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}