Angle between the orientation of a moving object and its velocity

No, just most of us have better things to do than watch you repeat the same multiple times debunked nonsense. Trippy's summary is bang on. No one has ever argued that in the axle rest frame the wheel doesn't give a Doppler shift. However, in any other frame some parts of the wheel do. I even explicitly did the algebra for you in the 2+1 dimensional case. It amounts to the same example I gave with the flat mirror, a boost not in the plane of the mirror leads to it giving a velocity in that direction. Really, it comes down to that! Mirror extended in x direction moves in x direction. Boost in y direction. Mirror now moves in both the x' and y' directions but is extended in x' direction. Your claim is demonstrated false. Do I need to do the transform or can you realise that boosting in the y direction alters the velocity in the direction?

I'm beginning to wonder if you do understand it because it's very basic stuff.

Mirror extended in x direction has velocity \(\mathbf{v} = (v_{x},0)\) in Frame S. Apply a Lorentz boost in direction \((0,v_{y})\). What's the velocity of the mirror in the new frame? Does it have non-zero y' component? If so your claim is false. Can you compute the velocity yourself or do you need to be provided with a circle of paper and a crayon?

 

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Prove it. You have asserted this several times, yet you never backed it up.

Which was obviously wrong since you failed to understand that each portion of the wheel moves parallel with its tangential speed. I have just given you the formal proof that this is the case. Now, if you understood the references (Pauli, Bateman), you would have understood that this is the classical case of zero Doppler effect.

Yes, I am wondering the same thing about you.

You repeat the same error over and over. If the mirror moves in the direction of its own plane (i.e. NO component along its normal) then, there is NO Doppler effect.(see references) This is the case of a rolling circular rim.

References

1. W. Pauli "Theory of relativity", Pergamon Press, p. 95, 1958(!)
2. H. Bateman , Phil. Mag, 18, (1908!!!)

 

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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.

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.

 

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Lets make this simpler. A mirror is extended in the x direction. Its velocity is in the x direction. I boost in the y direction. Is it now extended in only the x' direction?Does it have nonzero y' velocity component. Yes or no answers suffice.

 

Yes, it does have a velocity in the y direction
And yes, your example is totally irrelevant for the case in study.
Now that I answered your question, you answer mine please: you bounce a ray of light off the mirror in your example, while the mirror is moving between the source and the receiver. The mirror moves in its plane in the frame of the receiver-emitter (i.e. there is no velocity component aligned with the normal to the mirror surface). Does the receiver receive the same frequency as emitted from the source? Yes or no?

 

Basic geometry shows that the two of them have the same exact unit vector for every point on the circumference; both vectors rotate together, glued to each other , at all times and for all points on the circumference. The explanation is about the transformation of zero angles between directions. Time varying vectors are used to specify directions.

 

Look, the observer O, in the center of the wheel.
A second observer O', is in the frame of the source+receiver.
You need to do a Lorentz transformation from O's frame to the frame of O'. This has already been done for you, in post 145, all it is left is for you to try to understand the math. I cannot do that for you, you need to try doing it by yourself. What I can tell you about your picture is that the red line, the tangent to the circle, you need to draw one for each of your points, \(P_1, P_2,....\) just like you did for \(P_0\). In other words, you measuring \(P_1S_1, P_2S_2,...\) has no bearing whatsoever in determining the Doppler effect. Once you understand post 145, there is a better chance that you understand how the problem is actually solved.

 

This is precisely what I am telling you, in order for you to understand the explanation, it needs to be.

As long as you stubbornly insist on this, you will not understand how the problem is solved.

 

The math in post 145 contradicts your claim.

No, all points on the wheel are observed just the same.

Distance has nothing to do with the Doppler effect, you are now making the same error as Neddy Bate.

 

In the axle frame, yes.
But the opening posts show that the orientation of a surface transforms differently to a velocity vector.

In the ground frame, for example, the orientation of the tangent plane is tangent to the ellipse of the wheel, while its velocity is tangent to the cycloid trajectory.

 

Not when the tangent to the surface is a time-varying vector. You chose a surface fixed in time in your OP.

In the ground frame, the velocity is the Lorentz transform of the velocity vector in the axle frame.
In the ground frame, the tangent vector is the Lorentz transform of the tangent vector in the axle frame.
The points on the rim describe a tube of cycloids, not only one cycloid. The incremental positions of the ellipse, also describe a smeared band. It is impossible (at least for me) to figure out which tangent matches with what velocity vector, the Lorentz transforms keep things clean and clear: we have two time varying vectors that are coincident in the axle frame, they rotate together, one needs to Lorentz - transform both of them , through the same Lorentz boost, in the ground frame. They pass through the same transformation, not through two different kinds of transforms.

 

This rotating-mirror discussion might seem like futile trivia, but there is an important lesson to be learnt, to do with psychology. Tach is so utterly convinced that he's right that no amount of evidence or logic or anything else is going to make a blind bit of difference. He cannot supply a convincing argument to support his conviction, and is utterly immune to rock-solid counter-arguments. You might then be tempted to think That's Tach for you, but don't. This trait is commoner than you think. Look in the mirror and ask yourself this: do I do this sort of thing? And if you think the answer is no, don't be too sure of it. Because people who suffer from conviction never think that they do.

 

Do you look in the mirror also Farsight? How many arguments have you made without justification, evidence, rigour or mathematics?