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- TL;DR
- I evaluate the total energy of a relativistic rotating disk that consists of non-interacting dust particles, all of which undergo uniform circular motion, by including the centripetal potential energy that binds them to circles.
(This post is inspired by: KE of rotating disk.)
A Single Particle in Uniform Circular Motion
In an inertial frame, I begin by examining a single dust particle of mass ##m## moving in a plane with instantaneous position vector ##\vec{r}\equiv r\hat{r}\left(t\right)##. Time-differentiating gives the velocity vector ##\vec{v}\equiv d\vec{r}/dt\equiv\dot{\vec{r}}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta}##, so the Lorentz-factor for the moving particle is ##\gamma\equiv\left(1-\vec{v}\cdot\vec{v}/c^{2}\right)^{-1/2}=\left(1-\left(\dot{r}^{2}+r^{2}\dot{\theta}^{2}\right)/c^{2}\right)^{-1/2}##. The relativistic dynamics of that particle as it moves in an external 4-scalar potential ##U## is governed by the general Lagrangian ##L=-\gamma^{-1}mc^{2}+U\left(\vec{r}\right)##, here restricted to planar motion in a centripetal, power-law potential:$$L=-mc^{2}\sqrt{1-\left(\frac{\dot{r}^{2}+r^{2}\dot{\theta}^{2}}{c^{2}}\right)}+kr^{n}\tag{1}$$From this I obtain two Euler-Lagrange (EL) equations, as well as the Hamiltonian (energy):$$0=\frac{\partial L}{\partial\vec{r}}-\frac{d}{dt}\left(\frac{\partial L}{\partial\vec{v}}\right)\;,\quad H=\vec{v}\cdot\frac{\partial L}{\partial\vec{v}}-L\tag{2a,b}$$Rather than boring the reader by writing out eqs.(2a,b) for general ##r(t),\theta(t)##, I proceed directly to evaluating them for uniform circular motion by setting ##\dot{r}=0,\theta=\omega t##, which simplifies the Lorentz-factor to ##\gamma=\left(1-\omega^{2}r^{2}/c^{2}\right)^{-1/2}##. For this simple motion, the ##\theta## EL equation degenerates to ##0=0##, while the ##r## EL equation becomes purely algebraic, resulting in the two equations that characterize the circling particle:$$0=\frac{mr\omega^{2}}{\sqrt{1-\frac{\omega^{2}r^{2}}{c^{2}}}}-knr^{n-1}\:,\quad H=\frac{mc^{2}}{\sqrt{1-\frac{\omega^{2}r^{2}}{c^{2}}}}+kr^{n}\tag{3a,b}$$The ##r## EL equation (3a) serves to determine the orbital radius ##r## in terms of the constants ##c,m,\omega,k,n##. By rewriting it as:$$kr^{n}=\frac{m\omega^{2}r^{2}}{n\sqrt{1-\omega^{2}r^{2}/c^{2}}}=\frac{mc^{2}}{n}\left(\gamma-\frac{1}{\gamma}\right)\tag{4}$$it's evident that as ##\gamma\rightarrow\infty## the radius ##r## either dilates or contracts, depending on the sign of the power ##n##:$$
\lim_{\gamma\rightarrow\infty}r=\begin{cases}
\infty & \left(n>0\right)\\
0 & \left(n<0\right)
\end{cases}
\tag{5}
$$In other words, in the ultrarelativistic limit the orbits of a single particle become either infinite or infinitesimal!
Now substituting eq.(4) into (3b), I arrive at the simple final form for the total relativistic energy of a particle in uniform circular motion:$$H=\frac{mc^{2}}{\sqrt{1-\frac{\omega^{2}r^{2}}{c^{2}}}}+\frac{mr^{2}\omega^{2}}{n\sqrt{1-\frac{\omega^{2}r^{2}}{c^{2}}}}=\frac{mc^{2}}{n}\left(\left(n+1\right)\gamma-\frac{1}{\gamma}\right)\tag{6}$$Observe that the centripetal potential energy can be either positive or negative based on the sign of ##n##.
The Low and High Speed Limits
The nonrelativistic limit of ##H## is completely sensible:$$H^{NR}\equiv\lim_{c\rightarrow\infty}H=mc^{2}+\frac{1}{2}mr^{2}\omega^{2}+\frac{1}{n}mr^{2}\omega^{2}\tag{7}$$because it exhibits the well-known combination of classical kinetic and potential energies. For example, if the centripetal force obeys Hooke's Law, then ##n=2## and the total energy is positive and evenly divided between kinetic and potential, just like the time-averaged energies of a harmonic oscillator. Or if ##n=-1##, the particle is bound by an inverse-square force (like a satellite in circular orbit) and the sum of the two energies is precisely the negative of the kinetic energy, so assigning less total energy to the orbiting mass than to the stationary mass. Similarly, as expected for the ultrarelativistic limit, the total energy diverges for almost every form of centripetal force as the particle approaches light-speed:$$
H^{UR}\equiv\lim_{\gamma\rightarrow\infty}H=\begin{cases}
\infty & \left(n\neq-1\right)\\
0 & \left(n=-1\right)
\end{cases}
\tag{8}$$The sole exception is the inverse-square force ##n=-1##: in that case, by eq.(6) the total energy of the particle vanishes as it approaches the speed of light. And what a peculiar system it is at that limit: a particle with finite mass traveling at the speed of light around a circle of zero radius (per eq.(5)) and carrying zero total energy!
The Rotating Disk
I construct a rotating disk by uniformly distributing identical particles in a plane, all of them circling around a common center, such that the resulting disk has radius ##R##, total mass ##M## and uniform mass density ##\rho\equiv M/\pi R^{2}##. In the continuum limit, an infinitesimal piece of the disk has an energy ##dH_{disk}## found by substituting ##m\rightarrow dm## into eq.(6) to get:$$dH_{disk}=\frac{c^{2}\left(n+\frac{\omega^{2}r^{2}}{c^{2}}\right)}{n\,\sqrt{1-\frac{\omega^{2}r^{2}}{c^{2}}}}dm\tag{9}$$where ##dm\equiv\rho dA=\left(M/\pi R^{2}\right)rdrd\theta##. Integrating this over the area of the disk, I obtain the total relativistic energy in a rotating disk of non-interacting dust:\begin{align}
H_{disk} & =\left(\frac{Mc^{2}}{\pi R^{2}}\right)\intop_{0}^{2\pi}d\theta\intop_{0}^{R}rdr\left(\frac{n+\frac{\omega^{2}r^{2}}{c^{2}}}{n\,\sqrt{1-\frac{\omega^{2}r^{2}}{c^{2}}}}\right) \nonumber \\
& = \frac{2}{3}Mc^{2}\left(\frac{c^{2}\left(3n+2\right)-\left(c^{2}\left(3n+2\right)+\omega^{2}R^{2}\right)\sqrt{1-\frac{\omega^{2}R^{2}}{c^{2}}}}{n\omega^{2}R^{2}}\right) \nonumber\tag{10}
\end{align}The nonrelativistic limit of eq.(10) is clearly satisfactory:$$H_{disk}^{NR}\equiv\lim_{c\rightarrow\infty}H_{disk}=Mc^{2}+\frac{1}{4}M\omega^{2}R^{2}+\frac{1}{2n}M\omega^{2}R^{2}=Mc^{2}+\frac{1}{2}I\omega^{2}+\frac{1}{2n}M\omega^{2}R^{2}\tag{11}$$since ##I\equiv\frac{1}{2}MR^{2}## is indeed the classical moment-of-inertia for a disk of uniform density. At the other extreme is the ultrarelativistic limit, wherein the disk edge moves at the speed-of-light:$$H_{disk}^{UR}\equiv\lim_{\omega R\rightarrow c}H_{disk}=\frac{2}{3}Mc^{2}\left(\frac{3n+2}{n}\right)\tag{12}$$ In this limit, the total energy of a rotating dust-disk of finite radius ##R## is clearly non-zero finite over a wide range of choices for ##n## and hence its better behaved in that regard than an isolated dust particle traveling in a circle near light-speed. But the disk energy can still be made to take arbitrarily large positive or negative values by picking centripetal potentials having ##n## near enough to zero.
Of course, this is all merely a theoretical exercise as I know of no physics that permits the sculpting of centripetal potentials or implements experiments with relativistically rotating dust. But hopefully I've at least given some readers some ideas to somewhat think about sometimes.
A Single Particle in Uniform Circular Motion
In an inertial frame, I begin by examining a single dust particle of mass ##m## moving in a plane with instantaneous position vector ##\vec{r}\equiv r\hat{r}\left(t\right)##. Time-differentiating gives the velocity vector ##\vec{v}\equiv d\vec{r}/dt\equiv\dot{\vec{r}}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta}##, so the Lorentz-factor for the moving particle is ##\gamma\equiv\left(1-\vec{v}\cdot\vec{v}/c^{2}\right)^{-1/2}=\left(1-\left(\dot{r}^{2}+r^{2}\dot{\theta}^{2}\right)/c^{2}\right)^{-1/2}##. The relativistic dynamics of that particle as it moves in an external 4-scalar potential ##U## is governed by the general Lagrangian ##L=-\gamma^{-1}mc^{2}+U\left(\vec{r}\right)##, here restricted to planar motion in a centripetal, power-law potential:$$L=-mc^{2}\sqrt{1-\left(\frac{\dot{r}^{2}+r^{2}\dot{\theta}^{2}}{c^{2}}\right)}+kr^{n}\tag{1}$$From this I obtain two Euler-Lagrange (EL) equations, as well as the Hamiltonian (energy):$$0=\frac{\partial L}{\partial\vec{r}}-\frac{d}{dt}\left(\frac{\partial L}{\partial\vec{v}}\right)\;,\quad H=\vec{v}\cdot\frac{\partial L}{\partial\vec{v}}-L\tag{2a,b}$$Rather than boring the reader by writing out eqs.(2a,b) for general ##r(t),\theta(t)##, I proceed directly to evaluating them for uniform circular motion by setting ##\dot{r}=0,\theta=\omega t##, which simplifies the Lorentz-factor to ##\gamma=\left(1-\omega^{2}r^{2}/c^{2}\right)^{-1/2}##. For this simple motion, the ##\theta## EL equation degenerates to ##0=0##, while the ##r## EL equation becomes purely algebraic, resulting in the two equations that characterize the circling particle:$$0=\frac{mr\omega^{2}}{\sqrt{1-\frac{\omega^{2}r^{2}}{c^{2}}}}-knr^{n-1}\:,\quad H=\frac{mc^{2}}{\sqrt{1-\frac{\omega^{2}r^{2}}{c^{2}}}}+kr^{n}\tag{3a,b}$$The ##r## EL equation (3a) serves to determine the orbital radius ##r## in terms of the constants ##c,m,\omega,k,n##. By rewriting it as:$$kr^{n}=\frac{m\omega^{2}r^{2}}{n\sqrt{1-\omega^{2}r^{2}/c^{2}}}=\frac{mc^{2}}{n}\left(\gamma-\frac{1}{\gamma}\right)\tag{4}$$it's evident that as ##\gamma\rightarrow\infty## the radius ##r## either dilates or contracts, depending on the sign of the power ##n##:$$
\lim_{\gamma\rightarrow\infty}r=\begin{cases}
\infty & \left(n>0\right)\\
0 & \left(n<0\right)
\end{cases}
\tag{5}
$$In other words, in the ultrarelativistic limit the orbits of a single particle become either infinite or infinitesimal!
Now substituting eq.(4) into (3b), I arrive at the simple final form for the total relativistic energy of a particle in uniform circular motion:$$H=\frac{mc^{2}}{\sqrt{1-\frac{\omega^{2}r^{2}}{c^{2}}}}+\frac{mr^{2}\omega^{2}}{n\sqrt{1-\frac{\omega^{2}r^{2}}{c^{2}}}}=\frac{mc^{2}}{n}\left(\left(n+1\right)\gamma-\frac{1}{\gamma}\right)\tag{6}$$Observe that the centripetal potential energy can be either positive or negative based on the sign of ##n##.
The Low and High Speed Limits
The nonrelativistic limit of ##H## is completely sensible:$$H^{NR}\equiv\lim_{c\rightarrow\infty}H=mc^{2}+\frac{1}{2}mr^{2}\omega^{2}+\frac{1}{n}mr^{2}\omega^{2}\tag{7}$$because it exhibits the well-known combination of classical kinetic and potential energies. For example, if the centripetal force obeys Hooke's Law, then ##n=2## and the total energy is positive and evenly divided between kinetic and potential, just like the time-averaged energies of a harmonic oscillator. Or if ##n=-1##, the particle is bound by an inverse-square force (like a satellite in circular orbit) and the sum of the two energies is precisely the negative of the kinetic energy, so assigning less total energy to the orbiting mass than to the stationary mass. Similarly, as expected for the ultrarelativistic limit, the total energy diverges for almost every form of centripetal force as the particle approaches light-speed:$$
H^{UR}\equiv\lim_{\gamma\rightarrow\infty}H=\begin{cases}
\infty & \left(n\neq-1\right)\\
0 & \left(n=-1\right)
\end{cases}
\tag{8}$$The sole exception is the inverse-square force ##n=-1##: in that case, by eq.(6) the total energy of the particle vanishes as it approaches the speed of light. And what a peculiar system it is at that limit: a particle with finite mass traveling at the speed of light around a circle of zero radius (per eq.(5)) and carrying zero total energy!
The Rotating Disk
I construct a rotating disk by uniformly distributing identical particles in a plane, all of them circling around a common center, such that the resulting disk has radius ##R##, total mass ##M## and uniform mass density ##\rho\equiv M/\pi R^{2}##. In the continuum limit, an infinitesimal piece of the disk has an energy ##dH_{disk}## found by substituting ##m\rightarrow dm## into eq.(6) to get:$$dH_{disk}=\frac{c^{2}\left(n+\frac{\omega^{2}r^{2}}{c^{2}}\right)}{n\,\sqrt{1-\frac{\omega^{2}r^{2}}{c^{2}}}}dm\tag{9}$$where ##dm\equiv\rho dA=\left(M/\pi R^{2}\right)rdrd\theta##. Integrating this over the area of the disk, I obtain the total relativistic energy in a rotating disk of non-interacting dust:\begin{align}
H_{disk} & =\left(\frac{Mc^{2}}{\pi R^{2}}\right)\intop_{0}^{2\pi}d\theta\intop_{0}^{R}rdr\left(\frac{n+\frac{\omega^{2}r^{2}}{c^{2}}}{n\,\sqrt{1-\frac{\omega^{2}r^{2}}{c^{2}}}}\right) \nonumber \\
& = \frac{2}{3}Mc^{2}\left(\frac{c^{2}\left(3n+2\right)-\left(c^{2}\left(3n+2\right)+\omega^{2}R^{2}\right)\sqrt{1-\frac{\omega^{2}R^{2}}{c^{2}}}}{n\omega^{2}R^{2}}\right) \nonumber\tag{10}
\end{align}The nonrelativistic limit of eq.(10) is clearly satisfactory:$$H_{disk}^{NR}\equiv\lim_{c\rightarrow\infty}H_{disk}=Mc^{2}+\frac{1}{4}M\omega^{2}R^{2}+\frac{1}{2n}M\omega^{2}R^{2}=Mc^{2}+\frac{1}{2}I\omega^{2}+\frac{1}{2n}M\omega^{2}R^{2}\tag{11}$$since ##I\equiv\frac{1}{2}MR^{2}## is indeed the classical moment-of-inertia for a disk of uniform density. At the other extreme is the ultrarelativistic limit, wherein the disk edge moves at the speed-of-light:$$H_{disk}^{UR}\equiv\lim_{\omega R\rightarrow c}H_{disk}=\frac{2}{3}Mc^{2}\left(\frac{3n+2}{n}\right)\tag{12}$$ In this limit, the total energy of a rotating dust-disk of finite radius ##R## is clearly non-zero finite over a wide range of choices for ##n## and hence its better behaved in that regard than an isolated dust particle traveling in a circle near light-speed. But the disk energy can still be made to take arbitrarily large positive or negative values by picking centripetal potentials having ##n## near enough to zero.
Of course, this is all merely a theoretical exercise as I know of no physics that permits the sculpting of centripetal potentials or implements experiments with relativistically rotating dust. But hopefully I've at least given some readers some ideas to somewhat think about sometimes.