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Old Apr29-07, 04:37 PM       Last edited by pervect; Apr29-07 at 11:15 PM..            #1
pervect

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Stress-energy tensor of a wire under stress

[Edit] 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)

LaTeX Code: \\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) = LaTeX Code: \\frac{1+W}{(1+d/L)(AA/A)}

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

[add]
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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Old Apr29-07, 07:03 PM                  #2
pmb_phy

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Originally Posted by pervect View Post
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

LaTeX Code: \\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) = LaTeX Code: \\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.

[add]
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
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Old Apr29-07, 11:19 PM                  #3
pervect

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Originally Posted by pmb_phy View Post
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).
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Old Apr30-07, 07:13 PM                  #4
pmb_phy

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Originally Posted by pervect View Post
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
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Old May1-07, 03:15 PM                  #5
pervect

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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.

LaTeX Code: T^{ab}{}_{;b}=0
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Old May1-07, 05:56 PM                  #6
Chris Hillman

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Arrow 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.ne...lasticity.html
http://gregegan.customer.netspace.ne...erHorizon.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!
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Old May1-07, 06:47 PM       Last edited by pervect; May1-07 at 06:56 PM..            #7
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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,LaTeX Code: \\theta , z.

Define some new coordiantes t,r,LaTeX Code: \\Theta ,z where LaTeX Code: \\Theta = LaTeX Code: \\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 = LaTeX Code: \\sqrt{1 - \\omega^2 r^2} dt + \\frac{r^2 \\omega}{\\sqrt{1 - r^2 \\omega^2}} d \\Theta
w2 = dr
w3 = LaTeX Code: \\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 LaTeX Code: \\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,LaTeX Code: \\Theta ,z coordinates and convert it back to the t,r,LaTeX Code: \\theta ,z coordinates via the tensor transformation rules.

The t,r,LaTeX Code: \\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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Old May1-07, 07:00 PM                  #8
Chris Hillman

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Question

Originally Posted by pervect View Post
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!

Originally Posted by pervect View Post
Start with Minkowski cylindrical coordinates t,r,LaTeX Code: \\theta , z.

Define some new coordiantes t,r,LaTeX Code: \\Theta ,t, where LaTeX Code: \\Theta = LaTeX Code: \\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?...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)
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Old May1-07, 08:04 PM                  #9
pervect

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Originally Posted by Chris Hillman View Post
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?...oldid=53957524
This made my life a lot simpler, I didn't need to introduce that awkward second coordinate system (involving LaTeX Code: \\Theta ) with this approach.

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

LaTeX Code: T^{00} = \\rho(1+\\omega^2*r^2)
LaTeX Code: 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(_);
>
>
>
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Old May1-07, 08:27 PM       Last edited by pervect; May1-07 at 08:30 PM..            #10
pervect

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For those not fortunate enough to have GRtensor

What T looks like in the "frame field"

LaTeX Code: <BR>\\left[ \\begin {array}{cccc} \\rho&0&0&0\\\\\\noalign{\\medskip}0&0&0&0\\\\\\noalign{\\m  edskip}0&0&-{\\omega}^{2}\\rho\\,{r}^{2}&0<BR>\\\\\\noalign{\\medskip}0&0&0&0\\end {array} \\right] <BR>

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

T as it appears in cylindrical coordinates

LaTeX Code: <BR>\\left[ \\begin {array}{cccc}  \\left( {\\omega}^{2}{r}^{2}+1 \\right) \\rho&0&\\rho\\,\\omega&0\\\\\\noalign{\\medskip}0&0&0&0\\\\  \\noalign{\\medskip}<BR>\\rho\\,\\omega&0&0&0\\\\\\noalign{\\medskip}0&0&0&0\\end {array} \\right] <BR>
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Old May1-07, 09:19 PM       Last edited by Chris Hillman; May2-07 at 12:15 PM.. Reason: Better choice of dummy indices            #11
Chris Hillman

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Exclamation Frame components or coordinate basis components?

Originally Posted by pervect View Post
The final result I'm getting is that for (t,r,LaTeX Code: \\theta ,z) with t being represented by the 0 subscript, r by the 1 subscript, etc.

LaTeX Code: T^{00} = \\rho(1+\\omega^2*r^2)
LaTeX Code: 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 LaTeX Code: \\partial_r, \\, \\partial_z . That would make sense in terms of centrifugal "force" assuming your third frame vector is something like LaTeX Code: 1/r \\, \\partial_\\phi

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

Originally Posted by pervect View Post
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 velocityLaTeX Code: \\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.

For example, you could start with
LaTeX Code: ds^2 = -(dt-w\\, d\\phi)^2 + \\exp(2v) (dz^2+dr^2) + r^2 \\, d\\phi^2
LaTeX Code:  \\hspace{0.5in}<BR>= -dt^2 + 2\\, w \\, dt \\, d\\phi + \\exp(2v) (dz^2+r^2) + (r^2-w^2) \\, \\dphi^2,
LaTeX Code:  -\\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 LaTeX Code: \\partial_t which we naively assume has unit length, that is we assume that LaTeX Code: \\| \\partial_t \\|^2 = -1 . The spacelike Killing vector field is LaTeX Code: \\partial_\\phi and it has length LaTeX Code: \\| \\partial_\\phi \\|^2 = w^2-r^2 , where the function r is defined by LaTeX Code:  \\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 LaTeX Code: \\| \\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.

Read off the coframe
LaTeX Code: <BR>\\sigma^0 = -(dt-w\\,d\\phi), \\; <BR>\\sigma^1 = \\exp(v) \\, dz,  \\; <BR>\\sigma^2 = \\exp(v) \\, dr, \\;<BR>\\sigma^3 = r \\, d\\phi<BR>
Take the dual frame
LaTeX Code: <BR>\\vec{e}_0 = \\partial_t,\\; <BR>\\vec{e}_1 = \\exp(-v) \\, \\partial_z, \\;<BR>\\vec{e}_2 = \\exp(-v) \\, \\partial_r, \\;<BR>\\vec{e}_3 = \\frac{1}{r} \\; \\left( \\partial_\\phi + w \\, \\partial_t \\right)<BR>
and boost it by an undetermined amount (depending only on z,r) in the LaTeX Code: \\partial_\\phi direction. It is convenient to write
LaTeX Code: 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
LaTeX Code: <BR>\\vec{f}_0 = \\left( p + \\frac{\\sqrt{p^2-1}}{r} \\, w \\right) \\, \\partial_t<BR>+ \\frac{\\sqrt{p^2-1}}{r} \\, \\partial_\\phi, <BR>
LaTeX Code: <BR>\\vec{f}_1 = \\exp(-v) \\, \\partial_z, \\;<BR>\\vec{f}_2 = \\exp(-v) \\, \\partial_r, <BR>
LaTeX Code: <BR>\\vec{f}_3 = \\left( \\frac{p}{r} \\, w + \\frac{\\sqrt{p^2-1}}{r} \\right) \\, \\partial_t<BR>+ \\frac{p}{r} \\, \\partial_\\phi<BR>
Now LaTeX Code: \\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 LaTeX Code: \\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, LaTeX Code: c_0, \\, c_1, <BR>\\, c_2 \\, c_3 obey the unit contraint LaTeX Code: -c_0^2+c_1^2+c_2^2+c_3^2=-1 . For example LaTeX Code: \\vec{f}_0 = p \\, \\vec{e}_0 + (p^2-1) \\, \\vec{e}_3 above, and LaTeX Code: -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 LaTeX Code: 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 LaTeX Code: \\vec{f}_3 (and automatically orthogonal to LaTeX Code: \\vec{f}_0 since LaTeX Code: \\vec{e}_0 is a unit vector). The matter tensor has the form
LaTeX Code: <BR>T^{\\hat{m} \\hat{n}} = \\left[ \\begin{array}{cccc}<BR>e & 0 & 0 & 0 \\\\<BR>0 & a & f & 0 \\\\<BR>0 & f & b & 0 \\\\<BR>0 & 0 & 0 & c<BR>\\end{array} \\right]<BR>
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, LaTeX Code: a=f=b=0 , which is not sufficiently general for a fluid or for most elastic solids.

The assumption above that LaTeX Code: \\| \\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.)

(Additional: I'll try to add a bit more reconstruction):

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 LaTeX Code:  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, LaTeX Code: \\| \\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 LaTeX Code: 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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Old May1-07, 10:15 PM                  #12
Wallace

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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?
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Old May1-07, 10:26 PM       Last edited by Chris Hillman; May1-07 at 10:32 PM..            #13
Chris Hillman

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Hi, Wallace, see http://www.math.ucr.edu/home/baez/ph...igid_disk.html
Then see a tightly related quartet of WP articles I wrote:
http://en.wikipedia.org/w/index.php?...oldid=53957524
http://en.wikipedia.org/w/index.php?...oldid=51749949
http://en.wikipedia.org/w/index.php?...oldid=57888610
http://en.wikipedia.org/w/index.php?...oldid=58681705
See also
http://en.wikipedia.org/w/index.php?...oldid=45119659
http://en.wikipedia.org/w/index.php?...oldid=39874076
http://en.wikipedia.org/w/index.php?...oldid=45646663
See also Greg Egan's website.
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Old May1-07, 11:12 PM                  #14
Wallace

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Thanks Chris, lots of reading to keep me busy!
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Old May2-07, 12:38 AM       Last edited by pervect; May2-07 at 12:58 AM..            #15
pervect

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Originally Posted by Wallace View Post
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/physic...igid_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

LaTeX Code: \\int T^{00} dV , where dV is the volume element, in this case
dV = LaTeX Code: 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
LaTeX Code: 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 LaTeX Code: \\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 LaTeX Code: \\rho also stays constant.

We can then identify LaTeX Code: 2 \\pi \\, r dr \\, dz as the volume V of the wire, and LaTeX Code: \\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 + LaTeX Code: \\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 LaTeX Code: \\omega including SR corrections.
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Old May2-07, 12:07 PM       Last edited by Chris Hillman; May2-07 at 12:29 PM.. Reason: Add a caveat            #16
Chris Hillman

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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 LaTeX Code: \\omega \\, r rather than LaTeX Code: \\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/ph...igid_disk.html
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