<p>The whole exercise illustrates the usefulness of vectors in addressing problems.</p>

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

CC BY-SA 4.0
2019-12-15T19:23:00.833Z
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?)
ygtozc
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2019-12-15T19:24:54.527Z
@ygtozc because the force due to a pressure gradient acts in the opposite direction to the pressure gradient.
ProfRob
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2019-12-15T19:36:39.803Z
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.
ygtozc
CC BY-SA 4.0
2019-12-15T21:47:04.077Z
You've been a great help, thanks.
ygtozc
CC BY-SA 4.0
2019-12-15T09:54:37.057Z
2019-12-15T09:54:37.057Z
ProfRob
<p>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 <em>Introduction to Modern Astrophysics</em> (2nd ed. p. 286):</p>

<p>The authors consider an infinitesimal cylinder of gas (dimensions <span class="math-container">$A, dr$</span>) aligned on an r-axis pointing away from the center of the star. Equilibrium demands that acceleration be zero, so we get </p>

<p><span class="math-container">$$F_{bottom}  + F_{top} + F_G = 0,$$</span></p>

<p>Where the first two terms are the forces, taken normal to the surface, due to pressure, and <span class="math-container">$F_G$</span> is the gravitational force on the cylinder. The top of the cylinder is away from the center.</p>

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

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

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

<p>But if <span class="math-container">$dF$</span> is negative, then the way the (second) equation is written makes no sense. Wouldn't it make more sense to say <span class="math-container">$F_t = -F_b + dF$</span> ? </p>

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

<p>My main issue is that <strong>I do not understand why trying to define a positive <span class="math-container">$dF$</span> fails to produce the correct hydrostatic equilibrium equation.</strong> Furthermore, if we view <span class="math-container">$dF$</span> 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.</p>

<p>Can we only take <span class="math-container">$dF$</span> 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?</p>

<p>Edit:</p>

<p>I think I can illustrate my dilemma a bit more succinctly.</p>

<p>Considering the equation <span class="math-container">$F_t = -(F_b + dF)$</span>, we can try out 2 different possibilities: <span class="math-container">$dF$</span> being positive or negative. </p>

<p>If <span class="math-container">$dF &gt; 0$</span> then <span class="math-container">$F_t$</span> is bigger in magnitude than <span class="math-container">$F_b$</span>, in which case I don't understand how the pressure gradient creates an outward force because <span class="math-container">$|F_t|&gt;|F_b|$</span> implies the downward force is stronger.</p>

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

CC BY-SA 4.0
2019-12-15T09:10:08.630Z
Isn't all due to a mixing of vectorial and modulus notations?
Alchimista
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2019-12-15T09:44:10.160Z
Huh? $F_G$ is $- GMm/r^2$. Use vectors.
ProfRob
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2019-12-15T19:13:01.970Z
@Alchimista I don't understand what you mean by 'modulus notation'. (My apologies.)
ygtozc
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2019-12-15T19:18:47.793Z
@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 $\hat{r}$.
ygtozc
CC BY-SA 4.0
2019-12-15T21:59:55.440Z
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.
ygtozc
<hydrostatic-equilibrium>
Using a differential force in the derivation of hydrostatic equilibrium in a star
ygtozc
# Using a differential force in the derivation of hydrostatic equilibrium in a star

(Asked by: username_0 on 2019-12-15)

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?)

**Tags:** hydrostatic-equilibrium

> **Comments:**
> **username_1**: Isn't all due to a mixing of vectorial and modulus notations?  
> **username_2**: Huh? $F_G$ is $- GMm/r^2$. Use vectors.  
> **username_0**: @username_1 I don't understand what you mean by 'modulus notation'. (My apologies.)  
> **username_0**: @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 $\hat{r}$.  
> **username_0**: 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.  

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## 1 answers

### Answer by username_2 on 2019-12-15. Score: 2 | Accepted answer

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 $\vec{g}= -g\hat{r}$.

> **Comments:**
> **username_0**: 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?)  
> **username_2**: @username_0 because the force due to a pressure gradient acts in the opposite direction to the pressure gradient.  
> **username_0**: 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.  
> **username_0**: You've been a great help, thanks.  

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