### Faster Than Light

BackReaction has a post on special relativity which includes a nice $2 \times 2$ circulant matrix $\Lambda = (\gamma , - \gamma \beta )$, where $\gamma$ and $\beta$ are what you think they are.

Let's see how relativity likes playing with numbers. First observe that since the eigenvalues of a 2d circulant $(A,B)$ are always $A \pm B$, the eigenvalues of $\Lambda$ are given by

$\lambda \in \{ \frac{\sqrt{1 - v/c}}{\sqrt{1 + v/c}} , \frac{\sqrt{1 + v/c}}{\sqrt{1 - v/c}} \}$

and $\lambda \rightarrow 1$ as $c \rightarrow \infty$. Alternatively, we could have simplified $\lambda$ to

$\lambda = \frac{1 \pm v/c}{\sqrt{1 - (v/c)^{2}}}$

Now for $v/c = \sqrt{3}$, which is the speed of an observed preon, the two expressions for the eigenvalues lead to the basic numerical identity

$\sqrt{2} = (\sqrt{3} - 1) \cdot (\frac{\sqrt{3} + 1}{\sqrt{3} - 1})^{1/2} = ((\sqrt{3} + 1)(\sqrt{3} - 1))^{1/2}$

which is easily verified, even on my lousy calculator. This expression for $\sqrt{2}$ just amounts to the difference of squares $3 - 1$, but it's cute. Note also that the same expression applies for any pair $(n, n+1)$ of finite ordinals. That is, all square roots $\sqrt{n}$ are simply expressible in terms of $\sqrt{n + 1}$. For the ordinal $3$, or any $n$ such that $n+1$ is a square, it turns out to be a tautology, because $2 = \sqrt{4}$ trivialises the expression.

Let's see how relativity likes playing with numbers. First observe that since the eigenvalues of a 2d circulant $(A,B)$ are always $A \pm B$, the eigenvalues of $\Lambda$ are given by

$\lambda \in \{ \frac{\sqrt{1 - v/c}}{\sqrt{1 + v/c}} , \frac{\sqrt{1 + v/c}}{\sqrt{1 - v/c}} \}$

and $\lambda \rightarrow 1$ as $c \rightarrow \infty$. Alternatively, we could have simplified $\lambda$ to

$\lambda = \frac{1 \pm v/c}{\sqrt{1 - (v/c)^{2}}}$

Now for $v/c = \sqrt{3}$, which is the speed of an observed preon, the two expressions for the eigenvalues lead to the basic numerical identity

$\sqrt{2} = (\sqrt{3} - 1) \cdot (\frac{\sqrt{3} + 1}{\sqrt{3} - 1})^{1/2} = ((\sqrt{3} + 1)(\sqrt{3} - 1))^{1/2}$

which is easily verified, even on my lousy calculator. This expression for $\sqrt{2}$ just amounts to the difference of squares $3 - 1$, but it's cute. Note also that the same expression applies for any pair $(n, n+1)$ of finite ordinals. That is, all square roots $\sqrt{n}$ are simply expressible in terms of $\sqrt{n + 1}$. For the ordinal $3$, or any $n$ such that $n+1$ is a square, it turns out to be a tautology, because $2 = \sqrt{4}$ trivialises the expression.

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