# Homework Help: Expanding spherical shell of gas

1. Nov 15, 2009

### cepheid

Staff Emeritus
1. The problem statement, all variables and given/known data

I'm considering an expanding spherical shell of gas in the thin shell approximation, in which all of the mass can be considered to be at radius R. This shell encloses a bubble that has been blown out (by a supernova or something) and is expanding into the ambient interstellar medium (ISM), picking up mass along the way, like a snow plow. The simplest approximation would be that the shell mass M(R) at a given radius R is given by the total ambient mass of ISM material that was once enclosed by that radius. However, I do not think that I am supposed to assume any specific functional form for M(R) for the first part of this problem. I'm told also that the shell has some total force on it (which can consist of both inward and outward forces) given by F(R), which is some known function. The problem asks me to :

"Use the fact that F is the rate of change of the shell momentum to derive v(R) = dR/dt. Then show how R(t) can be derived from v(R). You do not have to plug in the actual v(R) in the second step."

2. Relevant equations

$$\textbf{F} = \frac{d\textbf{p}}{dt}$$ ​

3. The attempt at a solution

So far I have written:

$$F(R) = \frac{d}{dt}[ M(R) v(R) ] = M(R)\frac{dv(R)}{dt} + v(R)\frac{dM(R)}{dt}$$

$$F(R) = M(R)v^{\prime}(R)\dot{R} + v(R)M^{\prime}(R)\dot{R}$$

$$F(R) = M(R)v^{\prime}(R)v(R) + M^{\prime}(R)v^2(R)$$ ​

A differential equation which I have no idea how to solve. I'm thinking that I may be wrong about the expression for the total shell momentum just being Mv. After all momentum is a vector, and the total shell momentum should be the sum of the momenta of the individual particles comprising it. My thinking is that, in the absence of F(R), the total momentum is zero (in the rest frame of the object that exploded), and this can be shown simply by pointing out that for every bit of mass that is flying outward radially in some direction, there is another bit of the same mass that is flying radially outward in the opposite direction (by symmetry). However, another side issue that is also confusing me is that, using spherical coordinates, the particles of the gas all have momenta in the $+\hat{\textbf{r}}$ direction, which would lead one to state that there is some net momentum in the "radially outward" direction. I understand that the $\hat{\textbf{r}}$ vector changes direction, but I'm still confused. What is wrong with that statement?

Any help on both the problem itself and the side issue I identified would be much appreciated!

2. Nov 15, 2009

### cepheid

Staff Emeritus
Okay, I gave some more thought as to how to actually derive the shell momentum:

$$d\textbf{p} = dm\textbf{v}$$ ​

An element of mass on the shell is given by:

$$dm = \sigma dA$$ ​

where $\sigma$ is the surface mass density. I'm going to assume the shell is uniform so that:

$$\sigma = \frac{M(R)}{4\pi R^2}$$ ​

The velocity at any point on the shell is given by:

$$\textbf{v} = v(R)\hat{\textbf{r}}$$ ​

And so:

$$\textbf{p} = \int_A d\textbf{p} = \int_A dm\textbf{v}$$

$$\textbf{p} = \int_A \sigma v(R)\hat{\textbf{r}} dA$$

$$\textbf{p} = \int_0^{\pi} \int_0^{2\pi} \frac{M(R)}{4\pi R^2} v(R)\hat{\textbf{r}} R^2 \sin\theta \, d\phi d\theta$$ ​

Now, if you can take out the $\hat{\textbf{r}}$, then everything will cancel nicely and you'll get $\textbf{p} = M(R)v(R)\hat{\textbf{r}}$. But I don't think that the unit vector can come out of the integral. After all, by the same argument I gave before, not all of those individual infinitesimal momentum vectors are in the same direction! r-hat changes with theta and phi, and I feel like the integral of the unit normal vector over the surface of a sphere should come out to zero. I just don't know how to show it, and even if I do, I'm not sure how it will help me with my problem.

3. Nov 15, 2009

fun problem! I haven't solved it myself, but I might as well put down some thoughts.

It sounds like the problem is wanting you to derive something similar to the Taylor-Sedov solution (http://en.wikipedia.org/wiki/Blast_wave" [Broken] the Wikipedia page).

The simplistic way to derive the Taylor-Sedov solution with a given F(R) is to look at energy conservation:

$$E=\int^{\infty}_{0}{F(R)dR}=1/2M(R)\left(\frac{dR}{dt}\right)^2$$

Also, it might be okay to assume that all of the mass in the interstellar medium is swept up in the shock wave
$$\frac{dM}{dt}=4\pi R^2v\rho$$
and
$$M=4\pi R^3\rho/3$$.

However, I recognize this isn't the method the problem statement wants. Still, if you could try and mold your equations into the above form, then the solution would be easy (given on the wikipedia page). You might start by integrating your final F(R) equation in your first post and seeing what happens. You might have make some assumptions and play around with integration by parts and such also.

Last edited by a moderator: May 4, 2017
4. Nov 15, 2009

Okay, never mind with my taylor-sedov nonsense. Using the above expressions for M(R) and dM/dt are enough to allow you to solve the differential equation you set out in your first post.

5. Nov 16, 2009

### cepheid

Staff Emeritus
Okay, as an update to my post #2, I figured out how to evaluate the integral and found the expected result that:

$$\int_{\Omega} \hat{\textbf{r}}\, d\Omega = 0$$ ​

by expressing the unit vector in Cartesian coordinates:

$$\hat{\textbf{r}} = \hat{\textbf{x}} \sin \theta \cos \phi + \hat{\textbf{y}} \sin \theta \sin \phi + \hat{\textbf{z}} \cos \theta$$ ​

But if p = 0 by symmetry, then doesn't this suggest that dp/dt is always zero as well? Yet, F is non-zero. I think that the problem lies with my model. We're talking about a spherical shell of gas plowing through the ISM, not a spherical shell expanding in a vacuum. Yet, I don't know how to correct my model to account for this. Can anyone tell me where I am going wrong conceptually?

6. Nov 16, 2009

### gabbagabbahey

Why are you (implicitly) assuming that the velocity of the shell doesn't change?

EDIT: Nevermind; after reading your attempt again, I see that you are just calculating the momentum due to an infinitesimal piece of shell mass $dm$ here.

Last edited: Nov 16, 2009
7. Nov 16, 2009

### cepheid

Staff Emeritus
Hmmm....because I'm not too bright? So, should it be (using a physicist's abused notation)?

$$d\textbf{p} = d(m\textbf{v}) = \textbf{v}dm + md\textbf{v}$$ ​

8. Nov 16, 2009

### gabbagabbahey

There are two assumptions you've made that need justification:

(1) You are assuming that the shell has no angular momentum (doesn't rotate)

(2) You are assuming that the shell both starts with a spherically symmetric mass distribution, and accretes mass spherically symmetrically

9. Nov 16, 2009

Those are some interesting questions. I still think that what you are doing in your first post is the right way to look at this stuff. Here's why.

Your absolutely right that the total momentum of the system is a constant = 0 (in the frame of the supernova). However, that doesn't prevent a bubble from expanding or contracting when a radial force is applied to the bubble. Normally we talk about expansion being due to a pressure difference, $\Delta P$, between the inside and the outside of the bubble. However, you can also speak of the force on the bubble, $F(R)=4 \pi R^2 \Delta P$.

You can then look at the force, $\delta F=F d\Omega/4\pi$, applied to one small region of the bubble, $dA=R^2d\Omega$, of mass, $\delta M=\sigma R^2 d\Omega$. Then if you assume that mass of the bubble layer to be the swept up mass (i.e. $M = 4/3 \pi R^2 \rho$, you'll find that the force equation you write down (with the $\delta$'s) will be equivalent to the force equations you wrote in your first post.

However, once you write down the force equation with the proper M(R), it's difficult to see how to solve the equation! At least I haven't seen how to solve it yet.

10. Nov 16, 2009

### gabbagabbahey

No, I think you were right the firs time. (you must have missed my edit)

11. Nov 16, 2009

### gabbagabbahey

Just to add to bombadil's post, it seems as though you have some radial force $dF(R)=f(R)d\Omega=F(R)d\Omega/4\pi R^2$ on each piece of the shell and the "total force" they give you is $F(R)=\int_{\Omega}f(R)d\Omega$. Calling this quantity the total force is not strictly correct, as the net force on the shell is actually $\textbf{F}(R)= \int_{\Omega}f(R)\hat{\mathbf{r}}d\Omega=0$, but I gather it is still fairly common nomenclature in astrophysics (I'm not an astrophysicist).

Last edited: Nov 16, 2009
12. Nov 16, 2009

gabbagabbahey,

Right, the net force must be zero on the bubble as it's center of mass remains stationary. In this context, it's probably more fruitful to think of the force as pressure which is artificially being regulated (which is why F it is an arbitrary function of R).

13. Nov 16, 2009

### cepheid

Staff Emeritus
Thank you both, especially for post #'s 9 and 11. That really helps. I am going to give the problem a shot and report back.

gabbagabbahey - I did miss your edit in #6 at the time, but I see it now. As for the assumptions in post #8, I'm sort of okay with going with them...even if only for the sake of having a shot at solving this problem.

bombadil - Your remark in post #9 is ominous, but I have a differential equations textbook in hand and I will see if I can get anywhere!

14. Nov 16, 2009

### gabbagabbahey

The ODE looks fairly simple as long as you recognize that $\frac{d}{dR}\left(v(R)^2\right)=2v(R)v'(R)$

15. Nov 17, 2009

### cepheid

Staff Emeritus
But I don't have that factor of 2 in the second term. So, it doesn't look like that side of the ODE is an exact differential, because it's not quite equal to d[Mv^2]/dR or d[(Mv^2)/2]/dR.

16. Nov 17, 2009

### gabbagabbahey

You don't need the factor of two....it might help you see what I mean if you make the substitution $u(R)=v^2$

17. Nov 17, 2009

Okay, so here's the ode I get that looks hairy and scary:

$$F=\frac{dp}{dt}=\frac{d}{dt}\left(\frac{4}{3}\pi R^3 \rho \frac{dR}{dt}\right)$$

and then assuming (as cepheid did in the first post) that $dv/dt=vdv/dR$, I get:

$$F(R)=4 \pi R^2 \rho v^2+\frac{4}{3}\pi R^3 \rho v \frac{dv}{dR}$$

Am I wrong to be scared of such a non-linear beast?

gabbagabbahey: and what will the substitution $v^2=u$ do except to give you another nonlinear ode?

18. Nov 17, 2009

### gabbagabbahey

What makes you think the ODE is non-linear?

It's actually a linear 1st order separable ODE....

19. Nov 17, 2009

Your solving for $v(R)$ and the ode has a $v^2$ term.

Wait a second. I think your right about that substitution. It does make the ode [STRIKE]seperable and[/STRIKE] first order. very nice.

.

Last edited: Nov 17, 2009
20. Nov 17, 2009

### cepheid

Staff Emeritus
Here is what I have so far:

$$v^\prime v M + v^2 M^\prime = F$$

$$u = v^2 \Longrightarrow u^\prime (R) = u^\prime (v)v^\prime (R) = 2vv^\prime$$

$$\therefore \, \, \frac{1}{2}u^\prime M + uM^\prime = F$$

$$u^\prime + 2u\frac{M^\prime}{M} = 2\frac{F}{M}$$​

To make the left hand side an exact differential, multiply both sides by the following integrating factor:

$$\exp\left[2 \int \frac{M^\prime}{M} \, dR \right] = \exp \left[2\int \frac{d}{dR}(\ln M) \, dR\right] = \exp[2\ln M] = M^2$$​

With the result that:

$$u^\prime M^2 + 2u M^2 \frac{M^\prime}{M} = 2M^2 \frac{F}{M}$$

$$u^\prime M^2 + 2u M M^\prime = 2MF$$

$$[uM^2]^\prime = 2MF$$

$$uM^2 = 2\int MF \, dR$$

$$u(R) = \frac{2}{[M(R)]^2}\int M(R)F(R) \, dR$$

$$v(R) = \frac{\sqrt{2}}{M(R)}\left( \int M(R)F(R) \, dR \right)^{\frac{1}{2}}$$​

Am I right so far?

21. Nov 17, 2009

### gabbagabbahey

Looks good to me.

22. Nov 17, 2009

Great! I'm such an ignoramus when it comes to ODE, I'll have to file away that integrating factor trick. Just for fun, did you notice how you final result looked suspiciously like a conservation of energy? FYI here's how I solved the problem using conservation of energy:

When you have a distributed mass that is being worked on, the relevant quantities are always the displacement and velocity of the center of mass. This might seem weird in the context of a bubble as the center of mass of the system stays at the center for all time. However, you can analyse the infinitesimal work done along some ray, starting at $r=0$ and going out to some radius, $R$. Such an analysis will equivalent to this:
$$E=\int^R_0{Fdr_{cm}}=1/2 M(R)v_{cm}^2$$
where cm stands for center of mass of the ray. So, again, you can think of this as a one dimensional problem with an infinitesmal force applied along the radial direction ($\delta F=F d\Omega/4\pi$). Doing the usual analysis to find the location of the center of mass of the ray you find that $r_{cm}=1/4r^4/R^3+3/4R$ so that $dr_{cm}=r^3/R^3dr$ and $v_{cm}=r^3/R^3 v$. Plugging these results into the first equation gives:
$$\frac{1}{R^3}\int^R_0{r^3Fdr}=1/2M(R)\left(\frac{R^3}{R^3}v^2\right)^2$$
Remember that the right side of the equation represents the kinetic energy at $r=R$, so all the quantities are evaluated at r=R, hence $v_{cm}(r=R)=v$. Solving for v we get:
$$v=\left[\frac{2}{M(R)R^3}\int^R_0{r^3Fdr}\right]^{1/2}$$
If you multiply the denominator and the numerator in the square root with $4/3 \pi \rho$ you will recover the same result you got by solving the ODE:
$$v=\frac{\sqrt{2}}{M(R)}\left(\int^R_0{M(r)Fdr}\right)^{1/2}$$

Sorry if that bored you, but I love seeing two seemingly completely different ways of solving a problem produce the same answer.

Last edited: Nov 17, 2009
23. Nov 18, 2009

### cepheid

Staff Emeritus
Hey guys,

Thanks for the help so far. The second part of the question says to, "show how R(t) can be derived from v(R); (you do not have to plug in the actual v(R)..." So far all I've got is the following method:

Assume that an inverse function t(R), exists. If so, then it is true that:

dt/dR = [dR/dt]-1 = 1/v(R) ​

From which it follows that:

$$t(R) = \int_0^R \frac{dr}{v(r)}$$ ​

However, I'm wondering if that is a sensible way to proceed, or if there is a better way of deriving R(t) from v(R) that I'm missing (it becomes relevant in the next part of the question, in which we are given functional forms for M(R) and F(R) and asked to solve for R(t)).

24. Nov 19, 2009

### gabbagabbahey

I don't think you really have to assume that an inverse function exists to say that

$$\frac{dR}{dt}=v(R)\implies dt=\frac{dR}{v(R)}\implies t=\int_{R_0}^R\frac{dr}{v(r)}$$

($R_0$ represents tha radius of the shell at whichever point in time you take to be $t=0$)