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.(adsbygoogle = window.adsbygoogle || []).push({});

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

[itex]

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

[/itex]

For this to be true:

[itex]

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

[/itex]

[itex]

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

[/itex]

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

[itex]

x=ct'\sinh{\Psi}

[/itex]

[itex]

ct=ct'\cosh{\Psi}

[/itex]

and dividing the equations yields:

[itex]

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

[/itex]

But the speed [itex]V[/itex] is given by [itex]V=\frac{x}{t}[/itex] so:

[itex]

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

[/itex]

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

[itex]

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

[/itex]

[itex]

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

[/itex]

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

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# Hyperbolic Triangles

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