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Yes, the reason for this is that in maths $\nabla$ is often used as the vector differential opertator. This is a vector. When typesetting the convention to denote a vector is bold text e.g. $\bf{x}$. However for handwriting you can't really write bold font so other conventions are needed. Common ones are putting an arrow over as in your example or ...

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Yes, there are sometime different conventions for indicating vectors in hand-writing and printing. Yes, overset arrows in handwriting and boldface in printing is one of those conventions. No, it is not the only convention. Yes, you should familiarize yourself with the most common conventions in your sub-discipline. Yes, you should read the section on ...

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If $L(t)$ were a scalar valued function of time then, by the product rule, we have $$\frac{d}{dt}L^2(t) = \frac{d}{dt}\left(L(t) \cdot L(t) \right) = \frac{dL}{dt} \cdot L(t)+ L(t) \cdot\frac{dL}{dt} = 2L(t)\cdot\frac{dL}{dt}$$ Due to linearity, this holds for ordinary vector valued functions of time since $$L^2 = \vec L \cdot \vec L$$ Thus ...

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This is true for any vector quantity from a finite-dimensioned vector space that uses the standard definition of the inner product. Let $\mathbf u$ and $\mathbf v$ be elements of the vector space $\mathbb R^N$ with inner product $\mathbf u \cdot \mathbf v = \sum_{i=1}^N u_i v_i$. Then $\mathbf u \cdot \mathbf u = \sum_{i=1}^N {u_i}^2$. Using the standard ...

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It is the easiest to think of this problem in the component form. Let's say the vector $\vec L$ is in a 3 dimensional space (that is usually what we use). So in component form, i.e., written as a 3 by 1 matrix, $\vec L=(L_1,L_2,L_3)$. What is the left-hand-side of your expression? It is $\vec L\cdot\frac{\mathrm d\vec L}{\mathrm dt}=L_1\cdot\frac{\mathrm ... 1 From math and the power rule:$\dfrac{d(x^2)}{dx} = 2x$And we assume that L is a function of time:$\vec{L} = \vec{L(t)}$. To refresh you on the chain rule: if x were a function of time, then$\dfrac{d(f(x))}{dt} = \dfrac{d(f(x))}{dx} * \dfrac{dx}{dt}$. Back to math:$\dfrac{d(x^2)}{dx}$is actually$2x\dfrac{dx}{dx}$if you apply said chain rule. ... 1 The part that I don't get is mathematically how to get from one to the other. $$\frac{d}{dt}U_L(t) = p_L(t) = v_L(t) \cdot i_L(t) = L \cdot i(t) \cdot \frac{di}{dt}$$ but $$\frac{d}{dt}\left\{\frac{1}{2}Li^2(t)\right\} = L \cdot i(t) \cdot \frac{di}{dt}$$ thus $$U_L(t) = \frac{1}{2}Li^2(t)$$ 6 What Goldstein means by$\nabla_iV_i$is $$\nabla_iV_i=\left(\frac{\partial}{\partial x_{1,i}}\hat{x}_{1,i}+\frac{\partial}{\partial x_{2,i}}\hat{x}_{2,i}+\frac{\partial}{\partial x_{3,i}}\hat{x}_{3,i}\right)V_i$$ which is indeed a vector. Here,$\mathbf r_i=(x_{1,i},\,x_{2,i},\,x_{3,i})$is the position of the$i$th particle (with respect to the origin), ... 5 This is typical to see in situations where the potential is a function of the coordinates of more than one particle: $$V=V(\mathbf r_1,\ldots,\mathbf r_N)=V(x_1,y_1,z_1,\ldots,x_N,y_N,z_N).$$ The force produced by such a potential on the$i$th particle is the gradient of this function with respect to that particle's coordinates, while holding all the other ... 1 Consider an$n$-manifold$M$. A curve is simply a continuous map$\gamma : \mathbb{R} \to M$. For simplicity, suppose$M$is covered by a single coordinate chart (diffeomorphism)$\varphi : M \to \mathbb{R}^n$. Putting these together, we have \begin{eqnarray} \mathbb{R} & \stackrel{\gamma}{\longrightarrow} & M & ... 2$Q\$ is not a functional, but a linear operator. Since it is linear, there are no problems in using the chain rule. I will pretend there are no domain problems here, and that you can exchange limits and integrals as you wish (e.g. I'm supposing you can use the dominated convergence theorem), and obviously that each quantity is differentiable. ...

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