The second law of thermodynamics states that the total entropy of any isolated thermodynamic system tends to increase over time, approaching a maximum value (not an infinite value!). But the higher the entropy, the lower the free, useable energy of a system. Therefore it is perhabs easier to formulate the second law of thermodynamics the other way round (Feynman, 1967): the free, useable energy of any isolated thermodynamic system tends to decrease over time, approaching a minimum value (zero?!). In this sense we can also understand Penrose (Penrose, 1989). He pointed out, that a closed self-gravitating system will collapse to a black hole, which is the state with the maximum entropy of the system. But this state is not a very unordered state as one often imagines states with high entropy. But it is the state where no free energy is left, the potential energy of the system is zero and so no directed kinetic energy can be produced any more. Thats why the ”ordered state” of a black hole has the biggest entropy.

Newtonan gravity is often expressed in form of the poisson equation

$\mathrm{\Delta }\varphi =4\pi G\rho$

(Q.1)

Nevertheless this is equivalent to ³⁷³⁷K is a constant with units $\mathrm{kg}\mathrm{s}{\mathrm{m}}^{-2}$. The role of $K$ is similar to the role of $ϵ$ in the electromagnetic theory.

	$\nabla \cdot 𝒈=-4\pi G\rho$		(Q.2)
	$\nabla \times (K𝒈)=\mathrm{𝟎},$		(Q.3)

because $\nabla \times (-\nabla \varphi )=\mathrm{𝟎}$ and therefore we are able to express the gravitational field as a gradient of a potential like $𝒈=-\nabla \varphi$. As we can see these equations are similar to Maxwells equations for the electric field in the electrostatic case ($\frac{\partial 𝑩}{\partial t}=0$)

	$\nabla \cdot (ϵ𝑬)=-{\displaystyle \frac{1}{{ϵ}_0}}\rho$		(Q.4)
	$\nabla \times 𝑬=\mathrm{𝟎}$		(Q.5)

and therefore we might call the equations (Q.2) and (Q.3) Maxwells equations of gravity.

The following is just speculation:

One cannot derive the continuity equation from these equations

	${\displaystyle \frac{\partial \rho }{\partial t}}+\nabla \cdot \left(𝒗\rho \right)$	$=0$
	$\iff -{\displaystyle \frac{1}{4\pi G}}{\displaystyle \frac{\partial }{\partial t}}\left(\nabla \cdot 𝒈\right)+\nabla \cdot \left(𝒗\rho \right)$	$=0$
	$\iff -{\displaystyle \frac{1}{4\pi G}}\nabla \cdot \dot{𝒈}+\nabla \cdot \left(𝒗\rho \right)$	$=0$
	$\iff \nabla \cdot \left(𝒗\rho -{\displaystyle \frac{1}{4\pi G}}\dot{𝒈}\right)$	$=0$

But if we fixed equation (Q.3) in analogy to Maxwell by writing ³⁸³⁸The units of the left hand side and the right hand side are the same, because of our choice of units for $K$.

$\nabla \times (K𝒈)=𝒗\rho -{\displaystyle \frac{1}{4\pi G}}\dot{𝒈},$

(Q.6)

we could derive the continuity equation from our systems of equations like

	$\nabla \cdot \left(\nabla \times (K𝒈)\right)$	$=\nabla \cdot \left(𝒗\rho -{\displaystyle \frac{1}{4\pi G}}\dot{𝒈}\right)$
	$\iff 0$	$=\nabla \cdot \left(𝒗\rho \right)+{\displaystyle \frac{\partial }{\partial t}}\left(-{\displaystyle \frac{1}{4\pi G}}\nabla \cdot 𝒈\right)$
	$\iff 0$	$=\nabla \cdot \left(𝒗\rho \right)+{\displaystyle \frac{\partial }{\partial t}}\rho$

So instead of (Q.2) and (Q.3) we could have

	$\nabla \cdot 𝒈=-4\pi G\rho ,$		(Q.7)
	$\nabla \times (K𝒈)=𝒗\rho -{\displaystyle \frac{1}{4\pi G}}\dot{𝒈}.$		(Q.8)

In vacuum ($\rho =0$) we would have

	$\nabla \cdot 𝒈=0,$		(Q.9)
	$\nabla \times (K𝒈)=-{\displaystyle \frac{1}{4\pi G}}\dot{𝒈}.$		(Q.10)

From these vacuum equations we can try to derive a wave equation in analogy to the wave equation for the electromagnetic field like

$\nabla \times \left(\nabla \times (K𝒈)\right)=-{\displaystyle \frac{1}{4\pi G}}\left(\nabla \times \dot{𝒈}\right)=-{\displaystyle \frac{1}{4\pi G}}{\displaystyle \frac{\partial }{\partial t}}\left(\nabla \times 𝒈\right)=+{\displaystyle \frac{1}{{(4\pi G)}^{2}K}}{\displaystyle \frac{{\partial }^2}{\partial t^2}}𝒈$

(Q.11)

and

$\nabla \times \left(\nabla \times (K𝒈)\right)=K\nabla \left(\nabla \cdot 𝒈\right)-K\mathrm{\Delta }𝒈$

(Q.12)

With $\nabla \cdot 𝒈=0$ it follows that

${\displaystyle \frac{{\partial }^2}{\partial t^2}}𝒈+{(4\pi GK)}^2\mathrm{\Delta }𝒈=0$

(Q.13)

This could be interpreted as a wave equation with complex velocity $c=i\cdot 4\pi GK$.

At first we should mention that the limit $\nu ⟶0$ is not equivalent to the limit $Re⟶\mathrm{\infty }$. The definition of the Reynoldsnumber is³⁹³⁹Just from looking at the units we could also write $Re=\frac{\text{Area per time}}{nu}$ or $Re=\frac{\rho LV}{\eta }=\frac{\rho V^{2}t}{\eta }=\frac{\text{Action density}}{\eta }$.

$Re={\displaystyle \frac{LV}{\nu }}$

(Q.23)

where $L$ is a characteristic length and $V$ is a characteristic velocity of the system (whatever that means). Therefore the limit $Re⟶\mathrm{\infty }$ can mean $\nu ⟶0$, but also $L⟶\mathrm{\infty }$ or $V⟶\mathrm{\infty }$. So if we would make an experiment with the same fluid and the same setup, but would only increase the (characteristic) speed of the fluid, we would measure effects for higher and higher Reynoldsnumbers. But this has nothing to do with the limit $\nu ⟶0$, because we not only have a higher Reynoldsnumber, but also have a higher Machnumber. This would mean, that we would also measure more and more effects due to the the increasing compressibility of our fluid. Therefore when doing an experiment, that should give insights into the regime of vanishing viscosity we cannot simply increase the Reynoldsnumber by increasing the characteristic velocity. We have to take care that all the other characteristic numbers (Mach-Number, Froude-Number …) stay the same. Otherwise we measure effects that have nothing to with the interesting limit $\nu ⟶0$. Sadly the limit $\nu ⟶0$ is often mixed up with $Re⟶\mathrm{\infty }$ in the literature.

For understanding the limit $\nu ⟶0$ Feynman (1964) investigates the vorticity equation for a newtonian incompressible fluid

${\displaystyle \frac{\partial }{\partial t}}{\omega }_g-{ϵ}_{ghi}{\displaystyle \frac{\partial }{\partial r_h}}{ϵ}_{ijm}{v}_j{\omega }_m=\nu {\displaystyle \frac{{\partial }^2}{\partial r_j^2}}{\omega }_g$

(Q.24)

First he mentions that for the case of very high viscosity

$\nu {\displaystyle \frac{{\partial }^2}{\partial r_j^2}}{\omega }_g\gg {\displaystyle \frac{\partial }{\partial t}}{\omega }_g-{ϵ}_{ghi}{\displaystyle \frac{\partial }{\partial r_h}}{ϵ}_{ijm}{v}_j{\omega }_m.$

(Q.25)

Therefore the left side of the vorticity equation can be neglected and the problem describing a fluid with high viscosity can be simplified to solving the so called Stokes equation⁴⁰⁴⁰$0_g$ is a component of the zero vector.

${\displaystyle \frac{{\partial }^2}{\partial r_j^2}}{\omega }_g=0_g$

(Q.26)

But what happens for very low viscosity? Feynman (1964) says decreasing the viscosity of a fluid leads to an increase of the velocity fluctuations and so the increasing factor $\frac{{\partial }^2}{\partial r_j^2}{\omega }_g$ compensates the smallness of the viscosity. The product of viscosity and $\frac{{\partial }^2}{\partial r_j^2}{\omega }_g$ doesn’t go to the limit $0_g$, which we would expect from the equation of vorticity we get for an ideal fluid

${\displaystyle \frac{\partial }{\partial t}}{\omega }_g-{ϵ}_{ghi}{\displaystyle \frac{\partial }{\partial r_h}}{ϵ}_{ijm}{v}_j{\omega }_m=0_g$

(Q.27)

So the equations for an ideal fluid do not(!) yield the right limit for vanishing viscosity. Can we find other equations that do give the right limit for vanishing viscosity?

One idea might be the following: Feynman (1964) said that the limit of $\nu \frac{{\partial }^2}{\partial r_j^2}{\omega }_g$ is not $0_g$, but what is it then? The easiest alternative would be a constant vector $C_g\ne 0$! This leads us to the equations

${\displaystyle \frac{\partial }{\partial t}}{\omega }_g-{ϵ}_{ghi}{\displaystyle \frac{\partial }{\partial r_h}}{ϵ}_{ijm}{v}_j{\omega }_m=C_g$

(Q.28)

and(!)

${\displaystyle \frac{{\partial }^2}{\partial r_j^2}}{\omega }_g={\displaystyle \frac{1}{\nu }}{C}_g$

(Q.29)

If we ignore the second term on the left hand side of equation (Q.28) for a moment, we see that $C_g$ stand for a dissipation of vorticity independent of $\nu$. Something we might expect for a fluid with low viscosity. The second equation (Q.29) is more confusing, since it is a second order partial differential equation, dependent on viscosity and shows no dependency on time ⁴¹⁴¹If $𝑪=𝑪(x,t)$, then it would also depend on time. Maybe we can intepret the existence of the two equations in the sense that equation (Q.28) gives the vorticity for $\nu =0$ and therefore must be independent of $\nu$ and equation (Q.29) gives the vorticity for a very small but not zero viscosity and therefore is still dependent on the viscosity.

Nevertheless it is interesting that we can solve equation (Q.29) if we know $𝑪(𝒙)$, because it is a vector-poisson equation well know from electrodynamics. Written in vector notation equation (Q.29) is

$\mathrm{\Delta }𝝎={\displaystyle \frac{1}{\nu }}𝑪(𝒙)$

(Q.30)

The solution to this equation is

$𝝎={\displaystyle \frac{1}{4\pi \nu }}{\displaystyle \int \frac{𝑪({𝒙}^{\prime })}{\left|𝒙-{𝒙}^{\prime }\right|}𝑑V^{\prime }}$

(Q.31)

From this we can compute the velocity field because we know

$\nabla \cdot 𝒗=0,$

$\nabla \times 𝒗=𝝎$

(Q.32)

So the velocity is a so called pure curl field, because the divergence of the velocity is zero. Following Bronstein, p. 665 we make the ansatz

$𝒗=\nabla \times 𝑨,$

$\nabla \cdot 𝑨=0$

(Q.33)

That means

	$\nabla \times (\nabla \times 𝑨)$	$=𝝎$		(Q.34)
	$\iff \nabla (\nabla \cdot 𝑨)-\mathrm{\Delta }𝑨$	$=𝝎$		(Q.35)
	$\Rightarrow \mathrm{\Delta }𝑨$	$=-𝝎$		(Q.36)

So again this leads to a vector-poisson equation for our vectorfield $𝑨$. The complete solution for equation (Q.29) is then

$𝒗=\nabla \times 𝑨$

(Q.37)

with

	$𝑨$	$={\displaystyle \frac{1}{4\pi }}{\displaystyle \int \frac{𝝎({𝒙}^{\prime })}{\left|𝒙-{𝒙}^{\prime }\right|}𝑑V^{\prime }}$		(Q.38)
		$={\displaystyle \frac{1}{4\pi }}{\displaystyle \int \frac{-\frac{1}{4\pi \nu }\int \frac{𝑪({𝒙}^{\prime \prime })}{\left|{𝒙}^{\mathbf{\prime }}-{𝒙}^{\prime \prime }\right|}𝑑V^{\prime \prime }}{\left|𝒙-{𝒙}^{\prime }\right|}𝑑V^{\prime }}$		(Q.39)
		$=-{\displaystyle \frac{1}{16{\pi }^2\nu }}{\displaystyle \int \frac{1}{\left|𝒙-{𝒙}^{\prime }\right|}\int \frac{𝑪({𝒙}^{\prime \prime })}{\left|{𝒙}^{\mathbf{\prime }}-{𝒙}^{\prime \prime }\right|}𝑑V^{\prime \prime }𝑑V^{\prime }}$		(Q.40)