The derivation of the divergence $\nabla \cdot v$ in spherical coordinates begins with the general tensor calculus formula, $\nabla \cdot v=\frac{1}{\sqrt{g}} \partial_a\left(\sqrt{g} v^a\right)$. The crucial geometric factor for this coordinate system is the square root of the metric determinant, $\sqrt{ g }= r ^{ 2 } \sin (\theta)$. Substituting this into the formula and simplifying yields the divergence in terms of the contravariant components $\left(v^a\right): \nabla \cdot v= \frac{1}{r^2} \partial_r\left(r^2 v^r\right)+\frac{1}{\sin (\theta)} \partial_\theta\left(\sin (\theta) v^\theta\right)+\partial_{\varphi} v^{\varphi}$. To verify this result against the standard physics expression, the contravariant components were converted to the physical components ( $\tilde{v}_a$ ) using the relationship $\tilde{v}a=\sqrt{\text { g }{a a}} v^a$, which introduces specific scaling factors like $1 / r$ and $1 /(r \sin (\theta))$ for the $\theta$ and $\varphi$ components, confirming the tensor-based derivation is consistent with the traditional vector analysis formula.
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$\complement\cdots$Counselor
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$$ \nabla \cdot v=\frac{1}{\sqrt{g}} \partial_a\left(\sqrt{g} v^a\right) $$
$$ \sqrt{g}= r ^2 \sin (\theta) $$
$$ \nabla \cdot v=\frac{1}{r^2} \partial_r\left(r^2 v^r\right)+\frac{1}{\sin (\theta)} \partial_\theta\left(\sin (\theta) v^\theta\right)+\partial_{\varphi} v^{\varphi} $$
$$ v^\theta=\frac{1}{ r } \tilde{v}\theta \quad \text { and } \quad v^{\varphi}=\frac{1}{ r \sin (\theta)} \tilde{v}{\varphi} $$
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