Derivative of Lagrangian with respect to velocity

Question

My question revolves around this lecture notes on page $109$ equation $(5.1.10)$.

Let's stick to $\mathbb{R}^3$ and consider a particle in $3$-space with position vector $\mathbf{x} = (x, y, z)$. Denote its velocity by $\mathbf{v} = \dot{\mathbf{x}} = (\dot{x}, \dot{y}, \dot{z}).$

Basically, we have the Lagrangian describing the particle:

\begin{equation} L = -mc^2 \sqrt{1 - \frac{v^2}{c^2}}, \end{equation}

where $m$ is the mass of the particle, $c$ is the speed of light constant and $v = |\mathbf{v}| = \sqrt{\dot{x}^2 + \dot{y}^2 + \dot{z}^2}$ is the speed of the particle. Then the author derived in the notes:

\begin{equation} \frac{\partial L}{\partial \mathbf{v}} = -mc^2\left(-\frac{\mathbf{v}}{c^2}\right)\frac{1}{\sqrt{1 - v^2/c^2}} = \frac{m \mathbf{v}}{\sqrt{1 - v^2/c^2}}. \end{equation}

My question is, how did the author get the first equality in the derivation?

I know that I can just do it by computing this:

\begin{equation} \frac{\partial L}{\partial \mathbf{v}} = \left(\frac{\partial L}{\partial \dot{x}}, \frac{\partial L}{\partial \dot{y}}, \frac{\partial L}{\partial \dot{z}}\right). \end{equation}

But my question is more specific: how did the author get the first equality that fast? Is there a trick I'm missing here?

Answer 1

Component-wise, $\boldsymbol{p}=\partial L/\partial \boldsymbol{v}$ means $p_i=\partial L/\partial v_i$ for $i=1,2,3$. Then e.g. \begin{equation} \frac{\partial L}{\partial v_1} = \cdots = \gamma mv_1\,, \end{equation} yields the first momentum component. Notice that the differentiation is exactly the same for $i=2,3$. So you can imagine taking a shortcut by doing all three derivatives at the same time: write $v^2\rightarrow \boldsymbol{v}^2$, then perform the derivative w.r.t. $\boldsymbol{v}$ like it's a scalar.

Answer 2

Use the chain rule and the fact that $v^2 = \mathbf v\cdot\mathbf v$ whence $$\frac{\partial v^2}{\partial\mathbf v} = 2 \mathbf v.$$