# Debate: Lorentz invariance of certain zero angles

Discussion in 'Formal debates' started by Pete, Nov 25, 2011.

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1. ### TachBannedBanned

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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:

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

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

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.

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.

Last edited: Feb 8, 2012

3. ### PeteIt's not rocket surgeryRegistered Senior Member

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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 ignoring what I point out.
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.

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).

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?

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.

5. ### TachBannedBanned

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5,265
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.

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.

Last edited: Feb 8, 2012

7. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,167
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?

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?

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.

8. ### TachBannedBanned

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5,265
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.
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.

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.

Last edited: Feb 8, 2012
9. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,167
Is there any more rigorous way of determining whether two variables are independent?
It seems that you're working on intuition rather than rigor.

So far, the same as my post.

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.

Great! Let's see it.

10. ### TachBannedBanned

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5,265
I just gave you another proof. Should be rigorous enough.

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

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

11. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,167
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.
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}

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.

12. ### TachBannedBanned

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5,265
What in the concept of a 2D rotation transformation applied to a circle (expressed in polar coordinates) don't you understand?

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.

13. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,167
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.

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}

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.

Last edited: Feb 9, 2012
14. ### TachBannedBanned

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5,265
...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.

Last edited: Feb 9, 2012
15. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,167
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.

I think this step is incorrect:
This would be correct:
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?

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.

Last edited: Feb 9, 2012
16. ### TachBannedBanned

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5,265
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".

Last edited: Feb 10, 2012
17. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,167
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.

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 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.

18. ### TachBannedBanned

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5,265
...which is precisely what is expressed in the basic math that you seem totally unable to follow.

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.

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

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

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

19. ### PeteIt's not rocket surgeryRegistered Senior Member

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

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.
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?

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.

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.

Last edited: Feb 10, 2012
20. ### TachBannedBanned

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5,265
...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:

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

Last edited: Feb 10, 2012
21. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,167
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

22. ### TachBannedBanned

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5,265
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.

23. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,167
Here's the post that began this sidetrack: