Lorentz Boost Help: Why Use Hyperbolic Functions?

[fys iks!]

I was reading a section on lorentz boosts and i need some help understanding what they did:

the book starts off by defining the line element dS where:

(dS)^2 = -(CΔt)^2 + dx^2 + dy^2 + dz^2

then they say: "consider the analogs of rotations in the (ct) plane. These transformations leave y and z unchanged but mix ct and x. The transformations with this character that leave the analogies of rotations of (3.9) but with trig functions replaced by hyperbolic functions because of the non euclidean nature of space time. Specifically

ct= [cosh(theta)]*[ct] - [sinh(theta)]*x
x = [sinh(theta)]*[ct] + {cosh(theta)]*x
y= y
z = z

and 3.9 was

x = cos(gamma)*x - sin(gamma)*y
y = sin(gamma)*x + cos(gamma) * y


So what i don't understand is why did they decide to use the hyperbolic functions in 4 dimensions?

 

[bcrowell]

So what i don't understand is why did they decide to use the hyperbolic functions in 4 dimensions?

If you use a transformation with an ordinary sin and cos in it, you get a rotation. If you keep on rotating the x-t plane, at some point you'll have rotated it so it's 180 degrees upside-down. At that point you've reversed the direction of time. This violates causality.

 

[fys iks!]

Thanks! That helped out a lot.

So by using hyperbolic sin/cos they were able to create the same shift without violating causality.

Do you have any links that further explain this in more detail?

 

[Mentz114]

So what i don't understand is why did they decide to use the hyperbolic functions in 4 dimensions?

You have not been paying attention, have you. An ordinary rotation in 3D space leaves the length of a vector unchanged, where the length element is $ds^2=dx^2+dy^2+dz^2$. Rotating this by $\theta$ gives in the xy plane gives

$$ds'^2= (cos(\theta)dx+sin(\theta)dy)^2+(sin(\theta)dx-cos(\theta)dy)^2+dz^2$$
and because $cos(\theta)^2+sin(\theta)^2=1$ it follows that $ds'=ds$.

In the ct-plane, we wish to preserve $ds^2=dt^2-dx^2-dy^2-dz^2$. If you do the calculation as above with cosh and sinh instead, you find $ds'=ds$ because $cosh(\theta)^2-sinh(\theta)^2=1$. The geometry of the cx plane is not Euclidean.

[posted simultaneously with bcrowell]

 

[bcrowell]

Thanks! That helped out a lot.

So by using hyperbolic sin/cos they were able to create the same shift without violating causality.

Do you have any links that further explain this in more detail?

http://www.lightandmatter.com/area1book6.html

 

[resaypi]

In 4 dimensional spacetime intervals of constant length are hyperbolas. Also it is sometimes easier to work with hyperbolic functions

$$tanh(\theta) = v/c$$

The addition of velocities is reduced to adding hyperbolic functions.