Mathematical prerequisites

PREREQUISITE This page supports Unit 1. Read a section when a chapter links to it. Polar acceleration, conic geometry, and vector identities are developed where they enter the later chapters.

P1. Vectors, position, and motion

A scalar has magnitude; a vector has magnitude and direction. In fixed Cartesian axes, position is $\mathbf r=x\hat{\mathbf x}+y\hat{\mathbf y}+z\hat{\mathbf z}$. The hats denote unit vectors of length one. Velocity is $\mathbf v=d\mathbf r/dt$ and acceleration is $\mathbf a=d\mathbf v/dt$. With fixed axes, differentiate each component separately.

For $\mathbf r(t)=(2t,t^2,0)$ in SI units, $\mathbf v=(2,2t,0)$ and $\mathbf a=(0,2,0)$. At $t=1\,\mathrm s$, speed is $|\mathbf v|=\sqrt{2^2+2^2}=2\sqrt2\,\mathrm{m\,s^{-1}}$. Speed is the length of the velocity vector, not the vector itself.

The dot product is $\mathbf A\cdot\mathbf B=A_xB_x+A_yB_y+A_zB_z=|\mathbf A||\mathbf B|\cos\gamma$, where $\gamma$ is the angle between them. It measures the component of one vector along the other. Nonzero perpendicular vectors have zero dot product. That is why a normal reaction has no work along a tangent displacement.

The cross product $\mathbf A\times\mathbf B$ is perpendicular to both vectors, with magnitude $AB\sin\gamma$ and direction given by the right-hand rule. Torque about the origin is $\boldsymbol\tau=\mathbf r\times\mathbf F$. Angular momentum is $\mathbf L=\mathbf r\times m\mathbf v$. Unit 1 mainly uses the dot product; central-force motion will use the cross product extensively.

Check: for $\mathbf A=(1,2)$ and $\mathbf B=(2,-1)$, the dot product is $2-2=0$. They are perpendicular. Now explain why zero dot product alone does not mean either vector is zero.

P2. Newton’s law, work, and potential

In an inertial frame and for constant mass, Newton's second law is $\mathbf F_{\mathrm{total}}=m\mathbf a$. The total force includes constraint reactions unless you have a justified projection that removes them. At static equilibrium the total force is zero, but individual forces may remain nonzero.

An infinitesimal actual displacement $d\mathbf r$ receives work $dW=\mathbf F\cdot d\mathbf r$. Integrating along a path gives total work. Dot Newton's law with $\mathbf v$:

The last equality is the derivative of $\mathbf v\cdot\mathbf v$: its two product-rule terms are equal. Thus force power changes kinetic energy $T=\tfrac12mv^2$.

A conservative force can be expressed in terms of potential energy. In one dimension, $F_x=-dV/dx$. In three dimensions, $\mathbf F=-\nabla V$, where $\nabla V=(\partial V/\partial x,\partial V/\partial y,\partial V/\partial z)$. The minus sign means force points toward decreasing potential.

For gravity near Earth's surface, choose upward $y$; then $V=mgy$ and $F_y=-mg$. For a spring with extension $x$, $V=\tfrac12kx^2$ gives $F_x=-kx$. Adding a constant to $V$ changes no force because the constant's derivative is zero.

Check: if $V=3x^2$ joules when $x$ is measured in metres, $F_x=-6x$ newtons. For $x>0$ the force points toward the origin; for $x<0$ it again points toward the origin. This is a restoring force.

P3. Functions, partial derivatives, and the chain rule

A multivariable function depends on more than one independent argument. For $f(q,t)=q^2+3t$, the partial derivatives are $\partial f/\partial q=2q$ with time held fixed, and $\partial f/\partial t=3$ with $q$ held fixed.

Along a trajectory $q=q(t)$, both arguments can change. The total derivative adds both effects:

If $q=t^2$, direct substitution gives $f=t^4+3t$, whose derivative $4t^3+3$ agrees with the chain rule. This is the difference between explicit time dependence (the $3t$ term) and implicit time dependence (the change in $q(t)$).

With several coordinates,

The same rule applies component by component to vectors. The total differential is $df=\sum_i f_{,i}dq_i+f_{,t}dt$. A partial derivative is not a quotient of arbitrary small changes; it is defined by changing one argument while holding the others fixed.

In analytical mechanics, a function such as $T(q,\dot q,t)$ regards coordinate and velocity as separate arguments for partial differentiation. For $T=\tfrac12m\dot q^2+mq\dot q$, $\partial T/\partial\dot q=m\dot q+mq$, whereas $\partial T/\partial q=m\dot q$. Along a real trajectory the total derivative would include changes of both arguments.

Check: for $f(q_1,q_2,t)=q_1q_2+tq_1$, obtain $df/dt=(q_2+t)\dot q_1+q_1\dot q_2+q_1$. Each term has a specific source; none should be silently omitted.

P4. Trigonometry and polar directions

For a planar position, $x=r\cos\theta$ and $y=r\sin\theta$. The radial unit vector is $\hat{\mathbf e}_r=(\cos\theta,\sin\theta)$ and the increasing-angle unit vector is $\hat{\mathbf e}_\theta=(-\sin\theta,\cos\theta)$. Their dot product is zero, and each has length one.

Using derivatives of sine and cosine,

Thus $\mathbf r=r\hat{\mathbf e}_r$ gives velocity $\mathbf v=\dot r\hat{\mathbf e}_r+r\dot\theta\hat{\mathbf e}_\theta$. These basis vectors move with the particle; they are not fixed Cartesian directions. Chapter 4 also derives the velocity directly in Cartesian components.

The identity $\sin^2\theta+\cos^2\theta=1$ will repeatedly simplify speeds. The derivative formulas assume radians. For small angles, $\sin\theta\approx\theta$ and $\cos\theta\approx1-\theta^2/2$. These are approximations near zero, not exact identities.

P5. Differential equations and initial conditions

An equation of motion such as $\ddot x+\omega^2x=0$ relates an unknown function to its derivatives. Its general solution is $x=A\cos\omega t+B\sin\omega t$. Differentiate twice: $\ddot x=-\omega^2A\cos\omega t-\omega^2B\sin\omega t=-\omega^2x$, verifying the equation.

Initial data determine the constants: $x(0)=A$ and $\dot x(0)=B\omega$. Therefore

For constant acceleration $\ddot q=a_0$, integration gives $\dot q=v_0+a_0t$, then $q=q_0+v_0t+\tfrac12a_0t^2$. Two constants are needed for one second-order equation. They specify one position and one velocity, not two configuration degrees of freedom.

P6. Stationary points and elementary stability

At a stationary point $q_0$ of potential, $V'(q_0)=0$. For a small displacement $\eta=q-q_0$, Taylor expansion gives

There is no linear term because $V'(q_0)=0$. If $V''(q_0)>0$, nearby configurations have higher potential and the force tends to restore the system. If $V''(q_0)<0$, nearby positions have lower potential and the force tends to carry it away. If $V''(q_0)=0$, this test decides nothing: $V=q^4$ and $V=-q^4$ have opposite behaviour despite identical zero second derivative at the origin.

This is sufficient background for the equilibrium comments in Unit 1. Later chapters will derive the small-perturbation equations and state stability conditions more fully.