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per second (which they would abbreviate to something like 10 g cm/s). It is again going to be well worth our while to be a little bit more abstract and introduce placeholders rather than commit to any particular mass or speed. As ever, we certainly do not wish to transmogrify into the school mathematics teachers of our youth. But . . . if Δx is a placeholder for the length of the arrow, Δt is the time interval, and m is the mass of the ball (Δx = 2 centimeters, Δt = 2 seconds, and m = 10 grams in the example), then the momentum vector has a length equal to mΔx/Δt. It is common in physics to use the Greek symbol Δ(pronounced “delta”) to represent “difference,” and in that spirit Δt stands for the difference in time or the time interval between two things, and Δx stands for the length of something, in this case the distance in space between the start and the end of our measurement of the ball’s position.

We have succeeded in constructing the momentum vector of a ball in three-dimensional space, although it is hardly the most exciting thing we have done. We’re now going to make the bold step of trying to build a momentum vector in spacetime, and we will do it in an entirely analogous way to the three-dimensional case. The only constraint is that we will use only objects that are universal in spacetime.

FIGURE 12

Again we shall start with an arrow, this time pointing in four-dimensional spacetime, as illustrated in Figure 12. One end of the arrow specifies where our ball is at one instant and the other end specifies where it is some time later. The length of the arrow must be determined by Minkowski’s formula for the distance in spacetime, and it is therefore specified by (Δs) 2 = (cΔt)2 - (Δx)2. Remember that Δs is the only length that everyone in the universe can agree upon (something that most definitely cannot be said for Δx and Δt separately), and as such it is the distance measurement we must use, taking the place of Δx in the three-dimensional definition of momentum. But what is to take the place of the time interval Δt? (Remember, we are trying to find a four-dimensional replacement for mΔx/Δt). Here comes the crunch: We cannot use Δt because it is not a spacetime invariant. Not everyone agrees on time intervals, as we have emphasized again and again, and therefore we must not use time intervals in our quest for the four-dimensional momentum. What are our choices? By what could we possibly divide the length of the arrow by to determine the ball’s speed through spacetime?

We want to construct something that is an improvement over the old three-dimensional momentum. If we are dealing with objects moving around at speeds that are slow compared to the speed of light, then we should find that the new momentum is at least approximately equivalent to the old one. If that is to happen, we must divide the length of our arrow in spacetime Δs by some quantity that is of the same type as an interval in time. Otherwise the new four-dimensional momentum will be an entirely different beast from the old three-dimensional momentum. Intervals of time can be measured in seconds, so we would also like something that can be measured in seconds. Starting from our invariant spacetime quantities, the speed of light c and the distance Δs, there is only one viable combination: It is the number we obtain upon dividing the length of the arrow (Δs) by the speed c. In other words, if Δs is measured in meters, and the speed c is measured in meters per second, then Δs/c is measured in seconds. This must be the number we need to divide the length of our arrow by, since it is the only invariant thing we have at our disposal that is measured in the correct currency. So let us go ahead and divide Δs by the time Δs/c. The answer is simply c (for much the same reason that 1 divided by ½ is equal to 2). In other words, the four-dimensional analogue of the speed in our three-dimensional momentum formula is the universal speed limit c.

This all might feel rather familiar, and that is because it should be familiar. All we have done is to calculate