Minor typo correction, plus question

[keflavich]

I think in eqn 6.86, f_g should have a \rho out in front.

Also, the definition of f_g should either not include a negative or the sign should be flipped from 6.86 -> 6.87

[keflavich]

Question: to get from 6.88 -> 6.89, we have to assume \frac{\partial v}{\partial t} = 0. Why is this justified? Can this just be by assertion, that we're evaluating the spatial structure of v at a fixed time, noting that there is no explicit time dependence on the right side of the equation? If so, I think it would be best to spell this out in the text more explicitly.

I guess that's what's implied by "which is constant".

[markkrumholz]

OK, a couple of separate issues:

1. There is an inconsistency in the sign convention for f_g between equation 6.86 and the text right above 6.87 -- in equation 6.86 I used a sign convention whereby the value of f_g is taken to be a positive number, and thus it appears with a minus sign in the equation. However, I think inconsitently said that f_g = -G M_r / r^2 in the text, when to be consistent with the sign convention it should have been f_g = G M_r / r^2. Note that the sign is correct in 6.87; it's only in the text where it is inconsistent. I can fix this just by changing the text line above 6.87. Do you agree?

2. I also agree that there is a missing factor of rho in 6.86. Again, this is correct in 6.87, so it's just 6.86 that needs correction.

3. We do not need to assume that \partial v / \partial t = 0 to get from 6.88 to 6.89, provided that we understand v here to be defined as the velocity of a particular Lagrangian fluid element, not the velocity at a fixed (Eulerian) point in space. You seem to be thinking that we need to expand the Lagrangian derivative Dv/Dt into the Eulerian derivative dv/dt + v dv/dr. This would be the case if by v we meant the velocity at a particular position. However, this is not what we mean; we mean the velocity of a particular fluid element. In that case, one can get from 6.88 to 6.89 simply by recognizing that v = Dr/Dt = \dot{r}, and then integrating both sides with respect to t.

Does this satisfactorily resolve the issues? If so, I can go ahead and merge in the changes.

[keflavich]

On 1 and 2, I agree. For (1), I think either choice of where to put the negative sign is acceptable as long as it's consistent.

For (3), I don't follow still. If we start from 6.89 and take the derivative (w.r.t. r) of both sides, we get

If instead we square both sides, then take the derivative (w.r.t. r), we get:

The latter resembles the Eulerian expansion of the Lagrangian derivative, setting dv/dt = 0.

I am unclear how we would otherwise integrate both sides w.r.t. t; r is not a function of t, right? So integrating both sides w.r.t. t would simply yield

which isn't very useful. But I think here I'm ignorant of the properties of the Lagrangian. I think I last dealt with Lagrangians as an undergrad, so I don't discount this possibility.

[markkrumholz]

Hi Adam,

You're over-thinking it. Because we're working in Lagrangian coordinates, it doesn't matter that the thing we're following is a shell; we can treat this exactly like the equation of motion for a point mass, so the system reduces to just an ODE:

Dv/Dt = -G M / r^2 --> d^2r / dt^2 = -G M / r^2.

You can immediately verify just by plugging in that the solution I wrote down is a valid solution to this ODE. However, if you want to construct the solution directly, you can do so by using the chain rule to make a change from t to r as the independent variable:

d^2r / dt^2 = d/dt (dr/dt) = (dr/dt) d/dr (dr/dt) = v (dv/dr).

At this point the equation is immediately integrable by separation of variables:

v (dv/dr) = -G M / r^2 --> v dv = -G M dr / r^2 (1/2) (v^2 - v_0^2) = G M (1/r - 1/r_0),

where r_0 and v_0 are the starting position and velocity.

Would it help the clarity of this section if I went through the chain rule-based calculation explicitly?

[markkrumholz]

Just an additional comment: I think I realise the source of your confusion. It's this sentence:

"I am unclear how we would otherwise integrate both sides w.r.t. t; r is not a function of t, right?"

Yes, r is absolutely a function of t if we're working in Lagrangian as opposed to Eulerian coordinates. The variable r represents the radial position of a particular mass shell, and as this shell falls inward, r changes.

I'd welcome suggestions on how to make this clearer -- if I have succeeded in confusing you, then I am sure that I have been equally successful at confusing students!

[keflavich]

Thanks Mark, that's helpful. The change of variables trick with the chain rule is the piece I was missing. If that was ever in my math toolkit, it's grown very rusty. I'm getting ready to teach this material to grads, so I'm sweating all the details. It would indeed be helpful to spell that out

And yes, it makes sense that r is a function of t in Lagrangian coordinates. Thanks for clarifying.

You can immediately verify just by plugging in that the solution I wrote down is a valid solution to this ODE.

I got a bit stuck here. If I just take $d/dt$ of equ 6.89, I get the first messy equation I posted. It turns out that I needed one extra substitution to get back to equation 6.88. The other way to prove this was to square both sides, then take $d/dt$. So, yes, I confirm that that is a solution, but it wasn't trivial. Maybe I'm just out of practice!

[markkrumholz]

OK, given that the changes will be slightly more than what you checked in, I will decline this pull request, then make the required changes to both equations 6.86 / 6.87 and the added text explaining the chain rule trick. Will take care of that in the next couple of days, as family time permits.