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While searching for some nice and simple problems in Newton's theory of gravity, I've made this trivial calculation for which a proper physical interpretation is lacking. Some pieces are missing and I'm puzzled by the result, so here it is.

Consider a self-gravitating sphere of uniform density, of conserved mass $M$ and initial radius $R_0$. The sphere is initialy rotating with angular velocity $\omega_0$. The total mechanical energy is thus \begin{equation}\tag{1} E_0 = K_{\text{rot}} + U_{\text{sphere}} = \frac{1}{2} \, I_0 \, \omega_0^2 - \frac{3 G M^2}{5 R_0}, \end{equation} where $I_0 = \frac{2}{5} \, M R_0^2$ is the initial moment of inertia of the uniform sphere. The spin angular momentum of the sphere is simply \begin{equation}\tag{2} S_0 = I_0 \, \omega_0. \end{equation}

For some reason, the sphere change its radius and angular velocity (imploding or exploding). The radius is now $R \ne R_0$ and angular velocity $\omega \ne \omega_0$. The end mechanical energy and angular momentum are now these (the sphere is still of uniform density for simplicity) : \begin{align} E &= \frac{1}{2} \, I \, \omega^2 - \frac{3 G M^2}{5 R}, \tag{3} \\[12pt] S &= I \, \omega. \tag{4} \end{align}

If I impose strict conservation of BOTH mechanical energy and angular momentum : $E = E_0$ and $S = S_0$, I then get two constraints for two variables : $R$ and $\omega$. Solving these is trivial and give this : \begin{align} R &= \frac{R_0}{\frac{3 G M}{\omega_0^2 \, R_0^3} - 1}, \tag{5} \\[12pt] \omega &= \frac{R_0^2}{R^2} \, \omega_0 = \Big( \frac{3 G M}{\omega_0^2 \, R_0^3} - 1 \Big)^2 \, \omega_0. \tag{6} \end{align} Since $R > 0$, this solution is possible only if \begin{equation}\tag{7} \omega_0 < \sqrt{\frac{3 G M}{R_0^3}}, \end{equation} which is the same as saying $E < 0$. We can distinguish two cases for the final radius of the sphere, according to (5) : \begin{align} &R < R_0 \quad \text{if} \quad \omega_0 < \sqrt{\frac{3 G M}{2 R_0^3}}, \\[12pt] &R > R_0 \quad \text{if} \quad \sqrt{\frac{3 G M}{2 R_0^3}} < \omega_0 < \sqrt{\frac{3 G M}{R_0^3}}. \end{align}

This result is puzzling to me. It says that if (for some unspecified reason) a rotating uniform sphere change its size, conservation of energy and conservation of spin angular momentum gives a unique solution for its final state. If I put in the actual values for the Sun, for $M$, $R_0$ and $\omega_0$, I get a size $R$ and angular velocity $\omega$ of the same order of magnitude as a pulsar (very roughly).

The change must be brutal (no gradual or continuous change of $R$ and $\omega$), which is hard to accept !

In this case, my maths are okay but my physics is terrible (as Einstein once told Lemaître, but the other way around !)

So what is the physical interpretation of all this ? What am I missing ? Where are the mistakes in this reasoning ? What the maths are actually saying ? I don't get it and I'm confused !

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the fundamental flaw in the derivation according to me is the conservation of energy.You have to supply work when you contract or expand the sphere.Take the example of a ballerina dancing .Suppose she spreads her arms such that the her angular momentum is constant.You can instantly see that her total energy $1/2Iw^2$ is not conserved because she had to do work to spread her arms but those forces were internal and cannot change her net angular momentum.

Basically, you would also have to take into account the radial velocity of the mass distribution in your question.You can write that $a_{grav}\hat r=(\ddot r-rw^2) \hat r+(2\dot r\dot \theta+r\ddot \theta) \hat\theta $.You can calculate the accelearation due to gravity on a mass on hollow sphere by simple integration that i will leave as an exercise for you. Now, the theta component =0 will give you conservation of angular momentum and the other part could be solved for $\dot r$.Then ,in the question you would have to put in the additional kinetic energy part for the conservation.Assuming that the sphere is free to move,there would be a tug of war between centrifugal force and gravity .Centrifugal force would most probably win.The initial conditions for this question's are $r(0)=r_0$,$\dot r(0)=0,\dot{\theta}(0)=w_0$ and $\theta(0)$ is arbitrary.If you don't take into account the full motion of the sphere normally,you would have to provide an additional force,which will be your case.But then,you cannot apply the conservation of energy without this force work added.

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My physics is less terrible this morning. I have a solution to my own question :

Yes, energy as defined by (3) cannot be conserved. If the sphere changed its radius, it's because something changed internally (obviously !). Some internal force is resisting gravity, until it changed for some reason. So there are at least 2 ways to show this.

First solution (mechanical) : Suppose there's a kind of repulsive internal force (pressure, springs, wathever) that counter-balance gravity. Then its energy potential should be added, if it can be described by a potential. I'm using a "spring" energy potential as an example : \begin{equation}\tag{1} E = \frac{1}{2} \, I \, \omega^2 - \frac{3 G M^2}{5 R} + \frac{1}{2} \, k R^2. \end{equation} If the spring constant $k$ continuousy change for some reason : $k_0 \Rightarrow k_1$, then the sphere need to collapse/expand to adapt to the new situation. Energy (1) is conserved. Susbsituting the conservation of angular momentum to eliminate $\omega$ (and put in $\omega_0$), then we can find the final radius $R_1$ as a complicated function of $R_0$, $\omega_0$, $k_1$ and $k_0$ (the algebraic equation is of the fourth order !). Then the radius can change continuously since there's a new variable that enters the game (the constant $k$), that can also change continuously.

Second solution (thermodynamical) :

In place of an internal potential energy, we can simply state that some "heat" is produced by the change of state of the sphere. Keeping mechanical energy (3) from my question, we simply have to write \begin{equation}\tag{2} E_1 = E_0 + Q, \end{equation} where $Q$ is a new continuous variable that enters the game. That heat may stay inside the sphere after the state change, or be released outside, it doesn't matter. We then can solve the energy conservation (which is now an algebraic equation of the second order, if $Q$ is independant of $R$) : \begin{equation}\tag{3} E_1 = \frac{1}{2} \, \Big( \frac{2}{5} \, M R_1^2 \Big) \Big( \frac{R_0^2}{R_1^2} \, \omega_0 \Big)^2 - \frac{3 G M^2}{5 R_1} = \frac{1}{2} \, \Big( \frac{2}{5} \, M R_0^2 \Big) \, \omega_0^2 - \frac{3 G M^2}{5 R_0} + Q. \end{equation} I'll define the following dimensionless parameters to simplify things : \begin{align} p &= \frac{3 G M}{\omega_0^2 \, R_0^3}, \tag{4} \\[12pt] q &= \frac{5 Q}{M \omega_0^2 R_0^2}. \tag{5} \end{align} Parameter $q$ can be negative or positive, depending of the case. Then, the solution to (3) is this : \begin{equation}\tag{6} R_1 = \frac{p \pm \sqrt{p^2 - 4 (p - q - 1)}}{2 (p - q - 1)} \, R_0. \end{equation} Since $q$ can take any value, the final radius $R_1$ is a continuous function of $R_0$, $\omega_0$ and $Q$. If the dimensionless parameter $q$ is small, then to first order : \begin{equation}\tag{7} R_1 \approx \Big( \frac{1}{p - 1} - \frac{q}{(p - 2)(p - 1)^2} \Big) \, R_0. \end{equation}

In the simpler case of a non-rotating sphere : $\omega_0 \rightarrow 0$, then the solution to (3) is simply this : \begin{equation}\tag{8} R_1 = \frac{R_0}{1 - \frac{q}{p}} = \frac{R_0}{1 - \frac{5Q R_0}{3 G M^2}}, \end{equation} which can take any value, depending of $Q$ (positive or negative heat).

I think this solution solves all the issues. It even has great pedagogical values, in my opinion.

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  • $\begingroup$ Nice work on solving it.You can even make it more interesting if you consider the dynamics of the system.See,the point you found out is the point when the energy change of the system just compensates the amount of heat added or removed.If you consider the system,suppose you remove heat(more precisely work ) from it and it start's contracting and it kept on contracting till it reach equillibruim but now it will have some velocity.it will overshoot that point and the centrifugal force will now dominate.It may even start oscillating.the radius you found is not the equillibruim in 2nd approach. $\endgroup$ – Rishabh Jain Jun 14 '17 at 13:51

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