Theoretical Mechanics IPSP

In Section 2.10 we worked out the positions of masses for a mobile where all masses are the same and where all sticks are straight. It is worth while to revisit this problem from a more advanced mathematical perspective.

The mobile is at rest when its center of mass does not move, $\dot{\boldsymbol q} = \mathbf 0$, and when it has not spin. It remains at rest, when it does not experience a total force that will induce a motion of the center of mass, and no torque that induces spin. According to Equation 5.4.4 the center of mass can remain at rest when the total force $\mathbf F_{\text{tot}}$ vanishes. This implies that the force $\mathbf F_s$ at the suspension point of the mobile must balance the total weight of the mobile $M\, \mathbf g$, \begin{align*} \mathbf 0 = \mathbf F_{\text{tot}} = \mathbf F_s + M\, \mathbf g \quad \Rightarrow \quad \mathbf F_s = - M\, \mathbf g \end{align*} According to Equation 5.4.6 the mobile will not topple (i.e., pick up or change its spin) when $\mathbf M = \mathbf 0$, and in Section 5.4.3 we pointed out that the gravitational force does can not change the spin. Hence, we are left with the force $\mathbf F_s$ at the suspension point $\mathbf q_s = \mathbf Q + \mathbf r_s$, \begin{align*} \mathbf M = \mathbf r_s \times \mathbf F_s \end{align*} It vanishes iff the force $\mathbf F_s$ acts parallel to the direction $\mathbf r_s$ from the position of the center of mass to the suspension point. Since $\mathbf F_s$ acts antiparallel to gravity this entails that the center of mass of the mobile must either be located directly below or above the suspension point, irrespective of the shape of the arms or distribution of the masses.

The mobile is not a stiff body. Rather its arms can move with respect to each other. We assume again that the mass of the arms may be neglected. The mass of the mobile is concentrated in its $N$ weights that reside at the positions $\mathbf q_\nu$, $\nu=1, \cdots, N$. The position of the suspension will be denoted as $\mathbf q_0$. Let us attach the mobile to a spring so that we can explicitly measure the suspension force. Clearly the forces on the mobile are conservative, such that there is a potential $\Phi( \mathbf q_0, \mathbf q_1, \mathbf q_2, \cdots, \mathbf q_N )$. The force $\mathbf F^{(\nu)}$ acting on particle $\nu$ (or on the suspension) can then be calculated by taking the derivatives with respect to the coordinates $\mathbf q_\nu = ( q_{\nu,x}, q_{\nu,y}, q_{\nu,z} )$, of the particle \begin{align*} \mathbf F^{(\nu)} = - \nabla_{\mathbf q_\nu} \Phi = \begin{pmatrix} - \partial q_{\nu,x} \Phi \\ - \partial q_{\nu,y} \Phi \\ - \partial q_{\nu,z} \Phi \end{pmatrix} \end{align*} When the coordinates are collected into a single vector $\mathbf q = ( \mathbf q_0, \mathbf q_1, \mathbf q_2, \cdots, \mathbf q_N )$ then the mobile is in equilibrium when the $\mathbf q$-gradient of $\Phi( \mathbf q )$ vanishes, $\mathbf 0 = \nabla_{\mathbf q} \Phi( \mathbf q )$. However, when taking the partial derivatives one has to keep in mind that one must not fix the values of the other coordinates but rather keep in mind the constraints of motion of the mobile (recall Example 3.7). Alternatively, one can account for the elasiticity of the cords and bars in the mobile, and the resulting restoring forces to pulling and bending. When also all these forces are accounted for the stationary point can be found by a variation principle.

An more elegant way to deal with this problem will be presented in Chapter 6. Here we already note that the condition on $\Phi$ can be interpreted as a multi-dimensional requirement for a stationary point. The force will be zero, even when $\Phi( \mathbf q)$ takes a maximum. However, in that case small fluctuations will induce forces that drive the system away from the stationary point. The mobile will stay put when $\Phi( \mathbf q)$ takes a minimum. Small perturbations will then only lead to some wiggling close to the minimum. The mobile can slowly move because there are perturbations to its shape where all masses stay exactly at the same height. In terms of the potential this amounts to neutral directions where the potential is flat. In order to formally underpin this intuition we introduce multidimensional Taylor expansions.

We consider a scalar function $\Phi : \mathbb R^D \to \mathbb R$ that assigns a real value to its arguments $\mathbf x \in R^D$. For instance this may be the potential energy assigned to a configuration of masses characterized by a state vector $\mathbf x$. We select a reference point $\mathbf x_0$ and explore how $\Phi( \mathbf x )$ deviates from $\Phi( \mathbf x )$ for a small changes of the configuration, $\mathbf x = \mathbf x_0 + \mathbf\epsilon$, i.e., for a small change $\epsilon\in \mathbb R^D$ of the configuration. The multidimensional Taylor expansions states that \begin{align*} \Phi( \mathbf x ) = \Phi( \mathbf x_0 ) + (\epsilon_i \, \partial_i) \Phi( \mathbf x_0 ) & + \frac{1}{2} \: (\epsilon_i \, \partial_i) \: (\epsilon_j \, \partial_j) \Phi( \mathbf x_0 ) \\ & + \frac{1}{3!} \; (\epsilon_i \, \partial_i) \: (\epsilon_j \, \partial_j) \: (\epsilon_k \, \partial_k) \Phi( \mathbf x_0 ) + \dots \end{align*} Here, $\epsilon_i$ denotes the $i$-component of the vector ${\boldsymbol\varepsilon}$ with respect to an orthonormal basis $\hat{\mathbf e}_i$, and $\partial_i$ is the partial derivative with respect to the according coordinate $x_i$ of $\mathbf x$. Moreover, we use the Einstein convention that requires summation over repeated indices,i.e., $\epsilon_i \, \partial_i$ is an abbreviation for $\epsilon_i \, \partial_i = \sum_i \epsilon_i \, \partial_i$ where $i$ runs of the set of indices labeling the base vectors, and analogous statement hold for $(\epsilon_j \, \partial_j)$ and $(\epsilon_k \, \partial_k)$.

Remark 5.9. $\partial_j \Phi( \mathbf x_0 )$ should be interpreted as \begin{align*} \partial_j \Phi( \mathbf x_0 ) = \left. \frac{\partial}{\partial x_j} \Phi( x_1, \dots, x_j, \dots ) \right\rvert_{\mathbf x = \mathbf x_0} \, . \end{align*}

For scalar arguments $x \in \mathbb R$ the expression for the multi-dimensional Taylor expansion reduces to the one for real functions that we have discussed before.

Proof. For a one-dimensional function $f(x)$ the Taylor expansion around $x_0$ with $x=x_0+\epsilon$ the expression $(\epsilon_j \, \partial_j)$ reduces to $\epsilon \, \frac{\mathrm{d}}{\mathrm{d} x}$. Consequently, \begin{align*} f( x ) &= f( \mathbf x_0 ) + \left( \epsilon \,\frac{\mathrm{d}}{\mathrm{d} x} \right) f( \mathbf x_0 ) + \frac{1}{2} \: \left( \epsilon \,\frac{\mathrm{d}}{\mathrm{d} x} \right) \: \left( \epsilon \,\frac{\mathrm{d}}{\mathrm{d} x} \right) \: f( x_0 ) \\ &+ \frac{1}{3!} \;\left( \epsilon \,\frac{\mathrm{d}}{\mathrm{d} x} \right) \: \left( \epsilon \,\frac{\mathrm{d}}{\mathrm{d} x} \right) \: \left( \epsilon \,\frac{\mathrm{d}}{\mathrm{d} x} \right) \: f( x_0 ) + \dots \\ &= f( \mathbf x_0 ) + \epsilon \, f'( \mathbf x_0 ) + \frac{1}{2} \: \epsilon^2 \: f''( x_0 ) + \frac{1}{3!} \: \epsilon^3 \: f'''( x_0 ) + \dots \end{align*} These are the first terms of the 1D Taylor expansion.
qed

The first terms of the Taylor expansion can also be written in the form \begin{align*} \Phi( \mathbf x ) = \Phi( \mathbf x_0 ) + ({\boldsymbol\varepsilon} \cdot \nabla) \Phi( \mathbf x_0 ) + \frac{1}{2} \: {\boldsymbol\varepsilon}^T \, \mathsf C(\mathbf x_0) \: {\boldsymbol\varepsilon} + \dots \end{align*} where the matrix $\mathsf C(\mathbf x_0)$ has the components $c_{ij}(\mathbf x_0) = \partial_i \partial_j \Phi( \mathbf x_0 )$.

Proof. For the first-order term we have \begin{align*} ({\boldsymbol\varepsilon} \cdot \nabla) \Phi( \mathbf x_0 ) = \left( \sum_j \epsilon_j \, \partial_j \right) \, \Phi( \mathbf x_0 ) = \left( \epsilon_j \, \partial_j \right) \, \Phi( \mathbf x_0 ) \end{align*} where the second equality amounts to the simplification of notation achieved by the Einstein convention. For the second-order term we have \begin{align*} {\boldsymbol\varepsilon}^T \, \mathsf C(\mathbf x_0) \: {\boldsymbol\varepsilon} &= \begin{pmatrix} \epsilon_1 \, , \epsilon_2 \, , \epsilon_3 \, , \dots , \end{pmatrix} \begin{pmatrix} \partial_1^2 \Phi(x_0) & \partial_1 \, \partial_2 \Phi(x_0) & \partial_1 \, \partial_3 \Phi(x_0) & \dots \\ \partial_2 \, \partial_1 \Phi(x_0) & \partial_2^2 \Phi(x_0) & \partial_2 \, \partial_3 \Phi(x_0) & \dots \\ \partial_3 \, \partial_1 \Phi(x_0) & \partial_3 \, \partial_2 \Phi(x_0) & \partial_3^2 \Phi(x_0) & \dots \\ \vdots & \ddots \end{pmatrix} \: \begin{pmatrix} \epsilon_1 \\ \epsilon_2 \\ \epsilon_3 \\ \vdots \end{pmatrix} \\ &= \sum_{jk} \epsilon_j \, \partial_j \, \partial_k \Phi(x_0) \, \epsilon_k = \sum_{jk} ( \epsilon_j \, \partial_j ) \: ( \epsilon_k \, \partial_k ) \: \Phi(x_0) \\ &= \left( \sum_j \epsilon_j \, \partial_j \right) \: \left( \sum_k \epsilon_k \, \partial_k \right) \: \Phi(x_0) = \left( \epsilon_j \, \partial_j \right) \: \left( \epsilon_k \, \partial_k \right) \: \Phi(x_0) \end{align*} where the last equality amounts to the simplification of notation achieved by the Einstein convention.
qed

For scalar arguments the condition that $\nabla\Phi (\mathbf x_0) = \mathbf 0$ amount to the requirement that the slope vanishes at an extremum. When $\Phi(\mathbf x)$ is a potential then the requirement $\nabla\Phi (\mathbf x_0) = \mathbf 0$ amounts to the requirement that the force $\mathbf F(\mathbf x)$ vanishes at the position $x_0$, \begin{align*} \mathbf F (\mathbf x_0) = -\nabla\Phi (\mathbf x_0) = \mathbf 0 \end{align*} Hence, we say that the function $\Phi(\mathbf x)$ has a stationary point at $\mathbf x_0$ when $\nabla\Phi (\mathbf x_0) = \mathbf 0$. This underpins the heuristic discussion of the potential energy of the mobile that we gave above. In particular $\Phi(\mathbf x)$ has a minimum at $\mathbf x_0$ iff

• $\nabla\Phi (\mathbf x_0) = \mathbf 0$, and
• all eigenvalues of $\mathsf C(\mathbf x_0)$ are positive.

Proof. We explore how $\Phi(\mathbf x_0)$ changes when one considers a point $\mathbf x = \mathbf x_0 + \mathbf\varepsilon$ in the vicinity of $\mathbf x_0$, where we express the deviation in the orthonormal basis spanned by the eigenvectors $\hat{\mathbf e}_i$ of $\mathsf C$. Adopting Einstein notation we have \begin{align*} \mathbf\varepsilon &= \epsilon_i \: \hat{\mathbf e}_i \\ \Rightarrow\quad \mathbf\varepsilon^T \: \mathsf C \: \mathbf\varepsilon &= \epsilon_i \: \hat{\mathbf e}_i \cdot \bigl( \mathsf C \epsilon_k \: \hat{\mathbf e}_k \bigr) = \epsilon_i \: \hat{\mathbf e}_i \cdot \bigl( \lambda_k \epsilon_k \: \hat{\mathbf e}_k \bigr) \\ &= \lambda_k \epsilon_k \:\epsilon_i \: \hat{\mathbf e}_i \cdot \hat{\mathbf e}_k = \lambda_k \epsilon_k \:\epsilon_i \: \delta_{ik} = \lambda_k \epsilon_k \:\epsilon_k \end{align*} such that \begin{align*} \Phi( \mathbf x_0 + \mathbf \epsilon ) = \Phi( \mathbf x_0 ) + \epsilon_k \cdot \partial_k \Phi( \mathbf x_0 ) + \frac{1}{2} \lambda_k \epsilon_k \:\epsilon_k \end{align*}

1. Assume that $\partial_k \Phi( \mathbf x_0 ) \neq 0$ for some coordinate $k$. We will then choose the orientation of the associated unit vector such that $ \partial_k \Phi( \mathbf x_0 ) = m > 0 $ and consider a displacement $\mathbf\varepsilon = \epsilon \: \hat{\mathbf e}_k$. The change of the value of $\Phi( \mathbf x_0 )$ amounts then to \begin{align*} \Phi( \mathbf x_0 + \epsilon \: \hat{\mathbf e}_k ) - \Phi( \mathbf x_0 ) = m \, \epsilon + \frac{ \lambda_k }{2} \epsilon^2 + \dots = \epsilon \; \left( m + \frac{ \lambda_k }{2} \epsilon + \dots \right) \end{align*} For $| \epsilon | < 2m/|\lambda_k| $ the expression in the bracket takes a positive value, such that $\Phi( \mathbf x_0 + \epsilon \: \hat{\mathbf e}_k ) < \Phi( \mathbf x_0 )$ for small negative values of $\epsilon$. Consequently, $\Phi( \mathbf x_0 )$ can only be a minimum when $\nabla\Phi( \mathbf x_0 ) = \mathbf 0$.

2. Assume that $\nabla \Phi( \mathbf x_0 ) = \mathbf 0$ and that all eigenvalue $\lambda_k > 0$. For small $\mathbf\varepsilon$ the change of the value of $\Phi( \mathbf x_0 )$ amounts then to \begin{align*} \Phi( \mathbf x_0 + \epsilon \: \hat{\mathbf e}_k ) - \Phi( \mathbf x_0 ) \simeq \frac{ 1 }{2} \: \lambda_k \: \epsilon_k^2 > 0 \end{align*} such that that the function takes values larger than $\Phi( \mathbf x_0 )$ for all positions $\mathbf x$ in the vicinity of $\mathbf x_0$.
qed

Analogously, to the discussion of the minimum one shows that $\Phi( \mathbf x_0 )$ takes a maximum when the gradient vanishes, $\nabla \Phi( \mathbf x_0 ) = \mathbf 0$ and when all eigenvalue $\lambda_k$ take negative values. The function $\Phi$ takes a saddle at $\mathbf x_0$ when there are positive and negative eigenvalues and when $\nabla \Phi( \mathbf x_0 ) = \mathbf 0$. When some eigenvalues vanish and all others are positive (negative), then higher-order contributions of the Taylor expansion must be considered to determine if $\Phi$ takes a minimum (maximum).

Problem 5.9: Symmetry properties of the second-order contributions
Verify that the left and the right eigenvectors of $\mathsf C$ are identical, up to transposition.
Why does this imply that the normalized eigenvectors span an orthonormal basis?

Problem 5.10: Equipotential lines for a 2D potential
Consider a potential $\Phi( \mathbf x )$ with $\mathbf x \in \mathbb R^2$. Sketch the contour lines of the potential for the following situations

a) $\nabla\Phi(\mathbf x) = (1,1)$ and $\mathsf C(\mathbf x) = \mathsf 0$ for all positions $\mathbf x$.

b) $\nabla\Phi(1,2) = \mathbf 0$ and $\mathsf C(1,2) = \begin{pmatrix} 1 & b \\ b & 1 \end{pmatrix}$ with
1. $b>1$,
2. $1 > b > -1$,
3. $b < -1$,
4. $b=1$.