### Special Relativity and the Mercator Projection

#### Posted by John Baez

When you look at an object zipping past you at nearly the speed of light, it looks not squashed but *rotated*.

This phenomenon is well known: it’s called Terrell rotation. But this paper puts a new spin on it:

- Jack Morava, On the visual appearance of relativistic objects.

The abstract is pretty funny:

Abstract.It has been almost half a century since the realization [1,3] that an object moving at relativistic speeds (and observed by light reflected from some point source) is seen not as squashed (as a naive interpretation of the Lorentz-Fitzgerald contraction would suggest) but rather as rotated, through an angle dependent on its velocity and direction of motion. I will try to show here that this subject continues to be worth exploring.The author asserts his moral right to remuneration for any applications of these ideas to computer games or big-budget sci-fi movies.

But here’s the idea.

Suppose a rocket is moving at velocity v in units where $c = 1$. Suppose it’s in the upper half of the $x y$ plane, moving from left to right, parallel to the $x$ axis. Suppose you are at the origin.

If the rocket’s position is $(r,\theta)$ in polar coordinates, let

$\psi = \theta - \frac{\pi}{2}$

so $\psi = 0$ when rocket is ‘just passing you’. Let’s call $\psi$ the **line of sight angle**.

The rocket looks rotated counterclockwise by an angle $\phi$, where

$\cos(\phi + \psi) = \frac{\cos \psi - v}{1 - v cos \psi}$

But Morava found a nicer way to write this formula!

As we were all once forced to learn, the integral of $\sec x$ is

$\lambda(x) = \ln\vert\tan x + \sec x\vert$

Now you’re finally going to get some good out of this knowledge.

The reason why people figured out this integral in the first place is that in a Mercator projection map, a city with latitude $x$ lies on a line with $y$ coordinate $\lambda(x)$.

Now, define ‘Mercator addition’ by

$\lambda (x \oplus y) = \lambda (x) + \lambda (y)$

So, do the Mercator sum of angles $x \oplus y$ you think of them as latitudes, you convert them into heights up from the equator on a Mercator map, then add those heights, then convert the answer back into a latitude.

Suppose a rocket whizzing by at speed $v$ has light of sight angle $\psi$. Then Morava shows it looks rotated by an angle $\phi$, where

$\psi + \phi = \psi \oplus \arcsin v$

I’ll admit I haven’t checked his computation.

I don’t understand the deep inner meaning of this. So let me just say it in words!

To get the apparent rotation angle $\phi$ of a moving object, you take the Mercator sum of its line-of-sight angle $\psi$ and the arcsine of its velocity, then subtract off $\psi$ using ordinary subtraction.

What does it really mean? Why should the Mercator projection be related to special relativity? Can you help me out here?

I’m sure Jack Morava was interested in this because Mercator addition is an example of a formal group law, and formal group laws are important in homotopy theory.

One other weird and appealing fact is that the function $\lambda$, which you hated so much in calculus class, is almost its own inverse! To be precise,

$\lambda^{-1}(x) = - i\lambda (i x)$

Morava also gets some mileage out of this… but read his paper for details.

## Re: Special Relativity and the Mercator Projection

Dear John,

Srsly this is reminiscent of Wick rotation, but I don’t know how to make that intuition even vaguely precise. As

Private Eyesays, I think we should be told… I hope someone can make sense of this (assuming it’s not just false). Thanks, hopefully…