Angle between the orientation of a moving object and its velocity

[Neddy Bate]

Vectors are coordinate-independent entities, you get the same results independent of any system of coordinates you choose.

Hmmm, let's see. Velocity is a vector. So you are claiming I should get the same velocity for a bullet fired out of a gun whether I choose a coordinate system attached to the gun, or whether I choose a coordinate system attached to an airplane that happens to be flying by. That is what you are claiming, right?

 

[Tach]

We had previously agreed that demonstrating that the motion of the mirror has a component perpendicular to the surface of the mirror was sufficient to prove that doppler shifting would be observed in light reflected by the mirror.

Therein lies your repeated error.IF you insist in transforming the velocity of the mirror into the "camera frame" (for reasons I cannot fathom), THEN you also need to transform the normal to the surface. Only AFTER that you may compute the angle between the velocity and the normal. You are transforming only the velocity and you forgot to transform the normal in rushing to declare that the mirror velocity has a component along the facet normal as viewed from your "camera frame". Turns out that it doesn't.

Since you like Galilean relativity, since it is simpler, I will use an explanation that works in Euclidian space, for subrelativistic speeds. You have often resorted to this simplification, so I hope that you will not object. In Euclidian space, angles are frame invariant, I will follow a proof that you can also find in Rindler (Relativity, Special, General and Cosmological, second edition, page 96).

Let's consider two arbitrary vectors, $$\vec{a}$$ and $$\vec{b}$$. Their scalar product is an invariant. Indeed:

$$||\vec{a}+\vec{b}||^2=a^2+b^2+2 \vec{a} \vec{b}$$

But, in Euclidian geometry, the norm is an invariant, so:

$$2 \vec{a} \vec{b}=||\vec{a}+\vec{b}||^2-a^2-b^2$$

is an invariant, where $$a=|| \vec{a}||$$ . QED.

But there is more, the angle of the two vectors is also an invariant since:

$$cos(\theta)=\frac{\vec{a} \vec{b}}{ab}$$

Now, what I am asking, is whether or not this is equivalent to demonstrating that the camera has a component of motion perpendicular to the surface of the mirror.

Not, it doesn't. See above, angles are frame invariant. A 90 degree angle transforms into a 90 degree angle.

 

[Tach]

Hmmm, let's see. Velocity is a vector. So you are claiming I should get the same velocity for a bullet fired out of a gun whether I choose a coordinate system attached to the gun, or whether I choose a coordinate system attached to an airplane that happens to be flying by. That is what you are claiming, right?

Not. I really do not see how you manage to mangle everything I say so often and so badly.

 

[Neddy Bate]

Hmmm, let's see. Velocity is a vector. So you are claiming I should get the same velocity for a bullet fired out of a gun whether I choose a coordinate system attached to the gun, or whether I choose a coordinate system attached to an airplane that happens to be flying by. That is what you are claiming, right?

Not. I really do not see how you manage to mangle everything I say so often and so badly.

Well, you might not realize it, but you have been claiming that the velocity of the rim of a rolling wheel is frame-invariant. That false claim is the source of argument in this thread, as well as your debate with Pete. Now that you realize that velocity vectors are not frame-invariant, you might want to rethink your positions in these two threads.

 

[Tach]

Well, you might not realize it, but you have been claiming that the velocity of the rim of a rolling wheel is frame-invariant.

You are trolling now by making false claim after false claim. Cease and desist.

 

[Trippy]

But it doesn't , it is moving parallel to itself, so you are simply repeating the same error you made 117 posts ago.

That isn't what I said.
I didn't say the mirror is moving perpendicular to itself, I ASKED the question "IF the mirror can be shown to be moving perpendicular to itself THEN is that sufficient to prove doppler shifting." And you understood that at the time, because your reply was "Yes, but you will not be able to do so, because the mirror has no motion perpendicular to itself."

Please, try and keep to the questions that are actually being asked.

I know that you made this claim, it is just as wrong now as it was then. You calimed that you have additional math to prove your case, where is it?

I also said that I would only be forthcoming with such once we had agreed upon these basic assumptions on what follows from Pauli's condition as corollaries when we apply the principles of relativity and equivalence to it.

You received all the explanations to your ongoing misconceptions, so where is the beef? where is your math that completes[/b] my allegedly incomplete math?

As I stated above I said I would provide it when I felt we were done with this discussion. We are not yet done, because although you have replied, you have replied without answering.

I say you have not answered, because every time we have tried to discuss it you become distracted by talking about what you think that the mirror is doing, and what you think is relevant, rather than answering a straightforward yes or no question.

Now, I am going to ask again, and I am not interested in your opinion on the relevance of the question, nor am I interested in you repeating what the mirror is doing according to your formalism. That's not what the question is.

Pauli's criteria for doppler shift from the reflection off a mirror was that the mirror should have some component of velocity perpendicular to itself, and that there would be no doppler shift if the movement was paralell to itself. Now, setting aside for the moment that Pauli's setup implies a lightsource (or rather a source of photons) (without which there can be no doppler shifting or specular reflection), which in itself implies a restframe within which the mirror is moving paralell to itself...

Your experimental setup includes a camera.
That camera has a rest frame.
Your experimental setup dictates that the wheel should be moving in the Cameras rest frame.
You have agreed previously that demonstrating that the mirror has a component of velocity perpendicular to its own surface in the is sufficient to prove that Pauli's criterion for doppler shifting has been met. Now if we apply the principles of relativity and equivalence to your experimental setup, and how we have previously agreed Pauli's criterion applies to your experimental setup. Does it then follow that this is equivalent to proving that the camera has some velocity component perpedicular to the surface of the mirror, and consequently proving Pauli's criterion?

If not, why not?

 

[Trippy]

Not, it doesn't. See above, angles are frame invariant. A 90 degree angle transforms into a 90 degree angle.

So now you're trying to convince me that because a 90 degree angle transforms into a 90 degree angle, motion of the mirror toward the camera is not equivalent to motion of the camera towards the mirror?

Because that's what I'm asking.

Again, my question, that I have repeated several times now is this:
"IF the mirror is moving perpendicular to itself in the rest frame, is that equivalent to the camera moving perpendicular to itself in the mirrors rest frame."

And you've just answered that question with "No, angles are frame invariant, a 90 degree angle transforms into a 90 degree angle".

But it seems to me that if angles are frame invariant, and a 90 degree angle transforms into a 90 degree angle, then the answer must be yes, rather than no.

 

[Tach]

Does it then follow that this is equivalent to proving that the camera has some velocity component perpedicular to the surface of the mirror, and consequently proving Pauli's criterion?

If not, why not?

For the 6-th time. Not. Because you keep making the same mistake over and over and because you are transforming only the "camera" velocity and you are failing to transform the normal to the surface. had you read the piece I put together for you about angle invariance, you would not have asked this question again. So, you have made some errors, why not admit to them and move on? You have also made some claims about some math you have, I think that by now you probably realize that whatever you have , it is wrong.

 

[Tach]

So now you're trying to convince me that because a 90 degree angle transforms into a 90 degree angle, motion of the mirror toward the camera is not equivalent to motion of the camera towards the mirror?.

No, this is not what I am trying to explain to you. What does angle invariance mean? That a 90 degree angle in one frame transforms into a 90 degree angle in any other frame. So, your "camera" does not miraculously gain a non-zero component along the normal to the surface through a boost.

"IF the mirror is moving perpendicular to itself in the rest frame,

But it is NOT. It is moving PARALLEL to itself in its own frame. See , again, post 233:

-the camera is at rest wrt the source
-the mirror moves perpendicular on its normal , (i.e. parallel to its own plane) , hence your repeated attempts at boosting the camera along the normal are in error
-the mirror moves between the source and the camera

And you've just answered that question with "No, angles are frame invariant, a 90 degree angle transforms into a 90 degree angle".

But it seems to me that if angles are frame invariant, and a 90 degree angle transforms into a 90 degree angle, then the answer must be yes, rather than no.

Err, if the velocity makes a 90 degree angle with the normal, does it have a non-zero component along the normal? This is getting really bad, how long do you plan to keep claiming that you are right?.

 

[OnlyMe]

Tach, give me a little slack and tell me if there is any motion of the camera or the mirror that is perpendicular to the plane of the mirror, that would result in a Doppler effect?

You are only discussing motion that is parallel, to the plane of the mirror?

 

[Trippy]

No, this is not what I am trying to explain to you. What does angle invariance mean? That a 90 degree angle in one frame transforms into a 90 degree angle in any other frame.

I understand that Tach, this is the basis of my argument - that if it is 90 degrees in one frame, it must be 90 degrees in the other. The problem is that you either have not understood the question, or your jumping the gun and moving on to the next part of the discussion, which I'm not interested in discussing until we've established the validity of this part.

So, your "camera" does not miraculously gain a non-zero component along the normal to the surface through a boost.

This is not my argument, nor is it the question I am asking.

But it is NOT. It is moving PARALLEL to itself in its own frame.

Again, this is not the question I am asking.

See , again, post 233:

-the camera is at rest wrt the source
-the mirror moves perpendicular on its normal , (i.e. parallel to its own plane) , hence your repeated attempts at boosting the camera along the normal are in error
-the mirror moves between the source and the camera

Still not addressing the question I am asking. Look, my question is this simple.

Consider this diagram:

In A we have the mirror moving perpendicular to itself in the rest frame of the observer.

In B we have the Observer moving perpendicular to the plain of the mirror, in the rest frame of the observer.

My question is A equivalent to B?

Because the only thing I have asserted so far is that they are.

Err, if the velocity makes a 90 degree angle with the normal, does it have a non-zero component along the normal? This is getting really bad, how long do you plan to keep claiming that you are right?.

Not the question I asked, not the argument I have presented. See above.

 

[Tach]

Consider this diagram:

In A we have the mirror moving perpendicular to itself in the rest frame of the observer.

There is NO mirror moving perpendicular to itself in the scenario. The mirror is moving PARALLEL to itself, as in the microfacets tangent to the circular wheel. Turn $$\vec{V}$$ by 90 degrees in your picture. THIS is the scenario in discussion. You have modified the scenario rather than admitting that you have been wrong all along.

 

[Trippy]

There is NO mirror moving perpendicular to itself in the scenario.

That's not the question I asked. Stop trying to second guess me, and answer the question I am actually asking.
Is the scenario depicted by A equivalent to the scenario depicted by B, yes or no?

The mirror is moving PARALLEL to itself, as in the microfacets tangent to the circular wheel.

That's not the question I'm asking.
Is the scenario depicted by A equivalent to the scenario depicted by B, yes or no?

Turn $$\vec{V}$$ by 90 degrees in your picture. THIS is the scenario in discussion.

We've already agreed on that scenario. That's not the question I am asking. The question I am asking is this:
Is the scenario depicted by A equivalent to the scenario depicted by B, yes or no?
Even if you dispute its relevance, the least you could do is provide a yes or no answer to a yes or no question.

You have modified the scenario rather than admitting that you have been wrong all along.

I have done no such thing.
I am asking about a corollary of something that you agreed to 117 posts ago.
You agreed 117 posts ago, and have subsequently restated that moving the mirror perpendicular to itself will produce a doppler shift, and I have agreed to that. I am arguing that a corollary of that is the observer is moving perpendicular to the mirror in the mirrors rest frame, then the principles of relativity and equivalence predict that a doppler shift will occur.

I have made no effort, as yet, to demonstrate or debate whether or not this is actually the case. There's no point in my doing so until we've agreed wether or not it meets Pauli's criteria. I contend it does, you, so far have avoided answering the question.

So I ask you again. Regardless of whether or not you think its happening in any rest frame in your scenario, are A and B equivalent. yes, or no?

 

[Tach]

That's not the question I asked. Stop trying to second guess me, and answer the question I am actually asking.

I am not second guessing you, I am stopping you from the intellectual dishonesty of changing the scenario rather than admitting that you were wrong all along.
You know damn well that if you move the mirror in the direction of its normal it will produce a Doppler effect, this is why you are trying now desperately to change the rules of the game. You are trying to say that the mirror is moving perpendicular to itself in the frame of the ground. According to your logic, that would mean that it is moving perpendicular to itself in the frame of the axle. This is clearly false, so you should stop playing your game with the "equivalence principle", oops! "relativity principle".

On the other hand , you have never answered to the other error that I pointed out in your thinking. You are saying: "look, I can find a microfacet that is moving perpendicular to itself in the frame of the camera". And I have been telling you several times already: "no, it is not, you are failing to transform both the velocity and the normal of the microfacet, this is why you have been making this false claim". So, you do not even have a reason to claim that the microfacet is moving to itself in the "camera frame". Your whole "proof" is predicated on drawings that support your misconceptions but you have no math to back up you drawings A and B.

We used to have a very good math teacher that, when presented with incorrectly posed problems, would say "Yes, and if your grandmother had wheels, she would have been a trolleybus".

 

[Trippy]

I am not second guessing you...

sec·ond–guess verb \ˌse-kəŋ-ˈges, -kən(d)-\
2: : to seek to anticipate or predict

You are precisely second guessing me.

I am stopping you from the intellectual dishonesty of changing the scenario rather than admitting that you were wrong all along.

You false accusations are unfounded and misplaced.

You know damn well that if you move the mirror in the direction of its normal it will produce a Doppler effect...

So then you agree that A and B are equivalent, and that proving that B exists would be sufficient to prove that Pauli's criterion for doppler shifting exists, and we can move on, yes?

...this is why you are trying now desperately to change the rules of the game.

I'm not trying desperately to change anything. I can assure you that any proof that I present will be using your scenario, and you scenario alone.

You are trying to say that the mirror is moving perpendicular to itself in the frame of the ground. According to your logic, that would mean that it is moving perpendicular to itself in the frame of the axle. This is clearly false, so you should stop playing your game with the "equivalence principle", oops! "relativity principle"

Unless you're second guessing me, you don't know what I am trying to say, because I haven't said it yet. The only thing I have asked is if two scenarios, illustrated as A and B are equivalent, and whether or not proving B is sufficient to prove Pauli's criterion for doppler shifting from a mirror.

the outcome of any local, nongravitational test experiment is independent of the experimental apparatus’ velocity relative to the gravitational field and is independent of where and when in the gravitational field the experiment is performed..

-Einstein, 1907, more or less.

Are you suggesting that changing the velocity of the laboratory relative to (earths) gravitational field, or changing when and where we test to see whether A and B are equivalent will change the outcome of the experiment?

Because I think that in all circumstances the result will be the same, and that A and B are equivalent inertial frames.

But that's beside the point.

 

[Trippy]

And once again Tach substantialy alters his post.

On the other hand , you have never answered to the other error that I pointed out in your thinking. You are saying: "look, I can find a microfacet that is moving perpendicular to itself in the frame of the camera".

This is not what I have said. This is the result of you second guessing me and making assumptions about where this line of questioning.

And I have been telling you several times already: "no, it is not, you are failing to transform both the velocity and the normal of the microfacet, this is why you have been making this false claim".

And I keep telling you that this isn't what I'm asking or saying, and then asking you to address the question I am actually asking.

So, you do not even have a reason to claim that the microfacet is moving to itself in the "camera frame". Your whole "proof" is predicated on drawings that support your misconceptions but...

You don't know what my proof is, all you have at this point is your second guessing of my questions.

In the last 24hrs, I have asked you two questions.
Is a mirror moving paralell to itself equivalent to a camera moving paralell to the mirror?
And:
Is a mirror moving perpendicular to itself equivalent to a camera moving perpendicular to it?

That's all I've asked, nothing more. And in response to that I have had to put up with trite snippy responses, and irrelevant distractions.

...you have no math to back up you drawings A and B.

You want me to prove mathmatically that A and B are equivalent? Because that's all I've asked you at this point.

 

[Tach]

In the last 24hrs, I have asked you two questions.
Is a mirror moving paralell to itself equivalent to a camera moving paralell to the mirror?

No, it isn't.

And:
Is a mirror moving perpendicular to itself equivalent to a camera moving perpendicular to it?

No, it isn't.

You want me to prove mathmatically that A and B are equivalent? Because that's all I've asked you at this point.

I would like you to do the honest thing , admit that you are wrong all along and stop making up fake scenarios.

 

[Trippy]

And:

That's all I've asked, nothing more. And in response to that I have had to put up with trite snippy responses, and irrelevant distractions.


You want me to prove mathmatically that A and B are equivalent? Because that's all I've asked you at this point.[/QUOTE]

Is a mirror moving paralell to itself equivalent to a camera moving paralell to the mirror?

No, it isn't.

Is a mirror moving perpendicular to itself equivalent to a camera moving perpendicular to it?

No, it isn't.

So then it is your assertion that in this diagram:

A and B are not equivalent?

And that if we rotate The mirror 90 degrees in both cases, A and B are still not equivalent?

 

[Tach]

And that if we rotate The mirror 90 degrees in both cases, A and B are still not equivalent?

Please stop playing games. Rotate ONLY $$\vec{V}$$ without rotating the plane of the mirror.

 

[Trippy]

Please stop playing games. Rotate ONLY $$\vec{V}$$ without rotating the plane of the mirror.

I'm not playing games.

Please answer the question.