- #1
nonequilibrium
- 1,439
- 2
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].
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?
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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