Debate: Lorentz invariance of certain zero angles

[Tach]

Like I said, you're treating t as a constant in that derivative.

Please stop telling me what I am doing. It is very simple, really: t is NOT a function of $$\theta$$.

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

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

You're getting $$\frac{\partial t'}{\partial \theta}$$, not $$\frac{dt'}{d\theta}$$.
If a and b are independent of each other, then $$\frac{da}{db}$$ is undefined.

Nonsense, see above. Please look up the definition of total derivative.

Are you simply declaring that t and $$\theta$$ are independent?

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.

Can you derive $$\frac{\partial y'}{\partial \theta'}\frac{\partial \theta'}{\partial x'}$$?

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.

 

[Pete]

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.

 

[Tach]

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 $$\theta$$.

Nonsense, t' is a function of t and x whereby x IS a function of $$\theta$$ and t is NOT a function of $$\theta$$. Read the example, is self-explanatory.

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

Err, wrong. $$\theta$$ happens to be the angle between the positional vector and the x axis. Surely you should have recognized the equation of a circle in polar coordinates by now.

 

[Pete]

Unanswered questions:
What general rule or definition are you using to conclude that t and $$\theta$$ are independent?
What is it about the wheel equations that make $$\theta$$ and t independent?
Would you object if I opened a thread in Physics and Maths about partial differentiation and the independence of variables in this situation?

Nonsense, t' is a function of t and x whereby x IS a function of $$\theta$$ and t is NOT a function of $$\theta$$. Read the example, is self-explanatory.

Sorry, I can't make out your intended meaning.
Yes, we have t' as a function of t and x.
Yes, we have x as a function of $$\theta$$ (and t).
Yes, in this context t is independent of $$\theta$$.

That means you can't take a total derivative $$dt'/d\theta$$, because $$dt/d\theta$$ is undefined:

$$\begin{align}
\frac{dt'}{d\theta} &= \frac{\partial t'}{\partial x}\frac{dx}{d\theta} + \frac{\partial t'}{\partial t}\frac{dt}{d\theta} \\
&= \frac{\partial t'}{\partial x}(\frac{\partial x}{\partial \theta} + \frac{\partial x}{\partial t}\frac{dt}{d\theta}) + \frac{\partial t'}{\partial t}\frac{dt}{d\theta}
\end{align}
$$​

What example are you referring to?

Err, wrong. $$\theta$$ happens to be the angle between the positional vector and the x axis.

At t=0 only, like I said in the next sentence. Or if you prefer, it's the angle between the positional vector of a point on the circle and the positional vector of the point which was at (r,0) 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 corresponds to a unique physical wheel element, and that the smaller the difference in value of between two wheel elements, the closer together those wheel elements are.

 

[Tach]

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

Look, we need to break state, this is not constructive and it is leading nowhere.

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

I answered this several times, let's see if this explanation registers.
The stationary wheel is described by:

$$x=r cos (\theta)$$
$$y=r sin (\theta)$$

When the wheel starts rotating with angular speed $$\omega$$, each angle changes from $$\theta$$ to $$\omega t +\theta$$. There is nothing connecting the angle $$\theta$$ t the time $$t$$, hence the two variables are independent.
As an alternative, you can start with the equation of the stationary circle. When the circle rotates wrt frame S, the rotation is expressed as :

$$x_s=x cos (\omega t)+ y sin (\omega t)= r cos (-\omega t + \theta)$$
$$y_s=-x sin (\omega t)+ y cos (\omega t)= r sin (-\omega t + \theta)$$

Now, for the calculus part that you seem not to get either:

$$f=f(t,x(\theta))$$

Therefore:

$$\frac{df}{d \theta}=\frac{\partial f}{\partial t} \frac{dt}{d \theta}+\frac{\partial f}{\partial x} \frac{dx}{d \theta}=\frac{\partial f}{\partial t} 0+\frac{\partial f}{\partial x} \frac{dx}{d \theta}$$

Contrary to what you might think, $$\frac{dt}{d \theta}=0$$

Now, having said that, I do not plan to waste any more time on this basic stuff, it is irrelevant any way, since what is really needed, is not $$\frac{df}{d \theta}$$ but $$\frac{df}{d \theta'}$$ and I have just figured the way $$\theta'$$ transforms between frames.
If you want to do something constructive, we can talk about that.

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

Feel free to do so but count me out, I am not interested in wasting my time. You can start with the example that I have just given you above.

 

[Pete]

Look, we need to break state, this is not constructive and it is leading nowhere.

I answered this several times, let's see if this explanation registers.
The stationary wheel is described by:

$$x=r cos (\theta)$$
$$y=r sin (\theta)$$

When the wheel starts rotating with angular speed $$\omega$$, each angle changes from $$\theta$$ to $$\omega t +\theta$$. There is nothing connecting the angle $$\theta$$ t the time $$t$$, hence the two variables are independent.

Is there any more rigorous way of determining whether two variables are independent?
It seems that you're working on intuition rather than rigor.

Now, for the calculus part that you seem not to get either:

$$f=f(t,x(\theta))$$

Therefore:

$$\frac{df}{d \theta}=\frac{\partial f}{\partial t} \frac{dt}{d \theta}+\frac{\partial f}{\partial x} \frac{dx}{d \theta}$$

So far, the same as my post.

Contrary to what you might think, $$\frac{dt}{d \theta}=0$$

Whoops, wrong.
$$\frac{dt}{d \theta}=0$$
implies that t is a constant.

If t and $$\theta$$ are independent, then $$\frac{dt}{d \theta}$$ is undefined.

Now, having said that, I do not plan to waste any more time on this basic stuff, it is irrelevant any way, since what is really needed, is not $$\frac{df}{d \theta}$$ but $$\frac{df}{d \theta'}$$ and I have just figured the way $$\theta'$$ transforms between frames.

Great! Let's see it.

 

[Tach]

Is there any more rigorous way of determining whether two variables are independent?
It seems that you're working on intuition rather than rigor.

I just gave you another proof. Should be rigorous enough.

Whoops, wrong.
$$\frac{dt}{d \theta}=0$$
implies that t is a constant.

Nonsense, it implies that t is not a function of $$\theta$$

If t and $$\theta$$ are independent, then $$\frac{dt}{d \theta}$$ is undefined.

I don't know where you get this thing but it is wrong.

 

[Pete]

I just gave you another proof. Should be rigorous enough.

You said "There is nothing connecting the angle $$\theta$$ to the time t, hence the two variables are independent."

I don't call that rigorous, but let's explore the concept anyway.

What if we wrote an equation that added something to connect $$\theta$$ to t?
In that context, they would be dependent.
And what if we then derived a new equation that took the connection away?
In that context, they would be independent.

Well, that's what we did with t'.
It is t that connects t' to $$\theta$$.
With that connection removed, t' and $$\theta$$ are independent.
Look at this equation:

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

In that equation, there is nothing connecting the angle $$\theta$$ to the time t', hence the two variables are independent.

$$\frac{dt}{d\theta}=0$$ implies that t is a constant

Nonsense, it implies that t is not a function of $$\theta$$

Really?
What happens if you integrate it?

$$\begin{align}
\frac{dt}{d\theta} &= 0 \\
\int \frac{dt}{d\theta} d\theta &= \int 0 d\theta \\
t + C_1 &= C_2 \\
t &= C
\end{align}$$

If t and $$\theta$$ are independent, then $$dt/d\theta$$ is undefined.

I don't know where you get this thing but it is wrong.

I really don't understand how you conclude that it should be zero.
Is it something you read in a text?
It doesn't work out from the definition of the derivative - the definition implicitly requires one variable to be dependent on the other.

 

[Tach]

You said "There is nothing connecting the angle $$\theta$$ to the time t, hence the two variables are independent."

I don't call that rigorous, but let's explore the concept anyway.

What in the concept of a 2D rotation transformation applied to a circle (expressed in polar coordinates) don't you understand?

Really?

Really. I'll give it on more shot:

$$f=f(t,x(\theta))$$

Then, by definition, $$\frac{df}{d \theta}=\frac{\partial f}{\partial x} \frac{dx}{d \theta}$$

This is the last time I answer this basic stuff for you, I see that you have a whole thread going. PM me when you are ready to discuss the $$\theta$$ transformation.

 

[Pete]

What in the concept of a 2D rotation transformation applied to a circle (expressed in polar coordinates) don't you understand?

One, you're being rude.
Two, I understand it fine.
You don't seem to understand what I'm getting at.
Yes, I agree that it is intuitively obvious that t and $$\theta$$ are independent.
But I also think it's intuitively obvious that t' and $$\theta$$ are independent.
So, I'm trying to explore whether we can rigorously say whether two variables are independent.

Really. I'll give it on more shot:

$$f=f(t,x(\theta))$$

Then, by definition, $$\frac{df}{d \theta}=\frac{\partial f}{\partial x} \frac{dx}{d \theta}$$

By what definition?
And why did you ignore this proof:
$$\begin{align}
\frac{dt}{d\theta} &= 0 \\
\int \frac{dt}{d\theta} d\theta &= \int 0 d\theta \\
t + C_1 &= C_2 \\
t &= C
\end{align}$$

This is the last time I answer this basic stuff for you

Again, rude.
We're supposed to be engaged in mutual exploration, but you don't seem interested in exploring... please, don't let your pride hold you back.

 

[Tach]

And why did you ignore this proof:
$$\begin{align}
\frac{dt}{d\theta} &= 0 \\
\int \frac{dt}{d\theta} d\theta &= \int 0 d\theta \\
t + C_1 &= C_2 \\
t &= C
\end{align}$$

...because it is incorrect, you are trying to generalize incorrectly a conclusion from univariate functions to multivariate ones. The correct generalization is:

$$\begin{align}
\frac{dt}{d\theta} &= 0 \\
t=t(u,v,....) \\
\int \frac{dt}{d\theta} d\theta &= \int 0 d\theta \\
t + C_1(u,v,....) &= C_2 \\
t &= C(u,v,...)
\end{align}$$

This might remove your confusion on the subject of total derivatives. It is a rigorous treatment of total derivatives from base principles. I apologize if I came across as rude but so did you, every explanation that I gave you was met with "wrong" , though it was correct.

 

[Pete]

...because it is incorrect, you are trying to generalize incorrectly a conclusion from univariate functions to multivariate ones.

My understanding is that the expression $$\frac{dt}{d\theta}$$ only makes sense in a context where t depends only on $$\theta$$ (directly or indirectly).
It seems to me that you are trying to derive a total derivative in a context where you can only derive a partial derivative.

$$\begin{align}
\frac{dt}{d\theta} &= 0 \\
t=t(u,v,....) \\
\int \frac{dt}{d\theta} d\theta &= \int 0 d\theta \\
t + C_1(u,v,....) &= C_2 \\
t &= C(u,v,...)
\end{align}$$

I think this step is incorrect:

$$\int \frac{dt}{d\theta} d\theta &= t + C_1(u,v,....)$$

This would be correct:

\int \frac{\partial t}{\partial \theta} d\theta &= t + C_1(u,v,....)

This might remove your confusion on the subject of total derivatives. It is a rigorous treatment of total derivatives from base principles.

It seems to me that you are at least as confused as I on this topic.

Case 2 starts with t and $$\theta$$ as independent variables, then treats t as a function of $$\theta$$, such that:
$$t(\theta) = t(\theta + \Delta \theta)$$

I don't think that is a valid step. In effect, you are treating t as a constant, not a variable.
Can you show me a similar step in a maths text?

I apologize if I came across as rude but so did you, every explanation that I gave you was met with "wrong" , though it was correct.

The thing is that you presented essentially the same explanation in several different ways, without exploring my objections. Even now, you're still assuming that your explanations were correct and I just misunderstood them.
You're still using the same assumption that I objected to many posts ago, that if two variables are independent, then one can be treated as a constant function of the other.
You haven't tried to support that assumption, you've just presented it again in a different way.

 

[Tach]

Case 2 starts with t and $$\theta$$ as independent variables, then treats t as a function of $$\theta$$, such that:
$$t(\theta) = t(\theta + \Delta \theta)$$

You managed to get this just as wrong as the rest. Case 2 says clearly that t is INDEPENDENT of $$\theta$$. That means that one can use the general formalism developed for Case 1 and use the fact that $$t(\theta+\delta \theta)=t(\theta)=t$$ (i.e, that t does NOT depend on $$\theta$$).


This is precisely why I don't want to waste my time trying to correct your misconceptions. I prepared a rigorous presentation, you did not understand it , yet you keep claiming "wrong".

 

[Pete]

You managed to get this just as wrong as the rest.

That's what I'm talking about, Tach. Not on.

Case 2 says clearly that t is INDEPENDENT of $$\theta$$.

As I said.

That means that one can use the general formalism developed for Case 1 and use the fact that $$t(\theta+\delta \theta)=t(\theta)=t$$

No, that makes no sense.
In Case 2, t is not a function of $$\theta$$. $$t(\theta)$$ is meaningless.

This is precisely why I don't want to waste my time trying to correct your misconceptions.

I suggest that your time would be more productively spent on testing your own understanding.
I'm tested my own in the [thread=112351]Independent variables and partial differentiation[/thread] thread, and I'm growing more and more suspicious that you just don't know what you're talking about.

I prepared a rigorous presentation, you did not understand it , yet you keep claiming "wrong".

I suggest for your consideration the possibility that you may, in fact, be wrong.

-------
This discussion hasn't progressed for a while.
Perhaps we could change tack and examine the low velocity case, i.e. consider $$\frac{\partial y'}{partial \theta}\frac{\partial \theta}{\partial x'}$$ and $$\frac{\partial y'}{partial t'}\frac{\partial t'}{\partial x'}$$ under the transforms:
$$\begin{align}x' &= x - vt \\
y' &= y \\
t' &= t \\
\theta' &= \theta
\end{align}$$

I haven't worked it through yet, but I think we should be able to do it while avoiding the problems that we a bogged down in now.

 

[Tach]

That's what I'm talking about, Tach. Not on.

As I said.

No, that makes no sense.
In Case 2, t is not a function of $$\theta$$. $$t(\theta)$$ is meaningless.

...which is precisely what is expressed in the basic math that you seem totally unable to follow.

I suggest that your time would be more productively spent on testing your own understanding.
I'm tested my own in the [thread=112351]Independent variables and partial differentiation[/thread] thread, and I'm growing more and more suspicious that you just don't know what you're talking about.

I can see that people tried to set you straight and you still don't get it. So, I made an even simpler version , from base principles. If you manage to misunderstand this, we are done, there is no point in continuing.

I suggest for your consideration the possibility that you may, in fact, be wrong.

This is basic stuff, Pete, if you don't get this, then there is no point in continuing.

I haven't worked it through yet, but I think we should be able to do it while avoiding the problems that we a bogged down in now.

Not if you don't understand how multivariate functions are differentiated.

$$\theta' = \theta$$

$$\theta' \ne \theta$$, I already told you that.

 

[Pete]

...which is precisely what is expressed in the basic math that you seem totally unable to follow.

You seem to be agreeing that $$t(\theta)$$ is meaningless.
Is that what you meant?

I can see that people tried to set you straight and you still don't get it.

I'm not convinced you've actually read the thread.
James had some of the same thoughts as you, but wasn't definite about it. Trippy made a contribution, but again didn't pretend to be authoritative.
I'm inclined to give most weight to temur, who I believe has the best maths background, but I might be biased because what he says matches what I've been saying to you.

So, I made an even simpler version , from base principles. If you manage to misunderstand this, we are done, there is no point in continuing.

This:

By definition:
$$\frac{df}{d\theta} = \lim_{\Delta\theta \rightarrow 0}\frac{f(x(\theta+\Delta\theta),t) - f(x(\theta),t)}{\Delta\theta}$$

...is actually using the definition of the partial derivative:

$$\frac{\partial f}{\partial x_m} = \lim_{h\rightarrow0}\frac{f(x_1,...,x_m+h,...,x_n)-f(x_1,...,x_m,...,x_n)}{h}$$

(use $$x_1 = \theta, x_2 = t$$)
In other words, you're setting t constant and varying $$\theta$$ - that's a partial derivative.

This is basic stuff, Pete, if you don't get this, then there is no point in continuing.

I wouldn't call multivariate calculus basic. There's no shame in learning something new.
Look, you obviously are beyond considering that I could be right... but have you tried checking your understanding against a textbook, or with a professor, or a tutor, or any of the maths people on sciforums?
Or are you so convinced that you're right that you don't feel the need to double check?

Not if you don't understand how multivariate functions are differentiated.

We agree on enough for the simpler case, because there is no problem with whether t' is independent - we can use t in both cases.

$$\theta' \ne \theta$$, I already told you that.

In the low velocity limit, $$\theta'=\theta$$.
Look:

$$\begin{align}
x'&=x-vt \\
y'&=y \\
t'&=t \\
\\
r_x &= r \cos (\omega t + \theta) \\
r_y &= r \sin (\omega t + \theta) \\
\\
r_x' &= r \cos (\omega t + \theta) - vt \\
r_y' &= r \sin (\omega t + \theta)
\end{align}
$$

In both frames, $$\theta$$ is a true angle around the wheel axis.

 

[Tach]

You seem to be agreeing that $$t(\theta)$$ is meaningless.
Is that what you meant?


I'm not convinced you've actually read the thread.
James had some of the same thoughts as you, but wasn't definite about it. Trippy made a contribution, but again didn't pretend to be authoritative.
I'm inclined to give most weight to temur, who I believe has the best maths background, but I might be biased because what he says matches what I've been saying to you.

This:

...is actually using the definition of the partial derivative:

(use $$x_1 = \theta, x_2 = t$$)
In other words, you're setting t constant and varying $$\theta$$ - that's a partial derivative.

...which is the same exact thing as the total derivative in the case when t does not depend of $$\theta$$. Since you seem to refuse this simple explanation, I gave you the slightly more complicated one, where I took the total derivative while observing that $$t(\theta+\Delta \theta)=t(\theta)=t$$. You rejected that explanation as well, so I give up, there is no way of convincing you that you are wrong.

Maybe this will work:

$$f=f(u(\theta),v,w,...)$$

where $$v,w,...$$ are not functions of $$\theta$$.

Then, the total derivative of $$f$$ wrt $$\theta$$ is:

$$\frac{df}{d \theta}=\frac{\partial f}{\partial u} \frac{du}{d \theta}$$

It makes no sense to talk about the partial derivative of $$f$$ wrt $$\theta$$ (as you've been trying) since $$f$$ does not depend explicitly of $$\theta$$.

Now, if $$f=f(u(\theta),v (\theta),w,...)$$

then:

$$\frac{df}{d \theta}=\frac{\partial f}{\partial u} \frac{du}{d \theta}+\frac{\partial f}{\partial v} \frac{dv}{d \theta}$$

and so on.

By contrast, if: $$f=f(\theta, u(\theta),v,w,...)$$

then:


The total derivative wrt $$\theta$$is:


$$\frac{df}{d \theta}=\frac{\partial f}{\partial \theta}+\frac{\partial f}{\partial u} \frac{du}{d \theta}$$

while the partial derivative wrt $$\theta$$ is:

$$\frac{\partial f}{\partial \theta}$$

 

[Pete]

It seems we are at an impasse in this section.
I suggest we put it on hold for now, and instead consider (your choice):

• A differential approach to the low velocity case, as suggested in post 196, or
• Rindler's proof of angle invariance, last addressed in post 135, or
• My approach to my method, which (as agreed) uses the rod T1 to define the tangent to the wheel, or
• Your frequency measurement method

 

[Tach]

It seems we are at an impasse in this section.

We are at an impasse because you repeatedly rejected the textbook definition of the Differentiation of Composite Functions. Why don't we wait and see what you have learned on the subject?

Once this is cleared, we can proceed with the correct way of doing the differentiation , not only at low speeds but also at any speed. There is no need to use galilean transforms.

 

[Pete]

Here's the post that began this sidetrack:

$$
\begin{align}
x &= r \cos(\omega t + \theta) \\
y &= r \sin(\omega t + \theta) \\
\end{align}
$$​

Transform to S'...

$$
\begin{align}
t &= \gamma(t' + vx'/c^2) \\
x' &= \gamma(x - vt) \\
&= \gamma r\cos(\omega\gamma t' + \omega\gamma vx'/c^2 + \theta) - v\gamma^2t' - v^2\gamma^2x'/c^2 \\
y' &= y \\
&= r\sin(\omega\gamma t' + \omega\gamma vx'/c^2 + \theta) \\
\end{align}
$$​

Differentiate...

$$
\begin{align}\frac{\partial x'}{\partial t'} &= -\gamma r(\gamma \omega + \frac{\gamma\omega v}{c^2}\frac{\partial x'}{\partial t'})\sin(\omega\gamma t' + \omega\gamm vx'/c^2 + \theta) - v\gamma^2 - \frac{v^2\gamma^2}{c^2}\frac{\partial x'}{\partial t'} \\
\frac{\partial x'}{\partial \theta} &= -r\gamma(\frac{\gamma\omega v}{c^2}\frac{\partial x'}{\partial \theta} + 1)\sin(\omega \gamma t' + \omega \gamma vx'/c^2 + \theta) - \frac{v^2\gamma^2}{c^2}\frac{\partial x'}{\partial \theta} \\
\frac{\partial y'}{\partial t'} &= r(\omega\gamma + \frac{\omega\gamma v}{c^2}\frac{\partial x'}{\partial t'})\cos(\omega\gamma t' + \omega\gamma vx'/c^2 + \theta) \\
\frac{\partial y'}{\partial \theta} &= r(\frac{\omega\gamma v}{c^2}\frac{\partial x'}{\partial \theta} + 1)\cos(\omega\gamma t' + \omega\gamma vx'/c^2 + \theta) \\
\end{align}
$$​

What is your answer for those partial derivatives?