# 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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It's what you responded to:
It makes it very hard to hold a discussion if we don't pay read each other's posts carefully.

Because we're both supposed to work through each methodology.
I've given you enough information so that you can work through mine, you haven't given me enough to work through yours.
I'm not asking for the detailed math description, I'm just asking for the approach you take.

What factors do you plan to use to calculate the doppler shift at O'?
The obvious thing to do is to transform the ion velocity to S', then use the transformed velocity to calculate the doppler shift.
But that makes calculating the doppler shift redundant, because the ion velocity angle in S' is your end goal.

3. ### TachBannedBanned

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Yes.

No, I plan to use the frequencies at O, reflected towards O', in an extension of the Hasselkamp setup, wait a day or two and I will show you, I need to do some drawings that aren't that easy.
BTW, the info that you gave me about using markers and protractors doesn't help me at all but I have decided to make a run at your methodology on my own.

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

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That's methodology, not calculations.
I'll wait to see your drawings so I understand your methodology before I get too far into calculations.

The markers and protractors aren't necessary. As I said in post 69, it should be enough to simply calculate the angle between the defined vectors - practical measurement techniques don't add anything useful to the debate.
But if it's not enough, the coordinates of events on the vectors of interest can be directly measured by standard rods and clocks reference frames.

7. ### TachBannedBanned

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Ok, so you can think in these simple terms: equal frequencies detected as coming from positionally-coinciding sources moving with identical velocities in S' result into coinciding angles between the line of sight and said velocities.

Actually, they do because, if we do not agree on the theoretical predictions, the tiebreaker will be running the respective experiments. I know I am repeating this....

None of the experiments that I've done in remote measurements employed rods. Most, if not all of them, were radar based telemetry.

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

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We haven't yet established what's going on with "the frequencies at O, reflected towards O'."

We're not going to actually run any experiments. If anything, we can only theoretically predict what the result of experiments would be.

That's nice, but I don't see the relevance since we're not actually conducting experiments.
And I find it difficult to believe that you've never used some kind of ruler in the setup of an experiment.

9. ### TachBannedBanned

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Actually, the way I put it together, it is fully runnable.

Nope, never. Too imprecise.

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

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Again, not relevant, because we're not actually conducting experiments.
The only thing we can do in this thread is theoretical predictions.
Right?

11. ### TachBannedBanned

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If we end up disagreeing on the theoretical predictions, then the tiebreaker will be an experiment. So, I fashioned my theoretical proposal as a realizable experiment. It would be nice if you did the same with your proposal.

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

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No. We're not going to do an actual experiment.

If we disagree on the theoretical predictions, then we will burrow down as far as we need to to figure out where the mistake is.

It's maths. There should be no disagreement.

13. ### TachBannedBanned

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My experience is that people will go at great extends to prove their point and that only when confronted by experiment will admit to error, in many instances , not even then. Anyways, do as you wish, I am describing my formalism in a detail that would lend itself to experimental verification. We don't need to argue about this, it isn't even important.

Last edited: Dec 14, 2011
14. ### PeteIt's not rocket surgeryRegistered Senior Member

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In this debate thread, we're not going to do an experiment.

I'm rapidly losing interest. This thread is ruining my vacation.
If you want to explore empirical results, can you please do so where it isn't taking up my time?

15. ### TachBannedBanned

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I think you misunderstand, intent to present a full mathematical formalism, the fact that I intend to include the experimental setup should be of no worry to you. It will not take any of your time.

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

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So be it.

Can you please specify the labels you will use for the ion velocity vectors and any other variables not already defined, so that we are consistent in our writeup.

17. ### TachBannedBanned

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Sure:

Ion gun velocity: $\vec{v}$.
Ions velocity $\vec{u}$

18. ### TachBannedBanned

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I am ready, LMK when you are ready and I will post the link to the calculations.

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

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My calcs are good to go for my method.
I've noticed that I have used two different symbols for the unit vector tangent to P.
For consistency, I will no longer use $\hat{D_p}(t)$, and will use $\hat{P_t}(t)$ instead.

I still can't make enough sense of your method to make calculations.
At best, it appears to measure angles between velocities, and makes no measurement of orientation.

20. ### TachBannedBanned

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Isn't the invariance of the zero angle what we have been talking about? What "measurement of orientation" is included in this debate?

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

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By orientation, I mean the "angle of the microfacet" in the OP.

I said I'm good to go.
Notation:
$\vec{P}(t)$ is the position vector of wheel element P at time t in S.
$\vec{P'}(t')$ is the position vector of wheel element P at time t' in S'.
$\hat{P_t}(t)$ is the unit displacement vector tangent to wheel element P at time t in S.
$\hat{P_t}'(t')$ is the unit displacement vector tangent to wheel element P at time t' in S'.
$\vec{v_p}(t)$ is the velocity vector of wheel element P at time t in S.
$\vec{v_p}'(t')$ is the velocity vector of wheel element P at time t' in S'.
$\vec{T_1}(t,l)$ is the position vectors of the elements of rod T1 at time t in S. $l$ is distance along the rod.

Let $v$ denote the magnitude of the velocity of S' relative to S (this velocity is in the direction of the positive x-axis).
Let $t'_0$ denote the time in S' when the rod T1 is touching wheel element P.

Pete's method recap:
Determine whether $\vec{v_p}(t=0)$ is a scalar multiple of (ie is parallel to) $\hat{P_t}(t=0)$.
Determine whether $\vec{v_p}'(t'=t'_0)$ is a scalar multiple of (ie is parallel to) $\hat{P_t}'(t'=t'_0)$.

$\vec{v_p}(t=0)$ and $\vec{v_p}'(t'=t'_0)$ can be derived from $\vec{P}(t)$, the worldline of P.
$\hat{P_t}(t=0)$ is obvious, but can be formally derived from the definition the wheel at t=0 if necessary.
Rod T1 is defined in terms of $\hat{P_t}(t=0)$.
$\hat{P_t}'(t'=t'_0)$ can be derived from the transformed T1.

Calculations:
1. Finding $\vec{v_p}(t=0)$
This is trivial, as are much of the following calculations, but spelling it out points toward a general solution for times other than t=0.

\begin{align} \vec{P}(t) &= \begin{pmatrix} r\cos(\omega t) \\ r\sin(\omega t)\end{pmatrix} \\ \vec{v_p}(t) &= \frac{d}{dt} \vec{P}(t) \\ &= \begin{pmatrix} -r\omega\sin(\omega t) \\ r\omega\cos(\omega t) \end{pmatrix} \\ \vec{v_p}(t=0) &= \begin{pmatrix} 0 \\ r\omega \end{pmatrix} \end{align}
2. Finding $\vec{v_p}'(t'=t'_0)$
Transforming $\vec{P}(t)$ gives...
$\vec{P'}(t) = \begin{pmatrix}\gamma(r\cos(\omega t) - vt) \\ r\sin(\omega t) \end{pmatrix}$
Note that this is expressed as a function of t rather than t'. Given that
$t = \frac{t'}{\gamma} - \frac{v}{c^2}r\cos(\omega t)$
, I can't find a general closed form expression for $\vec{P'}(t')$. The same applies to the derivative:​
\begin{align} \vec{v_p}'(t) &= \frac{d}{dt'} \vec{P'}(t) \\ &= \begin{pmatrix}\frac{-r\omega\sin(\omega t) - v}{1 + \frac{vr\omega}{c^2}\sin(\omega t)} \\ \frac{r\omega\cos(\omega t)}{\gamma(1 + \frac{vr\omega}{c^2}\sin(\omega t))} \end{pmatrix} \end{align}
But, we can still calculate the result, since we know that $t=0$ at $t'=t'_0$ on the worldine of P.
\begin{align} \vec{v_p}'(t'=t'_0) &= \vec{v_p}'(t=0) \\ &= \begin{pmatrix}-v \\ r\omega / \gamma \end{pmatrix} \end{align}
3. Finding $\hat{P_t}(t=0)$
Trivially, $\hat{P_t}(t=0) = \begin{pmatrix}0 \\ 1 \end{pmatrix}$
If necessary, this can be formally derived by considering the direction of the displacement vector between P and a simultaneous wheel element a short distance away, and taking the limit as the distance from P approaches zero.
4. 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} \vec{T_1}'(t'=t'_0,l=l_0) &= \begin{pmatrix} \frac{r}{\gamma} - vt'_0 \\ l_0 + r\omega(\frac{t'_0}{\gamma} +\frac{vr}{c^2})\end{pmatrix} \\ \vec{T_1}'(t'=t'_0,l=l_1) &= \begin{pmatrix} \frac{r}{\gamma} - vt'_0 \\ l_1 + r\omega(\frac{t'_0}{\gamma} +\frac{vr}{c^2})\end{pmatrix} \\ \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}
$\hat{P_t}'(t'=t'_0) = \begin{pmatrix}0 \\ 1 \end{pmatrix}$

Results
$\vec{v_p}(t=0) = \begin{pmatrix} 0 \\ r\omega \end{pmatrix}$
...is parallel to...
$\hat{P_t}(t=0) = \begin{pmatrix}0 \\ 1 \end{pmatrix}$
i.e., in S at t=0, the microfacet at P is parallel to the velocity of P.

$\vec{v_p}'(t'=t'_0) = \begin{pmatrix}-v \\ \frac{r\omega}{\gamma} \end{pmatrix}$
...is not parallel to...
$\hat{P_t}'(t'=t'_0) = \begin{pmatrix}0 \\ 1 \end{pmatrix}$
i.e. in S' at t'=t'_0, the microfacet at P is not parallel to the velocity of P.

Last edited: Jan 25, 2012
22. ### TachBannedBanned

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My calculations can be found here. Several different methods, including the two we agreed on at the beginning of this debate. Pay special attention to the general method in Euclidian space. Any correct method in Minkowski space must reduce in the limit to the Euclidian one at non-relativistic speeds.
If I look at your solution, it is quite clear that at non-relativistic speeds ($\gamma=1$) your solution disagrees with Rindler's because $v'_p$ is still not parallel (according to your solution) to the microfacet. Indeed, according to the above:

$\vec{v_p}'(t'=t'_0) = \begin{pmatrix}-v \\ r\omega \end{pmatrix}$
...is not parallel to...
$\hat{P_t}'(t'=t'_0) = \begin{pmatrix}0 \\ 1 \end{pmatrix}$
i.e. in S' at t'=t'_0, the microfacet at P is not parallel to the velocity of P. This means that your method cannot be right.

Last edited: Jan 25, 2012
23. ### PeteIt's not rocket surgeryRegistered Senior Member

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Your post raises a number of issues, which we will address one at a time (see tracking list in the next post).
First, the most fundamental. The other issues might become obsolete when this is resolved.

3.1.1 Lorentz transformation of vectors
In your document you use three different transformations for vectors under a Lorentz boost.

The second transformation you use is this:
I agree that this is the correct lorentz transformation for a velocity vector.

On page 2, you have this:
I agree that this is the correct Lorentz transformation of a displacement 3-vector.

But I don't understand the first transformation you use:
Questions:
What is the source of this equation?
What does $dt$ represent in that equation?
What is $d\vec{r}$? You seem to imply that it is a displacement vector, perhaps $\hat{P_t}(0)$. Is that correct?

I think you have also incorrectly applied the equation to the parallel and perpendicular components.
Plugging in $d\vec{r_\perp}$ gives:
$d\vec{r_\perp} - \gamma\vec{V}dt$
Plugging in $d\vec{r_\parallel}$ gives:
$\gamma(d\vec{r_\parallel} - \vec{V}dt)$
$d\vec{r}' = d\vec{r_\perp} + \gamma(d\vec{r_\parallel} - 2\vec{V}dt)$