Unanswered questions:

What general rule or definition are you using to conclude that t and $$\theta$$ are independent?

Would you object if I opened a thread in Physics and Maths about partial differentiation and the independence of variables in this situation?

Please stop telling me what I am doing.

Please stop ignoring what I point out.

It is very simple, really: t is NOT a function of $$\theta$$.

Only if you leave t' out of the equation.

If you include t', then t can potentially be expressed as a function of t' and $$\theta$$.

Similarly for t'.

In this equation:

$$t'=\gamma(t-vr\cos(\omega t + \theta)/c^2)$$

t' is dependent on t and $$\theta$$.

In this equation:

$$x' = \gamma r\cos(\omega\gamma t' + \omega\gamma vx'/c^2 + \theta) - v\gamma^2t' - v^2\gamma^2x'/c^2$$

t' and $$\theta$$ are independent.

For any value of $$\theta$$, t' can have any value, and vice versa.

$$t'=\gamma(t-vx/c^2)$$ implies:

$$\frac{dt'}{d \theta}=\gamma( \frac{\partial t}{\partial \theta}-v/c^2 \frac {\partial x}{\partial \theta})=-\gamma vr/c^2 sin (\omega t +\theta)$$

Wrong.

You have t' as a function of variables t and x, where (it is implied) x is a function of t and $$\theta$$.

You can't get the

total derivative $$\frac{dt'}{d \theta}$$, unless you define t as a function of $$\theta$$.

If t and $$\theta$$ are independent, then the expression $$\frac{dt'}{d\theta}$$ just doesn't make sense.

You can get the

partial derivative by treating t as a constant:

$$\begin{align}

\frac{\partial t'}{\partial \theta} &= -\gamma v/c^2 \frac {\partial x}{\partial \theta} \\

&= -\gamma vr/c^2 sin (\omega t +\theta)

\end{align}$$

Note that $$\frac{dt}{d\theta}$$ and $$\frac{\partial t}{\partial \theta}$$ are undefined, unless t is considered constant or a function of $$\theta$$.

As I said before, if a and b are independent, then $$da/db$$ is undefined (it is

*not* zero).

I am not "simply declaring", they are independent variables by virtue of how the rotating wheel is described mathematically. I explained that several times already.

You've

*said * it several times, but you haven't explained your logic.

What is it about the wheel equations that make $$\theta$$ and t independent?

No, I cannot, since there is no definition of how $$\theta'$$ transforms between frames.

Determining this transform is equivalent to the problem that we are trying to solve in this debate.

The angle we're chasing at the core of the debate is the angle between the tangent vector and the velocity vector.

But that's different.

$$\theta$$ is simply a continuous parameter that distinguishes different points around the wheel.

In

**S**, it corresponds to the angle between the x-axis and a point on the wheel at t=0, but that particular definition is incidental to its purpose of acting as a tangent-defining parameter - all that really matters is that a particular value of $$\theta$$ corresponds to a unique physical wheel element, and that the smaller the difference in value of $$\theta$$ between two wheel elements, the closer together those wheel elements are.

In

**S'**, it can serve the same purpose - a small change in $$\theta$$ still corresponds to a small change along the wheel, so it still works for the purpose of defining the tangent.