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- 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.
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.
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