Let's say there are two inertial reference frames, F0 and F1 with relative velocity V=c*sqrt(3)/2 along the x axis. In frame F0, there is a large circular ring center at (0,0). The diameter of this ring is D. Per relativity, F1 observers measure the distance from top to bottom (in the y direction) to be D, and in the x direction to be D/2. Now if this ring is rotating with angular velocity V around the center as measured in F0 instead of having no rotation about the center, why do F1 observers measure that the shape of the rotating ring is identical to the shape of the non-rotating ring?
F1 observers measure that the instantaneous velocity along the top of the rotating ring is zero, and the instantaneous velocity along the bottom of the rotating ring is much greater than V. So why doesn't length contraction cause F1 observers to see a difference in the shape of a stationary ring versus the shape of a ring rotating with velocity V?
David Seppala
Bastrop TX

sepp623@yahoo.com wrote:
> Let's say there are two inertial reference frames, F0 and F1 with relative velocity V=c*sqrt(3)/2 along the x axis. In frame F0, there is a large circular ring center at (0,0). The diameter of this ring is D. Per relativity, F1 observers measure the distance from top to bottom (in the y direction) to be D, and in the x direction to be D/2. Now if this ring is rotating with angular velocity V around the center as measured in F0 instead of having no rotation about the center, why do F1 observers measure that the shape of the rotating ring is identical to the shape of the non-rotating ring?
> F1 observers measure that the instantaneous velocity along the top of the rotating ring is zero, and the instantaneous velocity along the bottom of the rotating ring is much greater than V. So why doesn't length contraction cause F1 observers to see a difference in the shape of a stationary ring versus the shape of a ring rotating with velocity V?
> David Seppala
> Bastrop TX

David, why are you doing this? Why are you doing this for DECADES?

On Tuesday, November 9, 2021 at 8:17:17 AM UTC-8, sep...@yahoo.com wrote:
> Let's say there are two inertial reference frames, F0 and F1 with relative velocity V=c*sqrt(3)/2 along the x axis. In frame F0, there is a large circular ring center at (0,0). The diameter of this ring is D. Per relativity, F1 observers measure the distance from top to bottom (in the y direction) to be D, and in the x direction to be D/2.

Correct

>Now if this ring is rotating with angular velocity V around the center as measured in F0 instead of having no rotation about the center, why do F1 observers measure that the shape of the rotating ring is identical to the shape of the non-rotating ring?

You are incorrect, the circle appears as an ellipse, rotated around the vertical axis. There is a lot of literature on the subject that explains the issue in great detail. Google "Penrose-Terrell effect". It is true that a SPHERE will appear as uncontracted and there are good reasons for that: https://www.google.com/search?client=firefox-b-1-d&q=Penrose+Terrell+effect

> F1 observers measure that the instantaneous velocity along the top of the rotating ring is zero, and the instantaneous velocity along the bottom of the rotating ring is much greater than V. So why doesn't length contraction cause F1 observers to see a difference in the shape of a stationary ring versus the shape of a ring rotating with velocity V?

Google "Penrose-Terrell effect".

On Tuesday, 9 November 2021 at 17:28:47 UTC+1, Python wrote:
> sep...@yahoo.com wrote:
> > Let's say there are two inertial reference frames, F0 and F1 with relative velocity V=c*sqrt(3)/2 along the x axis. In frame F0, there is a large circular ring center at (0,0). The diameter of this ring is D. Per relativity, F1 observers measure the distance from top to bottom (in the y direction) to be D, and in the x direction to be D/2. Now if this ring is rotating with angular velocity V around the center as measured in F0 instead of having no rotation about the center, why do F1 observers measure that the shape of the rotating ring is identical to the shape of the non-rotating ring?
> > F1 observers measure that the instantaneous velocity along the top of the rotating ring is zero, and the instantaneous velocity along the bottom of the rotating ring is much greater than V. So why doesn't length contraction cause F1 observers to see a difference in the shape of a stationary ring versus the shape of a ring rotating with velocity V?
> > David Seppala
> > Bastrop TX
> David, why are you doing this? Why are you doing this for DECADES?

You, on the other hand - for the same decades are spitting
and barking at the people opposing your moronic ideology;
but your reason are obvious - that's what your Shit has
trained you for.

Maciej Wozniak wrote:

>> David, why are you doing this? Why are you doing this for DECADES?
>
> You, on the other hand - for the same decades are spitting and barking
> at the people opposing your moronic ideology; but your reason are
> obvious - that's what your Shit has trained you for.

the Poles hate the Russians with a passion but still do not mind begging
for cheap gas from them. They have no shame whatsoever.

On Tuesday, November 9, 2021 at 8:17:17 AM UTC-8, sep...@yahoo.com wrote:
> [Usual erroneous expressions and confusions... leading to:]
> Why doesn't length contraction cause F1 observers to see a difference in the shape of
> a stationary ring versus the shape of a ring rotating with velocity V?

This is yet another standard question posed by countless clueless newbies. The answer is that the density distribution around the rim is non-uniform and is dependent on the speed of rotation of the ring. If you imagine spokes emanating from the hub to the rim, those spokes are not linear in terms of F1 if the wheel is rotating, and they yield the requisite distribution around the rim, consistent with the Lorentz transformation. This (like all your other questions) is Relativity 101. (It is not related to Terrell-Penrose optical image effects.)

Special Relativity: 844 ... Barnpole Dave: 0

Al Coe wrote:

> This is yet another standard question posed by countless clueless
> newbies.
> The answer is that the density distribution around the rim is
> non-uniform
> and is dependent on the speed of rotation of the ring. If you imagine
> spokes emanating from the hub to the rim, those spokes are not linear in
> terms of F1 if the wheel is rotating, and they yield the requisite
> distribution around the rim, consistent with the Lorentz transformation.
> This (like all your other questions) is Relativity 101. (It is not
> related to Terrell-Penrose optical image effects.)
>
> Special Relativity: 844 ... Barnpole Dave: 0

ohh my, relativity wouldn't care about your density distribution. It's
about the spacetime and whatever things in it.

On Tuesday, November 9, 2021 at 11:53:05 AM UTC-6, Al Coe wrote:
> On Tuesday, November 9, 2021 at 8:17:17 AM UTC-8, sep...@yahoo.com wrote:
> > [Usual erroneous expressions and confusions... leading to:]
> > Why doesn't length contraction cause F1 observers to see a difference in the shape of
> > a stationary ring versus the shape of a ring rotating with velocity V?
> This is yet another standard question posed by countless clueless newbies.. The answer is that the density distribution around the rim is non-uniform and is dependent on the speed of rotation of the ring. If you imagine spokes emanating from the hub to the rim, those spokes are not linear in terms of F1 if the wheel is rotating, and they yield the requisite distribution around the rim, consistent with the Lorentz transformation. This (like all your other questions) is Relativity 101. (It is not related to Terrell-Penrose optical image effects.)
>
> Special Relativity: 844 ... Barnpole Dave: 0
So you agree that the rotating ring and the non-rotating ring have the identical shapes. When you say the answer is in regard to the density distribution, please describe the differences in the following. If the rotating ring has a huge, huge diameter and it is rotating at a angular velocity V as measured in F0, and a bar is made of the same material, and it is traveling in a straight line with velocity V relative to F0, with this huge, huge diameter of the ring there is an extremely small centripetal force on the ring, please explain how the density of the rotating ring differs from the density of the bar traveling in a straight line.
David Seppala
Bastrop TX

On Tuesday, 9 November 2021 at 18:53:05 UTC+1, Al Coe wrote:
> On Tuesday, November 9, 2021 at 8:17:17 AM UTC-8, sep...@yahoo.com wrote:
> > [Usual erroneous expressions and confusions... leading to:]
> > Why doesn't length contraction cause F1 observers to see a difference in the shape of
> > a stationary ring versus the shape of a ring rotating with velocity V?
> This is yet another standard question posed by countless clueless newbies.. The answer is that the density distribution around the rim is non-uniform and is dependent on the speed of rotation of the ring. If you imagine spokes emanating from the hub to the rim, those spokes are not linear in terms of F1 if the wheel is rotating, and they yield the requisite distribution around the rim, consistent with the Lorentz transformation. This (like all your other questions) is Relativity 101. (It is not related to Terrell-Penrose optical image effects.)

In the meantime in the real world, however, forbidden
by your moronic religion GPS clocks keep measuring
t'=t, just like all serious clocks always did.

On Tuesday, November 9, 2021 at 10:04:05 AM UTC-8, sep...@yahoo.com wrote:
> If the rotating ring has a huge, huge diameter and it is rotating at a angular velocity V...

The correct term for what you mean is circumferential speed. The phrase "angular velocity" refers to the quantity dtheta/dt, i.e., it is a rate of change of an angle.

> as measured in F0, and a bar is made of the same material, and it is traveling in a
> straight line with velocity V relative to F0... please explain how the density of the
> rotating ring differs from the density of the bar traveling in a straight line.

Not at all... provided the material in the ring is unstressed. Needless to say, if you construct the ring while not rotating, and then set it into rotation, the circumference and radius would shrink if not constrained. But you can contrive to have a rotating ring of essentially unstressed material, which by definition would match a co-moving bar for each segment. This was entailed by the previous answer. The density corresponds to the speed, in accord with relativistic length contraction. Again, this is all Relativity 101. Understand?