In the Flow of dry water chapter of Feynman lecture, this following point is written (see here):
The law of hydrostatics, therefore, is that the stresses are always normal to any surface inside the fluid. The normal force per unit area is called pressure. From the fact that there is no shear in a static fluid, it follows that the pressure stress is the same in all directions (Fig. 40-1). We will let you entertain yourself by proving that if there is no shear on any plane in a fluid, the pressure must be the same in any direction.
I am trying to figure out how to prove the statement "if there is no shear on any plane in a fluid, the pressure must be the same in any direction."
What I've found so far:
The premise he puts itself is confusing for me because according to Wikipedia, the pressure is a scalar field rather than a vector field and hence it should have no associated direction.
Anyways, I assumed maybe that since pressure’s direction is determined by the area which it may act on and hence any area element below a given height from the top of the container would have the same pressure acting throughout it.
After some searching on stack exchange, I found that the explanation of pressure being the same in all direction is given by Pascal's law (See Here), but on seeing the Wikipedia page for pascal's law (here), I formed the impression that Pascal's law may be related to pressure transmission rather than what direction pressure acts in.
This led me to search for proof for pascals law.
I found this answer in which the author proves that pressure change in any point in a fluid is transmitted throughout the whole fluid undiminished(pascal's law) but I think that the statement doesn't really explain the isotropic nature of pressure.
Hence this leads me to two main questions:
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How is pascal's law related to the isotropic nature of pressure?
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How do you prove the isotropic nature of pressure using pascal's law or however another way?
Best Answer
The stress tensor is the physical quantity which has units of pressure and describes the force per unit area in different directions. It is a tensor which you can think of as a matrix. When you multiply a matrix by a vector you get another vector. In the case of the stress tensor you multiply the stress by an area vector and you get a force vector, the force acting on that area. This allows us to look at the force in all directions.
Since the stress tensor is a symmetric tensor it has three real eigenvalues: $\sigma_1$, $\sigma_2$, $\sigma_3$. These are the diagonal elements of the tensor in coordinates rotated so that all of the off diagonal elements are 0. These are called the principal stresses and the axes are called the principal axes. In these coordinates the stress tensor is: $$ \left( \begin{array}{ccc} \sigma_1 & 0 & 0 \\ 0 & \sigma_2 & 0 \\ 0 & 0 & \sigma_3 \\ \end{array} \right) $$
Let us order the eigenvalues from largest to smallest so $\sigma_1 \ge \sigma_2 \ge \sigma_3$. The maximum normal stress (pressure in a direction) is equal the largest principle stress, $\sigma_1$, and the minimum normal stress is equal to the smallest principle stress, $\sigma_3$. Pressure being the same in all directions implies that the normal stress is the same in all directions, which is true if and only if the maximum normal stress is equal to the minimum normal stress. So the proof reduces to proving that $\sigma_1=\sigma_3$
In a coordinate system aligned with the principal axes, the shear stress, $\sigma_s$, on a plane defined by its unit normal vector, $ \hat n=(n_1,n_2,n_3) $, where $ \hat n \cdot \hat n = 1$, is given by $$\sigma_s^2=(\sigma_1^2 n_1^2+\sigma_2^2 n_2^2+\sigma_3^2 n_3^2)-(\sigma_1 n_1^2+\sigma_2 n_2^2+\sigma_3 n_3^2)^2$$ We can find the extrema by solving $$\frac{\partial(\sigma_s^2)}{\partial n_i}=0$$ This has local minima of $\sigma_s^2=0$ for $\hat n=(\pm 1,0,0)$, $\hat n=(0,\pm 1,0)$, $\hat n=(0,0,\pm 1)$.
This has local maxima of
$\sigma_s^2=\frac{1}{4}(\sigma_1-\sigma_2)^2$ for $\hat n = (\pm 1/\sqrt{2},\pm 1/\sqrt{2},0)$
$\sigma_s^2=\frac{1}{4}(\sigma_2-\sigma_3)^2$ for $\hat n = (0,\pm 1/\sqrt{2},\pm 1/\sqrt{2})$
$\sigma_s^2=\frac{1}{4}(\sigma_1-\sigma_3)^2$ for $\hat n = (\pm 1/\sqrt{2},0,\pm 1/\sqrt{2})$
Out of these the global maximum is the last one $\sigma_s^2=\frac{1}{4}(\sigma_1-\sigma_3)^2$. Since a fluid in equilibrium has no shear stress we know that the maximum is zero (as is the minimum) so we have $\sigma_s^2=\frac{1}{4}(\sigma_1-\sigma_3)^2=0$ which implies $\sigma_1=\sigma_3$ which proves that the pressure is isotropic.