# Why does the solenoidal term vanishes in a barotropic fluid?

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?

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:

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

Thus:

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

• 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. Nov 19, 2019 at 16:14
• @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.
– 2b-t
Nov 20, 2019 at 0:30
• 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. Nov 20, 2019 at 8:40
• @Onturenio Oh, I see... I understand... I can't think of another way to be honest...
– 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 $$dH= dU+V dP + PdV= dU +\frac 1 \rho dP+ PdV= \frac 1 \rho dP.$$

• Wait, how comes that barotropic means $dS=0$. Barotropic means $\rho(p)$. Nothing is said about entropy. I'm puzzled now... Nov 17, 2019 at 23:49
• 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. Nov 18, 2019 at 15:56
• I don't think that is necessarily true. The flow of a barotropic fluid could also be isothermal.
– 2b-t
Nov 18, 2019 at 21:04
• 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 :) Nov 19, 2019 at 0:17
• @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.
– 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 $$\left(\frac{\partial p}{\partial x}\right)=\frac{1}{R_d T}\left(\frac{\partial \rho}{\partial x} \right), \quad \left(\frac{\partial p}{\partial y}\right)=\frac{1}{R_d T}\left(\frac{\partial \rho}{\partial y} \right)\Rightarrow \vec{\nabla p} = \frac{1}{R_dT}\vec{\nabla \rho} = -\frac{\rho^2}{R_d T}\vec{\nabla \alpha}$$