# Transformations and Rapidity

1. Sep 8, 2009

### Piamedes

1. The problem statement, all variables and given/known data

I'm not sure if this belongs in this section or in one of the physics homework sections. If it has been misposted please move it to the proper area.

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

$$x^{'} = \frac{x-vt}{\sqrt{1-v^2}}$$

$$t^{'} = \frac{t-vx}{\sqrt{1-v^2}}$$

Verfiy that s, the space-time interval is same for both:

$$s^2 = t^2 - x^2 = t^{'}^2 - x^{'}^2 = s^{'}$$

Show that if we parametrize the transformation terms of the rapidity $$\theta$$,

$$x^{'} = x\cosh{\theta} - t\sinh{\theta}$$
$$t^{'} = t\cosh{\theta} - x\sinh{\theta}$$

the space-time interval will be automatically invariant under this transformation thanks to an identity satisfied by hyperbolic functions. Relate $$\tanh{\theta}$$ to the velocity.

The question has three more parts, but they all just build on this aspect. My major problem here is that I do not understand what the term "rapidity" means. I solved the first part of the question by just substituting the prime values of x and t into the difference of squares equation and showed their equality. However for this second part I don't even know where to start.

I tried plugging in the equations relating rapidity to the remaining variables and came up with nothing. If someone could perhaps explain what the concept of rapidity is I would be most grateful.

I had considered taking a derivative of the rapidity equations, but didn't know if that were possible because I don't know if $$\theta$$ varies with regard to x or t

Thanks for any help

2. Sep 8, 2009

### Piamedes

Sorry for the double post, but after some more thought I think I may have answered the second part of the question.

By squaring both rapidity equations, one for x prime and the one for t prime, and then subtracting the two if reduces to the original differences of squares equation. I think this answer that part of the question, but since I still do not understand what rapidity is, it may just be a shot in the dark.

After doing a little more research on the internet I discovered that rapidity is defined as $$\arctanh{\frac{v}{c}}$$. So with that in mind I looked back at the transformation equations and compared them with the original equations. By assuming the independence of x and t I got these two equations:

$$\frac{x}{\sqrt{1-v^2}} = x\cosh{\theta}$$

$$\frac{vt}{\sqrt{1-v^2}} = t\sinh{\theta}$$

Which reduced to:

$$\cosh{\theta} = \frac{1}{\sqrt{1-v^2}}$$
and
$$\sinh{\theta} = \frac{v}{\sqrt{1-v^2}}$$

Dividing those two equations gave me

$$\tanh{\theta} = v$$

does this all make sense? Or am I completely off?

Last edited: Sep 8, 2009
3. Sep 8, 2009

### gabbagabbahey

Looks good to me!

4. Sep 8, 2009

thanks