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$$\dfrac{\partial\dot{r}_i^T\dot{r}_i}{\partial\dot{q}_j}=\dfrac{\partial\dot{r}_i^T\dot{r}_i}{\partial\dot{r}_i}\dfrac{\partial\dot{r}_i}{\partial\dot{q}_j}=2\dot{r}_i^T\dfrac{\partial\dot{r}_i}{\partial\dot{q}_j}$$ (A simple chain rule, while bearing in mind that $\dot{r}_i^T\dot{r}_i$ is essentially $\dot{r}_i^2$. You just need a transposed vector on the ...


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The definition of divergence is $$\textrm{div}\,\vec{F} = \lim_{V \to p}\iint_{S(V)} \frac{\vec{F}\cdot\vec{n}}{|V|}dS, \qquad [1]$$ where $\vec{F}$ is the vector field, $V$ is the volume surrounding the point $p$ where the divergence is calculated, $\vec{n}$ is a unit-length normal vector of the surface, $S(V)$, of the volume, and $|V|$ is the total volume. ...


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The difference is that $dW$ is an infinitesimal ''quantity'', whilst $W$ is not. I assume the context here is thermodynamics, which make use of calculus. In calculus there is the concept of the infinitesimal. I suggest, for you, to concern yourself with the structure of calculus if you are to tackle thermodynamics.


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The Lie derivative has a geometrical meaning: it measures the change of a tensor field (including scalar function, vector field and one-form), along the flow of another vector field. For example, the Lie derivative of the metric tensor along a Killing vector is zero (this defines the Killing vector equation). It means that a tensor (for example the metric) ...


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The function $A_z$ is only a function of $r=\sqrt{x^2+y^2+z^2}$ and $\theta$ (the integral over $z'$ is not a function of $z$, but does depend on $\theta$). I.e., $$ A_z=\frac{\mu e^{-iBr}}{4\pi r}{\tt C(\theta)} $$ Since only (apparently) $A_z$ is non-zero we only need to be able to take the derivative w.r.t. $z$ to get $\nabla \cdot A_z$. $$ \nabla \cdot ...



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