# Kinetic energy in Lagrangian formalism

In reading Goldstein's Classical Mechanics (2nd edition) I came across a confusing derivation. Goldstein (Eq. 1-71) derives the total kinetic energy of a system of (classical) particles as:

$$T = \sum_i \frac{1}{2}m_iv_i^2 = \sum_i \frac{1}{2}m_i \left( \sum_j \frac{\partial \mathbf{r}_i}{\partial q_j} \dot{q}_j + \frac{\partial \mathbf{r}_i}{\partial t}\right)^2$$

Where the $q_i$ are the generalized coordinates. He then expands the square to obtain three terms:

$$T = M_0 + \sum_j M_j\dot{q}_j + \frac{1}{2}\sum_{j,k} M_{jk} \dot{q}_j\dot{q}_k$$

Where $M_0$ only carries the time dependency of $\mathbf{r}$ on $t$, $M_i$'s carry linear dependence on $\frac{\partial \mathbf{r}_i}{\partial q_k}$, and $M_{i,k}$ carry the quadratic dependence.

Goldstein then claims that if the transformation equations do not contain the time explicitly, then only the last (third) term survives.

I don't understand this. The first two terms consist of $\frac{\partial \mathbf{r}_i}{\partial t}$'s, which are not necessarily zero. Nowhere in this definition of $T$ do I see something like $\frac{\partial q_i}{\partial t}$, which would be zero by Goldstein's assumption. How do the first two terms vanish?

$$\mathbf{r}_i~=~\mathbf{r}_i(q_1,\ldots, q_n ,t) \tag{1.38}$$
do not contain the time explicitly, then the explicit time derivative $\frac{\partial \mathbf{r}_i}{\partial t}=0$ is zero. Note that the position $\mathbf{r}_i$ of the $i$'th point particle also depends implicitly on time $t$ through the generalized positions $q_1,\ldots, q_n$.
• But that is not one of the assumptions.... if we assumed that $\frac{\partial r}{\partial t} = 0$ then the system would be stationary? Instead we assume that the $q_i$ (as functions of $r_i$) do not depend on time explicitly. Commented Jul 4, 2014 at 21:13
• @alexvas: One must distinguish between explicit time differentiation $\frac{\partial \mathbf{r}_i}{\partial t}$ and total time differentiation $\mathbf{v}_i\equiv\frac{d \mathbf{r}_i}{d t}$. If the latter is zero, the velocity is zero, and the particle would then be at rest. This is not necessarily the case if the former is zero. Commented Jul 4, 2014 at 21:20