Pentcho Valev wrote:

>

One of the deepest insights Einstein ever had can be found at the end of ch. 23 in his 1920 "Relativity". An observer who is sitting eccentrically" on the periphery of a rotating disc (K') measures the circumference of the disc with his measuring rod, then measures the diameter, divides and obtains a ratio greater than pi.

Yes.

>	However, although it is the observer in K' that does all this, the result is valid for the frame K which is at rest with respect to the rotation.

No. It is QUITE CLEAR that if an observer in K makes the same measurement and takes the same ratio he will obtain pi (the center of rotation is at rest in K, which is an inertial frame).

Conclusion: non-inertial systems are DIFFERENT from inertial ones. No surprise to anyone who understands the underlying geometry of SR.

For instance, if at every application of the ruler to the circumference the K' observer also E-syncs clocks at the two ends ofthe ruler, by the time he gets back to where he started the first and last clocks will NOT be in synch. If the observer in K does this they will be in synch.

Note in all cases the ruler used must be very short compared to the radius of the disc.

Tom Roberts tjroberts@lucent.com

"Tom Roberts" wrote in message news:cf7ui3$3q3@netnews.proxy.lucent.com...

>	Pentcho Valev wrote:

> >

One of the deepest insights Einstein ever had can be found at the end of ch. 23 in his 1920 "Relativity". An observer who is sitting eccentrically" on the periphery of a rotating disc (K') measures the circumference of the disc with his measuring rod, then measures the diameter, divides and obtains a ratio greater than pi.

>

Yes.

> >	However, although it is the observer in K' that does all this, the result is valid for the frame K which is at rest with respect to the rotation.

>	No. It is QUITE CLEAR that if an observer in K makes the same measurement and takes the same ratio he will obtain pi (the center of rotation is at rest in K, which is an inertial frame).

Why is it clear, Tom? Oh, right, SR makes the prediction.

>	Conclusion: non-inertial systems are DIFFERENT from inertial ones. No surprise to anyone who understands the underlying geometry of SR.

A wonderfully circular argument for the circular disc. :)

-- greywolf42
ubi dubium ibi libertas {remove planet for e-mail}

Tom Roberts wrote in message news:...

>	Pentcho Valev wrote:

> >

One of the deepest insights Einstein ever had can be found at the end of ch. 23 in his 1920 "Relativity". An observer who is sitting eccentrically" on the periphery of a rotating disc (K') measures the circumference of the disc with his measuring rod, then measures the diameter, divides and obtains a ratio greater than pi.

>

Yes.

> >	However, although it is the observer in K' that does all this, the result is valid for the frame K which is at rest with respect to the rotation.

>	No. It is QUITE CLEAR that if an observer in K makes the same measurement and takes the same ratio he will obtain pi (the center of rotation is at rest in K, which is an inertial frame).

Not "quite clear" at all. Rather, it is "quite clear" that, if we take for granted Einstein's main result, the observer in K must obtain a ratio SMALLER than pi. Here is this main result (quotation):

"If the observer [in K'] applies his standard measuring-rod (a rod which is short as compared with the radius of the disc) tangentially to the edge of the disc, then, as judged from the Galileian system [K], the length of this rod will be less than 1..."

The implication is that, as judged from K, the length of the ROTATING periphery will be L/gamma, where L is the length of a non-rotating periphery. Is the implication correct? If it is, you should be very sad from now on. Not only because the ratio is smaller than pi.

Pentcho Valev

"Tom Roberts" schreef in bericht news:cf7ui3$3q3@netnews.proxy.lucent.com...

>

Conclusion: non-inertial systems are DIFFERENT from inertial ones. No surprise to anyone who understands the underlying geometry of SR. For instance, if at every application of the ruler to the circumference the K' observer also E-syncs clocks at the two ends ofthe ruler, by the time he gets back to where he started the first and last clocks will NOT be in synch. If the observer in K does this they will be in synch. Note in all cases the ruler used must be very short compared to the radius of the disc.

It is not clear to me why you need clocks. Suppose I have a disc at rest with a Radius of R = 100 and instead of 1 I use 628 rulers each with a length l = 1. Do you agree that I nicely can place those rulers (front to end) against the circumference and that there will be a small space between no 1 and no 628 ? (not enough to insert ruler 629)

Now I start to rotate the disc. Do you agree that If I keep each ruler fixed at one place along the circumference that there will be a small space between each ruler ?
Or to state this different. In order to keep the 628 rulers in contact which each other, such that front meets end, you have to move the rulers along the circumference, resulting in a larger cap between no 1 and no 628. Meaning you can place more rulers along the circumference. (The more the faster you rotate the disc ?)

If you agree than how do you prove that this is true ?

Nicolaas Vroom
https://www.nicvroom.be/

>	Tom Roberts tjroberts@lucent.com

Nicolaas Vroom wrote:

>	"Tom Roberts" schreef in bericht news:cf7ui3$3q3@netnews.proxy.lucent.com...

>>

Conclusion: non-inertial systems are DIFFERENT from inertial ones. No surprise to anyone who understands the underlying geometry of SR. For instance, if at every application of the ruler to the circumference the K' observer also E-syncs clocks at the two ends ofthe ruler, by the time he gets back to where he started the first and last clocks will NOT be in synch. If the observer in K does this they will be in synch. Note in all cases the ruler used must be very short compared to the radius of the disc.

>	It is not clear to me why you need clocks.

You don't "need" clocks. This was just an illustration of another way the rotating system is different from an inertial system.

>

Suppose I have a disc at rest with a Radius of R = 100 and instead of 1 I use 628 rulers each with a length l = 1. Do you agree that I nicely can place those rulers (front to end) against the circumference and that there will be a small space between no 1 and no 628 ? (not enough to insert ruler 629)

By "at rest" I assume you mean at rest in an inertial frame. Then yes.

>	Now I start to rotate the disc.

OK. I assume its center remains at rest in the inertial frame you started with.

>

Do you agree that If I keep each ruler fixed at one place along the circumference that there will be a small space between each ruler ? Or to state this different. In order to keep the 628 rulers in contact which each other, such that front meets end, you have to move the rulers along the circumference, resulting in a larger cap between no 1 and no 628. Meaning you can place more rulers along the circumference. (The more the faster you rotate the disc ?)

Yes.

>	If you agree than how do you prove that this is true ?

Because of the limitations of materials and the smallness of the effect, this is not a feasible experiment.

Tom Roberts tjroberts@lucent.com

"Tom Roberts" schreef in bericht news:cfdf3u$e6s@netnews.proxy.lucent.com...

>	Nicolaas Vroom wrote:

> >

Do you agree that If I keep each ruler fixed at one place along the circumference that there will be a small space between each ruler ? Or to state this different. In order to keep the 628 rulers in contact which each other, such that front meets end, you have to move the rulers along the circumference, resulting in a larger cap between no 1 and no 628. Meaning you can place more rulers along the circumference. (The more the faster you rotate the disc ?)

>

Yes.

In fact there are two versions of this experiment. In the above version the disc is rotated such that the rulers along the circumference have a certain speed v.
In that case for example you can place 700 rulers front to end, fixed at one place with the disc. (Compared with 628 when v=0) (This also means that the observer obtains a ratio greater than pi.)

In a second version the disc is not rotated but only the rulers along the circumference.
The question if at the same speed v can you again place 700 rulers front to end ? (Or more or less ?)

If the answer is Yes, then is it not strange that for example the shape or the size of the ROTATED disc is not effected (changed) but only the length of the rulers ?

Nicolaas Vroom
https://www.nicvroom.be/

"Nicolaas Vroom" schreef in bericht news:ejHSc.211506$0x4.10430514@phobos.telenet-ops.be...

>

In fact there are two versions of this experiment. In the above version the disc is rotated such that the rulers along the circumference have a certain speed v. In that case for example you can place 700 rulers front to end, fixed at one place with the disc. (Compared with 628 when v=0) (This also means that the observer obtains a ratio greater than pi.) In a second version the disc is not rotated but only the rulers along the circumference. The question if at the same speed v can you again place 700 rulers front to end ? (Or more or less ?) If the answer is Yes, then is it not strange that for example the shape or the size of the ROTATED disc is not effected (changed) but only the length of the rulers ?

No reply. One reason could be that in reality we can not perform such an experiment.
In principle may be the only change that happens of a rotating disc is a change in the radius R, which results in the length of the circumference, but not in the number of rulers i.e. the ratio of pi does not change.

Nicolaas Vroom

>	https://www.nicvroom.be/

Nicolaas Vroom wrote:

>	Nicolaas Vroom wrote:

>>	is it not strange that for example the shape or the size of the ROTATED disc is not effected (changed) but only the length of the rulers ?

It's not strange at all -- the disc has internal circumferential stresses, while the rulers do not.

>	One reason could be that in reality we can not perform such an experiment.

For a real measurement, the tangential stress in a rotating disk is negligible compared to the radial stress. We have no hope of measuring it.

>	In principle may be the only change that happens of a rotating disc is a change in the radius R, which results in the length of the circumference, but not in the number of rulers i.e. the ratio of pi does not change.

The radial stresses will most dfinitely change the radius of a real disk. But don't go there -- instead consider the locus of points at radius R, and arrange the circumferential rulers to be placed there. In the rotating system the ratio of circumference/radius must change, for a measurement using small rulers lined up and at rest in the rotating system.

Tom Roberts tjroberts@lucent.com

Nicolaas Vroom
https://www.nicvroom.be/

"Pentcho Valev" wrote: in message news:bdf02d35.0408090218.222097f9@posting.google.com...

>	One of the deepest insights Einstein ever had can be found at the end of ch. 23 in his 1920 "Relativity". An observer who is sitting eccentrically" on the periphery of a rotating disc (K') measures the circumference of the disc with his measuring rod,

How do you measure circumference (curved) with a rod (straight)? Surely he should've taken some measuring string - or tape. He'll have to crawl around the disk to use it I guess. How big is the disk? He'll maybe have to crawl across it to get the diameter measured too.

Maybe he needs one of those disks on a stick that clicks every meter - that'd save his knees.

"Martin Stone" wrote in message news:cf7v27$1u1$1@rdel.co.uk...

>	"Pentcho Valev" wrote in message news:bdf02d35.0408090218.222097f9@posting.google.com...

> >	One of the deepest insights Einstein ever had can be found at the end of ch. 23 in his 1920 "Relativity". An observer who is sitting eccentrically" on the periphery of a rotating disc (K') measures the circumference of the disc with his measuring rod,

>	How do you measure circumference (curved) with a rod (straight)?

Have you ever studied calculus? The idea is for infinitesimal sized straight lines (rods) you can lay then end to end around a curve and get a very good approximation of the length of the curve - the smaller the line (rod) the better the accuracy. What the thought experiment shows is that since rods on the rotating disk are shortened more of then are required to measure the length of the circumference hence pi is bigger.

Bill

>	Surely he should've taken some measuring string - or tape. He'll have to crawl around the disk to use it I guess. How big is the disk? He'll maybe have to crawl across it to get the diameter measured too. Maybe he needs one of those disks on a stick that clicks every meter - that'd save his knees.

"Bill Hobba" wrote in message news:T3URc.43369$K53.3685@news-server.bigpond.net.au...

>	"Martin Stone" wrote in message news:cf7v27$1u1$1@rdel.co.uk...

> >	"Pentcho Valev" wrote in message news:bdf02d35.0408090218.222097f9@posting.google.com...

> > >

One of the deepest insights Einstein ever had can be found at the end of ch. 23 in his 1920 "Relativity". An observer who is sitting eccentrically" on the periphery of a rotating disc (K') measures the circumference of the disc with his measuring rod,

> >	How do you measure circumference (curved) with a rod (straight)?

>	Have you ever studied calculus? The idea is for infinitesimal sized straight lines (rods) you can lay then end to end around a curve and get a very good approximation of the length of the curve - the smaller the line (rod) the better the accuracy.

Yeah - that's not disputed (by me). I kinda imagine a guy on a table sized disk with a metre ruler, then waded in. Laying out those infinitesimal rods might take a while though. ;o) And actually, now I think about it you could easily measure round a curve with a straight edge - imagine rolling the disk along the ruler, but do the measurement by moving the ruler and not the disk. If ya see.

>	What the thought experiment shows is that since rods on the rotating disk are shortened more of then are required to measure the length of the circumference hence pi is bigger.

If we define pi as "the ratio of a circle's circumference to its diameter", OK. And for this thought experiment I'm not sure I'd take a measuring tape after all. Say we laid the tape round the disk right at the very edge and then started spinning and for some reason the tape didn't just fly off - would the ends of the tape just move apart or would they stay together while the whole length of the tape gets further from the edge of the disk?

Rods it is. :o)

>

Bill

> >	Surely he should've taken some measuring string - or tape. He'll have to crawl around the disk to use it I guess. How big is the disk? He'll maybe have to crawl across it to get the diameter measured too. Maybe he needs one of those disks on a stick that clicks every meter - that'd save his knees.

>

Martin Stone wrote:

>	"Bill Hobba" wrote in message news:T3URc.43369$K53.3685@news-server.bigpond.net.au...

>>	What the thought experiment shows is that since rods on the rotating disk are shortened more of then are required to measure the length of the circumference hence pi is bigger.

>	If we define pi as "the ratio of a circle's circumference to its diameter",

NO! That only holds in Euclidean space. The 3-space at rest on a rotating platform is not Euclidean. That's the point!

There are numerous mathematical definitions of pi, ALL of which must hold. This implies that the ratio of a circle's circumference to its diameter is NOT pi in a non-Euclidean space. That's well known.

>

OK. And for this thought experiment I'm not sure I'd take a measuring tape after all. Say we laid the tape round the disk right at the very edge and then started spinning and for some reason the tape didn't just fly off - would the ends of the tape just move apart or would they stay together while the whole length of the tape gets further from the edge of the disk?

That depends on the internal strength of the tape, its response to internal stress, and precisely how you "nail it to the rotating disk". Note the centripetal (radial inward) force required to keep it on the circumference of the rotating disk is MUCH larger than the force of the stress induced around the circumference.

In principle, if the tape always retains its intrinsic length no matter what, and the tape is held radially to the edge of the rotating disk, then yes the ends must move apart. And if the ends are held together no matter what, and the tape retains its intrinsic length no matter what, then the tape must move inward from the edge of the rotating disk.

In practice, of course, there are no materials with those properties.

Tom Roberts tjroberts@lucent.com

"Tom Roberts" wrote in message news:cfd6b5$af2@netnews.proxy.lucent.com...

>	Martin Stone wrote:

> >	"Bill Hobba" wrote in message news:T3URc.43369$K53.3685@news-server.bigpond.net.au...

> > >

What the thought experiment shows is that since rods on the rotating disk are shortened more of then are required to measure the length of the circumference hence pi is bigger.

> >	If we define pi as "the ratio of a circle's circumference to its diameter",

>	NO! That only holds in Euclidean space. The 3-space at rest on a rotating platform is not Euclidean. That's the point!

'kay... but he said ... hmmph - so pi isn't bigger then?

>

There are numerous mathematical definitions of pi, ALL of which must hold. This implies that the ratio of a circle's circumference to its diameter is NOT pi in a non-Euclidean space. That's well known. Note the centripetal (radial inward) force required to keep it on the circumference of the rotating disk is MUCH larger than the force of the stress induced around the circumference.

hence my: "for some reason the tape didn't just fly off "

>

In principle, if the tape always retains its intrinsic length no matter what, and the tape is held radially to the edge of the rotating disk, then yes the ends must move apart. And if the ends are held together no matter what, and the tape retains its intrinsic length no matter what, then the tape must move inward from the edge of the rotating disk.

And if you joined the ends, and placed the tape REALLY carefully, you wouldn't need to attach it to the disk at all. But, what if I joined the ends of the tape together and it ran snug round the edge of the disk - like a very slim "tyre" to the disks "wheel"?

Now I have an issue where by following a few steps I know that the tape "shrinks", so it should burst. But why does the tape shrink when the rim of the disk (doing pretty much exactly what the tape is doing) doesn't?

Yes, I will think about it on my own - no I don't think I've disproved relativity - thought I'd put it out there.

>	In practice, of course, there are no materials with those properties.

Good job it's just a though experiment. I'm enjoying this, hope it's not boring the rest of ya who've no doubt seen and solved this very "dilemma" a million times over. :o)

>	Tom Roberts tjroberts@lucent.com

Martin Stone wrote:

>

[tape around the edge of a rotating disk] But, what if I joined the ends of the tape together and it ran snug round the edge of the disk - like a very slim "tyre" to the disks "wheel"? Now I have an issue where by following a few steps I know that the tape "shrinks", so it should burst. But why does the tape shrink when the rim of the disk (doing pretty much exactly what the tape is doing) doesn't?

But, of course, the rim does indeed contract. This induces tangential stress in the disk (wheel). But the description calls it a disk, which implies that somehow the (gedanken) material of the disk can withstand this stress and remain a disk. In practice, of course, no real material could withstand the stresses involved -- but the disk will fail radially at a MUCH lower rotation rate than the rate for which the circumferential stresses become important.

This implicitly assumes that the disk was constructed in some inertial frame, and was engineered to be stress-free during construction. That means that there is the correct amount of material throughout the disk in Euclidean space. But when rotated the "rest 3-space" of the disk becomes non-Euclidean, requiring a different amount of material to be stress-free, but there is no mechanism to add material to the disk.

If you don't want to think of internal stresses, imagine the original "disk" is made up of a large number of radial fibers attached at the center and aligned into a disk, and imagine the fibers are strong enough radially to withstand the radial stress without deforming (the fibers of course withstand no circumferential stress at all). As you speed up its rotation, the fiber ends will separate from each other more and more.

Seen from the inertial frame of the center this is obvious. Seen by an observer standing on one fiber at the edge of the disk, the fiber remains its usual width (measured by a ruler comoving with both observer and fiber). The non-Euclidean nature of the rotating system dictates that spaces develop between fibers.

Tom Roberts tjroberts@lucent.com

"Tom Roberts" wrote in message news:cfdgra$evh@netnews.proxy.lucent.com...

>

But, of course, the rim does indeed contract. This induces tangential stress in the disk (wheel). But the description calls it a disk, which implies that somehow the (gedanken) material of the disk can withstand this stress and remain a disk. In practice, of course, no real material could withstand the stresses involved -- but the disk will fail radially at a MUCH lower rotation rate than the rate for which the circumferential stresses become important.

For a laboratory-sized disk that would be true. If we were willing to consider disks of astronomically large radius, then the radial stresses would remain small even when the rim has relativistic speeds. Then, the circumferential stresses within the disk would dominate over the radial stresses. Maybe this is a minor point, but I think it's helpful to realize that if the disk is large enough, then rim riders would not feel any appreciable 'centrifugal' or coriolis forces. They could comfortably carry out measurements in their neighborhood and they would find that these 'local' measurements essentially agree with measurements made in a comoving inertial frame.

>

This implicitly assumes that the disk was constructed in some inertial frame, and was engineered to be stress-free during construction. That means that there is the correct amount of material throughout the disk in Euclidean space. But when rotated the "rest 3-space" of the disk becomes non-Euclidean, requiring a different amount of material to be stress-free, but there is no mechanism to add material to the disk. If you don't want to think of internal stresses, imagine the original "disk" is made up of a large number of radial fibers attached at the center and aligned into a disk, and imagine the fibers are strong enough radially to withstand the radial stress without deforming (the fibers of course withstand no circumferential stress at all). As you speed up its rotation, the fiber ends will separate from each other more and more. Seen from the inertial frame of the center this is obvious. Seen by an observer standing on one fiber at the edge of the disk, the fiber remains its usual width (measured by a ruler comoving with both observer and fiber). The non-Euclidean nature of the rotating system dictates that spaces develop between fibers.

Sometimes I like to imagine replacing the disk with a hollow circular ring (hoola hoop). Start with a bunch of ball bearings sitting at rest and equally spaced within the hoop. The hoop itself remains at rest in some inertial frame while the balls within the hoop all accelerate up to some relativistic speed such that their spacing around the loop remains the same as measured in the inertial frame of the hoop. Riders moving with the balls will measure the spacing between consecutive balls to be greater than the spacing as measured by observers in the inertial frame. Thus, the non-inertial, circulating observers will conclude that the circumference is larger than the circumference as measured in the inertial frame. This avoids having to worry about stresses.

JJ

"J.J. Simplicio" wrote in message news:KGsSc.131311$eM2.29111@attbi_s51...

>	"Tom Roberts" wrote in message news:cfdgra$evh@netnews.proxy.lucent.com...

> >

But, of course, the rim does indeed contract. This induces tangential stress in the disk (wheel). But the description calls it a disk, which implies that somehow the (gedanken) material of the disk can withstand this stress and remain a disk. In practice, of course, no real material could withstand the stresses involved -- but the disk will fail radially at a MUCH lower rotation rate than the rate for which the circumferential stresses become important.

>

For a laboratory-sized disk that would be true. If we were willing to consider disks of astronomically large radius, then the radial stresses would remain small even when the rim has relativistic speeds. Then, the circumferential stresses within the disk would dominate over the radial stresses. Maybe this is a minor point, but I think it's helpful to realize that if the disk is large enough, then rim riders would not feel any appreciable 'centrifugal' or coriolis forces. They could comfortably carry out measurements in their neighborhood and they would find that these 'local' measurements essentially agree with measurements made in a comoving inertial frame.

Either you are crazy or I am. The further out you move from the center of a rotating disk (eg a space station) the greater the centripetal forces you measure eg the more you are flung against the 'rim' of the space station eg the greater the perceived 'gravitational' force.

Bill

>

> >

This implicitly assumes that the disk was constructed in some inertial frame, and was engineered to be stress-free during construction. That means that there is the correct amount of material throughout the disk in Euclidean space. But when rotated the "rest 3-space" of the disk becomes non-Euclidean, requiring a different amount of material to be stress-free, but there is no mechanism to add material to the disk. If you don't want to think of internal stresses, imagine the original "disk" is made up of a large number of radial fibers attached at the center and aligned into a disk, and imagine the fibers are strong enough radially to withstand the radial stress without deforming (the fibers of course withstand no circumferential stress at all). As you speed up its rotation, the fiber ends will separate from each other more and more. Seen from the inertial frame of the center this is obvious. Seen by an observer standing on one fiber at the edge of the disk, the fiber remains its usual width (measured by a ruler comoving with both observer and fiber). The non-Euclidean nature of the rotating system dictates that spaces develop between fibers.

>

Sometimes I like to imagine replacing the disk with a hollow circular ring (hoola hoop). Start with a bunch of ball bearings sitting at rest and equally spaced within the hoop. The hoop itself remains at rest in some inertial frame while the balls within the hoop all accelerate up to some relativistic speed such that their spacing around the loop remains the same as measured in the inertial frame of the hoop. Riders moving with the balls will measure the spacing between consecutive balls to be greater than the spacing as measured by observers in the inertial frame. Thus, the non-inertial, circulating observers will conclude that the circumference is larger than the circumference as measured in the inertial frame. This avoids having to worry about stresses. JJ

"Bill Hobba" wrote in message news:EUzSc.52809$K53.16304@news-server.bigpond.net.au...

>	"J.J. Simplicio" wrote in message news:KGsSc.131311$eM2.29111@attbi_s51...

> >	"Tom Roberts" wrote in message news:cfdgra$evh@netnews.proxy.lucent.com...

> > >

But, of course, the rim does indeed contract. This induces tangential stress in the disk (wheel). But the description calls it a disk, which implies that somehow the (gedanken) material of the disk can withstand this stress and remain a disk. In practice, of course, no real material could withstand the stresses involved -- but the disk will fail radially at a MUCH lower rotation rate than the rate for which the circumferential stresses become important.

> >

For a laboratory-sized disk that would be true. If we were willing to consider disks of astronomically large radius, then the radial stresses would remain small even when the rim has relativistic speeds. Then, the circumferential stresses within the disk would dominate over the radial stresses. Maybe this is a minor point, but I think it's helpful to realize that if the disk is large enough, then rim riders would not feel any appreciable 'centrifugal' or coriolis forces. They could comfortably carry out measurements in their neighborhood and they would find that these 'local' measurements essentially agree with measurements made in a comoving inertial frame.

>	Either you are crazy or I am. The further out you move from the center of a rotating disk (eg a space station) the greater the centripetal forces you measure eg the more you are flung against the 'rim' of the space station eg the greater the perceived 'gravitational' force. Bill

I'll vounteer to be the crazy one if you wish :-)

The point is that we want the rim of the wheel to move at some specified relativistic speed, say v = 0.99c. Now the centripetal acceleration at the rim will be v^2/r. For a radius of one light-year and v = 0.99c at the rim, the acceleration at the rim would be about 1 g (and, as you point out, it would be even less if you moved inward toward the center). For a radius of 10 light-years the acceleration at the rim would only be about 0.1 g, etc.

JJ

"J.J. Simplicio" wrote in message news:wxASc.244608$IQ4.69476@attbi_s02...

>	"Bill Hobba" wrote in message news:EUzSc.52809$K53.16304@news-server.bigpond.net.au...

> >	"J.J. Simplicio" wrote in message news:KGsSc.131311$eM2.29111@attbi_s51...

> > >

"Tom Roberts" wrote in message news:cfdgra$evh@netnews.proxy.lucent.com...

> > > >

But, of course, the rim does indeed contract. This induces tangential stress in the disk (wheel). But the description calls it a disk, which implies that somehow the (gedanken) material of the disk can withstand this stress and remain a disk. In practice, of course, no real material could withstand the stresses involved -- but the disk will fail radially at a MUCH lower rotation rate than the rate for which the circumferential stresses become important.

> > >

For a laboratory-sized disk that would be true. If we were willing to consider disks of astronomically large radius, then the radial stresses would remain small even when the rim has relativistic speeds. Then, the circumferential stresses within the disk would dominate over the radial stresses. Maybe this is a minor point, but I think it's helpful to realize that if the disk is large enough, then rim riders would not feel any appreciable 'centrifugal' or coriolis forces. They could comfortably carry out measurements in their neighborhood and they would find that these 'local' measurements essentially agree with measurements made in a comoving inertial frame.

> >	Either you are crazy or I am. The further out you move from the center of a rotating disk (eg a space station) the greater the centripetal forces you measure eg the more you are flung against the 'rim' of the space station eg the greater the perceived 'gravitational' force. Bill

>

I'll vounteer to be the crazy one if you wish :-) The point is that we want the rim of the wheel to move at some specified relativistic speed, say v = 0.99c. Now the centripetal acceleration at the rim will be v^2/r. For a radius of one light-year and v = 0.99c at the rim, the acceleration at the rim would be about 1 g (and, as you point out, it would be even less if you moved inward toward the center). For a radius of 10 light-years the acceleration at the rim would only be about 0.1 g, etc.

Without actually doing the calculations may I suggest that if you were traveling at the rim of a space station traveling at .99c and a radius of one light year the normal equations of classical mechanics would not apply? - see http://www.smcm.edu/nsm/physics/SMP03S/KeatingB.doc.pdf and its conclusion:

'Like most relativistic paradoxes, the Ehrenfest paradox arises due to ambiguities in defining simultaneity. It is clear that most of the physicists who have previously considered the rotating disk implicitly assumed that the circumference of the disk is a well-defined geometric entity. However, by contemplating rather simple Minkowski diagrams, one comes to appreciate that a self-consistent, natural definition of simultaneity is not possible for a rapidly rotating frame. One can force an extended splitting of space-time, but the results will not necessarily coincide with any experimentally observable feature of the system (indeed, this is how the curvature calculated in section 4 appeared).
The best way to view the paradoxes of the rotating disk is as a variant on the twin paradox. It is in the changing from inertial frame to inertial frame that time is "lost." In the words of Rizzi and Tartaglia [5], ".a rotating disk does not admit a well defined `proper frame'; rather, it should be regarded as a class of an infinite number of local proper frames, considered in different points at different times, and glued together according to some convention."

Thanks Bill

>

JJ

Bill Hobba wrote:

>

Without actually doing the calculations may I suggest that if you were traveling at the rim of a space station traveling at .99c and a radius of one light year the normal equations of classical mechanics would not apply? - see http://www.smcm.edu/nsm/physics/SMP03S/KeatingB.doc.pdf and its conclusion: 'Like most relativistic paradoxes, the Ehrenfest paradox arises due to ambiguities in defining simultaneity. It is clear that most of the physicists who have previously considered the rotating disk implicitly assumed that the circumference of the disk is a well-defined geometric entity. However, by contemplating rather simple Minkowski diagrams, one comes to appreciate that a self-consistent, natural definition of simultaneity is not possible for a rapidly rotating frame. One can force an extended splitting of space-time, but the results will not necessarily coincide with any experimentally observable feature of the system (indeed, this is how the curvature calculated in section 4 appeared). The best way to view the paradoxes of the rotating disk is as a variant on the twin paradox. It is in the changing from inertial frame to inertial frame that time is "lost." In the words of Rizzi and Tartaglia [5], ".a rotating disk does not admit a well defined `proper frame'; rather, it should be regarded as a class of an infinite number of local proper frames, considered in different points at different times, and glued together according to some convention."

That's a nice paper. Thanks, Bill.

---Tim Shuba---

shuba wrote in message news:...

>	Bill Hobba wrote:

> >

Without actually doing the calculations may I suggest that if you were traveling at the rim of a space station traveling at .99c and a radius of one light year the normal equations of classical mechanics would not apply? - see http://www.smcm.edu/nsm/physics/SMP03S/KeatingB.doc.pdf and its conclusion: 'Like most relativistic paradoxes, the Ehrenfest paradox arises due to ambiguities in defining simultaneity. It is clear that most of the physicists who have previously considered the rotating disk implicitly assumed that the circumference of the disk is a well-defined geometric entity. However, by contemplating rather simple Minkowski diagrams, one comes to appreciate that a self-consistent, natural definition of simultaneity is not possible for a rapidly rotating frame. One can force an extended splitting of space-time, but the results will not necessarily coincide with any experimentally observable feature of the system (indeed, this is how the curvature calculated in section 4 appeared). The best way to view the paradoxes of the rotating disk is as a variant on the twin paradox. It is in the changing from inertial frame to inertial frame that time is "lost." In the words of Rizzi and Tartaglia [5], ".a rotating disk does not admit a well defined `proper frame'; rather, it should be regarded as a class of an infinite number of local proper frames, considered in different points at different times, and glued together according to some convention."

>	That's a nice paper. Thanks, Bill.

Very nice indeed. There can't be anything nicer. Was it officially published? If yes, please give the reference.

Pentcho Valev

"Pentcho Valev" schreef in bericht news:bdf02d35.0408170016.14921552@posting.google.com...

>	shuba wrote in message news:...

> >	Bill Hobba wrote:

> > >

Without actually doing the calculations may I suggest that if you were traveling at the rim of a space station traveling at .99c and a radius of one light year the normal equations of classical mechanics would not apply? - see http://www.smcm.edu/nsm/physics/SMP03S/KeatingB.doc.pdf and its conclusion:

> >	That's a nice paper. Thanks, Bill.

>	Very nice indeed. There can't be anything nicer. Was it officially published? If yes, please give the reference.

Is that document really that good? It does not explain who is right Ehrenfest (circumference < 2pi*R), Einstein (ration > pi) or Strauss.
About the last they write: "Other physicists, such as Strauss, argued that if the measuring rods were contracted, then so were the distances they were measuring, so ratio C/D would still be pi"

IMO the following is a better document to study:: http://arxiv.org/PS_cache/physics/pdf/0404/0404027.pdf

But I think this one is even better: http://edu.supereva.it/solciclos/gron_d.pdf

This document gives an excellent overview of opinions of many people: Ehrenfest, Einstein and Strauss At the same time it also informs you why those opinions could be wrong For example at page 4 Max Planck discusses the Ehrenfest paradox
page 5 Einstein argues that the Ehrenfest paradox is wrong
page 7 Becquerel argues why Einstein is wrong
page 20 Eddington comments on Ehrenfest paradox
page 20 Lorentz reports on Eddington
page 37 Gron argues why Strauss is wrong. and many more.....

Many opinions.... but how do you know who is right ?

Fig 6 page 38 shows a disc with n measuring rods at rest. I assume that the rods are made of the same material as of the disc.
Fig 7 page 39 shows the same disc with n rods and angular velocity omega. The rods are Lorentz contracted. The radius of the disc of fig 6 and 7 is the same.

Again how do you know if this is right?

The standard answer is to do an experiment and test it. The problem is that such an experiment can not be performed with enough accuracy.

Consider "fig 8" a disc like fig 6 but now with a smaller radius but such that the rods are of the same length as of fig 7.

Is this the correct situation ?

Draw a second circle at radius R/2 inside fig 6 with the same number n of rods. Mark the rods in both circles from 1 to n. Draw the rods such that they are all "in line" like the spokes of a wheel.

Do the the same inside fig 8. However draw the rods at the inner circle slightly off line such that the spokes are slightly bended. (Can you envision this ?)

Is this the correct situation ?

There should be something like a Solvay Conference to decide what is right or at least what is wrong.

Nicolaas Vroom
https://www.nicvroom.be/

Nicolaas Vroom wrote:

[re: http://www.smcm.edu/nsm/physics/SMP03S/KeatingB.doc.pdf ]

>	Is that document really that good? It does not explain who is right Ehrenfest (circumference < 2pi*R), Einstein (ration > pi) or Strauss.

Read the conclusion. The resolution is that the "circumference" is not a well-defined geometric quantity.

---Tim Shuba---

Nicolaas Vroom wrote:

>	Many opinions.... but how do you know who is right ?

As I keep stressing: in an accelerated system (like a rotating disk), geometry is INHERENTLY AMBIGUOUS. You can get whatever answer you want by defining differently what you mean by "circumference of the rotating disk".

But that's OK -- you are attempting to discuss an abstract and insufficiently-well-described quantity. Specify what you are talking about well enough to correspond to ACTUAL MEASUREMENTS, and the ambiguities disappear.

Of course historically this was not always known, and early investigators made mistakes....

Tom Roberts tjroberts@lucent.com

"Nicolaas Vroom" wrote in message news:<1JbXc.225766$e54.11082887@phobos.telenet-ops.be>...

>	"Pentcho Valev" schreef in bericht news:bdf02d35.0408170016.14921552@posting.google.com...

> >	shuba wrote in message

>

news:...

> > >

Bill Hobba wrote:

> > > >

Without actually doing the calculations may I suggest that if you were traveling at the rim of a space station traveling at .99c and a radius of one light year the normal equations of classical mechanics would not apply? - see http://www.smcm.edu/nsm/physics/SMP03S/KeatingB.doc.pdf and its conclusion:

> > >

That's a nice paper. Thanks, Bill.

> >	Very nice indeed. There can't be anything nicer. Was it officially published? If yes, please give the reference.

>

Is that document really that good? It does not explain who is right Ehrenfest (circumference < 2pi*R), Einstein (ration > pi) or Strauss. About the last they write: "Other physicists, such as Strauss, argued that if the measuring rods were contracted, then so were the distances they were measuring, so ratio C/D would still be pi"

Under different circumstances, (circumpference < 2pi*R) would be universally accepted by relativists. However, in the present case, this result contradicts others and, if accepted, would destroy relativity. In such cases the Juggler is particularly creative, produces superabsurdities and so paralyses any possible criticism. See the end of ch. 23 in his 1920 "Relativity". It is the observer on the disc who does the experiment, then the validity of the result is assumed relative to the non-rotating frame and in the end the circumference proves greater than 2pi*R. Of course, this result is also contradictory but people have a limited ability to criticise absurdities. A science with too many absurdities becomes Divine Science immune to any criticism. In fact, this is the real discovery of Einstein.

Pentcho Valev

"Pentcho Valev" wrote in message news:bdf02d35.0408260616.41e6ee9f@posting.google.com...

[snip]

>

Under different circumstances, (circumpference < 2pi*R) would be universally accepted by relativists. However, in the present case, this result contradicts others and, if accepted, would destroy relativity. In such cases the Juggler is particularly creative, produces superabsurdities and so paralyses any possible criticism. See the end of ch. 23 in his 1920 "Relativity". It is the observer on the disc who does the experiment, then the validity of the result is assumed relative to the non-rotating frame and in the end the circumference proves greater than 2pi*R. Of course, this result is also contradictory but people have a limited ability to criticise absurdities. A science with too many absurdities becomes Divine Science immune to any criticism. In fact, this is the real discovery of Einstein.

In fact, the real discovery of Einstein, is a bunch of clueless idiots who can't read a text and then feel compelled to show how stupid they are by failing to understand and even reproduce parts of the texts to begin with.

Dirk Vdm

Tom Roberts wrote in message news:...

>	Nicolaas Vroom wrote:

> >	Many opinions.... but how do you know who is right ?

>

As I keep stressing: in an accelerated system (like a rotating disk), geometry is INHERENTLY AMBIGUOUS. You can get whatever answer you want by defining differently what you mean by "circumference of the rotating disk". But that's OK -- you are attempting to discuss an abstract and insufficiently-well-described quantity. Specify what you are talking about well enough to correspond to ACTUAL MEASUREMENTS, and the ambiguities disappear. Of course historically this was not always known, and early investigators made mistakes....

Poor early investigators! Why did not they recognize their mistakes later? Anyway, apart from the length contraction which has proved so inherently ambiguous, the same early investigators predicted that a clock on the rotating disc runs slow by a factor of 1/gamma. Then the early investigators based the rest of relativity on that result. Now the question is: is this time dilation independent of the inherently ambiguous length contraction? If it is not...

Pentcho Valev

"Tom Roberts" schreef in bericht news:cgkqm2$h8f@netnews.proxy.lucent.com...

>	Nicolaas Vroom wrote:

> >	Many opinions.... but how do you know who is right ?

>

As I keep stressing: in an accelerated system (like a rotating disk), geometry is INHERENTLY AMBIGUOUS. You can get whatever answer you want by defining differently what you mean by "circumference of the rotating disk". But that's OK -- you are attempting to discuss an abstract and insufficiently-well-described quantity. Specify what you are talking about well enough to correspond to ACTUAL MEASUREMENTS, and the ambiguities disappear.

In the first half I try to abstract if there is anything they agree about. And that is very difficult.

>	Of course historically this was not always known, and early investigators made mistakes....

And who is right ? Einstein, Lorentz, Strauss, Eddington, Ehrenfest ?

And what is your opinion about the 8 conclusions of this document: http://edu.supereva.it/solciclos/gron_d.pdf ?

IMO the whole purpose of this exercise is to answer the following questions:
1) What is the behavior of a rotating disc (compared to a disc at rest)
2) Is length contraction involved.
3) If yes is the amount in agreement with SR.

What makes this such a tricky issue because the concept of rigid and Born rigid are introduced.

In order for me to understand the issues involved I raised two questions:
1. Of a disc with R =100 with measuring rods of l =1 with speed v of circumference. How many rods can be placed (fixed at 1 point) on the circumference:
1) 628 2) more than 628 3) less than 628 4) impossible to answer.

2. Of a disc with R =100 with measuring rods of l =1 with speed 0 of circumference. How many rods with speed v can be placed on the circumference:
1) 628 2) more than 628 3) less than 628 4) impossible to answer.

Apparently (based on the number of answers) those questions are too difficult to answer.

I have a different questions: Are there readers of this newsgroup who were involved in an examination about physics were the rotating disc was discussed (investigated) ?
What were the questions ? What were the correct answers ?

I hope the questions were not like: What is the opinion of xyz about the rotating disc.

Nicolaas Vroom
https://www.nicvroom.be/

>	Tom Roberts tjroberts@lucent.com

"Tom Roberts" wrote in message news:cgkqm2$h8f@netnews.proxy.lucent.com...

>	Nicolaas Vroom wrote:

> >	Many opinions.... but how do you know who is right ?

>	As I keep stressing: in an accelerated system (like a rotating disk), geometry is INHERENTLY AMBIGUOUS. You can get whatever answer you want by defining differently what you mean by "circumference of the rotating disk".

That's the standard Relativist approach, all right. Define all problems away.....

>	But that's OK -- you are attempting to discuss an abstract and insufficiently-well-described quantity. Specify what you are talking about well enough to correspond to ACTUAL MEASUREMENTS, and the ambiguities disappear.

Only when you get to redefine your measurements after they're made.

>	Of course historically this was not always known, and early investigators made mistakes....

LOL!

-- greywolf42
ubi dubium ibi libertas
{remove planet for e-mail}

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