# Hyperbolic Triangles

1. May 16, 2014

### electricspit

Hello, I'm going through Landau and Lifshitz "The Classical Theory of Fields" this summer with a friend and in section 4 I've come to a bit of a math problem.

Assume you have an inertial frame $K'$ moving at speed $V$ relative to an inertial frame $K$ in the $x$-direction. In order for invariant intervals we require:

$(ct)^2 - x^2 = (ct')^2 - (x')^2$

For this to be true:

$x = x'\cosh{\Psi}+ct'\sinh{\Psi}$
$ct=x'\sinh{\Psi}+ct'\cosh{\Psi}$

Where $\Psi$ is the angle of rotation in the $xt$ plane. Which makes sense. Now if we just look at the origin of the $K'$ frame moving these can be reduced to:

$x=ct'\sinh{\Psi}$
$ct=ct'\cosh{\Psi}$

and dividing the equations yields:

$\tanh{\Psi}=\frac{x}{ct}$

But the speed $V$ is given by $V=\frac{x}{t}$ so:

$\tanh{\Psi}=\frac{V}{c}$

The next part is what is a bit confusing. If this were regular trigonometry to find both $\sin{\Psi}$ and $\cos{\Psi}$ would just require the construction of a triangle. In this case however, they end up with very, very similar results except where the hypoteneuse would be they end up with something like this:

$\sinh{\Psi}=\frac{\frac{V}{c}}{\sqrt{1-({\frac{V}{c}})^2}}$
$\cosh{\Psi}=\frac{1}{\sqrt{1-({\frac{V}{c}})^2}}$

If anyone can point me in the right direction of deriving these relationships either algebraically or using geometry that would be extremely helpful!

Last edited: May 16, 2014
2. May 16, 2014

### Staff: Mentor

3. May 16, 2014

### xox

$$\sinh{\Psi}=\frac{\tanh{\Psi}}{\sqrt{1-\tanh^2{\Psi}}}$$
$$\cosh{\Psi}=\frac{1}{\sqrt{1-\tanh^2{\Psi}}}$$

4. May 16, 2014

### electricspit

Okay hmm, assuming the Wikipedia section on "Comparison with circular functions" is correct then this is what I make of it all:

http://imgur.com/4eL9du1

Sorry for the mess, I've been working all day and am too tired to be perfect.