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jdstokes
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This problem comes from the GR problem book, question 5.1 (b).
The setup is a ring of N identical masses rotating in the x-y plane at a distance [itex]a[/itex] from the origin. Assume that N is large enough so the ring may be treated as continuous. Ignore the stress energies keeping the particles in orbit.
According to the solution, the energy density of the ring is
[itex]T^{00} = \frac{Nm\delta(r - a)\delta(z)}{2\pi a\gamma}[/itex].
I'm having trouble understanding why this is the case. It is certainly true that for a stationary ring the rest-energy density is
[itex]\rho_0 = \frac{Nm\delta(r-a)\delta(z)}{2\pi a}[/itex]
which can be easily seen by integrating over all of space in cylindrical coordinates. Alternatively it can be shown using the fact that [itex]\rho_0 2\pi r dr dz = Nm[/itex] and then dividing through by the infinitesimal [itex]2\pi r dr dz[/itex].
Why would it be that the energy density decreases as the angular velocity is increased?
The setup is a ring of N identical masses rotating in the x-y plane at a distance [itex]a[/itex] from the origin. Assume that N is large enough so the ring may be treated as continuous. Ignore the stress energies keeping the particles in orbit.
According to the solution, the energy density of the ring is
[itex]T^{00} = \frac{Nm\delta(r - a)\delta(z)}{2\pi a\gamma}[/itex].
I'm having trouble understanding why this is the case. It is certainly true that for a stationary ring the rest-energy density is
[itex]\rho_0 = \frac{Nm\delta(r-a)\delta(z)}{2\pi a}[/itex]
which can be easily seen by integrating over all of space in cylindrical coordinates. Alternatively it can be shown using the fact that [itex]\rho_0 2\pi r dr dz = Nm[/itex] and then dividing through by the infinitesimal [itex]2\pi r dr dz[/itex].
Why would it be that the energy density decreases as the angular velocity is increased?
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