You didn't even try. Stop dodging.
Angle between the orientation of a moving object and its velocity
- Thread starter Pete
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You didn't even try. Stop dodging.
What is the relevance? You are being asked to transform zero angles, nothing to do with "v approaching c". This is very straightforward stuff.
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Frankly, I am not interested in your bed time stories, if you can't write proper math, there is no way of falsifying your statements.
Yes there is. It's pretty trivial actually.
Ok,
I had a look, it is all nonsense. What you are being asked to figure out is what happens to an infinitely small portion of the tangent plane to the circumference (the piece that we call a microfacet) when the wheel rolls an infinitely small amount of time $$dt$$. Since the motion of the mirror happens to coincide with the tangential velocity in all frames, according to the theory of the Doppler effect, there will be a null shift.
What you did is totally irrelevant, you turned the wheel by $$2 \pi$$ and you claimed that the point under study advanced by $$2r$$ in the direction of the motion of the axle. From that, you conclude that there is an x component to the velocity (many others have noticed this before you).
There is also a y component to velocity. As long as the velocity and the tangent to the surface are colinear, there is zero Doppler effect.
Good job! But this is not how the microfacet moved, this is how a point on the circumference moves after a complete rotation.
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Tach appears to be claiming that Pete's surface A in post #2 of this thread makes the same angle with the x and x' axes, where the primed frame moves parallel to the x axis.
If so, clock this one up as another basic error by Tach.
Also, why is Tach apparently incapable of answering "Yes" or "No" to these two questions:
I don't think Tach knows what he agrees about. He's lost in his own mathematical maze, as usual.
If so, clock this one up as another basic error by Tach.
Also, why is Tach apparently incapable of answering "Yes" or "No" to these two questions:
Pete said:
I don't think Tach knows what he agrees about. He's lost in his own mathematical maze, as usual.
Nope, not at all. Clock this one as another redface for JamesR.
Why don't you let me discuss things with pete without butting in? At least, he understands the issue and he is capable of putting his thoughts into a mathematical formalism. His errors are much, much smaller than yours.
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Before we proceed further, I reproduce my post unedited and in full:
Yes, I'm aware of that.
Well, that's your term for it, I can think of better, less ambiguous terms.
Again, I'm aware of that.
No, this is only true in one reference frame, the non-rotating, co-moving reference frame of the axel.
No, that is not what I did. This is what I did:
Remind me. What direction is the axel moving in within its own restframe??
I didn't deal with the x component of the velocity. I dealt with the component of the velocity of a point on the circumference of your wheel, that is perpendicular to the plane of the tangent at that point. I then demonstrated how it was non zero in the rest frame of the axel, thus demonstrating according to Pauli's reasoning, which you have accepted, that it must show some doppler shift.
I dealt with neither the x component, nor the y component.
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.
Never mind, I thought that you were doing something decent, like calculating in the ground frame.
You are completely predictable Tach.., and absolutely hilarious!
Perhaps that you could contribute something of value instead of trolling. On second thoughts, scratch that, it isn't realistic. Could you, at least, refrain from trolling? You are wasting space and bandwidth.
Tach, I know I'm just trolling......, but doesn't something seem a little contractictory above?
Isn't an infinitely small portion, kind of like the same thing as a point? And wouldn't the point on the circumference and the associated point on the tangential plane, differ by less that a point? Can you even get less than a point difference?
Yes, but I'm still struggling to make out a completely consistent theme in what you're saying.
Are you making a distinction between the (flat) ground, and a tangent to the ground?
When you say tangent to the ground, I'm thinking a displacement vector between two simultaneous events on the ground.
Is that what you mean?
Or do you mean something different, such as a tangent lightlike worldine?
It seems to me that you are saying that the ground itself is not horizontal in the elevator frame, but then you imply that this is not what you're saying.
Maybe this has just been a huge case of mistaken understanding.
I don't believe you have ever addressed my derivation of $$theta'_A$$. If I'm wrong, I apologize, and can you please quote what you said.
Is there a mistake in the following? If so, where?
A is defined by:
$$y = x \tan(\theta_A)$$
Transforming A, we find:
$$y' = \gamma\tan(\theta_A) (x' + vt')$$
The angle between A' and the x'-axis is therefore:
$$\tan(\theta'_A) = \gamma\tan(\theta_A)$$
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.
I explained that you have no reason for treating $$\theta'_A$$ any different from $$\theta'_P$$. The error in your final result comes exactly from treating them via two different approaches. 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. 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 the velocity of the surface wrt to the ground when we pass from the axle frame to the ground frame. The final result must be consistent with the aberration formula, i.e. it must transform zero angles into zero angles.
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:
$$\begin{align}
x &= t |\vec{V_p}| \cos(\theta_P) \\
y &= t |\vec{V_p}| \sin(\theta_P)
\end{align}$$
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.
All I can see in your link there is the trivial claim that the tangent to a surface lies in the plane of the surface in any frame. That doesn't start to address the questions Pete has put to you in this thread.