How do you prove that pressure is same in all directions (i.e: isotropic)?

Question

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."

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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.

I assumed maybe that since pressure’s direction is determined by the area which it may act on and hence any area element about a point 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:

1. How is pascal's law related to the isotropic nature of pressure?

2. How do you prove the isotropic nature of pressure using pascal's law in lack of shear forces or however another way?

Answer 1

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.

Answer 2

Fluids can't sustain a shear stress and will rearrange their shape correspondingly. This can't be possible in the ideal hydrostatic case at steady state; otherwise, we could use this persistent flow to drive an engine to cool a fluid-containing isolated heat reservoir, in violation of the Second Law.

There's only one stress state that involves no shear in any orientation: equitrixial normal stress, or

$$\boldsymbol{\sigma}=\left[\begin{array}{ccc} \sigma & 0 & 0\\ 0 & \sigma & 0\\ 0 & 0 & \sigma \end{array}\right].$$

The compressive version ($\sigma=-P=-\rho g h$, generally corresponding to a gravity-based body load for density $\rho$, gravity $g$, and depth $h$) is known as pressure.

Answer 3

Suppose some volume in the fluid. The total force acting on this volume is equal to the integral of the pressure, take over the surface bounding the volume. $$-\oint p d\mathbf{f}$$

Transforming it to a volume integral, we have $$-\oint p d\mathbf{f}=-\int \nabla p dV$$ Hence we see that the fluid surrounding any volume element $dV$ exerts on that element a force $-dV \nabla p$. In other words, we can say that a force $-\nabla p$ acts on the unit volume of the fluid.

If there is no external force then $$\nabla p =0$$

i.e. $p = \mathrm{constant}$; the pressure is the same at every point in the fluid.

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Edit: For a fluid at rest in a uniform gravitational field, We have $$\nabla p =\rho g$$ This can be integrated immediately if the density of the fluid may be supposed constant throughout its volume i.e. if there is no significant compression of the fluid under action of the external force. Taking the z-axis vertically upward, we have $$\frac{\partial p}{\partial x}=\frac{\partial p}{\partial y}=0$$ $$\frac{\partial p}{\partial z}=-\rho g$$ Hence $$p=-\rho gz+\mathrm{constant}$$

If the fluid at rest has a free surface at height $h$, to which an external pressure $p_0$, the same at every poinnt, is applied, this surface must be the horizonal plane $z=h$. From the condtion $p=p_0$ for $z=h$, we find that the constant is $p_0+\rho gh$, so that $$p=p_0+\rho g(h-z)$$

Answer 4

The fact that different principal stresses leads to shear can be visualized in the picture below. The max and min. principal stresses are shown. Suppose there is a weak plane at the diagonal of the cube. It is clear the potential direction of displacement, corresponding to the shear components of the loading.

As showed in the Dale's answer, for an angle of $45^{\circ}$ there is a maximum of the shear stresses, that are proportional to the difference between the principal stresses.

The only way to avoid shear stresses is if all principal normal stresses are equal. In this case the projections at $45^{\circ}$ cancel.

Answer 5

Start with the stress tensor where all diagonal elements are equal.

$$\mathbf A= \left[ \begin {array}{ccc} \sigma_3&0&0\\ 0&\sigma_3&0 \\ 0&0&\sigma_3\end {array} \right] $$

with the eigenvalues $~\lambda_i=\sigma_3~,i=1..3~$ you obtain the Jordan transformation matrix (eigen vectors $~\mathbf n_i$) $$\mathbf T =\bigg[\mathbf n_1~,\mathbf n_2~,\mathbf n_3\bigg]=\left[ \begin {array}{ccc} 1&0&0\\ 1&1&0 \\ 1&1&1\end {array} \right] $$

with Gram-Schmidt method you get out of the matrix $~\mathbf T~$ orthogonal transformation matrix

$$\mathbf T_G=\left[ \begin {array}{ccc} 0&-1/3\,\sqrt {6}&1/3\,\sqrt {3} \\ -1/2\,\sqrt {2}&1/6\,\sqrt {6}&1/3\,\sqrt {3} \\ \,\sqrt {2}&1/6\,\sqrt {6}&1/3\,\sqrt {3} \end {array} \right] $$

where $~\mathbf T_G^T\,\mathbf A\,\mathbf T_G=\mathbf A$

if you apply this transformation matrix to transformed a stress tensor where the diagonal elements are not equal $$\mathbf B= \left[ \begin {array}{ccc} \sigma_1&0&0\\ 0&\sigma_2&0 \\ 0&0&\sigma_3\end {array} \right] $$

you obtain $$\mathbf Z=\mathbf T_G^T\,\mathbf B\,\mathbf T_G=\left[ \begin {array}{ccc} 1/2\,\sigma_{{2}}+1/2\,\sigma_{{3}}&-1/6\, \sqrt {3}\sigma_{{2}}+1/6\,\sqrt {3}\sigma_{{3}}&1/6\,\sqrt {2}\sqrt { 3} \left( \sigma_{{3}}-\sigma_{{2}} \right) \\ -1/6 \,\sqrt {3}\sigma_{{2}}+1/6\,\sqrt {3}\sigma_{{3}}&2/3\,\sigma_{{1}}+1 /6\,\sigma_{{2}}+1/6\,\sigma_{{3}}&1/6\,\sqrt {2} \left( \sigma_{{2}}+ \sigma_{{3}}-2\,\sigma_{{1}} \right) \\ 1/6\,\sqrt { 2}\sqrt {3} \left( \sigma_{{3}}-\sigma_{{2}} \right) &1/6\,\sqrt {2} \left( \sigma_{{2}}+\sigma_{{3}}-2\,\sigma_{{1}} \right) &1/3\,\sigma _{{1}}+1/3\,\sigma_{{2}}+1/3\,\sigma_{{3}}\end {array} \right] $$

from requirements no shear stress you get:

$$\sigma_2=\sigma_3\\ \sigma_1=\frac 12\,(\sigma_2+\sigma_3)=\sigma_3$$

substitute in $~\mathbf Z$

$$\mathbf Z= \left[ \begin {array}{ccc} \sigma_3&0&0\\ 0&\sigma_3&0 \\ 0&0&\sigma_3\end {array} \right] $$

so only if $\sigma_1=\sigma_2=\sigma_3$ there is no shear elements is the stress tensor

Answer 6

So lets start here:

Remember that there is no shear stress for a fluid with no force fields. So we get $$p_xw\Delta z-p_nb\Delta s \sin(\theta)=0$$ $$p_zb\Delta x -p_nb\Delta s \cos(\theta)-\rho b (\frac{1}{2} \Delta x \Delta z)=0$$

Now using $\Delta z=\Delta s \sin(\theta)$ and viz. And setting $\Delta x \to 0 $ $\Delta z \to 0$ which gives us our isotropic condition $$p_x=p_y=p_n$$ Note that $\hat n(\theta)$ is arbitrary. Now say a net force acts in z direction to a cubical element(gravity). Let's assume a pressure $p_0$ acts in upward direction at the bottom of the element $$p_{top}=p_0+\frac{d p}{dz} \Delta z$$ Now force acting on the element in z direction is $$F=-\frac{dp}{dz}dxdydz$$ Now force due to its weight is $$F_{weight}=-\rho g dxdydz$$ Condition for static equilibrium gives $$\frac{dp}{dz}=-\rho g$$