# Stress-energy tensor of a wire under stress

1. Apr 29, 2007

### 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.

Last edited: Apr 29, 2007
2. Apr 29, 2007

### pmb_phy

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

3. Apr 29, 2007

### pervect

Staff Emeritus
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).

4. Apr 30, 2007

### pmb_phy

My appologies for missing that.

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

5. May 1, 2007

### pervect

Staff Emeritus
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.

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

6. May 1, 2007

### Chris Hillman

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!

7. May 1, 2007

### pervect

Staff Emeritus
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.

Last edited: May 1, 2007
8. May 1, 2007

### Chris Hillman

I insist upon a Newtonian computation in a Galilei frame!

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)

9. May 1, 2007

### pervect

Staff Emeritus
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.

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(_);
>
>
>

10. May 1, 2007

### pervect

Staff Emeritus
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]

Last edited: May 1, 2007
11. May 1, 2007

### Chris Hillman

Frame components or coordinate basis components?

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.)

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 (Text):

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 (Text):

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 (Text):

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, gotta 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!

Last edited: May 2, 2007
12. May 1, 2007

### Wallace

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?

13. May 1, 2007

### Chris Hillman

14. May 1, 2007

### Wallace

Thanks Chris, lots of reading to keep me busy!

15. May 1, 2007

### pervect

Staff Emeritus
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.

Last edited: May 1, 2007
16. May 2, 2007

### Chris Hillman

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

Last edited: May 2, 2007
17. May 2, 2007

### pervect

Staff Emeritus
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.

Last edited: May 2, 2007
18. May 2, 2007

### Chris Hillman

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.

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 $1/L^2 \cdot L = 1/L$.

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".

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.

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.

19. May 3, 2007

### pervect

Staff Emeritus
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 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).

Last edited: May 3, 2007
20. May 3, 2007

### pervect

Staff Emeritus
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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Last edited: May 3, 2007