Work performed by hydrostatic pressure

Question

One should be able to show mathematically that the hydrostatic work done by an environment on an object undergoing a volume change $\Delta v$ should be $p \Delta v$, where $p$ is the (constant) pressure of the environment. I am interested in a derivation using mechanical principles, not thermodynamical principles.

Conceptually we imagine at each moment $t$ in the evolution of the object's shape an approximation the object's surface by a fine mesh of flat polygons, i.e. a tesselation. We can also imagine a "compatible tesselation" of the object's surface at a nearby moment $t+\delta t$ so that if we join each vertex of the tesselation at time $t$ to its associated vertex at time $t + \delta t$ we obtain for each polygon pair a small volume region capped by the polygon pair with no two volume regions intersecting. Then for each polygon there is associated a differential work $\delta W = \mathbf{\delta F} \cdot \mathbf{\delta x}$ performed by the environment in the evolution from $t$ to $t + \delta t$, where

• $\mathbf{\delta F}$ is the force applied on the polygon by the environment, where $\mathbf{\delta F} = p \mathbf{\delta A}$ with $\mathbf{\delta A} = \mathbf{n} \delta A$ a vector of magnitude $\delta A$ equal to the polygon's area and direction $\mathbf{n}$ perpendicular to the polygon pointing into the object, and
• $\mathbf{\delta x}$ is a vector starting at the centroid of the polygon at $t$ and ending at the centroid of the polygon at $t + \delta t$.

Taking the limit $\delta A, \delta t \to 0$ and summing up the differential works $\delta W$ for each polygon in an interval $t \to \delta t$ and then summing over all intervals, we should obtain the total work $W$ performed by the environment for the total transformation $t = t_o \to t'$, which should be equal to $p \Delta v$, where $\Delta v$ should be the difference $v(t_o) - v(t')$ in the initial and final object volumes.

I am able to derive (see my answer below) the desired relationship using somewhat sophisticated (at least for most physicists) differential geometry techniques. Should this elementary relation $W = p \Delta v$ have a more elementary proof?

Answer 1

With a little less math, the power of a stress distribution $\mathbf{t}_n$ over the boundary of a volume $V$ is

$$P(t) = \oint_{\partial V} \mathbf{t}_n \cdot \mathbf{u} \ ,$$

being $\mathbf{u}$ the velocity of the boundary. If stress is due only to pressure, $\mathbf{t}_n = - p \mathbf{\hat{n}}$. If pressure is uniform $p(\mathbf{r},t) = \overline{p}(t)$ and the power becomes

$$P(t) = - \overline{p}(t) \oint_{\partial V} \mathbf{\hat{n}} \cdot \mathbf{u} \ .$$

The integral is the rate of change $\frac{d V}{d t}$ of the volume $V$, as it can be proved with Reynolds theorem

$$\frac{d}{dt} \int_{V(t)} f = \int_{V(t)} \frac{\partial f}{\partial t} + \oint_{\partial V(t)} f \mathbf{u} \cdot \mathbf{\hat{n}} \ ,$$

with $f(\mathbf{r},t) = 1$ (remember that $\int_{V(t)} 1 = V(t)$),

$$\frac{d}{dt} V(t) = \oint_{\partial V(t)} \mathbf{u} \cdot \mathbf{\hat{n}} \ .$$

Thus, power done on the system reads

$$P(t) = - \overline{p}(t) \frac{dV}{dt}(t) \ ,$$

and the work

$$W = - \int_{t_0}^{t_1} \overline{p}(t) \frac{dV}{dt}(t) dt \ ,$$

that, if pressure is constant in time, becomes

$$W = - \overline{p} \int_{t_0}^{t_1} \frac{dV}{dt}(t) dt = - \overline{p} (V(t_1) - V(t_0)) = -\overline{p} \Delta V \ .$$

Please, note that the work done on the system is positive if the volume shrinks, $\Delta V < 0$.

Answer 2

My intuition is that we are looking at a special case of Stokes' theorem, since we are relating surface integrals to volume changes. To this end, it seems clear enough to me that the differential calculus formulation of the total work is the following: $$ W = p \int_{t_o}^{t'} dt \oint_{S(t)} \mathbf{f}(t) \cdot \left(-\mathbf{n}\right) dS = p \int_{t'}^{t_o} dt \oint_{S(t)} \mathbf{f}(t) \cdot \mathbf{n} dS $$ where $S(t)$ is the surface of the object at a time $t$, $\mathbf{n}$ is a unit vector normal to the surface element $dS$, and $$ \mathbf{f}\left(\mathbf{r};\tau\right) = \frac{d}{dt}\Bigr|_{t=\tau}\phi_t\left(\mathbf{r_o}\right) $$ where $\mathbf{\phi}_t\left(\mathbf{r_o}\right) = \mathbf{r}$ and $\mathbf{\phi}_t$ is a smooth function taking values from the set of points $V(t_o) \subset \mathbb{R}^3$ corresponding to the initial object geometry and returning points from the set $V(t) \subset \mathbb{R}^3$ corresponding to the object geometry at a moment $t$. We require also that $\phi_0\left(\mathbf{r_o}\right) = \mathbf{r_o}$ for all $\mathbf{r_o} \in V(t_o)$.

From Gauss' theorem we have for any vector field $\mathbf{W}$ that $\int_V \left(\nabla \cdot \mathbf{W}\right) dV = \oint_S \left( \mathbf{W} \cdot \mathbf{n} \right) dS$ so that in our case we have $$ \oint_{S(t)} \mathbf{f}(t) \cdot \mathbf{n} dS = \int_{V(t)} \left(\nabla \cdot \mathbf{f}\right) dV $$ At this point I need to leave the world of concrete vector calculus and resort to differential geometry tactics. We start by rewriting $\int_{V(t)} \left(\nabla \cdot \mathbf{f}\right) dV$ as a integral over $V(t)$ of the volume form $\left(\nabla \cdot \mathbf{f}\right) \omega$, where $\omega$ is the standard volume form $dx \wedge dy \wedge dz$ on $\mathbb{R}^3$: $$ \int_{V(t)} \left(\nabla \cdot \mathbf{f}\right) dV = \int_{V(t)} \left(\nabla \cdot \mathbf{f}\right) \omega $$ From here we apply the identity $$ \left(\nabla \cdot \mathbf{f}\right) \omega' = \mathcal{L}_\mathbf{f} \omega'\,\,\,, $$ true for any volume form $\omega'$, where $\mathcal{L}_{\mathbf{W}}$ denotes the Lie derivative of vector field $\mathbf{W}$, so that we now have $$ \int_{V(t)} \left(\nabla \cdot \mathbf{f}\right) dV = \int_{V(t)} \mathcal{L}_\mathbf{f} \omega $$ Now since $\mathbf{\phi}_t$ maps $V(t_o) \to V(t)$, we have, for any volume form $\omega'$ that $$ \int_{V(t)} \omega' = \int_{V(t_o)} \mathbf{\phi}^*_t \omega' $$ where $\psi^*$ denotes pullback of a function $\psi$. We further have for any vector field $\mathbf{W}$ that $$ \mathcal{L}_\mathbf{W} \omega' = \lim_{\epsilon \to 0} \frac{\left( \mathbf{\psi}^*_\epsilon - 1 \right)}{\epsilon} \omega' $$ where $\psi_\tau$ is the flow generated by the vector field $\mathbf{W}$. For our vector field $\mathbf{f}(t)$ we have $\psi_\tau = \phi_{t+\tau}\circ\phi_t^{-1}$, so that we can say $$ \int_{V(t)} \left(\nabla \cdot \mathbf{f}\right) dV = \int_{V(t_o)} \mathbf{\phi}^*_t \lim_{\epsilon \to 0} \frac{\mathbf{\psi}^*_\epsilon -1}{\epsilon} \omega $$ However, since $f^*g^*=\left( g \circ f \right)^*$, this also means that $$ \int_{V(t)} \left(\nabla \cdot \mathbf{f}\right) dV = \int_{V(t_o)} \lim_{\epsilon \to 0} \left( \phi_{t+\epsilon}^* - \phi_t^* \right) \omega = \frac{d}{d \tau}\Bigr|_{\tau=t} \int_{V(\tau)} \omega $$ The expression $\int_{V(t)} \omega$ is however simply $\int_{V(t)}dV$, i.e. the volume $v(t)$ of the solid at the moment $t$. Plugging this in to our expression for the work, we get $$ W = p \int_{t'}^{t_o} dt \frac{d}{dt} v(t) = p \left( v(t_o) - v(t') \right) = p \Delta v $$ as desired. Of course the mathematical result does not change if we modify the "speed" at which the transformation takes place. That is, the result holds should we replace $\phi_t$ by $\phi_{\sigma(t)}$ where $\sigma(t)$ is some continuous function satisfying $\sigma(t_o) = t_o$ and $\sigma(t') = t'$.