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I've often read that, for a mechanical system which can be described by $n$ generalized coordinates $q_1,...,q_n$, a point $\mathbf{Q}=(Q_1,...,Q_n)$ is a point of equilibrium if and only if the potential energy $U$ is stationary at that point, i.e. iff $\frac{\partial U}{\partial q_i}(\mathbf{Q})=0$ for all $i$. I've only seen a proof for the unidimensional case $n=1$, that uses the Lagrangian of the system to show that if the system is in $q=Q$, with $\dot q=0$, then $\ddot{q}=0 \iff \frac{\partial U}{\partial q}(Q)=0$ (in this case it is also easy to obtain the stable equilibrium condition $\partial ^2 U/\partial q^2>0$).

Following this example I was trying to prove the statement for the general case, however probably some steps aren't correct for the general case. The Lagrangian is (using Einstein's notation for summations $a_ib_i\dot{=}\sum _i a_ib_i$): $$L (\mathbf{q},\mathbf{\dot q})= T(\mathbf{q},\mathbf{\dot q})-U(\mathbf{q})=\frac{1}{2}\dot {q_i} A_{ij} \dot {q_j}-U(\mathbf{q}),$$ where I've expressed the kinetical energy as a quadratic form of the generalized velocities and $A_{ij}$ are functions of $\mathbf{q}$ (as I said, I don't know if this is generally possible). Now we have:$$\frac{d}{dt}\frac{\partial \cal L}{\partial \dot{q_i}}=\frac{d}{dt}[A_{ij}\dot q_j]=\dot{q_k}\frac{\partial A_{ij}}{\partial q_k}\dot{q_j}+A_{ij}\ddot{q_j},$$ and $$\frac{\partial \cal{L}}{\partial{q_i}}=\frac{1}{2}\dot{q_k}\frac{\partial A_{kj}}{\partial q_i}\dot{q_j}-\frac{\partial U}{\partial q_i}.$$ So, if all velocities $\dot{q_j}=0$, the equations of motion are:$$A_{ij}\ddot{q_j}=-\frac{\partial U}{\partial q_i},\qquad i=1,...,n.$$ Now, I see that if the system is in equilibrium in $\mathbf{Q}$, then, for all $j$, $\ddot {q_j}=0$ and it is required that $\partial U / \partial q_j = 0$. What about the vice-versa? Does $\partial U / \partial q_j = 0 \ \forall j \implies \ddot{q_j} = 0 \ \forall j$? I see that this is equivalent to require $\det A\neq0$, where $A$'s elements are $A_{ij}$, so is there a reason why it must be so? Thank you for your time.

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Throwing a monkey in the wrench, let us mention that the case $\det A=0$ is the starting point of the topic of constrained dynamics. Also there exist velocity dependent potentials $U(\mathbf{q},\mathbf{\dot q})$. –  Qmechanic Jul 5 '13 at 19:11
    
Hi @Qmechanic, I'm not sure I understand, can you give a simple example where $\det A=0$? For example if the system is composed by $n$ particles, the $A_{ij}$ terms are given by $\sum _k m_k (\frac{\partial\mathbf r _k}{\partial q_i})\cdot(\frac{\partial \mathbf r_k}{\partial q_j})$. Is in this case possible that $\det A=0$? –  pppqqq Jul 5 '13 at 20:06
    
... maybe I should say explicitly that I'm assuming a one-to-one correspondence beetween $\mathbf q$ and $\mathbf r _1 , \mathbf r _2 ,...,\mathbf r _n$, not dependent from time. –  pppqqq Jul 5 '13 at 20:24
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2 Answers 2

up vote 2 down vote accepted

Notice that $A$ is a linear operator on $\mathbb R^n$. Suppose that $A$ is singular, namely $\det A= 0$, then the kernel of $A$ is nontrivial. In other words, there exists some nonzero $v\in\mathbb R^n$ for which $Av=0$. It follows that the kinetic energy vanishes for this nonzero $v$.

There's nothing "wrong" with this mathematically speaking, but it's physically pathological because the kinetic energy represents energy due to the magnitude of the motion of the object, and we therefore expect that any state of the object for which $\dot q_i\neq 0$ for some $i$ should be assigned a nonzero kinetic energy.

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Thank you, this makes a lot of sense. –  pppqqq Jul 5 '13 at 17:31
    
@Kazz8 Sure thing. Cool question. –  joshphysics Jul 5 '13 at 17:49
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Your question actually is one of the most important questions in analytic mechanics. This is because, when you explicitly write the Eulero-Lagrange equations for any constrained system with $n$ degrees of freedom and Lagrangian of the form:

$$L(t, {\bf q},\dot{\bf q}) = T(t, {\bf q},\dot{\bf q}) - U(t, {\bf q},\dot{\bf q})$$

where $T$ is quadratic in $\dot{\bf q}$ and $U$ is at most linear in $\dot{\bf q}$, you have a set of equations of the form:

$$\sum_{j=1}^n A(t, {\bf q})_{ij} \frac{d^2 q^j}{dt^2} = G_i\left(t, {\bf q},\frac{d q^j}{dt}\right)\quad i= 1,2,\ldots, n\:.\qquad (1)$$

As is known from the general theory of differential equations, if the system of differential equations can be re-written as:

$$ \frac{d^2 q^i}{dt^2} = \sum_{j=1}^n A^{-1}(t, {\bf q})_{ij} G_j\left(t, {\bf q},\frac{d q^j}{dt}\right)\quad i= 1,2,\ldots, n \qquad(2)\:.$$

that is in normal form (only the highest order derivative appears in the left-hand side), then the system admits a solution in a neighborhood of any fixed $t_0$ and it is uniquely determined by initial conditions

$$({\bf q}(t_0),\dot{\bf q}(t_0)) = ({\bf q}_0,\dot{\bf q}_0)\:.$$ Actually, it is true when the right-hand side of (2) is sufficiently regular: $C^1$ jointly in all variables is OK (more weakly, continuity and the validity local Lipschitz condition in $({\bf q}_0,\dot{\bf q}_0)$ actually would also be enough). To pass from (1) to (2), it is necessary that $$\det A(t, {\bf q}) \neq 0\quad \mbox{everywhere in $(t, {\bf q})$.}$$

Let us show that, for a physical system of constrained material points, this condition is always verified under suitable hypotheses on the constraints.

Assume that the system is made of $N$, with $3N > n$, material points with masses $m_k>0$ and positions ${\bf r}_k$ in a given reference system. In this case the constraints are $c=3N-n$ requirements of the form:

$$f_l(t, {\bf r}_1, \ldots, {\bf r}_N)=0\quad l=1,2,\ldots, c \qquad (3)\:.$$

It is also assumed that the functions $f_l$ are sufficiently regular (for our computation $C^2$ is sufficient) and that the constraints are functionally independent. It means that, exactly on the set of points $(t, {\bf q})$ where (3) holds, it must also hold that the Jacobian matrix of elements (the 3N $x_k$ are all of the Cartesian components of all ${\bf r}_i$ labeled in any order, since it is not relevant here)

$$\frac{\partial f_l}{\partial x_k}$$

has $c$ linearly independent rows (or equivalently columns). These requirements assure that the set of allowed positions is a $n=3N-c$ manifold for every time $t$ and that, locally, there are $n=3N-c$ free coordinates, $q^1,\ldots, q^2$, on that manifold and in a neighborhood of any fixed time $t$.

It is worth remarking that these $n$ coordinates can always picked out among the $3N$ coordinates $x_i$, even if this choice is not necessary.

As a consequence of the existence of $n$ local coordinates on the manifold of admissible configurations, the local relations hold: ${\bf r}_k={\bf r}_k(t,q^1,\ldots, q^n)$. As you probably know, and the computation is straightforward, the matrix $A$ takes this form:

$$A_{ij} = \sum_{s=1}^N m_s \frac{\partial {\bf r}_s}{\partial q^i} \cdot \frac{\partial {\bf r}_s}{\partial q^j}\:.\qquad (4)$$

Now I wish to show that $\det A \neq 0$ when all the said hypotheses are true. It is obvious that it is sufficient to prove this fact only in one local coordinate system $q^1,\ldots, q^n$. Indeed, changing local coordinates and passing to the coordinates $Q^1,\ldots, Q^n$, we obtain a new matrix $A'$ related to the previous one by:

$$A'_{rs} = \sum_{i,j=1}^n\frac{\partial q^i}{\partial Q^r}\frac{\partial q^j}{\partial Q^s} A_{ij}\:.$$

Since the Jacobian matrix of elements $\frac{\partial q^i}{\partial Q^r}$ must be non singular, $\det A \neq 0$ is equivalent to $\det A' \neq 0$.

It is convenient to choose the free coordinates $q^1,\ldots, q^n$ as $n$ coordinates $x_k$ among the original components of the vectors ${\bf r}_j$ as said above. For the sake of simplicity we can suppose that $q^1=x_1, q^2= x_2,\ldots , q^n = x_n$.

With this choice of the coordinates, assume that $\det A=0$. Consequently, there is a vector $v \in R^n -\{0\}$ such that $Av=0$ and, consequently, $v^t A v =0$. Exploiting (4):

$$0=\sum_{ij}v_iA_{ij} v_j = \sum_{s=1}^N m_s \sum_{i,j}v_i\frac{\partial {\bf r}_s}{\partial q^i} \cdot v_j\frac{\partial {\bf r}_s}{\partial q^j} = \sum_{s=1}^N m_s \left| \sum_{i}v_i\frac{\partial {\bf r}_s}{\partial q^i}\right|^2\:.$$

Since $m_s>0$, in turn it implies:

$$\sum_{i}v_i\frac{\partial {\bf r}_s}{\partial q^i} =0 \quad \mbox{for $r=1,2,\ldots, N$}\:.$$ Passing to the components of the ${\bf r}_s$:

$$\sum_{i}v_i\frac{\partial x_l}{\partial q^i} =0 \quad \mbox{for $l=1,2,\ldots, 3N$}\:.\quad (5)$$

Finally, remember that $x_l=q^l$, for $l=1,2, \ldots, n$. Choosing $l=1$, the requirements (5) produce:

$$v_1 =0$$

Choosing $l=2$, the requirements (5) produce:

$$v_2 =0$$

and so on. Finally we get $v=0$. This is not possible since $v \in R^n -\{0\}$. The existence of such $v$ was a consequence of $\det A=0$ that, consequently is untenable.

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Sorry to bother for an algebraic fact, but I don't see how $\det A\neq0$ implies $ \det A'\neq 0$. The elements of $A'$ are of the form $a'_{ij}=B_i A B_j^T$ where, $B_k$ are the rows of a nonsingular matrix $B$. OK, why $\det B\neq 0$ implies $\det A' \neq 0$? –  pppqqq Dec 15 '13 at 19:18
    
Apart from this, thanks, this is really an improving of the accepted answer (that I still like because it gives a good physical insight about the matter). Truly, when I wrote this question I was thinking about a special case of a special case, so I'm happy to see that it brings out interesting discussions. –  pppqqq Dec 15 '13 at 19:23
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The matrix $B$ is non singular, so that $\det B \neq 0$, remember also that $\det B^t = \det B$ since it is useful shortly. On the other hand $A' = BAB^t$ implies $\det A' = \det B \det A \det B$, namely $\det A' = (\det B)^2 \det A$. Since $\det B \neq 0$, $\det A$ and $\det A'$ vanish simultaneously or do not vanish simultaneously. –  V. Moretti Dec 15 '13 at 19:38
    
Oh, I just didn't realize that $A'$ was of the form $BAB^T$. It is clear, then. Thanks. –  pppqqq Dec 15 '13 at 20:15
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OT (and even in Italian!) Dai allora un'occhiata alle mie dispense di meccanica anlitica sulla mia pagina web (sul mio profilo), potrebbero interessarti. Ciao, V. –  V. Moretti Dec 16 '13 at 18:21
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