Einstein on the rotating disc
Pentcho Valev
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. 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.
There have been several contests organized within zombiism for the
prize "Most confusing argument". This discovery of Einstein always
won. A few zombis have been expelled from the organization since they
claimed that, in K, the ratio of the periphery of the rotating disc
and the diameter is smaller than pi.
Pentcho Valev
Tom Roberts
> 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 tjro...@lucent.com
Martin Stone
"Pentcho Valev" <pva...@yahoo.com> wrote in message
news:bdf02d35.04080...@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.
davidoff404
The definitive review of this topic is contained in General Relativity
and Gravitation, Volume 1: "Einstein and the Rigidly Rotating Disk",
John Stachel, A. Held (Editor), ISBN 0-306-40265-3 (v. 1).
greywolf42
news:cf7ui3$3...@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}
Bill Hobba
"Martin Stone" <dont...@me.com> wrote in message
news:cf7v27$1u1$1...@rdel.co.uk...
>
> "Pentcho Valev" <pva...@yahoo.com> wrote in message
> news:bdf02d35.04080...@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
Pentcho Valev
> 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
Harry
news:<cf7ui3$3...@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).
>
>
> 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..."
Yes indeed - that is due to Lorentz contraction of the moving rod (along its
length ).
> 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?
Not according to Lorentz:
"If v is the velocity at the rim, the radius will be shortened in the ratio
of 1 to 1-1/8*v^2/c^2. The circumference changing to the same extent, its
decrease is seen to be exactly one-fourth of that of a rod moving with the
same velocity in the direction of its length." -Lorentz, Nature, feb.17
1921, p.793.
Your conclusion should be approximately correct however for a bicycle
wheel - not accounting for centrifugal (inertial) forces of course!.
Harald
Nicolaas Vroom
"Tom Roberts" <tjro...@lucent.com> schreef in bericht
news:cf7ui3$3...@netnews.proxy.lucent.com...
>
> Conclusion: non-inertial systems are DIFFERENT from inertial ones. No
> surprise to anyone who understands the underlying geometry of SR.
>
> For instance, if at every application of the ruler to the
> circumference the K' observer also E-syncs clocks at the
> two ends ofthe ruler, by the time he gets back to where he
> started the first and last clocks will NOT be in synch. If
> the observer in K does this they will be in synch.
>
> Note in all cases the ruler used must be very short compared
> to the radius of the disc.
It is not clear to me why you need clocks.
Suppose I have a disc at rest with a Radius of R = 100
and instead of 1 I use 628 rulers each with a length l = 1.
Do you agree that I nicely can place those rulers
(front to end)
against the circumference and that there will be a small
space between no 1 and no 628 ?
(not enough to insert ruler 629)
Now I start to rotate the disc.
Do you agree that If I keep each ruler fixed at one place
along the circumference that there will be a small
space between each ruler ?
Or to state this different.
In order to keep the 628 rulers in contact which each other,
such that front meets end, you have to move the rulers
along the circumference, resulting in a larger cap
between no 1 and no 628. Meaning you can place more
rulers along the circumference.
(The more the faster you rotate the disc ?)
If you agree than
how do you prove that this is true ?
Nicolaas Vroom
http://users.pandora.be/nicvroom/
>
> Tom Roberts tjro...@lucent.com
Martin Stone
"Bill Hobba" <bho...@rubbish.net.au> wrote in message
news:T3URc.43369$K53....@news-server.bigpond.net.au...
>
> "Martin Stone" <dont...@me.com> wrote in message
> news:cf7v27$1u1$1...@rdel.co.uk...
> >
> > "Pentcho Valev" <pva...@yahoo.com> wrote in message
> > news:bdf02d35.04080...@posting.google.com...
> > > One of the deepest insights Einstein ever had can be found at the end
> > > of ch. 23 in his 1920 "Relativity". An observer who is sitting
> > > eccentrically" on the periphery of a rotating disc (K') measures the
> > > circumference of the disc with his measuring rod,
> >
> > How do you measure circumference (curved) with a rod (straight)?
>
> Have you ever studied calculus? The idea is for infinitesimal sized
> straight lines (rods) you can lay then end to end around a curve and get a
> very good approximation of the length of the curve - the smaller the line
> (rod) the better the accuracy.
Yeah - that's not disputed (by me). I kinda imagine a guy on a table sized
disk with a metre ruler, then waded in. Laying out those infinitesimal rods
might take a while though. ;o) And actually, now I think about it you
could easily measure round a curve with a straight edge - imagine rolling
the disk along the ruler, but do the measurement by moving the ruler and not
the disk. If ya see.
> What the thought experiment shows is that
> since rods on the rotating disk are shortened more of then are required to
> measure the length of the circumference hence pi is bigger.
If we define pi as "the ratio of a circle's circumference to its diameter",
OK. And for this thought experiment I'm not sure I'd take a measuring tape
after all. Say we laid the tape round the disk right at the very edge and
then started spinning and for some reason the tape didn't just fly off -
would the ends of the tape just move apart or would they stay together while
the whole length of the tape gets further from the edge of the disk?
Rods it is. :o)
Tom Roberts
> "Bill Hobba" <bho...@rubbish.net.au> wrote in message
> news:T3URc.43369$K53....@news-server.bigpond.net.au...
>>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 tjro...@lucent.com
Martin Stone
> Martin Stone wrote:
> > "Bill Hobba" <bho...@rubbish.net.au> wrote in message
> > news:T3URc.43369$K53....@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 tjro...@lucent.com
Tom Roberts
> "Tom Roberts" <tjro...@lucent.com> schreef in bericht
> news:cf7ui3$3...@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 tjro...@lucent.com
Tom Roberts
> Tom Roberts <tjro...@lucent.com> wrote in message news:<cf7ui3$3...@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).
>
> Not "quite clear" at all.
Sure it is -- that's what we mean by an inertial frame: the 3-space of
such a frame is Euclidean, and all circles have pi as the ratio of
circumference to diameter.
BTW by "the same measurement" to an observer in K, I meant
measuring the circumference and diameter of the edge of the
disk using rulers at rest in K -- that's naturally how an
observer in K would do this. The fact that the disk is rotating
wrt K is irrelevant to measuring its circumference in K.
> 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.
It is the observer in K' that obtains a ratio larger than pi. The
observer in K obtains pi. The observer in K naturally uses rulers at
rest in K.
> 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..."
Everything I have said is consistent with that quote.
> 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?
No. Since they occupy the same locations, the circumference of the
rotating disk's edge, and the circumference of the same non-rotating
circular line in K, MUST be the same to the observer in K -- the
observer in K naturally uses rulers at rest in K to make all measurements.
Any attempt to apply Euclidean geometry to the ROTATING system will
fail. In the non-rotating (inertial) system K, Euclidean geometry is valid.
Tom Roberts tjro...@lucent.com
JM Albuquerque
"Tom Roberts" <tjro...@lucent.com> wrote:
> 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.
>
>
> > 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.
There is a way to do the experiment, which is by means of light
in the Sagnac experiment.
Sagnac Experiment not only shows that speed of light is not
independent of observer's speed, but also can prove there is no
contraction at all in the disk perimeter.
Why do you think that Sagnac experiment is not a true valid prove?
(I'm sure that you know perfectly well the Sagnac experiment).
Tom Roberts
> [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 tjro...@lucent.com
J.J. Simplicio
> 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
Bill Hobba
"J.J. Simplicio" <a...@nospam.com> wrote in message
news:KGsSc.131311$eM2.29111@attbi_s51...
>
> "Tom Roberts" <tjro...@lucent.com> wrote in message
> news:cfdgra$e...@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
Bill Hobba
That is the definition in Euclidean space. If experiment shows it is not
that then guess what - we no longer have Euclidean space.
> 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?
What is your point? If you used a tape and attached it to the rim then it
would be part of the rim and subject to its some effects so you can not
trust its measurement.
>
> Rods it is. :o)
>
In that case you are left with the inevitable conclusion space is not
Euclidian.
Bill
J.J. Simplicio
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
Bill Hobba
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
>
Martin Stone
> > 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:
I think the point being made was just that if we want to acheive a
particular speed then using a bigger disk will mean the lab technician
sitting on it experiences less g. The figures were done with near light
speeds while assuming everything was nicely, er, Newtonian(?) so they're
wrong but the point stands.
The technician on my table sized disk probably wouldn't hold on for long at
all. ;o)
Pentcho Valev
So you come to a contradiction and, if you did not belong to that
particular cult, you would reject the theory, as logic requires. On
one hand, as judged from K, any small segment of the rotating
periphery is length-contracted and one must conclude that the sum of
all such segments, that is, the periphery, is also length contracted.
On the other, as you say, "since they occupy the same locations, the
circumference of the rotating disk's edge, and the circumference of
the same non-rotating circular line in K, MUST be the same to the
observer in K".
Of course, the contradiction has other expressions and I suggest that
you look for them. For instance, place clocks at small distances on
the rotating periphery and see what would happen as a clock belonging
to K consecutively meets them. Don't say that the problem is
unsolvable since the rotating clocks cannot be synchronized etc. This
is part of Einstein's camouflage which he carefully built around this
extremely vulnerable issue. He was not very clever but his intuition
was strong indeed.
Pentcho Valev
Nicolaas Vroom
> 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
http://users.pandora.be/nicvroom/
J.J. Simplicio
> > 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?
Good point. I'll use the term ''lab frame'' for the inertial frame in which
the center of the disk is at rest and the disk is spinning such that the rim
is moving at a relativistic speed v. Relative to the lab frame, particles
on the rim of the disk would in fact have a centripetal acceleration of
v^2/r since this formula is derivable strictly from the definition of
acceleration and is valid whether v is relativistic or not.
However, you are correct to question the use of this formula in determining
the radial stresses in the disk. To decide if the disk will burst apart
from ''centrifugal'' forces we need to look at the ''proper acceleration''
of a point on the rim; i.e., the acceleration that a point on the rim would
actually ''feel'' due to its non-inertial motion. This proper acceleration
of a point on the rim is the acceleration as observed in an inertial frame
that is instantaneously comoving with the point and it turns out to be
*larger* than v^2/r by a factor of gamma^2.
Therefore, I should have been considering the formula v^2/(1-v^2/c^2) *
(1/r) rather than just v^2/r. Note, however, that we still have the inverse
proportionality with r. So, for a given specified speed of the rim, the
proper acceleration (and therefore the radial stress) will decrease
inversely with r.
For a rim speed (relative to the lab) of 0.99c, the proper acceleration at
the rim will be less than 1 g if r is larger than about 50 light years.
Astronomical indeed.
JJ
Tom Roberts
> [about rulers laid out around the circumference of a rotating disk]
> There is a way to do the experiment, which is by means of light
> in the Sagnac experiment.
Not really. The Sagnac experiment is related, but most definitely not
the same.
> Sagnac Experiment not only shows that speed of light is not
> independent of observer's speed, but also can prove there is no
> contraction at all in the disk perimeter.
No. All the Sagnac experiment can show is that there is a rotation-rate
dependent phase shift between the two directions around a rotating light
path.
Note that the results for actual Sagnac experiments are fully consistent
with the predictions of SR.
> Why do you think that Sagnac experiment is not a true valid prove?
Because it does not involve laying rulers down around the circumference
of a rotating disk. <shrug>
> (I'm sure that you know perfectly well the Sagnac experiment).
Yes I do. It seems you do not. Or you read a lot more into the results
than is really there.
Tom Roberts tjro...@lucent.com
Bill Hobba
Provided we also slow angular velocity - keeping it the same means as you
move further out from the center of rotation you experience more force.
>The figures were done with near light
> speeds while assuming everything was nicely, er, Newtonian(?) so they're
> wrong but the point stands.
>
> The technician on my table sized disk probably wouldn't hold on for long
at
> all. ;o)
Sure - if that was the case. As the link I gave showed as the length
increased and the angular velocity decreases it tends towards an inertial
frame - with this being a very interesting thing to consider:
'When ? = 0, the worldlines of the points on the edge of the disk are
straight lines, and the locus of events simultaneous to the flash is a
closed spacelike curve. But as soon as the disk is set into rotation the
spacelike path changes its topology and becomes an open curve, so that the
definition of simultaneity becomes a matter of
convention; it depends on where one starts the integration carried out in
section 5.'
Thanks
Bill
Bill Hobba
You have been repeatidly told it has been mathematically proven SR is as
consistent as Euclidian Geometry. No contradictions exist.
Bill
shuba
> 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---
Pentcho Valev
Very nice indeed. There can't be anything nicer. Was it officially
published? If yes, please give the reference.
Pentcho Valev
shuba
[Re: http://www.smcm.edu/nsm/physics/SMP03S/KeatingB.doc.pdf ]
> Very nice indeed. There can't be anything nicer.
Hmm. Is that what you really think?
> Was it officially
> published? If yes, please give the reference.
I don't know if it was published. As someone else mentioned,
there are more definitive treatments of the relativistic rotating
disk. I like the clarity of this presentation. There has been a
lot of nonsense written in this group about the Sagnac effect,
and this piece should clear most of it up. The author may be the
same Brian Keating who has several published papers about the
Cosmic Microwave Background, but that's just a guess on my part.
---Tim Shuba---
Pentcho Valev
> Pentcho Valev wrote:
>
> [Re: http://www.smcm.edu/nsm/physics/SMP03S/KeatingB.doc.pdf ]
>
> > Very nice indeed. There can't be anything nicer.
>
> Hmm. Is that what you really think?
Yes. The paper can be regarded as paradigmatic for the relativistic
approach. If C is the length of the rotating periphery and Co that of
a non-rotating one, then, as judged from the non-rotating frame,
1. C > Co, according to Einstein and other prominent physicists (p.
6).
2. C = Co, according to Strauss and other physicists (p. 2).
3. C < Co, according to the author of the paper (p. 8).
The reader is free to choose the result that suits him/her best.
Another essential point is that "Einstein did not consider it to be a
significant problem, however" (p. 6).
Pentcho Valev
shuba
> If C is the length of the rotating periphery
False premise.
---Tim Shuba---
Dirk Van de moortel
"shuba" <tim....@eudoramail.com> wrote in message news:tim.shuba-27F6D8.07530218082004@cp...
> Pentcho Valev wrote:
>
> > If C is the length of the rotating periphery
>
> False premise.
Actually, it depends a bit on the page :-)
Apart from his inability to
- read,
- interpret equations,
- behave,
he seems to be another victim of the dreaded
Darn-Minute-Syndrome:
http://users.pandora.be/vdmoortel/dirk/Stuff/PentchoAtSchool.gif
Dirk Vdm
Ken S. Tucker
Very interesting subject Mr. Valev, a real kook magnet,
so I'll reply.
Speaking of magnets, what if your disc was electrostatically
charged at rest, then spun? Make it easier, charge a loop and
spin it, (that looks like a simplified solenoid).
According to Lorentz and SR,
Force(magnetic) = q(V x B)
Using the charge q as a test charge, and supposing it
is at rest relative to the magnetic field B and therefore
also the spinning loop, then V=0, and the magnetic force
is zero, hence Strauss is correct (above) and C=Co, no
difference can be detected by the test charge.
That can be simplified. Suppose a disc is rotating about
the z-axis, in the x-y plane, and a observer is suspended
above the disc on the z-axis. The relative motion of any
point on the disc is zero, because the change in the
distance of the point relative to the observer is zero,
ie. dr/dt=0, hence v=0 in (1-v^2) and Lorentz contraction
doesn't apply, and no *Ehrenfest paradox* occurs.
((Poor Ehrenfest, he shot his young son, then himself,
that's the true paradox))
The complete understanding of a test charge moving
relatively to a charged spinning disc requires a
solution in accord with General Relativity - due to
centrifugal force, and Quantum Theory - due to the
photon exchange required to convey ElectroMagnetic
Force. The theory covering that is not well known
and is certainly a digression from the subject.
Regards
Ken S. Tucker
Nicolaas Vroom
"Nicolaas Vroom" <nicolaa...@pandora.be> schreef in bericht
news:ejHSc.211506$0x4.10...@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
Tom Roberts
>>Iis 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 tjro...@lucent.com
Nicolaas Vroom
1) When the disc is at rest and R = 100
and when the length of the rulers is 1
2)You can place 100 rulers along the radius
3)You can place 628 rulers along the circumference
4) and the ratio = 6.28
When the disc rotates and when the speed of the
circumference = v
5) The length of the radius does not change
(compared with a disc at rest)
6) The number of rulers comoving with the
circumference and all touching front to end is n
and larger than 628.
7) The ratio is larger than 6.28
8) When the disc is at rest but the rulers are moving
with a speed v along the circumference
than the number of rulers you can place
along the circumference is also n.
9)The length contraction of the rulers is in agreement
with lorentz transformations
10)It is not possible to perform an actual experiment
to prove that the above is true.
Nicolaas Vroom
http://users.pandora.be/nicvroom/
Nicolaas Vroom
"Pentcho Valev" <pva...@yahoo.com> schreef in bericht
news:bdf02d35.04081...@posting.google.com...
> shuba <tim....@eudoramail.com> wrote in message
news:<tim.shuba-085173.08585816082004@cp>...
> > 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:
> > >
> >
>
> 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
http://users.pandora.be/nicvroom/
shuba
[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---
Tom Roberts
> 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 tjro...@lucent.com
Pentcho Valev
> "Pentcho Valev" <pva...@yahoo.com> schreef in bericht
> news:bdf02d35.04081...@posting.google.com...
> > shuba <tim....@eudoramail.com> wrote in message
> news:<tim.shuba-085173.08585816082004@cp>...
> > > 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
Dirk Van de moortel
"Pentcho Valev" <pva...@yahoo.com> wrote in message news:bdf02d35.04082...@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
Pentcho Valev
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
Nicolaas Vroom
> 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.
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
http://users.pandora.be/nicvroom/
> Tom Roberts tjro...@lucent.com
greywolf42
> 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}