$$ collisions, roughly as claimed.

## Application to scenario 2

For this scenario, use

• $$R\sim 3\times 10^{-2}$$ meter

• $$\ell\sim 1$$ meter

• $$v\sim 1$$ m/s (https://billiards.colostate.edu/faq/speed/typical)

• $$m\sim 10^{2}$$ kg

• $$\Delta r\sim 1$$ meter

• $$r\sim 1$$ meter

• $$G\sim 10^{-10}$$ m$$^3$$ kg$$^{-1}$$ s$$^{-2}$$

Using these inputs in the preceding equations gives $$\Delta\theta_N\sim (30)^N \times 10^{-8}$$ so we get $$\Delta\theta_N\sim 1$$ ($$\sim 90^\circ$$) after $$N\sim 6$$ collisions, as claimed.