# Debate:Lorentz invariance of certain zero angles

Status
Not open for further replies.
We have:
\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}​
Agreed?

The question is whether $$\vec{v_1}'$$ and $$\vec{v_2}'$$ are parallel.
In your document, you determine the angle of vectors from the ratio of their components (note that this is not length dependent):
$$\tan(\theta(\vec{v_1}')) = \frac{-v_1}{\gamma V} \\ \tan(\theta(\vec{v_2}')) = \frac{-v_2}{\gamma V}$$​
If the angles are not the same, the vectors are not parallel. Obvious.

Now you suddenly prefer a new definition of parallelism:
Tach said:
two vectors are parallel if their unit vectors (versors) are identical.
Let's apply that definition to $$\vec{v_1}'$$ and $$\vec{v_2}'$$:
\begin{align} \vec{v_1}' &= (-V,v_1/\gamma) \\ \vec{v_2}' &= (-V,v_2/\gamma) \\ \hat{v_1}' &= \frac{\vec{v_1}'}{\|\vec{v_1}\|} \\ &= \left(\frac{-V}{\sqrt{V^2 +(v_1/\gamma)^2}}, \frac{v_1}{\sqrt{(\gamma V)^2 +v_1^2}}\right) \hat{v_2}' &= \frac{\vec{v_2}'}{\|\vec{v_2}\|} \\ &= \left(\frac{-V}{\sqrt{V^2 +(v_2/\gamma)^2}}, \frac{v_2}{\sqrt{(\gamma V)^2 +v_2^2}}\right) \\ v_1 &\ne \, v_2 \\ \hat{v_1}' &\ne \, \hat{v_2}' \end{align}​

Therefore by your unnecessary definition, $$\vec{v_1'}$$ and $$\vec{v_2}'$$ are still not parallel.

Tach, do you really want to argue this? It seems like you're letting your desired result get in the way of clear reasoning.

We have:
\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}​
Agreed?

In the previous post I explained why it makes no sense to talk about the "parallelism" of vectors $$\vec{v_1}'$$, $$\vec{v_2}'$$ unless $$\vec{v_1}$$, $$\vec{v_2}$$ are normalized to unit length. The notion of "parallelism" (more correctly, the Lorentz-invariance of parallelism) should not depend on length, should depend on the direction only. When you normalize the (parallel) vectors $$\vec{v_1}$$, $$\vec{v_2}$$ (in S), their transforms in any ther frame preserve parallelism.

In your document, you determine the angle of vectors from the ratio of their components (note that this is not length dependent):

Sure it is, as shown by the example in the previous post. Once you normalize the vectors, the frame invariance of parallelism is preserved.

Let's apply that definition to $$\vec{v_1}'$$ and $$\vec{v_2}'$$:
$$\vec{v_1}'= (-V,v_1/\gamma) \\ \vec{v_2}'= (-V,v_2/\gamma) \\$$

I'm afraid that you are misinterpreting what I said in the previous post. Your $$\vec{v_1}', \vec{v_2}'$$ do NOT have the same unit vector the way you chose them (they don't even have the same direction) . Nor do $$\vec{v_1}, \vec{v_2}$$, so you are violating the definition and you are obtaining a contradiction.

It is very simple:
-in frame S
$$\vec{v_1}=v_1 \vec{u}$$
$$\vec{v_2}=v_2 \vec{u}$$

Obviously, the vectors are parallel in S.
The transform in S' is $$\Lambda (\vec{u})$$ for both vectors (instead of the standard $$\Lambda ( v_1 \vec{u})$$, $$\Lambda ( v_2 \vec{u})$$ ) so , parallelism is preserved (by removing the dependency of the notion of parallelism on length).
Turns out that you can replace the above with an even weaker condition, i.e. both vectors have the same length, i.e.

$$\vec{v_1}=v_1 \vec{u}$$
$$\vec{v_2}=v_2 \vec{u} \frac{v_1}{v_2}$$

This means that you leave alone the first specimen (the ion velocity representing $$\vec{v_P(0)$$, in my example) and that you scale the ion velocity representing the tangent to the wheel by an appropriate amount to make the norms of the two velocities equal. This is perfectly legit since there is no specific requirement set on the length of the vector tangent.

Last edited:
Let's apply that definition to $$\vec{v_1}'$$ and $$\vec{v_2}'$$:
\begin{align} \vec{v_1}' &= (-V,v_1/\gamma) \\ \vec{v_2}' &= (-V,v_2/\gamma) \\ \hat{v_1}' &= \frac{\vec{v_1}'}{\|\vec{v_1}\|} \\ &= \left(\frac{-V}{\sqrt{V^2 +(v_1/\gamma)^2}}, \frac{v_1}{\sqrt{(\gamma V)^2 +v_1^2}}\right) \hat{v_2}' &= \frac{\vec{v_2}'}{\|\vec{v_2}\|} \\ &= \left(\frac{-V}{\sqrt{V^2 +(v_2/\gamma)^2}}, \frac{v_2}{\sqrt{(\gamma V)^2 +v_2^2}}\right) \\ v_1 &\ne \, v_2 \\ \hat{v_1}' &\ne \, \hat{v_2}' \end{align}​

Therefore by your unnecessary definition, $$\vec{v_1'}$$ and $$\vec{v_2}'$$ are still not parallel.

You are misapplying the definition in more than one way. For example, it is clear that your starting point is $$\vec{v_1}=(0,v_1)$$, $$\vec{v_1}=(0,v_2)$$. To do it correctly, apply the above again for $$v_1=v_2$$

Last edited:
Do you agree that these transformations are correct?
\begin{align} \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}​

You seem to be treating S as a privileged frame.
How does an observer in S' determine whether $$\vec{v_1}'$$ and $$\vec{v_2}'$$ are parallel?
Do they have to transform them back to S, normalise them in S, then transform the normalised vectors to S' before comparing them?

Do you agree that these transformations are correct?
\begin{align} \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}​

You seem to be treating S as a privileged frame.

No, not at all.

How does an observer in S' determine whether $$\vec{v_1}'$$ and $$\vec{v_2}'$$ are parallel?

As per the definition, the two vectors have the same unit vector.

Do you agree that these transformations are correct?
\begin{align} \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}​

Do you agree that these transformations are correct?
\begin{align} \vec{v_1} &= (0,v_1) \\ \vec{v_2} &= (0,v_2) \\ \end{align}​

I have already told you that this is not correct, three times. I have also explained to you why it isn't correct and how to make it correct: make $$v_2=v_1$$

Have you changed your mind about the lorentz transformation of a velocity vector?
The transformation below is a direct quote from your document.
Lorentz transformation of a velocity vector
$$\vec{v_p}' = \left(\frac{v_{p\parallel} - V}{1-v_{p\parallel}V/c^2} \ , \ \frac{v_{p\perp}}{\gamma(1-v_{p\parallel}V/c^2)}\right)$$​
We're both satisfied that this is the correct lorentz transformation for a velocity vector. Right?
Right.
([post=2896280]Post 128[/post])

Have you changed your mind about the lorentz transformation of a velocity vector?

It is not the transformation that you are doing wrong, the transformation is fine, it is the fact that you don't follow the part that says that the norm (length) of a vector should play no part in determining if two vectors are parallel. There are about 4 posts now on this subject, they are all saying the same thing, post 242 is the most explicit on the subject.

So you do agree that these transformations are correct?
\begin{align} \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}​

So you do agree that these transformations are correct?
\begin{align} \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}​

Yes, the transform is obviously correct . Now, try redoing the transform while you satisfy the condition that the vector norm plays no role in determining vector parallelism, i.e. make $$v_1=v_2$$.

OK, so the S' observer considers these two vectors:
\begin{align} \vec{v_1}' &= (-V,v_1/\gamma) \\ \vec{v_2}' &= (-V,v_2/\gamma) \end{align}​

How does the S' observer determine whether they are parallel? You said before that they would test if they have the same unit vector, which I agree is valid.

So, what does the S' observer determine the unit vectors to be?

OK, so the S' observer considers these two vectors:
\begin{align} \vec{v_1}' &= (-V,v_1/\gamma) \\ \vec{v_2}' &= (-V,v_2/\gamma) \end{align}​

How does the S' observer determine whether they are parallel? You said before that they would test if they have the same unit vector, which I agree is valid.

So, what does the S' observer determine the unit vectors to be?

He observes them to be different since the vectors did not satisfy the condition $$v_2=v_1$$ in S. If you continue to insist generating input that does not satisfy the condition of parallelism in S, why are you surprised that the condition is not satisfied in S'?

Are you saying that
\begin{align} \vec{v_1} &= (0,v_1) \\ \vec{v_2} &= (0,v_2) \end{align}​
Are not parallel?
An S observer determines the unit vectors to be identical, which satisfies your condition for parallelism.

Are you saying that
\begin{align} \vec{v_1} &= (0,v_1) \\ \vec{v_2} &= (0,v_2) \end{align}​
Are not parallel?

The convention says that they are parallel if:

\begin{align} \vec{v_1} &= (0,1) \\ \vec{v_2} &= (0,1) \end{align}​

The fact that their norms are different should play no role. Therefore, you need to consider only the unit vectors, since they are equal (and equal to (0,1)), the vectors are parallel. The above condition, means that they are parallel in all inertial frames:

\begin{align} \vec{v_1}' &= (V,1/ \gamma) \\ \vec{v_2}' &= (V,1/ \gamma) \end{align}​

I have shown you this before.

An S observer determines the unit vectors to be identical, which satisfies your condition for parallelism.

Correct. You only need to apply the Lorentz transform to the unit vector, this is the part you keep missing. When you do that, you satisfy the condition of parallelism in all frames, not just in S.

Last edited:
The convention says that they are parallel if:

\begin{align} \vec{v_1} &= (0,1) \\ \vec{v_2} &= (0,1) \end{align}​

The fact that their norms are different should play no role. Therefore, you need to consider only the unit vectors, since they are equal, the vectors are parallel.
I can't quite be certain of your meaning.
Please be patient while I make sure I understand you.

The S observer is considering the velocity vectors:
\begin{align} \vec{v_1} &= (0,v_1) \\ \vec{v_2} &= (0,v_2) \end{align}​
The units vectors of $$\vec{v_1}$$ and $$\vec{v_2}$$ are identical, so the S observer concludes that $$\vec{v_1}$$ and $$\vec{v_2}$$ are parallel (the fact that their norms are different plays no role.)

Right?

I can't quite be certain of your meaning.
Please be patient while I make sure I understand you.

The S observer is considering the velocity vectors:
\begin{align} \vec{v_1} &= (0,v_1) \\ \vec{v_2} &= (0,v_2) \end{align}​
The units vectors of $$\vec{v_1}$$ and $$\vec{v_2}$$ are identical, so the S observer concludes that $$\vec{v_1}$$ and $$\vec{v_2}$$ are parallel (the fact that their norms are different plays no role.)

Right?

Almost but not quite. The unit vector for both $$\vec{v_1}$$ and $$\vec{v_2}$$ is $$(0,1)$$. It is the transform of this particular unit vector that you need to put through the Lorentz transform (and not
\begin{align} \vec{v_1} &= (0,v_1) \\ \vec{v_2} &= (0,v_2) \end{align}​
) in order to determine that the condition of parallelism is satisfied in S'. Otherwise, you allow the norms of the two vectors to influence the parallelism condition. The parallelism should not be dependent on the length of the vectors, should only depend on their direction.

My previous post said nothing about transformations, Tach, it's purely about the S frame.
Do you agree with what I wrote?

My previous post said nothing about transformations, Tach, it's purely about the S frame.
Do you agree with what I wrote?

Almost but not quite, stop using $$(0,v_1)$$ and start using $$(0,1)$$

Last edited:
No, $$\vec{v_1}$$ and $$\vec{v_2}$$ are not unit vectors, they are of arbitrary length.

The S observer is considering the velocity vectors:
\begin{align} \vec{v_1} &= (0,v_1) \\ \vec{v_2} &= (0,v_2) \end{align}​
The units vectors of $$\vec{v_1}$$ and $$\vec{v_2}$$ are identical, so the S observer concludes that $$\vec{v_1}$$ and $$\vec{v_2}$$ are parallel.

Is that correct?

Status
Not open for further replies.