I vaguely understand the continuity equation (at least its integral form), but I don't really understand the differential form of the continuity equation. I'm having trouble understanding how to switch back and forth between the forms using the divergence theorem.


The intuitive way to think about this is to consider a gas inside a glass container (that cannot expand). If the gas expands, then what must happen as a result? The gas leaks out of the container. Similarly, if try I put more gas into the container, then the gas compresses.

The vector field $\mathbf F$ is what we use to describe the flow of a fluid. The divergence of this field describes the expansion or compression of a gas. What the divergence theorem says, is that the total expansion (or compression) of the gas in some volume $V$ is equal to the flux of the fluid out (or in) of the boundary (i.e., how much stuff is leaving (or entering) the surface $S$). Mathematically, this is $$ \int_V\nabla\cdot\mathbf F\,dV=\int_{\partial V}\mathbf F\cdot d\mathbf S $$ where $\mathbf F\cdot d\mathbf S$ represents the amount normal to the surface.

So for a volume of mass $m$, the time-rate-of-change of the mass is equal to the above (assuming that there aren't any other sources of matter): $$ \frac{\partial m}{\partial t}=-\int_V\nabla\cdot\mathbf F\,dV=-\int_{\partial V}\mathbf F\cdot d\mathbf S $$ which is our integral formulation of the continuity equation.

Since we know that $m=\int\rho\,dV$, then the above is $$ \frac{\partial}{\partial t}\int\rho\,dV=-\int_V\nabla\cdot\mathbf F\,dV=-\int_{\partial V}\mathbf F\cdot d\mathbf S $$ And since temporal and spatial coordinates are orthogonal, we can swap them to get $$ \int\frac{\partial\rho}{\partial t}\,dV=-\int_V\nabla\cdot\mathbf F\,dV=-\int_{\partial V}\mathbf F\cdot d\mathbf S $$ And lastly, since the volume is arbitrary, then the left two terms above must be $$ \frac{\partial\rho}{\partial t}=-\nabla\cdot\mathbf F $$ which is our differential form of the continuity equation.


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