So Reynold's transport theorem states that [itex]\frac{\mathrm d}{\mathrm d t} \int_{V(t)} f \; \mathrm d V = \int_{V(t)} \partial_t f \; \mathrm d V + \int_{V(t)} \nabla \cdot \left( f \mathbf v \right) \; \mathrm d V[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

Now I would expect (on basis of conceptual reasoning) that if I were to apply this to an infinitesimal element around [itex]\mathbf r(t)[/itex], I should get the well-known (the so-called convective derivative) result [itex]\frac{\mathrm d}{\mathrm d t} f(\mathbf r(t)) = \partial_t f + \left( \mathbf v \cdot \nabla \right) f[/itex]

However, it's straight-forward to see that one gets [itex]\frac{\mathrm d}{\mathrm d t} f(\mathbf r(t)) = \partial_t f + \nabla \cdot \left( f \mathbf v \right)[/itex]

I get a term [itex]f \; \nabla \cdot \mathbf v[/itex] too much. What gives?

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# [fluid dynamics] applying Reynold's transport theorem to an infinitesimal element (?)

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