In fluid dynamics, and in particular in atmospheric dynamics, the so-called solenoidal term is the line integral:

$$\oint \frac{\vec{\nabla p}}{\rho}\cdot d\vec l$$

where $p$ and $\rho$ are the pressure and density, respectively, related through the state equation $p=\rho R_d T$.

A barotropic fluid is a simplification of some fluids in which the density is assumed to be a function of pressure alone, i.e. $\rho\equiv \rho(p)$. This contrasts with a baroclinic fluid, where $\rho\equiv \rho(p,T)$.

Ok, so my questions is: Why the solenoidal term vanishes in a barotropic fluid? This is written in most text books as something obvious, but I could not work out a formal demonstration. I'm interested in a mathematically correct demonstration. I do not mean I want it demonstrated as a theorem, making explicit all required theorems and hypotheses involved, but I'd like to know the mathematical machinery of calculus that is behind the fact that $\rho(p)$ implies necessarily that this line integral vanishes.

I guess that at the very core of my doubt, the problem is that I do not mathematically understand what it's meant by having $\rho\equiv\rho(p)$. After all, $\rho = \frac{p}{R_dT}$, so density does seem to depend of $T$ in any case... right?


3 Answers 3


As $\vec{\nabla p}\cdot d\vec l$ = $dp$:

$$\oint \frac{\vec{\nabla p}}{\rho}\cdot d\vec l = \oint \frac{1}{\rho} dp$$

For a barotropic fluid $\frac{1}{\rho}$ = $f(p)$, and therefore:

$$\oint \frac{1}{\rho} dp = \oint f(p) dp$$

From the Second Fundamental theorem of Calculus:

$$ \int_a^b f(p)dp = F(b) - F(a) $$

where $F$ is the antiderivative of $f$ in the closed interval $[a,b]$.

If $a = b$:

$$ \oint f(p)dp = F(b) - F(b) = 0 $$

In your case: $$\oint \frac{1}{\rho} dp = \oint R_dT \frac{dp}{p} = \oint R_dT\ d(ln(p))$$

In an ideal gas, the barotropic assumption is valid when:

$$\rho = \frac{p}{R_dT} = f(p) \iff R_dT=constant$$


$$ \oint R_dT\ d(ln(p)) = R_dT \oint \ d(ln(p)) = R_dT\ [ln(p_0)-ln(p_0)]=0 $$

  • $\begingroup$ What I do not think it is clear in this demonstration is the fact that $R_dT$ is assumed to be constant and how it is related to barotropic fluid. Even for a barotropic fluid, $T$ is not necessarily constant through the closed loop, as it does not necessarily lie within an isothermal surface. Therefore it does not necessarily can be taken out of the integrate. $\endgroup$
    – Pythonist
    Nov 19, 2019 at 16:14
  • 1
    $\begingroup$ @Onturenio The integration was carried over to pressure space (meaning you would actually also have to replace the integration limits) and is not anymore in spatial coordinates. A closed loop in pressure space doesn't imply an isothermal but instead only a density that is independent of temperature: Whatever you do to a function $f(p)$ will only be influenced by pressure. $\endgroup$
    – 2b-t
    Nov 20, 2019 at 0:30
  • $\begingroup$ OK I see the argument, and I trust that it solves the issue. But again, it somehow hides with a clever notation the actual math "behind the scenes". In a nutshell, what I asked for is a solution that does not rely on the notation $\vec{\nabla p}\cdot \vec{dl}= dp$ (which is precisely the one used in the book I'm studying). But it's OK, I'll mark this as the correct answer. $\endgroup$
    – Pythonist
    Nov 20, 2019 at 8:40
  • $\begingroup$ @Onturenio Oh, I see... I understand... I can't think of another way to be honest... $\endgroup$
    – 2b-t
    Nov 20, 2019 at 10:50

For a barotropic fluid $$ \frac{1}{\rho}\nabla P= \nabla h $$ where $h$ is the specific enthalpy, i.e. $H=U+PV$ per unit mass. The line integral vanishes because it is the closed-path integral of the gradient of a globally defined function.

To see why $\frac 1 \rho dP=dH$ note that for a unit mass $\rho =1/V$ and for barotropic fluids $ dU= TdS -PdV$ becomes $dU=-PdV$ --- because barotropic means that $dS$ is zero. Therefore $$ dH= dU+V dP + PdV= dU +\frac 1 \rho dP+ PdV= \frac 1 \rho dP. $$

  • 1
    $\begingroup$ Wait, how comes that barotropic means $dS=0$. Barotropic means $\rho(p)$. Nothing is said about entropy. I'm puzzled now... $\endgroup$
    – Pythonist
    Nov 17, 2019 at 23:49
  • $\begingroup$ Usually $P$ is a function of $\rho$ and $S$. To make it a function of $\rho$ only you must stop heat flowing into and out of the fluid. This mean setttig $dS=0$, so, for example, an ideal gas has $P\propto \rho^\gamma$, where the constant of proportionaity depends on $S$. So to be barotropic $S$ has to stay fixed. $\endgroup$
    – mike stone
    Nov 18, 2019 at 15:56
  • $\begingroup$ I don't think that is necessarily true. The flow of a barotropic fluid could also be isothermal. $\endgroup$
    – 2b-t
    Nov 18, 2019 at 21:04
  • $\begingroup$ Yes. I think Yyou are correct, I have always taken the basic definition to be that the internal energy depends only on $\rho$ and that I think requires $dS=0$. All practical applications I know have $dS=0$ but a very slow change of pressure that is able to stay in equilibrium with a heat sink/source would also give $P(\rho)$.The answer by Fernando is also correct, (and so is the deleted answer by 2b-t) and more general than mine. You should mark him correct and give him some points :) $\endgroup$
    – mike stone
    Nov 19, 2019 at 0:17
  • $\begingroup$ @mikestone I have deleted my answer as Fernando's answer is almost identical to mine and the first version of my answer was misleading. I think his answer should be marked as the solution although your approach is also very interesting. $\endgroup$
    – 2b-t
    Nov 19, 2019 at 12:54

I found an alternative demonstration that does not rely on the "too clever" notation $\vec{\nabla p} \cdot \vec{dl}=dp$. For the sake of completeness, I'll post it here for future references.

Using Stokes theorem, and the notation $\alpha =1/\rho$, the integral can be converted into a surface integral:

$$\oint \vec{\alpha \nabla p}\cdot \vec{dl} = \int_\Sigma \nabla\times \alpha \vec{\nabla p} \cdot \vec{ds} = \int_\Sigma \vec{\nabla\alpha}\times\vec{\nabla p}\cdot ds$$

This a general expression, valid for any fluid.

Now, if $\rho(p)$, then we can demonstrate that $\vec{\nabla p}$ must be parallel to $\vec{\nabla \rho}$, and therefore antiparallel to $\vec{\nabla \alpha}$. Thus, the vectorial product within the integral vanishes, and the integral must be zero.

To demonstrate that $\rho(p)$ implies that both gradients are parallel:

$dp = \frac{1}{R_d T}d\rho = \left(\frac{\partial p}{\partial x} \right)dx + \left(\frac{\partial p}{\partial x} \right) dy = \frac{1}{R_d T}\left(\frac{\partial }{\partial x} \right)dx + \frac{1}{R_d T}\left(\frac{\partial \rho}{\partial x}\right)$

which can only be satisfied if $$\left(\frac{\partial p}{\partial x}\right)=\frac{1}{R_d T}\left(\frac{\partial \rho}{\partial x} \right), \quad \left(\frac{\partial p}{\partial y}\right)=\frac{1}{R_d T}\left(\frac{\partial \rho}{\partial y} \right)\Rightarrow \vec{\nabla p} = \frac{1}{R_dT}\vec{\nabla \rho} = -\frac{\rho^2}{R_d T}\vec{\nabla \alpha}$$


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