D'Alembert to Lagrange

From the D'Alembert principle we now derive Lagrange's equations.

Before this we clarify that D'Alembert principle does not require that the forces of constraints are time independent.
The forces of constraint do no work if they are time independent, but D'Alembert principle does not require that this
be so. There could well be work done. Either due to constraint forces or due to external time-dependent ones, entering through
the potential.
Once again D'Alembert principle is:

(1)
\begin{align} \sum_{i=1}^N (m_i {\ddot {\mathbf r}}_i -\mathbf{F}_i){\mathbf \cdot} \mathbf{\tau}_i=0 \end{align}

Where $\tau_i$ are $N$ arbitrary vectors that satisfy the condition that

(2)
\begin{align} \sum_{i=1}^N \tau_i \cdot \nabla_i f_k =0, \;\;\; k=1,\ldots,K. \end{align}

If the constraints $f_k=0$ depend on time then these vectors also depend on time.

Recall that the generalized coordinates are such that

(3)
\begin{align} {\mathbf r_i}={\mathbf r_i}(q_1,q_2,\ldots,q_{3N},t). \end{align}

Choosing the $\tau_i$

We can see that if

(4)
\begin{align} {\mathbf \tau_i}= \sum_{\alpha=1}^n \epsilon_{\alpha} \frac{\partial {\mathbf r_i}}{\partial q_{\alpha}} \end{align}

where $\epsilon_{\alpha}$ are arbitrary constants then these satisfy Eq.(2). Recall that $n=3N-K$ is the number of d.o.f..
Proof:

(5)
\begin{align} \sum_{i=1}^N \tau_i \cdot \nabla_i f_k =\sum_{i=1}^N \sum_{\alpha=1}^n \epsilon_{\alpha} \frac{\partial {\mathbf r_i}}{\partial q_{\alpha}}{\mathbf \cdot} \nabla_i f_k = \sum_{\alpha=1}^n \epsilon_{\alpha} \frac{\partial f_k}{\partial q_{\alpha}}=0. \end{align}

The last equality follows because the constraints depend only on the last $K$ generalized coordinates (see generalized-coordinates).
Note that $\tau_i$ have a total of $n=3N-K$ arbitrary components and these are the $\epsilon$.

Then D'Alembert gives

(6)
\begin{align} \sum_{\alpha=1}^n \epsilon_{\alpha} \sum_{i=1}^N (m_i {\ddot {\mathbf r}}_i -\mathbf{F}_i){\mathbf \cdot} \frac{\partial {\mathbf r_i}}{\partial q_{\alpha}}=0 \end{align}

Since $\epsilon_{\alpha}$ are independent and arbitrary we must have

(7)
\begin{align} \sum_{i=1}^N (m_i {\ddot {\mathbf r}}_i -\mathbf{F}_i){\mathbf \cdot} \frac{\partial {\mathbf r_i}}{\partial q_{\alpha}}=0,\;\;\; \alpha=1,2,\ldots,n. \end{align}

Evaluating $F\cdot \tau$

Using the above we now evaluate one piece of the D'Alembert equality. We are going to assume that the force on each particle
can be derived from a potential function that may or may not be explicitly dependent on time. That is

(8)
\begin{align} {\mathbf F_i} = -{\mathbf \nabla_i} V(\mathbf{r_1,r_2,\ldots,r_N},t). \end{align}

Then

(9)
\begin{align} \sum_{i=1}^N \mathbf{F_i \cdot }\frac{\partial {\mathbf r_i}}{\partial q_{\alpha}} = -\sum_{i=1}^{N} {\mathbf \nabla_i }V {\mathbf \cdot }\frac{\partial {\mathbf r_i}}{\partial q_{\alpha}}= -\sum_{i=1}^{N} {\mathbf \nabla_i }V {\mathbf \cdot} \frac{\partial {\mathbf r_i}}{\partial q_{\alpha}}= -\frac{\partial V}{\partial q_{\alpha}} \end{align}
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