Vectors

The bench launches a ball at 10 m/s, 45° above the ground, and two laboratories measure the same flight. A has put the origin of its axes at the launch point. B has put its own 8 m further along and 2 m higher, at a spot that can be changed by dragging the magenta circle. Half a second after the launch, A records the ball at (3.54, 2.31) m and B at (−4.46, 0.31) m, and both are right without agreeing on a single number.

What they have in common turns up in the differences. Between two consecutive rows of the tables, 0.25 s apart, the ball moves 1.77 m horizontally for A and 1.77 m for B, and the same happens vertically, because B’s origin enters both positions and drops out of the subtraction. Divided by the interval, these differences give the average velocity over each stretch, which is therefore the same in both laboratories. The Δy shrinks by about 0.61 m from one row to the next, in both tables, and that is the mark of gravity, an acceleration the two also measure alike.

With the same acceleration, the same mass and the same forces, the law \(F = ma\) that holds for A holds for B in the same form, only with other letters for the coordinates. No mechanical experiment could single out a privileged origin, then, since any point serves as well as any other; the secondary-school lesson on frames of reference relies on this freedom whenever it puts the zero wherever the arithmetic is simplest.

laboratory A   laboratory B   trajectory   velocity

Physicists call this a symmetry, in a sense Feynman borrows from Hermann Weyl. Something counts as symmetric when it comes out of an operation and nobody could tell that the operation had been carried out. The operation on this bench is taking the origin to another point, a translation, and what cannot tell the difference is the laws of motion. One then says that the laws of mechanics are invariant under translation.

The symmetry belongs to the laws, not to things. A building on top of a hill is not the same as a building by the sea, and a ball thrown from the edge of a cliff falls further than the one on the bench, because the ground is no longer where it was. For an experiment to come out the same somewhere else, everything that acts on it has to be taken along, and for the launch on the bench that includes the ground and the Earth that pulls on the ball.

In Feynman: §11-1 Symmetry in physics ↗ · §11-2 Translations ↗

Now the origin stays put and it is the axes that move. The red arrow on the bench is a 5 N force, with components of 4 N and 3 N on the green axes, the original ones. On the magenta axes, turned by 30°, the same force has components of 4.96 N and 0.60 N. The dashed lines show where these numbers come from: each component is the projection of the arrow onto one axis, and turning the axis changes the projection.

The calculation that takes one pair of numbers to the other needs only a cosine and a sine of the angle, and it is the same for the coordinates of a point and for the components of a force, a velocity or any other arrow. What does not change is the length. The squares of 4 and 3 add up to 25, so do those of 4.96 and 0.60, and the Length box stays still while the angle is turned, because turning the axes neither stretches nor shrinks anything. Dragging the tip of the force changes its length, but the new value is again the same in both systems.

original axes   rotated axes   force or gravity   trajectory

The Launch scenario puts the same turn into a problem in mechanics. A ball leaves at 10 m/s, and gravity, which on the green axes points straight down, has two components on the magenta axes turned by 30°: −4.90 m/s² along x′ and −8.49 m/s² along y′. Someone who used only the rotated axes would see a ball accelerated in both directions. Applying the same launch law to each component, they would find at every instant the position one gets by rotating the one in the green system, as the bench’s message checks. The secondary-school lesson on projectiles puts the y axis vertical to save arithmetic, and the bench shows that any other choice describes the same trajectory.

None of this means that every device works the same at every tilt. A pendulum clock mounted askew keeps time differently, or stops, and mechanics is not to blame: the pull on the pendulum comes from the Earth, which stayed where it was when the clock was tilted. Invariance under rotation promises the same result when everything that produces the forces is turned. Turning the whole laboratory, Earth included, amounts to what the bench does, looking at the same experiment through axes turned the other way. The clock example is Feynman’s.

In Feynman: §11-3 Rotations ↗

The bench starts with two arrows leaving the same origin, \(\vec a\) with components 3 and 4 and \(\vec b\) with −1 and 2, and each can be dragged by its tip. Adding them means adding the components one at a time, 3 − 1 = 2 in x and 4 + 2 = 6 in y, and the red arrow ending at (2, 6) is \(\vec a + \vec b\). The same point appears with no arithmetic at all when \(\vec b\) is redrawn to start where \(\vec a\) ends, because walking 3 and 4 and then −1 and 2 takes you where walking 2 and 6 in one go does. Starting with \(\vec b\) gets you there too, and the two routes close the dashed parallelogram on the bench.

The Rotate axes button swaps the green axes for the magenta ones, tilted by 30° as in step 2. On the new axes \(\vec a\) has components 4.60 and 1.96 and \(\vec b\) has 0.13 and 2.23, and the sums, about 4.73 and 4.20, are what the formulas of step 2 give for the point (2, 6); the small discrepancy in the second decimal place comes only from adding terms that were already rounded. Adding on the green axes or on the magenta ones produces the same red arrow, described by other numbers, and its length of 6.32 is unaltered.

The rotated axes hold the key to what a vector is. In space it has three components, and in the plane of the bench, two; what makes it a vector is the way they respond when the axes turn, by the same rule of cosines and sines that governs the coordinates of a point. Position and force pass this test. The sum passes too, because the rule only multiplies components by fixed numbers and adds the products, so rotating and then adding gives the same as adding and then rotating.

\(\vec a\)   \(\vec b\)   result   original axes   rotated axes

Difference relies on the same sum. Multiplying \(\vec b\) by −1 turns the arrow round, and adding this opposite to \(\vec a\) gives \(\vec a - \vec b\), which on the bench is (4, 2), with a length of 4.47. Scale multiplies \(\vec a\) by a number \(k\). With \(k = 1.5\) the arrow grows to (4.5, 6) and a length of 7.5 without leaving its line, and with a negative \(k\) it points the opposite way. Multiplying by a number does not depend on the axes either, and so \(k\vec a\) and \(\vec a - \vec b\) are vectors like the others. The secondary-school lesson on adding velocities uses these rules to cross a river.

The payoff comes when a law is written down. Newton’s second law in the form \(\vec F = m\vec a\) packs into one line the equalities \(F_x = m a_x\) and \(F_y = m a_y\), and packs them for any axes, because both sides change by the same rule when the axes turn. If it holds on the green axes, it holds on the magenta ones, and there is no need to redo the calculation of step 2 for every new law. Whatever is written with vectors alone, using only the operations on this bench, comes with rotation invariance built in.

In Feynman: §11-4 Vectors ↗ · §11-5 Vector algebra ↗

The bench puts a car on a circular track of 5 m radius, setting off at 2 m/s and gaining 0.5 m/s every second. The position arrow runs from the centre of the track to the car. It changes from instant to instant, and the rate of that change, the time derivative of the arrow, is the velocity, the blue arrow, always tangent to the track. The derivative is taken component by component, and since turning the axes only combines components with fixed cosines and sines, differentiating and then rotating gives the same as rotating and then differentiating, just as with the sum in step 3; velocity is therefore a vector. Acceleration is the derivative of velocity, and a vector for the same reason.

In step 2 of the chapter 9 lesson the point went round at constant speed, and the acceleration always pointed to the centre. Here the speedometer climbs, and the red arrow no longer points to the centre: with the Time control at 4 s and the car at 4 m/s, it leans about 9° forwards. The orange and magenta arrows split it into two parts at right angles, one along the velocity and the other towards the centre of the track.

velocity   acceleration and \(\Delta\vec v\)   tangential part   normal part

The velocity panel shows where the two parts come from. It copies the velocity of half a second earlier and the current one to a common origin, and \(\Delta\vec v\) is the red arrow joining their tips. The dashed arc marks how far the new arrow would reach if it kept its old size, and \(\Delta\vec v\) is the sum of two pieces: the magenta chord between the tips on the arc, which only turns the arrow, and the orange stretch along the new velocity, which only lengthens it. At t = 4 s, \(|\Delta\vec v|\) is 1.51 m/s, yet the speed has grown by only 0.25 m/s in that half second, so nearly all of \(\Delta\vec v\) is turning. With half a second between the two the chord is not yet perpendicular to the new velocity: at t = 4 s the arrow has turned through 22°, and the chord makes 90° minus half of that with it, 79°. It becomes perpendicular only in the limit of ever shorter intervals.

Divided by the interval, and with the interval shrinking, the two pieces become the two components of the acceleration. The tangential one is the rate at which the speed changes, 0.5 m/s² at every instant of this run. The normal one is the one from chapter 9, \(v^2/R\), which with 4 m/s and a 5 m radius comes to 3.2 m/s² and grows with the square of the speed. For the car to do this, the ground has to push on the tyres with the force \(m\vec a\), tilted like the red arrow, and the equality \(\vec F = m\vec a\) holds component by component along any pair of perpendicular directions, those of the x and y axes or these two, which travel with the car. The secondary-school lesson on uniformly accelerated circular motion reaches the same two parts through the angular velocity.

In Feynman: §11-6 Newton’s laws in vector notation ↗

The bench starts with two arrows at the same origin, \(\vec a\) with components 4 and 2 and \(\vec b\) with 1 and 3. Multiplying the components with the same name and adding the products, 4 × 1 + 2 × 3, gives the number 10. On the magenta axes, turned 30° clockwise, the components become 2.46 and 3.73 for \(\vec a\) and −0.63 and 3.10 for \(\vec b\); the products are now −1.56 and 11.56, and the sum is 10 once more. The Axis rotation control can go from −90° to 90° without moving that number, while the four numbers that produce it change all the time.

This number is the scalar product, written \(\vec a\cdot\vec b\). A scalar, here, is a plain number, one that has the same value for whoever uses the green axes and whoever uses the magenta ones, unlike the components, of which each system has its own. Step 2 already had a number of this kind, the length. The sum of the squares of the components is the scalar product of an arrow with itself, \(\vec a\cdot\vec a = |\vec a|^2\), and the Length box on that bench stayed still for the same reason that the product box stays still on this one.

The reason shows once each arrow is written through its length and the angle it makes with the x axis. If \(\vec a\) makes an angle \(\alpha\) and \(\vec b\) an angle \(\beta\), the sum of the products becomes \(|\vec a|\,|\vec b|\,(\cos\alpha\cos\beta + \sin\alpha\sin\beta)\), which is \(|\vec a|\,|\vec b|\cos(\beta - \alpha)\). Turning the axes takes the same angle off \(\alpha\) and \(\beta\) and leaves their difference intact, and that difference is the angle \(\theta\) between the arrows. On the bench, \(\vec a\) makes 26.6° with the green x and \(\vec b\) makes 71.6°; measured from the magenta x′, the angles become 56.6° and 101.6°, and the difference is still 45°. The product of the lengths, 4.47 × 3.16, is 14.14, and multiplied by cos 45° it gives the same 10.

\(\vec a\) or \(\vec d\)   \(\vec b\) or \(\vec F\)   shadow on \(\vec a\) or \(\vec d\)   original axes   rotated axes

The form with the cosine says what the number measures. \(|\vec b|\cos\theta\) is the size of the shadow \(\vec b\) casts on the line of \(\vec a\), the orange band on the bench, and the scalar product is the length of \(\vec a\) times that shadow, 4.47 × 2.24. The shadow shrinks as the angle opens, vanishes when the arrows are at right angles and, beyond 90°, falls on the other side of the origin, and the product turns negative.

The Work scenario carries the shadow over to mechanics. A 10 N force, tilted 60° above the ground, accompanies a block that slides 3 m horizontally. Only the shadow of the force on the displacement, 5 N, pushes the block forwards, and the rest pulls upwards, in a direction in which the block does not move. The work done by the force is \(\vec F\cdot\vec d\), 5 N × 3 m = 15 J, and the form with the cosine, 10 × 3 × cos 60°, gives the same. The block’s weight is perpendicular to the ground and does no work. On the slope in step 3 of the chapter 4 lesson, weight times height was this very product with its sign reversed: along the path up the slope, the weight sees only the vertical part, and its work is negative while the block rises.

Kinetic energy belongs to the same family. The \(v^2\) in \(\tfrac12 m v^2\) is \(\vec v\cdot\vec v\), and a 2 kg body with velocity components of 3 and 4 m/s has \(\vec v\cdot\vec v = 25\) m²/s² and 25 J of energy on any axes. This also sheds light on step 4. The normal part of the acceleration is perpendicular to the velocity, has a zero scalar product with it and so leaves the speed alone; step 6 of the chapter 4 lesson shows that \(K\) changes at the rate \(\vec F\cdot\vec v\). Both uses, work and kinetic energy, also appear in Feynman.

In Feynman: §11-7 Scalar product of vectors ↗