# Stress-energy tensor of a wire under stress

• pervect

#### pervect

Staff Emeritus
 switch to consistent geometric units and get rid of factors of 'c'.

I've been thinking about the relativistic hoop/disk again, and as a preliminary step I decided to look at what happens to the stress-energy tensor of a wire when we apply a load to it.

Suppose we have a wire, of mass m, length L, and cross sectional area A.

Then if the wire is not under stress, T^00 will (edit: in geometric units) just be m/(LA). Other components of the stress-energy tensor will be zero.

Suppose we apply a tension, T, gradually, and that the wire is within its elastic limit.

Then the wire will elongate from L to (L+d). This will require some amount of work W. To find the exact amount of work required would mean knowing the relationship between stress and strain, but if we use Hooke's law, it will be just

W = .5 K d^2

where K is the spring constant.

The wire may change its area when put under load, depending on it's Poisson's ratio http://en.wikipedia.org/wiki/Poisson's_ratio

Call the new area AA

What I'm interested in is the value for T^00 of the wire under load. This should be, by the conservation of energy, edit (in geometric units)

$$\frac{m + W}{(L+d)(AA)}$$

If we take the ratio of T^00 under load to the initial value, we get

T^00 (loaded) / T^00(initial) = $$\frac{1+W}{(1+d/L)(AA/A)}$$

The other component of the stress energy tensor will be the strain T in the wire.
T^11 = -T

I think this analysis should be OK even if the stresses exceed the elastic limit, as long as any temperature rise the wire may experience due to the non-reversible stretching (which will increase entropy) isn't allowed to radiate away and is uniformly distributed.

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I've been thinking about the relativistic hoop/disk again, and as a preliminary step I decided to look at what happens to the stress-energy tensor of a wire when we apply a load to it.

Suppose we have a wire, of mass m, length L, and cross sectional area A.

Then if the wire is not under stress, T^00 will just be m/(LA). Other components of the stress-energy tensor will be zero.
Note: tension = - stress
Suppose we apply a tension, T, gradually, and that the wire is within its elastic limit.

Then the wire will elongate from L to (L+d). This will require some amount of work W. To find the exact amount of work required would mean knowing the relationship between stress and strain, but if we use Hooke's law, it will be just

W = .5 K d^2

where K is the spring constant.

The wire may change its area when put under load, depending on it's Poisson's ratio http://en.wikipedia.org/wiki/Poisson's_ratio

Call the new area AA

What I'm interested in is the value for T^00 of the wire under load. This should be, by the conservation of energy

$$\frac{m + W/c^2}{(L+d)(AA)}$$

If we take the ratio of T^00 under load to the initial value, we get

T^00 (loaded) / T^00(initial) = $$\frac{1+W/mc^2}{(1+d/L)(AA/A)}$$

The other component of the stress energy tensor will be the strain T in the wire.

I think this analysis should be OK even if the stresses exceed the elastic limit, as long as any temperature rise the wire may experience due to the non-reversible stretching (which will increase entropy) isn't allowed to radiate away and is uniformly distributed.
Note: If the wire is under tension then the component of T which represents the tension (e.g. T^zz) will be less than zero.

Pete

Note: tension = - stress
Note: If the wire is under tension then the component of T which represents the tension (e.g. T^zz) will be less than zero.

Pete

Yes, of course. I've added a small note to that effect - I've also switched to geometric units.

Assuming this is OK (and I don't see much to argue with), it ought to be possible to use this to find the stress energy tensor and total energy of a relativistically rotating wire (for SR, i.e. with Minkowskian metric coefficients).

Yes, of course. I've added a small note to that effect ..
My appologies for missing that.

Assuming this is OK (and I don't see much to argue with), it ought to be possible to use this to find the stress energy tensor and total energy of a relativistically rotating wire (for SR, i.e. with Minkowskian metric coefficients).
I don't see why you needed to do that for stress/tension since the stress/tension that counts is the component parallell to the direction of motion. If there is a component of stress/tension perpendicular to the direction of motion then the stress/tension does not affect the energy density.

Pete

While I am still working on the transformations, the point is that in a frame comoving with the wire, there is stress along the length of the wire. Interestingly enough, I get the result there is no stress in the laboratory frame in the sense of the word that GR uses (which is different than the engineering usage).

I believe one can work this out from the continuity equations.

$$T^{ab}{}_{;b}=0$$

Emphasize a good web reference

Hi, pervect,

Did you do a Newtonian computation of the stress tensor of a unloaded and loaded wire first? In a static frame and also a (Galilei) "boosted" frame?

In passing to a relativistic analysis, be careful about assuming Hooke's law since this is not consistent with Lorentzian manifold structure!

A good web reference is Greg Egan's analysis at

http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/SimpleElasticity.html
http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/RindlerHorizon.html

Depending upon your ultimate intentions, I may have some suggestions about suitable frame fields for constructing gtr models and analyzing the physical experience of appropriate observers. If you seek exact solutions you will almost certainly need to carefully formulate your thought experiments in a stationary spacetime.

Everyone--- watch out, not everyone at PF and WP know as much as pervect does. Alas, there are a lot of mistaken/wrong eprints in the arXiv from non-relativists on "paradoxes" (often "rediscovering" mistakes which were cleared up long ago--- not everyone who writes arXiv eprints, it seems, reads the literature, or understands what they read). There are also some very good papers, such as the Ph.D. thesis cited by Egan, but anyone who reads that should make sure not to confuse the strain transformations undeformed -> deformed with a change of coordinates, despite the formal resemblance!

The Egan reference was very helpful - I remember looking at a different one that was related, but this one was a lot better than the one I recall looking at.

I did basically a Newtonian calculation of the stress-energy tensor in a Minkowski frame. The point is that the stress-energy tensor in this Minkowski frame should represent the stress-energy tensor of a small section of the wire in the wire's frame field.

The continuity equations for the flat Minkowski metric should basically guarantee that the amount of work done in stretching the wire goes into the stress energy tensor. Doing the analysis in the Minkowski frame makes life easier.

This sidesteps some issues, like finding the correct radius for the rotating hoop, that will have to be resolved later.

As far as the frame field goes, the approach I'm taking very briefly goes like this:

Start with Minkowski cylindrical coordinates t,r,$\theta$, z.

Define some new coordiantes t,r,$\Theta$,z where $\Theta$ = $\theta + \omega t$

Compute the metric in the new coordinate system. This coordinate chart is no longer diagonal.

Find an ONB of one-forms from the metric (this may have been a mistake!)

Currently, I have the following rather messy set of one-forms

w1 = $$\sqrt{1 - \omega^2 r^2} dt + \frac{r^2 \omega}{\sqrt{1 - r^2 \omega^2}} d \Theta$$
w2 = dr
w3 = $$\frac{r}{\sqrt{1 - \omega^2 r^2}} d \Theta$$
w4 = dz

The metric is right, but I just noticed that e3 (the dual of w3) doesn't point in the $\Theta$ direction according to GRTensorII. Which is of course related to the metric tensor not being diagonal.

So I need to get this issue resolved, then I can go on to find the stress-energy tensor in the t,r,$\Theta$,z coordinates and convert it back to the t,r,$\theta$,z coordinates via the tensor transformation rules.

The t,r,$\theta$,z is a flat Minkowski space, it should be easy to find the total energy once I get this far.

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I did basically a Newtonian calculation of the stress-energy tensor in a Minkowski frame.

I insist upon a Newtonian computation in a Galilei frame!

Start with Minkowski cylindrical coordinates t,r,$\theta$, z.

Define some new coordiantes t,r,$\Theta$,t, where $\Theta$ = $\theta + \omega t$

Come again?

I am pretty sure you are trying to derive the Langevin frame for Born chart on Minkowski spacetime--- see http://en.wikipedia.org/w/index.php?title=Born_coordinates&oldid=53957524
A good grasp of this is essential for the type of chart and frame field I was going to suggest for working on rotating hyperelastic disks in the context of gtr (having recourse to weak-field theory, if only to compare with the Newtonian analysis)

I insist upon a Newtonian computation in a Galilei frame!

As long as you agree that work = force*distance when we slowly stretch a wire, and that this work gets added to the total stress energy tensor of the wire.

I am pretty sure you are trying to derive the Langevin frame for Born chart on Minkowski spacetime--- see http://en.wikipedia.org/w/index.php?title=Born_coordinates&oldid=53957524

This made my life a lot simpler, I didn't need to introduce that awkward second coordinate system (involving $\Theta$) with this approach.

The final result I'm getting is that for (t,r,$\theta$,z) with t being represented by the 0 subscript, r by the 1 subscript, etc.

$$T^{00} = \rho(1+\omega^2*r^2)$$
$$T^{02} = T^{20} = \rho w$$

All other terms are zero

What's puzzling me is how this satisfies the Newtonian limit for a slowly rotating disk. Maybe I'm making a stupid mistake, but it seems like it's saying the kinetic energy in is mv^2, not .5 m v^2.

In case you want to see the Maple worksheet, here it is

> makeg(lang);
> 3;
> [t,r,theta,z];
> 2;
> [1/sqrt(1-omega^2*r^2),0,omega/sqrt(1-omega^2*r^2),0];
> [0,1,0,0];
> [omega*r/sqrt(1-omega^2*r^2),0,1/(r*sqrt(1-omega^2*r^2)),0];
> [0,0,0,1];
> 1;
> -1;
> 1;
> 1;
> 1;
> ;
> 1;
> 1
>
>
> grcalc(g(dn,dn));
> grdisplay(_);
> # define a stress energy tensor. In the frame field of the wire, there is a
> # term for the density of the wire, and a term for the stress in the wire
>
> grdef(T{^(a) ^(b)} := rho*kdelta{^a $t}*kdelta{^b$t} + P_th*kdelta{^a $theta}*kdelta{^b$theta});
>
> grcalc(T(bup,bup));
> grdisplay(_);
>
> grcalc(T(up,up));
> grdisplay(_);
>
> # find the tension by the continuity equation
> grdef(J{^a} := T{^a ^b ;b});
> grcalc(J(up));
>
> grdisplay(_);
> P_th := -omega^2*rho*r^2;
> grdisplay(_);
>
> gralter(T(up,up));
> 1;
> grdisplay(_);
>
>
>

For those not fortunate enough to have GRtensor

What T looks like in the "frame field"

\left[ \begin {array}{cccc} \rho&0&0&0\\\noalign{\medskip}0&0&0&0\\\noalign{\medskip}0&0&-{\omega}^{2}\rho\,{r}^{2}&0 \\\noalign{\medskip}0&0&0&0\end {array} \right]

The value for tension, $$-\rho \, \omega^2 \, r^2$$ can be calculated by the continuity equation.

T as it appears in cylindrical coordinates

\left[ \begin {array}{cccc} \left( {\omega}^{2}{r}^{2}+1 \right) \rho&0&\rho\,\omega&0\\\noalign{\medskip}0&0&0&0\\\noalign{\medskip} \rho\,\omega&0&0&0\\\noalign{\medskip}0&0&0&0\end {array} \right]

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Frame components or coordinate basis components?

The final result I'm getting is that for (t,r,$\theta$,z) with t being represented by the 0 subscript, r by the 1 subscript, etc.

$$T^{00} = \rho(1+\omega^2*r^2)$$
$$T^{02} = T^{20} = \rho w$$

All other terms are zero

For other readers: the components with respect to the coordinate basis in general have no physical meaning. The components wrt a frame, OTH, are the components the observer with said frame would actually measure.

You can get the frame components from the coordinate components computed via index gymnastics by using the expression for the frame vectors (or dual coframe covectors) in terms of the coordinate vectors (or coordinate covectors).

As for the interpretation: we know from Newtonian analysis that our wire should be under tension, i.e. there should be a nonzero diagonal component in the stress tensor computed in the frame. Since we aer using a frame comoving with the matter, the momentum components should vanish. And they do. The sole surviving component of the stress tensor is negative because this is a tension.

A tricky point in comparing with Newtonian analysis: sometimes a frame is actually "spinning". To find out, compute the Fermi derivatives of the spatial frame vector fields along the timelike frame vector field. If these vanish (after projection orthogonal to the timelike frame vector), the frame is nonspinning.

Since we are accelerating our wire (as I recall), we don't expect our frame to be inertial. We can compute the acceleration vector as the covariant derivative of the timelike frame vector along itself.

BTW, pervect, I am a bit confused about what you are doing here. Since you used the Langevin frame for the Born chart (right?) your wire is, I guess, a circular wire which we have set rotating about the axis of symmetry, so that we have a stationary axisymmetric scenario. You never wrote out your line element or your frame, but I guess you found a tension along the length of the wire (orthogonal to $\partial_r, \, \partial_z$. That would make sense in terms of centrifugal "force" assuming your third frame vector is something like $1/r \, \partial_\phi$

(Additional: sorry, pervect; I see now that you did say exactly this, so all is well.)

What's puzzling me is how this satisfies the Newtonian limit for a slowly rotating disk.

Well, first of all, if you expand to first order in the angular velocity$\omega = V/r$, then you are ignoring the tension. To make a comparision with Newton you will probably need to use a weak-field analysis but you'll need to carry out the computations to higher order in the velocity.

Also, to obtain a reasonable model incorporating elasticity you will need to use something more elaborate so that you can try to match the RHS of the EFE to a matter tensor suitable for "hyperelastic" matter.

Here is the short version of how you might begin a more careful analysis for matter which is rotating rigidly:

Start with a chart for a stationary axisymmetric spacetime; that is, a Lorentzian manifold with two commuting Killing vector fields, one timelike and the other spacelike. We don't want a static spacetime so the timelike Killing vector should not be hypersurface orthogonal, i.e. should have nonzero vorticity.

To be safe, you can take the Weyl canonical chart, although that is awkward to interpret geometrically. So you can play with making some simplifying assumptions.

$$ds^2 = -(dt-w\, d\phi)^2 + \exp(2v) (dz^2+dr^2) + r^2 \, d\phi^2$$
$$\hspace{0.5in} = -dt^2 + 2\, w \, dt \, d\phi + \exp(2v) (dz^2+r^2) + (r^2-w^2) \, \dphi^2,$$
$$-\infty < t, \, z < \infty, \; w < r < \infty, \; -\pi < \phi < \pi$$
where w,v are functions of z,r only. Here the timelike Killing vector field is $\partial_t$ which we naively assume has unit length, that is we assume that $\| \partial_t \|^2 = -1$. The spacelike Killing vector field is $\partial_\phi$ and it has length $\| \partial_\phi \|^2 = w^2-r^2$, where the function r is defined by $\partial_t \cdot \partial_\phi = r$. So the radial coordinate has a known geometric interpretation, as do the time and angular coordinates. Finally we choose z so that $\| \partial_z \| = \| \partial_r \| = \exp(-v)$. Turning this around, I have somewhat described how one could try to derive this chart starting from purely coordinate-free considerations, which of course guarantees that all the metric functions appearing in our line element have coordinate-free interpretations.

$$\sigma^0 = -(dt-w\,d\phi), \; \sigma^1 = \exp(v) \, dz, \; \sigma^2 = \exp(v) \, dr, \; \sigma^3 = r \, d\phi$$
Take the dual frame
$$\vec{e}_0 = \partial_t,\; \vec{e}_1 = \exp(-v) \, \partial_z, \; \vec{e}_2 = \exp(-v) \, \partial_r, \; \vec{e}_3 = \frac{1}{r} \; \left( \partial_\phi + w \, \partial_t \right)$$
and boost it by an undetermined amount (depending only on z,r) in the $\partial_\phi[/tex] direction. It is convenient to write $$1/\sqrt{1-V^2} = p(z,r), \; V/\sqrt{1-V^2} = \sqrt{p(z,r)^2-1}$$ Then the new frame field is $$\vec{f}_0 = \left( p + \frac{\sqrt{p^2-1}}{r} \, w \right) \, \partial_t + \frac{\sqrt{p^2-1}}{r} \, \partial_\phi,$$ $$\vec{f}_1 = \exp(-v) \, \partial_z, \; \vec{f}_2 = \exp(-v) \, \partial_r,$$ $$\vec{f}_3 = \left( \frac{p}{r} \, w + \frac{\sqrt{p^2-1}}{r} \right) \, \partial_t + \frac{p}{r} \, \partial_\phi$$ Now [itex]\vec{f}_0$ already has vanishing expansion scalar, so require that the shear tensor also vanish. In GRtensor speak:
Code:
casesplit([seq(seq(grarray(expv(bdn,bdn))[j,k], j=1..4),k=1..4)] );
where expv(bdn,bdn) is the expansion tensor of $\vec{f}_0$
Code:
grdef(expv({((a) (b))} := p{(a) ^(m)}*p{(b) ^(n)}*(v{(m) ;(n)}+v{(n) ;(m)})/2):
where p(dn,dn) is the projection orthogonal to the timelike unit vector field whose expansion tensor is to be computed, v(dn). We can cause GRTensor to ask this vector field to be input (as a linear combination of the frame vector fields) by
Code:
grdef(v{ ^(a) }):
Just make sure the coefficients in your linear combination, $c_0, \, c_1, \, c_2 \, c_3$ obey the unit contraint $-c_0^2+c_1^2+c_2^2+c_3^2=-1$. For example $\vec{f}_0 = p \, \vec{e}_0 + (p^2-1) \, \vec{e}_3$ above, and $-p^2+(p^2-1)=-1$.

(Additional: my post seems to have been accidently truncated at this point. I'll try to reconstruct it.)

The condition that the expansion tensor vanish (i.e. that the shear tensor vanish) turns out to give w in terms of p via quadrature.

Next, we want our frame field to be comoving with the matter, so require that the momentum components vanish. Two already do so, and the condition $G^{\hat{0} \hat{3}} = 0$ gives an equation in p only. We should check that these conditions are not mutually inconsistent, although this is pretty obvious in this case. The easiest way to do that with GRTensor is to casesplit the three equations and let maple do its differential ring magic.

The acceleration vector of our matter is orthogonal to $\vec{f}_3$ (and automatically orthogonal to $\vec{f}_0$ since $\vec{e}_0$ is a unit vector). The matter tensor has the form
$$T^{\hat{m} \hat{n}} = \left[ \begin{array}{cccc} e & 0 & 0 & 0 \\ 0 & a & f & 0 \\ 0 & f & b & 0 \\ 0 & 0 & 0 & c \end{array} \right]$$
where the hats are often used to signal that an expression refers to components with respect to an ONB or frame, rather than a coordinate basis.

At this point, you can start to try use some elasticity theory to further determine the precise form of the matter tensor; so far we only know that some components vanish as above.

Note that pervect's computation, $a=f=b=0$, which is not sufficiently general for a fluid or for most elastic solids.

The assumption above that $\| \partial_t \|^2 = -1$ simplifies the analysis, but is rather artificial and inconsistent with dust, for example. So if the above doesn't work out, go back and start over using a more general stationary axisymmetric line element (at worst, you can use the Weyl canonical chart).

(Oh darn, got to go. I said a lot more but I won't be able to reconstruct it after all.)

The idea here is to use elasticity to put constraints on the form of the matter tensor, perhaps eventually coming up with sufficiently simply equations to yield a solution with a reasonable interpretation (including a good understanding of what assumptions enter into the derivation). So for example for uniaxial tension in an homogeneous isotropic elastic body you'd expect the stress tensor to be diagonal with form $f \, \operatorname{diag}( k, 1,1)$ where f is some function and k is a constant (simply related to standard characterizations of isotropic homogeneous elastic materials such as the two Lame constants), and you'd expect to relate f to the energy density. Note one tricky point: many textbooks on elasticity theory provide tensor equations which are only valid in a Cartesian chart and thus are not "tensorial" in the sense students of gtr might expect. There is nothing wrong with this practice, of course, one just has to be aware of it and to make adjustments. This will be relevant if you consider uniaxial torsion (as in spinning up a disk by applying a torque) rather than unaxial tension (as in pulling on a rod).

By the way, one of the many advantages of using frame fields is that this is by far the conceptually and computationally easiest way to make connections with other theories.

Don't forget that the particular metric I chose above involved an assumption, $\| \partial_t \|^2 = -1$ which certainly doesn't hold for the Schwarzschild vacuum or Schwarzschild fluid, or indeed for very many stationary axisymmetric solutions! It was chosen for mathematical convenience. If you play around you'll soon appreciate that using fewer metric functions of fewer variables helps alot, but there is actually another reason why my choice is convenient: the integrability condition for the quadrature (which gives $w_z, \, w_r$ as a function of p and its partials) is precisely the equation in p alone which results from the demand that the momentum vanish, i.e. that our new frame be comoving with the matter. This is in fact very typical and similar statements will hold for good analyses starting from more general line elements, so when playing around be sure to look out for this! See the Maple help for casesplit for using this command to find the integrability condition for two quadrature equations of the type we found above.

Those of you who are interested in solitons will want to learn about the Lax pair formulation, which involves basically the same phenomenon, in which we obtain a second order PDE of interest as the integrability condition for two first order equations giving some function of two variables in terms of another by quadrature. This is in turn related to defining a certain connection, whose curvature--- well, see the review paper I've cited elsewhere!

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Apologies for butting in, I have nothing to add to the discussion but am enjoying following in and learning a lot.

I have one question for you though Pervect, I am curious to know to what end you are attempting this interesting calculation?

Thanks Chris, lots of reading to keep me busy!

Apologies for butting in, I have nothing to add to the discussion but am enjoying following in and learning a lot.

I have one question for you though Pervect, I am curious to know to what end you are attempting this interesting calculation?

What I'm trying to do is simple enough - it's to find the total energy (in the SR sense) of a rotating wire, ignoring gravity - i.e. assuming a Minkowskian metric.

I then want to compare it to the expected Newtonian result.

Chris Hillman in post #11talking about a rather more elaborate computation, which involves a gravitating system. My computation is ignoring the self-gravity (or gravitational binding energy) of the wire - it's a purely SR computation in a flat space-time.

This will hopefully fill in some holes in http://math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html

which I've never found very satisfying. (I'm working a simpler problem than that url.)

Unfortunately, I'm not quite happy with the results yet.

In the coordinate basis (the second result for the stress energy tensor) we can find the total energy (in the SR sense) easily enough by simply taking

$$\int T^{00} dV$$, where dV is the volume element, in this case
dV = $$r \, dr \, d\theta \, dz$$

Because we have a flat metric and this is SR, the energy adds, so we can just integrate the energy per unit volume to get the total energy.

Doing this we get
$$2 \pi r \rho \left( 1+r^2 \omega^2 \right) dr dz$$

We've not integrated over r, because dr is tiny, and r is essentially constant.

(Example: R dR $\approx$ .5 r^2 evaluated at r=R+dR - .5 r^2 evaluated at r=R)

If we imagine slow velocities, and a rigid wire, for the purposes of a sanity check by comparing with the Newtonian result, the wire won't stretch and there won't be any Lorentz contraction. We expect that r will be unchanged before and after we start it rotating. Since the wire doesn't stretch, so no work is done on the wire and its volume doesn't change either, thus $\rho$ also stays constant.

We can then identify $$2 \pi \, r dr \, dz$$ as the volume V of the wire, and $\rho \, V$ as the rest energy of the wire (also its mass, since I'm assuming geometric units where c=1).

So this expression boils down to E = M(1 + $\omega^2 r^2$ )

where M is the rest mass / rest energy.

Unfortunately, this seems to have twice the kinetic energy that it should have in this simple Newtonian limit, which is why I'm not happy yet, i.e

E = M + Mv^2 :-(.

The intent is to get a result that works in general - taking the Newtonian limit is just a check, a check that appears to have failed :-(.

Also of some interest is getting an expression for the angular momentum of the rotating wire vs $\omega$ including SR corrections.

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Suggestion: start with a purely Newtonian analysis, understood in terms of applying a body force to obtain a Galilei boost. As you know, to obtain "rigid" linear acceleration in str, you need to apply carefully and rather artificially chosen body forces, accelerating harder at trailing points than at leading points along the wire.

Indeed, I suggest backing off from your rotating circular wire for a moment and studying a linearly accelerated bit of straight wire. I just noticed that you appear to be using $\omega \, r$ rather than $\frac{\omega \,r}{\sqrt{1-\omega^2 \, r^2}}$, so maybe you are already following my advice to consider Galilei boosts first. (Try going back to the definition of the stress-energy tensor if this point isn't clear!)

When you return to the rotating wire (or disk), watch out for a tricky point in integrating in the case of a rotating circular wire: you should choose coordinates which are explicitly comoving, and part of the point of the "paradox" is that if you naively draw "boosted" axes on a cylinder, you run into a problem at branch cut, so to speak. You need to avoid "double counting" and you need to address a serious conceptual issue: the Langevin congruence of rigidly rotating observers is not hypersurface orthogonal , so there is no hope of finding well-defined "spaces at a time" for "spatial integrations".

I can now see that I was getting way ahead of your post in my previous reply, and I agree that being systematic is a very good idea! In fact, I think that by following my suggestion just above you should be able to resolve the difficulty you mentioned.

Eventually, you want to consider torsion. Again I'd suggest first purely Newtonian analysis, then careful construction of a simple but consistent str model. The goal should be relating the different components of the matter tensor as expressed wrt a suitable frame field in Minkowski spacetime. Then you can apply your results to seek an exact solution by symmetry Ansatz, modeling a rotating disk as an elastic material treated in gtr, which is what Michael Weiss was calling for in http://www.math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html

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While I'm not quite ready to tackle in detail your more advanced analysis yet, as I am having significant problems with a much simpler one, looking at it was interesting. There's a few simple things I want to revisit at some later time (techniques I want to find more out about eventually), but I digress.

If we apply the analysis to the accelerated wire, an equivalent analysis would be what happens if we accelerate a wire for a while, then stop. The acceleration can be done at either end of the wire, which will have some stress during the acceleration, details of the motion are worked out in for instance some of the Nikolic papers.

I can pretty much tell you what I expect:

$T^{00}$ is going to get multiplied by $$\frac{1}{1-v^2}$$

the wire is going to span a length L in it's comoving frame, but a contracted length $\sqrt{1-v^2} L$ in the inertial frame.

So the stress energy tensor increases by a factor of gamma^2, the volume decreases by a factor of gamma, the total energy of the wire increases by gamma, which is the correct result.

Going back to the hoop, note that I'm not doing the integration in the frame-field, for the reasons you mention - finding a coherent notion of what constitutes the hoop would be a mess.

The logical choice is to do the integration in the lab frame. Then we know exactly what set of points constitues the hoop at a given time:

r varies from R to R+dR
t=0
theta varies from 0 to 2pi
z varies from Z to Z+dZ

So the plan was to find the stress energy tensor in the lab frame, and integrate it. What could be more straightforward?

One thing to look at is to see if I have the right value for tension. If we leave the pressure in the Langevian frame-field as a variable named P_th (we expect P_th to be a negative number), the expression for T^00 in the lab frame becomes

$${\frac {\rho+{\omega}^{2}{r}^{2}{\it P\_th}}{1-{\omega}^{2}{r}^{2}}}$$

The value I computed for P_th by the continuity equation was $$-\rho \omega^2 r^2$$ - when simplified, this gave the earlier result.

Rambling a bit, we see:

The density is going up as expected because of the velocity

theta varies from 0 to 2*pi, there is no change in the boundary conditions as there is for the linear wire

If nothing else happened, the energy would go up by gamma^2, which is too much.

But something else does happen - the tension transforms to reduce the energy density in the lab frame.

Unfortunately, I don't see any error in the value for the tension, or what plausible value for it would make things balance out.

As far as acceleration goes, we see that while the wire is accelerating, the acceleration is doing no work. So we don't have some of the bookeeping issues that arise with the relativity of simultaneity and forces on a distributed body that do work.

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The acceleration can be done at either end of the wire, which will have some stress during the acceleration, details of the motion are worked out in for instance some of the Nikolic papers.

For other readers: I repeat that even for linear acceleration and even in str, one cannot rigidly accelerate a rod by pushing or pulling at one end; one has to apply carefully calibrated forces at each point, which is not physically realistic. Hence the interest in more elaborate models in which we push or pull a rod made of an elastic material.

So [typical components of] the stress energy tensor increase by a factor of gamma^2, the volume decreases by a factor of gamma, the total energy of the wire increases by gamma, which is the correct result.

Right, in relativistic units energy density, momentum flux, and pressure/stress all have the units of 1/L^2, i.e. units of curvature, so they scale as you said. Then $1/L^2 \cdot L = 1/L$.

Going back to the hoop, note that I'm not doing the integration in the frame-field, for the reasons you mention - finding a coherent notion of what constitutes the hoop would be a mess.

The logical choice is to do the integration in the lab frame. Then we know exactly what set of points constitues the hoop at a given time:

One of the elementary points which people who should know better (authors of bad arXiv eprints) often miss: because the Langevin congruence and other rotating congruences have nonzero vorticity, the world lines (integral curves of the congruence) are not hypersurface orthogonal, i.e. there exists no orthogonal hypersurfaces we can call "space at a time".

So the plan was to find the stress energy tensor in the lab frame, and integrate it. What could be more straightforward?

I think you are saying that you still hope to integrate over a rotating hoop in a comoving chart (such as the Born chart, which is comoving with the Langevin observers). However, one still has to deal with the inconsistency about what time it is at the branch cut, e.g. $\phi=0, 2\, \pi$! See the picture in my WP article on the Ehrenfest paradox.

One thing to look at is to see if I have the right value for tension.

I say again you should start with a careful Newtonian analysis, if for no other reason so that you can compare with any str expression after expanding in powers if v/c.

I think you are saying that you still hope to integrate over a rotating hoop in a comoving chart (such as the Born chart, which is comoving with the Langevin observers). However, one still has to deal with the inconsistency about what time it is at the branch cut, e.g. $\phi=0, 2\, \pi$! See the picture in my WP article on the Ehrenfest paradox.

Nope. To put it in terms of frame-fields, since you don't seem to like the coordinate formulation

The frame-field I want to use to integrate the energy is just the lab frame-field, i.e.

$$\frac{\partial}{\partial t},\frac{\partial}{\partial r},\frac{1}{r} \frac{\partial}{\partial \theta} , \frac{\partial}{\partial z}$$

So we just convert from the Langevian frame-field to the above. The relationship between the two frame-fields is just a Lorentz boost. This is actually convenient, because of what's below.

Unpacking and going through some of my papers, Rindler for instance gets the following result for a Lorentz boost of the stress-energy tensor in "Introdouction to Special Relativity". (I don't own this, but I photocopied some of the pages as a result of some past arguments). I've taken the liberty of removing the factors of 'c' from Rindler's results

pg 132 eq 45.8

$$\rho = \gamma^2 \left( \rho_0 + u^2 t_0^{11} \right)$$

Here u is the 4-velocity, $\rho$ is the 0,0 component of the stress-energy tensor in the boosted frame, $\rho_0$ is the 0,0 component of the stress-energy tensor in the original frame, and $t_0^{11}$ is the pressure in the original frame.

So this gives in my notation
$$T^{00} = \rho + v^2 P$$

where $T^{00}$ is the energy density in the lab frame-field rield, $\rho$ is the 0,0 component of the stress-energy tensor in the Langevian frame field, and P is the 1,1 component of the stress-energy tensor in the Langevian frame field.

Also v = $r \omega$

This is the same result I got earlier.

The volume and notion of simultaneity in the lab-frame field is well defined, and that's where we want our results, ultimately.

I've also done a Newtonian analysis that seems to confirm the expression for P by drawing a free-body diagram. This is

$$P = -\rho v^2$$

Putting this together we get

$$T^{00} = \gamma^2 (\rho - v^4 \rho) = \frac{1-v^4}{1-v^2} \rho = (1+v^2) \rho$$

Which is exactly where I started.

Note that Rindler also gives an expression for momentum density and pressure

momentum = $$u \gamma^2 (\rho_0 + t_0^{11} )$$

pressure = $$\gamma^2 (t_0^{11} + u^2 \rho)$$

This also gives the same results I got earlier, the pressure for instance is zero in the lab frame-field.

But I still don't see how this could be the right answer, because it seems to have twice the value for kinetic energy that it should in the Newtonian limit of small v and a stiff wire.

I say again you should start with a careful Newtonian analysis, if for no other reason so that you can compare with any str expression after expanding in powers if v/c.

I don't think I quite understand what you are proposing. Since I seem to be hitting a dead end with the current approach, it probably is time to start trying to view the problem from a different angle. (No, that's not supposed to be a pun).

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Here's the free-body diagram

$x = d \theta / 2$

Then if the total tension force in the wire is T (a positive number), the force towards the center is

$$2 \sin \frac{d \theta}{2} T \approx T d \theta$$

Let the mass of the element in the arc $d \theta$ be dm.

Then
$$dm \frac{v^2}{r} = T d\theta$$

Now
$$dm = \rho A r d\theta$$

and
$$T = -P A$$

where A is the cross-sectional area of the wire.

Then

$$\rho A r d\theta \frac{v^2}{r} = -P A d \theta$$

or
$$\rho v^2 = -P$$

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The frame-field I want to use to integrate the energy is just the lab frame-field, i.e.

Let's give it a name and write it like this:
$$\vec{e}_1 = \frac{\partial}{\partial t}, \; \vec{e}_2 = \frac{\partial}{\partial z}, \; \vec{e}_3 = \frac{\partial}{\partial r}, \; \vec{e}_4 = \frac{1}{r} \frac{\partial}{\partial \phi}$$
This is just the obvious frame to use in our cylindrical chart, the one which is dual to the coframe read right off the line element
$$ds^2 = -dt^2 + dz^2 + dr^2 + r^2 \, d\phi^2, \; -\infty < t, \, z < \infty, \, 0 < r < \infty, \; -\pi < \phi < \pi$$
namely
$$\sigma^1 = -dt, \; \sigma^2 = dz, \; \sigma^3 = dr, \, \sigma^4 = r \, d\phi$$
The Langevin frame field is obtained by boosting each of these by some amount in the $\partial_\phi$ direction:
$$\vec{f}_1 = \cosh(p) \, \vec{e}_1 + \sinh(p) \, \vec{e}_4,$$
$$\vec{f}_2 = \vec{e}_2, \; \vec{f}_3 = \vec{e}_3,$$
$$\vec{f}_4 = \sinh(p) \, \vec{e}_1 + \cosh(p) \, \vec{e}_4$$
Here, p is a function of r; requiring the expansion tensor of the Langevin congruence (the integral curves of $\vec{f}_1$) to vanish gives p up to a constant, the angular velocity.

So we just convert from the Langevian frame-field to the above. The relationship between the two frame-fields is just a Lorentz boost.

A different boost in the different tangent spaces at the various events.
Locally, frames are related by choice of some global section in a trivial SO(1,3)-bundle.

Have you seen the discussion of the stress-energy tensor in the textbook by Schutz? I think you should find that very helpful here!

The volume and notion of simultaneity in the lab-frame field is well defined, and that's where we want our results, ultimately.

Why? I don't see why you think the problems I mentioned are cured simply by integrating in the original inertial frame.

Here u is the 4-velocity,
What is u^2 doing in an equation relating components, then? Surely that isn't $\| \vec{e} \|^2 = -1$.

(EDIT: it was v^2 as in the expressions from Schutz below.)

This also gives the same results I got earlier, the pressure for instance is zero in the lab frame-field.

Uh oh, you are backsliding a bit! The component wrt the coordinate basis vanishes, but it has no physical meaning. The pressure (or rather, tension) which would actually be measured by a comoving (Langevin) observer is the component computed with respect to the Langevin frame.

But I still don't see how this could be the right answer, because it seems to have twice the value for kinetic energy that it should in the Newtonian limit of small v and a stiff wire.

I think Schutz should help here. See his (4.21), which gives the frame components of the stress-energy tensor for dust (in flat spacetime) in terms of bits of matter in the body:
$$T^{\hat{0}\hat{0}} = \rho \, U^{\hat{0}} \, U^{\hat{0}} = \frac{\rho}{1-v^2}$$
$$T^{\hat{0} \hat{j}} = \rho \, U^{\hat{0}} \, U^{\hat{j}} = \frac{ \rho \, v^{\hat{j}} }{1-v^2}$$
$$T^{\hat{j} \hat{k}} = \rho \, U^{\hat{j}} \, U^{\hat{k}} = \frac{ \rho \, v^{\hat{j}} \, v^{\hat{k}} }{1-v^2}$$
Here,
$$\vec{U} = \frac{1}{\sqrt{1-v^2}} \; \left( \partial_t + v^1 \, \partial_x + v^2 \, \partial_y + v^3 \, \partial_z \right)$$
defines the matter congruence, i.e. this is the unit tangent vector field generated by the world lines of the bits of matter in the body, and $\rho$ is the density in the MCRF (momentarily comoving reference frame), which corresponds in our discussion to the Langevin frame, so the above gives the expressions in the inertial frame. Always using the frame components, of course!

(Edit: Sorry, I see I failed to explain my suggestion: treat the the Langevin congruence as the world lines of a dust, with comoving density a specified function of r, and compute the stress-energy tensor in terms of str formalism. The result can be used as a reality check for the desired stress-energy tensor of an accelerated circular wire--- at least, I think that's what we're discussing!)

Dropping the hats, the first of these says
$$T^{00} = \frac{1}{1-v^2} \, T^{0' 0'} = (1 + v^2) \, T^{0'0'} + O(v^4)$$
(primed indices for Langevin, unprimed for inertial lab frame), or
$$T^{0' 0'} = (1-v^2) \, T^{00}$$

Similarly, the boost along one of the spatial vectors in some frame of the matter tensor of a perfect fluid $\operatorname{diag}(\rho,p,p,p)$ is
$$T^{ab} = \left[ \begin{array}{cccc} c^2 \, \rho+s^2 \, p & -c \, s \, (\rho + p) & 0 & 0 \\ -c \, s \, (\rho + p) & c^2 \, p+s^2 \, \rho & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \end{array} \right]$$
where $c = \cosh(\tau), \; s = \sinh(\tau)$. In our context, this is valid eventwise if you interpret the velocities in the appropriate way to represent the relationship between the two frames at that particular event.

pg 132 eq 45.8

$$\rho = \gamma^2 \left( \rho_0 + u^2 t_0^{11} \right)$$

Here u is the 4-velocity, $\rho$ is the 0,0 component of the stress-energy tensor in the boosted frame, $\rho_0$ is the 0,0 component of the stress-energy tensor in the original frame, and $t_0^{11}$ is the pressure in the original frame.

So this gives in my notation
$$T^{00} = \rho + v^2 P$$

where $T^{00}$ is the energy density in the lab frame-field rield, $\rho$ is the 0,0 component of the stress-energy tensor in the Langevian frame field, and P is the 1,1 component of the stress-energy tensor in the Langevian frame field.

Also v = $r \omega$

This is the same result I got earlier.

And agrees with my trigonometry if you decode Rindler's expression :-/

Similarly, for a matter tensor which in the comoving frame looks like
$$T^{ab} = \left[ \begin{array}{cccc} \rho & 0 & 0 & 0 \\ 0 & u & v & 0 \\ 0 & v & w & 0 \\ 0 & 0 & 0 & p \end{array} \right]$$
the boost along $\vec{e}_4[/tex] gives in the boosted frame $$T^{a'b'} = \left[ \begin{array}{cccc} c^2 \, \rho + s^2 \, p & 0 & 0 & -c\,s (\rho + p) \\ 0 & u & v & 0 \\ 0 & v & w & 0 \\ -c\,s \, ( \rho + p ) & 0 & 0 & c^2 \, p + s^2 \, \rho \end{array} \right]$$ For other readers: we won't obtain precisely Newtonian expressions except as an approximation, of course. Relativity matters! Writing $$c^2 = \frac{1}{1-V^2} = 1 + V^2 + O(V^4), \; s^2 = \frac{V^2}{1-V^2} = V^2 + O(V^4), \; c s = \frac{V}{1-V^2} = V + O(V^3)$$ we obtain up to O(V^2) $$T^{a'b'} \approx \left[ \begin{array}{cccc} (1+V^2) \, \rho + V^2 \, p & 0 & 0 & -V (\rho + p) \\ 0 & u & v & 0 \\ 0 & v & w & 0 \\ -V \, (\rho + p) & 0 & 0 & (1+V^2) \, p + V^2 \, \rho \end{array} \right]$$ In many cases we can probably neglect terms [itex]O(p V)$, so this becomes
$$T^{a'b'} \approx \left[ \begin{array}{cccc} (1+V^2) \, \rho & 0 & 0 & -V \, \rho \\ 0 & u & v & 0 \\ 0 & v & w & 0 \\ -V \, \rho & 0 & 0 & p + V^2 \, \rho \end{array} \right]$$

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A different boost in the different tangent spaces at the various events.
Locally, frames are related by choice of some global section in a trivial SO(1,3)-bundle.

Have you seen the discussion of the stress-energy tensor in the textbook by Schutz? I think you should find that very helpful here!

I don't have Schutz. Does he actually work out this particular problem? If he does, I'll go to the library and order it via interlibrary loan. If not, I'm not sure that I would find it that much more helpful than MTW, or the section of Rindler that I photocopied.

As far as the vectors pointing in different directions go, any directional issues don't matter to the energy, which is a scalar. When we get around to computing the momentum, we can argue about it if needed, but I think it's pretty obvious that the total momentum is zero, and the angular momentum is r x volume x $T_{\hat{t}\hat{\theta}}$

Why? I don't see why you think the problems I mentioned are cured simply by integrating in the original inertial frame.

I don't see why you think there is a problem in the lab frame. Energy is a frame-dependent concept. And what we are trying to find is the energy of the rotating disk in the lab frame. Unless you are arguing that a spinning disk actually doesn't have an energy in the lab frame?

One of the whole point of tensors is that they allow one to switch between frames. So we have the tensor in the Langevian frame, and convert it to the lab frame, and integrate it, where a) we want the answer and b) the integral is trivial.

What is u^2 doing in an equation relating components, then? Surely that isn't $\| \vec{e} \|^2 = -1$.

I should have said that u is the velocity.

Uh oh, you are backsliding a bit! The component wrt the coordinate basis vanishes, but it has no physical meaning.

The pressure (or rather, tension) which would actually be measured by a comoving (Langevin) observer is the component computed with respect to the Langevin frame.

I think Schutz should help here.

I'm not actually saying that the pressure in the lab frame is "physically significant". In fact, it's pretty irrelevant to anything physical, except that it could be interesting if one was trying to find the Komar energy. The fact that the pressure is zero means that the Komar energy (in this flat manifold, just $$\int \left( \rho + P_{\hat{r}\hat{r}} + P_{\hat{\theta}\hat{\theta}} + P_{\hat{z}\hat{z}} \right) dV$$ agrees with the SR energy, which is what we expect for a closed system.

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I don't have Schutz. Does he actually work out this particular problem? If he does, I'll go to the library and order it via interlibrary loan. If not, I'm not sure that I would find it that much more helpful than MTW, or the section of Rindler that I photocopied.

I just wrote out the relevant bit while you were replying :-/

As far as the vectors pointing in different directions go,

I don't see why you think there is a problem in the lab frame. Energy is a frame-dependent concept.

Locally. Globally--- well that's always tricky.

One of the whole points of [tensor algebra] is that it allows one to switch between frames.

Locally.

So we have the tensor in the Langevian frame, and convert it to the lab
frame, and integrate it, where a) we want the answer and b) the integral is trivial.

Think about the integral. Draw a picture of what submanifold you are integrating over. Think about the endpoints of the integration, and notice an awkward global feature of a helical "axis" (integral curve of spatial vectors from the Langevin frame) wrapping around a cylinder.

A zillion eons ago, Nathan Urban, Tom Roberts, myself, and some others discussed this very carefully in sci.physics.relativity. (In those days, we managed to keep the cranks more or less under control.) A year or so ago I noticed an arXiv eprint making many of the same points (but I've forgotten the citation).

Schutz's expressions agree with Rinder's IF one assumes $T_0^{11}$ is zero. I would therefore conclude that Schutz did not include the effects of tension in his analysis, which makes Rindler's analysis (which I have) more suitable.

I don't expect to get the Newtonian result, but I do expect to get a result that is the same as the Newtonian result to order v^2, i.e. a series expansion of the energy around v=0 should give M + .5Mv^2, where M is the mass of the disk and v=r omega, for a "stiff" disk. A disk is stiff if the amount of work done in stretching the wire is small compared to it's rotational kinetic energy.

Aha - I think I've got the answer.

I was incorrectly assuming that for a rigid wire, the radius 'r' of the hoop did not change when it was spun up.

Actually, for the assumed rigid wire, the radius of the hoop must contract by sqrt(1-v^2). I was ignoring this because it was a "small effect" - but it is of order v^2, therefore it needs to be included.

If we paint marks on the wire, and the wire is rigid, the distance between marks must not change in the frame-field of the wire, what we've been calling the Langevian frame. If we assume that r is constant, the distance between marks would increase. Therfore, r must decrease. (There must be a more rigorous way of deriving this).

This corrected value for r gives the corrected value for the energy as

M (1+v^2) sqrt(1-v^2)

The density of the wire goes up by (1+v^2), but the volume of the wire goes down, multiplied by a factor of sqrt(1-v^2), due to the radius shrinking. The cross section of the wire doesn't change under the simple assumptions made.

A series expansion of sqrt(1-v^2) (1+v^2) is 1 + v^2/2 - O(v^4), so this yields the correct Newtonian behavior.

What's interesting is that the relativistic rotating wire has less energy and also less angular momentum that it's Newtonian counterpart, assuming that this is (finally) the correct answer.

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Sorry for the confusion!

Schutz's expressions agree with Rinder's IF one assumes $T_0^{11}$ is zero. I would therefore conclude that Schutz did not include the effects of tension in his analysis, which makes Rindler's analysis (which I have) more suitable.

Right, I got distracted and failed to complete the description of my suggestion (I just added another edit to the post in question clarifying this). And you are of course completely correct--- Schutz was discussing dust before discussing perfect fluids.

I don't expect to get the Newtonian result, but I do expect to get a result that is the same as the Newtonian result to order v^2, i.e. a series expansion of the energy around v=0 should give M + .5Mv^2, where M is the mass of the disk and v=r omega, for a "stiff" disk. A disk is stiff if the amount of work done in stretching the wire is small compared to it's rotational kinetic energy.

Agreed.

World's shortest sketch of some major Ehrenfest paradox pitfalls

For the assumed rigid wire, the radius of the hoop must contract by sqrt(1-v^2).

I know what you mean, but it's not nearly that simple!

First of all, anyone trying to follow along needs to be told that this computation is much trickier than first appears, largely because--- as some well known history of physics shows--- people who attempt it tend to make all kinds of hidden assumptions which invalidate their reasoning. I urge everyone to read (or re-read!) some WP articles I wrote (the last four were left unfinished when I was driven out of WP, and the fourth benefited from significant contributions from another WikiProject Physics member, Peter Jacobi):

http://en.wikipedia.org/w/index.php?title=Frame_fields_in_general_relativity&oldid=42117350
http://en.wikipedia.org/w/index.php?title=Congruence_(general_relativity)&oldid=43337327

http://en.wikipedia.org/w/index.php?title=Rindler_coordinates&oldid=51749949
http://en.wikipedia.org/w/index.php?title=Born_coordinates&oldid=53957524

See also the review paper by Gron cited in the last article. The book by Poisson cited at http://math.ucr.edu/home/baez/RelWWW/HTML/reading.html#gtrsupplement [Broken] should be very helpful for learning some of the mathematical techniques which make life much easier in studying questions like this: "geometrically, without getting confused by coordinate effects, how can I tell whether these two observers in this scenario moving apart, or not?"

I can't emphasize this too strongly: a book could be written simply on the tangled history of the simple-sounding problem of treating a rotating disk, and that poor-quality eprints continue to appear which have been written by authors who are unaware of this history and who have simply (independently) repeated much earlier mistakes. Books could also be written on the philosophical issues involved here! Indeed, in a sense at least one has been--- see the citations in the above article.

See also some other articles in the versions listed at http://en.wikipedia.org/wiki/User:Hillman/Archive Please note that my attempt to outline the history and the philosphical, physical, and mathematical issues involved in the problem of rotating relativistic matter at WP was left incomplete at the time I was driven out of WP. These resources (and indeed the review paper by Gron) are only starting points for discussion, which I think everyone interested in this stuff should study carefully to "get up to speed". I don't claim they provide an adequate discussion--- I am convinced that only a book, or actually several books, could provide that!

Just a few points to look for in studying the above cited background resources, which I see have already arisen:

1. A rod can be accelerated rigidly, but only by applying carefully calibrated "body forces" to each bit of matter (basically, for fundamental geometric reasons, trailing bits have to be accelerated harder than leading bits). It might seem from this that a disk can be "spun up" rigidly, but only by applying carefully calibrated "body forces" so that the world lines look like a Langevin congruence with time variable omega. However, things aren't so simple! In fact, a disk can rotate rigidly with constant angular velocity, but we should not expect be "spin it up" from a nonrotating initial state while "maintaining rigidity", in the global sense that observers comoving with the axis of rotation aver that the disk (or hoop) maintains it's shape. To see why not, note that I already sketched a "physicist's proof" that the Langevin congruence arises if we ask for observers riding a rigidly rotating cylinder (with constant angular velocity!), then consider a generalized congruence in which omega is a function of time, and compute the new expansion tensor of the new congruence! The result shows that global rigidity is incompatible with local rigidity, in the sense of vanishing expansion tensor.

2. In any case, such scenarios are highly artificial and basically impossible to achieve in practice. Hence the interest in studying non-rigid linear and angular acceleration of matter, which for simplicity one naturally wishes to treat as the simplest kind of nonrigid matter known from Newtonian physics, a homogeneous isotropic material.

3. But the theory of elasticity ultimately rests upon Hooke's law, which is strictly speaking not compatible with relativity. This problem can be fixed up, but things quickly get tricky.

4. This is yet another of those areas where the local to global transition is highly problematical. Decades of early work suffered from lack of adequately addressing this point (see the review paper by Gron).

5. Two methods of dimensional reduction are taking hyperslices and forming a quotient manifold. For example, in a static spacetime we can project geodesics onto curves in a constant time slice (hyperslice). But rotating congruences have nonzero vorticity, so by the Frobenius integrability criterion they admit no hyperslices everywhere orthogonal to the world lines. If the rotation is stationary as for the Langevin congruence, one can form a quotient by the congruence. But one needs to be very careful about glib assumptions concerning metrical relations. One of the important points here which is often overlooked is that the "distances" measured in some thought experiment by two or more accelerating observers depend upon the method of defining and measuring distance, even in flat spacetime.

6. This is yet another area where good pictures help enormously, and good enough pictures are easy enough to draw, but it is time consuming to reformat and upload them to forums like PF. Thus, it is best to discuss tricky stuff in a facutly office standing at the whiteboard, as human scholars say.

7. When discussing a topic which is known to be chock full of pitfalls (historical, philosophical, physical, mathematical), it pays to be very systematic, and to start with simple stuff, to continually check results against previous knowledge, simpler cases, and so on. I hope everyone sees that pervect has been trying to do just this! I just wanted to try to briefly assure lurkers that there is good reason for trying to be so painstaking about absolutely everything said here.

Here are some more details of the computation I suggested:

In the cylindrical chart for Minkowski vacuum,
$$ds^2 = -dt^2 + dz^2 + dr^2 + r^2 \, d\phi^2, \; -\infty < t, \, z < \infty, 0 < r < \infty, -\pi < \phi < \pi$$
the Langevin frame is
$$\vec{f}_1 = \frac{1}{\sqrt{1-\omega^2 r^2}} \; \left( \partial_t + \omega \, \partial_\phi \right), \; \vec{f}_2 = \partial_z, \; \vec{f}_3 = \partial_r, \; \vec{f}_4 = \frac{1}{\sqrt{1-\omega^2 r^2}} \; \left( \frac{1}{r} \partial_\phi + \omega \, r \, \partial_t \right)$$
(in the first vector, a coefficient of r/r = 1 appears, or rather, doesn't appear after simplication!). As an exercise, anyone who hasn't already done this should transform these vectors to see what they look like in the Born chart. Remember that rougly speaking the cylindrical chart is comoving with inertial observers comoving with the axis of the disk, whereas the Born chart is nondiagonal and is comoving with the bits of rotating matter.

Letting $\omega$ become a function of t, let's evaluate some quantities of interest. (All components refer to the Langevin frame I wrote out above, not to the coordinate basis.) Those of you who use GRTensorII can use this input:
Code:
Ndim_ := 4:
x1_ := t:
x2_ := z:
x3_ := r:
x4_ := phi:
eta11_ := -1:
eta22_ :=  1:
eta33_ :=  1:
eta44_ :=  1:
b11_ := 1/sqrt(1-omega(t)^2*r^2):
b14_ := omega(t)/sqrt(1-omega(t)^2*r^2):
b22_ := 1:
b33_ := 1:
b41_ := omega(t)*r/sqrt(1-omega(t)^2*r^2):
b44_ := 1/r/sqrt(1-omega(t)^2*r^2):
Info_ := Minkowski vacuum (cyl chart; Langevin frame; time varying omega):
# Chart covers -infty < t,z < infty, 0 < r < infty, -pi < u < pi
# However, this frame (Langevin 1933) is only valid on 0 < r < 1/omega.
# It is obtained by boosting original static frame in f_4 direction
# with velocity v = omega*r where omega is a function of t only.
We find that the acceleration vector of the Langevin observers is given by
$$\nabla_{\vec{f}_1} \vec{f}_1 = \frac{r}{1-\omega^2 r^2} \; \left( \frac{\omega_t}{\sqrt{1-\omega^2 r^2}} \; \vec{f}_4 - \omega^2 \; \vec{f}_3 \right)$$
Notice that this is precisely the acceleration needed to make them move in circular orbits (in track sense), or to make their world lines look like helices (in spacetime, and with "variable pitch" when $\omega$ is slowly varying). In the case of constant $\omega$, this acceleration is purely radial; otherwise it acquires a $\vec{f}_4$ component.

The vorticity tensor has only nonzero component (up to algebraic symmetries):
$$\Omega[\vec{f}_1]_{34} = \frac{-\omega}{1-\omega^2 r^2}$$
Equivalently, the vorticity vector is
$$\vec{\Omega}[\vec{f}_1] = \frac{\omega}{1-\omega^2 r^2} \, \vec{f}_2$$
The expansion tensor has only one nonzero component:
$$\Theta[\vec{f}_1]_{44} = \frac{ r^2 \, \omega \, \omega_t}{(1-\omega^2 r^2)^{3/2}}$$
When $\omega_t > 0$, this shows an expansion along $\vec{f}_4$, which in an elastic solid would be resisted by a tension along the circular wire, just as pervect found.

What can we conclude from this simple computation? Well, the assumption here is that the world lines of bits of matter in our hoop (or more ambitiously, our disk or rotating solid cylinder) look like variable pitch helices winding around a fixed coordinate cylinder in our chart, i.e. we are asking what happens infinitesimally if in some sense we insist that inertial observers comoving with the centroid of the hoop declare that according to a suitable method of measurement, the hoop, considered setwise rather than pointwise, neither expands nor contracts--- globally speaking. Or IOW, we tried to spin up the hoop while keeping it "globally rigid" in the sense that its shape is not deformed, according to inertial observers comoving with the centroid (the ones who are static in our cylindrical chart). We found that infinitesimally, observers riding on the hoop must measure an expansion along the hoop. Just to be perfectly clear: all notions of distance agree infinitesimally, and according to any physically reasonable notion of distance in the large, in our scenario, two initially "nearby" hoop-riding observers really will declare that they are moving apart from each other, despite what the inertial observers say about the "rigid" shape of the hoop as a whole. This phenomenon (which was predicted in the very early days of relativity by Einstein himself, who was correcting a mistake by another physicist) has been well illustrated in the figures Peter Jacobi made for the beginning of the version of "Ehrenfest paradox" which I cited above.

Of course, real materials subject to forces--- clearly, spinning up a body produces centrifugal forces which will act within the body--- respond by deforming slightly in a manner dependent upon the properties of the material.
Generally speaking, to a first approximation, locally speaking, we have two counteracting tendecies: "centrifugal force" versus "tension" within the hoop. (Note that in my posts "infinitesimally", "locally", and "globally" refer to phenomena which arise respectively at the level of jet spaces [generalizations of tangent spaces], local neighborhoods, and, well, globally. Many authors use sloppy terminology which confuses these levels of structures, a failing I try to avoid, particularly in discussing such a delicate topic as rotating relavitistic matter.) The most careful analyses to date show that we should expect a real hoop made of elastic material to expand when it is spun up, initially for the obvious Newtonian reasons. This is just what we should expect. Eventually relativistic effects will come into play (the effect pervect has in mind would oppose the centrifugal effect); the problem is to describe these effects correctly and clearly. But the details will clearly depend upon our material model.

Note that computing the Fermi derivatives of
$\vec{f}_2, \; \vec{f}_3, \; \vec{f}_4$ along $\vec{f}_1$ shows that this frame is spinning about $\vec{f}_2$ with respect to a gyrostabilized frame; in the case of constant angular velocity this is of course just the well-known Thomas precession. To compute it, we can introduce a rotation by $\theta = p \, t$ about $\vec{f}_2$, where p is an undetermined function of r, and we demand that the new frame (the third discussed in this post!) should be nonspinning. The result is
Code:
Ndim_ := 4:
x1_ := t:
x2_ := z:
x3_ := r:
x4_ := phi:
eta11_ := -1:
eta22_ :=  1:
eta33_ :=  1:
eta44_ :=  1:
b11_ :=  1/sqrt(1-omega^2*r^2):
b14_ :=  omega/sqrt(1-omega^2*r^2):
b22_ :=  1:
b31_ := -sin(f(r)*t)*omega*r/sqrt(1-omega^2*r^2):
b33_ :=  cos(f(r)*t):
b34_ := -sin(f(r)*t)/r/sqrt(1-omega^2*r^2):
b41_ :=  cos(f(r)*t)*omega*r/sqrt(1-omega^2*r^2):
b43_ :=  sin(f(r)*t):
b44_ :=  cos(f(r)*t)/r/sqrt(1-omega^2*r^2):
constraint_ := {
f(r) = omega/sqrt(1-omega^2*r^2)
}:
Info_ := Minkowski vacuum (cyl chart; despun Langevin frame):
# Chart covers -infty < t,z < infty, 0 < r < infty, -pi < u < pi
Similarly, we introduce a fourth frame whose spatial vectors are "modelocked" with a distant observer. Comparing these we can deduce the precession of the gyrostabilized spatial frame carried by one of our comoving observers which would be observed by said distant observer.

In the generalization to time varying $\omega$, we let p be an undetermined function of t,r and we find that the nonspinning intertial frame obtained from our time varying omega Langevin frame has $\vec{n}_1 = \vec{f}_1, \, \vec{n}_2 = \vec{f}_2$ but sets $\vec{f}_3, \; \vec{g}_4$ spinning about $\vec{f}_2$ at the angular rate
$$p \, t = \int \frac{\omega \, dt}{\sqrt{1-\omega^2 r^2}}$$
In the new frame, the expansion scalar and vorticity tensor components are unchanged, but now the expansion tensor "rotates" with respect to the "principal axis" or diagonalized form we encountered in the spinning frame.

I have shown elsewhere that the Thomas precession can also be computed rather more easily in the velocity space, using the curvature tensor of hyperbolic 3-space, and as Theodor Kaluza was the first to observe, to a good first approximation, the geometry of the Landau-Lifschitz-Langevin metric for the quotient manifold is hyperbolic.

Note also that when you speak of "radius" or "distance" in this scenario (even with constant $\omega$), you need to be very careful because there are multiple physically meaningful notions of "distance in the large" which apply here. Failure to recognize this is the root cause of many bootless arguments about the Ehrenfest paradox, which is unfortunate since the multiplicity of notions of distance is so easy to verify and to understand. This doesn't mean you can't discuss "distance in the large", but it does mean you have to be very careful.

In some of my posts in this thread, I was getting way ahead of the story by proposing to discuss a general relativistic treatment taking account of the effects of gravitation. This would be of potential interest to astrophysicists in connection with creating convenient and accurate models of rotating relavistic objects such as neutron stars or black hole accretion disks. (Needless to say, much is already known about these topics!) The simplest possible models of rotating matter configurations are exact dust solutions. Two simple examples which illustrate some of the new wrinkles which appear (like closed causal curves) are described in two more WP articles I wrote or rewrote:

http://en.wikipedia.org/w/index.php?title=Van_Stockum_dust&oldid=39874076
http://en.wikipedia.org/w/index.php?title=Gödel_metric&oldid=45646663

From dust solutions it might seem to be a simple matter to generalize to rotating perfect fluid solutions. But this turns out to be far trickier than one might readily anticipate. For example, no perfect fluid solution is known which can act as a source for important stationary axisymmetric vacuum solutions such as the Kerr vacuum! Contrast the situtation for perfect fluid sources for the Schwarzschild vacuum, which are now rather well understood: http://arxiv.org/abs/gr-qc/0609088 For a rotating disk which is being spun up, we probably want to use the relativistic theory of elasticity, which (suprisingly enough) has only recently achieved attention from more than a handful of researchers (Robin Carter of Kerr vacuum fame is one of them). See http://arxiv.org/abs/gr-qc/0605025

I think that public discussion (public in the sense of "world readable", like the op-ed pages of the New York Times, not in the sense of an "open mike forum" like PF or WP) by knowledgeable parties of an interesting and philosophically and technically challenging--- yet seemingly very simple--- problem would be very instructive for students and interested laypersons, but only if all participants share sufficient common background, scholarly and scientific ethos, and so on! Thus, what is sorely needed is a something like a closed edit blog where only pervect, myself, and other individuals known to be reasonable in discussions can participate. But PF is not such a "world-readable idgit-uneditable forum", so my further input will have to be made elsewhere.

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A couple of (ok, a bunch of) comments:

Born rigidity (Born rigid motion) to be carried out exactly requires the body forces you mention. But as Nikolic aka Demystifier points out in

http://arxiv.org/abs/physics/9810017

if the acceleration is gentle enough a physical rod will move in essentially the same manner as a Born rigid rod.

I'll agree that it's not possible to spin up a disk rigidly - this is mentioned with appropriate references in the sci.physics.faq, this was first pointed out by Ehrenfest. But I don't believe there is any problem in spinning up a hoop, i.e. a disk of zero thickness.

To quote the FAQ:
Integrating tau out of Born's condition, we see that infinitesimally close particles must keep the same proper distance. So in the original rest frame, they suffer Lorentz contraction in the transverse direction but none in the radial direction. The circumference contracts but the radius doesn't. But in the original rest frame, the circumference is a circle, sitting in a spatial slice (t=constant) of ordinary flat Minkowski spacetime. In other words, we would have a "non_Euclidean circle" sitting in ordinary Euclidean space. This is a contradiction.

But by making the radius zero and considering a hoop of zero thickness, we avoid this problem.

The notion of nearby observers maintaining the same distance is the notion of what I was looking for in "rigidly" spinning up the hoop, it's the standard notion of Born rigidity.

I'll agree that if you keep the radius of the hoop fixed in the lab frame that different observers must move apart. This is an argument that the radius does not stay constant. We simply require that the radius adjust in such a manner that a pair of observers, an infinitesimal distance apart, keep the same infinitesimal distance apart.

If we have an inertial frame of reference, the lab frame, with a Minkowskian metric, we can talk about distances in the lab frame without any confusion even over large intervals. Thus there isn't any ambiguity in defining what we mean by the radius of the disk in the lab frame. [add]This is probably different from what you're used to, for instance in your calculation including gravitational self-energy, there is not a Minkowski metric, and the usual problems with defining distance in a general GR metric arise. However, in the context of SR, the only thing we need to beware of is that distance is frame-dependent, in any given frame there's only one definition.

While we do need to consider distance in the moving frame(s), we don't need to consider "large" distances, we only need to consider short distances, so there isn't any problem there, either.

So putting this together, if we spin up a hoop of zero thickness (rather than a disk), and we do it "gently enough", we will achieve an arbitrarily close approximation to Born rigid motion. And this Born-rigid hoop should contract, changing its radius.

I definitely have to read more of the literature, especially including Gron, however, which has been on my list for some time.

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As Nikolic aka Demystifier points out in
http://arxiv.org/abs/physics/9810017
if the acceleration is gentle enough a physical rod will mone in essentially the same manner as a Born rigid rod.

Well, of course that just says that if you look at the expressions I derived in the article, under some conditions you can neglect, over a sufficiently brief period of time, the expansion in the Bell congruence (or the nonequal accelerations in the Rindler congruence).

I'll agree that it's not possible to spin up a disk rigidly - this is mentioned with appropriate references in the sci.physics.faq, this was first pointed out by Ehrenfest. But I don't believe there is any problem in spinning up a hoop, i.e. a disk of zero thickness.

To quote the FAQ:
But by making the radius zero and considering a hoop of zero thickness, we avoid this problem.

I think this confuses the "big" and "small" radii of a circular torus in analytic geometry :-/

The notion of nearby observers maintaining the same distance is the notion of what I was looking for in "rigidly" spinning up the hoop, it's the standard notion of Born rigidity.

This is exactly the condition that the expansion tensor must vanish.

I'll agree that if you keep the radius of the hoop fixed in the lab frame that different observers must move apart. This is an argument that the radius does not stay constant.

I just tried rather hard to carefully explain that statements like that last DMS! Because "distance in the large" requires qualification for accelerating observers, even in flat spacetime. And of course, the Langevin observers are accelerating, even when the angular velocity is constant.

If we have an inertial frame of reference, the lab frame, with a Minkowskian metric, we can talk about distances in the lab frame without any confusion even over large intervals. Thus there isn't any ambiguity in defining what we mean by the radius of the disk in the lab frame.

So I think we agree that qualifications are neccessary.

While we do need to consider distance in the moving frame, we don't need to consider "large" distances, we only need to consider short distances, so there isn't any problem there, either.

Well, when we talked about preserving the shape of the hoop, we of course were talking about distance in the large. Since that distance was referred to an inertial frame in flat spacetime, minimal qualification was neccessary.

So putting this together, if we spin up a hoop of zero thickness (rather than a disk), and we do it "gently enough", we will achieve an arbitrarily close approximation to Born rigid motion.

I think you might still be missing the point that if we don't find that Newtonian effects dominate for small angular velocities, we are in trouble. IOW, Nikolic's statement makes qualified sense, but is little help in discussing any realistic material.

I definitely have to read more of the literature, especially including Gron, however, which has been on my list for some time.

Go to, go to! It's very helpful, as are some of the better of the hundred or so papers he cites. I collected and studied some three dozen when I was writing the WP articles; some of these were pretty awful, some expressed valuable insights. Please carefuly reread my previous post after reading Gron and working some kinematic decomposition exercises from Poisson, plus verifying the computations I offered in this thread. Also, I finished some thoughts in my post #21.

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As far as energy goes, the relativistically spinning disk appears to me now to be very much like the linearly accelerated one.

The correction factor due to tension is of order v^4. We have a volume reduction of sqrt(1-v^2) which is of order (1-v^2/2), and we have a density increase of exactly (1+v^2), which is different from the linearlly accelerated result of 1/(1-v^2) but is the same to order v^2.

The net result is a kinetic energy that agrees with the Newtonian result for both cases, i.e. a kinetic energy of m v^2/2 to second order.

The effect of the tension terms on the angular momentum density are much more significant, though. This is an effect of order v^2, and exactly cancels out the increase in density, i.e. the angular momentum density is just rho v r.

However, the volume decreases as noted before, resulting in an order v^2 decrease in angular momentum, quite different than one might expect. This can be attributed to the shrinking radius of the hoop.

Summarizing (?) the discussion so far

Hi, pervect,

Do you agree with my understanding of your current approach?:

1. You are working in flat spacetime with the stress-energy tensor of a model of a rotating hoop.

2. You are trying to compute the total mass-energy of a rotating hoop (constant nonzero angular velocity omega) as described by the inertial observer comoving with the centroid of the hoop. The idea is to compare with the total mass-energy of an identical but non-rotating hoop.

3. The world lines of the matter in the "before" hoop should be straight lines in Minkowski spacetime (vertical coordinate lines in the cylindrical chart). The world lines of the matter in the "after" hoop should agree with world lines belonging to one of the unique family, parameterized by omega, of Langevin congruences in Minkowski spacetime, which are stationary cylindrically symmetric with vanishing expansion tensor, i.e. they describe the world lines of a cylindrically symmetric configuration of rigidly rotating observers.

4. Given the global issues with integrating in the comoving Born chart, you are trying to integrate the kinetic energy (plus stored energy from the tension) along the hoop in the cylindrical chart (your "lab frame"). Specifically, you hope to find the mass-energy in the (constant omega) hoop in the obvious inertial frame and integrate over the hoop in a constant time slice. This should give the total mass-energy as described by the inertial observer comoving with the centroid of the hoop, which is stationary as described by this observer.

5. As far as I can see, you are not (yet) actually using any stress-strain relationship or indeed anything from the theory of elasticity. Rather, you are trying to impose "local rigidity" (vanishing expansion tensor of the world lines of the matter in the hoop).

6. As far as I can see, you avoided trying to model spinup phase and tried to guess what rigidly rotating hoop is equivalent to a given nonrotating hoop. As I understand it, you simply assumed that it will be possible to maintain rigidity during spin up (presumably via a complicated but unique congruence), and then deduced, based on the Lorentz contraction, that the diameter of the hoop, as described by the inertial observer, must have decreased by a specific ammount. IOW, your rotating hoop is supposed to have the same circumference as measured by comoving observers as the nonrotating hoop (since local rigidity is assumed to have been maintained throughout). You deduce that the mass-energy and angular momentum of a rotating hoop (with constant nonzero omega), as described by the inertial observer comoving with the centroid, are both smaller than that of a nonrotating hoop (zero omega).

As far as energy goes, the relativistically spinning disk appears to me now to be very much like the linearly accelerated one.

My objections to your current approach (as I understand it) are these:

1. My intuition tells me that your result can't be right: the total mass-energy of a rotating hoop should not be smaller than that of an "equivalent" nonrotating hoop, in any physically reasonable sense of "equivalent"!

2. To make the desired comparison between the two hoops, you need to be able to set up a physically reasonable equivalence between "before" and "after", i.e. between a nonrotating hoop (no doubt what this means!) and an equivalent but rotating hoop (nonzero omega, nonzero tension along the hoop). I maintain that the only "safe" way I see of doing this is to set up a simple material model and to try to model the spin-up phase. You maintain that you can evade this by insisting that local rigidity (vanishing expansion tensor of congruence of world lines of the matter in the hoop) must be maintained during spin-up, but it's not clear to me that this is possible. I showed that the congruence you imagine would have to be rather complicated.

3. You assumed the radius of the "equivalent" rotating hoop (constant nonzero omega) must be smaller by an amount deduced from the Lorentz contraction factor, but measuring circumference is problematical for the comoving observers. I think you are thinking of integrating length of a spacelike curve in a "constant time slice" in the comoving Born chart to compute this circumference, but there is no such slice. If we try to compute C as measured by comoving observers in the cylindrical chart, we run into global inconsistencies (draw the picture of alleged "space at a time" for hoop matter). So I maintain that you need to worry about measuring "distance in the large", and to confront the fact that theory shows clearly that results will depend upon the method of measurement used by the comoving observers (possibly one, possibly many acting together) and the definition they use to compute a distance from these measurements.

The correction factor due to tension is of order v^4. We have a volume reduction of sqrt(1-v^2) which is of order (1-v^2/2), and we have a density increase of exactly (1+v^2), which is different from the linearlly accelerated result of 1/(1-v^2) but is the same to order v^2.

The net result is a kinetic energy that agrees with the Newtonian result for both cases, i.e. a kinetic energy of m v^2/2 to second order.

I suggest backing off from a rotating hoop and starting over for a short rod which is linearly accelerated along the axis of the rod. Can you find a reasonable expression for the mass-energy of such a rod as described by inertial observers?

As we know from Rindler versus Bell congruence, such a rod can remain rigid only if we assume very special accelerations.

At his website, Greg Egan models a rod being accelerated by being tugged at one endpoint. He focuses on deriving the displacement of an elastic rod under these conditions, but he does give expressions for the stress-energy tensor in a frame comoving with the matter in the rod. His $\vec{u}=\vec{e}_1 is the timelike unit vector in this frame, and his [itex]\vec{w}=\vec{e}_2$ is the spacelike unit vector pointing along the axis of the rod. He writes down expressions for the stored energy and the tension (from a material model), then writes down the stress-energy tensor in our frame, then takes a (flat spacetime) divergence to obtain an equation which can be rearranged (as I understand it) to give either (a) an equation for the displacement, or (b) equations for the frame written in an ordinary cylindrical or cartesian chart for Minkowski spacetime. He finds an exact solution for (a) by assuming boost invariance. This models a rod which is static in the Rindler chart, so with trailing points being accelerated harder. I guess the implicit claim is that for a suitable elastic type material model, the tensions in the rod distribute themselves to make this possible.

(BTW, I am not sure I understand his claims correctly, but I am seeking clarification from him.)

The effect of the tension terms on the angular momentum density are much more significant, though. This is an effect of order v^2, and exactly cancels out the increase in density, i.e. the angular momentum density is just rho v r.

However, the volume decreases as noted before, resulting in an order v^2 decrease in angular momentum, quite different than one might expect. This can be attributed to the shrinking radius of the hoop.

At the moment, I doubt that you are comparing the right pair of hoops. Indeed, I doubt that your proposed notion of setting up an equivalence between nonrotating hoops of radius R and stationary rotating hoops (of some different radius) makes sense.

If I am right, it seems to me that one cannot evade postulating a material model. If so, the obvious choice would be to try to create a suitable model of an elastic material. Then, for small omega Newtonian effects should dominate. The question becomes: how do relativistic effects alter Newtonian predictions as omega increases? For reasonable elastic constants I think this question should be answerable, both for linearly accelerated short rods and for spinning hoops.

I think there would be considerable interest in simple but reasonable ("elastic material") models in gtr of the interior of (a) a tugged rod (constant acceleration of leading endpoint) (b) a rotating hoop (constant omega, so stationary spacetime), particularly if the latter could be matched to an exact stationary axisymmetric vacuum solution.

I agree with your understandings expressed in 1-5.

[add]Ooops, I was hasty in agreeing. Some of the fine print doesn't quite work.

6 is correct in that I don't have a detailed model of spin-up currently. I thought that you agreed that the congruence in 3) was unique, however?

I agree that the derivation needs to be improved in this area.

I definitely do NOT think that the energy decreases when you spin it up. My result for the energy was expressed in terms of v, the radial velocity of the hoop.

$$E = m_0 \left(1 + v^2\right) \sqrt(1-v^2)$$

here m_0 is the initial mass of the hoop.

where $r_0$ is the intial radius of the hoop.

The energy increases with v, the term of order v^2 is .5 m v^2

My observations related to "lower energy" were observations that coefficients of v^n of order 4 and higher in the expression for energy written as a polynomial in v are negative. I've actually only verified that the coefficients are negative for 4<=n<=48, howver, I haven't computed all of them :-).

We can find v as a function of $\omega$ by solving the relationship

$$v = \sqrt{1-v^2} r_0 \omega$$

this gives

$$v := \frac{{\it r_0 \omega}}{\sqrt { \left( {{\it r_0}}^{2}{\omega}^{2}+1 \right) }}$$

The series expansion in terms of $\omega$ turns out to be
$$(m+1/2\,m{{\it r\_0}}^{2}{\omega}^{2}-{\frac {9}{8}}\,m{{\it r\_0}}^{4 }{\omega}^{4}+{\frac {25}{16}}\,m{{\it r\_0}}^{6}{\omega}^{6}-{\frac { 245}{128}}\,m{{\it r\_0}}^{8}{\omega}^{8}+O \left( {\omega}^{10} \right) )$$

so unlike the series expansion for v, not all of the higher order terms are negative.

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OK, let's define the congruence more precisely.

We initially have a point at r=r0, and some angle theta. Through this point there must be a uniqie curve in the congruence. We parameterize this curve using lab-frame coordinates

(r,theta) -> (r(t), theta + phi(t))

We require that r initially be at r0, however.

A feature of this congruence is that the difference in angles, theta, in the lab frame between two worldlines is independent of lab time.

in order to have rigid motion, the separation of worldlines (orthogonal to the worldlines) must be constant.

We assume that we can pick phi(t) in such a manner that dr/dt is as small as desired, that in the limit dr/dt is zero, therefore we can neglect it. This assumption can be checked

In this case, the velocity of the worldline will just be

v = r(t) * d(phi)/dt

Doing a Lorentz boost, it is intuitively obvious that the separation between worldlines passing through points (t,theta) and (t,theta+dtheta) is

r(t) dtheta / sqrt(1-v^2), where v = r(t) dphi/dt.

To do better than this bold statement, I'd have to draw some space-time diagrams. (add), but basically, all that is happening is that the separation between worldlines is the proper distance, and r*dtheta is the Lorentz contracted distance in the lab frame.

Because dtheta is invariant as a function of time, in order for the separation between worldlines to remain constant r(t) must equal sqrt(1-v^2).

Let's check the assumptions

Let's let r_0 = 1 for simplicity. Let $$\omega = \frac{d \phi}{dt}$$. Then we have using some of the results from earlier posts

$$v = \frac{\omega}{\sqrt{1+\omega^2}}$$
$$r = \sqrt{1-v^2} = \frac{1}{\sqrt{1+\omega^2}}$$

so
$$\frac{dr}{dt} = \frac{dr}{d\omega} \frac{d \omega}{dt} = -\frac{\omega}{\left( 1 + \omega^2 \right)^\frac{3}{2}} \frac{d \omega}{dt}$$

so by making $$\frac{d\omega}{dt}$$ very small (quasi-stationary), we can make dr/dt small as well.

I suppose at this point I ought to learn how to compute the expansion scalar of this congruence to check that it's zero, at least in some limiting sense as $d\omega/dt$ -> 0

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Trying to clarify my objections

I agree with your understandings expressed in 1-5.

6 is correct in that I don't have a detailed model of spin-up currently. I thought that you agreed that the congruence in 3) was unique, however?

The Langevin congruence is the unique rigid "helical" congruence in Minkowski vacuum. That is, if we write down the frame for a general congruence in which the world lines form helices of fixed pitch depending upon r (i.e. the orbits are circles and observers move uniformly in each circle, but we don't specify how quickly a given observer moves around his circle), and if we then demand that the expansion tensor vanish, we obtain a differential equation, and solving this, we arrive at the Langevin congruence with fixed omega.

If we allow omega to vary with time, the expansion tensor acquires a nonzero component $\Theta[\vec{e}_1]_{44}$ which is positive for $\omega_t > 0$, i.e. our observers are moving apart along $\vec{e}_4$, so the "comoving circumference" is increasing in this sense when we are increasing the angular velocity. Naturally this means that the Langevin congruence is no longer region when omega is conconstant. And each world line remains on a fixed cylinder of constant radius in the cylindrical chart, but their pitch varies. We can call this the variable pitch Langevin congruence.

You were turning the expansion on its head and arguing that, without modeling the spinup, we can imagine a spinup phase which is infinitesimally rigid throughout, i.e. the expansion tensor is zero throughout. I said it's not clear (at least not to me) that this is possible and provided some evidence that it might not be. But under your assumption, you said the comoving circumference should be the same, so the diameter measured by an inertial observer comoving with the centroid must be smaller, i.e. infinitesmal rigidity implies global contraction. In this case, during the spinup phase, each quasihelical world line stays on a quasicylinrical surface which shrinks in radius during the spinup phase, but before spin up they still look like vertical rays and after spinup they look like helices of constant pitch.

But I said your notion of circumference measured by the observers comoving with the hoop is incompletely specified, so your assumptions are invalid on at least two counts.

I definitely do NOT think that the energy decreases when you spin it up.

I asked you in a PM if you meant that the expression you found means the energy decreases for sufficiently small V, and I thought you said you did. I must have misunderstood your answer.

OK, now that we agree on this point, I know what your answer is and I see how you are coming up with it, but I don't think your answer is physically reasonable. Or at least, I don't think you have provided convincing reasoning that your assumptions about spinup are plausible. Or if you like, your assumptions about the equivalence between rotating hoops of radius $R(\omega)$ as described by inertial observers comoving with the centroid and nonrotating hoops of radius $R_0$.

I said that it is clear that you need some way to establish such an equivalence. If you wish to avoid postulating a material model, I understand that your idea is to try to argue that there's some criterion like "keep expansion tensor zero throughout" which establishes the necessary equivalence. But then you'd have to show that spinups satisfying this condition exist and you'd have to argue that the implied material properties are not physically absurd (e.g. speed of sound smaller than speed of light).

So what about the angular momentum? Do you still say that rotating hoops which are spun up rigidly (assuming that's even possible) have decreasing angular momentum?

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Clarification

by making $$\frac{d\omega}{dt}$$ very small (quasi-stationary), we can make dr/dt small as well.

I think this is besides the point; it seems to me that all my objections still stand. I never said anything which involves the size of r_t, only the (varying) size of r.

I suppose at this point I ought to learn how to compute the expansion scalar of this congruence to check that it's zero, at least in some limiting sense as $d\omega/dt$ -> 0

It's easy. In fact computing the expansion tensor for a given vector field is built into GRTensorII. When you installed that you should have obtained a (well written) manual which explains how to do this. I included the "spacetime definitions" (actually, frame definitions, which is at a higher level of structure!) I used so that you could quickly run your own computations as well as checking by hand (since these are simple examples, this is fortunately very easy, but you want to have confidence GRTensorII is doing what you expect).

I myself don't use the built in definition, I use grdef to define the expansion tensor directly. (In fact, I read in a whole bunch of useful definitions from a file before using GRTensorII, and read in more definitions from other files as needed.) In index gymnastics, if we write the velocity vector as a covector $\vec{u} = u_j \partial_{x^j}$, the coordinate basis components of the expansion tensor are $u_{(j;k)}$ (symmetrize the covariant derivative). Now you can find the frame components. Since u is a unit vector, the expansion tensor is in fact a three dimensional symmetric tensor living in the hyperplane element spanned by the spatial vectors in the frame. See the book by Poisson, A Relativist's Toolkit, which offers a fine discussion of the kinematic decomposition of a vector field into acceleration vector, expansion tensor, and vorticity vector.

(You can also compute the expansion tensor directly in the frame by hand-- GRTensorII also has a built in command to compute covariant derivatives of a tensor wrt a frame vector , but you probably want to stick close to what you are most comfortable with for now!)

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