Pressure in Fluids: Conceptual Q & Maths Explained

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SUMMARY

This discussion centers on the concept of pressure in fluids, specifically addressing the isotropy of pressure in hydrostatics. It is established that pressure in a fluid at rest is equal in all directions at a given depth, as stated in the textbook. The conversation highlights that while vertical pressure differs due to height, horizontal pressure must remain equal in opposing directions to prevent unbalanced forces. The isotropy condition is crucial, indicating that pressure is a scalar quantity and does not depend on the orientation of the surface element.

PREREQUISITES
  • Understanding of hydrostatics and fluid mechanics
  • Familiarity with the concept of pressure as a scalar
  • Knowledge of isotropy in physical systems
  • Basic principles of statistical mechanics related to particle interactions
NEXT STEPS
  • Study the derivation of hydrostatic pressure equations in fluids
  • Explore the implications of the isotropy condition in fluid dynamics
  • Investigate the statistical mechanics of fluids and particle distribution
  • Learn about the mathematical representation of pressure as a scalar field
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Students and professionals in physics, particularly those focusing on fluid mechanics, hydrostatics, and statistical mechanics, will benefit from this discussion.

crazy student
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I have a conceptual question about fluid...
we derive the equations about pressure in fluid by considering the mass above a particular point in fluid.
and for fluid at rest, the pressure above equals pressure from below.
how about on the perpendicular plane? is the pressure on the sides equal to pressure from above and below also?
my textbook says that 'At any point in a fluid at rest, the pressure is the same in all directions at a given depth',
is it an experimental statement or can it be shown mathematically?
 
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If the pressure was not the same on either side of your fluid "element", there would be an unbalanced (net) lateral force on it, and it would move. The fact that we're considering hydrostatics means that must not happen.
 
crazy student said:
is the pressure on the sides equal to pressure from above and below also?

To tackle this specific point, the whole point of the analysis was to show that pressure varies only with height. Since "above", "below", and "on the sides", generally describes three different heights, I don't see why the pressure would be the same at any of these points. For instance, you already know that the pressure below is higher than the pressure above, so how could the pressure on the sides be equal to the pressure "above and below"?!
 
cepheid said:
If the pressure was not the same on either side of your fluid "element", there would be an unbalanced (net) lateral force on it, and it would move. The fact that we're considering hydrostatics means that must not happen.

But this is not sufficient to demonstrate that the horizontal pressure is the same as the vertical pressure. It only requires that horizontal pressure be the same in two sets of opposite directions and that vertical pressure be the same in both directions, not that they all be equal. The equality in all directions is unique to fluids. It is certainly not a requirement for solids.

I think you would have to look microscopically at the nature of fluids and get into the statistical mechanics of ensembles of particles to justify the statement of equal pressure in all directions. The pressure is the same in all directions because the kinetic energy distribution of the particles in a fluid is isotropic. The interaction of an infinitesimal volume of fluid with its surroundings will involve collisions of energetic particles with isotropic momentum distribution, etc etc etc
 
OlderDan said:
But this is not sufficient to demonstrate that the horizontal pressure is the same as the vertical pressure.
Pressure is a scalar. It has no direction, so there's no such things as horizontal and vertical pressure.
 
OldeDan is correct, what we have to show is that the scalar pressure, p=p(x,y,z,t)
at most, and not p(x,y,z,t,\vec{n}) where \vec{n} is the normal to the surface element we're looking at, at the point (x,y,z) at time t
This is not completely trivial, since forces would balance on either side of a surface, as long as p were an even function of \vec{n}
Since, then:
p(\vec{x},t,\vec{n})\vec{n}dA+p(\vec{x},t,-\vec{n})(-\vec{n})dA=\vec{0}
The condition that the pressure does not depend on the orientation of the local surface element is called the isotropy condition.
 
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