In the previous section we saw that Newton's laws can be expressed as equations relating the second derivative of the position of a particle to the forces acting on the particle. The forces are determined as part of setting up the physical model. Subsequently, determining the time dependence of the position is a mathematical problem. Often it can be solved by finding constraints on the solution that must hold for all times. Such a constraint is called a

It provides us with an opportunity to take a closer look at the expressions that emerge when taking derivatives of functions with arguments that are vectors. In order to evaluate the time derivative of $\mathcal C$ we write $\mathbf q = (q_1 , ....., q_D)$, and apply the chain rule \begin{align} \frac{\mathrm{d}}{\mathrm{d} t} \mathcal{C} (\mathbf q(t) , \dot{\mathbf q}(t) ,t) &= \frac{\mathrm{d}}{\mathrm{d} t} \mathcal{C} (q_1(t), \cdots, q_D(t) , \dot q_1(t), \cdots, \dot q_D(t) ,t) \nonumber \\ &= \sum_{i=1}^{D} \frac{\mathrm{d} q_i}{\mathrm{d} t} \frac{\partial \mathcal C}{\partial q_i} + \sum_{i=1}^{D} \frac{\mathrm{d} \dot q_i}{\mathrm{d} t} \; \frac{\partial \mathcal{C}}{\partial \dot q_i} + \frac{\partial \mathcal{C}}{\partial t} \tag{3.4.1} \end{align} In this expression the operation $\partial$ is called `partial', and the derivative $\partial \mathcal{C} / \partial q_i$ is denoted as partial derivative of $\mathcal C$ with respect to $q_i$. For the purpose of calculating the partial derivative, we consider $\mathcal{C}$ to be a function of only the single argument $q_i$. For sake of a more compact notation we also write $\partial_{q_i} \mathcal{C}$ rather than $\partial \mathcal{C} / \partial q_i$. Moreover, when it is not clear from the context which conditions are adopted, they can explicitly be stated as subscript of a vertical bar to the right of the derivative (or even square brackets).

A compact notation that allows us to state the expression of Equation 3.4.1 in a more transparent way is achieved as follows: We observe that the expressions in the sums amount to writing out in components a scalar product of $\mathbf q$ and $\dot{\mathbf q}$ with vectors that are obtained by the partial derivatives. These vectors are denoted gradients with respect to $\mathbf q$ and $\dot{\mathbf q}$, and they will be written as ¹⁾ \begin{align} \nabla\!_{\mathbf q}\, \mathcal C = \begin{pmatrix} \partial_{q_1} \mathcal C \\ \vdots \\ \partial_{q_D} \mathcal C \end{pmatrix} \qquad \text{and} \qquad \nabla\!_{\dot{\mathbf q}}\, \mathcal C = \begin{pmatrix} \partial_{\dot q_1} \mathcal C \\ \vdots \\ \partial_{\dot q_D} \mathcal C \end{pmatrix} \tag{3.4.2} \end{align} such that \begin{align*} \frac{\mathrm{d}}{\mathrm{d} t} \mathcal{C} (\mathbf q(t) , \dot{\mathbf q}(t) ,t) &= \dot{\mathbf q} \cdot \nabla\!_{\mathbf q}\, \mathcal C + \ddot{\mathbf q} \cdot \nabla\!_{\dot{\mathbf q}}\, \mathcal C + \frac{\partial \mathcal{C}}{\partial t} \end{align*} In terms of the phase-space coordinates $\boldsymbol\Gamma = (\mathbf q, \dot{\mathbf q})$ one can also adopt the even more compact notation \begin{align*} \frac{\mathrm{d}}{\mathrm{d} t} \mathcal{C} (\mathbf q(t) , \dot{\mathbf q}(t) ,t) &= \dot{\boldsymbol\Gamma} \cdot \nabla\!_{\boldsymbol\Gamma}\, \mathcal C + \frac{\partial \mathcal{C}}{\partial t} \end{align*} or even \begin{align*} \frac{\mathrm{d}}{\mathrm{d} t} \mathcal{C} (\mathbf\Gamma(t) ,t) &= \dot{\boldsymbol\Gamma} \cdot \nabla\, \mathcal C (\mathbf\Gamma(t) ,t) + \frac{\partial \mathcal{C}}{\partial t} (\mathbf\Gamma(t) ,t) \end{align*} where the index of the nabla operator has been dropped with the understanding that it is clear from the context what the operator refers to. We make use of these derivatives while introducing some important physical quantities that are constants of the motion in specific settings.

When no forces are acting on a particle, $\mathbf F_{\text{tot}} = \mathbf 0$, it moves with constant velocity. All functions that depend only on the velocity will then be constant. In particular this holds for the kinetic energy, $T$, that will play a very important role in the following.

Proof.
\begin{align*} \frac{\mathrm{d}}{\mathrm{d} t} T & = \frac{m}{2} \; \frac{\mathrm{d}}{\mathrm{d} t}\; \sum_i \dot q_i \cdot \dot q_i = m \; \sum_i \dot q_i \cdot \ddot q_i \\ & = m \dot{ \mathbf q } \cdot \ddot {\mathbf q} = \dot{\mathbf q} \cdot (m \: \ddot{\mathbf q}) = \dot{\mathbf q} \cdot \mathbf F_{\text{tot}} = 0 \end{align*} In the last two steps we used Newton's 2nd law, and the assumption that $\mathbf F_{\text{tot}} = \mathbf 0$.
qed

From a physics perspective, work is performed when a body is moved in the presence of an external force.

• When the force $\mathbf F$ is constant along a straight path of displacement $\mathbf s = \mathbf q_E - \mathbf q_I$, from a position $\mathbf q_I$ to the position $\mathbf q_E$, then the work $W$ amounts to the scalar product $W = \mathbf F \cdot \mathbf s$.
• When the force depends on the position along the path, we parameterize the motion along the path by time, $\mathbf q(t)$, with $\mathbf q(t_I) = \mathbf q_I$ and $\mathbf q(t_E) = \mathbf q_E$ and break it into sufficiently small pieces $\mathbf s_i = \mathbf q(t_i) - \mathbf q(t_{i} - \Delta t)$ where the force $\mathbf F_i = \mathbf F( t_i )$ and the velocity of the particle $\dot q(t_i)$ may be assumed to be constant, such that $\dot{\mathbf q}(t_i) = \bigl( \mathbf q(t_i) - \mathbf q(t_{i-1}) \bigr) / \Delta t$. Then

\begin{align*} W = \sum_i \mathbf F_i \cdot \mathbf s_i =\lim_{\Delta t \to 0} \sum_i \mathbf F_i \cdot \dot{\mathbf q} \; \Delta t = \int_{t_0}^{t_1} \mathbf F (t) \cdot \dot{\mathbf q} (t) \; \mathrm{d} t = \int_{\mathbf q (t)} \mathbf F \cdot \mathrm{d}\mathbf q \end{align*} The last equality should be understood here as a definition of the final expression that is interpreted here in the spirit of the substitution rule of integration. Moreover, $\mathbf F (t)$ denotes here the force acting on the particle at time $t$, where the particle is at the position $\mathbf q (t)$ and moving with velocity $\dot{\mathbf q} (t)$, i.e., one may also have $\mathbf F \bigl( \mathbf q(t) \bigr)$, or $\mathbf F \bigl( \mathbf q(t), \dot{\mathbf q} (t) \bigr)$, or an explicit time dependence on top of the dependence on the particle position.

Remark 3.5. Here, $\mathbf F (t)$ denotes the force that is acting on the particle at time $t$, irrespective of how it emerges. Specifically, the time dependence of $\mathbf F (t)$ accounts for changes of the position $\mathbf q(t)$, the velocity $\dot{\mathbf q}(t)$, and any additional time dependences due to external incluences (if applicable).

Remark 3.6. The scalar product $\mathbf F \cdot \mathrm{d}\mathbf q$ or $P(t) = \mathbf F (t) \cdot \dot{\mathbf q} (t)$ singles out only the action of the force parallel to the trajectory. The perpendicular components do not perform work. Hence, a force that is always acting perpendicular to the velocity, i.e., perpendicular to the path of the particle, does not perform any work, \[ W = \int \mathbf F (t) \cdot \dot{\mathbf q} (t) \; \mathrm{d} t = \int 0 \; \mathrm{d} t = 0 \] It only changes the direction of motion.

Remark 3.7. The result of the integral does not rely on the parameterization of the path by time. For instance mathematicians prefer to use the length $\ell$ of the path. The speed of the particle is then $\dot \ell(t) = \left\lvert \dot{\mathbf q} (t) \right\rvert$ and one finds \[ W = \int \mathbf F (t) \cdot \mathrm{d}\mathbf q = \int \mathbf F (t(\ell)) \cdot \dot{\mathbf q} (t(\ell)) \; \frac{\mathrm{d} \ell}{\dot\ell} = \int \mathbf F (\ell) \cdot \frac{\mathrm{d}\mathbf q(\ell)}{\mathrm{d} \ell} \; \mathrm{d} \ell \] where $\mathrm{d}{\boldsymbol q} / \mathrm{d}\ell$ is a unit vector pointing in the direction of the trajectory.

Remark 3.8. Line integrals are also used to determine the length, $L$, of a path in space. After all, the length amounts to the time integral of the speed, $\dot\ell(t)$, and one has \begin{align*} % \label{eq:} L = \int\mathrm{d}\ell = \int \mathrm{d} t \: \dot\ell(t) = \int \mathrm{d} t \: \frac{\dot{\mathbf q}^2(t)}{\dot\ell(t)} = \int \mathrm{d} \vec q \cdot \frac{\dot{\mathbf q}(t)}{\dot\ell(t)} = \int \mathrm{d} \vec q \cdot \frac{\mathrm{d} \vec q(\ell)}{\mathrm{d}\ell} \end{align*} This is further illustrated in Problem 3.22.

The calculation of work simplifies dramatically when the force can be written as gradient of another function, $\Phi$.

Remark 3.9. Conservative forces only depend on position, $\mathbf F = \mathbf F (\mathbf q)$. They neither explicitly depend on time nor on the velocity $\mathbf q$.

Proof.
\begin{align*} W &= \int_{t_0}^{t_1} \mathbf F \cdot \dot{\mathbf q} \; \mathrm{d} t = - \int_{t_0}^{t_1} \nabla \Phi \cdot\dot{\mathbf q} \; \mathrm{d} t \\ &= - \int_{t_0}^{t_1} \sum_i \frac{\partial \Phi} {\partial q_i}\; \frac{\partial q_i}{\partial t} \; \mathrm{d} t = -\int_{t_0}^{t_1} \frac{\mathrm{d}\ \Phi}{\mathrm{d} t}\; \mathrm{d} t \\ & = - \bigl( \Phi( \mathbf q(t_1)) - \Phi( \mathbf q(t_0)) \bigr) = \Phi (\mathbf q_0) - \Phi( \mathbf q_1) \\ \end{align*} qed

Remark 3.10. The work performed along a closed path vanishes for conservative forces. After all, in that case $\mathbf q_1 = \mathbf q_0$ such that $W = \Phi (\mathbf q_0) - \Phi( \mathbf q_1) = 0$.

Remark 3.11. The potential in itself is not an observable. ²⁾ One can only observe the work, which is the potential difference between two positions, and the force, which is the negative gradient of the potential. Therefore, the potential is only defined up to adding a constant.

Remark 3.12. According to Theorem 3.3 differences of the value of the potential between two positions amount to the work performed in the potential. Different approaches to calculate the value of this scalar observable must yield identical results. Therefore, the functional dependence of the potential must not depend on the choice of the coordinate system. This invariance requires that the potential can always be expressed in terms of scalar products. For the potentials in Example 3.9 this is achieved by writing \begin{align*} \Phi_1( \mathbf q ) &= m \, \mathbf g \cdot \mathbf q \quad\text{ with }\quad \mathbf g = (0,0,-g)
\Phi_2 ( \mathbf q ) &= - G \, M_E \, m / \sqrt{ \mathbf q \cdot \mathbf q }
\end{align*}

Conservative forces are called conservative forces because motion in such a potential conserves the sum of the potential energy and the kinetic energy.

Proof.
\begin{align*} \frac{\mathrm{d} E}{\mathrm{d} t} & = \frac{\mathrm{d} T}{\mathrm{d} t} + \frac{\mathrm{d} \Phi}{\mathrm{d} t} = m\ \dot{\mathbf q} \cdot \ddot{\mathbf q}\ + \nabla\Phi \cdot \dot{\mathbf q} = \dot{\mathbf q} \cdot \underset{=\ \mathbf 0}{\underbrace{\big( m \ddot{\mathbf q} - \mathbf F \big)}} =\ \mathbf 0 \end{align*} In the third equality we used that the force is conservative, and in the final step, we used Newton's second law which states that $m \ddot{\mathbf q} = \mathbf F$.
qed

Proof. The time derivative of the total momentum is $\displaystyle \frac{\mathrm{d}}{\mathrm{d} t} \mathbf P = \sum_{i=1}^N m_i \ddot{\mathbf q}_i(t)$ where $m_i \ddot{\mathbf q}_i(t)$ amounts to the force on particle $i$. This force amounts to the sum of an external force $\mathbf F_i$ on particle $i$ and the forces $\mathbf f_{ji}$ exerted by other particles $j$ on $i$. The net force amounts to the sum of the external forces, $\mathbf 0 = \mathbf F_{\text{tot}} = \sum_i \mathbf F_i$. Newton's third law requires that $\mathbf f_{ji} = -\mathbf f_{ij}$, and we will set $\mathbf f_{ii} = \mathbf 0$ to simplify notations of indices in the sums. Consequently, \begin{align*} \frac{\mathrm{d}}{\mathrm{d} t} \mathbf P = \sum_{i=1}^N \left( \mathbf F_i + \sum_{j=1}^N \mathbf f_{ji} \right) &= \sum_{i=1}^N \mathbf F_i + \sum_{i=1}^N \sum_{j=1}^N \mathbf f_{ij} \\ &= \mathbf F_{\text{tot}} + \frac{1}{2} \; \sum_{i=1}^N \sum_{j=1}^N \bigl( \mathbf f_{ij} + \mathbf f_{ji} \bigr) = \mathbf 0 \end{align*} qed

In the immediate vicinity of the collisions the balls in Newton's cradle perform a motion along a horizontal line, as discussed in Example 3.12. However, during the excursions to the left and right they follow a circular track where the chains act as arms and their suspension as fulcrum of the circular motion. In such settings it is often desirable to also consider the evolution of the angular momentum.

Proof.
\begin{align*} \frac{\mathrm{d}}{\mathrm{d} t} \mathbf L &= \sum_{i=1}^N m_i \Bigl( \dot{\mathbf q}_i(t) \times \dot{\mathbf q}_i(t) + \mathbf q_i(t) \times \ddot{\mathbf q}_i(t) \Bigr) \\ &= \sum_{i<j} \Bigl( \mathbf q_i(t) \times \mathbf f_{ij} + \mathbf q_j(t) \times \mathbf f_{ji} \Bigr) \\ &= \sum_{i<j} \Bigl( \mathbf q_i(t) - \mathbf q_j(t) \Bigr) \times \mathbf f_{ij} = \mathbf 0 \end{align*} where we used that $\mathbf f_{ij} = - \mathbf f_{ji}$ due to Newton's third law, and that $(\mathbf q_i(t) - \mathbf q_j(t) )$ is parallel to $\mathbf f_{ij}$ by assumption on the particle interactions.
qed

Problem 3.7: Derivatives of common composite expressions
Evaluate the following derivatives.

1. $\frac{\mathrm{d}}{\mathrm{d} x} ( a + x )^b$
   
2. $\frac{\partial}{\partial x} ( x + b \, y )^2$
   
3. $\frac{\mathrm{d}}{\mathrm{d} x} ( x + y(x) )^2$
4. $\frac{\mathrm{d}}{\mathrm{d} t} \sin \theta(t)$
5. $\frac{\mathrm{d}}{\mathrm{d} t} \bigl( \sin\theta(t) \; \cos\theta(t) \bigr)$
6. $\frac{\mathrm{d}}{\mathrm{d} t} \sin\bigl( 2 \theta(t) \bigr)$
7. $\frac{\mathrm{d}}{\mathrm{d} z} \sqrt{a + b \, z^2}$
8. $\frac{\partial}{\partial x_3} \left[ \sum_{j=1}^6 x_j^2 \right]^{-1/2}$
9. $\frac{\partial}{\partial y_1} \ln( \mathbf x \cdot \mathbf y )$

In these expressions $a$ and $b$ are real constants, and $\mathbf x$ and $\mathbf y$ are 6-dimensional vectors.

Problem 3.8: Running mothers
Demonstrate that \begin{align*} I = \dot x_1(t) \: \dot x_2(t) + \omega^2 \; x_1(t) \: x_2(t) \end{align*} is a constant of motion of a two-dimensional harmonic oscillator with equation of motion \begin{align*} \ddot{\mathbf x} (t) = - \omega^2 \: \mathbf x (t) \qquad \text{with } \quad \omega \in \mathbb R \quad \text{and} \quad \mathbf x (t) = \bigl( x_1(t), x_2(t) \bigr) \in \mathbb R^2 \end{align*}

Problem 3.9: Anvil shooting

Anvil shooting is a tradition in some US communities to celebrate St.Clement's Day, honoring Pope Clement I, the patron saint of blacksmiths and metalworkers. Typical anvils have a mass of about $150 \, \text{kg}$ and they are shot up to a height of $60 \, \text{m}$. Which energy must the gun powder release to the anvil for such a feat?

Problem 3.10: Running mothers
In the Clara Zetkin Park one regularly encounters blessings³⁾ of dozens of mothers jogging in the park while pushing baby carriages. Troops of kangaroo mothers rather carry their youngs in pouches.

1. Estimate the energy consumption spend in pushing the carriages as opposed to carrying the newborn.
   The carriages suffer from friction. Let the friction coefficient be $\gamma=0.3$.
   When carrying the baby the kangaroo must lift it up in every jump and the associated potential energy is dissipated.
2. How does the running speed matter in this discussion?
3. How does the mass of the babies/youngs make a difference?

Problem 3.11: The sledgehammer experiment
In his magnificent book “Thinking Physics” Lewis Carroll (Epstein (2009)) sets out a class room experiment that he used to perform in his physics class: He placed an anvil on his chest and asked a student from the audience to hit the anvil with a sledge hammer as hard as he could manage. What will happen?⁴⁾

Problem 3.12: The rotating chair experiment
The spin increases when an ice dancer pulls inwards arms and legs. This is illustrated in the picture of Yuko Kawaguti in the margin, and the physical principle has beautifully been demonstrated in a wikimedia movie by Oliver Zajkov from the Physics Institute at the University of Skopje.

1. Assume that a less careful experimenter starts his motion with a spin of $1 \, \text{Hz}$, holding $5 \, \text{kg}$ barbells with stretched-out arms $1 \, \text{m}$ away from the rotation axis.
   Estimate his spin rate when he pulls in his arms till the barbells reach a distance of $20 \, \text{cm}$ from the rotation axis.
2. Which trajectory will they take when the careless experimenter gets dizzy and looses hold on the barbells?