• A

## Main Question or Discussion Point

Consider the following flow: x = (1+ct)x0. Let the density rho(t) = rho0/(1+ct) so that the flow conserves mass. Physically, this is just a bunch of fluid elements on the positive x0-axis each given initial velocities that are proportional to their initial positions. Each fluid element should therefore continue with its initial speed, not accelerate, and have no forces acting on it while the density decays asymptotically to zero. Plugging this solution into Euler's equation, however, gives a pressure (1/2)rho0c2x2/(1+ct)3 which is proportional to the non-zero partial derivative of the density with respect to time times the fluid speed. Is this pressure to be regarded as fictitious/ non-physical?

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vanhees71
Gold Member
2019 Award
Let's whether really everything is really fulfilled. From your flow we get
$$v=\dot{x}=c x_0=\frac{c x}{1+c t}.$$
$$\dot{\rho}=-\partial_x (\rho v).$$
This is indeed fulfilled by your equation. Also the pressure looks right. So why should it be fictitious?

The last sentence is wrong. In checking the Euler equation I've wrongly put the density under the time differentiation, but it must be as @Chestermiller writes in #4, i.e.,
$$\rho (\partial_t \vec{v}+(\vec{v} \cdot \vec{\nabla}) \vec{v})=-\partial_x p.$$
And this vanishes as shown also in #4.

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If you think of this fluid as a continuum limit of a bunch of point masses evenly spread out at time t=0, they would each continue to move according to x = (1+ct)x0 only if there are no forces acting on them - so, physically, the pressure should be zero. Literally imagine setting up a flow of point particles with these initial conditions in the lab - and notice there had better not be any pressure anywhere. Conversely, the calculated pressure could be obtained if you e.g. put a continuum of strainers that would remove a certain fraction of the particles from each volume element per unit time - but then mass would be physically removed from the flow and the continuity equation would not hold. Specifically, the mass per unit volume per unit time that would have to be removed from the flow and brought to zero velocity would be rho0c/(1+ct) to give the calculated pressure. These pressures could be recorded at each of strainers in the lab.

Chestermiller
Mentor
This is funny, because I get: $$\frac{\partial v}{\partial t}=-\frac{c^2x}{(1+ct)^2}$$
$$\frac{\partial v}{\partial x}=\frac{c}{(1+ct)}$$
$$v\frac{\partial v}{\partial x}=+\frac{c^2x}{(1+ct)^2}$$
So, $$\frac{\rho_0}{(1+ct)}\left(\frac{\partial v}{\partial t}+v\frac{\partial v}{\partial x}\right)=\frac{\rho_0}{(1+ct)} \left(-\frac{c^2x}{(1+ct)^2}+\frac{c^2x}{(1+ct)^2}\right)=0=-\frac{\partial p}{\partial x}$$

• vanhees71
vanhees71
Gold Member
2019 Award
Hm, I made a mistake with Euler's equation in #2. @Chestermiller is, of course, right. The pressure must be constant along the $x$ axis and thus there's no force acting and the fluid elements are moving uniformly along the $x$ axis. I'll put a not in #2.

I have two ways to set up this flow:

If you set up a continuum filter in the flow with a differential cross section of dx/x, the calculated pressure is felt on the filter - though technically not every particle in the flow obeys x=(1+ct)x0 nor is the mass conserved globally IN the flow; ct/(1+ct) of the original mass gets stuck in the filter.

If you set up the flow with no filters, then every particle obeys x=(1+ct)x0, the mass IN the flow is conserved globally (with those that were stuck on the filter now sailing forward out of volume elements and reducing the density that way instead), but the calculated Euler pressure is indeed fictitious - no real physical forces are felt.

Let's whether really everything is really fulfilled. From your flow we get
$$v=\dot{x}=c x_0=\frac{c x}{1+c t}.$$
$$\dot{\rho}=-\partial_x (\rho v).$$
This is indeed fulfilled by your equation. Also the pressure looks right. So why should it be fictitious?

The last sentence is wrong. In checking the Euler equation I've wrongly put the density under the time differentiation, but it must be as @Chestermiller writes in #4, i.e.,
$$\rho (\partial_t \vec{v}+(\vec{v} \cdot \vec{\nabla}) \vec{v})=-\partial_x p.$$
And this vanishes as shown also in #4.
I think for Euler's equation, you want to take the convective derivative of the product of the density and the velocity. The convective derivative of the velocity is indeed zero - as there are no accelerations in the (unfiltered) flow. The convective derivative of the density, on the other hand, does not vanish - for mass conserving flows, it is equal to the negative of the density times the divergence of the flow - and this flow is diverging. This enables one to calculate a non-zero Euler pressure.

This is funny, because I get: $$\frac{\partial v}{\partial t}=-\frac{c^2x}{(1+ct)^2}$$
$$\frac{\partial v}{\partial x}=\frac{c}{(1+ct)}$$
$$v\frac{\partial v}{\partial x}=+\frac{c^2x}{(1+ct)^2}$$
So, $$\frac{\rho_0}{(1+ct)}\left(\frac{\partial v}{\partial t}+v\frac{\partial v}{\partial x}\right)=\frac{\rho_0}{(1+ct)} \left(-\frac{c^2x}{(1+ct)^2}+\frac{c^2x}{(1+ct)^2}\right)=0=-\frac{\partial p}{\partial x}$$
Calculations are correct! But I think there is also a (convective derivative of the density)xV on the left hand side of the bottom equation.

Chestermiller
Mentor
Calculations are correct! But I think there is also a (convective derivative of the density)xV on the left hand side of the bottom equation.
I think you mean the following (correct me if I'm wrong):
$$\frac{\partial (\rho v)}{\partial t}+\frac{\partial (\rho v^2)}{\partial x}=-\frac{\partial p}{\partial x}$$or, differentiating using the product rule,
$$\rho\frac{\partial v}{\partial t}+\rho v\frac{\partial v}{\partial x}+v(\frac{\partial \rho}{\partial t}+v\frac{\partial \rho}{\partial x}+\rho\frac{\partial v}{\partial x})=-\frac{\partial p}{\partial x}$$
Is this what you meant?

• spin2he2
vanhees71
Gold Member
2019 Award
The worst sort of files you can post to a forum where (theoretical) physicists like to participate is Word. I can open it with libreoffice, but it's so distorted that it's unreadable. It's very easy to use LaTeX in this forum, and that's the right way to communicate. I cannot read your derivations. As I repeatedly said, have a look at Landau, Lifshits, Course of Theoretical Physics, vol VI. It's the most clear exposition of fluid mechanics, I've ever seen.

The correct local expression of the momentum-conservation law for ideal fluids reads
$$\partial_t (\rho v_j)=-\partial_j p - \partial_k (\rho v_j v_k),$$
where on the right-hand side you have the negative divergence of the momentum-flow tensor,
$$\Pi_{jk}=p \delta_{jk} + \rho v_j v_k.$$

$$\partial_t \left (\frac{\rho}{2} \vec{v}^2+\rho \epsilon \right)=-\vec{\nabla} \cdot \left [\vec{v} \left (\frac{\rho}{2} \vec{v}^2 + \rho h \right) \right ].$$
Note that on the left-hand side in the bracket you have the total energy density (energy per fluid volume), and thus the correct thermodynamics quantity is the internal energy per volume, $\rho \epsilon$, where $\epsilon$ is the internal energy per fluid mass. The expression in the square bracket on the right-hand side is the energy current, and there the right thermodynamic quantitiy to use is the enthalpy density $\rho h$, where $h$ is the enthalpy per fluid mass. This is understandable, because the total energy corrent through a surface element on the one hand consists of the amount of energy per unit time transported through the surface element by the flow of the fluid, which is
$$\mathrm{d}^2 \vec{f} \cdot \vec{v} \left (\frac{\rho}{2} \vec{v}^2 + \rho \epsilon \right)$$
and by the work done per unit time on the fluid element at this surface element by the pressure, which is
$$\mathrm{d}^2 \vec{f} \cdot \vec{v} p.$$
So together you have for the energy current
$$\vec{v} \left [\frac{\rho}{2} \vec{v}^2 + \rho \left (\epsilon+\frac{p}{\rho} \right ) \right],$$
but
$$\epsilon+\frac{p}{\rho}=h.$$