Proof
Consider an infinitesimal element of fluid in the form of a triangular prism containing a point PP of interest as shown in the figure below.
Let the pressures P_xPx along xx axis, P_yPy along yy axis, and P_sPs normal to any plane inclined at any angle thetaθ to the horizontal at this point be acting as shown.
There can be no shearing forces for a fluid in equilibrium. As such the sum of the forces in any direction must be zero. The forces acting are due to the pressures on the surrounding and the gravity.
Knowing the definition of pressure and that x and yxandy axes are orthogonal to each other, sum of forces in the xx-direction is
P_x xxdeltayxxdeltaz +( - P_sxxdeltazxxdeltas sin theta) = 0Px×δy×δz+(−Ps×δz×δssinθ)=0
=>P_x xxdeltayxxdeltaz-Psxxdeltasxxdeltazxx(deltay)/(deltas)=0⇒Px×δy×δz−Ps×δs×δz×δyδs=0
=>P_x xxdeltayxxdeltaz-Psxxdeltazxxdeltay=0⇒Px×δy×δz−Ps×δz×δy=0
=>P_x = P_s⇒Px=Ps .....(1)
Sum of forces in the yy-direction including weight of the fluid element is
P_y xxdeltax xxdeltaz +( - P_sxxdeltazxxdeltas cos theta) +(-"specific gravity"xx "Volume"xxg)= 0Py×δx×δz+(−Ps×δz×δscosθ)+(−specific gravity×Volume×g)=0
=>P_y xxdeltax xxdeltaz - P_sxxdeltazxxdeltas (deltax)/(deltas)⇒Py×δx×δz−Ps×δz×δsδxδs
-rhoxx 1/2deltax xxdeltayxxdeltazxxg= 0−ρ×12δx×δy×δz×g=0
Since deltax, deltay and deltazδx,δyandδz are infinitesimal; therefore, the last term which is product of three infinitesimals can be ignored in comparison to first two terms which is product of two infinitesimals. We have
P_y xxdeltax xxdeltaz - P_sxxdeltazxxdeltax = 0Py×δx×δz−Ps×δz×δx=0
=>P_y = P_s⇒Py=Ps .....(2)
From (1) and (2) we get
P_x = P_y=P_sPx=Py=Ps .....(3)
By selecting orientation of coordinate system appropriately the above expression could be extended to include the zz-axis.
:.P_x = P_y=P_z
In the limiting case the element can be considered a point so the derived expression indicates that pressure at any point is the same in all directions.
