Notes on Classical Mechanics by John R. Taylor.

$$ \newcommand{\xhat}{\vec{e_x}} \newcommand{\yhat}{\vec{e_y}} \newcommand{\rhat}{\vec{e_r}} \newcommand{\phihat}{\vec{e_\phi}} \newcommand{\r}{\vec{r}} \newcommand{\v}{\vec{v}} \newcommand{\p}{\vec{p}} \newcommand{\a}{\vec{a}} \newcommand{\F}{\vec{F}} \newcommand{\vector}[1]{\begin{bmatrix}#1\end{bmatrix}} $$

Chapter 1 - Newton's Laws of Motion

Basics

The basic object of interest is a moving particle. Its position at time \(t\) is \(\r\). It has that arrow over it because it is a vector. A vector is something that specifies a direction and a magnitude. Think of \(\r\) as an arrow from the origin pointing to the current position. Don't think of \(\r\) yet as a column vector containing numbers, because we haven't said what coordinate system we're using. Regardless of what coordinate system we use, \(\r\) is always a vector pointing from the origin to the current position.

The particle is moving, i.e. the position changes over time. So instead of just writing \(\r\), we write \(\r(t)\) which says that it's a function of time. Think of that as giving the answer to a question: "At a given time \(t\), what is the position?". The answer (position) is a vector, so we can say that this is a "vector-valued function" (i.e. whatever output it gives, it's always a vector).

Its velocity is a function \(\v(t)\) whose value is also a vector (at time \(t\) it's going at some speed in some direction). The velocity function \(\v(t)\) is the derivative with respect to time of the position function \(\r(t)\). That sounds very familiar, but what exactly is the derivative of a vector-valued function?

In normal, non-vector, calculus we imagine some curve like \(y = x^2\). So \(y\) is a function of \(x\). The value of that function is not a vector; it's just a number (a scalar). The derivative of that function with respect to \(x\) is saying: at a particular point along the x-axis, if I start advancing \(x\) a tiny bit, how fast is \(y\) changing? So, it's the slope of the curve at that point (also just a number, not a vector).

In vector calculus, the derivative of \(\r(t)\) with respect to \(t\) is saying: at some particular time \(t\), if I start advancing time a tiny bit, where is the position going and how fast is it going there? So the derivative of a vector-valued function is a vector -- an arrow with direction and magnitude (speed).

Coordinate systems

Thinking of \(\r(t)\) as an arrow with direction and magnitude is correct but a bit abstract. How specifically do we use numbers to represent position? The chapter covers two main coordinate systems. Let's say the particle is moving in 2D space for now.

  • Cartesian coordinates: we write down how far the particle currently is in the x-direction, \(x(t)\), and how far it currently is in the y-direction, \(y(t)\).

  • Polar coordinates: we write down how far the particle currently is, \(r(t)\), in the current direction to the particle.

Note that \(x(t)\), \(y(t)\), and \(r(t)\) were not written with arrows. They are just numbers, saying how far the particle is in some direction. The "in some direction" part corresponds to the concept of a unit vector. A "unit vector" is basically a vector where the direction is of interest, but the magnitude is just set to 1 for convenience.

Cartesian coordinates use two directions to specify the position. We'll write these directions as the unit vectors \(\xhat\) and \(\yhat\). So in Cartesian coordinates, the position is

$$\r(t) = x(t)\xhat + y(t)\yhat$$ $$\mathrm{Go~} x(t) \mathrm{~units~in~the~} \xhat \mathrm{~direction} \mathrm{~and~} y(t) \mathrm{~units~in~the~} \yhat \mathrm{~direction}$$

In contrast, polar coordinates just use one direction to specify the position: the direction of a direct line to the particle's current position. This direction is the unit vector \(\rhat(t)\). So in polar coordinates, the position is

$$\r(t) = r(t)\rhat(t)$$ $$\mathrm{Go~} r(t) \mathrm{~units~in~the~} \rhat(t) \mathrm{~direction}$$

Notice (and this is pretty important; it's basically the reason the chapter is covering polar coordinates) that in polar coordinates the unit vector \(\rhat(t)\) is a function of time (its direction changes as the particle moves); in contrast, in Cartesian coordinates, \(\xhat\) and \(\yhat\) are constant; they always point in the same direction. The polar unit vector is a function of time because it is the direction to wherever-the-particle-currently-is. The Cartesian unit vectors are not functions of time because they are just the x-axis direction and the y-axis direction and these do not change.

Velocity

We can now differentiate these position functions to get the velocity. Recall that the answer is going to be a vector because it is the derivative of a vector-valued function.

Cartesian coordinates

Because \(\xhat\) and \(\yhat\) are not functions of time, differentiating is straightforward:

$$\v(t) = \frac{d}{dt}\bigg(x(t)\xhat + y(t)\yhat\bigg) = \frac{d x(t)}{dt}\xhat + \frac{d y(t)}{dt} \yhat$$

Physicists use a dot to represent derivative-with-respect-to-time. So they might write this as

$$\v(t) = \dot x(t) \xhat + \dot y(t) \yhat$$

Either way, what this is saying is that in Cartesian coordinates, the velocity function is a vector comprised of current x-speed in the x-direction and current y-speed in the y-direction. In other words, it's what you expect.

Polar coordinates

$$\v(t) = \frac{d}{dt}\bigg(r(t)\rhat(t)\bigg)$$

That's a product of two things that are both a function of time, so we use the "product rule"1 to differentiate it:

$$\frac{d}{dt}\bigg(r(t)\rhat(t)\bigg) = \dot r(t) \rhat(t) + r(t)\frac{d \rhat(t)}{dt}$$

There's quite a few \(r\)s there and it's important at this stage not to get lost in the symbols. We know that the answer (velocity) is a vector. That means we can write it as a bunch of things added together, where each thing is a number times some unit vector. And we're using polar coordinates, so the unit vectors are going to be the polar unit vectors. So the thing on the left \(\dot r(t) \rhat(t)\) is fine: that's saying that the velocity has one component which is the current radial speed (a number \(\dot r(t)\)) in the current radial direction (the unit vector \(\rhat(t)\)).

What about the thing on the right? It's the current radial distance times the current derivative of the unit vector function. We've said that in polar coordinates the unit vector \(\rhat(t)\) changes over time, so it does make sense that we could ask what its derivative with respect to time is. So what is it? The answer is that it's a vector-valued function whose current value always points at right-angles to the current radial direction, but that requires explaining:

Going back to the informal definition of derivatives above, we're at some point \(t\) in time, and we imagine starting to advance time a tiny bit, and we look at the change in where the unit vector points, after this infinitesimally small amount of time passes. A unit vector always has length 1, so it can't grow in length. There's only one thing it can do: it can point in a slightly different direction. What direction has it gone in? It's basically like the hand of a clock. It's not too hard to see that if the hand of a clock changes just a tiny bit, then the tip moves in a direction that's almost a tangent to the circle. Change "tiny" to "infinitesimally small" and the "almost" goes away: so the time derivative of the radial unit vector is a vector pointing at right angles to the radial vector. This unit vector in that direction is called \(\phihat\), because it points in the direction that you go in when you increase the angle \(\phi\), as opposed to \(\rhat\) which points in the direction you go in if you increase the radius \(r\). How fast does the radial unit vector move in the \(\phihat\) direction? The answer is that it moves at the speed that the angle is increasing, so \(\dot \phi\)2. In other words, the time derivative of the radial unit vector is \(\dot \phi(t) \phihat(t)\)

The conclusion of all that is that in polar coordinates, the velocity vector is

$$\v(t) = \dot r(t) \rhat(t) + r(t) \dot \phi(t) \phihat(t)$$

Compare this with the expression for velocity in Cartesian coordinates

$$\v(t) = \dot x(t) \xhat + \dot y(t) \yhat$$

and we see it's a bit more complicated in polar coordinates.

I understand the polar coordinates version as follows. At time \(t\) the particle might be moving radially, and its angle might also be changing. The velocity vector has two components, one in the radial direction, and one in the tangent direction. In the radial direction, it's moving at whatever speed the radius is changing with. In the tangent direction it's moving at the speed that the angle is changing, multiplied by the current radius. That multiplication by radius makes sense informally, because if you are further out from the center of a circle, and the circle rotates by a few degrees, then you move further in space than if you were closer in to the center.

Acceleration

The acceleration function is the derivative of the velocity function with respect to time. Therefore, it is also a vector: at time \(t\) the particle is accelerating by some amount, in some direction.

Cartesian coordinates

Again, because the unit vectors do not change with time, it's as you expect: there's an x-acceleration in the x-direction, and a y-acceleration in the y-direction.

$$\a(t) = \ddot x(t) \xhat + \ddot y(t) \yhat$$

Polar coordinates

Above we saw that because, in polar coordinates, the directions of the coordinate system change with time, the function for velocity was more complicated than when using Cartesian coordinates. For acceleration, we differentiate the velocity expression and of course it gets even more complicated. But basically the answer is still a function of the form

$$\a(t) = \bigg( \text{Some function of } t \bigg) \rhat(t) + \bigg( \text{Another function of } t \bigg) \phihat(t)$$

The functions of \(t\) involve the current radius length, the speed and acceleration in the current radius direction, and the speed and acceleration of the angle parameter \(\phi\). The full expression is in the footnote3.

Newton's second law as a differential equation

A key point seems to be: view Newton's second law \(\F = m\a\) as a differential equation4:

$$m \ddot \r(t) = \F$$

I'm understanding this as follows: You know what forces are acting on the body in question. You want to know how the position of the body will evolve through time: \(\r(t)\). This is a function satisfying the following differential equation: the second derivative with respect to time of \(\r(t)\), times \(m\), is equal to the net force acting on the body.

In practice: in a typical problem you have some expression for \(\F\) derived from consideration of a diagram showing forces acting on the body. You might be able to discover \(\r(t)\) by finding a function whose second derivative is \(\F\).

Example problems

Cartesian coordinates

1.37 A student kicks a frictionless puck with initial speed \(v_0\), so that it slides up a plane that is inclined at an angle \(\theta\) above the horizontal. (a) Write down Newton's second law for the puck and solve to give its position as a function of time.

This is a simple example of using the Second Law as a differential equation. We write down the forces acting on the particle, set them equal to \(m\ddot r(t)\) and integrate twice to get position.

The only force acting on the puck is its weight, i.e. its mass times acceleration due to gravity: \(mg\). The puck can only move along the surface of the plane, so we are only interested in the component of the force that acts parallel to the plane. This component is \(-mg sin(\theta)\). So taking \(x\) as the direction up the plane, Newton's second law is

$$ m\ddot x(t) = -mgsin(\theta)$$

Integrating once gives velocity

$$ \dot x(t) = -g sin(\theta) t + v_0$$

Integrating again gives position

$$ x(t) = -\frac{1}{2} g sin(\theta) t^2 + v_0t + x_0$$

and \(x_0=0\) since we start measuring from its starting position.

(b) How long will the puck take to return to its starting point?

The puck is at its starting point whenever \(x = 0\):

$$0 = t\bigg(-\frac{1}{2} g sin(\theta) t + v_0\bigg)$$

The solutions of that are either \(t=0\) (which we already knew) or (the solution we want)

\(t = \frac{2v_0}{g sin(\theta)}\)

Polar coordinates

A "halfpipe" at a skateboard park consists of a concrete trough with a semicircular cross section of radius \(R = 5m\). I hold a frictionless skateboard on the side of the trough pointing down toward the bottom and release it. Discuss the subsequent motion using Newton's second law. In particular, if I release the skateboard just a short way from the bottom, how long will it take to come back to the point of release?

Conceptually, we do the same thing as for the problem using Cartesian coordinates: we write down Newton's second law resolved into two orthogonal directions. It's just that with polar coordinates, these orthogonal directions are constantly changing.

The weight of the skateboard acts downwards. This results in a tangent force causing the skateboard to move along the halfpipe, and also presses the skateboard into the halfpipe a bit, with an associated reaction force. We ignore the force/reaction force between the skateboard and the pipe and focus only on the tangent force: \(-mg sin(\phi)\).

The equation for acceleration says that, at time \(t\), acceleration in the current tangent direction is \(R\ddot \phi(t)\) (halfpipe radius times current angular acceleration5). So Newton's second law in this context is the differential equation

$$mR \ddot \phi(t) = -mg sin(\phi(t))$$

We read this as saying:

We don't know how the angle is changing over time \(\phi(t)\) -- that is precisely what we want to know. But what we do know is that whatever that function is, its second derivative at time \(t\) is equal to the sin of the current angle (times \(g/R\) and with a minus sign because the way we've defined the angle it gets smaller as the weight force takes the skateboard towards the bottom).

Once we've got to that point, finding the angle function \(\phi(t)\) is just math. It turns out that the only function for which it is true that the second derivative has this property6 is

$$\phi(t) = \phi_0 cos\bigg(\sqrt\frac{g}{R}t\bigg)$$

where \(\phi_0\) is the angle that the skateboard was released at at time \(t=0\). This is the "solution" of the differential equation: a function matching the criteria that the differential equation specified.

So we have our answer: the forces acting on the skateboard imply (via Newton's second law) that the way the angle of the skateboard changes is a cosine function of time. So the skateboard angle does what cosines do: it starts off at its maximum, decreases to zero, crosses zero and becomes negative for a while, starts turning back towards zero, crosses zero and becomes positive again and gets back to its maximum where it turns around again.

Conservation of momentum

Momentum is mass times velocity, \(\p(t) = m\dot \r(t)\), so another way of stating the second law is: rate of change of momentum is equal to force. In a multi-particle system the forces-and-reaction-forces of the third law cancel each other out when summing the rate of change of momentum of the whole system. So, total momentum doesn't change due to internal forces (but it does if there are external forces).

pp 21-23 show that conservation of momentum does not hold when considering magnetic and electrostatic forces between charged particles moving close to the speed of light. However I am unfamiliar with those forces and with the "right-hand rule" for fields/forces and I haven't understood this section.

  1. The product rule is the thing when you studied differentiation that says: when you're differentiating the product of two functions you differentiate one and keep the other as-is, then you differentiate the other while keeping the first as-is, and you add the two things together: \(\frac{d(f(t)g(t))}{dt} = \dot f(t) g(t) + f(t) \dot g(t)\) ↩
  2. You can prove this by writing the unit vector in Cartesian coordinates, \(cos(\phi) \xhat + sin(\phi) \yhat\), and then differentiating it to give \(\dot \phi\big(-sin(\phi)\xhat + cos(\phi)\yhat\big)\) which is \(\dot \phi\) times a vector orthogonal to the original one. ↩
  3. In polar coordinates, if you suppose that you know functions \(r(t)\) and \(\phi(t)\) giving the angle and distance at time \(t\), then the accelerations in the two orthogonal directions at time \(t\) are \(\a(t) = \bigg( \ddot r(t) - r(t) \dot\phi(t)^2 \bigg) \rhat(t) + \bigg( 2\dot r(t) \dot \phi(t) + r(t) \ddot \phi(t)\bigg) \phihat(t)\) ↩
  4. The dot means "differentiated with respect to time". So if \(r\) is position as a function of time then \(\dot r\) is velocity and \(\ddot r\) is acceleration. ↩
  5. To see this, start with the \(\phihat(t)\) (tangent direction) part of the full expression for acceleration and note that the radial distance of the skateboard is fixed by the presence of the half-pipe, so speed \(\dot r(t)\) (and acceleration) in the radial direction is zero. ↩
  6. Actually the solution is a function with second derivative having a different property, but one which is very similar to the desired property as long as we're restricting ourselves to the angle being fairly small. ↩