Debate: Lorentz invariance of certain zero angles

[Pete]

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)​

 

[Tach]

Here's the post that began this sidetrack:

What is your answer for those partial derivatives?

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?

 

[Pete]

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, $$\frac{\partial y'}{\partial \theta}$$?

 

[Tach]

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

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

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

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 $$\theta' \ne \theta$$

 

[Pete]

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

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.

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 $$\theta' \ne \theta$$

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.

 

[Tach]

No

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

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?

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.

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

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.

 

[Pete]

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.

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.
So show your results.

This is belief, not science. It is a prejudice that has been coloring all your calculations, there is no reason for it.

I've explained the reasons for that statement, and I judge them to be solid. I acknowledge that you disagree.

To wit, the transforms that tie $$\theta'$$ to $$\theta$$ are nonlinear, so your above claim has a very low probability of being true.

So let's see your transformation.

Anyways, at this stage it is just your prejudice about the solution.

Insults are against the rules.

 

[Tach]

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.
So show your results.

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.

Insults are against the rules.

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.

 

[Tach]

So let's see your transformation.

Certainly. As you can see, the transforms are non-linear, contradicting your intuition.

 

[Pete]

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

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.

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.

 

[Tach]

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?

Yes.

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.

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.

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.

No, see the explanation given above.

 

[Pete]

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.

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'?

 

[Tach]

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'

good.

How does that relate to the debate?

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

 

[Pete]

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

That's something you haven't demonstrated.

 

[Tach]

That's something you haven't demonstrated.

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.

 

[Pete]

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:

And how does the O' observer relate the velocity of the T1 ion beam to the orientation of T1?
Does he assume it is the same?

If so, you'll have to support that assumption at some stage?

I will , don't worry.

 

[Tach]

You have demonstrated that in S':

• the ion velocities are parallel to each other.

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

 

[Pete]

..meaning that $$\vec{v'_P(0)}$$ is parallel to $$\vec{T'_1}$$.

Can you spell out how you reach that conclusion?
It doesn't appear to follow from the given premise.

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

No, that's something you need to demonstrate.

 

[Tach]

Can you spell out how you reach that conclusion?
It doesn't appear to follow from the given premise.

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.

 

[Tach]

No, that's something you need to demonstrate.

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 $$\vec{v'_P(0)}$$,$$\vec{T'_1}$$.