A star is born

Classical Mechanics Level 4

A large, spherical, and very diffuse cloud of molecular hydrogen sits in empty space. The radius of the cloud is $r_0$ and the molecules are initially at rest. Let $t_0$ be the time it takes for this cloud to completely collapse, i.e. the time it takes for a molecule at $r_0$ to free-fall into the origin. Similarly, let $t_1$ be the time it takes for a cloud of radius $r_1=4r_0$ to collapse.

What is $\frac{t_1}{t_0}?$

Details and Assumptions:

The entire cloud is collapsing in this process.
Both clouds have the same mass.

The answer is 8.

4 solutions

Josh Silverman Staff
Nov 10, 2013

Because this problem asks for the relative magnitude of the collapse times, we can replace doing hard physics with the easy (sexy) physics of dimensional analysis. The least we can ask of a correct physics equation is that its units be consistent across the equals sign, i.e. if we're not comparing apples to apples, there is no hope of the equation being right.

As it turns out, a system that is as simple as this is quite constrained when it comes to finding a combination of the relevant parameters that gives us the unit of time. Here, the only possibilities for physical parameters that can affect the collapse of the cloud are the gravitational constant $G$ , the mass of the cloud $m$ , and the radius of the cloud $r$ . Let's try to find some combination of $G$ , $r$ , and $m$ that gives units of time.

Starting with $G$ , we have

$\displaystyle \left[G\right] = \frac{ML}{T^2}\frac{L^2}{M^2} = \frac{L^3}{T^2M}$

where $M$ , $T$ , and $L$ are the units of mass, time, and length, respectively.

This expression beckons us to cancel the units of length by introducing a factor of $\displaystyle \frac{1}{r^3}$ , bringing us to the manageable state of affairs:

$\displaystyle\left[\frac{G}{r^3}\right] = \frac{1}{T^2M}$ .

Try as we might, there isn't much to do with this but to introduce the last remaining variable at our disposal, the mass of the cloud, $m$ , to cancel the $M$ :

$\displaystyle\left[\frac{Gm}{r^3}\right] = \frac{1}{T^2}$

which gives us units of inverse time squared. Clearly, if we take the square root of the inverse, we'll arrive at units of time, as we desire.

This suggests that the collapse time of the cloud is given by $\displaystyle t\sim\sqrt{\frac{r^3}{Gm}}$

Now, the problem wording is a little ambiguous in that it is unclear whether the two clouds have the same mass or the same density. If the two clouds had the same density, then we'd have $m \sim \rho r^3$ and $t \sim \sqrt{\frac{1}{G}}$ so that the time to collapse was constant with respect to the radius. However, 1 is not being accepted as the correct answer, and so this assumption must be incorrect.

Let's suppose instead that the two clouds have the same amount of hydrogen, and so different densities.

In this case, $\displaystyle t \sim \sqrt{\frac{1}{G\rho}}$ .

Further, as the masses of the two clouds are equal, $\displaystyle \frac{\rho_0}{\rho_1} = \frac{r_1^3}{r_0^3} = 64$

and therefore, $\boxed{\displaystyle \frac{t_1}{t_0} = \sqrt{\frac{\rho_0}{\rho_1}} = \sqrt{64} = 8}$ .

One might object to this line of reasoning, after all we made absolutely no physical argument for the validity of the final result. Also, couldn't the equation be something more complex like a sum of terms or something? Sure, but each term would have to keep the same form as the one we arrived at because it is the only combination of the relevant variables that result in units of time.

You can read more about the usefulness of dimensional analysis for these kinds of scaling problems, and about the validity of the uniqueness claim at Buckingham's $\pi$ theorem .

Damn, I actually came up with a full-on classical derivation for the equation of motion.

Fantastic work, though, quite clear.

Guillermo Angeris - 7 years, 7 months ago

God ! I didn't know dimensional analysis could get you this far ! Just outta curiosity, did you choose the dimensional analysis method just like that, or is there a particular set of conditions for which it works in problems ?

Gaurav Simha - 7 years, 7 months ago

The dimensional analysis approach can be helpful in a wide variety of problems but in many cases, we have complicated tradeoffs that result in polynomials or transcendental equations or some such thing that can't be fully resolved by insisting that the units work out. In those cases we can sometimes still get the asymptotic behavior, i.e. "high Reynolds number ensures that only effect X has any relevance in this regime".

For this problem, it was clear that the collapse of the cloud only had one important effect (gravitation) and that the behavior should be smooth over the whole collapse of the cloud (neglecting any quantum effects the cloud is simply collapsing under the approximation of classical gravitation), so the dimensional analysis approach would work.

Also, obviously, the method can't tell us what numerical prefactors are required, so if a function goes like $f(t) \sim t$ we can't know if it's $t$ or if it's $5\pi t$ . However, when we ask how things scale, like we are here, it's clear that these unit-less numerical coefficients cancel out and we only care about the functional form of the quantity

Josh Silverman Staff - 7 years, 7 months ago

Why 1 is not an answer?

Nihhaar Chandra Routhu - 7 years, 7 months ago

There are two possible answers due to the ambiguity of the problem. By trial and error, 8 is the correct answer.

Si Yu How - 7 years, 7 months ago

great work

avijit batra - 7 years, 6 months ago

Maedhros 777
Jan 4, 2014

Without actually finding an equation for the time of collapse, we can use dimensional analysis to find the answer. The variables involved are the mass of the cloud (in units of $kg$ ), the radius of the cloud (in units of $m$ ), and the gravitational constant G (in units of $\frac{m^3}{kg\cdot s^2}$ ). Now the equation for the time of collapse will be of the form $T = M^a R^b G^c$ . Reducing this to units: $s = (kg)^a(m)^b(\frac{m^3}{kg\cdot s^2})^c = (kg)^{a-c}(m)^{b+3c}(s)^{-2c}$ $-2c = 1$ , $c = -\frac{1}{2}$

$b+3c = b - \frac{3}{2} = 0$ , $b = \frac{3}{2}$

$a-c=0$ , $a=c=-\frac{1}{2}$

Thus, $T = kM^{-\frac{1}{2}}R^{\frac{3}{2}}G^{-\frac{1}{2}}$ for some dimensionless constant k and so quadrupling $R$ will multiply $T$ by $4^{\frac{3}{2}} = 8$ . We now find that $\frac{t_1}{t_0} = \boxed{8}$ .

Lucas Tell Marchi
Dec 30, 2013

According to energy conservation $\frac{G m M}{r_{0}} = \frac{1}{2} (\frac{dx}{dt})^{2}$ Which solution is $t = \int_{0}^{S} \sqrt{\frac{r_{0}}{2 G M}} dx = \sqrt{\frac{r_{0}}{2 G M}} \times S$ Now if we choose $S$ as our starting position, we have the time $t_{0} = \sqrt{\frac{r_{0}}{2 G M}} \times r_{0}$ Doing the same for the cloud with $S = 4 r_{0}$ we have $t_{1} = 8 t_{0}$ Clearly $\frac{t_{1}}{t_{0}} = 8$ .

My friend has pointed out that I forgot the mass on the right side of the first equation. Sorry for that, I wish I could edit it.

Lucas Tell Marchi - 7 years, 5 months ago

Gaurav Simha
Nov 11, 2013

Firstly, to solve any problem, we need a good approach. Here, we need to find the ratio of two times taken by the particles to reach a certain displacement, so it suggests it may be related to kinematics. One more key idea here is that motion of the particles is caused by a force (without it, kinematics would be boring, and ultimately the universe would be boring). Now, we can say that forces are a more fundamental approach to this problem. So, we need to use Newton's Laws analyze the forces on the particles. Here, we find that the gravitational forces are the only forces acting (although inter molecular forces may be present, they are neglected here). However to apply these laws, we need to make an assumption, that mass of the core of the sphere is constant for both the cases.

Fortunately, we have an equation giving the gravitational force on the particle,

F= G[(mM)/(r^2)]; (here 'm' is the mass of the particle and 'M' is the mass of the core it is attracted to)

Also, we need to fuse the kinematic laws & Newton's laws to get the right answer. Thus, since acceleration is a common factor for both laws, we take it as a parameter.

Thus, from the above equation using and F=ma & cancelling the mass of the particle 'm' from both sides we get,

a=G(M/r^2)

Now, integrating twice with respect to time on both sides, and setting the constants of integration of velocity and displacement each to zero by default, we get displacement of the particle

s= \frac {G (t^2) M}{2*r^2},

In the first case, r 0 = \frac {G*(t 0)^2 M}{r^2 2}

In the second case, 4 r_0 = \frac {G (t 1)^2*M}{(4r 0)^2 2} \Rightarrow 64 r 0 = \frac {G*(t 1)^2 M}{(r_0)^2 2}

Now, getting t 0 and t 1 on the right side for both cases and dividing them (getting the ratio) , we cancel r_0, G & M, we get

\frac {t 1}{t 0} = \sqrt{\frac {128}{2}} = \sqrt{64} = \boxed{8}

I fiddled around with a straightforward solution to the problem, I think it goes like:

The time for collapse is equal to

$\displaystyle t_c = \int\limits_0^{t_c} dt = \int\limits_{r_0}^{0} \frac{dt(r)}{dr}dr$

So, if we can find an expression for $\displaystyle\frac{dt}{dr}$ , we're in business

Let's consider a particle that starts at the edge of the cloud. It is attracted to the mass on the interior of the sphere as if it's all concentrated at the center, i.e.

$\displaystyle \ddot{r} = -\frac{GM_{cloud}}{r^2}$

Multiply both sides by $\dot{r}$ and realize you've got some exact differentials

$\begin{aligned} \dot{r}\ddot{r} &= - GM\frac{\dot{r}}{r^2} \\ \frac12\frac{d}{dt}\left(\dot{r}^2\right) &= GM\frac{d}{dt}\left(\frac{1}{r}\right)\\ \frac12\int\limits_0^{t_f}d\left(\dot{r}^2\right) &= GM\int\limits_0^{t_f}d\left(\frac{1}{r}\right) \\ \frac12 \dot{r_f}^2 &= GM\left(\frac{1}{r_f} - \frac{1}{r_0}\right) \\ \dot{r_f} &= -\sqrt{2GM\left(\frac{1}{r_f} - \frac{1}{r_0}\right)} \end{aligned}$

Now you've got yourself a stew goin'!

From before: $\displaystyle t_c = \int\limits_{r_{initial}}^{0} \frac{dt(r)}{dr}dr$

So... $\displaystyle t_c = -\frac{1}{\sqrt{2GM}}\int\limits_{r_0}^{0} \left[\left(\frac{1}{r_f} - \frac{1}{r_0}\right)\right]^{-1/2}dr_f$

Now on the one hand, there is an integral. On the other hand, we don't have to do it. So we won't.

Let's consider what happens when we move the initial radius from $r_0$ to $ar_0$ . Let's insert a copy of the Heaviside function to remind ourselves that the integrand is undefined for $r_f > ar_0$

$\begin{aligned} I(a) &= -\int\limits_{ar_0}^{0} \left[\left(\frac{1}{r_f} - \frac{1}{ar_0}\right)\right]^{-1/2}\Theta\left(ar_0-r_f\right)dr_f \\ \end{aligned}$

Let $r_f' = r_f/a$ , we have

$\begin{aligned} I(a) &= -a\int\limits_{ar_0}^{0}\left[\left(\frac{1}{ar_f'} - \frac{1}{ar_0}\right)\right]^{-1/2}\Theta\left(ar_0-ar_f'\right)dr_f' \\ &= -a\int\limits_{ar_0}^{0}\left[\left(\frac{1}{ar_f'} - \frac{1}{ar_0}\right)\right]^{-1/2}\Theta\left(r_0-r_f'\right)dr_f' \\ &= -a\int\limits_{r_0}^{0}\left[\left(\frac{1}{ar_f'} - \frac{1}{ar_0}\right)\right]^{-1/2}dr_f' \\ &= -a^{3/2}\int\limits_{r_0}^{0}\left[\left(\frac{1}{r_f'} - \frac{1}{r_0}\right)\right]^{-1/2}dr_f' \end{aligned}$

which is just $a^{3/2}$ times the original integral, i.e.:

$\displaystyle I(a) = a^{3/2}I(1)$

From this it is obvious that the time of collapse scales as the cloud radius to the $3/2$ power and so for the dilation by a factor of 4, $\displaystyle t_{collapse}(4r_0) = 8t_{collapse}(r_0)$

Josh Silverman Staff - 7 years, 7 months ago

The subsitution $r_f = r_0\sin^2\theta$ makes calculating the integral a piece of cake.

Mark Hennings - 7 years, 6 months ago

Hi, your approach is incorrect, actually you are taking $r$ to be constant, but actually it is changing and its double derivative with respect to time is acceleration,

The actually equation would be : $\frac{d^2r}{dt^2} = \frac{GM}{r^2}$

jatin yadav - 7 years, 7 months ago

Here's a solution using Conservation of Energy.Let $v$ be the instantaneous velocity when the particle is at a distance $x$ from the center.

$\frac{1}{2}mv^2-\frac{GMm}{x}=-\frac{GMm}{r_0}$ $\rightarrow v=\frac{dx}{dt}=\sqrt{\frac{2GM}{r_0}\left(\frac{r_0-x}{x}\right)}$ $\rightarrow \int^t_0\,dt=\int^{r_0}_0\sqrt{\frac{r_0}{2GM}\left(\frac{x}{r_0-x}\right)}\,dx$

Substitute $x=r_0\sin\theta$

$t=\sqrt{\frac{r_0}{2GM}}\left(\frac{\pi r_0}{2}\right)$

Shaswata Roy - 7 years, 6 months ago

I'm actually new to latex, and i'm quite frankly struggling with it. Can anyone please tell me whether you see the equations in my question with or without latex ?

Gaurav Simha - 7 years, 7 months ago

I only see the $\LaTeX$ code but it isn't formatted. I think you need to wrap your mathematical statements in \ ( and \ ) (no space in between) to get them to format.

Josh Silverman Staff - 7 years, 7 months ago

A star is born

The answer is 8.

4 solutions

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