Angle between the orientation of a moving object and its velocity

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Tach, I think your math is correct in post 145, but your conclusions about it are misapplied. No one claims that the spinning motion of a mirrored-rim wheel would produce a Doppler effect, and your math proves that the tangent plane and circumferential point on the wheel share a velocity vector (basically by definition). However I don't see why this matters, because the Doppler effect is a function of imparting a net energy change to the incoming light, which is determined by orientation and velocity of the reflective surface. Why does it matter if a point on the rim and its tangent plane move together? They could BOTH provide the necessary orientation and velocity to impart a Doppler effect on the reflected light. To say otherwise is akin to saying that because a perpendicularly oriented, stationary mirror produces no such Doppler effect, then the same mirror moving towards the camera would also produce no such effect, which is trivially false.

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If you cannot "see" what I mean in your mind's eye I can do another analysis (with math, visuals, etc) but it will have to wait until tonight.

 

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It demonstrates that the claim a mirror moving in its plane in one frame is moving in its plane in all frames to be false so it is relevant.

Frame S : Mirror is extended in x direction. Velocity in x direction. Extension and velocity parallel.
Boost done in y direction
Frame S' : Mirror is extended in x' direction. Velocity in both x and y directions. Extension and velocity not parallel.

Don't even have to do any algebra. So the assertion the angles between extension and velocity don't change are false. Unless you think the mirror is now slanted? Please explain how this is not relevant to your claim when it specifically addresses it.

This was something I explicitly addressed in previous threads with you. I went through the algebra with you about how I accept that there's no Doppler shift in that case but if the observer/receiver is moving relative to the centre of mass frame of the wheel then there will be. I did the Lorentz transform explicitly, based on the accepted result that there's no Doppler shift in the wheel's centre of mass frame.

No shift in that frame necessarily lead to a shift in the frame I was considering. I gave you the algebra. Clearly the cycloid path shows the velocity vector of a point being non-parallel to the tangent of the circle at all by 2 places, the two places which have been highlighted in discussion. No one is denying that in specific cases what you're saying is correct but your carte blanche statements are not.

All you seem to be doing is saying "Pauli agrees with me!", when he doesn't and "That example is irrelevant" when it's very relevant. People are right, you've invested such a vast amount of emotion into this now, across so many threads, and basically insulted so many people that you are unable to let yourself admit a mistake. You struggle at the best of times, like the spheres vs balls incident showed but I think this is the largest "I'm right, everyone else shut up!" incident you've done, as far as I can remember. Everyone disagrees with you. You said that someone should look at themselves when Farsight and RJ are their only supporters. What does it say when everyone on both sides of the pseudoscience fence disagree with you? Something worse.

 

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You are not even able to cite my claims correctly, for the fourth time : ZERO angles are frame invariant (this has been shown to you in post 145, please go back and read it). NOT arbitrary angles. Exactly as in the case of the light aberration angles.

So, you aren't answering the question. It was a simple question, I will restart interacting with you when you give a straight answer and you learn to cite my simple claim correctly.

It says that you are on the same side of the fence with RJBeery,OnlyMe and Neddy Bate.

 

Please note that my example involves a zero angle.

Actually I did answer it. I know the first word after the close quote tags wasn't a yes or no but it was answered in the sentences which followed, if you cared to read it.

Tach, I answered it and my example involved a zero angle. I hardly think you're in a position to try to play the "I'll interact with you when you be a little more mature" card if you fail to read what I said twice in one short post. Are you reading them? I'm beginning to wonder because not only did you not realise I answered your question this time around but your question asks me something I not only made a post about in the debates forum but I even PM'd you a link to when you were still engaged in the debate.

Perhaps I'm reading too much into it but I always read your posts with the mental image (or sound rather) of you 'snapping' at the person you're replying to, as if they're just being ridiculous. Of course everyone now and again writes posts with that frame of mind but with you I get that feeling constantly. Like I just said in another thread, tone it down. It gets tiresome after a while and given the amount of people who've tried to set you right on these issues I would guess you're exasperating more people than just myself.

 

Then you got the wrong answer. You are passing two vectors that are identical through the same Lorentz boost. Are they identical after the boost? Yes or no?

 

Ah, I see your mistake. Let's go through it in steps, starting with the set up.

The mirror is extended in the x direction in a 2+1 dimensional space-time. Let's suppose it's an infinite mirror so we can define it by the y location it is at, say \(y=y_{0}\). It's moving in the x direction, say velocity \(\mathbf{v} = (v_{x},0)\).

To clarify, since the infinite nature might seem a little odd with the velocity if you put a dot on the mirror at time t=0, say at \(\mathbf{p}(0) = (p_{x},y_{0})\) then it is later at \(\mathbf{p}(t) = (p_{x},y_{0}) + t(v_{x},0) = (p_{x}+tv_{x},y_{0})\).

That's the setup in Frame S. We do a Lorentz boost in the y direction by \((0,V)\) to get Frame S'.

What is the new definition of the mirror, ie what is the Lorentz image of the equation \(y=y_{0}\)? Using the usual Lorentz transforms we get \(y' = \gamma(y_{0}-Vt)\). The mirror is now moving in the y' direction but it is still extended in the x' direction only. The mirror is not slanted, ie defined by an equation y=mx+c for non-zero m, it is still a horizontal line but one which moves up (or down, depending on the sign of V).

So that is the Lorentz image of the mirror as a whole, if you took a snap shot of it at some point in time in the new frame. But what about its original velocity in the x direction? What is it's Lorentz image? Well obviously nothing changes there, the boost was orthogonal to it. Thus we can compute the Lorentz image of \(\mathbf{p}(t)\). which by the above argument should have the new y' equation as it's y' component and the original expression as the x' component, which it indeed does.

So while a point on the mirror is moving diagonally, in both the x' and y' directions, the mirror's extension is still parallel to the x' axis, defined by a single value of y' at any given moment in time.

The velocity vector of the mirror in S was \((v_{x},0)\) and indeed the point in space corresponding to that value of distance (putting units aside etc) does indeed map to the same point but you forgot that that doesn't mean the mirror remains parallel to that vector. Instead the mirror's orientation is left unchanged and instead it is translated in the y direction. That's why I picked the y direction, this effect is nice and simple. A y direction boost on a y=constant line shifts it to a y = new constant + linear time term, which means it's still a y=constant line at any moment in time. The only way the mirror would remain in the same orientation to the velocity vector would be if it became slanted, ie satisfies y = linear equation in x (ie y = mx+c). It doesn't. All of the pieces of the mirror move diagonally but they do it in such a way as to laid out in a horizontal line and they remain that way.

You only considered a single point and assumed something incorrect about what that implied for the layout of the mirror. Do you understand what I have outlined or do you wish me to provide additional explanation (possibly a picture?), though it'll have to wait until morning for that (its 12.40am here).

 

First off, thank you for being polite this time and for trying to address the issue rather than launching into another personal attack.
Second off, the above is not the scenario in discussion since frame S is defined as the frame of the axle in the debate. In frame S, the velocity \(\vec{V_p}\) of any point on wheel circumference coincides with the velocity \(\vec{v_t}\) of any infinitesimally small region tangent to the wheel circumference. The coincidence is for any position and for any time, t: \(\vec{v_t}=\vec{V_p}\). Now pass the two vectors through the same boost in order to get from S to S' (ground frame).

 

Third off, if you wanted to study the much simpler , already solved case described much earlier in the debate (translating mirrors), then your boost should have been in the direction \((V,0)\), not in \((0,V)\). This is what the physics of the example dictates.

Actually, this is a misrepresentation of what I did.

Yes, I understood and no, that would not be necessary. Thanks for the offer, though.

 

That isn't how you should be thinking about it.

By that argument the cycloid should be a circle. You get a cycloid by doing a Galilean transform on the motion of a spinning wheel whose centre of mass is stationary, ie the velocity vector is parallel to the tangent of the wheel. In the new frame the velocity is obviously not tangent to the wheel. The velocity of the bit certainly is a vector and should transform as such but as the cycloid shows, the velocity vector at the 'front' of a rolling wheel isn't parallel to the tangent of the wheel.


The tangent line to the wheel isn't a vector as such, it's a set defined by the arrangement of the wheel's outer edge. We're considering the rolling wheel and the bit at the front. In the WCoM (wheel centre of mass) frame the velocity is parallel to the tangent line of the circle. This tangent line is vertical. Under a Lorentz transform in the x direction you get an ellipse. The ellipse is formed by squashing in the x direction, which leaves the vertical line vertical by precisely the argument I just used in my previous post about the infinite mirror (hence why it's directly relevant). Vertical line to vertical line, but the vector defined on that line is no longer parallel to the vertical line as we've boosted from a zero x component to a non-zero x' component.

That's your issue, you've been assuming you can work with the tangent vector when in fact you can't. As I just demonstrated with the mirror, vectors can go from horizontal to non-horizontal yet the layout of physical objects go from horizontal to horizontal, meaning their planes, tangents, normals etc don't change (other than x to x', y to y' etc).

My mirror example is directly applicable to the spinning wheel case, just you move right instead of up and down instead of right (ie rotate 90 degrees clockwise and you're looking at the bit of the mirrored wheel in question).

As I said, the simplest counter example to your claim is the cycloid itself. If your argument were valid then it wouldn't matter what transform you used as you claim you're only considering 1 vector for 2 things, so a Galilean transform is a valid counter example. Clearly the cycloid sweeps out a path where the velocity vector isn't tangent to the wheel's edge, yet it's obtained by doing a transform on the case where the velocity vector is tangent to the wheel's edge.

Anyway, I'm off to sleep. Hopefully you'll understand where you've made your mistake or at least see it is a mistake even if my less than perfect explanation doesn't give you understanding of it.

 

Yet, the physics of the problem is represented by the reflection of light off the moving infinitesimal section of the wheel tangent to the wheel. As such , the velocity of the infinitesimal section of the wheel is what enters the physics of reflection and not the tangent "line". So, you need to transform the velocity of the microfacet, not its tangent line. If what you are claiming is true, it would lead to a monumental failure of the light aberration formulas for the case when the angle between the mirror velocity and its direction of motion is zero. Anyways, I don't think that we will ever agree on this subject but thank you for being polite for a change.

 

Ok,

Let me try to explain better.

1. The light rays bounce of the infinitesimal portion of the rim. We (pete and I ) call this a "microfacet"
2. In the axle frame S, the microfacet describes a circular trajectory of equation:

\(x=r cos (\omega t)\)
\(y=r sin (\omega t)\) (eq.1)

3. In the ground frame S', the microfacet describes the Lorentz - transformed of eq (1), i.e. a different trajectory of equation :

\(x'=\gamma V (t+\frac{cos (\omega t)}{\omega})\)

\(y'=V \frac{sin (\omega t)}{\omega}\)

\(V=r \omega\)

(an elongated cycloid)
4. Point is, that you are trying to compare the microfacet trajectories in two different frames, a mistake commonly known in special relativity as "frame jumping".

5. As the result of your "frame jumping" you get the tangents not matching.

6. There is no reason for them to match since they represent the trajectories in different frames so any attempt at matching them is meaningless.

7. Now, in both frames, S and S', the microfacets move in their own plane. The proof is trivial in the axle frame, you can do it for yourself in the ground frame, it is not difficult.

8. As such, the Doppler shift is zero in the axle frame and it is also zero in the ground frame.

9. I appreciate you being so persistent in your beliefs that I am wrong and you are right, it gave me the opportunity not only to find the errors in your reasoning but to also develop yet another proof (about the 5-th to my counting)

10. BTW, the "ellipse" plays no role in the whole argumentation.

Yes, I understood all right. You are getting condescending again. This is too bad, especially given the fact that the error is yours.

 

Tach, you really don't seem to be reading what I'm saying.

I consider both frames. In the first frame I explain how the directions line up. I then go on to explain how applying a transform (it doesn't even matter what, be it Galilean or Lorentz) leads you to a situation where they don't line up. I am not frame mixing. I know the equations are different, I have been over this again and again.

As for the ellipse, your own equations describe the position of a point moving around an ellipse which is moving (ie the Lorentz image of the rolling wheel) so it is relevant. You don't even understand your own equations.

Tach, you really aren't in a position to complain about other people being condescending, you are one of the more ruder and abrasive posters here. You can't behave as you do and then turn around and complain when I offer to walk you through your mistakes, which you consider condescending, and not be a hypocrite.

There is a Doppler shift, I've previously given you the algebra which shows that for the front of the wheel with the observer with a torch in front of it. The zero angle between a mirror's plane of extension and it's velocity is not frame invariant either, I've walked you through that. I've also explained how it doesn't apply to a wheel either. You repeatedly say "That's irrelevant" or "It plays no role!" yet you've shown you don't read what I say, demanding I consider examples I literally just addressed.

I'm wondering if you're even reading half of what people say to you. You're certainly not thinking about it.

 

I've been looking in the mirror for five years. I haven't made any arguments without justification and evidence. Whilst I generally don't give any mathematics, do note that many of my arguments concern the meaning of mathematical terms, such as t. I can't use mathematics to address that.

 

The direction of the motion of a point on the circumference lines up with the direction of the motion of the microfacet in both the axle frame and the ground frame. Post 145 gives you the formalism explaining why this is true.

Actually, your explanation doesn't hold since you are trying to compare the tangent to the ellipse to the tangent to the cycloid, so you are doing worse than jumping frames as evidence by the quotes below:

The ellipse is simply the locus of the points (x',y') measured in S' for the same t. That is, the simultaneity is considered in S when the space coordinates are considered in S'. As such, the "ellipse" is just a snapshot, devoid of any physical meaning. Therefore, there is no meaning in your attempt at comparing the tangent to the ellipse (snapshot) with the tangent to the cycloid and concluding that they do not coincide.

 

I've only been following this thread casually, but is this what Tach is saying? I thought he was claiming a point on the rim and its tangent plane share a velocity vector, which I believe to be true but irrelevant. If your interpretation is correct I'd say the comment is quite curious indeed...

 

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

Again, see the answer at 160

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