Angle between the orientation of a moving object and its velocity

[Trippy]

I have already answered about your mistakes, see post 160.

Again, see the answer at 160

You answered nothing with that snippy trite crap. You replied in a condescending manner, without actually addressing the points I made.

...which is precisely what the equations of speed composition reflect. If you think otherwise, feel fee to post your own equations.

No they don't - the first set of equations is the motion of a micro facet around its axis, but it doesn't include a (-V,0) boosting for the motion of the camera in the axels frame.

Do you agree with this statement?
"If the wheel is boosted in the cameras frame by (V,0) then the Camera must be boosted by (-V,0) in the axels frame".


If you agree with that, then what part of this, precisely, deals with the motion of the camera relative to the tangent plane of the microfacet?

Ok,

Time to put this away for good.
1. In the frame co-moving with the axle, the trajectory of a point on the circumference is:

$$x=r cos(\omega t)$$
$$y=r sin(\omega t)$$

where $$\omega$$ is the angular speed of the wheel and r is its radius.

The tangential speed of a particle $$\vec{v_p}$$ is identical to the speed of the tangent plane in any point $$\vec{v_p}=\vec{v_t}=(-r \omega sin(\omega t), r \omega cos (\omega t))$$, a well known fact. So,the angle between $$\vec{v_p}$$ and $$\vec{v_t}$$ is ZERO everywhere in the axle frame.

Answer: None of it. That is the equation of motion for a point moving in a circular orbit around some central point.

There is nothing in that that considers the direction and magnitude of the motion of the camera relative to the plane of the tangent, but that is the neccessary value, according to Pauli, for determining whether or not doppler shifting occurs.

 

[Tach]

You answered nothing with that snippy trite crap. You replied in a condescending manner, without actually addressing the points I made.

Not really, I simply pointed out your errors.

No they don't - the first set of equations is the motion of a micro facet around its axis, but it doesn't include a (-V,0) boosting for the motion of the camera in the axels frame.

This is exactly what the second set of equations does.

Answer: None of it. That is the equation of motion for a point moving in a circular orbit around some central point.

...in the frame of the axle, you need to pay attention to the details. If you don't pay attention, you only have knee-jerk reactions that are so easy to prove wrong.

 

[Trippy]

Not really, I simply pointed out your errors.

No, you posted condescending snippy trite crap, that made no corrections, and offered no answers.

This is exactly what the second set of equations does.

...in the frame of the axle, you need to pay attention to the details. If you don't pay attention, you only have knee-jerk reactions that are so easy to prove wrong.

No it doesn't.

The second part of the post:

2. In the ground frame

The axle moves with speed $$\vec{V}=(V,0)$$ so:

$$\vec{v'_p}=\vec{v'_t}=(\frac {V-r \omega sin (\omega t)}{1-Vr \omega sin(\omega t) }, \frac{r \omega/ \gamma(V) cos(\omega t)}{1-Vr \omega sin(\omega t)})$$, c=1.

so, not only that the two speeds are identical (of course they are), their angle is ZERO in the ground frame AS WELL. This is true for both relativistic and "classical" case (pay attention, Trippy). It is also true whether the wheel motion has slippage or not. The "no slip" situation (much touted as being "key" by AN) is just a particular case of the above, $$r \omega=V$$:

$$\vec{v'_p}=\vec{v'_t}=(\frac{V(1- sin (\omega t))}{1-V^2 sin(\omega t) }, \frac{V/\gamma(V) cos(\omega t)}{1-V^2 sin(\omega t) })$$

Of course, the same holds, the zero angle in the axle frame transforms into a zero angle in the ground frame.

The END.

You open with a frame jump to the rest frame of the ground, and conclude that because the zero angle in axle frame transforms to a zero angle in the ground frame, that there is no doppler shift.

But, your conclusion is wrong because you didn't calculate the angle between the motion of the microfacet and the motion of the camera in the axles rest frame, you only calculated the motion of the microfacet.

 

[Tach]

You open with a frame jump to the rest frame of the ground,

Nope, the post opens at point 1, which is the frame of the axle.
From the frame of the axle, the equations are TRANSFORMED into the frame of the ground , at point 2.

But, your conclusion is wrong because you didn't calculate the angle between the motion of the microfacet and the motion of the camera in the axles rest frame, you only calculated the motion of the microfacet.

But what is needed for a zero Doppler effect off a moving mirror is a (zero) angle between the velocity of the mirror and the tangent plane of the mirror (microfacet). The angle wrt to the camera DOES NOT factor in the equation. So, your repeated demands to calculate the angle wrt the camera are groundless.

 

[Trippy]

Nope, the post opens at point 1, which is the frame of the axle.

Now go back and re-read what I actually said, because this isn't it.

What I said was that the second set of equations opens with that, not the post.

And what you have said is precisely my point - you start by calulating the motion of the microfacet in the axles frame, but did not calculate the angle between the motion of the microfacet and the motion of the camera in the axels frame.

From the frame of the axle, the equations are TRANSFORMED into the frame of the ground , at point 2.

Again, this is what I said, but it's beside the point that I'm making here. The point that I'm making is that in point 1 you don't consider the motion of the camera relative to the axle (you don't boost the cameras in the axles frame) but in point 2 you boost the axel in the cameras frame.

But what is needed for a zero Doppler effect off a moving mirror is a (zero) angle between the velocity of the mirror and the tangent plane of the mirror (microfacet). The angle wrt to the camera DOES NOT factor in the equation. So, your repeated demands to calculate the angle wrt the camera are groundless.

That's not what Pauli said.

Pauli said that if the mirror is moving paralell to itself in the frame of the observer, there would be no doppler shift. The corollary of Pauli's statement is that if the observer is moving paralell to the surface of the mirror there would be no doppler shift. Therefore, it is absolutely neccessary to consider the motion of the camerlative to the motion of the microfacet to establish whether or not Pauli's criterion applies to all observers in all frames and all microfacets.

So, according to Pauli the angle of the mirror wrt the angle of the motion of the camera is precisely neccessarily to know to determine in the axels rest frame whether or not doppler shifting occurs.

 

[Tach]

but did not calculate the angle between the motion of the microfacet and the motion of the camera in the axels frame.

Because, once again, the angle wrt the camera is IRRELEVANT in this type of exercise.

Again, this is what I said, but it's beside the point that I'm making here. The point that I'm making is that in point 1 you don't consider the motion of the camera relative to the axle (you don't boost the cameras in the axles frame) but in point 2 you boost the axel in the cameras frame.

...because the angle wrt to the camera is IRRELEVANT. What IS relevant is the angle between the tangent plane of the microfacet and its velocity.

That's not what Pauli said.

Pauli said that if the mirror is moving paralell to itself in the frame of the observer, there would be no doppler shift. The corollary of Pauli's statement is that if the observer is moving paralell to the surface of the mirror there would be no doppler shift.

I don't know what you are reading but page 95 in his book says textually : "If the mirror is moving parallel to its own plane, then $$\alpha'_2=2 \pi - \alpha'_1$$ and it also follows from (15) and (16) that $$\alpha_2=2 \pi - \alpha_1$$, $$\nu_2=\nu_1$$".
Now when you pug the above into the relativistic Doppler equations and you perform all the necessary calculations you get zero Doppler effect between the source, the mirror and the camera.

Therefore, it is absolutely neccessary to consider the motion of the camerlative to the motion of the microfacet to establish whether or not Pauli's criterion applies to all observers in all frames and all microfacets.

No, see above.

So, according to Pauli the angle of the mirror wrt the angle of the motion of the camera is precisely neccessarily to know to determine in the axels rest frame whether or not doppler shifting occurs.

The detailed math of the experiment shows otherwise. If you still belive that you are right , feel free to do the derivation on your own and post it.

 

[Trippy]

Because, once again, the angle wrt the camera is IRRELEVANT in this type of exercise.

wrong.

...because the angle wrt to the camera is IRRELEVANT. What IS relevant is the angle between the tangent plane of the microfacet and its velocity.

See above.

I don't know what you are reading but page 95 in his book says textually : "If the mirror is moving parallel to its own plane, then $$\alpha'_2=2 \pi - \alpha'_1$$ and it also follows from (15) and (16) that $$\alpha_2=2 \pi - \alpha_1$$, $$\nu_2=\nu_1$$".

And the corrolary of this is what, precisely?

If the Mirror is moving paralell to itself in the Camera's rest frame, what direction is the camera moving in the Mirrors rest frame (relative to the mirrors surface)?
1) Paralell?
2) Not paralell?

No, see above.

Yes, see above.

The detailed math of the experiment shows otherwise. If you still belive that you are right , feel free to do the derivation on your own and post it.

Your detailed math is wrong, because it is incomplete.

You use an incomplete proof and as a result of that come to a wrong conclusion.

 

[Tach]

If the Mirror is moving paralell to itself in the Camera's rest frame, what direction is the camera moving in the Mirrors rest frame?

All the frames of reference make sense (axle, camera, mirror) etc. What DOESN'T make sense is your insistence in demanding that I calculate the angle between the camera line of sight and the direction of the moving microfacet

1) Paralell?
2) Not paralell?

Parallel to its velocity, in all frames. I have been with pete over this in much greater detail, the short of this is the mirror is moving parallel to itself in the axle frame, there is no argument about that. When you transform the equations in the ground (camera) frame as I did, contrary to your claims that I didn't, you get that the microfacet is STILL moving parallel to itself. Pete and I are still having a disagreement over the latter and we will sort it out in the next few days.

Your detailed math is wrong, because it is incomplete.

You use an incomplete proof and as a result of that come to a wrong conclusion.

Your repeating the same mantra is not cutting it. Let's see you post the "complete" math.

 

[Trippy]

Look,

I have been with pete over this in much greater detail, the short of this is the mirror is moving parallel to itself in the axle frame, there is no argument about that.

That's not the question I'm asking.

Please try and stay focused and answer the question I am asking you.

If the Mirror is moving paralell to itself in the Camera's rest frame, what direction is the camera moving in the Mirrors rest frame (relative to the mirrors surface)?
1) Paralell?
2) Not paralell?

I'll even reword it for you, phrase it as a yes/no question.

Does the statement "The mirror is moving paralell to its own surface in the cameras rest frame" imply the statement "The camera is moving paralell to the surface of the mirror, in the mirrors rest frame" as a corollary.

Yes, or no.

When you transform the equations in the ground (camera) frame as I did, contrary to your claims that I didn't, you get that the microfacet is STILL moving parallel to itself.

We'll move onto that, soon enough.

Your repeating the same mantra is not cutting it. Let's see you post the "complete" math.

Let's see you give a straight yes or no answer to this question first eh? Then we'll negotiate further proofs of anything.

 

[Trippy]

And of course you've substantialy altered your post while I've been replying. Brilliant.

 

[Tach]

I'll even reword it for you, phrase it as a yes/no question.

Does the statement "The mirror is moving paralell to its own surface in the cameras rest frame" imply the statement "The camera is moving paralell to the surface of the mirror, in the mirrors rest frame" as a corollary.

Yes, or no.

No, it does not. Not only that it doesn't , it is a nonsensical question. You made the same statement a few posts ago:

But, your conclusion is wrong because you didn't calculate the angle between the motion of the microfacet and the motion of the camera in the axles rest frame, you only calculated the motion of the microfacet.

It is as wrong now, as it was a few posts ago.

Then we'll negotiate further proofs of anything.

You made the unsubstantiated claim that my math is incomplete. I posted my math, now post yours or retract the claim.

 

[Trippy]

No, it does not. Not only that it doesn't , it is a nonsensical question.

So... If I have a mirror in 3d space, that is oriented in the plain (X,Y,0).
And I have a Camera some height Z above the plain of the mirror.
And in the cameras rest frame I boost the mirror by (0,V,0).
You're arguing that the motion of the camera in the rest frame of the mirror is something other than (0,-V,0)?

Or, to put it another way. Are you arguing that Zero angles are not preserved by a lorentz transformation?

You made the unsubstantiated claim that my math is incomplete. I posted my math, now post yours or retract the claim.

I will, in good time, however, obviously, for what ever reason I consider establishing this neccessary before I tackle anything else.

 

[Tach]

Are you arguing that Zero angles are not preserved by a lorentz transformation?

No, I have been arguing exactly the opposite. many of the arguments have been with you, so how can you claim such a thing? It is you and the others that are arguing that zero angles are not preserved by the Lorentz transforms.

I will, in good time,

I answered your questions (even the nonsensical ones), you made an claim that my math is incomplete, back it up or withdraw it. No more questions, please back up your claim by posting your own math that allegedly would complete mine.

 

[Trippy]

No, I have been arguing exactly the opposite. many of the arguments have been with you, so how can you claim such a thing? It is you and the others that are arguing that zero angles are not preserved by the Lorentz transforms.

Right, so then you agree that because zero angles are preserved under a lorentz transformation, stating that that the mirror is moving paralell to itself in the rest frame of the camera implies that the camera is moving paralell to the surface of the mirror in the rest frame of the mirror?

I answered your questions (even the nonsensical ones), you made an claim that my math is incomplete, back it up or withdraw it. No more questions, please back up your claim by posting your own math that allegedly would complete mine.

I'm trying to back up my claim, but every time you refuse to answer the questions, you make it impossible for me to present any form of proof, because the questions I am asking are pertinent to the basic assumptions that form the frame work to the maths.

 

[Tach]

Right, so then you agree that because zero angles are preserved under a lorentz transformation, stating that that the mirror is moving paralell to itself in the rest frame of the camera implies that the camera is moving paralell to the surface of the mirror in the rest frame of the mirror?

We've been over this for weeks. The answer is "Yes". Has been "Yes" for the last 4 weeks. You argued the opposite , multiple times. Are you having a change of heart?
Now, would you please put up the math that you claimed that you had?

 

[Trippy]

Now, would you please put up the math that you claimed that you had?

We're not done yet. I will tell you when we are done, and when I am ready to put up the maths mmkay?

Right, so then you agree that because zero angles are preserved under a lorentz transformation, stating that that the mirror is moving paralell to itself in the rest frame of the camera implies that the camera is moving paralell to the surface of the mirror in the rest frame of the mirror?

We've been over this for weeks. The answer is "Yes". Has been "Yes" for the last 4 weeks.

So then when you said this:

Does the statement "The mirror is moving paralell to its own surface in the cameras rest frame" imply the statement "The camera is moving paralell to the surface of the mirror, in the mirrors rest frame" as a corollary.

Yes, or no.

No, it does not. Not only that it doesn't , it is a nonsensical question.

You were in error?

Good then progress.

On to the next question.

We have established that there will be no doppler shift if the motion of the observer is paralell to the surface of the mirror in the rest frame, because that is what is implied as a corollary of Pauli's statement that if a mirror is moving paralell to itself, an observer will see no doppler shift.

Now. Do you agree that, as pauli suggested if the mirror is moving perpendicular to itself, doppler shift will be observed, then by the same logic we have just agreed upon, if, in the mirrors rest frame, the observer has any motion perpendicular to the surface of the mirror, then the observer will measure a doppler shift?

 

[Tach]

We're not done yet. I will tell you when we are done, and when I am ready to put up the maths mmkay?


So then when you said this:


You were in error?

No, I was pointing to your error, I even redlined it, the answer has nothing to do with the "direction of the camera". This is the 4-th time I am correcting this misconception in your today's posts.

Now. Do you agree that, as pauli suggested if the mirror is moving perpendicular to itself, doppler shift will be observed, then by the same logic we have just agreed upon, if, in the mirrors rest frame, the observer has any motion perpendicular to the surface of the mirror, then the observer will measure a doppler shift?

Same mistake, phrased differently. Let me try to correct your errors for good: velocity of the mirror perpendicular to the normal to the mirror => zero Doppler shift. Velocity of the mirror has a non-zero component along the normal to the mirror => Doppler shift. NOTHING to do with the motion of the observer, ok?

No more questions, just show your math or admit that you don't have any.

 

[Trippy]

Same mistake, phrased differently. Let me try to correct your errors for good: velocity of the mirror perpendicular to the normal to the mirror => zero Doppler shift. Velocity of the mirror has a non-zero component along the normal to the mirror => Doppler shift. NOTHING to do with the motion of the observer, ok?

So your argument at this point is that the motion of an observer has nothing to do with the doppler shift of objects, as measured by that observer?

Or to put it another way, that Einsteins equivalence principle was wrong to suggest that the motion of the mirror wrt a stationary observer is equivalent to the motion of an observer wrt a stationary mirror?

No more questions, just show your math or admit that you don't have any.

I have stated that I have the math.
I have demonstrated more than once that not only can I do the math, but that I understand the math well enough to be able to present the equivalent scenario as a diagram, maths, or prose.
I have also stated that I regard these questions as neccessary before we can proceed with the math.

Now, are you going to answer these questions, or are you going to keep trolling with this trite crap?

 

[Fudge Muffin]

gotta love these hardcore maths battles.

 

[arfa brane]

It's going to be pitched shovels at dawn, I'll wager.