Debate: Lorentz invariance of certain zero angles

[Pete]

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

You reached that conclusion by assuming that:

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

I dispute that assumption, and ask you to demonstrate its validity.

 

[Tach]

You reached that conclusion by assuming that:

I dispute that assumption, and ask you to demonstrate its validity.

The experimental setup is constructed that the ion beam speeds are the prototypes for the vectors $$\vec{v_P(0)}$$,$$\vec{T_1}$$. IN ALL inertial FRAMES. This means S' just as well as it meant S.

 

[Pete]

The experimental setup is constructed that the ion beam speeds are the prototypes for the vectors $$\vec{v_P(0)}$$,$$\vec{T_1}$$. IN ALL inertial FRAMES.

No, you're setting it up in S, and assuming that the ion velocities transform so as to stay parallel to the transformed $$\vec{v_P(0)}$$ and the transformed rod.

I dispute that assumption.
Can you support it?

Note - repeating the assumption doesn't count as support.

 

[Tach]

No, you're setting it up in S,

No, frame S is not privileged in any way, the ion speeds are the prototype for the vectors in study in all frames.

and assuming that the ion velocities transform so as to stay parallel to the transformed $$\vec{v_P(0)}$$ and the transformed rod.

No, I am not assuming anything: there is no such thing as a "transformed rod", there is only its realization in the form of the ion speed. You are still thinking in terms of your setup. The ion speeds are the prototypes for the vectors in ALL frames. The ion velocity in S' IS the prototype for $$\vec{v'_P(0)}$$ . The ion velocity in S' IS the prototype for $$\vec{T'_1}$$ . This is the way of getting a measure for $$\vec{T'_1}$$ .

 

[Pete]

No, I am not assuming anything: there is no such thing as a "transformed rod", there is only its realization in the form of the ion speed.

The rod is a physical object, directly measurable in any reference frame.

The ion speeds are the prototypes for the vectors in ALL frames.

Repeating an assumption doesn't count as support.

I don't think that assumption even works for $$\vec{v_p(0)}$$.
While checking it out, I've found something I don't follow in your document:

The ion guns speed wrt the observer comoving with the axle is, by definition

$$\mathbf{v} = (0, \omega R)$$​

The speed of the ions with respect to the guns is

$$\mathbf{u} = (0,u)$$​

The speed of the ions wrt the axle is:

$$\mathbf{U} = \left(u\sqrt{1 - \frac{(\omega R)^2}{c^2}} \, , \, \frac{u + \omega R}{1 + u\omega R/c^2}\right)$$​

​

I don't understand how you derived the expression for the first component of U?
Should it be this?

$$\mathbf{U} = \left(0 \, , \, \frac{u + \omega R}{1 + u\omega R/c^2}\right)$$​

 

[Tach]

Rubbish. The rod is a physical object, directly measurable in any reference frame.

You are just being rude.

 

[Pete]

Sorry, I'll remove the word "rubbish"

 

[Tach]

I don't understand how you derived the expression for the first component of U?
Should it be this?

$$\mathbf{U} = \left(0 \, , \, \frac{u + \omega R}{1 + u\omega R/c^2}\right)$$​

You are right.Cut and paste typo, thank you for the catch. Doesn't change anything in the proof.

 

[Pete]

No problem.

The problem is that when you transform $$\vec{U}$$ and $$\vec{v_p}(0)$$ to to S', according to the velocity transformation we agreed on (eqn 1.3 in your document), you find that they are not parallel.

$$\begin{align}
\vec{v_p}(0) &= (0,\, v_p) \\
\vec{U} &= (0,\, U) \\
\vec{v_p}'(t=0) &= (-V ,\, v_p/\gamma\) \\
\vec{U}' &= (-V,\, U/\gamma\)
\end{align}$$​

 

[Tach]

No problem.

The problem is that when you transform $$\vec{U}$$ and $$\vec{v_p}(0)$$ to to S', according to the velocity transformation we agreed on (eqn 1.3 in your document), you find that they are not parallel.

$$\begin{align}
\vec{v_p}(0) &= (0,\, v_p) \\
\vec{U} &= (0,\, U) \\
\vec{v_p}'(t=0) &= (-V ,\, v_p/\gamma\) \\
\vec{U}' &= (-V,\, U/\gamma\)
\end{align}$$​

Formula (2.6) clearly and un-ambiguously shows you that only U' intervenes in the calculation of the frequencies in S'. U,U' are the only speeds relevant in calculating the angles.

 

[Pete]

I agree. I have no problem with your calculations of the ion velocity angles in S'.
The problem is that you've onlycalculated the ion velocity angle. You haven't calculated the angles that the debate is about.
The problem is that you assert without support a relationship between the ion velocity angle and the angles of $$\vec{v_p}'(0)$$ and the angle of the tangent:

The ion velocity in S' IS the prototype for $$\vec{v'_P(0)}$$.

I assume that by "the prototype for $$\vec{v_p}'(0)$$", you mean that the angle of the ion velocity in S' is the same as the angle of $$\vec{v_p}'(0)$$.
If not, then what do you mean?
If so, then it's wrong, as demonstrated.

The ion velocity in S' IS the prototype for $$\vec{T_1}'$$. This is the way of getting a measure for $$\vec{T_1}'$$ .

I'm assuming that by "the prototype for $$\vec{T_1}'$$," you mean that the angle of the ion velocity in S' is the same as the angle of rod T1.
If not, then what do you mean?
If so, then you need to support that statement, as I asked back in post 85.

 

[Tach]

No problem.

The problem is that when you transform $$\vec{U}$$ and $$\vec{v_p}(0)$$ to to S', according to the velocity transformation we agreed on (eqn 1.3 in your document), you find that they are not parallel.

$$\begin{align}
\vec{v_p}(0) &= (0,\, v_p) \\
\vec{U} &= (0,\, U) \\
\vec{v_p}'(t=0) &= (-V ,\, v_p/\gamma\) \\
\vec{U}' &= (-V,\, U/\gamma\)
\end{align}$$​

I see the issue, you are complaining that $$\vec{U}'$$ is not parallel with $$\vec{v'_p}(0)$$. This is easily fixable, since the ion speed $$u$$ gives us the degree of freedom that allows for $$U/\gamma=v_p/\gamma$$ . Fixed it in the writeup , thank you for the contention. Next.

 

[Pete]

I see the issue, you are complaining that $$\vec{U}'$$ is not parallel with $$\vec{v'_p}(0)$$.

Just to be clear,
Do you agree that you can have two velocities with a zero-angle between them in S that have a non-zero angle between them in S'?

This is easily fixable, since the ion speed $$u$$ gives us the degree of freedom that allows for $$U/\gamma=v_p/\gamma$$ . Fixed it in the writeup , thank you for the contention. Next.

For U' to be parallel to $$\vec{v_p}'(0)$$, you have to set u=0.

$$\begin{align}
u &= \frac{\vec{v_p}(0) - \omega R}{1 - (\vec{v_p}(0)\omega R)/c^2)} \\
\vec{v_p}(0) &= \omega R \\
u &= 0
\end{align}$$​

Next, you need to show that address the last part of the previous post.

The ion velocity in S' IS the prototype for $$\vec{T_1}'$$. This is the way of getting a measure for $$\vec{T_1}'$$ .

I'm assuming that by "the prototype for $$\vec{T_1}'$$," you mean that the angle of the ion velocity in S' is the same as the angle of rod T1.
If not, then what do you mean?
If so, then you need to support that statement, as I asked back in post 85.

 

[Tach]

Just to be clear,
Do you agree that you can have two velocities with a zero-angle between them in S that have a non-zero angle between them in S'?

Yes.

For U' to be parallel to $$\vec{v_p}'(0)$$, you have to set u=0.

$$\begin{align}
u &= \frac{\vec{v_p}(0) - \omega R}{1 - (\vec{v_p}(0)\omega R)/c^2)} \\
\vec{v_p}(0) &= \omega R \\
u &= 0
\end{align}$$​

Yes.

Next, you need to show that address the last part of the previous post.

I'm assuming that by "the prototype for $$\vec{T_1}'$$," you mean that the angle of the ion velocity in S' is the same as the angle of rod T1.

Yes, this was established early on when we agreed on the method. So, it is by design, there is no further justification for you to demand. There is no "rod", it has been replaced by the ion beam.

 

[Pete]

Do you agree that you can have two velocities with a zero-angle between them in S that have a non-zero angle between them in S'?

Yes.

So it follows that if those two velocities are parallel to the same rod in S, then they can not both be parallel to that rod in S'?

Yes, this was established early on when we agreed on the method. So, it is by design, there is no further justification for you to demand.

In post 85, I pointed out that you'd need to support that assumption, but you never addressed it.

Why do you assume that in S' the ion velocity is parallel to the rod T1?

 

[Tach]

So it follows that if those two velocities are parallel to the same rod in S, then they can not both be parallel to that rod in S'?

No, it doesn't. The logical inference is if those two velocities are NOT parallel to the same rod in S, then they can not both be parallel to that rod in S'. Totally irrelevant to the case we are studying.

In post 85, I pointed out that you'd need to support that assumption, but you never addressed it.

I know what you said in post 85 and I also know the answer, it is the same one I gave you then, the ion beams replace both the tangent facet and the tangential velocity. By the way the experiment has been constructed.

Why do you assume that in S' the ion velocity is parallel to the rod T1?

One last time, there is no "rod" anymore. The "rod" is part of your setup, there is no such thing in my setup.

 

[Pete]

No, it doesn't. The logical inference is if those two velocities are NOT parallel to the same rod in S, then they can not both be parallel to that rod in S'.

We started from the premise that two velocities are parallel in S[/b] and not parallel in S'.
If the velocities are parallel in S, then they can both be parallel to the same surface.
If the velocities are not parallel in S', then they can not both still be parallel to that surface.

Totally irrelevant to the case we are studying.

On the contrary, it is very relevant. But it is a side issue that we will return to later.

I know what you said in post 85 and I also know the answer, it is the same one I gave you then, the ion beams replace both the tangent facet and the tangential velocity.

In what post did you give that answer, and what support did you give for that statement?
It looks like the same unsupported statement that's been repeated and rejected already.

One last time, there is no "rod" anymore. The "rod" is part of your setup, there is no such thing in my setup. There is no "rod", it has been replaced by the ion beam.

You previous post:

I'm assuming that by "the prototype for $$\vec{T_1}$$," you mean that the angle of the ion velocity in S' is the same as the angle of rod T1.

Yes, this was established early on when we agreed on the method.

So if there is no rod T1, then what is $$\vec{T_1}$$?

I also notice you're talking about "ion beams" again, which is ambiguous.
I ask you to be careful to keep in mind the difference between the direction of the ion velocity and the orientation of the line between an emitted ion and the ion gun at some time after emission.

Your document refers only to the velocity of ions emitted at t=0, but when you say the rod has been replaced by the ion beam, you appear to be referring to the line between the emitted ion and the gun.

 

[Tach]

We started from the premise that two velocities are parallel in S[/b] and not parallel in S'.

No, you started with this premise. Not "we".

If the velocities are not parallel in S', then they can not both still be parallel to that surface.

This is your premise, the jury is out on its validity. All proofs I have shown prove the opposite, i.e. zero angles are Lorentz invariant.

So if there is no rod T1, then what is $$\vec{T_1}$$?

The vector tangent to the wheel , whose prototype is one of the two ion beams. There is no "rod" in my setup. We have been over this multiple times.

Your document refers only to the velocity of ions emitted at t=0, but when you say the rod has been replaced by the ion beam, you appear to be referring to the line between the emitted ion and the gun.

I refer to the velocity of the ion coming out of the gun as observed in frames S and S'. These are the velocity vectors $$\vec{U}, \vec{U'}$$ in the writeup. The document is quite clear.

 

[Pete]

All proofs I have shown prove the opposite, i.e. zero angles are Lorentz invariant.

We just agreed to a case where a zero angle is not invariant:

Do you agree that you can have two velocities with a zero-angle between them in S that have a non-zero angle between them in S'?

Yes.

e.g.:

$$\begin{align}
v_1 &\ne v_2 \\
\vec{v_1} &= (0,v_1) \\
\vec{v_2} &= (0,v_2) \\
\vec{v_1}' &= (-V,v_1/\gamma) \\
\vec{v_2}' &= (-V,v_2/\gamma)
\end{align}$$​

The vector tangent to the wheel , whose prototype is one of the two ion beams.

"ion beams" meaning "ion velocity vectors".
OK, so we're still stuck on the issue of whether the ion velocity vector is parallel to the vector tangent to the wheel at P in S'.
You claim that it is.
Do you claim that it is true regardless of the value of u, the ion speed relative to the gun?

I refer to the velocity of the ion coming out of the gun as observed in frames S and S'. These are the velocity vectors $$\vec{U}, \vec{U'}$$ in the writeup.

Good. I think it would be better to avoid the ambiguous term "ion beam". If you mean ion velocity, then say ion velocity.

 

[Tach]

We just agreed to a case where a zero angle is not invariant:

e.g.:

$$\begin{align}
v_1 &\ne v_2 \\
\vec{v_1} &= (0,v_1) \\
\vec{v_2} &= (0,v_2) \\
\vec{v_1}' &= (-V,v_1/\gamma) \\
\vec{v_2}' &= (-V,v_2/\gamma)
\end{align}$$​

This is a bad definition of "parallelism" because it makes the notion vector component-dependent (length-dependent). Two vectors that are parallel in one frame (S) are no longer parallel in another frame (S'), THOUGH their unit vectors are STILL parallel in S'. This results into the loss of Lorentz-invariance for no other reason than the fact that the above definition is coordinate-dependent.

Parallelism should be determined by direction, should not be sensitive to length:

$$\begin{align}
v_1 &\ne v_2 \\
\vec{V_1} &= (0,\frac{\vec{v_1}}{v_1}) \\
\vec{V_2} &= (0,\frac{\vec{v_2}}{v_2}) \\
\vec{V_1}' &= (-V,1/\gamma) \\
\vec{V_2}' &= (-V,1/\gamma)
\end{align}$$​

"ion beams" meaning "ion velocity vectors".
OK, so we're still stuck on the issue of whether the ion velocity vector is parallel to the vector tangent to the wheel at P in S'.
You claim that it is.
Do you claim that it is true regardless of the value of u, the ion speed relative to the gun?

Yes, it is , regardless of the value of u, provided you use the appropriate definition of parallelism (see above).

Good. I think it would be better to avoid the ambiguous term "ion beam". If you mean ion velocity, then say ion velocity.

I will. In the same vein, let's also use the proper way of defining parallelism: two vectors are parallel if their unit vectors (versors) are identical.