We want to roughly model the fluid flow of a star; consider the following proposition:

In the absence of viscosity, and heat conduction, the Navier-Stokes Equations for a steady, spherically symmetric, radial flow of an ideal gas, in the presence of a gravitating point mass situated ate the origin are: \begin{equation}\frac{d}{dr}(\rho u r^2)=0 \label{nscm1}\tag{1}\end{equation} \begin{equation}\rho u \frac{du}{dr}=-\frac{dp}{dr}-\frac{GM\rho}{r^2}=0\label{nscqdm1}\tag{2}\end{equation} \begin{equation}\frac{d}{dr}\left(\frac{1}{2}u^2+\frac{\gamma}{\gamma-1}\frac{p}{\rho}-\frac{GM}{r}\right)=0\label{nsce1}\tag{3}\end{equation} Where:

  • $\rho$ is the density
  • $p$ is the pressure
  • $\gamma$ is the ratio of the specific heats
  • $G$ is the gravitational constant
  • $M$ is the mass situated in the origin
  • $r$ is the radial coordinate

Or at least that's what is stated in my lecture notes. Now: equation (1) is simply conservation of mass, the continuity equation, no problem. Equation (2) is conservation of momentum, or you can think of it as Newton's second law of dynamics, no problem there either; sure whe have the density $\rho$ instead of the mass but that's because we were integrating on the volume, and then the integral got elided, and we are left with the density. But I have a problem with equation (3): this equation should simply be the conservation of energy, but it seem strange to me for a couple of reason:

  1. I sort of get that the first term on the left is the kinetic energy, and I also sort of get that the last term is gravitational potential energy, but what about the mid term? What kind of energy $\frac{\gamma}{\gamma-1}\frac{p}{\rho}$ represents? Also we are dealing with only an ideal gas here, and $\gamma$ is defined as a ratio between specific heats.. But a ratio between what specific heats exactly? I suspect this bit wants to represent some kind of thermal energy, but we are also in a situation where we assume absence of heat conduction.. I don't get what is going on.

  2. I have a problem with all this third equation in general.. Why do we derive all the left side by $r$? How can we prove that, in this context, equation (3) is the right equation for energy conservation? I would like a complete mathematical proof.

I don't have a lot of experience with Navier-Stokes equations. I only know that they should represent conservation of mass, momentum and energy. Indeed I suspect that my confusion regarding equation (3) is motivated by my inexperience. For this reason I would like an answer that does not assume possession of an excessive amount of knowledge about fluid flow and Navier-Stokes equations from the beginning.

  • $\begingroup$ The $p/rho$ term in your energy equation looks like the static pressure term of the bernoulli equation $\endgroup$ Commented Mar 4, 2021 at 15:56
  • $\begingroup$ if you multiply eq. 2 by $dr/\rho$ and integrate you obtain something like eq. 3 . $\endgroup$
    – Eli
    Commented Mar 4, 2021 at 17:04

1 Answer 1


As @Eli mentioned in a comment, you can derive equation 3 (Bernoulli's equation) from the Euler equation (2) (this is just a name for the subcase of the Navier-Stokes equations in which we assume adiabaticity and no viscosity, as is done here).

I will be denoting radial derivatives with a prime. After dividing through by the density $\rho$ in the Euler equation, two identifications are easy: $$ \begin{align} \left( \frac{u^2}{2} \right)' &= u u' \\ \frac{GM}{r^2} &= \left( - \frac{GM}{r}\right)', \end{align} $$ but it is less immediate to see that $$ \frac{P'}{\rho} \overset{?}{=} \left(\frac{\gamma}{\gamma - 1} \frac{P}{\rho} \right)'. $$

Here, the assumption of adiabaticity (no heat conduction) means that the equation of state is in the form $P = K \rho^\gamma$, with constant $\gamma$ (equal, as you mentioned, to the ratio of the specific heat indices) as well as constant $K$. This is known as a polytropic equation of state.

Substituting this in, the two sides of the equation read: $$ \frac{\gamma\rho^{\gamma -1}}{\rho} \overset{?}{=} \frac{\gamma}{\gamma - 1} (\rho^{\gamma-1})' = \frac{\gamma}{\gamma - 1} (\gamma-1) \rho^{\gamma-2}, $$ so all the three terms do indeed match.

As an aside, this is quite similar to how you might derive the energy conservation equation for the motion of a particle in one dimension ($x$) starting from $F = ma$, assuming that $F$ is conservative (so, expressible as $F = - \partial_x V$): $$ -\frac{\mathrm{d} V}{\mathrm{d} x} = m \frac{\mathrm{d} v}{\mathrm{d} t} = m \frac{\mathrm{d} v}{\mathrm{d} x} \frac{\mathrm{d} x}{\mathrm{d} t} = m v \frac{\mathrm{d} v}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \left( \frac{1}{2} m v^2 \right), $$ which means that $$ \frac{\mathrm{d} }{\mathrm{d} x} \left( V + \frac{1}{2} m v^2 \right) = 0. $$

Edit, to answer the comment. Why do we bother to write this new equation, if it can be derived from the Euler one? Well, they are both useful! Under the assumption of stationarity, we can integrate the Bernoulli equation and get a constant of motion, which will be the same for the whole fluid motion (without the assumption of stationarity, the integral could be time-dependent). This is very convenient: as your lecture notes will probably do further on, we can compute it at different radii in order to gather interesting information.

For other calculations, instead, it is useful to keep the differential formulation: this is the reason why one typically writes both.

The fact that energy and momentum conservation are related is not a problem: after all, if you think of a single particle with mass $m$ there is a one-to-one relation between its energy and its momentum, $E = p^2 / 2m$ (in the nonrelativistic limit).

  • $\begingroup$ I thought that equations (1),(2),(3) represent three distinct conservation laws. So how could it be that we can derive equation (3) (conservation of energy) from equation (1)? And also, if we can indeed derive (3) from (2) then why bother writing (3) at all? Seems redundant information to me, why not just write (1) and (2) if this is the case? And at last: is this the only derivation possible for (3), can we derive it from conservation of energy? $\endgroup$
    – Noumeno
    Commented Mar 9, 2021 at 11:13
  • $\begingroup$ I've tried to answer your first two points in an edit; as for being able to derive the Bernoulli equation from the conservation of energy equation: I believe it can be done, you will need to account for both thermodynamics and kinematics, since the energy at hand is has partly thermal and partly mechanical. $\endgroup$ Commented Mar 9, 2021 at 15:52
  • $\begingroup$ If it helps, the term $\gamma / (\gamma -1) P/\rho$ is the specific enthalpy of the fluid element. $\endgroup$ Commented Mar 9, 2021 at 15:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.