Question

Proof that pressure in a liquid is equal in all direction

Answer 1

Answer:

Explanation:

It is an interesting question, and I am sure there are various ways of looking at it. One view, that of similar to John (which is essentially a microscopic view-point), could be that when a fluid is at rest, one could fairly expect an isotropicity of the molecular motions giving rise to pressure at a point, that is equal in any direction.

But, if restrained to continuum mechanics, and then one tries to answer the above question, few continuum properties of fluids has to be invoked. Fluid, by its very definition, will deform continuously when shear force is applied to it. In other words, a fluid can not stay at rest when a shear force/stress is acting on it. Therefore, if any fluid is at rest, it can be inferred (with certainty) that shear stress vanishes everywhere in the fluid, otherwise the fluid would deform continuously in some fashion. Conclusively, because of the reasoning above, a fluid body at rest can only sustain normal stresses; compressive in most cases.

Given this, consider an infinitesimal two-dimensional, square, fluid element (a two-dimensional element is considered for the sake of simplicity but, the arguments also carry over to a more general three-dimensional scenario). In this orientation of the fluid element, only normal stresses would exist (given that the fluid is at rest). Say, [math]\sigma_x [/math]in the x-direction, and [math]\sigma_y [/math]in y-direction. For a graphical picture, one is referred to figure 4 of the following: Mohr's circle, and is also shown below. It is in this page, one can also find the equation of the Mohr circle. For convenience, I will reproduce the equations here:

[math]\sigma_n = \frac{1}{2} (\sigma_x+\sigma_y) + \frac{1}{2}(\sigma_x-\sigma_y)\cos 2\theta + \tau_{xy}\sin 2\theta[/math]

[math]\tau_n = -\frac{1}{2}(\sigma_x-\sigma_y)\sin 2\theta + \tau_{xy}\cos 2\theta[/math]