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:

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

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!

 

[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 $K'$ 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!

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

 

[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.

 

[sardar]

Hello there! I would like to provide a response to the content you have shared about hyperbolic triangles and deriving relationships algebraically or geometrically. First of all, I must commend you for taking on the challenge of studying Landau and Lifshitz's "The Classical Theory of Fields" this summer. It is a complex and advanced topic, and I am glad that you are seeking help when you encounter difficulties.

Now, let's address the specific problem you have mentioned. It seems like you are trying to understand the relationships between the x, ct, and ct' coordinates in the K and K' frames, and how they relate to the speed V. This is indeed a tricky problem, but it can be solved using both algebraic and geometric methods.

First, let's look at the algebraic approach. You have correctly derived the equations x = x'\cosh{\Psi}+ct'\sinh{\Psi} and ct=x'\sinh{\Psi}+ct'\cosh{\Psi}. These equations relate the x, ct, and ct' coordinates in the two frames using the angle of rotation \Psi. Now, if we rearrange these equations, we can obtain:

x' = (x-ct\tanh{\Psi})/\cosh{\Psi} and ct' = (ct-x\tanh{\Psi})/\sinh{\Psi}

Substituting these into the invariant interval equation (ct)^2 - x^2 = (ct')^2 - (x')^2, we get:

(ct)^2 - x^2 = \frac{(ct-x\tanh{\Psi})^2}{\sinh^2{\Psi}} - \frac{(x-ct\tanh{\Psi})^2}{\cosh^2{\Psi}}

Simplifying this equation, we get:

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

This shows that the invariant interval is indeed preserved, as it should be. Now, to find the relationships between \sinh{\Psi} and \cosh{\Psi}, we can use the definitions of hyperbolic sine and cosine functions:

\sinh{\Psi} = \frac{e^{\Psi} - e^{-\Psi}}{2} and \cosh{\Psi} = \frac{e^{\Psi} + e^{-\Psi}}{2}

Using these definitions, we can obtain the expressions