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I've been banging my head against this particular derivation of hydrostatic equilibrium in a star for the last few days, from Carroll and Ostlie's Introduction to Modern Astrophysics (2nd ed. p. 286):

The authors consider an infinitesimal cylinder of gas (dimensions $A, dr$) aligned on an r-axis pointing away from the center of the star. Equilibrium demands that acceleration be zero, so we get

$$F_{bottom} + F_{top} + F_G = 0,$$

Where the first two terms are the forces, taken normal to the surface, due to pressure, and $F_G$ is the gravitational force on the cylinder. The top of the cylinder is away from the center.

Alternative derivations I have found from this point onward make perfect sense- substitute $-\rho A dr$ for the cylinder mass $dm$ in the $F_G$ term, substitute $P_{bottom}$ and $P_{top}$, the rest is straightforward (this is the derivation on Wikipedia).

However, Carroll and Ostlie take an approach I haven't been able to understand. At this point they define $$F_{top} = -(F_{bottom} + dF_P)$$ where $dF_P$ -I'll write it $dF$ hereafter- is called the differential force caused by the change in pressure due to a change in $r$. Substitution into the first equation and a few other steps here ($dF = A dP$) lead to $dP/{dr} = -\rho g$ as needed.

I have had trouble with the following: $dF$ needs to be negative because otherwise $F_{top}$ is greater in magnitude than $F_{bottom}$ which makes no sense- if $|F_t| > |F_b| $ then pressure isn't acting against gravity, is it? Pressure itself would be pushing the cylinder down in that case. ($b$ is 'bottom', $t$ is 'top'.)

But if $dF$ is negative, then the way the (second) equation is written makes no sense. Wouldn't it make more sense to say $F_t = -F_b + dF$ ?

Partially motivated by this question and partially trying to approach the problem from a slightly different angle, I decided to define $dF = F_b + F_t$ as the net outward force due to the infinitesimal change in pressure between the top and bottom of the cylinder. Then, as $F_{net} = 0,$ we get $dF + F_G = 0$, but this leads to an incorrect equation, namely $$ dF + F_G = 0 \implies dP/dr = \rho g . $$

My main issue is that I do not understand why trying to define a positive $dF$ fails to produce the correct hydrostatic equilibrium equation. Furthermore, if we view $dF$ as an infinitesimal quantity, trying to ensure its sign is consistent with the directions of the forces (outward being positive), we once again get the wrong answer.

Can we only take $dF$ as an inward force? If so, why? If not, how can I establish a positive/outward differential force that will let me derive the correct hydrostatic equilibrium equation?

Edit:

I think I can illustrate my dilemma a bit more succinctly.

Considering the equation $F_t = -(F_b + dF)$, we can try out 2 different possibilities: $dF$ being positive or negative.

If $dF > 0$ then $F_t$ is bigger in magnitude than $F_b$, in which case I don't understand how the pressure gradient creates an outward force because $|F_t|>|F_b|$ implies the downward force is stronger.

If $dF < 0$ then $F_t = -F_b + -dF = -F_b + |dF|$ but as far as I can understand, this form cannot, like any of my attempts to define a positive $dF$, output the correct sign in the final equilibrium equation. So computing the signs (that is, saying $(-dF) = |dF|$) somehow makes the equation incorrect. (Additional edit: For me, this also begs the question, if $dF = A dP$ then is $dP$ negative, and what does that signify?)

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  • $\begingroup$ Isn't all due to a mixing of vectorial and modulus notations? $\endgroup$
    – Alchimista
    Dec 15, 2019 at 9:10
  • $\begingroup$ Huh? $F_G$ is $- GMm/r^2$. Use vectors. $\endgroup$
    – ProfRob
    Dec 15, 2019 at 9:44
  • $\begingroup$ @Alchimista I don't understand what you mean by 'modulus notation'. (My apologies.) $\endgroup$
    – ygtozc
    Dec 15, 2019 at 19:13
  • $\begingroup$ @RobJeffries Yes, $F_G$ is negative, but my point is then if you take a positive $dF$, you get the wrong answer, and I don't understand why. This problem is one-dimensional anyhow, so I see no need to use $\hat{r}$. $\endgroup$
    – ygtozc
    Dec 15, 2019 at 19:18
  • $\begingroup$ I think I have finally understood the way $dF$ was defined by the text: $dF$ is a negative force along a positive change in pressure $dP$, that is to say, the force is negative along a direction in which dP increases. $\endgroup$
    – ygtozc
    Dec 15, 2019 at 21:59

1 Answer 1

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The whole exercise illustrates the usefulness of vectors in addressing problems.

$$\vec{F_g} = -\frac{GM(r)\delta m}{r^2}\ \hat{r}$$ $$\vec{F_P} = -\left(\frac{dP}{dr}\right) \delta r \delta A\ \hat{r}$$ $$\vec{F_P} + \vec{F_g} = 0$$ leads to the scalar equation of hydrostatic equilibrium $$ \frac{dP}{dr}=- \rho g,$$ where $\vec{g}= -g\hat{r}$.

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  • $\begingroup$ Why does $\overrightarrow{F_P}$ have a negative sign? As far as I understand it describes the net force of pressure, so I don't understand where the negative sign comes in. (In hindisght, of course, $dP/dr < 0$ lets us say that but how can we claim the $-$ sign has to be there before deriving the formula?) $\endgroup$
    – ygtozc
    Dec 15, 2019 at 19:23
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    $\begingroup$ @ygtozc because the force due to a pressure gradient acts in the opposite direction to the pressure gradient. $\endgroup$
    – ProfRob
    Dec 15, 2019 at 19:24
  • $\begingroup$ Oh, I understand- intuitively at least. As pressure force is essentially 'pushing', increasing pressure means force increases (in magnitude) in the opposite direction. Then in terms of differential forces, I suppose $dF = -AdP$ (instead of $AdP$) would provide the correct answer. $\endgroup$
    – ygtozc
    Dec 15, 2019 at 19:36
  • $\begingroup$ You've been a great help, thanks. $\endgroup$
    – ygtozc
    Dec 15, 2019 at 21:47

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