And now a more formal approach.
Consider a flat surface
A in inertial motion in a 2+1 spacetime.
Choose an inertial reference frame
S(x,y,t), such that in
S:
- A is at rest.
- The angle between the surface A and the x-axis is $$\theta_A$$.
- A intersects (x,y,t)=(0,0,0)
A is thus defined by:
$$y = x \tan(\theta_A)$$
An inertial particle
P passes through (x,y,t)=(0,0,0) with velocity $$\vec{V_p}$$.
The angle between $$\vec{V_p}$$ and the x-axis is $$\theta_P$$.
The worldline of
P is thus defined by:
$$\begin{align}
x &= t |\vec{V_p}| \cos(\theta_P) \\
y &= t |\vec{V_p}| \sin(\theta_P)
\end{align}$$
A reference frame
S'(x',y',t') has velocity v in the x-direction relative to
S, with parallel axes and coincident origin.
$$\begin{align}
\beta &= v/c \\
\gamma &= 1/\sqrt{1-\beta^2}
\end{align}$$
Transforming
A, we find:
$$y' = \gamma\tan(\theta_A) (x' + vt')$$
The angle between
A' and the x'-axis is thus defined by:
$$\tan(\theta'_A) = \gamma\tan(\theta_A)$$
Here, I note that if $$\tan(\theta_A) = 0$$, then $$\tan(\theta'_A)=0$$ as well.
Transforming the worldine of
P, we find:
$$\begin{align}
x' &= t'\frac{|\vec{V_p}| \cos(\theta_P) - v}{1 - v|\vec{V_p}| \cos(\theta_P)/c^2} \\
y' &= \frac{t'}{\gamma}\frac{|\vec{V_p}| \sin(\theta_P)}{{1 - v|\vec{V_p}|\cos\theta_P/c^2}
\end{align}$$
The angle between $$\vec{V'_p}$$ and the x'-axis is thus given by:
$$\tan(\theta'_P) = \frac{|\vec{V_p}| \sin(\theta_P)}{\gamma(|\vec{V_p}| \cos(\theta_P) - v)}$$.
Again, I note that if $$\tan(\theta_P) = 0$$, then $$\tan(\theta'_P)=0$$ as well, and I also note that this transformation is not the same as for $$\theta_A$$.
The angle $$\phi$$ between
A and $$\vec{V_p}$$ is given by:
$$\begin{align}
\phi &= \theta_A - \theta_P \\
&= \frac{\tan(\theta_A) - \tan(\theta_P)}{1+\tan(\theta_A)\tan(\theta_P)}
\end{align}$$
The angle $$\phi'$$ between
A' and $$\vec{V'_p}$$ is given by:
$$\begin{align}
\phi' &= \theta'_A - \theta'_P \\
\tan\phi' &= \frac{\tan(\theta'_A) - \tan(\theta'_P)}{1+\tan(\theta'_A)\tan(\theta'_P)} \\
\end{align}$$
Note very well:
Due to the different transformations of $$\theta_A$$ and $$\theta_P$$, $$\tan\phi = 0$$ does
not imply that $$\tan(\phi')=0$$, except in special cases such as $$\theta_A = \theta_P = 0$$.
---------------------------
Notes:
Let me know if something needs clarification, doesn't seem to make sense, or is just wrong.
I have a more general case for a surface moving with arbitrary velocity, but this simpler case is sufficient.
No, I'm not going to do a (x,y,z,t) derivation. Too hard, no returns.
