I'm trying to resolve the following question: does the angular momentum
of a rapidly rotating object diverge as R\omega approaches c? It
should, right, since the linear momentum does.

The formula L=mrv\gamma seems to be the right relativistic expression
for the angular momentum of a point particle. I've done an integration
on that formula to get the angular momentum of a rigid disk, and that
expression does *not* seem to diverge. However, that formula is
obviously naive, I expect the binding energy of the disk to make a
contribution. How much the contribution is depends on what rigid
means, and I'm not too sure how to approach this question. I guess
Born has made some method for describing rigidity in relativity, but I
don't know how to use it.

I've found some allusions in my textbook to the fact that Lense and
Thirring derived the metric of a rapidly spinning rigid sphere of
uniform density, and that maybe this metric is the Kerr metric. The
Kerr metric contains a parameter denoted a, which is, according to
D'Inverno's textbook, related to the angular velocity, but it doesn't
say how. Then the angular momentum is ma. I can't find Lense and
Thirring's paper on the internet, there seems to be a 1984 translation
by Mashhoon, Hehl, and Theiss called "On the gravitational effects of
rotating masses: The Thirring-Lense Papers", I guess my uni doesn't
have a site license for Springerlink, cause I can't get it either. So
how is the rotation parameter a in the Kerr metric related to angular
velocity? Does the Kerr metric also describe rigid rotating spheres
(for points outside the sphere)?

Maybe a simpler approach would be to take two point masses on the end
of a spring revolving around its center. Can I write the stress tensor
for the energy stored in the spring? With that tensor, I could
calculate the total angular momentum, including the binding energy as
well as the rotation of the masses. But how do I write the stress
tensor for the spring?

After looking around on the internet, I found, through google books,
this collection of papers called "Gravity, Particles, and Space-Time"
edited by P. Pronin, G. Sardanashvily, which contains a paper called
"Rotating Boson Stars", by E. Mielke and F. Schunk. It says "Because
of the finite velocity of light and the infinite range of the scalar
matter within the boson star, our localized configuration can only
rotate..." at which point google books has cut it off. I feel like it
is about to tell me that the rotation rate must be bounded, which I
would take to be an equivalent assertion. Is there such a result for
rotating stars?

So can you help me resolve any or all of these four issues?

wrote in message
ups.com...
I'm trying to resolve the following question: does the angular momentum
of a rapidly rotating object diverge as R\omega approaches c? It
should, right, since the linear momentum does.

It will depend on the nature of the rotating object. Keep
in mind that a rotating object, where the tangent speed at
the farthest point from the axis of rotation has speed c,
nothing there actually travels at c and so the integral
calculating the total angular momentum may be finite.
A similar thing holds for the total rotational kinetic
energy of a relativistic rod. If you have a rod that is
anchored at one end and the mathematical point at its
other ends travels at some speed, v, the total rotational
kinetic energy is:

(mc^3/v)*Arcsin(v/c) - mc^2

When v = c this equation reduces to (mc^2)(Arcsin(1) - 1)
which is finite. In the limiting case, as v c, this reduces
to the classical expression mv^2/6. The rotational kinetic
energy of a rotating disk turns out to be (m rest mass):

(2mc^4/v^2)[1 - sqrt(1 - v^2/c^2)] - mc^2

and when v = c this reduces to just mc^2. And, when
v c, this reduces to the classical expression of
mv^2/4 where v is the tangential speed of the disk's
perimeter and m is the rest mass of the disk. Now, due
to length contraction the "disk" looks more like to two-
petal flower since the disk must break at least in two
places. All of these calculations assume an even rest
density in the disk or stick and, apart from black holes
and neutron stars, no material objects could rotate at
anything like a substantial fraction of the speed of light
without flying apart.

The formula L=mrv\gamma seems to be the right relativistic expression
for the angular momentum of a point particle. I've done an integration
on that formula to get the angular momentum of a rigid disk, and that
expression does *not* seem to diverge. However, that formula is
obviously naive, I expect the binding energy of the disk to make a
contribution. How much the contribution is depends on what rigid
means, and I'm not too sure how to approach this question. I guess
Born has made some method for describing rigidity in relativity, but I
don't know how to use it.

I've found some allusions in my textbook to the fact that Lense and
Thirring derived the metric of a rapidly spinning rigid sphere of
uniform density, and that maybe this metric is the Kerr metric. The
Kerr metric contains a parameter denoted a, which is, according to
D'Inverno's textbook, related to the angular velocity, but it doesn't
say how. Then the angular momentum is ma. I can't find Lense and
Thirring's paper on the internet, there seems to be a 1984 translation
by Mashhoon, Hehl, and Theiss called "On the gravitational effects of
rotating masses: The Thirring-Lense Papers", I guess my uni doesn't
have a site license for Springerlink, cause I can't get it either. So
how is the rotation parameter a in the Kerr metric related to angular
velocity? Does the Kerr metric also describe rigid rotating spheres
(for points outside the sphere)?

This question is answered on page 914 of the book
"Gravitation" by Charles W. Misner, Kip S. Thorne,
and John Archibald Wheeler. It gives angular speeds,
relativie to distant observers, in terms of Kerr metric
variables.

Maybe a simpler approach would be to take two point masses on the end
of a spring revolving around its center. Can I write the stress tensor
for the energy stored in the spring? With that tensor, I could
calculate the total angular momentum, including the binding energy as
well as the rotation of the masses. But how do I write the stress
tensor for the spring?

This is a fairly complex problem and if you are really
interested in solving it, check out the aforementioned book.
They do a really good job covering angular momentum densities.
Go to the index and look under `Angular momentum'.

After looking around on the internet, I found, through google books,
this collection of papers called "Gravity, Particles, and Space-Time"
edited by P. Pronin, G. Sardanashvily, which contains a paper called
"Rotating Boson Stars", by E. Mielke and F. Schunk. It says "Because
of the finite velocity of light and the infinite range of the scalar
matter within the boson star, our localized configuration can only
rotate..." at which point google books has cut it off. I feel like it
is about to tell me that the rotation rate must be bounded, which I
would take to be an equivalent assertion. Is there such a result for
rotating stars?

It depends on what you mean. Angular momentum, like energy and
linear momentum, is conserved in general relativity so there are
bounds in different situations. Frame dragging, for example, allows
very high rotation rates. As a rotating star shrinks, gravitational
time dilation tends to slow down its rotation rate, relative to distant
observers, whereas its shrinking and frame dragging tend to speed
it up. When all three effects are accounted for, you get the equation
on page 914 of "Gravitation".


R

wrote:
The formula L=mrv\gamma seems to be the right relativistic expression
for the angular momentum of a point particle.

No. In SR angular momentum must be represented as a 2-form, not as a
"pseudo 3-vector" (as one could in Newtonian mechanics).


Tom Roberts

RLG wrote:
wrote in message
ups.com...
I'm trying to resolve the following question: does the angular momentum
of a rapidly rotating object diverge as R\omega approaches c? It
should, right, since the linear momentum does.

It will depend on the nature of the rotating object. Keep
in mind that a rotating object, where the tangent speed at
the farthest point from the axis of rotation has speed c,
nothing there actually travels at c and so the integral
calculating the total angular momentum may be finite.

Indeed, I did such a calculation, and did find a finite angular
momentum.

The rotational kinetic
energy of a rotating disk turns out to be (m rest mass):

(2mc^4/v^2)[1 - sqrt(1 - v^2/c^2)] - mc^2

and when v = c this reduces to just mc^2.

But doesn't seem weird that a body made up of particles with infinite
linear momentum can add up to have a finite angular momentum? It's
certainly true that as the angular velocity increases, the internal
forces of the body have to grow. I was thinking that maybe the
increased internal energy would make the angular momentum go to
infinity. Does that expression you mention include internal forces?
Mine didn't.


So
how is the rotation parameter a in the Kerr metric related to angular
velocity? Does the Kerr metric also describe rigid rotating spheres
(for points outside the sphere)?

This question is answered on page 914 of the book
"Gravitation" by Charles W. Misner, Kip S. Thorne,
and John Archibald Wheeler. It gives angular speeds,
relativie to distant observers, in terms of Kerr metric
variables.


Hmm... I had gone through the index of MTW looking for anything they
said about angular velocity. But I somehow missed this page. Thank
you for pointing it out! That is helpful.

It says that the angular velocity of the Kerr black hole is
a/(r^2+a^2), where r =m + sqrt(m^2-a^2) is the horizon radius (assuming
no charge). These imply that the angular momentum becomes imaginary
when the radial velocity at the horizon radius becomes greater than one
half the speed of light. This is odd; I was expecting possibly a
divergence, not an imaginary number. I don't know what to make of
that.




Maybe a simpler approach would be to take two point masses on the end
of a spring revolving around its center. Can I write the stress tensor
for the energy stored in the spring? With that tensor, I could
calculate the total angular momentum, including the binding energy as
well as the rotation of the masses. But how do I write the stress
tensor for the spring?

This is a fairly complex problem and if you are really
interested in solving it, check out the aforementioned book.
They do a really good job covering angular momentum densities.
Go to the index and look under `Angular momentum'.

I'm OK with the angular momentum density tensor, and calculating total
angular momentum from it. That's not the complexity you're referring
to, is it? What I don't know is how to turn a spring into one of those
bad boys. Should I assume the spring has a finite volume, calculate
the energy and momentum density by dividing 1/2kx^2 by this volume?
MTW seems to have nothing helpful for this, at least not in their
chapter on angular momentum.

Tom Roberts wrote:
wrote:
The formula L=mrv\gamma seems to be the right relativistic expression
for the angular momentum of a point particle.

No. In SR angular momentum must be represented as a 2-form, not as a
"pseudo 3-vector" (as one could in Newtonian mechanics).


Tom Roberts


I find this wrong statement rather bizarre. Notation is arbitrary.
Saying that in SR angular momentum *must* be represented one way is
like saying that Newton's laws must be written in the Cyrillic
alphabet.

But if it makes you feel better, then you can pretend that the equation
above is for magnitude of the spatial Hodge dual of the spatial
pullback of the angular momentum two-form.

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

The rotational kinetic
energy of a rotating disk turns out to be (m rest mass):

(2mc^4/v^2)[1 - sqrt(1 - v^2/c^2)] - mc^2

and when v = c this reduces to just mc^2.

But doesn't seem weird that a body made up of particles with infinite
linear momentum can add up to have a finite angular momentum? It's
certainly true that as the angular velocity increases, the internal
forces of the body have to grow. I was thinking that maybe the
increased internal energy would make the angular momentum go to
infinity. Does that expression you mention include internal forces?
Mine didn't.

Ask yourself this question: do any of the particles in the disk
have an infinite linear momentum? If the very edge of the disk
has speed c, every material point in the disk has a speed less
than c. Think of the range of speeds that every point in the
disk has as the open set [0,c) or all speeds v such that
0 = v c. Although there are material points in the disk
that have every speed up to c, no material point in the disk
has speed c. The total integral is finite for the same reason
that the integral of (1/x^2) from 1 to infinity is the finite
number 1. Although there are rectangles of arbitrary
length, under the curve (1/x^2), their height becomes so
small that the total area integral is finite. An analagous thing
occurs in the rotational kinetic energy described above.
As far as internal forces go, those could vary depending
on the materials in the disk.

This question is answered on page 914 of the book
"Gravitation" by Charles W. Misner, Kip S. Thorne,
and John Archibald Wheeler. It gives angular speeds,
relativie to distant observers, in terms of Kerr metric
variables.

It says that the angular velocity of the Kerr black hole is
a/(r^2+a^2), where r =m + sqrt(m^2-a^2) is the horizon radius (assuming
no charge). These imply that the angular momentum becomes imaginary
when the radial velocity at the horizon radius becomes greater than one
half the speed of light. This is odd; I was expecting possibly a
divergence, not an imaginary number. I don't know what to make of
that.

Double check your calculations! You made a huge goof here.


Maybe a simpler approach would be to take two point masses on the end
of a spring revolving around its center. Can I write the stress
tensor
for the energy stored in the spring? With that tensor, I could
calculate the total angular momentum, including the binding energy as
well as the rotation of the masses. But how do I write the stress
tensor for the spring?

This is a fairly complex problem and if you are really
interested in solving it, check out the aforementioned book.
They do a really good job covering angular momentum densities.
Go to the index and look under `Angular momentum'.

I'm OK with the angular momentum density tensor, and calculating total
angular momentum from it. That's not the complexity you're referring
to, is it? What I don't know is how to turn a spring into one of those
bad boys. Should I assume the spring has a finite volume, calculate
the energy and momentum density by dividing 1/2kx^2 by this volume?
MTW seems to have nothing helpful for this, at least not in their
chapter on angular momentum.

It depends on the material properties of the spring. If you are trying to
do relativistic continuum mechanics, for example calculating the
relativistic equivalent of strain tensors and so on, I would start by first
reading chapter 8 of "Introduction to the Theory of Relativity" by
Peter Gabriel Bergmann.


R

RLG wrote:
wrote in message
oups.com...
RLG wrote:


Ask yourself this question: do any of the particles in the disk
have an infinite linear momentum? If the very edge of the disk
has speed c, every material point in the disk has a speed less
than c. Think of the range of speeds that every point in the
disk has as the open set [0,c) or all speeds v such that
0 = v c. Although there are material points in the disk
that have every speed up to c, no material point in the disk
has speed c. The total integral is finite for the same reason
that the integral of (1/x^2) from 1 to infinity is the finite
number 1. Although there are rectangles of arbitrary
length, under the curve (1/x^2), their height becomes so
small that the total area integral is finite. An analagous thing
occurs in the rotational kinetic energy described above.
As far as internal forces go, those could vary depending
on the materials in the disk.


OK, I'm pleased with this explanation of why it shouldn't be
counterintuitive that the angular momentum doesn't diverge. Thank you.
(Though I'd still like an explicit calculation including internal
energy).



It says that the angular velocity of the Kerr black hole is
a/(r^2+a^2), where r =m + sqrt(m^2-a^2) is the horizon radius (assuming
no charge). These imply that the angular momentum becomes imaginary
when the radial velocity at the horizon radius becomes greater than one
half the speed of light. This is odd; I was expecting possibly a
divergence, not an imaginary number. I don't know what to make of
that.

Double check your calculations! You made a huge goof here.

You're right about that! OK, now I think I get L=2m^2\beta for the
angular momentum, which is finite for all velocities. Insofar as we
can consider the horizon to be the edge of a black hole, that it can go
as fast as it wants. That's not my favorite solution, since it's
probably not very realistic to assume the solid body extends that far
(or indeed any finite radius after the singularity), but it's an
answer.



I'm OK with the angular momentum density tensor, and calculating total
angular momentum from it. That's not the complexity you're referring
to, is it? What I don't know is how to turn a spring into one of those
bad boys. Should I assume the spring has a finite volume, calculate
the energy and momentum density by dividing 1/2kx^2 by this volume?
MTW seems to have nothing helpful for this, at least not in their
chapter on angular momentum.

It depends on the material properties of the spring. If you are trying to
do relativistic continuum mechanics, for example calculating the
relativistic equivalent of strain tensors and so on, I would start by first
reading chapter 8 of "Introduction to the Theory of Relativity" by
Peter Gabriel Bergmann.

Thanks for the reference. I'll check it out.

-zig