# 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?

## 1 Answer

Hint: If the transformation equations

$$\mathbf{r}_i~=~\mathbf{r}_i(q_1,\ldots, q_n ,t) \tag{1.38}$$

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 $q_1,\ldots, q_n$.

References:

1. H. Goldstein, Classical Mechanics, Chapter 1.
• 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. – alexvas Jul 4 '14 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. – Qmechanic Jul 4 '14 at 21:20