Why Does Relativity Use Hyperbolic Functions for Space-Time Transformations?

[Mzachman]

I wasn't sure if this should go here or the relativity section, but it is for a "physics math" course, so I figured I would have the best luck here.

Homework Statement

According to the Theory of Relativity, if an event happens at a space-time point (x,t) according to an observer, another moving relative to him at speed v (measured in untis in which the velocity of light c = 1) will ascribe it to the coordinates

x' = (x-vt)/sqrt(1-v^2)
t' = (t-vx)/sqrt(1-v^2)

Verify that s, the space-time interval is the same for both: s^2 = (t^2 - x^2) = (t'^2 - x'^2) = s'^2. Show that if we parametrize the transformation terms of the rapidity θ,

x' = x*coshθ - t*sinhθ
t' = t*coshθ - x*coshθ

the space-time interval will be automatically invariant under this transformation thanks to an identity satisfied by hyperbolic functions. Relate tanhθ to the velocity. Suppose a third observer moves relative to the second with speed v', that is, with rapidity θ'. Relate his coordinates (x'', t'') to (x,t) going via (x', t'). Show that the rapidity parameter θ'' = θ' + θ in obvious notation. (You will need to derive a formula for tanh(A + B).) Thus it is the rapidity, not the velocity that really obeys a simple addition rule. Show that if v and v' are small (in units of c), that this reduces to the daily life rule for addition of velocities. (Use the Taylor series for tanhθ.) This is an example of how hyperbolic functions arise naturally in mathematical physics.

Homework Equations

x' = (x-vt)/sqrt(1-v^2)
t' = (t-vx)/sqrt(1-v^2)

x' = x*coshθ - t*sinhθ
t' = t*coshθ - x*coshθ

θ'' = θ' + θ

tan(A + B) = (tanh(A) + tanh(B))/(1 + tanh(A)*tanh(B))

The Attempt at a Solution

I have already done the first part, showing where s^2 = s'^2, but after that I am totally lost. From the first part about rapidity onwards... I have been trying to understand for over an hour now and I have absolutely no idea where to start even (internet searches haven't helped either...). Thanks very much for any help.

Also, this is a hint my professor posted about this problem specifically:
"This problem has 4 parts (though they are not explicitly marked as separate parts). In the third part (starting from the bottom of p.26 to the top 4 lines of p.27), they define θ" and v" as those that relate (x",t") directly to (x,t), while θ' and v' relate (x",t") to (x',t') and θ and v relate (x',t') to (x,t). In this part, it is useful to realize that:

cosh(θ+θ') = cosh θ cosh θ' + sinh θ sinh θ'

sinh(θ+θ') = sinh θ cosh θ' + cosh θ sinh θ'

Don't forget the fourth part where you will consider the limit of small v and v'."Thank you very much!

 

[lanedance]

$$x' = \frac{(x-vt)}{\sqrt{1-v^2}} = x \frac{1}{sqrt(1-v^2)} - t\frac{v}{\sqrt{1-v^2}} = x cosh(\theta) - t sinh(\theta)$$

$$t' = \frac{(t-vx)}{\sqrt{1-v^2}} = x \frac{1}{sqrt(1-v^2)} - t\frac{v}{\sqrt{1-v^2}} = t cosh(\theta) - x sinh(\theta)$$

which gives
$$cosh(\theta) = \frac{1}{sqrt(1-v^2)}$$
$$sinh(\theta) = \frac{v}{sqrt(1-v^2)}$$

i would try & derive theta form those equations and the porperteis of hyperbolic functions (hint sinh(x) + cosh(x) = ?)

if you want to see how to write tex, click on one of the expressions

 

[lanedance]

updated post #2