Confusion regarding the concepts and derivation of Hydrostatic Equilibrium for a star

First and foremost I am sorry for this; in order to make the questions clear I must first upload the lecture notes from my institution $^\zeta$ for the derivation of Hydrostatic Equilibrium:

$^\zeta$ Lecture notes courtesy of Imperial College London, Astrophysics dept, 2017-2018 edition.

Firstly, in equation $(3.1)$ why is there no negative sign for $F_g$? Last time I checked, Newton's law of gravitation told us that

$$F=-\frac{GMm}{r^2}$$

Nextly, I have a conceptual problem with equation $(3.2)$; when they use the term 'pressure' I am going to assume that they are referring to the radiation pressure that is released due to nuclear fusion reactions taking place in the sun. In that formula it must be the case that $P(r) \gt P(r+\delta r)$ if this is the case then $(3.2)$ should read $$F_p=-\left(\frac{dP}{dr}\right)\delta r\,\delta A$$

Correct me if I'm wrong but Hydrogen nuclei are fused together to generate the more stable Helium and hence lots of radiation pressure acting outwards from the centre of the star. If this is correct then $P(r+\delta r)=0$, since the outside face of that cylindrical volume element in Figure $3.1$ will not 'feel' any of this radiation pressure because this pressure only acts on the inner face. What am I not understanding here?

Moving on to equation $(3.3)$, let's suppose the sign of the gravitational force in $(3.1)$ really is neglected. Does this mean that the negative sign in $(3.3)$ is due to $P(r+\delta r) \lt P(r)$ and hence $$\frac{GM(r)\rho(r)}{r^2}=-\left(\frac{dP}{dr}\right)\,?$$ I know this is the same equation as in $(3.3)$ but I have written it with the negative sign next to the pressure differential as I would like to know where this minus sign originates from. Does anyone know?

My final concern is the sign in $(3.4)$ as last time I checked $$g=-\frac{GM}{r^2}$$ and not $$g=\frac{GM}{r^2}$$

Hence, equation $(3.4)$ should read $$\left(\frac{dP}{dr}\right)=\rho(r)\,g\,?$$

I forgot to mention that I have done some research and this is the closest derivation I could find, which, needless to say is no less helpful than the notes I have.

Equation 3.1 probably should have a minus sign in front of it. The gravitational force acts in the opposite direction to which $r$ increases. However, the text clearly states that it acts towards the centre and so long as we bear that in mind there is not necessarily any problem with what is written - it is merely the magnitude of the gravitational force.

The pressure is the pressure. There is no need to distinguish what the source of the pressure is (thermal, degeneracy, radiation, magnetic fields); hydrostatic equilibrium applies in all (equilibrium) situations. In a star like the Sun the radiation pressure is negligible, but it becomes much more important in more massive stars.

In all cases $P(r) > P(r + \delta r$), so $dP/dr$ is negative and indeed the force due to the pressure gradients should be written as $$F_P = -\left(\frac{dP}{dr}\right) \delta r \delta A,$$ if we define positive forces as acting towards increasing $r$. I think this is a problem with your notes, because the direction defined for $F_P$ is ambiguous.

The final piece of the puzzle is to sum $F_g$ and $F_P$ (as vectors) and equate to zero (for equilibrium). Since they both should have minus signs in front of them (IMO), you end up with equation 3.3. Your notes have fudged the issue and said $F_g = F_P$, which is incorrect as they are defined, and then inserted an unexplained additional minus sign to correct the mistake and you end up with the same thing.

Finally, $g$ is usually described as a positive quantity unless written as a vector $\vec{g}$, though ideally the forces might also have been written as vectors. In other words $g$ here is a scalar quantity, the magnitude of the gravitational acceleration. Thus we say the gravitational acceleration at the surface of the Earth for example, is 9.8 m/s$^2$ acting downwards, where $g = GM/r^2$.

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

• I had a query, and please bear in mind it is a high school student asking, so pardon me if it is a trivial question. If we integrate the function for dp we must get the pressure at a distance r from the centre. This pressure multiplied by the area of the shell elemeny that lies at r distance must equal the gravitational force (in magnitude) that is acting on the mass beyond the shell element, given it's an isolated system. Is my assertion correct? Commented Sep 26, 2021 at 19:26

Answer for the first point: it depends on the reference system. The only thing that says Newton's law is that there is a force directed toward the center of mass. For the second point: it depends again on the reference system. If you want, you can think that the sign of the equation (3.2) is inside the $(dP/dr)$ term.

The important thing is that $F_g$ and $F_P$ have the opposite sign, as in equation (3.3).

Finally, the eqaution (3.4) is written correctly, because as you have written the gravity and the pressure must be the opposite sign if your in an hydrostatic equilibrium.

• I thought that $g$ is defined to be $g=-\frac{GM}{r^2}$, this cannot possibly depend on the reference system. Also, what do you mean by "(3.2) is inside the derivate"? Is 'derivate' a typo? Commented Aug 22, 2018 at 18:00
• g is a vector, so the sign depends on the reference system. Yes it is a typo, I mean "the derivative of P" Commented Aug 22, 2018 at 20:46
• You keep telling me "the sign depends on the reference system". But how about showing me a reference system which is consistent with the sign of $g$ and of the $F_g$ and $F_p$? Now that would be very helpful Commented Aug 23, 2018 at 1:53
• Indeed $F_g$ and $F_p$ should point in opposite directions, but $dP/dr$ is a negative quantity, so in the notes above they do not! Commented Aug 23, 2018 at 8:04
• @AlessandroPeca $dP/dr$ is negative. The $r$ coordinate increases outwards, whilst $P$ decreases. Your answer and comments (especially the second one) are adding to the confusion. Commented Aug 23, 2018 at 12:01