Did I Solve the Radial Pulsation of Stars Problem Correctly?

[Boorglar]

I am trying to do the problems in this set:
http://farside.ph.utexas.edu/teaching/336k/lectures/node13.html

They seem quite hard, but I think I've managed to get to #4.
The statement of the problem is at Problem 4 in the above link.

My attempted solution is the following:

The gravitational field at a point inside the sphere can be calculated using Gauss' law for gravitation, so that 4πr²g = -4∏G ∫r^-αdV (V is a sphere of radius r)

= -4∏GMr^3-α/R^3-α. (M and R have the meanings given in the problem statement, and the integral is done in spherical coordinates).

so g = -GM/R^3-α*r^1-α, and its potential is GM/(2-α)*r^2-α/R^3-α since its negative gradient gives g.

By integrating the potential with respect to the density in the whole sphere, where density is proportional to r^-α, I get that U, the potential energy of the system, is :

4∏/(2-α) * GM/(5-2α) * R^2-α and its moment of inertia around the origin is 4∏R^5-α/(5-α) so that U = 1/(2-α) * (5-α)/(5-2α) * GM/R³ and plugging this in the Virial equation, and recognizing the second order linear differential equation, we find that the angular frequency should be

(2/(2-α) * (5 - α)/(5 - 2α) * GM/R³ )^1/2.

This has an additional factor of 2/(2-α) which is not in the problem statement, so I wonder whether I made a mistake or something is wrong.


Thank you for your help

 

[TSny]

so g = -GM/R^3-α*r^1-α, and its potential is GM/(2-α)*r^2-α/R^3-α since its negative gradient gives g.

By integrating the potential with respect to the density in the whole sphere, ...

The potential that you want to integrate with the density is the potential at the surface of a sphere of the material of radius r (with the material between r and R removed). Thus, V(r) = -Gm(r)/r where m(r) is the total mass contained within the sphere of radius r.

This is not the same as the anti-derivative of the acceleration g(r) at location r within a sphere of radius R.

It's as though you're building up the sphere layer by layer. When you bring in the next layer of mass, dm, from infinity and place it on the sphere of radius r, the work requred is dm*V(r) where
V(r) = -Gm(r)/r.

 

[Boorglar]

Yes, this makes sense, and I think the answer works that way. Thank you!

But then why isn't the acceleration vector the negative gradient of the potential? I thought this was one of the definitions of a potential... I'll have to think more about it.

EDIT:

since U will be negative and the equation is I'' = -2U + c, doesn't this mean that it will grow exponentially?

U = -(positive constant)I so I'' = 2(positive constant) I + c which is unbounded??

I also realize I made a few typos in the original post. But it's the question of U being negative that bothers me

 

[TSny]

why isn't the acceleration vector the negative gradient of the potential? I thought this was one of the definitions of a potential...

In finding the total U by building the sphere layer by layer, you only need the expression for V at the surface of the sphere as its radius grows. If $\bar{r}$ is the current radius during the build up, then V at the surface is $-Gm(\bar{r})/\bar{r}$. For a point outside the sphere at radius $r>\bar{r}$, the potential would be $V(r) = -Gm(\bar{r})/r$. The negative gradient of this function would be the acceleration for points outside the sphere.

since U will be negative and the equation is I'' = -2U + c, doesn't this mean that it will grow exponentially?

U = -(positive constant)I so I'' = 2(positive constant) I + c which is unbounded??

I don't think that U = -(positive constant)I. Let ##r_{i\}$be the equilibrium position of the$i^{th}## particle during the pulsation. For small radial oscillations in which the amplitude is proportional to the distance from the center of the star, you can write [itex]r_{i}(t)= r_{i\} + r_{i\} \epsilon \: cos(\omega\:t)[/itex] where $\epsilon << 1$ and $\epsilon$ is the same for all particles. You can then write I and U as functions of time to first order in $\epsilon$. See if that helps.

 

[Boorglar]

ri(t)=rio+rioϵcos(ωt) where ϵ<<1 and ϵ is the same for all particles.

But how do you know that?

I don't think that U = -(positive constant)I.

Here's my reasoning:

The moment of inertia around the origin that I found is

$$I = \int\int\int_{V} r^{2} d\rho = 2\pi\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{R}r^{2}r^{-\alpha}r^{2}\cos{\theta} dr d\theta = \frac{4\pi}{5-\alpha}R^{5-\alpha}$$

To calculate the potential energy, first find the mass inside a sphere of radius R:

$$m(R) = \int\int\int_{V} r^{-\alpha} dV = 2\pi\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{R}r^{-\alpha}r^{2}\cos{\theta} dr d\theta = \frac{4\pi}{3-\alpha}R^{3-\alpha}$$

then $$U = -\int_{0}^{R}\frac{G(\frac{4\pi}{3-\alpha}r^{3-\alpha})(4\pi r^{2}r^{-\alpha})dr}{r} = -\frac{16}{3-\alpha}\pi^{2}G\frac{R^{5-2\alpha}}{5-2\alpha}$$ (the potential is -Gm(r)/r where m(r) is the mass inside the sphere of radius r, and the mass of the next layer is 4∏r²*ρdr = 4∏r²*r^-αdr.)

so $$\frac{U}{I} = -\frac{4\pi}{3-\alpha}R^{3-\alpha}*\frac{5-\alpha}{5-2\alpha}\frac{G}{R^{3}} = -\frac{5-\alpha}{5-2\alpha} \frac{GM}{R^{3}}$$

Due to the fact that the oscillations have small amplitudes compared to the (very large for a star) equilibrium radius, R doesn't vary much and can be approximated by it. So U is -(positive constant)*I.

 

[Boorglar]

Although I must have made a mistake somewhere since a factor of 1/2 is missing to cancel with the 2 in the Virial equation...

As an additional clarification, when I say that m(r) = -4/(3-a)R^(3-a) (which seems to be a volume) it is understood that there is a unit of density attached to the constants (so that units work out) but it will cancel in the division of U with I anyways.

 

[TSny]

The equation [itex]r_{i}(t)= r_{i\} + r_{i\} \epsilon \: cos(\omega\:t)[/itex] is the way I think you would mathematically express “small amplitude radial pulsations of the star (in which the radial displacement is directly proportional to the radial distance from the center)”.

During the pulsations, $U$ and $I$ will be time dependent. It’s true that for the instants of time where the pulsation is passing through the equilibrium position you would have [itex]r_{i}(t)= r_{i\}[/itex] and $U_o = -KI_o$ where $K$ is the positive constant that you derived. But for arbitrary times during the pulsations, it will not be true that $U(t) = -KI(t)$.

However, using the expression [itex]r_{i}(t)= r_{i\} + r_{i\} \epsilon \: cos(\omega\:t)[/itex] I think you can show that to first order in $\epsilon$, the "distortions" from equilibrium of $U$ and $I$ satisfy $U(t) – U_o = +\frac{K}{2}(I-I_o)$.

 

[Boorglar]

I must admit I didn't quite understand what they meant by that phrase... I couldn't see the relevance to the question. But I'll try what you suggested.

Just to make sure: will it be true that U(t)=−K(t)I(t), where K now varies due to the R³(t) which is the time-varying radius of the star? I think that in the derivation I didn't assume it was at equilibrium.

EDIT:

Ok, here's what I get with r(t) = R(1+εcos(ωt)).

I(t) is proportional to R^5-α and U(t) is proportional to R^5-2α.

So I(t) = I₀(1+εcos(ωt))^5-α ≈ I₀[1+(5-α)εcos(ωt)].
and U(t) = U₀(1+εcos(ωt))^5-2α ≈ U₀[1+(5-2α)εcos(ωt)].

Plugging into the differential equation:
I"(t) = -ω²I₀(5-α)εcos(ωt) = -2U₀(5-2α)εcos(ωt) - 2U₀ + C

[2U₀(5-2α)-ω²I₀(5-α)]εcos(ωt) = C - 2U₀.

in order for the LHS to be constant, the coefficient before cos must be 0 (and even then the RHS might not be 0). Moreover, solving for ω gives a different answer.

 

[TSny]

Just to make sure: will it be true that U(t)=−K(t)I(t), where K now varies due to the R³(t) which is the time-varying radius of the star? I think that in the derivation I didn't assume it was at equilibrium.

Not only will R vary, but the mass density function will also change. Otherwise the total mass of the star would change.

Of course, you could always define a time varying K(t) such that U(t) =-K(t)I(t) by letting K(t) = -U(t)/I(t) (I know that's not what you meant.)

 

[Boorglar]

I just edited my previous post.

Not only will R vary, but the mass density function will also change. Otherwise the total mass of the star would change.

I didn't think of that! But then in what way is the density changing? It should be something like ρ(t) = k(t)*r^-α since the problem says the density is proportional to r^-α. So $k(t) = (\frac{R}{r(t)})^{3-\alpha}$ and $U(t) = -(\frac{R}{r(t)})^{3-\alpha}K(t) I(t)$ where K(t) is the "time varying constant" .

Wow this problem really got me stumped. And I don't see how this will lead me anywhere, since we still get I" = -2U + C and U being -(positive, time varying coefficient)*I so the system is still unbounded

 

[Boorglar]

You must be a genius if you can make sense of this, because seriously this problem confuses me the more I think about it.

 

[TSny]

OK, I think you were right about how $K(t)$ would vary with time: $K(t) = [(5-\alpha)/(5-2\alpha)]*GM/R(t)^3$.

I was thinking at first that maybe the pulsations would alter the mass density in such a way that $\alpha$ would also change. But you can show that if the density varies with $r$ as $\rho(r) = \rho_o r^{-\alpha}$ when there is no pulsation, then with the pulsations the density will vary with $r$ and $t$ as $\rho(r,t) = \rho_o(t) r^{-\alpha}$ where now $\rho_o$ varies with time but the exponent of $r$ is still the same. At any given time the ratio of $U(t)$ to $I(t)$ will not depend on $\rho_o(t)$, it will only depend on the total mass $M$, the total radius $R(t)$,and $\alpha$. So, the only time dependence of $K(t)$ will be due to the change in radius, as you were saying. Good.

So, back to the issue of deriving the expression for the frequency of pulsation. Were you able to express $U(t)$ and $I(t)$ in terms of $\epsilon\: cos(\omega t)$, where $r_i(t) = r_{i o} + r_{i o} \epsilon \: cos(\omega t) = (1+\epsilon\: cos(\omega t))r_{i o}$? Once you have that, it should be fairly easy to get the frequency.

 

[Boorglar]

Yes, I edited one of my previous posts:

EDIT:

Ok, here's what I get with r(t) = R(1+εcos(ωt)).

I(t) is proportional to R^5-α and U(t) is proportional to R^5-2α.

So I(t) = I₀(1+εcos(ωt))^5-α ≈ I₀[1+(5-α)εcos(ωt)].
and U(t) = U₀(1+εcos(ωt))^5-2α ≈ U₀[1+(5-2α)εcos(ωt)].

Plugging into the differential equation:
I"(t) = -ω²I₀(5-α)εcos(ωt) = -2U₀(5-2α)εcos(ωt) - 2U₀ + C

[2U₀(5-2α)-ω²I₀(5-α)]εcos(ωt) = C - 2U₀.

in order for the LHS to be constant, the coefficient before cos must be 0 (and even then the RHS might not be 0). Moreover, solving for ω gives a different answer.

Also, I think that the correction factor k(t) won't cancel since U is proportional to the square of density, while I is proportional to the 1st power only.

 

[TSny]

The easiest way to see how I(t) and U(t) vary with time is to go back to the basic expressions $I(t) = \sum m_i \: r_i(t)^2$ and $U(t) = \sum_{ij} G \: m_i \: m_j/|r_i(t)-r_j(t)|$ and use the fact that each $r_i(t) = r_{io}(1+ \epsilon \: cos(\omega t))$.

[ Hey, I just made my 1000 post! But, man, you've really made me sweat for it! Actually, I'm enjoying this problem ]

 

[TSny]

Ok, here's what I get with r(t) = R(1+εcos(ωt)).

I(t) is proportional to R^5-α and U(t) is proportional to R^5-2α.

So I(t) = I₀(1+εcos(ωt))^5-α ≈ I₀[1+(5-α)εcos(ωt)].

The reason that this is not giving the right answer is that it leaves out the time dependence of the density function.

$I(t)$ is proportional to $\rho_o(t)R(t)^{5-\alpha}$ where $\rho_o(t)$ is the time dependent factor in $\rho = \rho_0(t)r^{-\alpha}$.

You can eliminate the $\rho_o(t)$ factor in $I(t)$ using the the expression $M = \frac{4\pi}{3-\alpha}\rho_0(t)R^{3-\alpha}(t)$ to solve for $\rho_o(t)$ in terms of the constant mass $M$ and $R(t)$ and substitute into the expression for $I(t)$. Similar remarks for $U(t)$.

But you can avoid all of this by going back to the basic expressions for $I(t)$ and $U(t)$ in terms of summations over the particles.

 

[Boorglar]

When you speak of summation, you mean the integral over a volume?

$$I(t) = \int\int\int_{V(t)} r^{2}\rho(r,t)dV$$

and $$U(t) = \int\int\int_{V(t)} -\frac{Gm(r,t)}{r}\rho(r,t) dV$$

where V(t) is the time varying sphere of radius R(1+εcos(ωt)) and m(r,t) is the mass of a sphere of radius r at time t?

 

[TSny]

See what happens if you go back to $I(t) = \sum m_i \: r_i(t)^2$ and $U(t) = \sum_{ij} G \: m_i \: m_j/|r_i(t)-r_j(t)|$ and use the fact that each $r_i(t) = r_{io}(1+ \epsilon \: cos(\omega t))$.

You should be able to express $I(t)$ in a simple way in terms of $I_o$ and $\epsilon \: cos(\omega t)$ and likewise for $U(t)$. Then you can sub these into I'' = -2U +C.

 

[Boorglar]

Using this method I get
I(t) = I₀(1+εcos(ωt))² ≈ I₀(1+2εcos(ωt))
U(t) = U₀(1+εcos(ωt))^-1 ≈ U₀(1-εcos(ωt))

So I" = -2εω²I₀cos(ωt) = -2U₀(1-εcos(ωt)) + C

so -2ε(ω²I₀ + U₀)cos(ωt) = C - 2U₀

and ω = (-U₀/I₀)^1/2 !

This actually works! (Honestly, I would never have thought of that, the way the question was asked really confused me)

But what about the right hand side? If C - 2U₀ is not 0, then the star can't undergo the small amplitude harmonic oscillations we assumed it did? So it's like a condition on the initial kinetic and potential energy? (C is 4(K₀ + U₀) when the equation I" = -2U + C is derived).

And then from the derivation I made in post #5, we find that -U₀/I₀ is the constant K which is the answer to the second part of the problem.

 

[TSny]

So, the method I led you to (pushed you to?) gives the right answer. But I now think it’s bogus! (You can’t punch me because I’m not nearby.)

When I wrote $r_i(t) = r_{io}(1+cos(\omega t))$ I was thinking of $r_{io}$ as a constant, as though the motion of the $i^{th}$ particle is due solely to the pulsation. This is so wrong! The particles are also moving like mad chaotically to give the star its large internal kinetic energy. When there is no pulsation, there must still be a large total internal kinetic energy K to "oppose" gravitational collapse. Heck, that’s kind of what the virial theorem is all about: to relate K to U!

Rather than going back to summations over individual particles, I think we should just use the virial theorem directly. You were on the right track! By changing the summations to integrations using the given mass density function, I think you have essentially shown that

$I(t) = \frac{3-\alpha}{5-\alpha} M R(t)^2$

$U(t) = -\frac{(3-\alpha)GM^2}{(5-2\alpha)R(t)}$

We know $R(t) = R_o(1-\epsilon \: cos(\omega t))$. Using these expressions in the virial theorem should yield the result for the pulsation frequency. Moreover, we can understand the requirement that $C = 2U_o$ by considering the virial theorem applied to the star when it is not pulsating. The theorem says

$I_o’’(t) = 4K_o + 2U_o = 4E_o – 2U_o$ where the subscript refers to the star in equilibrium with no pulsation.

In equilibrium you can argue that $I_o(t) = constant$. (Actually, due to random motions of the particles, $I_o$ , $K_o$, and $U_o$ would fluctuate over small time scales, but we can assume we are only interested in their steady state average values.) So, see if setting $I_o’’(t) = 0$ solves the mystery of the constant C.

 

[Boorglar]

ha! Will see this. Is tried in right formula. Never worse than bed ned.

EDIT:

But it does lead to the same answer. Using the expressions for I and U in terms of alpha, M and R(t), and then writing R(t) as R(1+e*cos(wt)) and using the Bernouilli approximation in the powers, you get the same answer for the frequency of pulsation. So I guess we did solve the problem correctly, after all (well, mostly you since I followed your idea about R(1+e*cos(wt))...).