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Rotating Ring

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  Rotating Ring sepp623@yahoo.com - 2016-02-10 17:51 -0800
    Re: Rotating Ring Henry Wilson <HGW@home.com> - 2016-02-11 08:24 +0000
      Re: Rotating Ring JanPB <filmart@gmail.com> - 2016-02-11 01:30 -0800
        Re: Rotating Ring Muddy Shoes <shoes@muddyshoes.org> - 2016-02-11 13:46 +0000
        Re: Rotating Ring Henry Wilson <HGW@home.com> - 2016-02-11 15:51 +0000
    Re: Rotating Ring Sylvia Else <sylvia@not.at.this.address> - 2016-02-11 21:49 +1100
    Re: Rotating Ring Tom Roberts <tjroberts137@sbcglobal.net> - 2016-02-14 17:12 -0600

#376003 — Rotating Ring

In a rest frame there is a circular ring that is at rest.  If another inertial reference frame is moving with velocity V say along the x-axis of the rest frame, that moving frame measures this circular ring to be oval shaped due to Einstein's length contraction.

Now if this rest frame has a circular ring that has an angular rotation velocity of V, the moving frame has zero instantaneous velocity with with the top arc of the ring and has a velocity about 2V with respect to the bottom arc of the ring. If we make the ring diameter very, very large so that centrifical forces are negligible, using Einstein's length contraction concept why does the moving frame still measure the rotating ring to be the same symmetrical oval shape as when the ring is not rotating?  Why doesn't the moving frame measure a different length contraction at the top arc than it measures for the bottom arc resulting in a shape that is not symmetrical about the x-axis?

Thanks,
David Seppala
Bastrop TX

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#376012

On Wed, 10 Feb 2016 17:51:52 -0800, sepp623 wrote:

> In a rest frame there is a circular ring that is at rest.  If another
> inertial reference frame is moving with velocity V say along the x-axis
> of the rest frame, that moving frame measures this circular ring to be
> oval shaped due to Einstein's length contraction.
> 
> Now if this rest frame has a circular ring that has an angular rotation
> velocity of V, the moving frame has zero instantaneous velocity with
> with the top arc of the ring and has a velocity about 2V with respect to
> the bottom arc of the ring. If we make the ring diameter very, very
> large so that centrifical forces are negligible, using Einstein's length
> contraction concept why does the moving frame still measure the rotating
> ring to be the same symmetrical oval shape as when the ring is not
> rotating?  Why doesn't the moving frame measure a different length
> contraction at the top arc than it measures for the bottom arc resulting
> in a shape that is not symmetrical about the x-axis?

 The ring stays the same in all frames. Einstein's whole theory is crap! 

Thanks,
> David Seppala Bastrop TX

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#376015

On Thursday, February 11, 2016 at 12:24:12 AM UTC-8, Henry Wilson wrote:
> 
> Einstein's whole theory is crap! 

A strange sentiment to come from an adult who cannot calculate, or even set up,
simple Newtonian mechanics problems.

--
Jan

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#376024

JanPB wrote:

> On Thursday, February 11, 2016 at 12:24:12 AM UTC-8, Henry Wilson wrote:
>> 
>> Einstein's whole theory is crap!
> 
> A strange sentiment to come from an adult who cannot calculate, or even
> simple Newtonian mechanics problems.

Right. Because these are difficult, easy to be fooled by particular 
configurations. Not easy and simple as relativity. Relativity is the 
simplest theory today in circulation. They are struggling for decades to 
prove it right. 

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#376029

On Thu, 11 Feb 2016 01:30:23 -0800, JanPB wrote:

> On Thursday, February 11, 2016 at 12:24:12 AM UTC-8, Henry Wilson wrote:
>> 
>> Einstein's whole theory is crap!
> 
> A strange sentiment to come from an adult who cannot calculate, or even
> set up,
> simple Newtonian mechanics problems.

One very simple Newtonian calculation is that nothing happens to a 
spinning wheel no matter how many differently moving observers look at it.

Another is that all dingleberries are subhuman morons.

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#376017

On 11/02/2016 12:51 PM, sepp623@yahoo.com wrote:
> In a rest frame there is a circular ring that is at rest.  If another
> inertial reference frame is moving with velocity V say along the
> x-axis of the rest frame, that moving frame measures this circular
> ring to be oval shaped due to Einstein's length contraction.
>
> Now if this rest frame has a circular ring that has an angular
> rotation velocity of V, the moving frame has zero instantaneous
> velocity with with the top arc of the ring and has a velocity about
> 2V with respect to the bottom arc of the ring. If we make the ring
> diameter very, very large so that centrifical forces are negligible,
> using Einstein's length contraction concept why does the moving frame
> still measure the rotating ring to be the same symmetrical oval shape
> as when the ring is not rotating?  Why doesn't the moving frame
> measure a different length contraction at the top arc than it
> measures for the bottom arc resulting in a shape that is not
> symmetrical about the x-axis?

Oh goody, Seppala's back.

The rotating ring has other relative velocity components, which you're 
casually ignoring. Before you can find your hoped for contradiction[*], 
you have to show that taking those other velocities into account doesn't 
lead to the oval shape. Given your demonstrated lack of mathematical 
competence, I doubt you're capable of that.

So far, you have nothing.

Sylvia.

[*] This isn't stated, but it's always what Seppala is about.

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#376245

On 2/10/16 2/10/16   7:51 PM, sepp623@yahoo.com wrote:
> In a rest frame there is a circular ring that is at rest.  If another
> inertial reference frame is moving with velocity V say along the x-axis of
> the rest frame, that moving frame measures this circular ring to be oval
> shaped due to Einstein's length contraction.
>
> Now if this rest frame has a circular ring that has an angular rotation
> velocity of V, the moving frame has zero instantaneous velocity with with the
> top arc of the ring and has a velocity about 2V with respect to the bottom
> arc of the ring. If we make the ring diameter very, very large so that
> centrifical forces are negligible, using Einstein's length contraction
> concept why does the moving frame still measure the rotating ring to be the
> same symmetrical oval shape as when the ring is not rotating?

Because in setting up the physical situation you constrained the ring to be 
circular in its rest frame.


> Why doesn't
> the moving frame measure a different length contraction at the top arc than
> it measures for the bottom arc resulting in a shape that is not symmetrical
> about the x-axis?

Because in its rest frame the ring is a circle.

Remember that "length contraction" is just a geometrical projection onto 
different coordinates. The locus occupied by the (rotating or non-rotating) ring 
is a circle in its rest frame, which will be measured as a symmetric oval in a 
relatively moving frame. The rotation of the ring (relative to the rest frame of 
its center) has nothing to do with this BECAUSE THAT LOCUS IS INDEPENDENT OF ITS 
ROTATION (that's how you specified the physical situation).


Tom Roberts

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