Tom Roberts wrote:
[I just happened to see this.]

wrote:
On the other hand, given the complications that occur
with acceleration, I agree it's a bit surprising that
Born-rigid acceleration is possible at all, even in
theory.

This is not "surprising", it is natural:

Born rigid motion is the natural way a solid object accelerates, as long
as it is pushed or pulled from a single point, and the acceleration is
small enough that the inter-molecular bonds of the object remain on
average unstressed (i.e. the object does not break or deform). Those
bonds will try to keep the inter-molecular spacings constant, making
this as rigid as possible for an atomic object, and that's just what
Born rigid motion is.

Yes, I knew that. And yet, I managed to say some stupid
things about it in this thread -- so perhaps we should
still say it was surprising *to me*, even though it
should not have been, and hopefully won't be so in the
future. Surprising to the uninitiated, let us say.

I'm reviving a discussion we had a couple weeks ago...

Todd wrote:
wrote in message
oups.com...
Todd wrote:

[snip my rotating corkscrew paradox]

....which I posed as an ordinary corkscrew rotating along its
axis, and I made the claim that the corkscrew must be straight
in some frame. As Todd corrected me, this could only happen
if the corkscrew had a very long "pitch", or a very high rate
of rotation, or both.

But assuming we could construct such a corkscrew, there is
still a problem in the moving frame: we would have a perfectly
straight rod pulling itself along in a helical path by its
own bootstraps, as it were, in violation of our momentum
conservation laws.


I think the solution to this 'paradox' is that we need to include the
momentum contributed by the _stresses_ in the corkscrew. These stresses
arise because the corkscrew is rotating. In the frame in which the COM
is
screaming along at relativistic speeds, these contributions are large.

I guess your word "scream" is more appropriate than my "zip".
But in principle the issue exists at any speed.


Good point. The parameters are related in a nice way. Let v be the speed
between the rest frame of the rotating helix and the frame that sees the
helix unwound as a straight line. Let f be the frequency of rotation of the
helix in revolutions per second. Finally let p be the pitch of the helix as
measured in the rest frame in units of _light-seconds_. Then

(p)(v/c)(f) = 1

So, you can have any speed for v (less than c, of course) as long as you are
willing to adjust p and f appropriately. If you want v to be
nonrelativistic and f to be low enough so that the stresses aren't
ridiculously large, then the pitch will have to be enormous. For example,
if v/c = .01 and f = 100 Hz, then p = 1 light-second.


It must turn out that the total momentum (including the contributions
from
the stresses) is indeed conserved if no external forces are applied to
the
system. To actually check this, you would have to first work out the
stresses inside the rotating corkscrew in the frame in which the COM of
the
corkscrew is at rest (yuck), and then transform these components to the
moving frame (pretty easy).

A nice treatment of stress contributions to momentum is given in Tolman's
_Relativity, Thermodynamics, and Cosmology_ which is cheaply available
from
Dover Pub. A famous example of a paradox that arises if you don't
include
the stress momentum is the so-called Right-Angled Lever Paradox (see page
79
of Tolman ).

Thank you, Todd, that is just the sort of expert contribution
that I was hoping for. Similarity to the Right-Angled Lever
had actually occurred to me separately, but I've never seen
that one worked out. I've heard about Tolman's book, and now
I guess I'll have to get a copy.


It's certainly outdated in many ways, but it contains some interesting
topics that you generally can't find in the modern texts.

I now have it. And yes, you are spot-on about that book.
What you didn't mention, though, is that the writing is
wonderfully clear, a classic example of how one should talk
about relativity (albeit in 1934-style conventions, some of
which requiring translation into modern terms). In its
concise felicity of phrase it reminds me of some of the
posts Steve Carlip has made here.



Aside 1: A helix with an integral number of windings will have its COM
on
the symmetry axis of the helix. By 'symmetry axis', I mean the axis of
the
cylinder on which the helix may be imagined to be wound. However, I
don't
believe the symmetry axis of the helix will be one of the _principle
axes_
of the moment of inertia tensor of the helix. This means that if you
start
the helix rotating about its symmetry axis, it will wobble in the absence
of
any external forces. So, this is an added complication.

No doubt you're right about that. But actually I was only
trying for something easy to visualize; I don't think it's
in any way a crucial detail.


Right.

But on second thought, it does start to matter a lot when
you are talking about a helix of, say, one coil (or less).
Which indeed we are, in *some* frame. So I think it
probably *is* crucial to getting the proper theoretical
limit for a corkscrew of finite length.



Aside 2: If I'm not mistaken: In order for there to exist an inertial
frame in which the rotating helix appears straight, the helix must be
wound
such that in the helix's own rest frame the 'pitch' of the helix
(distance
between consecutive windings) is greater than 2*Pi*R where R is the
radius
of the helix. I think that's kind of interesting.

Doesn't it depend on omega? But I guess you mean that at
the theoretically limiting omega (c/R) this is the minimum
pitch, and therefore it is minimum overall. I think you
are right.


If p is measured in ordinary units rather than light-seconds then the
relation is

(p/c)(v/c)(f) = 1

If we let u be the tangential speed of a point of the helix as it rotates in
the rest frame, then f = u/(2*Pi*R). So, the relation may be rearranged as

p = (2*Pi*R)(c/v)(c/u)

Thus, p is always greater than 2*Pi*R and can only approach this value for
both v and u approaching c.

This injects a further dose of healthy reality into my
scenario of the ordinary steel corkscrew: unless the spin
is some ungodly rate (too much for steel) there won't be
much of an effect to measure.

I think that's right. For any real corkscrew rotating at any physically
allowable rate, it appears that even when v approaches c the screw will
appear only 'slightly unwound'.

And as you suggest, the stress energy in the metal will
provide the necessary momentum to compensate for any deviation
of the COM from the central axis, and also keep the principal
axis pointing where it should (wherever that may be ;-).

But now I must say, I do not think that even in theory we
can have a fully straightened corkscrew of the kind I've
posed. (And for finite-length corkscrews, I think the
problem must occur somewhere around one coil.) Otherwise
the stress momentum must lie *outside* the rod to avoid
the bootstrap problem I mentioned above. Yes, I know EM
fields could conceivably do that, but there simply aren't
any of note here, outside of the material itself.

Instead, I think it would be found on calculation (and a
hairy one it would be too, AFAICS) that a corkscrew could
not be rigid enough, even in theory, to maintain its required
high-pitch helical shape under a rotation of the magnitude
necessary to approach straightness in some inertial frame.
I think it would be found that the speed of sound in the
material must exceed c for that to happen.

wrote:
...which I posed as an ordinary corkscrew rotating along its
axis, and I made the claim that the corkscrew must be straight
in some frame. As Todd corrected me, this could only happen
if the corkscrew had a very long "pitch", or a very high rate
of rotation, or both.

A normal corkscrew won't do, you need one of fixed pitch, and it must
lie on a cone with apex at its point where it is driven to rotate. That
is, it is a helix with linearly varying radius, having zero radius at
its point (where it is driven).


But assuming we could construct such a corkscrew, there is
still a problem in the moving frame: we would have a perfectly
straight rod pulling itself along in a helical path by its
own bootstraps, as it were, in violation of our momentum
conservation laws.

No. The driving force in the initial frame will have tangential
components that must conserve momentum; they will make momentum be
conserved in any other frame, including the one in which the corkscrew
is a straight rod (angled relative to the axis around which it rotates,
of course, and driven from the point it intersects the axis).


Tom Roberts

Tom Roberts wrote:
wrote:
...which I posed as an ordinary corkscrew rotating along its
axis, and I made the claim that the corkscrew must be straight
in some frame. As Todd corrected me, this could only happen
if the corkscrew had a very long "pitch", or a very high rate
of rotation, or both.

A normal corkscrew won't do, you need one of fixed pitch, and it must
lie on a cone with apex at its point where it is driven to rotate. That
is, it is a helix with linearly varying radius, having zero radius at
its point (where it is driven).

Thanks, Tom, this is very interesting.

As you can see later in my post, I do reach the conclusion
that my scenario wouldn't fly, owing to too much stress
(without compensating strain) requiring the speed of sound
to exceed c in the material. Was this a correct statement,
albeit one I didn't attempt to justify by calculation? In
other words, what is it that forces the conical shape.

Also, I while do see why your conical helix would have to
spin mounted on some exceedingly massive base (to provide
the momentum necessary for a stable axis) I don't see why
it would have to be *driven* -- apart from the question of
gravitational radiation, of course. Is there some other
reason for energy to be lost here?



But assuming we could construct such a corkscrew, there is
still a problem in the moving frame: we would have a perfectly
straight rod pulling itself along in a helical path by its
own bootstraps, as it were, in violation of our momentum
conservation laws.

No. The driving force in the initial frame will have tangential
components that must conserve momentum; they will make momentum be
conserved in any other frame, including the one in which the corkscrew
is a straight rod (angled relative to the axis around which it rotates,
of course, and driven from the point it intersects the axis).

That much I see.

Thanks again for any further info you might suggest.

Some clarifications --

wrote:
Tom Roberts wrote:
wrote:
...which I posed as an ordinary corkscrew rotating along its
axis, and I made the claim that the corkscrew must be straight
in some frame. As Todd corrected me, this could only happen
if the corkscrew had a very long "pitch", or a very high rate
of rotation, or both.

A normal corkscrew won't do, you need one of fixed pitch, and it must
lie on a cone with apex at its point where it is driven to rotate. That
is, it is a helix with linearly varying radius, having zero radius at
its point (where it is driven).

Thanks, Tom, this is very interesting.

As you can see later in my post, I do reach the conclusion
that my scenario wouldn't fly, owing to too much stress
(without compensating strain) requiring the speed of sound

i.e. if we gave it enough strain to keep its Young's
modulus within theoretically allowed limits, it wouldn't
stay helical

to exceed c in the material. Was this a correct statement,
albeit one I didn't attempt to justify by calculation? In
other words, what is it that forces the conical shape.

Also, I while do see why your conical helix would have to

while I do

spin mounted on some exceedingly massive base (to provide
the momentum necessary for a stable axis) I don't see why
it would have to be *driven* -- apart from the question of
gravitational radiation, of course. Is there some other
reason for energy to be lost here?



But assuming we could construct such a corkscrew, there is
still a problem in the moving frame: we would have a perfectly
straight rod pulling itself along in a helical path by its
own bootstraps, as it were, in violation of our momentum
conservation laws.

No. The driving force in the initial frame will have tangential
components that must conserve momentum; they will make momentum be
conserved in any other frame, including the one in which the corkscrew
is a straight rod (angled relative to the axis around which it rotates,
of course, and driven from the point it intersects the axis).

That much I see.

Actually, what I see is that given a sufficiently massive
mounting (and a forgiving bearing) you can spin an angled
bar by one end so that it describes a cone, without violating
any conservation laws. But once it's spun up, this does not
require any tangential component of force within the bar or
the mounting, as far as I can see.


Thanks again for any further info you might suggest.

Particularly any modern references on this sort of thing,
as Tolman is, for all its virtues, old.

wrote in message
ups.com...
I'm reviving a discussion we had a couple weeks ago...

Todd wrote:
wrote in message
oups.com...
Todd wrote:

Aside 1: A helix with an integral number of windings will have its
COM
on
the symmetry axis of the helix. By 'symmetry axis', I mean the axis
of
the
cylinder on which the helix may be imagined to be wound. However, I
don't
believe the symmetry axis of the helix will be one of the _principle
axes_
of the moment of inertia tensor of the helix. This means that if you
start
the helix rotating about its symmetry axis, it will wobble in the
absence
of
any external forces. So, this is an added complication.

No doubt you're right about that. But actually I was only
trying for something easy to visualize; I don't think it's
in any way a crucial detail.


Right.

But on second thought, it does start to matter a lot when
you are talking about a helix of, say, one coil (or less).
Which indeed we are, in *some* frame. So I think it
probably *is* crucial to getting the proper theoretical
limit for a corkscrew of finite length.


I think it's possible to arrange for the helix to rotate freely about its
axis without any wobbling by choosing just the right amount of winding and
by also adding a point mass of just the right magnitude to the helix halfway
along the helix (bear with me). This is based on some calculations that I'll
summarize here. I would be grateful to anyone who bothers to check my
results. All the integrations are elementary.

I parameterized the helix as

(x, y, z) = (a*theta, R*cos(theta), R*sin(theta))

where theta is the parameter that runs from -thmax to +thmax. The helix
winds around the x-axis from x = -a*thmax to x = +a*thmax with radius R and
pitch 2*Pi*a. At x = 0, the helix passes through the positive y axis at y =
R.

Letting lambda be the mass per unit length, the total mass is found to be

M = 2*lambda*Sqrt(a^2+R^2)*thmax

For a general value of thmax, the center of mass of the helix is located at

(Xcm, Ycm, Zcm) = (0, R*sin(thmax)/thmax, 0).

For integral number of windings (thmax = 0, Pi, 2Pi, 3Pi, etc.) the center
of mass is located at the origin of the coordinate system. Otherwise, it is
located somewhere on the positive or negative y-axis.

In order for no wobbling to occur as the helix rotates about the x-axis, it
is necessary that the x-axis be one of the principle axes of the moment of
inertia tensor. This will be the case if all the off-diagonal elements
I_xy, I_yx, and I_xz of the tensor are zero. Recall that these are defined
as

I_xy = - integral [x*y*dm], etc.

where dm is an element of mass.

For the helix with an arbitrary value of thmax, I find

I_xy = I_yz = 0 (odd integrands)

I_xz = 2*lambda*a*R*Sqrt(a^2+R^2)*( sin(thmax) - thmax*cos(thmax) )

= M*a*R*( sin(thmax)/thmax - cos(thmax) )

Thus, we can force I_xz to vanish by choosing thmax to be a positive root of

tan(thmax) = thmax

Call these roots: th1, th2, th3, th4, etc. (in order of increasing
magnitude). These roots must be found numerically. For these values of
thmax the helix has a non-integral number of windings.

For these special values of thmax, all off-diagonal elements of the moment
of inertia tensor vanish. We need one further requirement in order for the
helix to be able to freely rotate about the x-axis without any wobble. We
need the center of mass of the helix to be located on the x-axis. From
above, we had Xcm=Zcm=0 and

Ycm = R*sin(thmax)/thmax

If thmax is a root of tan(thmax)=thmax, this may be written as

Ycm = R*cos(thmax).

It is easy to check that for th1, th3, th5, etc., the value of cos(thmax) is
negative and the center of mass is located on the negative y-axis. For th2,
th4, th6, etc. the value of cos(thmax) is positive and the center of mass is
located on the positive y-axis.

Thus, for th1, th3, th5, etc. we can shift the center of mass to the origin
by just adding a point mass m0 to the helix at (x, y, z) = (0, R, 0). For
example, if we choose thmax = th1, then m0 = M*cos(th1). For thmax = th3,
m0 = M*cos(th3), etc.

Thus, for thmax = th1, th3, th5, etc., we can arrange for the center of mass
to be at the origin. It is easy to check that adding the point mass does
not mess up the vanishing of the off diagonal elements of the inertia
tensor. Thus, the system should be able to rotate freely about the x-axis
without wobbling.

If the pitch of the helix is chosen large enough, then there will exist a
'moving' inertial frame in which the rotating helix appears to be completely
unwound into a straight line parallel to the x-axis (with the point mass m0
attached to the midpoint). The system orbits around the x-axis as it also
moves parallel to the x-axis. And it does this without any externally
applied forces! Strange.

Either something's wrong with this analysis or else such a strange motion is
actually allowed. Linear momentum and angular momentum do not *appear* to
be conserved for the system as viewed in the moving frame. However, it
could be that they are in fact conserved if the momentum contributions of
the internal stresses in the helix are taken into account.

Todd

Todd wrote:
wrote in message
ups.com...
I'm reviving a discussion we had a couple weeks ago...

Todd wrote:
wrote in message
oups.com...
Todd wrote:

Aside 1: A helix with an integral number of windings will have its
COM
on
the symmetry axis of the helix. By 'symmetry axis', I mean the axis
of
the
cylinder on which the helix may be imagined to be wound. However, I
don't
believe the symmetry axis of the helix will be one of the _principle
axes_
of the moment of inertia tensor of the helix. This means that if you
start
the helix rotating about its symmetry axis, it will wobble in the
absence
of
any external forces. So, this is an added complication.

No doubt you're right about that. But actually I was only
trying for something easy to visualize; I don't think it's
in any way a crucial detail.


Right.

But on second thought, it does start to matter a lot when
you are talking about a helix of, say, one coil (or less).
Which indeed we are, in *some* frame. So I think it
probably *is* crucial to getting the proper theoretical
limit for a corkscrew of finite length.


I think it's possible to arrange for the helix to rotate freely about its
axis without any wobbling by choosing just the right amount of winding and
by also adding a point mass of just the right magnitude to the helix halfway
along the helix (bear with me).

Like dynamically balancing a tire with one well-positioned
weight... seems plausible.

This is based on some calculations that I'll
summarize here. I would be grateful to anyone who bothers to check my
results. All the integrations are elementary.

I parameterized the helix as

(x, y, z) = (a*theta, R*cos(theta), R*sin(theta))

where theta is the parameter that runs from -thmax to +thmax. The helix
winds around the x-axis from x = -a*thmax to x = +a*thmax with radius R and
pitch 2*Pi*a. At x = 0, the helix passes through the positive y axis at y =
R.

Letting lambda be the mass per unit length, the total mass is found to be

M = 2*lambda*Sqrt(a^2+R^2)*thmax

For a general value of thmax, the center of mass of the helix is located at

(Xcm, Ycm, Zcm) = (0, R*sin(thmax)/thmax, 0).

For integral number of windings (thmax = 0, Pi, 2Pi, 3Pi, etc.) the center
of mass is located at the origin of the coordinate system. Otherwise, it is
located somewhere on the positive or negative y-axis.

In order for no wobbling to occur as the helix rotates about the x-axis, it
is necessary that the x-axis be one of the principle axes of the moment of
inertia tensor. This will be the case if all the off-diagonal elements
I_xy, I_yx, and I_xz of the tensor are zero. Recall that these are defined
as

I_xy = - integral [x*y*dm], etc.

where dm is an element of mass.

For the helix with an arbitrary value of thmax, I find

I_xy = I_yz = 0 (odd integrands)

I_xz = 2*lambda*a*R*Sqrt(a^2+R^2)*( sin(thmax) - thmax*cos(thmax) )

= M*a*R*( sin(thmax)/thmax - cos(thmax) )

Thus, we can force I_xz to vanish by choosing thmax to be a positive root of

tan(thmax) = thmax

Call these roots: th1, th2, th3, th4, etc. (in order of increasing
magnitude). These roots must be found numerically. For these values of
thmax the helix has a non-integral number of windings.

FWIW I get 4.49341, 7.72525, 10.90412 for the first three.
There are fortunately no solutions less than pi, which agrees
with the "obvious" requirement that there should be more than
one complete turn in the helix.


For these special values of thmax, all off-diagonal elements of the moment
of inertia tensor vanish. We need one further requirement in order for the
helix to be able to freely rotate about the x-axis without any wobble. We
need the center of mass of the helix to be located on the x-axis. From
above, we had Xcm=Zcm=0 and

Ycm = R*sin(thmax)/thmax

If thmax is a root of tan(thmax)=thmax, this may be written as

Ycm = R*cos(thmax).

It is easy to check that for th1, th3, th5, etc., the value of cos(thmax) is
negative and the center of mass is located on the negative y-axis. For th2,
th4, th6, etc. the value of cos(thmax) is positive and the center of mass is
located on the positive y-axis.

Thus, for th1, th3, th5, etc. we can shift the center of mass to the origin
by just adding a point mass m0 to the helix at (x, y, z) = (0, R, 0). For
example, if we choose thmax = th1, then m0 = M*cos(th1). For thmax = th3,
m0 = M*cos(th3), etc.

This looks and feels right to me. Note that as n increases,
thn gets closer and closer to (2n+1)pi/2, i.e. for more and
more turns the required mass gets closer and closer to zero.


Thus, for thmax = th1, th3, th5, etc., we can arrange for the center of mass
to be at the origin. It is easy to check that adding the point mass does
not mess up the vanishing of the off diagonal elements of the inertia
tensor. Thus, the system should be able to rotate freely about the x-axis
without wobbling.

If the pitch of the helix is chosen large enough, then there will exist a
'moving' inertial frame in which the rotating helix appears to be completely
unwound into a straight line parallel to the x-axis (with the point mass m0
attached to the midpoint). The system orbits around the x-axis as it also
moves parallel to the x-axis. And it does this without any externally
applied forces! Strange.

But we really need to do some calculations to see whether
this is physically possible from the standpoint of the
speed of sound in the material. I have speculated in my
earlier post that it is in fact not possible, but the
calculation does not appear easy even to set up. If I were
Feynman, perhaps I'd have worked it all out already, but as
anyone can see, I'm not. So for now, I can only talk about
what I think here, not what I know.

(OTOH the quickness and certainty with which Tom Roberts
answered this, suggests that it is given somewhere in the
literature. Unfortunately Roberts has not responded to my
followup questions, moreover there are some points in his
answer that don't seem quite right to me. Which could of
course be my problem, not his.)

Note that the corkscrew must be almost straight to begin
with (i.e. even 2piR is a pretty long pitch, and anything
close to that would require a relativistic spin) yet we
require it to provide, through its own torsional rigidity,
the centripetal force to accelerate a relativistic mass
through a small-radius arc. I think this requirement is
too great.


Either something's wrong with this analysis or else such a strange motion is
actually allowed. Linear momentum and angular momentum do not *appear* to
be conserved for the system as viewed in the moving frame. However, it
could be that they are in fact conserved if the momentum contributions of
the internal stresses in the helix are taken into account.

What I find a bit perplexing is that even if it unwinds
as little as would be required to put thmax at the next-
lower multiple of pi (i.e. not anywhere close to straight)
it seems you are still guaranteed to get a wobble. So, by
making the corkscrew really long, with a lot of turns, it
seems you can make it wobble even if the angular velocity
is quite small. OTOH the longer you make it, the more
demands you put on its rigidity to maintain the helical
shape, so I guess that doesn't necessarily lead to a
contradiction.

Assuming it unwinds only a very little bit, within what I am
calling the allowed range, does it make sense that the two
ends of the helix will gain in their contribution to the
moment of inertia enough to keep the rotation wobble-free?
This should be a much easier calculation but I'm not sure
how to do even that.


Todd

wrote in message
ups.com...
Todd wrote:
I think it's possible to arrange for the helix to rotate freely about its
axis without any wobbling by choosing just the right amount of winding
and
by also adding a point mass of just the right magnitude to the helix
halfway
along the helix (bear with me).

Like dynamically balancing a tire with one well-positioned
weight... seems plausible.

yes

This is based on some calculations that I'll
summarize here. I would be grateful to anyone who bothers to check my
results. All the integrations are elementary.

I parameterized the helix as

(x, y, z) = (a*theta, R*cos(theta), R*sin(theta))

where theta is the parameter that runs from -thmax to +thmax. The helix
winds around the x-axis from x = -a*thmax to x = +a*thmax with radius R
and
pitch 2*Pi*a. At x = 0, the helix passes through the positive y axis at
y =
R.

Letting lambda be the mass per unit length, the total mass is found to be

M = 2*lambda*Sqrt(a^2+R^2)*thmax

For a general value of thmax, the center of mass of the helix is located
at

(Xcm, Ycm, Zcm) = (0, R*sin(thmax)/thmax, 0).

For integral number of windings (thmax = 0, Pi, 2Pi, 3Pi, etc.) the
center
of mass is located at the origin of the coordinate system. Otherwise, it
is
located somewhere on the positive or negative y-axis.

In order for no wobbling to occur as the helix rotates about the x-axis,
it
is necessary that the x-axis be one of the principle axes of the moment
of
inertia tensor. This will be the case if all the off-diagonal elements
I_xy, I_yx, and I_xz of the tensor are zero. Recall that these are
defined
as

I_xy = - integral [x*y*dm], etc.

where dm is an element of mass.

For the helix with an arbitrary value of thmax, I find

I_xy = I_yz = 0 (odd integrands)

I_xz = 2*lambda*a*R*Sqrt(a^2+R^2)*( sin(thmax) - thmax*cos(thmax) )

= M*a*R*( sin(thmax)/thmax - cos(thmax) )


I left out the an overall negative sign, but that doesn't change anything.

Thus, we can force I_xz to vanish by choosing thmax to be a positive root
of

tan(thmax) = thmax

Call these roots: th1, th2, th3, th4, etc. (in order of increasing
magnitude). These roots must be found numerically. For these values of
thmax the helix has a non-integral number of windings.

FWIW I get 4.49341, 7.72525, 10.90412 for the first three.
There are fortunately no solutions less than pi, which agrees
with the "obvious" requirement that there should be more than
one complete turn in the helix.


Good, I get the same numerical values.


For these special values of thmax, all off-diagonal elements of the
moment
of inertia tensor vanish. We need one further requirement in order for
the
helix to be able to freely rotate about the x-axis without any wobble.
We
need the center of mass of the helix to be located on the x-axis. From
above, we had Xcm=Zcm=0 and

Ycm = R*sin(thmax)/thmax

If thmax is a root of tan(thmax)=thmax, this may be written as

Ycm = R*cos(thmax).

It is easy to check that for th1, th3, th5, etc., the value of cos(thmax)
is
negative and the center of mass is located on the negative y-axis. For
th2,
th4, th6, etc. the value of cos(thmax) is positive and the center of mass
is
located on the positive y-axis.

Thus, for th1, th3, th5, etc. we can shift the center of mass to the
origin
by just adding a point mass m0 to the helix at (x, y, z) = (0, R, 0).
For
example, if we choose thmax = th1, then m0 = M*cos(th1). For thmax =
th3,
m0 = M*cos(th3), etc.

This looks and feels right to me. Note that as n increases,
thn gets closer and closer to (2n+1)pi/2, i.e. for more and
more turns the required mass gets closer and closer to zero.


Yes


Thus, for thmax = th1, th3, th5, etc., we can arrange for the center of
mass
to be at the origin. It is easy to check that adding the point mass does
not mess up the vanishing of the off diagonal elements of the inertia
tensor. Thus, the system should be able to rotate freely about the
x-axis
without wobbling.

If the pitch of the helix is chosen large enough, then there will exist a
'moving' inertial frame in which the rotating helix appears to be
completely
unwound into a straight line parallel to the x-axis (with the point mass
m0
attached to the midpoint). The system orbits around the x-axis as it
also
moves parallel to the x-axis. And it does this without any externally
applied forces! Strange.

But we really need to do some calculations to see whether
this is physically possible from the standpoint of the
speed of sound in the material. I have speculated in my
earlier post that it is in fact not possible, but the
calculation does not appear easy even to set up. If I were
Feynman, perhaps I'd have worked it all out already, but as
anyone can see, I'm not. So for now, I can only talk about
what I think here, not what I know.

(OTOH the quickness and certainty with which Tom Roberts
answered this, suggests that it is given somewhere in the
literature. Unfortunately Roberts has not responded to my
followup questions, moreover there are some points in his
answer that don't seem quite right to me. Which could of
course be my problem, not his.)

Note that the corkscrew must be almost straight to begin
with (i.e. even 2piR is a pretty long pitch, and anything
close to that would require a relativistic spin) yet we
require it to provide, through its own torsional rigidity,
the centripetal force to accelerate a relativistic mass
through a small-radius arc. I think this requirement is
too great.


Well, as a 'thought experiment', I don't see why we can't consider a helix
of *very* large radius. Then the helix could rotate so that points of the
helix move at relativistic speed and yet have small centripetal
acceleration. So, the rigidity would not have to be great.


Either something's wrong with this analysis or else such a strange motion
is
actually allowed. Linear momentum and angular momentum do not *appear*
to
be conserved for the system as viewed in the moving frame. However, it
could be that they are in fact conserved if the momentum contributions of
the internal stresses in the helix are taken into account.

What I find a bit perplexing is that even if it unwinds
as little as would be required to put thmax at the next-
lower multiple of pi (i.e. not anywhere close to straight)
it seems you are still guaranteed to get a wobble. So, by
making the corkscrew really long, with a lot of turns, it
seems you can make it wobble even if the angular velocity
is quite small. OTOH the longer you make it, the more
demands you put on its rigidity to maintain the helical
shape, so I guess that doesn't necessarily lead to a
contradiction.

Assuming it unwinds only a very little bit, within what I am
calling the allowed range, does it make sense that the two
ends of the helix will gain in their contribution to the
moment of inertia enough to keep the rotation wobble-free?
This should be a much easier calculation but I'm not sure
how to do even that.


Good point. There is still an apparent paradox even if we just look at the
rotating helix in a 'moving frame' in which the helix is only slightly
unwound. And for this, we don't need to assume that the helix is rotating
at relativistic speeds.

For the slightly unwound helix, thmax would change such that the
condition tan(thmax) = thmax would no longer hold. So, it would *appear*
that the rotating helix would have to wobble in this reference frame. But
on the other hand, it is clear that if the rotating helix doesn't wobble in
the original 'rest frame' then it can't wobble when transforming to the
'moving frame' in which the helix is slightly unwound. So, there appears to
be a paradox. The thought experiment in which we arrange for the helix to
be completely unwound is just a more dramatic example of the same paradox.
I still think the solution probably lies in taking into account the momentum
contributions from the stresses in the helix.

Todd

Todd wrote:
wrote in message
ups.com...
Todd wrote:

[snip]

Note that the corkscrew must be almost straight to begin
with (i.e. even 2piR is a pretty long pitch, and anything
close to that would require a relativistic spin) yet we
require it to provide, through its own torsional rigidity,
the centripetal force to accelerate a relativistic mass
through a small-radius arc. I think this requirement is
too great.

I guess I should not have referred to the radius as
necessarily small.



Well, as a 'thought experiment', I don't see why we can't consider a helix
of *very* large radius. Then the helix could rotate so that points of the
helix move at relativistic speed and yet have small centripetal
acceleration. So, the rigidity would not have to be great.

Yes it would, because we would have miles and miles more
material in each turn. For a given thickness of material,
and constant angular velocity, the rigidity must increase
as R increases. It would of course be different if we were
dealing here with a closed hoop, but we're not.

[snip]

Assuming it unwinds only a very little bit, within what I am
calling the allowed range, does it make sense that the two
ends of the helix will gain in their contribution to the
moment of inertia enough to keep the rotation wobble-free?
This should be a much easier calculation but I'm not sure
how to do even that.


Good point. There is still an apparent paradox even if we just look at the
rotating helix in a 'moving frame' in which the helix is only slightly
unwound. And for this, we don't need to assume that the helix is rotating
at relativistic speeds.

For the slightly unwound helix, thmax would change such that the
condition tan(thmax) = thmax would no longer hold. So, it would *appear*
that the rotating helix would have to wobble in this reference frame. But
on the other hand, it is clear that if the rotating helix doesn't wobble in
the original 'rest frame' then it can't wobble when transforming to the
'moving frame' in which the helix is slightly unwound. So, there appears to
be a paradox.

Exactly my thinking. So I think it must be the case that
the two ends of the helix have a greater "weight" than the
middle, or perhaps there is even some more complicated
distribution of stress in the material to cause the
wobble to disappear when relativistic mass is accounted
for.

The thought experiment in which we arrange for the helix to
be completely unwound is just a more dramatic example of the same paradox.
I still think the solution probably lies in taking into account the momentum
contributions from the stresses in the helix.

I was trying to grope toward a quantitative statement of
the latter, but I don't have the expertise yet to see it.
I think it would make better sense for me to hone my skills
on some easier problems first.

wrote in message
oups.com...
Todd wrote:
wrote in message
ups.com...
Todd wrote:

[snip]

Note that the corkscrew must be almost straight to begin
with (i.e. even 2piR is a pretty long pitch, and anything
close to that would require a relativistic spin) yet we
require it to provide, through its own torsional rigidity,
the centripetal force to accelerate a relativistic mass
through a small-radius arc. I think this requirement is
too great.

I guess I should not have referred to the radius as
necessarily small.



Well, as a 'thought experiment', I don't see why we can't consider a
helix
of *very* large radius. Then the helix could rotate so that points of
the
helix move at relativistic speed and yet have small centripetal
acceleration. So, the rigidity would not have to be great.

Yes it would, because we would have miles and miles more
material in each turn. For a given thickness of material,
and constant angular velocity, the rigidity must increase
as R increases. It would of course be different if we were
dealing here with a closed hoop, but we're not.

[snip]

Yes, I goofed in thinking that increasing R would reduce the stress.
Instead I believe that the stress would be independent of R just as for a
closed hoop.

For a closed hoop spinning such that the material has a speed u, the tension
in the hoop will equal lambda*u^2 which is independent of R (at least for
nonrelativistic speeds).

For the helix I suspect that the tension will still be lambda*u^2 except for
an extra numerical factor that would depend on the ratio of the pitch to the
radius. As you scale up the helix with larger radius, you would also scale
up the pitch so that the ratio of pitch to radius would stay constant. (For
example, we might choose the pitch to equal 2*Pi*R) So, I think the tension
would be independent of R for a given rotational speed u and a given mass
per unit length lambda. I also think that the tension stress would be
essentially independent of the length (number of wraps) of the helix.

Anyway, I was wrong in believing that increasing R would help.
Nevertheless, I don't believe that the resolution of the paradox lies in
establishing that there are no real materials that are rigid enough to
withstand the rotations imagined in the thought experiment. For some
reason, that doesn't seem to me to get to the heart of the paradox. I could
be wrong of course. As in David's original post, we can consider a thought
experiment in which the pitch is astronomically long and then we can get by
with a slow rotation speed. I think the stress would then be small and we
wouldn't have to worry about the strength of the material.

Todd