If there are two inertial reference frames, F0 and F1, with a relative velocity of V in the x-direction, if there is a circle of radius R centered at (0,0) in F0, that circle is an oval shape for viewers in F1. If that circle is rotating around (0,0) at some angular velocity in F0, why do viewers in F1 see the shape as being identical to the non-rotating circle while they view the top of the circle moving at a different angular velocity than the bottom of the circle?
David Seppala
Bastrop TX

On Thursday, April 28, 2022 at 4:27:14 PM UTC-7, sep...@yahoo.com wrote:
> If there are two inertial reference frames, F0 and F1, with a relative velocity of V in
> the x-direction, if there is a circle of radius R centered at (0,0) in F0, that circle is
> an oval shape for viewers in F1.

Correction: It's spatial shape is an oval (an ellipse, to be exact) in terms of F1. Talking about "viewers" is misleading unless your intention is to discuss relativistic optics... which it obviously is not.

> If that circle is rotating around (0,0) at some angular velocity in F0...

Using the word "circle" for a rotating ring or disk is misleading. If you are meaning a solid object like a circular ring or disk, than say so. Note that, in general, setting an initially non-rotating ring or disk into rotation about its axis necessarily results in some deformation due to circumferential length contraction, so if your interest was in how the object deforms when it is set into rotation, you would need to specify the properties of the material and so on. Of course, you can posit a rotating ring with a certain radius, but you can't assume it is the same radius it would have if the rotation was stopped, etc. This is all just Relativity 101.

> Why do viewers [sic] in F1 see [sic] the shape as being identical to the non-rotating
> circle [sic] while they view the top of the circle [sic] moving at a different angular
> velocity than the bottom of the circle [sic]?

The subject of your post was "why no length contraction", but now you aren't asking why no length contraction (which would be a non sequitur), you are just asking why the description of the rotating ring or disk in terms of F1 exhibits the well-known and expected relativistic effects. The reason it exhibits the relativistic effects is because physics is locally Lorentz invariant. This is just Relativity 101.

Special Relativity: 868.... Barnpole Dave: 0

On Thursday, April 28, 2022 at 8:28:26 PM UTC-5, Al Coe wrote:
> On Thursday, April 28, 2022 at 4:27:14 PM UTC-7, sep...@yahoo.com wrote:
> > If there are two inertial reference frames, F0 and F1, with a relative velocity of V in
> > the x-direction, if there is a circle of radius R centered at (0,0) in F0, that circle is
> > an oval shape for viewers in F1.
> Correction: It's spatial shape is an oval (an ellipse, to be exact) in terms of F1. Talking about "viewers" is misleading unless your intention is to discuss relativistic optics... which it obviously is not.
>
> > If that circle is rotating around (0,0) at some angular velocity in F0....
>
> Using the word "circle" for a rotating ring or disk is misleading. If you are meaning a solid object like a circular ring or disk, than say so. Note that, in general, setting an initially non-rotating ring or disk into rotation about its axis necessarily results in some deformation due to circumferential length contraction, so if your interest was in how the object deforms when it is set into rotation, you would need to specify the properties of the material and so on. Of course, you can posit a rotating ring with a certain radius, but you can't assume it is the same radius it would have if the rotation was stopped, etc. This is all just Relativity 101.
>
> > Why do viewers [sic] in F1 see [sic] the shape as being identical to the non-rotating
> > circle [sic] while they view the top of the circle [sic] moving at a different angular
> > velocity than the bottom of the circle [sic]?
>
> The subject of your post was "why no length contraction", but now you aren't asking why no length contraction (which would be a non sequitur), you are just asking why the description of the rotating ring or disk in terms of F1 exhibits the well-known and expected relativistic effects. The reason it exhibits the relativistic effects is because physics is locally Lorentz invariant. This is just Relativity 101.
> op
> Special Relativity: 868.... Barnpole Dave: 0
If any object in F0 in the x-y plane is circular or any object has a circular hole in it, frame F1 observers moving with V in the x direction view the object as an elliptical shape or if it has a hole they view the hole as an elliptical shape. Now, if any of those objects rotate in the x-y plane around a stationary point in F0, observers in F1 view the object with the identical shape of the now rotating object even though the top is rotating in the opposite x direction than the bottom is rotating. So length contraction doesn't happen for some reason. This is true no matter what the object is composed of or its relative angular velocity. If it is viewed as a circular shape in F0, then F1 says the shape when rotating is identical to the object's shape when it is not rotating, or if one object is rotating clockwise and an identical object is rotating counter-clockwise they have the identical shape as viewed in F1.
David Seppala
Bastrop TX

On Friday, April 29, 2022 at 12:37:19 PM UTC-5, sep...@yahoo.com wrote:
> On Thursday, April 28, 2022 at 8:28:26 PM UTC-5, Al Coe wrote:
> > On Thursday, April 28, 2022 at 4:27:14 PM UTC-7, sep...@yahoo.com wrote:
> > > If there are two inertial reference frames, F0 and F1, with a relative velocity of V in
> > > the x-direction, if there is a circle of radius R centered at (0,0) in F0, that circle is
> > > an oval shape for viewers in F1.
> > Correction: It's spatial shape is an oval (an ellipse, to be exact) in terms of F1. Talking about "viewers" is misleading unless your intention is to discuss relativistic optics... which it obviously is not.
> >
> > > If that circle is rotating around (0,0) at some angular velocity in F0...
> >
> > Using the word "circle" for a rotating ring or disk is misleading. If you are meaning a solid object like a circular ring or disk, than say so. Note that, in general, setting an initially non-rotating ring or disk into rotation about its axis necessarily results in some deformation due to circumferential length contraction, so if your interest was in how the object deforms when it is set into rotation, you would need to specify the properties of the material and so on. Of course, you can posit a rotating ring with a certain radius, but you can't assume it is the same radius it would have if the rotation was stopped, etc. This is all just Relativity 101.
> >
> > > Why do viewers [sic] in F1 see [sic] the shape as being identical to the non-rotating
> > > circle [sic] while they view the top of the circle [sic] moving at a different angular
> > > velocity than the bottom of the circle [sic]?
> >
> > The subject of your post was "why no length contraction", but now you aren't asking why no length contraction (which would be a non sequitur), you are just asking why the description of the rotating ring or disk in terms of F1 exhibits the well-known and expected relativistic effects. The reason it exhibits the relativistic effects is because physics is locally Lorentz invariant. This is just Relativity 101.
> > op
> > Special Relativity: 868.... Barnpole Dave: 0
> If any object in F0 in the x-y plane is circular or any object has a circular hole in it, frame F1 observers moving with V in the x direction view the object as an elliptical shape or if it has a hole they view the hole as an elliptical shape. Now, if any of those objects rotate in the x-y plane around a stationary point in F0, observers in F1 view the object with the identical shape of the now rotating object even though the top is rotating in the opposite x direction than the bottom is rotating. So length contraction doesn't happen for some reason. This is true no matter what the object is composed of or its relative angular velocity. If it is viewed as a circular shape in F0, then F1 says the shape when rotating is identical to the object's shape when it is not rotating, or if one object is rotating clockwise and an identical object is rotating counter-clockwise they have the identical shape as viewed in F1.
> David Seppala
> Bastrop TX
There was a typo - it should read ... F1 view the object with the identical shape of the non- rotating object -- instead of now rotating

On Friday, April 29, 2022 at 10:37:19 AM UTC-7, sep...@yahoo.com wrote:
> > > If there are two inertial reference frames, F0 and F1, with a relative velocity of V in
> > > the x-direction, if there is a circle of radius R centered at (0,0) in F0, that circle is
> > > an oval shape for viewers in F1.
> > Correction: It's spatial shape is an oval (an ellipse, to be exact) in terms of F1. Talking about "viewers" is misleading unless your intention is to discuss relativistic optics... which it obviously is not.
> >
> > > If that circle is rotating around (0,0) at some angular velocity in F0...
> >
> > Using the word "circle" for a rotating ring or disk is misleading. If you are meaning a solid object like a circular ring or disk, than say so. Note that, in general, setting an initially non-rotating ring or disk into rotation about its axis necessarily results in some deformation due to circumferential length contraction, so if your interest was in how the object deforms when it is set into rotation, you would need to specify the properties of the material and so on. Of course, you can posit a rotating ring with a certain radius, but you can't assume it is the same radius it would have if the rotation was stopped, etc. This is all just Relativity 101.
> >
> > > Why do viewers [sic] in F1 see [sic] the shape as being identical to the non-rotating
> > > circle [sic] while they view the top of the circle [sic] moving at a different angular
> > > velocity than the bottom of the circle [sic]?
> >
> > The subject of your post was "why no length contraction", but now you aren't asking why no length contraction (which would be a non sequitur), you are just asking why the description of the rotating ring or disk in terms of F1 exhibits the well-known and expected relativistic effects. The reason it exhibits the relativistic effects is because physics is locally Lorentz invariant. This is just Relativity 101.
> >
> If...

As always, you ignore the explanation, and simply repeat your question.

> If any of those objects rotate in the x-y plane around a stationary point in F0,
> observers in F1 view the object with the identical shape of the non-rotating object...

No, your misconception about this was explained above. In general, setting an object into rotation will introduces stresses and strains that will cause it to either break or deform, and the precise results will depend on the material properties and structure of the object and so on. It is possible, in principle, to prepare a solid object that is rotating with a circular perimeter (in terms of the rest frame of the axis of rotation), but this object would have a different shape when its rotation is stopped. The point is that, due to the length contraction associated with the circumferential motion, the spatial shape of a solid put into rotation is different (in general) that it's shape when not rotating, and this is true for both F0 and F1. You see, you are confusing two different distinctions: rotating vs. non-rotating (active transformation), and F0 vs F1 (passive coordinate transformation). Understand?

> So length contraction doesn't happen for some reason.

That is false, as explained to you many times before. The length contraction associated with changing from non-rotating to rotating occurs, and the length contraction associated with the different systems of reference F0 and F1 also is present, all exactly in accord with elementary Relativity 101. Do you understand this?

> If it is viewed as a circular shape in F0...

It is wrong to say "viewed as". Say simply "If it's shape is circular in terms of F0..."

> then F1 says the shape when rotating is identical to the object's shape
> when it is not rotating...

Non-sequitur. See above. You're again confusing rotating-vs-non rotating, and F0 vs F1.

> > In terms of F1 (the inertial coordinates in which the traveler is at rest on the outbound leg) the traveler arrives at B (after a lapse of 10 seconds on his clock) simultaneous with the emitter reaching 5 seconds of elapsed time on its clock. This corresponds to the fact that in terms of F1 the source clock is running at half speed. Now, in terms of F2 (the inertial coordinates in which the traveler is at rest on the return leg), the turn-around event at B is simultaneous with the source event after 35 seconds have elapsed on its clock, so from that event to the reunion the source has 5 seconds of elapsed time, and the traveler has 10, which corresponds to the fact that in terms of F2 the source clock is running at half speed. What you stupidly call the "extra flashes" are emitted by the source clock between 5 sec and 35 sec on its clock.
>
> Tell me how many flashes the traveler says have been emitted by the flashing light
> when the traveler is at B.

The traveler says the same thing that everyone says, about everything, always. For example, everyone says that, in terms of F1, the number of flashes emitted at the source between the coordinate time when he departed and the coordinate time of event B (traveler reaching turn-around) is 5, because 5 seconds of the source's proper time has elapsed in that span of time. This was answered up above.

> Does the number of flashes the traveler says is emitted when he is at B
> change as he accelerates?

You must mean "have been emitted", not "is emitted". What you have stated in meaningless because you failed to specify coordinates. Again, everyone always agrees on all the objective facts. In terms of F1, the number of flashes emitted at the source between the coordinate time at the origin and the coordinate time of event B is 5. In terms of F2, the number of flashes emitted at the source between the coordinate time of the origin and the coordinate time of event B is 35. In terms of F0 the number of flashes between the coordinate time at the origin and the coordinate time of event B is 20. These are all objective facts, and none of them depend on the traveler's state of motion, given the time and place of event B. It isn't even necessary to have a traveler. This was answered up above.

Now do you understand?

On Friday, April 29, 2022 at 2:07:23 PM UTC-5, Al Coe wrote:
> On Friday, April 29, 2022 at 10:37:19 AM UTC-7, sep...@yahoo.com wrote:
> > > > If there are two inertial reference frames, F0 and F1, with a relative velocity of V in
> > > > the x-direction, if there is a circle of radius R centered at (0,0) in F0, that circle is
> > > > an oval shape for viewers in F1.
> > > Correction: It's spatial shape is an oval (an ellipse, to be exact) in terms of F1. Talking about "viewers" is misleading unless your intention is to discuss relativistic optics... which it obviously is not.
> > >
> > > > If that circle is rotating around (0,0) at some angular velocity in F0...
> > >
> > > Using the word "circle" for a rotating ring or disk is misleading. If you are meaning a solid object like a circular ring or disk, than say so. Note that, in general, setting an initially non-rotating ring or disk into rotation about its axis necessarily results in some deformation due to circumferential length contraction, so if your interest was in how the object deforms when it is set into rotation, you would need to specify the properties of the material and so on. Of course, you can posit a rotating ring with a certain radius, but you can't assume it is the same radius it would have if the rotation was stopped, etc. This is all just Relativity 101.
> > >
> > > > Why do viewers [sic] in F1 see [sic] the shape as being identical to the non-rotating
> > > > circle [sic] while they view the top of the circle [sic] moving at a different angular
> > > > velocity than the bottom of the circle [sic]?
> > >
> > > The subject of your post was "why no length contraction", but now you aren't asking why no length contraction (which would be a non sequitur), you are just asking why the description of the rotating ring or disk in terms of F1 exhibits the well-known and expected relativistic effects. The reason it exhibits the relativistic effects is because physics is locally Lorentz invariant. This is just Relativity 101.
> > >
> > If...
>
> As always, you ignore the explanation, and simply repeat your question.
>
> > If any of those objects rotate in the x-y plane around a stationary point in F0,
> > observers in F1 view the object with the identical shape of the non-rotating object...
>
> No, your misconception about this was explained above. In general, setting an object into rotation will introduces stresses and strains that will cause it to either break or deform, and the precise results will depend on the material properties and structure of the object and so on. It is possible, in principle, to prepare a solid object that is rotating with a circular perimeter (in terms of the rest frame of the axis of rotation), but this object would have a different shape when its rotation is stopped. The point is that, due to the length contraction associated with the circumferential motion, the spatial shape of a solid put into rotation is different (in general) that it's shape when not rotating, and this is true for both F0 and F1.. You see, you are confusing two different distinctions: rotating vs. non-rotating (active transformation), and F0 vs F1 (passive coordinate transformation). Understand?
> > So length contraction doesn't happen for some reason.
> That is false, as explained to you many times before. The length contraction associated with changing from non-rotating to rotating occurs, and the length contraction associated with the different systems of reference F0 and F1 also is present, all exactly in accord with elementary Relativity 101. Do you understand this?
>
> > If it is viewed as a circular shape in F0...
>
> It is wrong to say "viewed as". Say simply "If it's shape is circular in terms of F0..."
> > then F1 says the shape when rotating is identical to the object's shape
> > when it is not rotating...
>
> Non-sequitur. See above. You're again confusing rotating-vs-non rotating, and F0 vs F1.
>
> > > In terms of F1 (the inertial coordinates in which the traveler is at rest on the outbound leg) the traveler arrives at B (after a lapse of 10 seconds on his clock) simultaneous with the emitter reaching 5 seconds of elapsed time on its clock. This corresponds to the fact that in terms of F1 the source clock is running at half speed. Now, in terms of F2 (the inertial coordinates in which the traveler is at rest on the return leg), the turn-around event at B is simultaneous with the source event after 35 seconds have elapsed on its clock, so from that event to the reunion the source has 5 seconds of elapsed time, and the traveler has 10, which corresponds to the fact that in terms of F2 the source clock is running at half speed. What you stupidly call the "extra flashes" are emitted by the source clock between 5 sec and 35 sec on its clock.
> >
> > Tell me how many flashes the traveler says have been emitted by the flashing light
> > when the traveler is at B.
>
> The traveler says the same thing that everyone says, about everything, always. For example, everyone says that, in terms of F1, the number of flashes emitted at the source between the coordinate time when he departed and the coordinate time of event B (traveler reaching turn-around) is 5, because 5 seconds of the source's proper time has elapsed in that span of time. This was answered up above.
>
> > Does the number of flashes the traveler says is emitted when he is at B
> > change as he accelerates?
>
> You must mean "have been emitted", not "is emitted". What you have stated in meaningless because you failed to specify coordinates. Again, everyone always agrees on all the objective facts. In terms of F1, the number of flashes emitted at the source between the coordinate time at the origin and the coordinate time of event B is 5. In terms of F2, the number of flashes emitted at the source between the coordinate time of the origin and the coordinate time of event B is 35. In terms of F0 the number of flashes between the coordinate time at the origin and the coordinate time of event B is 20. These are all objective facts, and none of them depend on the traveler's state of motion, given the time and place of event B. It isn't even necessary to have a traveler. This was answered up above.
>
> Now do you understand?

If a steel wheel is at rest in the F0 inertial reference frame in the x-y plane and it is not rotating, it is a circle. If that wheel is rotating in the F0 inertial reference frame in the x-y plane isn't it still a circle in shape?
Correct?
David Seppala
Bastrop TX

On Friday, April 29, 2022 at 1:28:37 PM UTC-7, sep...@yahoo.com wrote:
> On Friday, April 29, 2022 at 2:07:23 PM UTC-5, Al Coe wrote:
> > On Friday, April 29, 2022 at 10:37:19 AM UTC-7, sep...@yahoo.com wrote:
> > > > > If there are two inertial reference frames, F0 and F1, with a relative velocity of V in
> > > > > the x-direction, if there is a circle of radius R centered at (0,0) in F0, that circle is
> > > > > an oval shape for viewers in F1.
> > > > Correction: It's spatial shape is an oval (an ellipse, to be exact) in terms of F1. Talking about "viewers" is misleading unless your intention is to discuss relativistic optics... which it obviously is not.
> > > >
> > > > > If that circle is rotating around (0,0) at some angular velocity in F0...
> > > >
> > > > Using the word "circle" for a rotating ring or disk is misleading. If you are meaning a solid object like a circular ring or disk, than say so.. Note that, in general, setting an initially non-rotating ring or disk into rotation about its axis necessarily results in some deformation due to circumferential length contraction, so if your interest was in how the object deforms when it is set into rotation, you would need to specify the properties of the material and so on. Of course, you can posit a rotating ring with a certain radius, but you can't assume it is the same radius it would have if the rotation was stopped, etc. This is all just Relativity 101.
> > > >
> > > > > Why do viewers [sic] in F1 see [sic] the shape as being identical to the non-rotating
> > > > > circle [sic] while they view the top of the circle [sic] moving at a different angular
> > > > > velocity than the bottom of the circle [sic]?
> > > >
> > > > The subject of your post was "why no length contraction", but now you aren't asking why no length contraction (which would be a non sequitur), you are just asking why the description of the rotating ring or disk in terms of F1 exhibits the well-known and expected relativistic effects. The reason it exhibits the relativistic effects is because physics is locally Lorentz invariant. This is just Relativity 101.
> > > >
> > > If...
> >
> > As always, you ignore the explanation, and simply repeat your question.
> >
> > > If any of those objects rotate in the x-y plane around a stationary point in F0,
> > > observers in F1 view the object with the identical shape of the non-rotating object...
> >
> > No, your misconception about this was explained above. In general, setting an object into rotation will introduces stresses and strains that will cause it to either break or deform, and the precise results will depend on the material properties and structure of the object and so on. It is possible, in principle, to prepare a solid object that is rotating with a circular perimeter (in terms of the rest frame of the axis of rotation), but this object would have a different shape when its rotation is stopped. The point is that, due to the length contraction associated with the circumferential motion, the spatial shape of a solid put into rotation is different (in general) that it's shape when not rotating, and this is true for both F0 and F1. You see, you are confusing two different distinctions: rotating vs. non-rotating (active transformation), and F0 vs F1 (passive coordinate transformation). Understand?
> > > So length contraction doesn't happen for some reason.
> > That is false, as explained to you many times before. The length contraction associated with changing from non-rotating to rotating occurs, and the length contraction associated with the different systems of reference F0 and F1 also is present, all exactly in accord with elementary Relativity 101. Do you understand this?
> >
> > > If it is viewed as a circular shape in F0...
> >
> > It is wrong to say "viewed as". Say simply "If it's shape is circular in terms of F0..."
> > > then F1 says the shape when rotating is identical to the object's shape
> > > when it is not rotating...
> >
> > Non-sequitur. See above. You're again confusing rotating-vs-non rotating, and F0 vs F1.
> >
> > > > In terms of F1 (the inertial coordinates in which the traveler is at rest on the outbound leg) the traveler arrives at B (after a lapse of 10 seconds on his clock) simultaneous with the emitter reaching 5 seconds of elapsed time on its clock. This corresponds to the fact that in terms of F1 the source clock is running at half speed. Now, in terms of F2 (the inertial coordinates in which the traveler is at rest on the return leg), the turn-around event at B is simultaneous with the source event after 35 seconds have elapsed on its clock, so from that event to the reunion the source has 5 seconds of elapsed time, and the traveler has 10, which corresponds to the fact that in terms of F2 the source clock is running at half speed. What you stupidly call the "extra flashes" are emitted by the source clock between 5 sec and 35 sec on its clock.
> > >
> > > Tell me how many flashes the traveler says have been emitted by the flashing light
> > > when the traveler is at B.
> >
> > The traveler says the same thing that everyone says, about everything, always. For example, everyone says that, in terms of F1, the number of flashes emitted at the source between the coordinate time when he departed and the coordinate time of event B (traveler reaching turn-around) is 5, because 5 seconds of the source's proper time has elapsed in that span of time. This was answered up above.
> >
> > > Does the number of flashes the traveler says is emitted when he is at B
> > > change as he accelerates?
> >
> > You must mean "have been emitted", not "is emitted". What you have stated in meaningless because you failed to specify coordinates. Again, everyone always agrees on all the objective facts. In terms of F1, the number of flashes emitted at the source between the coordinate time at the origin and the coordinate time of event B is 5. In terms of F2, the number of flashes emitted at the source between the coordinate time of the origin and the coordinate time of event B is 35. In terms of F0 the number of flashes between the coordinate time at the origin and the coordinate time of event B is 20.. These are all objective facts, and none of them depend on the traveler's state of motion, given the time and place of event B. It isn't even necessary to have a traveler. This was answered up above.
> >
> > Now do you understand?
>
> If a steel wheel is at rest in the F0 inertial reference frame in the x-y plane and it is not rotating, it is a circle. If that wheel is rotating in the F0 inertial reference frame in the x-y plane isn't it still a circle in shape?

That doesn't follow, even for shape, let alone for size, for the reason thoroughly explained *twice* up above. What is wrong with your brain? What part of the careful and thorough explanation was unclear to you?

And why have you run away from your remedial course on the twins scenario? Have you finally understand your misconceptions, and now understand what you've been told, and just forgot to say so?

On Friday, April 29, 2022 at 6:17:27 PM UTC-5, Al Coe wrote:
> On Friday, April 29, 2022 at 1:28:37 PM UTC-7, sep...@yahoo.com wrote:
> > On Friday, April 29, 2022 at 2:07:23 PM UTC-5, Al Coe wrote:
> > > On Friday, April 29, 2022 at 10:37:19 AM UTC-7, sep...@yahoo.com wrote:
> > > > > > If there are two inertial reference frames, F0 and F1, with a relative velocity of V in
> > > > > > the x-direction, if there is a circle of radius R centered at (0,0) in F0, that circle is
> > > > > > an oval shape for viewers in F1.
> > > > > Correction: It's spatial shape is an oval (an ellipse, to be exact) in terms of F1. Talking about "viewers" is misleading unless your intention is to discuss relativistic optics... which it obviously is not.
> > > > >
> > > > > > If that circle is rotating around (0,0) at some angular velocity in F0...
> > > > >
> > > > > Using the word "circle" for a rotating ring or disk is misleading.. If you are meaning a solid object like a circular ring or disk, than say so. Note that, in general, setting an initially non-rotating ring or disk into rotation about its axis necessarily results in some deformation due to circumferential length contraction, so if your interest was in how the object deforms when it is set into rotation, you would need to specify the properties of the material and so on. Of course, you can posit a rotating ring with a certain radius, but you can't assume it is the same radius it would have if the rotation was stopped, etc. This is all just Relativity 101.
> > > > >
> > > > > > Why do viewers [sic] in F1 see [sic] the shape as being identical to the non-rotating
> > > > > > circle [sic] while they view the top of the circle [sic] moving at a different angular
> > > > > > velocity than the bottom of the circle [sic]?
> > > > >
> > > > > The subject of your post was "why no length contraction", but now you aren't asking why no length contraction (which would be a non sequitur), you are just asking why the description of the rotating ring or disk in terms of F1 exhibits the well-known and expected relativistic effects. The reason it exhibits the relativistic effects is because physics is locally Lorentz invariant. This is just Relativity 101.
> > > > >
> > > > If...
> > >
> > > As always, you ignore the explanation, and simply repeat your question.
> > >
> > > > If any of those objects rotate in the x-y plane around a stationary point in F0,
> > > > observers in F1 view the object with the identical shape of the non-rotating object...
> > >
> > > No, your misconception about this was explained above. In general, setting an object into rotation will introduces stresses and strains that will cause it to either break or deform, and the precise results will depend on the material properties and structure of the object and so on. It is possible, in principle, to prepare a solid object that is rotating with a circular perimeter (in terms of the rest frame of the axis of rotation), but this object would have a different shape when its rotation is stopped. The point is that, due to the length contraction associated with the circumferential motion, the spatial shape of a solid put into rotation is different (in general) that it's shape when not rotating, and this is true for both F0 and F1. You see, you are confusing two different distinctions: rotating vs. non-rotating (active transformation), and F0 vs F1 (passive coordinate transformation). Understand?
> > > > So length contraction doesn't happen for some reason.
> > > That is false, as explained to you many times before. The length contraction associated with changing from non-rotating to rotating occurs, and the length contraction associated with the different systems of reference F0 and F1 also is present, all exactly in accord with elementary Relativity 101. Do you understand this?
> > >
> > > > If it is viewed as a circular shape in F0...
> > >
> > > It is wrong to say "viewed as". Say simply "If it's shape is circular in terms of F0..."
> > > > then F1 says the shape when rotating is identical to the object's shape
> > > > when it is not rotating...
> > >
> > > Non-sequitur. See above. You're again confusing rotating-vs-non rotating, and F0 vs F1.
> > >
> > > > > In terms of F1 (the inertial coordinates in which the traveler is at rest on the outbound leg) the traveler arrives at B (after a lapse of 10 seconds on his clock) simultaneous with the emitter reaching 5 seconds of elapsed time on its clock. This corresponds to the fact that in terms of F1 the source clock is running at half speed. Now, in terms of F2 (the inertial coordinates in which the traveler is at rest on the return leg), the turn-around event at B is simultaneous with the source event after 35 seconds have elapsed on its clock, so from that event to the reunion the source has 5 seconds of elapsed time, and the traveler has 10, which corresponds to the fact that in terms of F2 the source clock is running at half speed. What you stupidly call the "extra flashes" are emitted by the source clock between 5 sec and 35 sec on its clock.
> > > >
> > > > Tell me how many flashes the traveler says have been emitted by the flashing light
> > > > when the traveler is at B.
> > >
> > > The traveler says the same thing that everyone says, about everything, always. For example, everyone says that, in terms of F1, the number of flashes emitted at the source between the coordinate time when he departed and the coordinate time of event B (traveler reaching turn-around) is 5, because 5 seconds of the source's proper time has elapsed in that span of time. This was answered up above.
> > >
> > > > Does the number of flashes the traveler says is emitted when he is at B
> > > > change as he accelerates?
> > >
> > > You must mean "have been emitted", not "is emitted". What you have stated in meaningless because you failed to specify coordinates. Again, everyone always agrees on all the objective facts. In terms of F1, the number of flashes emitted at the source between the coordinate time at the origin and the coordinate time of event B is 5. In terms of F2, the number of flashes emitted at the source between the coordinate time of the origin and the coordinate time of event B is 35. In terms of F0 the number of flashes between the coordinate time at the origin and the coordinate time of event B is 20. These are all objective facts, and none of them depend on the traveler's state of motion, given the time and place of event B. It isn't even necessary to have a traveler. This was answered up above.
> > >
> > > Now do you understand?
> >
> > If a steel wheel is at rest in the F0 inertial reference frame in the x-y plane and it is not rotating, it is a circle. If that wheel is rotating in the F0 inertial reference frame in the x-y plane isn't it still a circle in shape?
> That doesn't follow, even for shape, let alone for size, for the reason thoroughly explained *twice* up above. What is wrong with your brain? What part of the careful and thorough explanation was unclear to you?
>
> And why have you run away from your remedial course on the twins scenario? Have you finally understand your misconceptions, and now understand what you've been told, and just forgot to say so?
So if a wheel is rotating in the x-y plane in F0, if you don't think it remains circular in shape, what shape do you think it is. I didn't say the radius was identical of a wheel at rest versus one that is rotating. But tell me what shape you think the spinning wheel is in F0 if it is not circular..
David Seppala
Bastrop TX

On Friday, April 29, 2022 at 9:08:13 PM UTC-7, sep...@yahoo.com wrote:
> If a wheel is rotating in the x-y plane in F0, if you don't think it remains circular
> in shape, what shape do you think it is?

Again, when an object is set into rotation, it is subjected to stresses and strains, due to (among other things, like centrifugal effects) relativistic length contraction, and the resulting shape of the object is indeterminate without specifying the material properties and precise structure of the object (down to the microscopic level), which together determine whether the object breaks into fragments or folds and buckles along arbitrary lines, or deforms more or less uniformly. And of course, in every case, the behavior of the object will be in perfect accord with local Lorentz invariance (special relativity), including the effects of length contraction. This is the third time I've explained this to you.

Now, if you wish to admit that you have no interest in the many ways in which a real solid object would deform when it is set rotating, and hence you have nothing to say about the active circumferential length contraction that accompanies such action, then you can simply stipulate that you have a rotating circular solid. In that case you aren't talking about active transformations at all (despite your erroneous statements about "looks the same rotating as no rotating), all you are talking about is the passive descriptions of a rotating circular disk in terms of inertial rest coordinates of its axis and in terms of a relatively moving system of coordinates. Needless to say, the descriptions of the rotating solid in terms of these two coordinate systems are perfectly consistent and correct, confirming that those systems are related by Lorentz transformation.

Now that you've conceded your silly conceptual mistakes in the other thread and run away to launch still dumber misconceptions in this thread, I ask again: If you see any contradiction, go ahead and point it out.

Special Relativity: 869 ..... Barnpole Dave: 0

On Friday, April 29, 2022 at 11:43:10 PM UTC-5, Al Coe wrote:
> On Friday, April 29, 2022 at 9:08:13 PM UTC-7, sep...@yahoo.com wrote:
> > If a wheel is rotating in the x-y plane in F0, if you don't think it remains circular
> > in shape, what shape do you think it is?
>
> Again, when an object is set into rotation, it is subjected to stresses and strains, due to (among other things, like centrifugal effects) relativistic length contraction, and the resulting shape of the object is indeterminate without specifying the material properties and precise structure of the object (down to the microscopic level), which together determine whether the object breaks into fragments or folds and buckles along arbitrary lines, or deforms more or less uniformly. And of course, in every case, the behavior of the object will be in perfect accord with local Lorentz invariance (special relativity), including the effects of length contraction. This is the third time I've explained this to you.
>
> Now, if you wish to admit that you have no interest in the many ways in which a real solid object would deform when it is set rotating, and hence you have nothing to say about the active circumferential length contraction that accompanies such action, then you can simply stipulate that you have a rotating circular solid. In that case you aren't talking about active transformations at all (despite your erroneous statements about "looks the same rotating as no rotating), all you are talking about is the passive descriptions of a rotating circular disk in terms of inertial rest coordinates of its axis and in terms of a relatively moving system of coordinates. Needless to say, the descriptions of the rotating solid in terms of these two coordinate systems are perfectly consistent and correct, confirming that those systems are related by Lorentz transformation.
>
> Now that you've conceded your silly conceptual mistakes in the other thread and run away to launch still dumber misconceptions in this thread, I ask again: If you see any contradiction, go ahead and point it out.
>
> Special Relativity: 869 ..... Barnpole Dave: 0

Let's say in F0 there is a spinning wooden disc of radius R+1 centered at (0,0) in the x,y plane. At (R,0,1) there is a laser knife that cuts through this wood. When the cutting is done, is the spinning disc circular? Or as you repeatedly state, in what way does it deform?
David Seppala
Bastrop TX

On Saturday, April 30, 2022 at 10:24:56 AM UTC-4, sep...@yahoo.com wrote:

> Let's say in F0 there is a spinning wooden disc of radius R+1 centered at (0,0) in the x,y plane.
> At (R,0,1) there is a laser knife that cuts through this wood. When the cutting is done, is the
> spinning disc circular? Or as you repeatedly state, in what way does it deform?

Don't you understand? What you had for breakfast is irrelevant to the problem.
How the spinning disc was created is irrelevant to the problem.

Just specify that you have a Spinning Disk of radius R.

This does not mean that as the disc is immobile that it's radius is R.
It means once you have set up it's spinning, it's stabilized to a final radius of R.
Again, your three lines sentence above can be simply stated as:

"Consider a Spinning Disk of radius R. "

Isn't that simple?

On Saturday, April 30, 2022 at 7:24:56 AM UTC-7, sep...@yahoo.com wrote:
> Let's say in F0 there is a spinning wooden disc of radius R+1 centered at (0,0) in the x,y plane.

As I explained (three times), if you stipulate a circular spinning disk in a specified system of coordinates, then you have successfully specified a circular spinning disk in a specified system of coordinates. Duh. You are disavowing any interest in the many ways in which a real solid object would deform/break when subjected to torque that sets it rotating, and hence you have nothing to say about the active circumferential length contraction that accompanies such action.

This establishes clearly that, contrary to your claims, you aren't talking about active transformations at all (despite your erroneous statements about "has the same shape rotating and not rotating"), all you are talking about is the passive descriptions of a rotating circular disk in terms of two different systems of inertial coordinates, exhibiting the relations of the Lorentz transformation, including the effects of length contraction, time dilation, and the relativity of simultaneity. You note that everything conforms perfectly to the relativistic predictions.

So, if your purpose was to affirm the correctness of special relativity (under a subject line that pretends to deny it), then mission accomplished. It's good that you've finally realized all your silly complaints were idiotic and fallacious. It only took 870 rounds.

Special Relativity: 870 ..... Barnpole Dave: 0

On Saturday, April 30, 2022 at 9:47:17 AM UTC-5, Al Coe wrote:
> On Saturday, April 30, 2022 at 7:24:56 AM UTC-7, sep...@yahoo.com wrote:
> > Let's say in F0 there is a spinning wooden disc of radius R+1 centered at (0,0) in the x,y plane.
> As I explained (three times), if you stipulate a circular spinning disk in a specified system of coordinates, then you have successfully specified a circular spinning disk in a specified system of coordinates. Duh. You are disavowing any interest in the many ways in which a real solid object would deform/break when subjected to torque that sets it rotating, and hence you have nothing to say about the active circumferential length contraction that accompanies such action.
>
> This establishes clearly that, contrary to your claims, you aren't talking about active transformations at all (despite your erroneous statements about "has the same shape rotating and not rotating"), all you are talking about is the passive descriptions of a rotating circular disk in terms of two different systems of inertial coordinates, exhibiting the relations of the Lorentz transformation, including the effects of length contraction, time dilation, and the relativity of simultaneity. You note that everything conforms perfectly to the relativistic predictions.
>
> So, if your purpose was to affirm the correctness of special relativity (under a subject line that pretends to deny it), then mission accomplished. It's good that you've finally realized all your silly complaints were idiotic and fallacious. It only took 870 rounds.
>
> Special Relativity: 870 ..... Barnpole Dave: 0

Let a steel disc of radius R+r spin with a fixed spin rate in F0 be centered at (0,0) in F0. Let a laser knife be positioned at (R,0). Let the laser-knife reduce the radius of the disc to R as measured in F0 and let R be large enough so that any force that you think effects the circumference or radius so the disc is not circular as measured in F0 be insignificant. Do you understand that is posssible?
David Seppala
Bastrop TX

On Saturday, April 30, 2022 at 10:56:07 AM UTC-7, sep...@yahoo.com wrote:
> Let a steel disc...

It's self-evident that, in terms of any inertial coordinate system, a circular disk rotating about a fixed normal axis through its center can be created... in infinitely many ways. The point about how solid objects change during active accelerations (Ehrenfest disk) is obviously beyond your comprehension, and evidently irrelevant to the fallacy that you are proposing. You always couch your fallacies inside larger fallacies (inside still larger fallacies...), and then you get confused and distracted when presented with a thorough debunking of all your nested fallacies, since you are really only interested in (and indeed conscious of) the most rudimentary level. So let's dumb this down to the most rudimental level...

In terms of inertial coordinate system S0, consider two thin superimposed circular disks of radius R in the xy plane with common fixed axis in the z direction, one stationary and the other rotating about the z axis at some uniform angular speed. Their circular perimeters always coincide. In terms of another inertial coordinate system S1, aligned and moving in the positive x direction at some uniform speed relative to S0, the perimeters obviously still coincide (since the mapping is one-to-one), but in a contracted elliptical shape due to relativistic effects. Note to the slow-witted student: The density of the constituent particles on the rim of the rotating disk is greater on the edge whose circumferential speed adds to the disk's translational speed, and lower on the opposite edge, consistent with the length contraction (and time dilation, and relativity of simultaneity) predicted by special relativity. Understand?

On Saturday, 30 April 2022 at 20:41:15 UTC+2, Al Coe wrote:
> On Saturday, April 30, 2022 at 10:56:07 AM UTC-7, sep...@yahoo.com wrote:
> > Let a steel disc...
>
> It's self-evident that, in terms of any inertial coordinate system, a circular disk rotating about a fixed normal axis through its center can be created... in infinitely many ways. The point about how solid objects change during active accelerations (Ehrenfest disk) is obviously beyond your comprehension, and evidently irrelevant to the fallacy that you are proposing. You always couch your fallacies inside larger fallacies (inside still larger fallacies...), and then you get confused and distracted when presented with a thorough debunking of all your nested fallacies, since you are really only interested in (and indeed conscious of) the most rudimentary level. So let's dumb this down to the most rudimental level...
>
> In terms of inertial coordinate system S0, consider two thin superimposed circular disks of radius R in the xy plane with common fixed axis in the z direction, one stationary and the other rotating about the z axis at some uniform angular speed. Their circular perimeters always coincide. In terms of another inertial coordinate system S1, aligned and moving in the positive x direction at some uniform speed relative to S0, the perimeters obviously still coincide (since the mapping is one-to-one), but in a contracted elliptical shape due to relativistic effects. Note to the slow-witted student: The density of the constituent particles on the rim of the rotating disk is greater on the edge whose circumferential speed adds to the disk's translational speed, and lower on the opposite edge, consistent with the length contraction (and time dilation, and relativity of simultaneity) predicted by special relativity. Understand?

Sure, it's easy to understand that the predictions of The Shit are
consistent with the predictions of The Shit; though, of course
there are exceptions.

On April 30, sep...@yahoo.com wrote:
> Let's say in F0 there is a spinning wooden disc of radius R+1 centered at (0,0) in the x,y plane.
> At (R,0,1) there is a laser knife that cuts through this wood. When the cutting is done, is the spinning
> disc circular? Or as you repeatedly state, in what way does it deform?

?
What is (R,0,1)?
The knife cuts... in which direction?
Does it spin with the disc?

It isn't enough to write, one must also write clearly.

--
Rich

On April 28, sep...@yahoo.com wrote:
> If there are two inertial reference frames, F0 and F1, with a relative velocity of V in the x-direction,
> if there is a circle of radius R centered at (0,0) in F0, that circle is an oval shape for viewers in F1.
>If that circle is rotating around (0,0) at some angular velocity in F0, why do viewers in F1 see the
> shape as being identical to the non-rotating circle while they view the top of the circle moving at
> a different angular velocity than the bottom of the circle?

Your question is mushy, but very interesting, if properly phrased.

Einstein mentioned nothing about angular velocity transformation
in the 1905 paper. Perhaps he thought it follows trivially from the
linear velocity formula. It doesn't.

ω = d(L/R)/dt, where L is the arc length of angle θ. In F1,
(dP/dt)', for any point P on L, is given by the velocity
transformation (from F0), so it looks trivial.

But dL' = R' dθ, so we have to consider R', in F1. Using your
construction, R' will vary with θ, as R's orientation varies
sinusoidally with respect to the velocity vector V along the x axis.

Thus we have a rotating solid object with non-uniform (quasi-sinusoidal)
variation of angular velocity, around its perimeter! Which is slightly
mind bending.
But doesn't refute relativity, despite your fantasies -

--
Rich

On Monday, May 2, 2022 at 5:04:43 PM UTC-5, RichD wrote:
> On April 30, sep...@yahoo.com wrote:
> > Let's say in F0 there is a spinning wooden disc of radius R+1 centered at (0,0) in the x,y plane.
> > At (R,0,1) there is a laser knife that cuts through this wood. When the cutting is done, is the spinning
> > disc circular? Or as you repeatedly state, in what way does it deform?
> ?
> What is (R,0,1)?
> The knife cuts... in which direction?
> Does it spin with the disc?
>
> It isn't enough to write, one must also write clearly.
>
> --
> Rich
I should clarify. The laser knife is stationary. It is located in F0 at x=R, y=0, and z = 1. The laser knife is pointed in the -z direction. The rotating object is centered at x=0, y=0,z=0 and spins in the x-y plane. The object could be a piece of wood, or steel, or whatever.
Thanks,
David Seppala
Bastrop TX

On Wednesday, May 4, 2022 at 8:56:56 AM UTC-7, sep...@yahoo.com wrote:
> I should clarify...

Your question was thoroughly answered, in detail up above. Again, in terms of inertial coordinate system S0, consider two thin superimposed circular disks of radius R in the xy plane with common fixed axis in the z direction, one stationary and the other rotating about the z axis at some uniform angular speed. Their circular perimeters always coincide. In terms of another inertial coordinate system S1, aligned and moving in the positive x direction at some uniform speed relative to S0, the perimeters obviously still coincide (since the mapping is one-to-one), but in a contracted elliptical shape due to relativistic effects.

Your confusion was due to your failure to realize that the density of the constituent particles on the rim of the rotating disk is greater on the edge whose circumferential speed adds to the disk's translational speed, and lower on the opposite edge, consistent with the length contraction (and time dilation, and relativity of simultaneity) predicted by special relativity.

If anything about this is still unclear to you, go ahead and point it out, so we can clarify it for you.

On Thursday, April 28, 2022 at 4:27:14 PM UTC-7, sep...@yahoo.com wrote:
> If there are two inertial reference frames, F0 and F1, with a relative velocity of V in the x-direction, if there is a circle of radius R centered at (0,0) in F0, that circle is an oval shape for viewers in F1. If that circle is rotating around (0,0) at some angular velocity in F0, why do viewers in F1 see the shape as being identical to the non-rotating circle while they view the top of the circle moving at a different angular velocity than the bottom of the circle?
> David Seppala
> Bastrop TX

falling into an event horizon length should contract to zero along one axis....