How to go from Lagrange equations to d'Alembert's principle?

Question

All sources I know show how to use d'Alembert's principle and/or Hamilton's principles to derive Lagrange equations. It is also common to use d'Alembert's principle to derive Hamilton's principle (see Lanczos "the variational principles of mechanics", p.112) But what about the opposite direction? If you only have Lagrange equations, how can we derive d'Alembert's principle?

Answer 1

On one hand Lagrange equations $$\frac{d}{dt} \frac{\partial T}{\partial \dot{q}^j} -\frac{\partial T}{\partial q^j}~=~Q_j,\qquad j~\in~\{1, \ldots, n\}, \tag{LE}$$ make sense in pretty much any setting, while on the other hand d'Alembert's principle $$ \sum_{i=1}^N \left(\dot{\bf p}_i-{\bf F}_i\right)\cdot \delta {\bf r}_i ~=~0\tag{DAP} $$ basically only makes sense within the context of Newtonian point mechanics$^1$. It also seems that information about the point particle positions $${\bf r}_i(q^1,\ldots, q^n,t), \qquad i\in\{1, \ldots, N\},$$ in terms of the generalized coordinates $(q^1,\ldots, q^n)$ and time $t$ is needed.

In that case the equivalence of DAP and LE now follows from the following key identity $$ \sum_{i=1}^N \left(\dot{\bf p}_i-{\bf F}_i\right)\cdot \delta {\bf r}_i ~=~ \sum_{j=1}^n \left(\frac{d}{dt} \frac{\partial T}{\partial \dot{q}^j} -\frac{\partial T}{\partial q^j}-Q_j\right) \delta q^j, $$ where the kinetic energy is $$ T~:=~\sum_{i=1}^N\frac{m_i}{2}v^2_i,$$ and the generalized forces are $$ Q_j~:=~\sum_{i=1}^N{\bf F}_i\cdot \frac{\partial {\bf r}_i}{\partial q^j},\qquad j~\in~\{1, \ldots, n\}, $$ cf. e.g. my related Phys.SE answer here.

References:

1. H. Goldstein, Classical Mechanics, Chapter 1.

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$^1$ A relativistic extension is possible.

Answer 2

Lets look at this example

Pendulum 2D space

the kinetic energy is

$$T=\frac m2 (\dot x^2+\dot y^2)$$

the potential energy is:

$$U=m\,g\,y+\lambda\,\underbrace{(x^2+y^2-l^2)}_{\text {holonomic constraint eq.}}$$

thus the equations of motion

$$\ddot x=-\lambda\,\frac{2\,x}{m}$$ $$\ddot y=-g-\lambda\,\frac{2\,y}{m}$$

how to eliminate the generalized constraint force $~\lambda~$ from the equations of motion ?

the EOM's

\begin{align*} \begin{bmatrix} \ddot{x} \\ \ddot{y}\\ \end{bmatrix}=-\frac\lambda m\,\underbrace{\begin{bmatrix} 2\,x \\ 2\,y \\ \end{bmatrix}}_{\mathbf{C}_c^T}\tag 1 \end{align*}

from the constraint equation

\begin{align*} &z=x^2+y^2-l^2=0\quad\Rightarrow\quad \dot z=2\,x\,\dot x+2\,y\dot y=0\\ & \dot z=\mathbf{C}_c\,\begin{bmatrix} \dot x \\ \dot y \\ \end{bmatrix}=0 \end{align*}

to eliminate the constraint force we multiply from the left equation (1) with

\begin{align*} &\mathbf{J}^T=\begin{bmatrix} -\frac yx & 1 \\ \end{bmatrix}\quad, \mathbf{J}^T\,\mathbf{C}_c^T=0 \end{align*} this is the "d'Alembert principle".

Fazit:

from the holonomic constraint equation you obtain the constraint matrix $~\mathbf{C}_c~$ and from here ein orthogonal matrix $~\mathbf J~$ .

notice that

\begin{align*} \begin{bmatrix} \dot{x} \\ \dot{y}\\ \end{bmatrix}\mapsto\mathbf{J}\,\dot{y} \end{align*} thus $~y~$ is the generalized coordinate , and you can transformed the kinetic and potential energy and obtain the EOM's with out the generalized constraint force

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How to obtain the Jacobi matrix $~\mathbf J~$

with:

\begin{align*} &x^2+y^2=l^2\quad\Rightarrow x^2=l^2-y^2\quad ,2\,x\dot{x}=-2\,y\dot{y}\\ &\dot{x}=-\frac{y}{x}\dot{y}\quad,\dot{y}:=\dot{y}\quad\Rightarrow \mathbf{J}=\begin{bmatrix} -\frac{y}{x} \\ 1 \\ \end{bmatrix}=\begin{bmatrix} -\frac{y}{\sqrt{l^2-y^2}}\\ 1 \\ \end{bmatrix} \end{align*} Newton equation of motion

the generalized coordinate if $~y~$ \begin{align*} &m\,\mathbf J^T\,\mathbf J\,\ddot{y}=\mathbf J^T\begin{bmatrix} 0 \\ -m\,g \\ \end{bmatrix} -m\,\,\mathbf J^T\,\frac{\partial \mathbf v}{\partial y}\,\dot{y}\quad, \text{where}~\mathbf v=\mathbf J\,\dot{y}\\ &\Rightarrow \end{align*} \begin{align*} & \ddot{y}=-\left({\frac {y{{\dot{y}}}^{2}}{{l}^{2}-{y}^{2}}}+{\frac { \left( {l}^{2}-{y} ^{2} \right) g}{{l}^{2}}}\right) \end{align*}