Rapidity and 4-velocity
snoopies622 said:
A long time ago I was taught that an angle is two rays with a common origin that is measured with a protractor, so I am having trouble with 'hyperbolic' angles. If a rocket ship moves away from me in the x direction at v=0.5c and I decide to draw this on a spacetime diagram (x vs. ct in my frame of reference) the angle the ship's world line makes with mine is
<br />
\theta = arctan (0.5) \approx 26.6 ^ {\circ}<br />
but the rapidity between us is
<br />
\phi = arctanh (0.5) \approx 31.5 ^ {\circ}<br />.
Is there a way to draw spacetime diagrams such that the angles that appear are the rapidities? Is a 'hyperbolic angle' really an angle?
"Hyperbolic angle" is not literally an angle in the same way that "spacetime interval"
ds is not literally a distance, but, nevertheless, we can mentally picture it as a distance and in many (but not all) ways it resembles a distance. You can't measure a spacetime interval directly using a ruler on a spacetime diagram, and similarly you can't measure a rapidity using a protractor either.
In 3D Euclidean geometry, distance, angle and scalar product are related by
\textbf{u} \cdot \textbf{v} = ||\textbf{u}|| \, ||\textbf{v}|| \, \cos\theta
In 4D Minkowski geometry, the spacetime interval, rapidity and metric are related by
\textbf{U} \cdot \textbf{V} = ||\textbf{U}|| \, ||\textbf{V}|| \, \cosh\phi
where \textbf{U} = (U_t, U_x, U_y, U_z) and
V are both timelike 4-vectors (e.g. 4-velocities),
\textbf{U} \cdot \textbf{V} = g_{\mu\nu}U^{\mu}V^{\nu} = U_tV_t - U_xV_x - U_yV_y - U_zV_z
and ||
U|| denotes the spacetime interval
||\textbf{U}|| = \sqrt{\textbf{U} \cdot \textbf{U}}
When a velocity is along the
x-axis (i.e.
U lies in the
tx-plane and the observer's
V lies along the
t-axis) we can also write
\textbf{U} = (c\cosh\phi, c\sinh\phi, 0, 0)
In 2D Euclidean geometry, rotation through an angle \theta is represented by the matrix
<br />
\left(\begin{array}{cc}<br />
\cos\theta & -\sin\theta \\<br />
\sin\theta & \cos\theta \\<br />
\end{array}\right) <br />
In 2D (1+1) Minkowski geometry, the Lorentz transform for velocity c\tanh\phi is represented by the matrix
<br />
\left(\begin{array}{cc}<br />
\cosh\phi & \sinh\phi \\<br />
\sinh\phi & \cosh\phi \\<br />
\end{array}\right) <br />
You can see the
resemblance between angles in Euclidean geometry (with trig functions) and rapidities in Minkowski geometry (with hyperbolic functions), which is why the term "hyperbolic angle" is used.
How embarassing. You are exactly correct. I simply meant to agree with snoopies662's math and re-phrase the math into english, but I got the math to english translation exactly backwards. Sorry for the confusion.