What is the “pressure on a point” precise definition?

Question

I'm reading Landau's Fluid Mechanics and in the first page is defined the pressure in every point and every time: $p=p(x,y,z,t)$. Here every "point" $(x,y,z)$ is really a tiny differential volume $dV$ e.g. a small rectangular box of dimensions $dx$, $dy$, $dz$ ($dV=dx dy dz$), which contains a lot of particles.

This pressure $p$, as a function, has the property that $\oint_S p\ dS$ is the total exterior force over any surface $S$, this suggests that the pressure is defined as the total exterior force over the surface of a tiny volume dV divided the value of its surface. For example, if we apply forces to every face of a box of dimensions $a, b, c$:

Then the pressure over this box is: \begin{equation} p=\frac{F_{x+}+F_{x-}+F_{y+}+F_{y-}+F_{z+}+F_{z-}}{2ab+2bc+2ca} \end{equation}

Now, for example, if I have a large box of dimensions $L$, $2L$, $2L$, and over this box are exterior forces $F_x$, $F_y$, $F_z$ trying to compress this box, and the box doesn't moves, then the total exterior force applied to the box is $2(F_x +F_y +F_z)$. Suppose that the forces are uniformly distributed over the faces.

Now let's calculate the integral of the pressure over the surface of this box (it must be $2(F_x +F_y +F_z)$). To do this, we can divide the box in little cubes of volume $L^3/n^3$. The force over each of the two faces orthogonal to the $x$ axis is $F_x/4n^2$, and the force over the faces orthogonal to $y$ axis is $F_y/2n^2$, similarly the force over the faces orthogonal to $z$ axis is $F_z/2n^2$.

Then the pressure over each tiny cube of volume $L^3/n^3$ is: \begin{equation} p_0=\frac{2\left(\frac{F_x}{4n^2}+\frac{F_y}{2n^2}+\frac{F_z}{2n^2} \right)}{6 L^2/n^2} \end{equation}

Ignoring the edges and vertices, we can estimate the pressure surface integral taking the total number of little cubes on the surface but the edges, and multiplying it by $p_0$. There are $(2n-2)^2$ such cubes on the two faces with surface $4L^2$, and $(2n-2)(n-2)$ on each of the four remaining faces of surface $2L^2$. Let $S$ be the surface of the large box. Let $\Delta S$ be the surface of the face of a little cube ($\Delta S = L^2/n^2$).

\begin{equation} \oint_S p\ dS\approx \left(2(n-2)^2 + 4(2n-2)(n-2) \right)p_0 \Delta S = \left(2(n-2)^2 + 4(2n-2)(n-2) \right)\frac{2\left(\frac{F_x}{4n^2}+\frac{F_y}{2n^2}+\frac{F_z}{2n^2} \right)}{6 L^2/n^2}\frac{L^2}{n^2}=\frac{4}{3}\frac{3n^2-8n+5}{n^2}\left(\frac{F_x}{4}+\frac{F_y}{2}+\frac{F_z}{2} \right) \end{equation}

Taking limit as $n\rightarrow \infty$, and considering that edges are negligible for the surface integration:

\begin{equation} \oint_S p\ dS = F_x + 2F_y + 2F_z \end{equation}

But this can't be correct, because the force over the surface is $2(F_x +F_y +F_z)$. I don't really understand what is wrong. Is it the definition of pressure? Or is it the integration?

Answer 1

In the book it's stated that the quantity $-\oint p\mathrm d \mathbf f$ is the total force. If you notice the the $\mathrm d \mathbf f$ in bold you can see it's a vector and it essentially means that the integral is done component by component so your calculations don't apply. So, for example: $$\int p_{x+}dS = \int \frac{F_{x+}}{bc}dS=F_{x+}\int \frac{dS}{bc}=F_{x+}$$ and similarly for the other components. In this example dS is not a vector. As you can see you always retrieve the original component.

As for the precise definition it is the constant of proportionality between the vectors $\mathrm d \mathbf F_n$, the normal component of $\mathrm d \mathbf F$ in the surface, and $\mathrm d \mathbf S$. Note it's defined infinitesimally as these vectors generally are functions of the position.

Answer 2

Your problems start when you start treating forces and areas as scalars.

Then the pressure over this box is:

\begin{equation} p=\frac{F_{x+}+F_{x-}+F_{y+}+F_{y-}+F_{z+}+F_{z-}}{2ab+2bc+2ca} \end{equation}

is incorrect.

You need to use the vector form of the equation which gives you the force on an area as described in the Wikipedia article on Pressure.