Stevemg, the Minkowski bilinear form g (I won't call it a scalar product, or an inner product, since it isn't positive definite) is defined by
g(x,y)=x^T\eta y
I usually take
\eta=\begin{pmatrix}-1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{pmatrix}
to be the definition of \eta, but you could choose the opposite sign if you want to. This wouldn't change any of the physics. It would just change a few signs here and there. The two different "norms" (they're not really norms, since they're not positive definite) are just different choices of what sign to use in the definition of \eta.
My definition above is more common in GR books, and the opposite sign is more common in QFT books. Some people prefer to write \langle x,y\rangle instead of g(x,y).
Since you have taken an interest in the concept of rapidity, you may find the notes I made for myself useful. Note that I'm using units such that c=1. Keeping that c around is like putting something sharp in one of your shoes before you go out running. It's a pain with no benefits. This is what I wrote in my notes:
In special relativity, we define the
rapidity \phi of a particle moving with velocity v by \tanh\phi=v. Note that the rapidity and the velocity are approximately the same when the velocity is small. To be more precise, in the limit \phi\rightarrow 0, we have v=\phi+O(\phi^2).
The concept of rapidity is useful because it's easier to "add" rapidities than velocities. If the velocity of frame B in frame A is v, and the velocity of a particle in frame B is v', then the velocity of the particle in frame A is
v\oplus v'=\frac{v+v'}{1+vv'}
If the rapidity of frame B in frame A is \phi, and the rapidity of a particle in frame B is \phi', then the rapidity of the particle in frame A is just \phi+\phi'. This follows from the velocity addition formula above, the definition of rapidity, and the identity
\tanh(\phi+\phi')=\frac{\tanh\phi+\tanh\phi'}{1+\tanh\phi\ \tanh\phi'}
The definition of rapidity also implies that
\cosh\phi=\gamma=\frac{dt}{d\tau}
\sinh\phi=\gamma v=\frac{dx}{d\tau}Proof:
v^2=\tanh^2\phi=\frac{\sinh^2\phi}{\cosh^2\phi}=\frac{\cosh^2\phi-1}{\cosh^2\phi}=1-\frac{1}{\cosh^2\phi}
\frac{1}{\cosh^2\phi}=1-v^2=\frac{1}{\gamma^2}
\cosh\phi=\gamma
\sinh^2\phi=\cosh^2-1=\gamma^2-1=\frac{1}{1-v^2}-\frac{1-v^2}{1-v^2}=\frac{v^2}{1-v^2}=\gamma^2v^2
\sinh\phi=\gamma v
\tau=\int\sqrt{dt^2-dx^2}=\int dt\sqrt{1-\dot x^2}=\int\frac{dt}{\gamma}
\frac{d\tau}{dt}=\frac 1 \gamma
\frac{dt}{d\tau}=\gamma=\cosh\phi
\frac{dx}{d\tau}=\frac{dt}{d\tau}\frac{dx}{dt}=\gamma v=\sinh\phi
These results enable us to express an arbitrary proper and orthochronous Lorentz transformation as a hyperbolic rotation by an "angle" equal to the rapidity.
\Lambda=\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}=\begin{pmatrix}\cosh\phi & -\sinh\phi\\ -\sinh\phi & \cosh\phi\end{pmatrix}
stevmg said:
Cyosis -
Would this be the correct representation with the hyperbolic functions?<br />
\begin{pmatrix}<br />
ct' \\ x'<br />
\end{pmatrix}<br />
<br />
=<br />
<br />
\begin{pmatrix}<br />
\cosh \(arctanh(v/c) & -\sinh\(arctanh(v/c) \\<br />
-\sinh\(arctanh(v/c) & \cosh\(arctanh(v/c)<br />
\end{pmatrix}<br />
<br />
\begin{pmatrix}<br />
ct \\ x<br />
\end{pmatrix}<br />
If this is the case, then all one would have to do is look up the arctanh of v/c or (\beta) and plug that into the cosh and sinh to get your matrix. That can be done on a scientific calculator directly.
Yes, but if you look at my last equation before the quote, I'm sure you'll see an easier way.
