Angle between the orientation of a moving object and its velocity

[Tach]

This is for Tach, who claims that if a surface is moving parallel to itself in one reference frame, it is move parallel to itself in all reference frames


First, the easy handwaving rebuttal:

An observer in a train looks at the ground, and notes that the ground is moving parallel to itself.

An observer in an elevator looks at the same ground, and notes that the ground is moving perpendicular to itself.

Therefore, Tach is wrong.

Ahh, you are making the same misrepresentation of my claim that AN did.
I thought that we were on the same page, you are clearly misrepresenting my claim.
What I claim is that a surface having its velocity colinear with its tangent in one frame F, has it colinear in any other (inertial) frame F' moving wrt F. Like for example the motion of the microfacets of a spinning wheel.
In discussion it is NOT the direction of surface motion wrt the observer(s) BUT the ANGLE between the surface tangent and its velocity.
I am really surprised about your above blunder, we have been discussing this subject for so long, I thought that it was very clear. Now, you come up with this nonsense?

Now, that I have answered this silly question you've been hounding me with, please answer mine.

 

[Trippy]

Didn't Pauli say that it was the motion perpendicular to the mirror that was the only motion that was important?

 

[Pete]

No, ir is your (mis)interpretation of special relativity

Wrong again.
You clearly don't know what I'm thinking.

First off, you never addressed my contentions, you simply pretended that they don't exist. Tell you what, address my contentions and I will answer your questions.

Still avoiding, I see.
Address the elementary scenario I posted in the OP, then I'll consider the different scenario you designed.

What do you think post 5 is?

A mistake.
You used:
$$\tan(\theta'_A) = \frac{|\vec{V_p}| \sin(\theta_A)}{\gamma(|\vec{V_p}| \cos(\theta_A) - v)}$$
Instead of:
$$\tan(\theta'_A) = \gamma\tan(\theta_A)$$

 

[Tach]

Didn't Pauli say that it was the motion perpendicular to the mirror that was the only motion that was important?

Pauli (pg 95-96) shows that:


-the component of the motion along the normal to the mirror surface produces Doppler effect.

-mirrors moving colinear with their tangent (as in a spinning wheel) produce no Doppler effect

 

[Tach]

Still avoiding, I see.
Address the elementary scenario I posted in the OP, then I'll consider the different scenario you designed.

I did. See post 41. It is another one of your blunders. I can't believe that you would make such a blunder after all that we have gone through.

A mistake.
You used:
$$\tan(\theta'_A) = \frac{|\vec{V_p}| \sin(\theta_A)}{\gamma(|\vec{V_p}| \cos(\theta_A) - v)}$$
Instead of:
$$\tan(\theta'_A) = \gamma\tan(\theta_A)$$

I know what you used, it is wrong. This is why you end up contradicting the results of the aberration formulas. So, your model is unphysical.

 

[Trippy]

Pauli (pg 95) shows that:


-the component of the motion along the normal to the mirror surface produces Doppler effect.

-mirrors moving colinear with their tangent (as in a spinning wheel) produce no Doppler effect

So then all one has to do to prove you wrong about rolling wheels is prove that an elementary point on a mirror has some component of motion that is perpendicular to the surface of the mirror, and you will admit you are in error right?

 

[Pete]

Ahh, you are making the same misrepresentation of my claim that AN did.
I thought that we were on the same page, you are clearly misrepresenting my claim.
What I claim is that a surface having its velocity colinear with its tangent in one frame F, has it colinear in any other (inertial) frame F' moving wrt F.

Right, that's what I said.

Call the the train frame F.
The surface (ie the ground) has its velocity colinear with its tangent.
Right?

Call the elevator frame F'
The surface (ie the ground) has its velocity perpendicular to its tangent.
Right?

In discussion it is NOT the direction of surface motion wrt the observer(s) BUT the ANGLE between the surface tangent and its velocity.

It's velocity with rest

Velocity with respect to what?

I thought that it was very clear.

It seems it's not clear at all.
You seem to be making a distinction between "the direction of surface motion wrt the observer/s" and the angle of "its velocity"surface velocity.

What do you mean?

 

[Tach]

Right, that's what I said.

What you said shows that you are not even understanding the basics of the discussion.

Call the elevator frame F'
The surface (ie the ground) has its velocity perpendicular to its tangent.
Right?

Wrong. The surface velocity and its tangent are still in the same plane. You are transforming two colinear vectors and you are claiming that they are no longer colinear in the boosted frame (the elevator). The vectors made a zero angle in frame F, they make a zero angle in frame F'. This is why your formalism does not recover the aberration formulas correctly.

 

[Tach]

So then all one has to do to prove you wrong about rolling wheels is prove that an elementary point on a mirror has some component of motion that is perpendicular to the surface of the mirror, and you will admit you are in error right?

Yes, this is the discussion with pete. You will not be able to prove that because you would be contradicting the formulas of light aberration (see post 4) that tell you that zero angles are frame invariant. Since the angle between the microfacets forming the circumference and their velocity is zero in the axle frame, it follows that they are also zero in all other frames in inertial motion wrt the axle. What trips people is that while the velocity is affected by the boost, so is the equation of the circumference so the colinearity is preserved.

 

[Trippy]

Yes, this is the discussion with pete. You will not be able to prove that because you would be contradicting the formulas of light aberration (see post 4) that tell you that zero angles are frame invariant. Since the angle between the microfacets forming the circumference and their velocity is zero in the axle frame, it follows that they are also zero in all other frames in inertial motion wrt the axle. What trips people is that while the velocity is affected by the boost, so is the equation of the circumference so the colinearity is preserved.

Not only will I prove it, but I will prove it using only the classical limit, so there can be no worming out, and no whining about boosts being mis-applied. Will you accept this?

 

[Tach]

Not only will I prove it, but I will prove it using only the classical limit,

I think that you are getting a false start. Did you not get the part about contradicting the relativistic aberration? How are you going to deal with this if you are using "classical" limit?

so there can be no worming out, and no whining about boosts being mis-applied. Will you accept this?

How about you do it right, rather than trying to hack it?

 

[Trippy]

I think that you are getting a false start. Did you not get the part about contradicting the relativistic aberration? How are you going to deal with this if you are using "classical" limit?

How about you do it right, rather than trying to hack it?

Because, as you have stated repeatedly, zero angles are preserved by relativistic aberation, but, IIRC as v approaches c, non-zero angles are still non zero - they may approach zero, but they are still non-zero.

So all I have to do is demonstrate it's existence in the classical limit, and if it exists in the classical limit, it will still be there in the relativistic limit.

Besides which, I can always perform the relativistic corrections at a later time when things are less chaotic at my end.

But hey, if you're too scared...

 

[Pete]

Wrong. The surface velocity and its tangent are still in the same plane.

Are you referring to a tangent to the velocity, or a tangent to the surface?

In the elevator frame, the ground (or the tangent to the ground, if you prefer) is horizontal. Right?

I know what you used, it is wrong.

So you say.

It looks fine to me:
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:
$$\tan(\theta'_A) = \gamma\tan(\theta_A)$$

If there's a mistake there, I'd be happy for you to point it out.

 

[James R]

An observer in a train watches the ground passing by, and notes that the ground's velocity is parallel to the ground itself.
Do you agree?

An observer in an elevator watches the ground falling away, and notes that the ground's velocity is perpendicular to the ground.
Do you agree?

Why is Tach apparently incapable of answering "Yes" or "No" to these two questions?

 

[Tach]

Because, as you have stated repeatedly, zero angles are preserved by relativistic aberation, but, IIRC as v approaches c, non-zero angles are still non zero - they may approach zero, but they are still non-zero.

This is false, the aberration formula transforms zero angles into exact zero angles.

But hey, if you're too scared...

Why would I be scared about your total lack of understanding of relativistic physics?

 

[Tach]

Are you referring to a tangent to the velocity, or a tangent to the surface?

Tangent to the surface. We have been talking about this for hundreds of posts.

In the elevator frame, the ground (or the tangent to the ground, if you prefer) is horizontal. Right?

This is where you go wrong, it is not. Go back to see how vectors transform (see the writeup I linked eralier for you):

$$r'=r+V(\frac{\gamma-1}{V^2}r.V+\gamma t)$$

$$dr'=dr+V(\frac{\gamma-1}{V^2}dr.V+\gamma dt)$$

Since the motion is transverse:

$$dr.V=0$$ so

$$dr'=dr+V \gamma dt$$

$$dr'_x=\gamma V dt$$

$$dr'_y=dr_y$$

So you say.

It looks fine to me:
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:
$$\tan(\theta'_A) = \gamma\tan(\theta_A)$$

If there's a mistake there, I'd be happy for you to point it out.

I already did, several times, starting with post 5. You insist that there isn't so we reached an impasse.

 

[Trippy]

Here's the prose version, Tach. Do try and follow it.

I have a mirrored pipe, of zero thickness.
Its C[sub]n[/sub] axis is stationary, but the pipe is rotating around it IE we are in the rest frame of the axis (but not co-rotating).
I pause the rotation, at some arbitrary point, and measure its position relative to the axle to be $$(r_1,\theta_1)$$ where r is the radius of our pipe, and $$\theta$$ is the angle of rotation.
I then rotate it through an angle of $$2\pi$$ radians, with an angular velocity of $$\omega$$, and measure it's position again $$(r_2,\theta_2)$$. Performing the neccessary vector addition, I find that the point I am interested in has moved with a displacement of 2r in a time of $$\frac{2\pi}{\omega}$$ seconds, along a vector that is oriented $$\theta_1 +\pi$$ radians from the vertical. I also find that this vector is perpendicular to the tangents representing the mirror plane at both points.

Q.E.D.

 

[OnlyMe]

Why is Tach apparently incapable of answering "Yes" or "No" to these two questions?

It would require a commitment. Yes or no, leaves no wiggle room.

 

[Trippy]

This is false, the aberration formula transforms zero angles into exact zero angles.

Why would I be scared about your total lack of understanding of relativistic physics?

Once again, your prose literacy, it seems, fails you.

Allow me to repeat myself:

Because, as you have stated repeatedly, zero angles are preserved by relativistic aberation, but, IIRC as v APPROACHES c, NON-zero angles are still NON zero - they may APPROACH zero, but they are still NON-zero.

I have hilighted the salient key words and phrases for you, seeing as how you seem to have missed them the first time around (again).

 

[Tach]

Here's the prose version, Tach. Do try and follow it.

Please try putting this in a mathematical form, so I can point out your errors.