Debate: Lorentz invariance of certain zero angles

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

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  1. Pete It's not rocket surgery Moderator

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    You reached that conclusion by assuming that:
    I dispute that assumption, and ask you to demonstrate its validity.
     
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  3. Tach Banned Banned

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    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.
     
    Last edited: Feb 15, 2012
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  5. Pete It's not rocket surgery Moderator

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    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.
     
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  7. Tach Banned Banned

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


    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 \(\vec{T'_1}\) .
     
    Last edited: Feb 15, 2012
  8. Pete It's not rocket surgery Moderator

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    The rod is a physical object, directly measurable in any reference frame.
    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?
    \(\mathbf{U} = \left(0 \, , \, \frac{u + \omega R}{1 + u\omega R/c^2}\right)\)​
     
  9. Tach Banned Banned

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    You are just being rude.
     
  10. Pete It's not rocket surgery Moderator

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    Sorry, I'll remove the word "rubbish"
     
  11. Tach Banned Banned

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    You are right.Cut and paste typo, thank you for the catch. Doesn't change anything in the proof.
     
  12. Pete It's not rocket surgery Moderator

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    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}\)​
     
    Last edited: Feb 15, 2012
  13. Tach Banned Banned

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    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.
     
    Last edited: Feb 15, 2012
  14. Pete It's not rocket surgery Moderator

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    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:
    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.
    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.
     
    Last edited: Feb 15, 2012
  15. Tach Banned Banned

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    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.
     
    Last edited: Feb 15, 2012
  16. Pete It's not rocket surgery Moderator

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    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'?
    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.
    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.
     
    Last edited: Feb 15, 2012
  17. Tach Banned Banned

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    Yes.

    Yes.


    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.
     
    Last edited: Feb 15, 2012
  18. Pete It's not rocket surgery Moderator

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    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'?
    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?
     
  19. Tach Banned Banned

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    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.

    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.

    One last time, there is no "rod" anymore. The "rod" is part of your setup, there is no such thing in my setup.
     
  20. Pete It's not rocket surgery Moderator

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

    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.

    You previous post:
    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.
     
  21. Tach Banned Banned

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

    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.





    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.



    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.
     
  22. Pete It's not rocket surgery Moderator

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    We just agreed to a case where a zero angle is not invariant:
    e.g.:
    \(\begin{align} v_1 &\ne v_2 \\ \vec{v_1} &= (0,v_1) \\ \vec{v_2} &= (0,v_2) \\ \vec{v_1}' &= (-V,v_1/\gamma) \\ \vec{v_2}' &= (-V,v_2/\gamma) \end{align}\)​

    "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?

    Good. I think it would be better to avoid the ambiguous term "ion beam". If you mean ion velocity, then say ion velocity.
     
    Last edited: Feb 16, 2012
  23. Tach Banned Banned

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    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:

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


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



    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.
     
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