Relativistic multiplication

A couple years ago I wrote about relativistic addition. Given two numbers in the interval (-c, c) you can define their relativistic sum by

x \oplus y \equiv \frac{x + y}{1 + \dfrac{xy}{c^2}}

We can set c = 1 without loss of generality.

Given this exotic definition of addition, what is multiplication?

We’d expect 2 ⊙ x to be the same as xx, and 3 ⊙ x to be the same as xxx, etc. If you stare at these examples you may come up with the proposal

n \odot x = \frac{(1+x)^n-(1-x)^n}{(1-x)^n+(1+x)^n}

It turns out this works for all positive integers n. We can replace n with any real number r and take this as the definition of multiplication by any real number. (Because x is between -1 and 1, the terms in parentheses are always positive and so there’s no difficulty defining real exponents.)

Now what kind of algebraic properties does this multiplication have? It’s not symmetric. The left argument of ⊙ can be any real number, while the right argument has to be in (-1, 1). And if the first argument r happens to be in (-1, 1), it’s not generally the case that rx equals xr.

What we have is scalar multiplication in a vector space. If we define vectors to be numbers in the interval (-1, 1), with vector addition defined by ⊕, and scalar multiplication defined by ⊙, all the axioms of a vector space are satisfied.

Reference: Michael A. Carchidi. Generating Exotic-Looking Vector Spaces. The College Mathematics Journal, Vol. 29, No. 4, pp. 304–308

One thought on “Relativistic multiplication

  1. As a proposal I would have thought of rapidity and reached for tanh( n atanh(x)), which of course turns out to be the same thing.

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