# Debate: Lorentz invariance of certain zero angles

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

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1. No, it isn't correct. You have a class of (parallel) vectors, $(0,v_i), 1<i<n$. None of these vectors is a unit vector. The representative of this class is the unit vector$(0,1)$. When judging vector parallelism, you need to stop using the members of the class and start using the class representative. This way, the notion of parallelism is independent of vector norm (length).

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

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Which bit?
So you agree with the conclusion, that v1 and v2 are parallel.
And you agree with how the conclusion was reached.
So what's the problem?

Last edited: Feb 17, 2012
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5. The problem is that you insist on using unnormalized vectors in your reasoning.
Normalized vectors retain their parallelism under Lorentz transformations, unnormalized ones don't (when the lengths of the vectors differ). The geometric notion of parallelism has nothing to do with vector length.

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

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Why is that a problem?
Trivially true, because normalised parallel vectors are identical.
I agree.
I agree.

Conclusion: Zero angles between velocity vectors are not lorentz invariant, except for the trivial special case of identical vectors.

8. Vectors that have the same unit vector aren't identical.

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

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They are when they're normalized.
If they're not, then (as you said in your previous post) they don't retain their parallelism under Lorentz transformations.

10. Well, I see that we are not seeing eye to eye in terms of what vector parallelism means in SR. This is not surprising since there is no definition for what an angle is in SR (otherwise we wouldn't have the dispute on the Lorentz invariance of zero angles). Let me try to explain the same part of my document we have been discussing a different way. The velocity of one of the ion beams models $\vec{v_P(0)}$ and the speed of the ions is , accordingly, $v_P(0)$. The velocity of the second ion beam models $\vec{T_1}$. Since $\vec{T_1}$ is a vector of any length, one can choose any value for its norm $T_1$, so there is nothing stopping the choice $T_1=v_P(0)=U$ in the writeup.

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

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Only because you are using your incorrect conclusion that parallelism is lorentz invariant as an assumption, which means you can't develop a self-consistent argument.

If you held to the obvious and simple meaning of parallelism that you were happy with before you found that it contradicted your side of the debate, there would be no problem.

Sure there is. You've used it yourself in your document. If two vectors form the same angle with the x-axis, they are parallel.
Or, use the dot product if you like.

The only problems arise when questions arise over exactly what vectors are being compared, but this is not an issue for velocity vectors because there is no confusion about how they transform between frames.

As we established earlier, the velocity of the $v_p(0)$ modelling ions must be equal to $v_p(0)$ (ie u=0).
So, the ions must have the same speed as the guns, making the guns redundant. The ions are actually a simple light source of known frequency attached to P, just as I suggested back in post 49.

Now, your T1 beam.
It is true that you can choose the S' angle of the ion velocity within certain limits by setting u, the ion velocity relative to the gun.

Now you just have to find what value of u (if any) makes the T1 ion velocity tangent to the wheel in S'.

12. You have agreed earlier with me that the notion of parallelism should not be dependent on vector norm (length). You have agreed that the parallelism of the unit vectors is frame-invariant. Yet, you insist in bringing back vector length in the argument....

...applied to unit vectors, such that it is independent of vector length, as you agreed earlier.

This is not the issue. The issue is what parallelism should be defined as.
We both agree that it should be defined as the identity of unit vectors.
You apply the above definition as "transform the vectors, normalize them and calculate their relative angle".
I apply the above definition as "transform the unit vectors and calculate their relative angle".

Yes, so what? Adding an extra degree of liberty (u) and finding out that it has to have a specific value (i.e. u=0) does not affect the problem in a negative way.

Err, no, you are missing the point explained in the previous post: I can chose the velocity of the ions such that the vector modeling $\vec{T_1}$ equals $\vec{v_P(0)}$ by choosing to make their lengths equal. This way, the two velocity vectors transform into identical vectors in S'.

Last edited: Feb 19, 2012
13. ### PeteIt's not rocket surgeryRegistered Senior Member

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Correct.
In any given reference frame, the length of two vectors does not affect whether they are parallel - ie the notion of parallelism does not depend on vector length.

However, that has no bearing on whether parallelism is frame-invariant.

Correct.
Yes, because you have agreed that the parallelism of vectors that have different lengths is not lorentz invariant.

No, you correctly applied the obvious and simple meaning of parallelism to arbitrary vectors. Parallelism doesn't depend on length, as we agreed, so there is simply no need to normalize.

No, I said that it could be defined as the identity of unit vectors, not should.
No, I apply the definition as "normalize the vectors $\vec{v_1}'$ and $\vec{v_2}'$, and calculate their relative angle".
Which tells you nothing about $\vec{v_1}'$ and $\vec{v_2}'$.

So, you wasted a lot of time faffing about with first lasers, then ion beams, instead of considering my suggestions. It would be appropriate for you to apologize.

There's only one tangent orientation. You need to model it correctly.

14. First off, I never agreed to that.
Second off, I have repeatedly pointed out that , in my opinion, the notion of parallelism has to be defined independent of vector length. So, you are not only contradicting yourself, you are also putting words in my mouth.

...which is precisely what I said you were doing. By doing so, you create the paradox that the transformed vectors are no longer parallel , since the unit vectors, as computed by your approach are no longer parallel because the result of your approach DEPENDS on the original length of the vectors.

You are not only wrong, you are also being rude.

...which is precisely what I have done.

Last edited: Feb 19, 2012
15. ### PeteIt's not rocket surgeryRegistered Senior Member

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Yes you did, just a couple of posts ago:
And I have repeatedly agreed. The problem is that you are confusing parallelism with transforms.
Parallelism is independent of length.
Transforming between frames is not.

Wrong.
When considering vectors in frame S', I do not start from S.
How do you decide if $\vec{v_1}'$ and $\vec{v_2}'$ are parallel?
Not by looking at $\vec{v_1}$ and $\vec{v_2}$, obviously.

What paradox?
Finding a contradicting to an assumption means the assumption is wrong. It's not a paradox unless you can't let it go.

You're treating frame S as privileged.
There's nothing "original" about frame S.

No Tach, you are wrong.
I politely pointed out very early in the thread that your lasers and ion beams were a waste of time, but you ignored me. Now you've discovered I was right, but you don't acknowledge it. I think you should apologize.

You have a vector with adjustable direction, and you think that no matter how you adjust it will always be tangent to the wheel?
Prove it.

Last edited: Feb 19, 2012
16. Not the way you are trying to apply it, that much is clear.

This is only because you are miss-applying the notion of parallelism by taking through the transformation two vectors of different lengths. Then you say "see, they are no longer parallel".

You are again putting words in my mouth, we aren't talking about vectors that "no matter how you adjust...", we are talking about vectors tangent to the wheel (at least in frame S). The fact that you see them no longer tangent in any other frame is your problem. See part 1 of my writeup. The part that you failed to understand earlier. The part that you keep insisting on making $dt=0$ without any justification.

Last edited: Feb 19, 2012
17. ### PeteIt's not rocket surgeryRegistered Senior Member

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The debate has stalled.
Prepare your summary.

18. I think that you may have learned by now how to calculate the partial derivatives, if not for multivariate functions, at least for univariate ones. Anyways, the error in the above stems from missing terms:

Since :
$t'=\gamma(t-vx/c^2)$ implies that you are missing the term:

$\frac{\partial t'}{\partial \theta}=-\gamma v/c^2 \frac {\partial x}{\partial \theta}=-\gamma vr/c^2 sin (\omega t +\theta)$

In other words, you are missing:

$\gamma^3 v^2r/c^2 sin (\omega t +\theta)$ from your failed attempt at calculating $\frac{\partial x'}{\partial \theta}$

BTW, this is not the only part you got wrong, the bit with $+1$ in $(\frac{\gamma\omega v}{c^2}\frac{\partial x'}{\partial \theta} + 1)$ is also wrong. Same mistake, you are missing $\frac{\partial t'}{\partial \theta}$

Last edited: Feb 20, 2012
19. ### PeteIt's not rocket surgeryRegistered Senior Member

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I'm satisfied with the correctness of my answer, as I'm sure you are with yours.

As the debate has stalled, no more posts may be made to this thread until we have prepared our summaries, in accordance with the agreed debate rules.

20. You don't decide unilaterally what happens next and most definitely you don't get to order me around. I will give you one more formal proof, then we will see where we go, by mutual agreement as the rules state. In the meanwhile, I pointed out where you went wrong in your calculations, perhaps you can take some time to look at it while I transcribe my third (and final proof). This should give you some time to correct your errors in the calculations of the partial derivatives.

Last edited: Feb 19, 2012
21. ### PeteIt's not rocket surgeryRegistered Senior Member

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My summary is complete. When yours is ready, we'll post them up and the thread can be closed.

22. The declared intent of the thread was to debate and learn. Here is another explanation. Before you attempt to shut down the conversation, may I suggest that you read it.

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

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Sorry Tach, I'm done wasting time with you.

Prepare your summary.

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