# Debate: Lorentz invariance of certain zero angles

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

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

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Updated tracking list

1.0 Scenario (Complete)
1.1 - Coordinate dependence vs. coordinate independence (Resolved)
1.2 - Definition of rods T1 and T2 (Resolved)
1.3 - Definition of points A and B (Obsolete)​

2.0 Methodology (Complete)
2.1 - Tach's proposed measurements (Complete)
2.1.1 - Transverse doppler effect (Complete)
2.2 - Pete's proposed measurements (Complete)
2.3 - Measuring remote events using background Rods and Clocks (Complete)

3.0 Calculations (Active)
3.1 Calculations for Pete's method (Active)
3.1.1 Lorentz transformation of vectors (Active)
3.1.1.1 Lorentz transformation of displacement vectors (Active)
3.1.1.1.1 Lorentz transformation of $\hat{P_t}$(Active)
3.1.2 Orientation of rod T1 (Pending)
3.1.3 Angle between surface and velocity in the low velocity limit (Galilean spacetime) (Pending)
3.1.3.1 Rindler's proof of angle invariance (Pending)
3.2 Calculations for Tach's method (Pending)

4. Summary and reflection (not started)​

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

Messages:
10,167
3.1.1.1.1 Lorentz transformation of $\hat{P_t}$
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.

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 dr is not $\hat{P_t}(0)$, and that dr' is not $\hat{P_t}'(t'_0)$.
In fact, I don't think that your using dr 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 dr/dt, which does not simply equate to dr.

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

5. But you are, whether you realize it or not:

Yes, of course. BOTH the velocity and the tangent are time-varying (not stationary) vectors. This is the root of our disagreement.

....which is the velocity of a point on the circumference

....which is the tangent to the circumference. The two aren't the same thing but are co-linear. This is the point.

We can try something very simple, in frame S, the equation of the wheel is:

$x=r \cos ({\omega t+\theta})$
$y=r \sin ({\omega t+\theta})$

with $0< \theta <2 \pi$

The equation of the velocity is:

$\frac{dx}{dt}=-r \omega \sin ({\omega t+\theta})$
$\frac{dy}{dt}=r \omega \cos ({\omega t+\theta})$

The equation of the tangent to the wheel (A NON-STATIONARY vector) is:

$\frac{dx}{d \theta}=-r \sin ({\omega t+\theta})$
$\frac{dy}{d \theta}=r \cos ({\omega t+\theta})$

Obviously, the velocity and the tangent differ in magnitude but are co-linear.

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

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10,167
You've bolded simultaneous, as if that implies I'm using differentiation, but I don't see the connection.
It's clear from my posted calculations that I don't use differentiation to get $\hat{P_t}$.
I could use it, as mentioned in post 118, but that would be with t held constant anyway.

Yes, they're both time-varying, but that really doesn't figure in your approach.

Look at $\vec{v_P}$. In your approach to my method, you derive $\vec{v_P}$ at a particular instant based on the time-varying nature of $\vec{P}$, but the time-varying nature of $\vec{v_P}$ doesn't come into your approach at all.
By definition, dr is the small change in r corresponding to a small change in some other value. Until that other value is specified, dr is not well defined.

Thanks... this is a good approach that we can try to follow rigorously. We should use partial differentials now.

It's something that I thought we could avoid by using the rod T1 as a surrogate for $\frac{\partial\vec{r}}{\partial \theta}$ and $\frac{\partial\vec{r}'}{\partial \theta}$, because I think the transforms will get ugly, but we'll see how we go.

I agree that:
$\frac{\partial y}{\partial \theta} \frac{\partial \theta}{\partial x} = \frac{\partial y}{\partial t} \frac{\partial t}{\partial x}$​

I agree that this means that in S, the velocity of a point on the wheel is parallel to the tangent to the wheel at that point.

But now transform to S', as equations of x', y', t' and $\theta$.
In S', if the velocity of a point on the wheel is parallel to the tangent to the wheel at that point, then we should find that:
$\frac{\partial y'}{\partial \theta} \frac{\partial \theta}{\partial x'} = \frac{\partial y'}{\partial t'} \frac{\partial t'}{\partial x'}$​

Right?

I haven't worked it through yet. I don't know if it will be doable.

Last edited: Feb 6, 2012
8. First of all, "simultaneous" means that one uses $dt=0$ (ot $dt'=0$, according to the frame of reference). Second, differentiation is nothing but taking the difference of two entities. So you are BOTH taking the difference AND making $dt'=0$ in your calculation of :

when you make $t'=t'_0$ for BOTH $\vec{T_1}'(t'=t'_0,l=l_0)$ AND $\vec{T_1}'(t'=t'_0,l=l_1)$.
In addition to that, if you pay attention, the tips of the vectors $\vec{T_1}'(t'=t'_0,l=l_0)$ and $\vec{T_1}'(t'=t'_0,l=l_1)$ belong to DIFFERENT , NON-INERTIAL (rotating) frames, so the notion of simultaneity doesn't even make sense for them. I can recommend a very good paper by Dieks that deals exactly with this issue. Anyways, the insistence of taking the endpoints as simultaneous in frame S' (when this notion doesn't even make sense) is what is causing the problem in your derivation.

Sure it does, as pointed out repeatedly. See that $dt \ne 0$?

Right.

Last edited: Feb 6, 2012
9. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,167
That's your approach. dt doesn't appear in my approach at all.
Subtraction is not the same as differentiation.

Yes I'm doing subtraction. No, that doesn't involve calculus.

I'm not sure what you mean.
The rod T1 is not rotating, it is moving inertially.

That relates to the time-varying nature of $\vec{P}$, not to the time-varying nature of the velocity or tangent vectors.

Now, finding:
$\frac{\partial y'}{\partial t'} \frac{\partial t'}{\partial x'}$​
is easy. It's the gradient of $\vec{v_P}'$, which we've agreed on already.

Finding:
$\frac{\partial y'}{\partial \theta} \frac{\partial \theta}{\partial x'}$​
Is harder. This is the tangent vector, parallel to $\hat{P_t}'$.
I think the easy way is to consider the gradient of rod T1 in S', which is what I posted pages ago.

In S', T1 is straight and tangent to the wheel at point P at some time t'=t'_0.
Therefore, it seems clear that the gradient of the displacement vector between two points on the rod T1 at t'=t'_0 is equal to:
$\frac{\partial y'}{\partial \theta} \frac{\partial \theta}{\partial x'} (t'=t'_0)$​

If this approach is not valid, then we'll have to try the hard way of figuring it out analytically.
I'll give it a shot later today, but I'm not confident of reaching a solution.
Can you help?

10. Yes, it appears but you make it go away by taking the the endpoints of the vectors at the same time.

The tangent follows the circumference of the rotating wheel.

The tangent vectors are time varying, just the same.

No, this approach is only valid for stationary vectors, the tangent is not stationary so you can't make $t'=t'_0$. Besides, the tangent (considered as a vector) in S' is $\frac{\partial\vec{r}'}{\partial \theta}$. I think that you are saying that the tangent to the x' axis has the angle
$arctan(\frac{\partial y'}{\partial \theta} \frac{\partial \theta}{\partial x'})$. I agree with this.​

Last edited: Feb 7, 2012
11. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,167
Yes, the ends of the displacement vector are simultaneous.
That doesn' t mean I'm differentiating anything.

In the scenario, we explicitly agreed to use rod T1 to model the tangent to the wheel at t=0, and rod T1 was explicitly defined as moving inertially.
We did consider using rod T2, which rotates with the wheel, but you said it wasn't necessary and we discarded it.

That's not correct.
Taking a partial derivative means holding other variables constant.
It doesn't imply that those other variables aren't subject to change.

So $\partial x/\partial \theta$, where x is a function of $\theta$ and t, for example, means the change in x over a small change in $\theta$, with t held constant.

The gradient of the tangent has a particular value at a particular instant.
We are interesting in that gradient at $t'=t'_0$ in S'.

Yes, that correct.
$\frac{\partial\vec{r}'}{\partial \theta}$ (where r' is a function of $\theta$ and t'), is by definition the rate of change in r' with respect to $\theta$, with t' held constant.

Yes, that's right, if x' and y' are functions of $/theta$ and t'.
If x' and y' are functions of t, then we're mixing frames and will get a different answer.

The problem is that I don't think that x' and y' can be expressed as neat functions of $\theta$ and t', so I don't know if we'll be able to solve it in that way.

Last edited: Feb 7, 2012
12. Sure you are but you don't realize it. It is very simple, really:

$r'=\gamma(r-vt)$

means:

$r'_1=\gamma(r_1-vt_1)$

$r'_2=\gamma(r_2-vt_2)$

Then:

$r'_2-r'_1=\gamma((r_2-r_1)-v(t_2-t_1))$

which is exactly the same thing as:

$dr'=\gamma(dr-vdt)$

Sure , but this is not what you have been doing. You have been calculating full differentials (see above) and you've been following by setting $dt=0$ despite my repeatedly pointing out that you have no justification to do so.

I have shown earlier, in another thread that the "cure" to this is to use the fact that:

$t=\gamma (t'+vx'/c^2)$

This makes $x'$ an implicit function.

Worth a shot.

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

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10,167
Subtraction isn't calculus.

I can't get a general solution, but I can get the special case for our point P as it crosses the x-axis:
\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}​

I haven't tried getting $\frac{\partial y'}{\partial t'}\frac{\partial t'}{\partial x'}$ or $\frac{\partial y'}{\partial \theta}\frac{\partial \theta}{\partial x'}$ from the above. Maybe everything cancels and it's easy, I don't know.
I just went on to the special case, by substituting in known values for point P crossing the x-axis:
\begin{align} \theta &= 0 \\ t &= 0 \\ x &= r \\ y &= 0 \\ t' &= -\gamma vr/c^2 \\ x' &= \gamma r \\ y' &= 0 \\ \end{align}​

It helps to do the constant expression in the trig functions first...
\begin{align} \omega\gamma t' + \omega\gamma vx'/c^2 + \theta &= -\omega\gamma^2 vr/c^2 + \omega\gamma^2 vr/c^2 + \theta \\ &= \theta \\ &= 0 \\ \end{align}​

Now it's just plug'n'chug, ending up with...
\begin{align} \frac{\partial x'}{\partial t'} &= -v \\ \frac{\partial y'}{\partial t'} &= \frac{r\omega}{\gamma} \\ \frac{\partial x'}{\partial \theta} &= 0 \\ \frac{\partial y'}{\partial \theta} &= r \end{align}​

Now, if I wrote up all that without making at least one mistake, I'll be surprised!

14. I never said it was, but in this case, it is. It always is in the case of linear combinations.

Last edited: Feb 7, 2012

15. I don't think it is that simple: $t$ is definitely independent of $\theta$ but $t'$ isn't by virtue of the fact that $t'=\gamma(t-vx(\theta)/c^2$, meaning that you need to account for that in calculating the derivatives, you are currently missing a term. I trust that you are using a symbolic package (like Mathematica) so, all the calculations are bound to come out right without much effort.

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

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How do you determine independence?
And $t = \gamma(t' + vx'(\theta)/c^2)$, but so what?
Once you take t out of the equations, t' is independent of $\theta$.
That is, setting $\theta$ to a particular value doesn't affect t' unless t is also set to a particular value. For any given value of $\theta$, t' can have any value at all.

Nope, hand coded.
Better check it twice.

17. Simple, for a stationary wheel, the equation is :

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

For a particle describing a circular trajectory, the equation is:

$x=r \cos (\omega t)$
$y=r \sin (\omega t)$

For a rotating wheel, the equation becomes (via superposition of effects):

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

Nope, while $t$ is independent of $\theta$ by virtue of the problem statement (see above), $t'$ is not independent of $\theta$ by virtue of the fact that it is dependent of both $t$ and $x(\theta)$ .

18. Come to think of it, it isn't clear why $\theta$ would not transform into $\theta'$ , after all, this is precisely what we have been after and there is no reason to consider apriori that $\theta$ is frame invariant.
So, the rigorous definition of the tangent in S' should be $(\frac{d x'}{d \theta'}, \frac{d y'}{d \theta'})$ and not $(\frac{d x'}{d \theta}, \frac{d y'}{d \theta})$ as in the above calculations.

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

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So what general rule are you applying?
It seems to be something like this:
If:
• x is a function of $\theta$ when t is fixed, and
• x is a function of t when $\theta$ is fixed, then
• t and $\theta$ are independent
Is that right?

We can write an equation in which t' is dependent on t and $\theta$.
We can also write an equation in which t is dependent on t' and $\theta$.
We can write equations in which t is independent of $\theta$, (but only if they don't contain t').
We can write equations in which t' is independent of $\theta$, (but only if they don't contain t).

It would make no difference to the direction of the vectors.
In S', the result of a small change in $\theta$ is obviously parallel to the result of a small change in $\theta'$.
In S' $theta$ can potentially be measured by physical markings on the wheel.
Note that this makes it obvious that t' and $\theta$ are independent.
But let's see what you come up with anyway.

20. x is a function of the independent variables $t$ AND $\theta$.

I don't think it is "obvious" at all, we can't accept statements based on faith, there should be a rigorous proof.

"Obvious"? Again? Simple math says that your above claim is incorrect:

$\frac{d t'}{d \theta}=\gamma \frac{v}{c^2} \frac{dx}{d \theta}=-\gamma r \frac{v}{c^2} sin (\omega t + \theta)$

Last edited: Feb 7, 2012
21. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,167
Yes, but you didn't answer the question.
What general rule are you applying to what premises to conclude that t is independent of $\theta$?
It is because x can be expressed as a function of t when $\theta$ is fixed, and vice versa, right?

OK.
How do you define $\theta'$?
Can you derive $\frac{\partial y'}{\partial \theta'}\frac{\partial \theta'}{\partial x'}$?

Yes, it's obvious.
Consider the wheel with physical markings at constant increments of $\theta$.

It's obvious that at every instant in S', there is an x and a y corresponding to every value of $\theta$, i.e. setting t' doesn't influence $\theta$, i.e. for any given value of t' there is some function $x' = f(\theta)$ valid over the full range of $\theta$, ie $\theta$ is independent of t'.

It's also obvious that for every value of $\theta$, there is an x' and a y' corresponding to every value of t', i.e. for any given value of $\theta$ there is some function $x' = g(t')$ valid over the full range of t', i.e. t' is independent of $\theta$.

Your simple math is incorrect, unless t is set constant.

It seems we're both out of our depth.
Perhaps we need a ruling from a qualified person.
Choose anyone on Sciforums with qualifications in mathematics or physics, and I'll trust their judgement on whether the partial derivative I posted is correct.

22. Nonsense, $t$ is a variable independent of $\theta$, so $\frac{dt}{d \theta}=0$.
I explained very clearly the meanings of the two variables, several times already, last time only a few posts ago.

x is a function of two INDEPENDENT variables, $t$ and $\theta$.

The transform in S' of the variable $\theta$ in S.

Last edited: Feb 7, 2012
23. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,167
Like I said, you're treating t as a constant in that derivative.
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.

Putting it in caps doesn't explain anything.
Are you simply declaring that t and $\theta$ are independent?
If not, then what general rule or definition are you using to conclude that they are independent?

I don't trust your judgement on this issue, and it's clear that you don't trust mine.
I would like to ask for an authority opinion.
Would you object if I opened a thread in Physics and Maths about partial differentiation and the independence of variables in this situation?

That much is obvious, but it doesn't serve to define $\theta'$.
Do you mean that $y'/x' = \tan(\theta')$?

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

Last edited: Feb 8, 2012