**3.1.1.1.1 Lorentz transformation of $$\hat{P_t}$$**
There is no requirement of events to be simultaneous in either definition.

From earlier in the thread:

A displacement vector is tangent to the wheel at a given point at a given instant if it is a scalar multiple of the displacement between two points on a straight rod tangent to the wheel at that point at that instant.

So,

- $$\hat{P_t}(0)$$ is defined as a unit vector between two points on rod
**T1** at t=0 in **S**.
- $$\hat{P_t}'(t'_0)$$ is defined as a unit vector between two points on rod
**T1** at $$t'=t'_0 \$$ in **S**.

Neither **tangent** vector is stationary so you have no reason to impose either $$dt=0$$ or $$dt'=0$$

I'm not saying anything about dt or dt' in relation to $$\hat{P_t}$$, because I'm not using differentials.

I am saying that d

**r** is not $$\hat{P_t}(0)$$, and that d

**r'** is not $$\hat{P_t}'(t'_0)$$.

In fact, I don't think that your using d

**r** appropriately anyway.

You seem to be using it to mean "a small change in

**r**", but it is implied that the change you mean is change

*over time*. In other words, you're talking specifically about a small change in the direction d

**r**/dt, which does not simply equate to d

**r**.

Now.

I do agree that $$d\vec{r}/dt$$ is tangent to the wheel at

**P** in

**S**.

But, I do

*not* agree that $$d\vec{r}'/dt'$$ is tangent to the wheel at

**P** in

**S'**.

That's the whole point of this debate.

In your approach to my method, you are in effect simply

*declaring* $$d\vec{r}'/dt'$$ to be a tangent to the wheel at

**P**, and you are ignoring the part of my method that describes how to prove tangency.

In my approach, I explicitly derive $$\hat{P_t}'(t'_0)$$ as tangent to the wheel at

**P** in

**S'** using

**T1**, which is a straight rod tangent to the wheel at the given point at the given time:

$$\begin{align}

\hat{P}(0) &= \begin{pmatrix}r \\ 0 \end{pmatrix} \\

\hat{P_t}(0) &= \begin{pmatrix}0 \\ 1 \end{pmatrix} \\

\vec{v_p}(0) &= \begin{pmatrix}0 \\ \omega r \end{pmatrix}

\end{align}$$

**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}

\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}$$

$$ \hat{P_t}'(t'=t'_0) = \begin{pmatrix}0 \\ 1 \end{pmatrix}$$