Stress-energy tensor of a rotating rod

[Ibix]

TL;DR Summary

Various threads recently set me thinking about the stress-energy tensor of a rotating rod. It seemed relatively straightforward to write down, up to integrals that depend on the material model. Is the following hopelessly naive?

A rod rotates freely (edit: about an axis perpendicular to its length) in empty space. Working in an inertial coordinate system where the rod rotates around a fixed point, the rod is straight, of length $2L$ in its spinning state, and its mass distribution is symmetric along its length. The rod's angular velocity is $\omega$.

We can choose to work in cylindrical polars $(t,r,\phi,z)$ such that the rod rotates in the $r-\phi$ plane, centred at the origin. The basis vectors are orthonormal, so it's possible to interpret the components of the stress-energy tensor, $T^{\mu\nu}$, as the components of four-momentum density (in the case of $T^{0\nu}$ and $T^{\mu 0}$) and of the Cauchy stress tensor (in the case of $T^{ij}$, where $i,j=1,2,3$), as measured by someone at rest in our coordinate system.

The four-momentum density is just the four-momentum divided by volume. In this case, that means that inside the rod $T^{00}=\gamma\rho$ and $T^{20}=T^{02}=\gamma\rho \omega r$, where $\gamma=(1-\omega^2r^2)^{-1/2}$. The only stress in the rod is a radial tension, which means that $T^{11}$ is non-zero. All other components are zero. So we have
$$T^{\mu\nu}=\left(
\begin{array}{cccc}
\gamma\rho&0&\gamma\rho \omega r&0\\
0&T^{11}&0&0\\
\gamma\rho\omega r&0&0&0\\
0&0&0&0
\end{array}\right)$$inside the rod. Obviously, $T^{\mu\nu}=0$ outside the rod.

The only question remaining is what is $T^{11}$? In Newtonian physics we'd simply note that the centripetal force per unit volume needed to keep an element of the rod at radius $r'$ circling is $\rho\omega^2 r'$, then integrate this over $r'>r$ to obtain the tension in the rod at $r$. I think the only difference in relativity is that the centripetal force is $\gamma^2\rho\omega^2r'$, and hence $$T^{11}=\omega^2\int_r^L\gamma^2(r')\rho(r')r'dr'$$where $L$ is the half-length of the rod. We can't do this integral without specifying a mass density for the rod, which would typically depend on its material properties since the rod is under stress. And hence we'd probably have to rewrite it as a differential equation because the mass density will depend on the stress via Hooke's law (or whatever material model we pick).

So there it is - if you specify a material model (or state a mass density by fiat) and the shape of the rod, the above is what I think is the stress-energy tensor for a rotating rod. I'd be interested in any comments.

 

[PeterDonis]

Working in an inertial coordinate system where the rod rotates around a fixed point

A fixed point or a fixed axis? It seems to be the latter.

 

[Ibix]

A fixed point or a fixed axis? It seems to be the latter.

Around a fixed axis, the z-axis, yes. I thought that was clear from the "rotates in the $r-\phi$ plane" comment (?).

 

[PeterDonis]

I thought that was clear from the "rotates in the $r-\phi$ plane" comment

Taken at face value, that comment would imply that we don't have a rod (cylinder) but a disk (plane). There isn't a single $r-\phi$ plane that the rod occupies; there is a whole stack of them.

 

[Ibix]

Taken at face value, that comment would imply that we don't have a rod (cylinder) but a disk (plane). There isn't a single $r-\phi$ plane that the rod occupies; there is a whole stack of them.

I'm not sure I see what you're getting at, but that's obviously coloured by the fact that I know exactly what I meant.

Anyway - I was intending to describe a rod whose ends are at $(t,r,\phi,z)=(t,L,\omega t,0)$ and $(t,L,\pi+\omega t,0)$. I think that's unambiguous...

 

[PeterDonis]

I was intending to describe a rod whose ends are at $(t,r,\phi,z)=(t,L,\omega t,0)$ and $(t,L,\pi+\omega t,0)$.

Ah, I see; I was imagining a different rotation than you are.

 

[Ibix]

Ah, I see; I was imagining a different rotation than you are.

Yes - I see that I didn't say the rod was rotating about an axis perpendicular to its length. I'll edit something in...

 

[pervect]

I don't think I'd get quite the same stress-energy tensor. I haven't had time to work it out in detail, but it's similar to the stress-energy tensor of a relativistic rotating hoop, which I have worked out in https://www.physicsforums.com/threads/relativistic-hoop-with-constant-circumference.976746/

Borrowing from this, we can define a line elelment and a set of orthonormal basis vectors in 2 d cylindrical coordinates as I did earlier, the Langevian basis vectors. See also the original post for the inspiration by Egan.

The line element is (using geometric units) is just

$$-dt^2 + dr^2 + r^2 d\phi^2$$

The basis vectors in terms of these coordinates

$$u = \frac{1}{\sqrt{1-r^2\omega^2}} \left( \partial_t + \omega \, \partial_\phi \right) \quad w = \frac{1}{\sqrt{1-r^2\omega^2}} \left( \omega \, r\,\partial_t + \frac{1}{r} \partial_\phi \right) \quad q = \partial_r$$

With respect to these basis vectors, we can write for the rod case

$$T = \rho \, u \otimes u + P \, q \otimes q $$

Basically, the tension is radial, the basis vector I called q in this example, where it's circumfrential (the basis vector I call w) in the rotating hoop.

We then need to solve for P, by solving $\nabla_a T^{ab}=0$ as before, where $\nabla_a$ is defined from our line element in the usual manner. I haven't done this, though.

 

[pervect]

I ran through some calculations based on the above.

In our orthonormal basis [u,q,w] the stress-energy tensor is just

$$\begin{bmatrix} \rho & 0 & 0 \\ 0 & P(r) & 0 \\ 0 & 0 & 0 \end{bmatrix} $$

This is just the matrix representation of

$$T = \rho \, u \otimes u + P(r) \, q \otimes q$$

In a coordinate basis $[\partial_t, \partial_r , \partial_\phi]$ our stress energy tensor has a matrix representation of

$$\begin{bmatrix} \frac{\rho}{1-r^2\omega^2} & 0 & \frac{\omega \rho}{{1-r^2\omega^2}} \\ 0 & P(r) & 0 \\ \frac{\omega \rho} {{1-r^2\omega^2}} & 0 & \frac{\omega^2 \rho}{1-r^2\omega^2} \end{bmatrix}$$

This is done simply by multiplication and our defintion of u,q, and w, which I gave in the previous post.

The basis vectors in terms of these coordinates

$$u = \frac{1}{\sqrt{1-r^2\omega^2}} \left( \partial_t + \omega \, \partial_\phi \right) \quad w = \frac{1}{\sqrt{1-r^2\omega^2}} \left( \omega \, r\,\partial_t + \frac{1}{r} \partial_\phi \right) \quad q = \partial_r$$

Note that wiki also has these basis vectors, referred to as the basis vectors of a Langevin observer, in https://en.wikipedia.org/w/index.ph...6#Langevin_observers_in_the_cylindrical_chart

The main reason for bothering to find the stress-energy tensor in a coordinate basis is so we can compute $\nabla_a T^{ab}$. To me it has little physical significance, the stress-energy tensor in the [u,q,w] basis is much more meaningful. But it's easiest to calculate the covariant divergence in a coordinate basis, so it is convenient to convert it.

The end result of this is the differential equation for P(r) that we need to solve, with the appropriate boundary conditions. I believe that it's appropriate to make P(r) = 0 at the end of the bar.

Doing this calculation of $\nabla_a T^{ab}$, I get for the differential equation

$$ P(r)/r + \partial_r P(r) - \rho \frac{r^2 \omega^2}{1-r^2\omega^2} = 0$$

I assumed that $\rho$ was constant. It'd be possible to redo the calculation to make $\rho$ a function of r, but I didn't do that.

I rushed this out, and I find I make many more errors nowadays :(. But these are the results I get, for comparison purposes.

 

[pervect]

A few comments on some of the differences with Ibix's results. As far as $T^{00}$ goes, Ibix has a factor of $\gamma$ where I have a factor of $\gamma^2$. Here $\gamma = \frac{1}{\sqrt{1-v^2}}$ and $v = r \omega$.

In the lab frame, Lorentz contraction would make the rotating rod "squash" in the direction of moment, so the cross section of the rod wouldn't be constant in the lab frame. If one wanted to use the tensor to calculate the total energy by integrating $T^{00}$, in my case one would have to take into account this changing cross-section.

The vector $\partial_\phi$ isn't of unit length. It's not really clear to me what basis vectors Ibix used. Perhaps he used a unit length vector in the lab frame, $\frac{1}{r} \partial_\phi$, rather than the coordinate basis vector? That difference probably accounts for the discrcepancy factor of "r" in the momentum term. I used the coordinate basis vectors, in spite of them not being unit length, as in my view they are mainly a calculational tool to get the continuity equation for P(r). We've already addressed the factor of $\gamma^2$ vs $\gamma$ discrepancy.

Finally, Ibix omitted the $T^{22}$ term that I had. I suspect is just an omission on his part, though it may not matter for many purposes (like finding the energy and angular momentum of the rod in the lab frame).

 

[pervect]

One more basis:

If we let $\hat{t} = \partial_t$, $\hat{r} = \partial_r$, and $\hat{\phi} = \frac{1}{r} \partial_\phi$, we have an orthonormal basis $[\hat{t}, \hat{r}, \hat{\phi}]$ in the lab frame.

This is a convenient basis for physical interpretation in the lab frame, because the vectors are othornormal. (But it's not a coordinate basis, so it's not convenient for calculating the divergence of the stress energy tensor).

Then in matrix notion we can write:

$$\begin{bmatrix} \frac{\rho}{1-r^2\omega^2} & 0 & \frac{r \omega \rho}{{1-r^2\omega^2}} \\ 0 & P(r) & 0 \\ \frac{r \omega \rho} {{1-r^2\omega^2}} & 0 & \frac{r^2 \omega^2 \rho}{1-r^2\omega^2} \end{bmatrix}$$

which we can write in index free notation as

$$ \frac{\rho}{1-r^2\omega^2} \hat{t} \otimes \hat{t} + \frac{r \omega \rho}{{1-r^2\omega^2}} \left( \hat{t} \otimes \hat{\phi} + \hat{\phi} \otimes \hat{t} \right) + \frac{r^2 \omega^2 \rho}{1-r^2\omega^2} \hat{\phi} \otimes \hat{\phi} + P(r) \hat{r} \otimes \hat{r}$$

This is closest to Ibix's result, though we still have the extra factor of $\gamma$, and the extra $T^{22}$ term.

 

[Ibix]

Thanks pervect. I need to read through that carefully before I respond properly.

I will say I don't immediately see why you've got a $T^{22}$. I see where it appeared in your maths, but I thought this would represent tangential stresses which I expected to be zero in the rod (obviously not in your hoop). So it's not an oversight that it's not there in my version. Maybe wrong, but not an oversight.

 

[pervect]

Thanks pervect. I need to read through that carefully before I respond properly.

I will say I don't immediately see why you've got a $T^{22}$. I see where it appeared in your maths, but I thought this would represent tangential stresses which I expected to be zero in the rod (obviously not in your hoop). So it's not an oversight that it's not there in my version. Maybe wrong, but not an oversight.

Try computing the SET of a moving mass. It's $\rho \, u \otimes u$, where u is the 4-velocity. You'll see a similar term.

You can also compute the SET via the change of basis formula: $T^{c'd'} = \Lambda^{c'}{}_a \Lambda^{d'}{}_b T^{ab}$, where $\Lambda$ is the Lorentz boost. In two dimensions:

$$\Lambda = \begin{bmatrix} \gamma & -\gamma \beta \\ -\gamma \beta & \gamma \end{bmatrix}$$

 

[PeterDonis]

I thought this would represent tangential stresses which I expected to be zero in the rod

The tangential stress is zero in the rod in the orthonormal frame, since that frame is moving with the rod (i.e., it is the rod's momentarily comoving rest frame). But the tangential stress is not zero in the coordinate frame, since the rod is moving in that frame, and therefore exerts pressure in the tangential direction.

 

[Ibix]

Try computing the SET of a moving mass. It's $\rho \, u \otimes u$, where u is the 4-velocity. You'll see a similar term.

The tangential stress is zero in the rod in the orthonormal frame,

Right. In pervect's example, which can describe dust, the term that emerges is the wind of passage felt by someone moving through. There's an analogous concept with the rod, which is the pressure you'll feel in your fingers if you try to stop it rotating.

I thought that was covered by the $02$ and $20$ terms, but apparently not.

 

[pervect]

I thought that was covered by the $02$ and $20$ terms, but apparently not.

Nope. To repeat my earlier point in more general language, the power of tensors is that if you know the value of the components in one coordinate system or frame of reference, you can compute the components in arbitrary frame of refrences.

Specifically, if you know the stress-energy tensor in some particular inertial frame of reference is

$$\begin{bmatrix} \rho & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

you can find out what the components of the stress-energy tensor are in anoter inertial frame that is moving relative to the first. This is sometimes called a "passive" transformation, or a passive boost.

There are various methods one might use to carry out the transform, I am basically suggesting picking one and working through the math. I've demonstratd a few ways I do it with algebra, but feel free to use whatever method you like, perhaps one that your favorite textbook uses.

The next two cases I'd suggest analyze how pure tension / stress/ pressuretransforms. I.e a stress-energy tensor like:

$$\begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & P & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}$$

There are two sub cases to analyze in this case - when the boost is in transverse to the tension (which is the case in the rotating rod, and is the simpler case), and when the boost is parallel to the tension (which is the case for the rotating hoop, the answers are not so trivial).

[add]
For the fisrt problem, one only needs 1 space + 1 time dimesion. The transverse case for the second problem needs a transvere spatial dimension, of course, so you need at least 2 spatial dimensions so that one can be transverse to the direction of motion.

 

[Ibix]

@pervect - thank you so much for taking the time to go through all of that. It's a huge help, and your analysis of my errors appears to be spot on. I'll just say that you don't need to sell tensor methods to me - I'm just apparently not very good (edit: or, being kinder to myself, just not very experienced) at actually applying them to analysing the SET.

I was indeed attempting to write down the SET in the orthonormal basis used by non-rotating observers, and failed to do so correctly. (Side note - I don't think I specified my basis clearly, which is the GR equivalent of the perennial "please specify which frame is measuring what" in SR undergrad problems, I think.) We've already discussed $T^{22}$, which I failed to include at all. And you are correct that when I wrote down $T^{0i}$ and $T^{i0}$ I wrote down the momentum of the elementary part of the rod, but neglected to correct for the length contraction of the volume element.

I tried to reproduce your maths, and agree except for two points. First, I think that, in the coordinate basis, your signs are wrong in the $T^{02}$ and $T^{20}$ elements given the Langevin basis you specify. The Langevin basis appears to be defined with an inverse Lorentz transform from the coordinate basis (there's a plus sign between the $\partial_t$ and $\partial_\phi$ terms in your definition) so the $t\phi$ components of the transform to the coordinate basis have negative signs, I think. I don't think this affects anything beyond the handedness of one of our bases.

Second, the differential equation I get is $$\partial_rP+\frac Pr-\rho\frac{r\omega^2}{1-r^2\omega^2}=0$$Edit: added missing $P$ to first term. You have an $r^2$ in the numerator of the last term. I think mine's correct on dimensional analysis grounds, but I've been known to make mistakes...

Again, thank you, and any cimments welcome.

 

[PeterDonis]

I think that, in the coordinate basis, your signs are wrong in the $T^{02}$ and $T^{20}$ elements given the Langevin basis you specify. The Langevin basis appears to be defined with an inverse Lorentz transform from the coordinate basis

The signs @pervect gave for $T^{02}$ and $T^{20}$ in the coordinate basis look right to me given the physics: in the non-rotating coordinates the rod is rotating in the positive $\phi$ direction, so its momentum and energy flux in that direction should be positive.

The expressions for the Langevin basis vectors in terms of the coordinate basis vectors are not the same as an inverse Lorentz transform. The plus signs in those expressions are correct given the physics I just described (since the rod is rotating in the positive $\phi$ direction, the expressions for the Langevin basis vectors in terms of the coordinate basis vectors should have positive $\partial_\phi$ components). But those are expressions for the basis vectors, not the coordinates. The transformations for the coordinates will have opposite signs.

As an exercise, try working this out for two inertial frames in Minkowski spacetime. You have a primed frame whose coordinates are given by the usual Lorentz transforms from the unprimed frame, with minus signs. Now work out the expressions for the primed basis vectors $\partial_{t^\prime}$ and $\partial_{x^\prime}$ in terms of the unprimed basis vectors $\partial_t$ and $\partial_x$. What signs do they have?

 

[pervect]

Second, the differential equation I get is $$\partial_r+\frac Pr-\rho\frac{r\omega^2}{1-r^2\omega^2}=0$$You have an $r^2$ in the numerator of the last term. I think mine's correct on dimensional analysis grounds, but I've been known to make mistakes...

Again, thank you, and any cimments welcome.

I'm not too worried about a possible sign error, as physically that just depends on the direction of rotation. However, the second error you point out is indeed an error. My worksheet has the same answer as you wrote, so it appears to be a typo on my part.

$$\partial_r+\frac Pr-\rho\frac{r\omega^2}{1-r^2\omega^2}=0$$

This is more or less what one expects from taking the Newtonian limit, with $r \omega << 1$.

 

[pervect]

I don't see a sign error either.

If one wants to work out the problem from first principles without using the derivation of the Langevin basis vectors, one simply defines the dimensonless velocity $\beta$and the associated $\gamma$ factor as:

$$ \beta = r \omega \quad \gamma = 1/\sqrt{1-\beta^2} $$

One needs to pick a basis, so we pick $\hat{t} = \partial_t$ as a unit vector in the t direction in the lab frame, $\hat{\phi} = \frac{1}{r} \partial_{\phi}$ as a unit vector in the $\phi$ direction, and $\hat{r} = \partial_r$ as a unit vector in the r direction for our basis.

Then one can write the 4 velocity u and the stress energy tensor T in the chosen basis knowing the 4-velocity u as:

$$u = \gamma \, \hat{t} + \gamma \beta \, \hat{\phi} \quad T = \rho \, u \otimes u + P(r) \, \hat{r} \otimes \hat{r} $$

which expands out via algebra into the stress energy tensor in the $\hat{t}, \hat{\phi}, \hat{r}$ basis as:

$$ T = \gamma^2 \rho \, \hat{t} \otimes \hat{t} + \beta \gamma^2 \rho \, \left( \hat{t} \otimes \hat{\phi} + \hat{\phi} \otimes \hat{t} \right) + \beta^2 \gamma^2 \rho \, \hat{\phi} \otimes \hat{\phi} + P(r) \, \hat{r} \otimes \hat{r}$$

If $\beta$ is defined as positive, then $T^{01}$ is positive, as one expects. This is true of the rod is rotating clockwise with the usual polar coordinate conventions.

 

[Ibix]

My worksheet has the same answer as you wrote, so it appears to be a typo on my part.

Maxima has a tex() function that converts a result into rather ugly TeX, which I usually cut and paste precisely to avoid this kind of thing (I didn't follow my own advice this time, which is why the first term in my version is just $\partial_r$ instead of $\partial_r P$, which is what it should be). I don't know if Maple or Mathematica have a similar function, but it wouldn't surprise me.

I'll take a look at the basis change - I'm obviously missing something. My guess is I'm somehow mixing up coordinate and basis changes, as Peter seems to be saying. Edit: Peter's physical argument is compelling - I just need to understand what bit I'm transforming wrong.

 

[Ibix]

Right - I see what I did. I am correct, I think, that $\partial_t$ and $\partial_\phi$ can be expressed in terms of the Langevin basis vectors $u$ and $w$ using a forward Lorentz transform (or at least, something algebraically identical). But what I'm interested in, the components of a tensor, must transform the opposite way to the basis in order that the tensor itself be invariant under the basis change.

All of which I knew from chapter 2 of Carroll, so I'm a little annoyed I didn't get it right. I agree that the sign error is mine.