# 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 (On hold)
3.1.1.1.1 Lorentz transformation of $\hat{P_t}$(On hold)
3.1.1.2 Tangent and velocity vectors by partial differentiation (On hold)
3.1.1.2.1 Partial differentiation, total differentiation, dependent and independent variable(Impasse)
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 (Active)
4. Summary and reflection (not started)​

Last edited: Feb 14, 2012

3. My objections were already formulated at post 172.
Additionally, we need to consider the derivatives wrt $t',\theta'$ meaning that we need to work out the transformation between $\theta$ and $\theta'$. I worked that out while you were trying to straighten out your knowledge on total differentiation. Do you agree with post 197 now or do you still have issues with it?

Last edited: Feb 13, 2012

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

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10,167
Yes, I know your objections. Do you have answers? Are you capable of doing the partial differentiation?

Given:
\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}​

What do you get for $\frac{\partial x'}{\partial t'}, \ \frac{\partial x'}{\partial \theta}, \ \frac{\partial y'}{\partial t'},$ and, $\frac{\partial y'}{\partial \theta}$?

7. Yes, I am . Do you agree that post 197 is correct? Yes or No?

You need to calculate $\frac{\partial x'}{\partial t'}, \ \frac{\partial x'}{\partial \theta'}, \ \frac{\partial y'}{\partial t'},$ and, $\frac{\partial y'}{\partial \theta'}$

since $\theta' \ne \theta$

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

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10,167
No. Like I said, we're at an impasse.
You insist (without support other than your own handwaving) that the total derivative df/dx is meaningful when some parameters of f do not depend on x.
I insist (backed up by the video you like, by temur, and by przyk) that it is not.
We're getting nowhere with that argument, so I suggest we move on.

So do you agree with with calculation for:
$\frac{\partial x'}{\partial t'}$ and $\frac{\partial y'}{\partial t'},$?
If not, then what do you get?

And I'm still asking what you get for $\frac{\partial x'}{\partial \theta}$ and, $\frac{\partial y'}{\partial \theta}$.
You claimed my calculation is wrong, but never presented your own.

I don't find $\theta'$ very interesting, since any reasonable transform will result in:
$\frac{\partial y'}{\partial \theta} \, \frac{\partial \theta}{\partial x'} \, = \, \frac{\partial y'}{\partial \theta'} \, \frac{\partial \theta'}{\partial x'}$

...but if you present your preferred transform for $\theta'$, I'll see if I can calculate $\frac{\partial x'}{\partial \theta'}$ and, $\frac{\partial y'}{\partial \theta'}$ anyway.

9. Then, we cannot go any further since the formalism requires the proper calculation of total derivatives.

I told you before that you are missing that $t'$ is a function of $\theta$, so, all the corresponding terms are missing in your derivation. This ties into the disagreement from the previous point. Until we solve that, we cannot proceed with this path.

This is belief, not science. It is a prejudice that has been coloring all your calculations, there is no reason for it.
To wit, the transforms that tie $\theta'$ to $\theta$ are nonlinear, so your above claim has a very low probability of being true. Anyways, at this stage it is just your prejudice about the solution.

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

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10,167
Nevertheless, I would like to see what you believe to be the correct calculations.
Can you please present what you judge to be the correct calculations for $\frac{\partial x'}{\partial t'}$, $\frac{\partial y'}{\partial t'}$, $\frac{\partial x'}{\partial \theta}$ and $\frac{\partial y'}{\partial \theta}$.
You said you could do it.
I've explained the reasons for that statement, and I judge them to be solid. I acknowledge that you disagree.

Insults are against the rules.

11. No, it doesn't work this way, Pete. If you do not agree on the way of taking total derivatives (and you spent a LOT of time on this diversion), there is no way of proceeding down this path. We need to shift gears and try to discuss my method using ion guns.

I am sorry, it is not meant as an insult, it is an observation: you are assuming your final answer to be true and you are constructing your derivations backwards from there.

Last edited: Feb 14, 2012
12. Certainly. As you can see, the transforms are non-linear, contradicting your intuition.

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

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10,167
We'll never know, unless you're willing to do the calculations.

Your judgments of my motivations are not observations.

If I suspect of you what you suspect of me, it would be insulting for me to say so. The only acceptable course of action is to point out the steps and assumptions in your arguments that I disagree with, and ask you to support them.

3.2 Calculations for Tach's method

I agree with your result that the velocities of the ions emitted at t=0 are parallel in S and in S'.
The term "ion beam" is not defined.
By "ion beam", do you mean the path over time of the ions emitted at t=0?
Or do you mean the collective positions at a given instant of all previously emitted ions?

You also haven't defined $\theta'_{v_p(0)}$ or $\theta'_{T_1}$.
You seem to imply two meanings for $\theta'_{v_p(0)}$:
• the angle with the x' axis of the velocity of the ion emitted from the gun at wheel element P at t=0.
• the angle in S' of $\vec{v_p}(0)$ (the velocity of wheel element P at t=0)
However, you need to prove that these two quantities are the same.

Similarly, you seem to imply two meanings for $\theta'_{T_1}$.
• the angle with the x' axis of the velocity of the ion emitted from the gun on rod T1 at t=0.
• the angle with the x' axis of rod T1 in S'
Again, you need to prove that these two quantities are the same.

14. Yes.

These are the angles that intervene in the relativistic Doppler effect formulas. So, there is only one meaning. They aren't the angles wrt to the x' axis, they are the angles the velocities make with the line of sight. This is in the text.

No, see the explanation given above.

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

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10,167
OK, so they are the angle with the line of sight of the velocity of the ions emitted at t=0.

I agree that those velocities (angle and direction) are equal in S, and equal in S'

How does that relate to the debate?
How does the ion velocity in S' relate to the velocity of P in S'?
How does the ion velocity in S' relate to the tangent to the wheel at P in S'?

16. good.

The velocities model $\vec{v_P(0)}$ and $\vec{T_1}$. The angles are the same in S' as in S .

Last edited: Feb 14, 2012
17. ### PeteIt's not rocket surgeryRegistered Senior Member

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That's something you haven't demonstrated.

18. 1. The angles are the same in S, by construction.
2. Therefore the frequencies due to transverse Doppler effect are the same in S.
3. The frequencies are the same in S' by virtue of the fact that they are the transformed of the identical frequencies in S.
4. If one assumes by absurd that the angles are different in S', one arrives to a contradiction, therefore the angles are the same in S'

All the above is shown in the writeup.

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

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10,167
We agree that in S:
• the ion velocities are parallel to each other,
• the ion velocities are parallel to $\vec{v_p}(0)$
• the ion velocities are parallel to the rod T1.

You have demonstrated that in S':
• the ion velocities are parallel to each other.

You claim, but have not demonstrated that in S':
• the ion velocities are parallel to $\vec{v_p}'(0)$
• the ion velocities are parallel to the rod T1

This was mentioned and acknowledged back in posts 85 and 86:

20. ..meaning that $\vec{v'_P(0)}$ is parallel to $\vec{T'_1}$.
You need to remember that the ion velocities are the prototypes for $\vec{v'_P(0)}$,$\vec{T'_1}$

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

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10,167
Can you spell out how you reach that conclusion?
It doesn't appear to follow from the given premise.
No, that's something you need to demonstrate.

22. The angles made by the vectors $\vec{v'_P(0)}$,$\vec{T'_1}$ with the SAME direction (OS') are the SAME. It is in the text.

23. In frame S that the ion velocities in S are the prototypes for $\vec{v_P(0)}$,$\vec{T_1}$. By construction.
In frame S' that the ion velocities in S' are the prototypes for $\vec{v'_P(0)}$,$\vec{T'_1}$.

Last edited: Feb 15, 2012