How does a pressure distribution keep the fluid from moving?

[Adesh]

Let’s say we have a unit volume of some fluid in a column on the Earth surface. Let $\mathbf F$ be the gravitational force that acts on the unit volume of the fluid.

Consider a small volume element $\Delta \tau$ in the fluid and let’s assume it to be a cuboid with dimensions $\Delta x$, $\Delta y$ and $\Delta z$. The seemingly backward face of $\Delta \tau$ have $x$ coordinate as $x$ and the forward face would have the $x+\Delta x$ as coordinate.
Now, forces acting in the $x-$direction: backwards face = $p(x) \Delta y \Delta z$
forwards face = $-p(x+\Delta x) \Delta y \Delta z$
on the whole fluid =$F_x \Delta \tau$. For equilibrium we must have $$ \left(
p(x+\Delta x) - p(x) \right) \Delta y \Delta z = F_x \Delta \tau \\
\frac{\partial p}{\partial x} \Delta x \Delta y \Delta z = F_x \Delta \tau \\
F_x = \frac{\partial p}{\partial x}$$

Similarly, for other directions and therefore we have $$ \mathbf F = \nabla p$$ .

My problem is that Mr. Arnold Sommerfeld is saying that this pressure distribution is keeping the fluid from moving. He gives the reason that gravity has a potential and can be written like $$\mathbf F = - \nabla U$$ And for this he says
Equilibrium is only possible if the external force has a potential.

My problem is how is potential or pressure distribution is keeping the fluid from moving? I think it is the lower wall and side walls of the column that is preventing the fluid from moving. I’m sceptical because later on he proves that if the force were to be magnetic, then fluid will start to flow in a circular motion. What is the significance of $\nabla P$ and $\nabla U$?

Please explain me what is he trying to emphasise.

 

[jbriggs444]

My problem is how is potential or pressure distribution is keeping the fluid from moving?

The pressure distribution is preventing the fluid from accelerating. If the fluid starts from rest, a lack of acceleration is sufficient to keep it at rest. A lack of acceleration is also required to keep the fluid at rest.

The author is expecting you to consider the interior of the fluid volume. He does not seem concerned here with the conditions at the boundaries of the volume.

 

[Adesh]

The pressure distribution is preventing the fluid from accelerating.

Wasn’t there any pressure in fluid when there was no gravitational force? I mean will there be a pressure when there is no external conservative force?

 

[vanhees71]

Take an ideal fluid. Then Euler's equation reads
$$\rho \mathrm{D}_t \vec{v}=\rho (\partial_t \vec{v} + (\vec{v} \cdot \vec{\nabla} \vec{v})=-\vec{\nabla} P+ \rho \vec{g},$$
were I assumed only the gravitational force close to Earth as an external force. In static situations the left-hand side vanishes and then
$$\vec{\nabla} P=\rho \vec{g}.$$

 

[jbriggs444]

Wasn’t there any pressure in fluid when there was no gravitational force? I mean will there be a pressure when there is no external conservative force?

The pressure gradient in such a case must be zero. As @vanhees71 showed above.

The pressure can take on any value. For instance, it depends on how much air you put in the tire.

 

[Adesh]

Will there be no pressure (or any value of pressure) when the fluid is subjected to a non-conservative force?

 

[jbriggs444]

Will there be no pressure (or any value of pressure) when the fluid is subjected to a non-conservative force?

You cannot have an equilibrium in such a case. There can be pressure, certainly. But the fluid will also be accelerating.

 

[Adesh]

Thank you so much.