In his derivation of the Euler-Lagrange equations from D'Alembert's principle, Goldstein uses the parametrization (equation 1.45')

$$\displaystyle{\vec{r_i}=\vec{r_i}(q_1,q_2, ..., q_n, t)}\tag{1.45'}$$

Where $\vec{r_i}$ are the old coordinates and $\vec{q_k}$ are the generalized coodinates.

Using the "chain rules" he arrives at the expression (equation 1.46)

$$\displaystyle{\vec{v_i}\equiv \frac{\mathrm{d}\vec{r_i} }{\mathrm{d} t}= \sum_{k}\frac{\partial \vec{r_i}}{\partial q_k} \dot{q_k}+\frac{\partial \vec{r_i}}{\partial t}}\tag{1.46}$$

I didn't understand, what is the difference between $\displaystyle{\frac{\mathrm{d}\vec{r_i} }{\mathrm{d} t}}$ and $\displaystyle{\frac{\partial \vec{r_i}}{\partial t}}$? I thought they were the same thing, but this way

$$\displaystyle{\sum_{k}\frac{\partial \vec{r_i}}{\partial q_k} \dot{q_k}}$$

vanishes. Is it correct? If it is, why does the sum vanish?


1 Answer 1


He is using the standard chain rule for partial differentiation from calculus. The partial derivative and the total derivative are not the same. See any calculus text such as one by Kaplan, Thomas, or Taylor.

$$\displaystyle{\sum_{k}\frac{\partial \vec{r_i}}{\partial \dot{q_k}}\cdot \vec{\dot{q_k}}}$$

does not vanish. $\vec{\dot{q_k}}$ is ${d\vec q_k \over dt}$.

Suggest you look at an example application of the Lagrangian approach to a real problem to understand this more clearly.


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