Hyperbolic Triangles: Deriving Relationships Algebraically or Geometrically

[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:

<br /> (ct)^2 - x^2 = (ct&#039;)^2 - (x&#039;)^2<br />

For this to be true:

<br /> x = x&#039;\cosh{\Psi}+ct&#039;\sinh{\Psi}<br />
<br /> ct=x&#039;\sinh{\Psi}+ct&#039;\cosh{\Psi}<br />

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&#039; frame moving these can be reduced to:

<br /> x=ct&#039;\sinh{\Psi}<br />
<br /> ct=ct&#039;\cosh{\Psi}<br />

and dividing the equations yields:

<br /> \tanh{\Psi}=\frac{x}{ct}<br />

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

<br /> \tanh{\Psi}=\frac{V}{c}<br />

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:

<br /> \sinh{\Psi}=\frac{\frac{V}{c}}{\sqrt{1-({\frac{V}{c}})^2}}<br />
<br /> \cosh{\Psi}=\frac{1}{\sqrt{1-({\frac{V}{c}})^2}}<br />

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

 

[PeterDonis]

Try the Wikipedia page on hyperbolic trig functions:

http://en.wikipedia.org/wiki/Hyperbolic_function

 

[xox]

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&#039; moving at speed V relative to an inertial frame K in the x-direction. In order for invariant intervals we require:

<br /> (ct)^2 - x^2 = (ct&#039;)^2 - (x&#039;)^2<br />

For this to be true:

<br /> x = x&#039;\cosh{\Psi}+ct&#039;\sinh{\Psi}<br />
<br /> ct=x&#039;\sinh{\Psi}+ct&#039;\cosh{\Psi}<br />

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&#039; frame moving these can be reduced to:

<br /> x=ct&#039;\sinh{\Psi}<br />
<br /> ct=ct&#039;\cosh{\Psi}<br />

and dividing the equations yields:

<br /> \tanh{\Psi}=\frac{x}{ct}<br />

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

<br /> \tanh{\Psi}=\frac{V}{c}<br />

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:

<br /> \sinh{\Psi}=\frac{\frac{V}{c}}{\sqrt{1-{\frac{V}{c}}^2}}<br />
<br /> \cosh{\Psi}=\frac{1}{\sqrt{1-{\frac{V}{c}}^2}}<br />

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

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

 

[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.