Isn't the invariance of the zero angle what we have been talking about? What "measurement of orientation" is included in this debate?
By orientation, I mean the "angle of the microfacet" in the OP.
Good, ready to point at the link where you uploaded them?
I said I'm good to go.
Notation:
$$\vec{P}(t)$$ is the position vector of wheel element
P at time t in
S.
$$\vec{P'}(t')$$ is the position vector of wheel element
P at time t' in
S'.
$$\hat{P_t}(t)$$ is the unit displacement vector tangent to wheel element
P at time t in
S.
$$\hat{P_t}'(t')$$ is the unit displacement vector tangent to wheel element
P at time t' in
S'.
$$\vec{v_p}(t)$$ is the velocity vector of wheel element
P at time t in
S.
$$\vec{v_p}'(t')$$ is the velocity vector of wheel element
P at time t' in
S'.
$$\vec{T_1}(t,l)$$ is the position vectors of the elements of rod
T1 at time t in
S. $$l$$ is distance along the rod.
Let $$v$$ denote the magnitude of the velocity of
S' relative to
S (this velocity is in the direction of the positive x-axis).
Let $$t'_0$$ denote the time in
S' when the rod
T1 is touching wheel element
P.
Pete's method recap:
Determine whether $$\vec{v_p}(t=0)$$ is a scalar multiple of (ie is parallel to) $$\hat{P_t}(t=0)$$.
Determine whether $$\vec{v_p}'(t'=t'_0)$$ is a scalar multiple of (ie is parallel to) $$\hat{P_t}'(t'=t'_0)$$.
$$\vec{v_p}(t=0)$$ and $$\vec{v_p}'(t'=t'_0)$$ can be derived from $$\vec{P}(t)$$, the worldline of
P.
$$\hat{P_t}(t=0)$$ is obvious, but can be formally derived from the definition the wheel at t=0 if necessary.
Rod
T1 is defined in terms of $$\hat{P_t}(t=0)$$.
$$\hat{P_t}'(t'=t'_0)$$ can be derived from the transformed
T1.
Calculations:
- Finding $$\vec{v_p}(t=0)$$
This is trivial, as are much of the following calculations, but spelling it out points toward a general solution for times other than t=0.
$$\begin{align}
\vec{P}(t) &= \begin{pmatrix} r\cos(\omega t) \\ r\sin(\omega t)\end{pmatrix} \\
\vec{v_p}(t) &= \frac{d}{dt} \vec{P}(t) \\
&= \begin{pmatrix} -r\omega\sin(\omega t) \\ r\omega\cos(\omega t) \end{pmatrix} \\
\vec{v_p}(t=0) &= \begin{pmatrix} 0 \\ r\omega \end{pmatrix}
\end{align}$$
- Finding $$\vec{v_p}'(t'=t'_0)$$
Transforming $$\vec{P}(t)$$ gives...
$$\vec{P'}(t) = \begin{pmatrix}\gamma(r\cos(\omega t) - vt) \\ r\sin(\omega t) \end{pmatrix}$$
Note that this is expressed as a function of t rather than t'. Given that
$$t = \frac{t'}{\gamma} - \frac{v}{c^2}r\cos(\omega t)$$
, I can't find a general closed form expression for $$\vec{P'}(t')$$. The same applies to the derivative:
$$\begin{align}
\vec{v_p}'(t) &= \frac{d}{dt'} \vec{P'}(t) \\
&= \begin{pmatrix}\frac{-r\omega\sin(\omega t) - v}{1 + \frac{vr\omega}{c^2}\sin(\omega t)} \\ \frac{r\omega\cos(\omega t)}{\gamma(1 + \frac{vr\omega}{c^2}\sin(\omega t))} \end{pmatrix}
\end{align}$$
But, we can still calculate the result, since we know that $$t=0$$ at $$t'=t'_0$$ on the worldine of P.
$$\begin{align}
\vec{v_p}'(t'=t'_0) &= \vec{v_p}'(t=0) \\
&= \begin{pmatrix}-v \\ r\omega / \gamma \end{pmatrix}
\end{align}$$
- Finding $$\hat{P_t}(t=0)$$
Trivially, $$\hat{P_t}(t=0) = \begin{pmatrix}0 \\ 1 \end{pmatrix}$$
If necessary, this can be formally derived by considering the direction of the displacement vector between P and a simultaneous wheel element a short distance away, and taking the limit as the distance from P approaches zero.
- Finding $$\hat{P_t}'(t'=t'_0)$$
Rod T1:
$$\begin {align}
\vec{T_1}(t,l) &= \vec{P}(0) + l\hat{P_t}(0) + t\vec{v_p}(0) \\
&= \begin{pmatrix} r \\ l + tr\omega\end{pmatrix}
\end{align}$$
Transforming to S':
$$\vec{T1'}(t',l) = \begin{pmatrix} \frac{r}{\gamma} - vt' \\ l + r\omega(\frac{t'}{\gamma} +\frac{vr}{c^2})\end{pmatrix}$$
$$\hat{P_t}'(t'=t'_0)$$ is parallel to the displacement vector between two points on the rod at $$t'=t'_0$$ with different values of $$l$$:
$$\begin{align}
\vec{T_1}'(t'=t'_0,l=l_0) &= \begin{pmatrix} \frac{r}{\gamma} - vt'_0 \\ l_0 + r\omega(\frac{t'_0}{\gamma} +\frac{vr}{c^2})\end{pmatrix} \\
\vec{T_1}'(t'=t'_0,l=l_1) &= \begin{pmatrix} \frac{r}{\gamma} - vt'_0 \\ l_1 + r\omega(\frac{t'_0}{\gamma} +\frac{vr}{c^2})\end{pmatrix} \\
\hat{P_t}'(t'=t'_0) &= \frac{\vec{T_1}'(t'=t'_0,l=l_0) - \vec{T_1}'(t'=t'_0,l=l_1)}{\left\|\vec{T_1}'(t'=t'_0,l=l_0) - \vec{T_1}'(t'=t'_0,l=l_1)\right\|
\end{align}$$
$$ \hat{P_t}'(t'=t'_0) = \begin{pmatrix}0 \\ 1 \end{pmatrix}$$
Results
$$\vec{v_p}(t=0) = \begin{pmatrix} 0 \\ r\omega \end{pmatrix}$$
...is parallel to...
$$\hat{P_t}(t=0) = \begin{pmatrix}0 \\ 1 \end{pmatrix}$$
i.e., in
S at t=0, the microfacet at
P is parallel to the velocity of
P.
$$\vec{v_p}'(t'=t'_0) = \begin{pmatrix}-v \\ \frac{r\omega}{\gamma} \end{pmatrix}$$
...is not parallel to...
$$\hat{P_t}'(t'=t'_0) = \begin{pmatrix}0 \\ 1 \end{pmatrix}$$
i.e. in
S' at t'=t'_0, the microfacet at
P is
not parallel to the velocity of
P.
