Angle between the orientation of a moving object and its velocity

[Trippy]

$$V_i$$ is the tangent to the cycloid, the trajectory of the point on the circumference calculated in the ground frame. What you have been asked to do is to calculate the angle between the velocity $$V_t$$ and the tangent to the surface , not between $$V_t$$ and $$V_i$$.

Haven't you figured it out yet?
The angle between V[sub]t[/sub] and V[sub]i[/sub] in my treatment is precisely the angle between the velocity of the point and the tangent plane, because V[sub]t[/sub] is the velocity of the point in the tangent plane, while V[sub]i[/sub] is the velocity of the point relative to the ground.

If you read my exchange with pete as to what is being calculated you would not have wasted your time. Besides, you would have learned the correct derivation, including the relativistic case.

My solution and Pete's solution are equivalent, as I demonstrated here:

And that this:

$$\psi=tan^{-1}(\frac{\omega r sin(\alpha)}{V+\omega r cos(\alpha)})$$

Which is just another way of saying this:
$$tan(\psi)=\frac{\omega r sin(\alpha)}{V+\omega r cos(\alpha)}$$
Is just the classical limit of this:

$$\tan(\theta'_A) = \frac{|\vec{V_p}| \sin(\theta_A)}{\gamma(|\vec{V_p}| \cos(\theta_A) - v)}$$

Because $$|\vec{V_p}|=\omega r$$
And $$\psi = \theta'_A$$
And $$\alpha = \theta_A$$

When I illustrated the equivalence. The only difference is that where Pete took the relativistic approach, I took the classical approach, and we arrived at the same conclusion.

I do not need to learn the relativistic approach, because I already understand the relativistic approach. I understand it well enough to recognize that it is un-neccessary and that a proof in the classical limit will suffice.
Why? because I understand what the equations describe.
I understand that in the rest frame as v→c
$$\psi$$→0
and that V[sub]i[/sub]→V[sub]t[/sub]
But that for ALL cases where V<c, V[sub]i[/sub]≠V[sub]t[/sub] $$\psi$$≠0 unless the condition where $$\alpha=0$$ or $$\alpha=\pi$$ radians are met.

 

[Tach]

My solution and Pete's solution are equivalent, as I demonstrated here:

Arrgh,

$$\tan(\theta'_A) = \frac{|\vec{V_A}| \sin(\theta_A)}{\gamma(|\vec{V_A}| \cos(\theta_A) - v)}$$

is my correction to pete's derivation. When you make $$\tan(\theta'_A) = \frac{|\vec{V_A}| \sin(\theta_A)}{\gamma(|\vec{V_A}| \cos(\theta_A) - v)}$$ you get the transformed angle to be ZERO. So, if you agree with that it means that you are agreeing with my correction to pete's formalism such that the ZERO angle between the particle speed and the tangent to the surface is preserved through the boost from the axle frame to the ground frame.
You need to start with a zero angle (which you obviously don't) and you need to apply a Lorentz boost (which you also do not)
in order to address this problem. (which, despite so many posts, you did NOT).

The angle between $$\vec{V'_p}$$ and the x'-axis is thus given by:
$$\tan(\theta'_P) = \frac{|\vec{V_p}| \sin(\theta_P)}{\gamma(|\vec{V_p}| \cos(\theta_P) - v)}$$.

...meaning that :

$$\tan(\theta'_A) = \frac{|\vec{V_A}| \sin(\theta_A)}{\gamma(|\vec{V_A}| \cos(\theta_A) - v)}$$

But, in the unprimed frame: $$\theta_A= \theta_P$$
$$\vec{V_p}=\vec{V_A}$$

so:

$$\tan(\theta'_A) =\tan(\theta'_P)$$

meaning that:

$$\tan(\phi') =0$$

 

[Pete]

What you have is the angle between a stationary flat surface and the velocity of a point on the rim.

No, in the opening posts I have the angle between a stationary flat surface and the velocity of an arbitrary moving point.

Therefore your derivation does not address the problem at all.

It's the problem I'm addressing, and it was clearly defined in the opening posts.
If you don't want to address it, then why are you in this thread?

Post 3 describes how it can be applied to moving surfaces and spinning wheels, but the specific problem in the opening posts stands on its own.

I think that you are backpeddalling now that your solution has been shown to be wrong. We have been talking about the microfacets making up the rim for quite awhile, now you are changing your story?

The "story" in this thread has been consistent from the opening post.

If you don't want to address the problem defined in this thread, then you shouldn't have jumped in saying my solution to that problem is wrong.

For me, the scenario was always the same, find out how the angle between tangent to moving microfacet and point velocity transforms from the axle frame in the ground frame.

And you have a thread for that already.
This thread is about a specific side issue, which is why I made it a separate thread.

 

[Trippy]

Arrgh,

$$\tan(\theta'_A) = \frac{|\vec{V_A}| \sin(\theta_A)}{\gamma(|\vec{V_A}| \cos(\theta_A) - v)}$$

Mea culpa. But that's worse for you, isn't it? You've just spent the last how many posts telling me how wrong I am when all along my derivation was equivalent to yours, while the worst I've done is mis-attirbute it.

is my correction to pete's derivation. When you make $$\tan(\theta'_A) = \frac{|\vec{V_A}| \sin(\theta_A)}{\gamma(|\vec{V_A}| \cos(\theta_A) - v)}$$ you get the transformed angle to be ZERO.

No. It is only zero when the angle of rotation is Zero (or $$\pi$$). In other words, it's only Zero for an observer on the radial line connecting the point in question to the axle. All other points must show some doppler shift, and all other observers must see some doppler shift.

So, if you agree with that it means that you are agreeing with my correction to pete's formalism such that the ZERO angle between the particle speed and the tangent to the surface is preserved through the boost from the axle frame to the ground frame.

In the special case that the observer is on the radial line connecting the micro facet with the axle of the wheel, sure, but for all other microfacets to the same observer, or all other observers of the same microfacet, doppler shift will be observed.

You need to start with a zero angle (which you obviously don't)...

No I don't, because I'm considering the general case - the whole wheel. You're restricting yourself to a very specific case that happens to be the only case that supports your example.

...and you need to apply a Lorentz boost (which you also do not)

No I don't, I only need to prove that doppler shifting exists in the classical limit. If it exists in the classical limit it must also exist in relativistic case, because you can not lorentz transform it out of existence. If it exists for one obsevrer, barring the previously stated special cases, it exists for all observers. What they will disagree on is the extent and the sign.

in order to address this problem. (which, despite so many posts, you did NOT).

I have addressed this problem precisely, and repeatedly. And continuously it has been demonstrated that the problem does not lie in my ability to understand the problem.

And yet you insist on investing emotional currency in being right with comments such as this.

 

[Pete]

Ok, open a debate between the two of us, it is long overdue.

[thread=111135]Proposal: A surface moving parallel to itself in one frame is not doing so in all frames[/thread]

 

[James R]

I see that you are still smarting from the arsing you got in the debate.

"Arsing"? How mature of you. Exactly how old are you? You sound about 19.

As I recall, I won our debate, despite your wimping out of it before it finished. I had already won by that time.

Also, don't think that you've managed to sidestep the issue of plagiarism. If you were actually involved in real research, you'd know what a serious case of academic misconduct plagiarism is.

 

[Tach]

Mea culpa. But that's worse for you, isn't it? You've just spent the last how many posts telling me how wrong I am when all along my derivation was equivalent to yours, while the worst I've done is mis-attirbute it.

No, it is not. Because if you read on, if you assumme my correction to pete formalism you end up with zero angle transforming into zero angle.

No. It is only zero when the angle of rotation is Zero (or $$\pi$$). In other words, it's only Zero for an observer on the radial line connecting the point in question to the axle. All other points must show some doppler shift, and all other observers must see some doppler shift.

Post 5 in this thread shows that your claim is false.

In the special case that the observer is on the radial line connecting the micro facet with the axle of the wheel, sure, but for all other microfacets to the same observer, or all other observers of the same microfacet, doppler shift will be observed.

Actually, post 5 shows this claim also false.

No I don't, I only need to prove that doppler shifting exists in the classical limit.

But you didn't do that, even in the classical limit. You "jumped frames" by calculating the angle between $$V_t$$ measured in the axle frame and $$V_i$$ measured in the ground frame. You simply repeated an earlier error made by pete.

If it exists in the classical limit it must also exist in relativistic case, because you can not lorentz transform it out of existence. If it exists for one obsevrer, barring the previously stated special cases, it exists for all observers. What they will disagree on is the extent and the sign.

If you jump frames (as you did) in the classical limit, you will be jumping frames in the relativistic limit as well.

I have addressed this problem precisely, and repeatedly. And continuously it has been demonstrated that the problem does not lie in my ability to understand the problem.

Yes, you made a different mistake every time. Every time I pointed it out to you, you devised a new method (without ever admitting to the error just pointed out) with new errors.

 

[Tach]

No, in the opening posts I have the angle between a stationary flat surface and the velocity of an arbitrary moving point.

...which is not the problem that we agreed to solve. This is the first time you admit that you have not solved the problem we agreed to solve.

Post 3 describes how it can be applied to moving surfaces and spinning wheels, but the specific problem in the opening posts stands on its own.

I already refuted your post 3 in my post 5.

If you don't want to address the problem defined in this thread, then you shouldn't have jumped in saying my solution to that problem is wrong.

I was on the impression that this thread was a continuation of the "Discussion" thread on the subject of the zero Doppler shift off a rolling wheel. Now, that I proved that you have not solved the problem, you are saying that it isn't the solution that we agreed to seek, the one that I posted a few days ago,

And you have a thread for that already.
This thread is about a specific side issue, which is why I made it a separate thread.

I will look at it, thanks for the late admission that you did not solve the problem that we agreed to work on. I will see to what extent your new solution addresses anything germaine to the zero Doppler shift.

 

[Tach]

"Arsing"? How mature of you. Exactly how old are you? You sound about 19.

As I recall, I won our debate, despite your wimping out of it before it finished.

Err, no you begged for extra "chancies" , yet you failed to produce an answer.

I had already won by that time.

Delusions, delusions.

Also, don't think that you've managed to sidestep the issue of plagiarism. If you were actually involved in real research, you'd know what a serious case of academic misconduct plagiarism is.

Still sour grapes? Grasping at straws?

 

[Trippy]

No, it is not. Because if you read on, if you assumme my correction to pete formalism you end up with zero angle transforming into zero angle.

Seperate issue.

Post 5 in this thread shows that your claim is false.

No it doesn't. The fact that you would claim it does suggests that maybe you should spend some time drawing "colourful pictures".

Actually, post 5 shows this claim also false.

No it doesn't, it proves it true, because the only time $$sin(\theta)$$ is 0 is when $$\theta$$ is zero.

But you didn't do that, even in the classical limit.

Yes I did, because...

You "jumped frames" by calculating the angle between $$V_t$$ measured in the axle frame and $$V_i$$ measured in the ground frame. You simply repeated an earlier error made by pete.

I did not jump frames.

The angle of V[sub]t[/sub] In the ground frame is the same as the angle of rotation of the points location in the ground frame, from the vertical in the ground frame. There is no frame jumping required, nor was there any performed.

If you jump frames (as you did) in the classical limit, you will be jumping frames in the relativistic limit as well.

Which would be relevant if I had jumped frames.

Yes, you made a different mistake every time. Every time I pointed it out to you, you devised a new method (without ever admitting to the error just pointed out) with new errors.

You have yet to demonstrate an error in anything that I have done (aside from one misattribution) that does not stem from your own understanding of what has been presented to you.

 

[RJBeery]

Please don't hijack this thread. Open another thread elsewhere and I'll be more than happy to point out your new mistakes.

Tach, I moved my explanation to a new thread as you requested. I know you seem to have made quite a few battles for yourself but I'm most curious to see what mistakes I've made in my analysis. Thanks

 

[Tach]

Seperate issue.

No, key issue as in showstopper.

I did not jump frames.

Sure you did, $$V_t$$ is measured in the axle frame (tangent to the circle, $$V_i$$ is measured in the ground frame (tangent to the cycloid). This is why you get a non-zero angle between them from the start. If you didn't jump frames, you would have gotten zero angle as the initial angle.

 

[Trippy]

Sure you did, $$V_t$$ is measured in the axle frame (tangent to the circle, $$V_i$$ is measured in the ground frame (tangent to the cycloid).

No.
It is measured in the frame of the ground.
The ground itself is proof of that, it just represents a 'special case' where $$\alpha=\pi$$ radians, and where $$|\vec{V_i}|=0$$

This is why you get a non-zero angle between them from the start. If you didn't jump frames, you would have gotten zero angle as the initial angle.

No, I get a non-zero angle from the start because the vector that represents the velocity in the ground frame combines with the vector that describes a linear component of the angular motion relative to the ground, in the ground frame, to describe V[sub]i[/sub].

There is no frame jumping involved.

The only information I need, is to know at two different times, as measured by a clock in my frame, where the center of the wheel is as measured with rods in my frame, and where the microfacet is as measured with roads in my frame. I can derive every other piece of information I need from those 5 pieces of information, all measured in my own frame.

 

[Tach]

No.
It is measured in the frame of the ground.

Nope, you see, $$\vec{V_t}=\vec{\omega}x\vec{r}$$ in the frame of the axle, not in the frame of the ground.
In the frame of the ground, things are more complicated.

 

[Trippy]

Nope, you see, $$\vec{V_t}=\vec{\omega}x\vec{r}$$ in the frame of the axle, not in the frame of the ground.
In the frame of the ground, things are more complicated.

Nope.

All. I repeat ALL measurements are made relative to the ground.

The only things I need to know are their positions at two seperate times, relative to some point on the ground. Every aspect can be measure using rods and clocks in the frame of the ground.

I do not need any information other than that, and I do not need to make any measurements in any frame other than the ground frame.

 

[Pete]

...which is not the problem that we agreed to solve. This is the first time you admit that you have not solved the problem we agreed to solve.

To what agreement do you refer?

I already refuted your post 3 in my post 5.

Post 5 doesn't even mention post 3.

I was on the impression that this thread was a continuation of the "Discussion" thread on the subject of the zero Doppler shift off a rolling wheel. Now, that I proved that you have not solved the problem, you are saying that it isn't the solution that we agreed to seek, the one that I posted a few days ago,

Perhaps you should have read the opening posts of this thread, instead of making assumptions.

I will see to what extent your new solution addresses anything germaine to the zero Doppler shift.

I would appreciate it if you would first address the validity of the analysis as it stands, before jumping ahead to doppler shift.

 

[Tach]

Nope.

All. I repeat ALL measurements are made relative to the ground.

This can't be since $$\vec{V_t}=\vec{\omega}x\vec{r}$$ in the frame of the axle, so you can't have the same expression in the frame of the ground.

 

[Pete]

To what agreement do you refer?

Oh yes, [post=2861705]this one[/post].
I said I would derive how to properly transform the angle between a surface and an arbitrary velocity vector.

Which is exactly what I did in post 2.

 

[Tach]

To what agreement do you refer?

that we are going to figure out whether or not the zero angle between surface tangent and tangential velocity transforms into a zero angle in the the ground frame.

Post 5 doesn't even mention post 3.

Sorry, it refutes your post 2.

Perhaps you should have read the opening posts of this thread, instead of making assumptions.

I did, I knew it was not true so I assumed that you are working in earnest on the case describing the spinning wheel in the axle frame being boosted in the ground frame. I was under the impression that you were dealing with the microfacets on the rim, as I did in the "Discussion" forum. So, now you want to open a debate thread on the SAME subject? This is a monumental waste of time, I thought that you were planning in earnest to solve the microfacet problem so we could put the zero Doppler effect to bed once and for all.

[qoute=Tach]I will see to what extent your new solution addresses anything germaine to the zero Doppler shift.[/quote]
I would appreciate it if you would first address the validity of the analysis as it stands, before jumping ahead to doppler shift.[/QUOTE]

It is totally useless for the purpose of the studying the Doppler shift. Now I see why we were having the disagreement, you were using a fixed surface while I solved the problem for the moving surfaces (microfacets).

 

[Tach]

Oh yes, [post=2861705]this one[/post].
I said I would derive how to properly transform the angle between a surface and an arbitrary velocity vector.

Which is exactly what I did in post 2.

Which is not only not the correct solution, as explained further down but also , as I have just learned, doesn't even address the case of surfaces comoving with the vectors, as the microfacets in the axle frame. We are going around in circles.