# Minkowski diagram - the angle between axes

1. Sep 24, 2011

### 3nTr0pY

I was reading through my textbook and it said that the angle between the axes of two inertial frames, one stationary and one moving at velocity v is supposed to be tan^-1(v/c). I assumed this would be easy to show, but after spending a couple of hours on this probably trivial problem, I can't for the life of me get the right answer.

I'm using the transformation:

ct' = ctcosh(a) -xsinh(a)
x' = -ctsinh(a) + xcosh(a)
y' = y
z' = z

where a is the 'rapidity function', tanh^-1(v/c) (not that I really understand what that is)

What I've done is choose a value of ct' and kept x' = 0 i.e. (0, ct'), because I'm supposed to be looking at the axes. I'm supposed to convert this into the other inertial frame, I think, but I'm really confused...

2. Sep 24, 2011

### 3nTr0pY

Ok, I solved it using bog standard Lorentz transformations. As (annoyingly) easy as that.

But I'm still really confused as to why I couldn't get the transformation I mentioned in my first post to work (the Lorentz boost). I don't really get how it works/what the point of it is, so any clarification on that would be helpful.

3. Sep 25, 2011

### Bill_K

The x' axis is where t'=0. So set t'=0 and solve for the slope of this line, ct/x, and I think you'll get the answer you're looking for.