Angle between the orientation of a moving object and its velocity

[Pete]

See the example that I gave you in terms of determining the flatness of the floor by shining a laser beam parallel with it. So, the tangent to the surface is the light ray coming from the laser mounted on the surface.

So you mean a light-like worldline tangent to the surface?
If so, I wish you'd spelled that out before.
Well, perhaps we can explore that next.

I explained that you have no reason for treating $$\theta'_A$$ any different from $$\theta'_P$$.

Yes I do, as clearly described in post 2.

The formula for calculating $$tan(\theta'_A)$$ is the root of our disagreement. The choice you made in calculating $$tan(\theta'_A)$$ has the undesirable result in your solution being incompatible with the aberration formulas.

What choice?
The equation for $$tan(\theta'_A)$$ was derived from the lorentz transform. There's no choice involved.

Think about it, in the aberration formulas you have exactly the same situation: in our problem we want to calculate the transformation of the angle between the tangent to the surface (represented by a light ray)

And there's the misunderstanding.
You want to represent the tangent to the surface with a light-like worldline.
I don't.
I'm plainly talking about the surface itself, its position in space at a particular instant.
From the post 2:

The angle between the surface A and the x-axis is $$\theta_A$$.

Do you understand what I mean?

 

[Tach]

And there's the misunderstanding.
You want to represent the tangent to the surface with a light-like worldline.
I don't.

Well, then you pay by getting a result that it is incompatible with the light aberration formalism.
There is no prescribed way of representing geometric elements in SR (tangents, normals, etc). I chose the way that ensures compatibility with the physical world, you did not <shrug> .

 

[Pete]

OK, let's try this, the point P is moving wrt your frame used for calculation. Therefore you calculated its wordline, I agree with the approach.

But so is the facet under observation. so, its points must also describe a wordline , exactly like P. Instead, you chose to do a simply geometric intersection," A intersects (x,y,t)=(0,0,0)" when , in reality, I contend that you should have calculated the wordline(s) corresponding to the surface tangent plane. By doing the calculations differently, you end up with a set of formulas that are incompatible with the formulas for light aberration.

The diagram is a step removed from the moving facet problem.
The surface is at rest, and is fully described by:
$$y = x \tan(\theta_A)$$

 

[Pete]

Well, then you pay by getting a result that it is incompatible with the light aberration formalism.

I get a result that doesn't apply to light aberration.
If I want to model light aberration, then clearly I'll model light rays.

I'm not modelling light aberration.
I'm modelling a surface.

 

[Tach]

I get a result that doesn't apply to light aberration.
If I want to model light aberration, then clearly I'll model light rays.

I'm not modelling light aberration.
I'm modelling a surface.

I already explained that there is no prescribed way of representing geometric elements like normals, tangents, bi-normals, etc. in SR. You chose one way, I chose a different way. The price you pay is that you get results that are incompatible with the existent body of physics.

 

[Pete]

There is no incompatibility here. There can't be, unless I've made a mistake in the derivation, which you've yet to point out.

I defined $$\theta_A$$ as the angle between the surface and the x-axis, not between a tangent light ray and the x-axis.
It is not the angle of a light ray, so it says nothing about light aberration. It is the angle of the surface.

Do you understand?

For example, do you agree that at any instant in the elevator rest frame, the ground (the ground itself, not a tangent light ray) is horizontal?

 

[Trippy]

This is much worse , the wheel is spinning in the frame of the axle, so there can be no "velocity of a point on the circumference of your wheel, that is perpendicular to the plane of the tangent at that point".

Would you please , please let me deal with pete only? At least he knows what he's talking about, his errors are much, much smaller than yours.

I assure you, I have made no errors.

Consider this diagram:

The wheel is grey.
The axel is black.
The yellow point, labled P is the point of interest - one of your 'microfacets'.
Now, if the curved path, shown in black, that connects P to P' by rotation, is the path that P, then it follows that, by definition, the vector shown in yellow, and labled S that also connects P to P' is the displacement of your microfacet in the non-rotating rest frame of the axel?
Does it not follow, that S is perpendicular to the tangential planes at both P and P' if the angle of rotation from P to P' is $$\pi$$ radians?
Does it not follow that because there is a displacement - S, and that this displacement has occured during some period of time dt, that there is a velocity associated with the motion that has a component that is paralell to the vector S?
Therefore, if S is perpendicular to the tangent, and the velocity is paralell to S, then the velocity must be perpendicular to the tangential planes, therefore meeting Pauli's criteria for causing a doppler shift.

Never mind, I thought that you were doing something decent, like calculating in the ground frame.

I'm fairly sure you're not in a position for such displays of arrogance.

 

[AlphaNumeric]

I have already answered this hundreds of posts ago in the "rolling whee" thread.

And you still seem to be labouring under the same problems.

Already answered this hundreds of posts ago. The full Lorentz transform, it is not very difficult.

Sure but doing it in explicit detail doesn't add anything to my point.

I have already addressed this issue with you long ago, the y-component is irrelevant, what is relevant is that the velocity in frame S' is colinear with the tangent to the wheel. This is precisely the point where pete and I have arrived. So, rather than having to explain this over to you, why don't you let us continue the debate?

The y' component is very relevant. If it is non-zero then you have motion which is not colinear with the tangent of the wheel or plane of the mirror.

The claim "Once colinear, always colinear" means to prove it wrong is to show there's a non-zero component in a direction orthogonal to the colinear direction. In my setup that's the y direction. There's motion in the y direction, thus it isn't colinear, thus your claim is wrong.

I really don't understand how you can so dismiss simple things. Everyone is disagreeing with you and providing the mathematics. Once again you're acting like Chinglu (and yes, I mean that as a very negative thing).

 

[Trippy]

I really don't understand how you can so dismiss simple things. Everyone is disagreeing with you and providing the mathematics. Once again you're acting like Chinglu (and yes, I mean that as a very negative thing).

What amazes me is that someone can miss something this obvious:

(approximately to scale)

And be so assured that the error is on the part of others.

 

[James R]

I'm beginning to think that Tach can't actually understand explanations and questions that are put to him. He can only understand the mathematics he generates himself. He seems quite incapable of analysing a problem.

Oh, and in the case of his wheel example I think he plagiarised an article by Sfarti, so he probably doesn't understand that problem either.

 

[Tach]

There is no incompatibility here. There can't be, unless I've made a mistake in the derivation, which you've yet to point out.

I defined $$\theta_A$$ as the angle between the surface and the x-axis, not between a tangent light ray and the x-axis.
It is not the angle of a light ray, so it says nothing about light aberration. It is the angle of the surface.

Which is precisely the error, the inconsistency in modeling the two vectors (velocity of the microfacet is modeled one way and tangent to the microface is modeled a different way). Instead of:

And now a more formal approach.

Consider a flat surface A in inertial motion in a 2+1 spacetime.
Choose an inertial reference frame S(x,y,t), such that in S:

• A is at rest.
• The angle between the surface A and the x-axis is $$\theta_A$$.
• A intersects (x,y,t)=(0,0,0)

A is thus defined by:
$$y = x \tan(\theta_A)$$

An inertial particle P passes through (x,y,t)=(0,0,0) with velocity $$\vec{V_p}$$.
The angle between $$\vec{V_p}$$ and the x-axis is $$\theta_P$$.
The worldline of P is thus defined by:
$$\begin{align}
x &= t |\vec{V_p}| \cos(\theta_P) \\
y &= t |\vec{V_p}| \sin(\theta_P)
\end{align}$$

you need to have:



The tangent surface points A pass through (x,y,t)=(0,0,0) with velocity $$\vec{V_A}$$:
$$x = t |\vec{V_A}| \cos(\theta_A)$$
$$y = t |\vec{V_A}| \sin(\theta_A)$$

An inertial particle P passes through (x,y,t)=(0,0,0) with velocity $$\vec{V_p}$$.
The angle between $$\vec{V_p}$$ and the x-axis is $$\theta_P$$.
The worldline of P is thus defined by:
$$\begin{align}
x &= t |\vec{V_p}| \cos(\theta_P) \\
y &= t |\vec{V_p}| \sin(\theta_P)
\end{align}$$

Basically what you did is you took two vectors that are colinear in one frame (the axle) and you made them divergent in the boosted frame of the ground through an obvious modeling error. So, your result is unphysical and can be easily corrected by correcting your modeling error.

can you open a debate in the debate section such that we can discuss this without all the background noise?

 

[Tach]

The y' component is very relevant. If it is non-zero then you have motion which is not colinear with the tangent of the wheel or plane of the mirror.

That would be true if the wheel were a circle both in the ground frame and in the axle frame. That is not the case.

The claim "Once colinear, always colinear" means to prove it wrong is to show there's a non-zero component in a direction orthogonal to the colinear direction. In my setup that's the y direction. There's motion in the y direction, thus it isn't colinear, thus your claim is wrong.

Yet, this is exactly the case of the light aberration formalism. You keep ignoring that. (see post 4). So, whatever formalism pete cooked up is incompatible with the physical reality. Think about it, what pete did is he took two vectors that are colinear in one frame, boosted the frame and voila, got them to be non-colinear.

I really don't understand how you can so dismiss simple things. Everyone is disagreeing with you and providing the mathematics.

If science were decided by popular vote....yet, it isn't.

Once again you're acting like Chinglu (and yes, I mean that as a very negative thing).

The person that acts like Chinglu is you. Superficial, incapable of putting together a complete formal argument. So, before you make this comparison again, look in the mirror.

 

[Tach]

What amazes me is that someone can miss something this obvious:

(approximately to scale)

And be so assured that the error is on the part of others.

You know, if proofs were based on drawings, you would get the prize. Unfortunately, they are based on hard math. I have no idea what your drawing is supposed to convey but I will guess that you want to prove that V_i and V_t (whatever they are in your mind) are not colinear. Can you produce the equations that prove that? The reason I ask, is that , by correcting pete's formalism I have proven exactly the opposite.

 

[Tach]

I'm beginning to think that Tach can't actually understand explanations and questions that are put to him. He can only understand the mathematics he generates himself. He seems quite incapable of analysing a problem.

Oh, and in the case of his wheel example I think he plagiarised an article by Sfarti, so he probably doesn't understand that problem either.

I see that you are still smarting from the arsing you got in the debate. If you are unable to contribute any science , as it appears from your repeated content-free posts, could you, at least, refrain from trolling?

 

[Tach]

I assure you, I have made no errors.

Consider this diagram:

The wheel is grey.
The axel is black.
The yellow point, labled P is the point of interest - one of your 'microfacets'.
Now, if the curved path, shown in black, that connects P to P' by rotation, is the path that P, then it follows that, by definition, the vector shown in yellow, and labled S that also connects P to P' is the displacement of your microfacet in the non-rotating rest frame of the axel?

OMG, you are way off. You definitely misunderstood how the microfacets move. Hint: in the axle frame they follow the tangent to the circle as the circle spins. In the frame of the ground they follow the tangent to a cycloid.

 

[Trippy]

OMG, you are way off. You definitely misunderstood how the microfacets move. Hint: in the axle frame they follow the tangent to the circle as the circle spins. In the frame of the ground they follow the tangent to a cycloid.

Your microfacets exist on the surface of the cylinder, and follow the surface of the cylinder as it rolls do they not?

 

[Tach]

Your microfacets exist on the surface of the cylinder, and follow the surface of the cylinder as it rolls do they not?

...meaning that your yellow line PP' makes absolutely no sense.

 

[Trippy]

...meaning that your yellow line PP' makes absolutely no sense.

The vector (note that it has an arrow head, and is described as such in previous posts) represents the displacement of the point P when it has moved to the point P'.

As a reminder:
dis·place·ment (ds-plsmnt)
n.
3. Physics
a. A vector or the magnitude of a vector from the initial position to a subsequent position assumed by a body.
The American Heritage® Dictionary of the English Language, Fourth Edition.

 

[Tach]

The vector (note that it has an arrow head, and is described as such in previous posts) represents the displacement of the point P when it has moved to the point P'.

As a reminder:
dis·place·ment (ds-plsmnt)
n.
3. Physics
a. A vector or the magnitude of a vector from the initial position to a subsequent position assumed by a body.
The American Heritage® Dictionary of the English Language, Fourth Edition.

The microfacet has traversed a succession of incremental positions from P to P' following:

-the circle in the axle frame
-the cycloid in the ground frame

In no frame did the microfacet move along PP'. As such, there is clearly no Doppler effect in the axle frame. The only subject in discussion is if there is any Doppler as measured in the ground frame. If, after all the explanations I have gone back and forth with pete you still think that there is Doppler effect, please post the calculations that evaluate such effect. Colored pictures will not cut it.

 

[Trippy]

The microfacet has traversed a succession of incremental positions from P to P' following:

-the circle in the axle frame
-the cycloid in the ground frame

In no frame did the microfacet move along PP'. As such, there is clearly no Doppler effect in the axle frame. The only subject in discussion is if there is any Doppler as measured in the ground frame.

Read the definition again.

Displacement is not dependent on the path taken.