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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 $a, b, c$:

tiny box and forces

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.

enter image description here

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?

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  • $\begingroup$ Pressure at a point should be a tensor that maps a differential vector area to a force. $\endgroup$ – velut luna Jan 3 '17 at 21:55
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    $\begingroup$ Two remarks: First, we are doing continuum mechanics here. When we say "pressure at a point" then that's exactly what we mean. There's no need to ever consider "tiny differential volumes containing a lot of particles". There are no particles. Second, I didn't spend any time going through the details (TL;DR), but apart from the conceptual mistake explained in the answer(s) below it is entirely clear that you should obtain a symmetric result that is invariant with respect to changes in your coordinate labels. The fact that you did not means you did something wrong. $\endgroup$ – Pirx Jan 4 '17 at 1:16
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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.

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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.

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