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While reading the section on Hamiltonian mechanics in Taylor's Classical mechanics, I realized that I didn't fully understand what he was saying when he was explaining why $$\nabla\cdot\vec{F(\vec{x_0})}=\frac{1}{V}\frac{dV}{dt}, $$for small volumes around $\vec{x_0}.$

$$\frac{dV}{dt}=\int_{\partial V}\vec{n}\cdot\vec{F}dA=\int_V\nabla\cdot\vec{F}dV.$$ Taking over a small enough volume, $\nabla\cdot\vec{F}$ is constant, factors out of the integral, and the result follows.

I'm not sure why $\nabla\cdot\vec{F}$ is locally constant. The first time I read this, I thought is was just a matter of assuming sufficient smoothness on the partial derivatives, but after thinking for a second, that can't imply locally constant. So I'm a little confused.

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    $\begingroup$ Let $U$ be a simply connected open subset of $\mathbb R^3$. Take any smooth function $F : U \to \mathbb R$. The author claims that as the volume $V_U$ of $U$ approach 0, the following approximate result holds: $$\int_U F dU \simeq V_U F$$ where $dU$ is the pullback of the usual $\mathbb R^3$ volume form by the inclusion. See if you can convince yourself of that. $\endgroup$
    – zzz
    Jun 26, 2015 at 23:46

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Think about it one more time. If $\vec{F}$ has continuous partial derivatives, then $$\vec\nabla\cdot\vec{F}=\sum_i \frac{\partial F_i}{\partial x_i}$$ is also continuous. If a function is continuous, it's approximately constant on sufficiently small volumes: that's pretty much the definition of continuity! So your original understanding was just fine.

Maybe your confusion is on what locally constant means? It doesn't mean that the function is actually constant on any given region, just that as the region gets smaller and smaller, the variation of the function over the region tends to zero.

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  • $\begingroup$ Yes, that was exactly my misunderstanding. Thank you! :) $\endgroup$
    – user153582
    Jun 27, 2015 at 9:25
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Divergence at a point measures the flux flowing out from an infinitesimal volume around that point.

Suppose you have a volume $v$ in vector field $\mathbf{h}$. Divide it into two parts: $$v = v_1 +v _2$$. Now the flux out of volume $v$ is given by : $\int \mathbf{h}\cdot da_v = \int \mathbf{h} \cdot da_{v_1} + \int \mathbf{h}\cdot da_{v_2}$. Start dividing the volume into $N$ parts so that $v = \sum_{i=1}^N v_i$. So, the net flux out of volume $v$ is $$\int \mathbf{h} \cdot da_v= \sum_{i=1}^N \int \mathbf{h} \cdot da_{v_i}$$. This is a macrosopic quantity. However, we want to find some microscopic property; in order to do this, we make $N \to \infty$ so that $da_{v_i} \to 0$. Let the flux out of such infinitesimal volume $v_i$ is $\int\mathbf{ h}\cdot da_i$. This quantity is surely approximating to $0$. But if we take the ratio of the flux divided by the volume it encloses, we can get a finite quantity. This is what we call divergence. $$\text{div} \mathbf{h}_i \equiv \lim_{v_i \to 0} \dfrac{1}{v_i} \int \mathbf{h}\cdot da_{v_i} $$.

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