# Using a differential force in the derivation of hydrostatic equilibrium in a star

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?)

• Isn't all due to a mixing of vectorial and modulus notations? Dec 15, 2019 at 9:10
• Huh? $F_G$ is $- GMm/r^2$. Use vectors. Dec 15, 2019 at 9:44
• @Alchimista I don't understand what you mean by 'modulus notation'. (My apologies.) Dec 15, 2019 at 19:13
• @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 $\hat{r}$. Dec 15, 2019 at 19:18
• 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. Dec 15, 2019 at 21:59

$$\vec{F_g} = -\frac{GM(r)\delta m}{r^2}\ \hat{r}$$ $$\vec{F_P} = -\left(\frac{dP}{dr}\right) \delta r \delta A\ \hat{r}$$ $$\vec{F_P} + \vec{F_g} = 0$$ leads to the scalar equation of hydrostatic equilibrium $$\frac{dP}{dr}=- \rho g,$$ where $$\vec{g}= -g\hat{r}$$.
• 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?) Dec 15, 2019 at 19:23
• 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. Dec 15, 2019 at 19:36