Why Does E=mc2_ - Brian Cox [51]
To finish our construction of the new spacetime momentum arrow, all we need to do is multiply the spacetime velocity vector by the mass m. It follows that our proposed momentum arrow always has a length equal to mc and points in the direction of travel of the object in spacetime. At first glance this new momentum arrow is a little boring because its length in spacetime is always the same. It seems we are hardly off to a good start. But we should not be deterred. It remains to be seen whether the spacetime momentum vector that we have just constructed bears any relation to the old-fashioned three-dimensional momentum or, for that matter, whether it will be of any use to us in our new spacetime world.
To delve a little deeper, we will now take a look at the portions of our new spacetime momentum vector that point in the space and time directions separately. To do this bit of delving, we need a bit of absolutely unavoidable mathematics. We can only apologize to the nonmathematical reader and promise that we will go very slowly. Remember, it is always an option to skim over the equations in search of the punch line. The mathematics makes the argument more convincing but it is okay to read on without following the details. Similarly, we must also apologize to the reader familiar with mathematics for laboring the point. We have a saying in Manchester: “You can’t have your cake and eat it.” This saying is perhaps harder to understand than the mathematics.
Recall that we arrived at an expression for the length of the momentum vector in three-dimensional space, mΔx/Δt. We have just argued that Δx should be replaced by Δs and Δt should be replaced by Δs/c to form the four-dimensional momentum vector, which has a seemingly rather uninteresting length of mc. Indulge us for one more paragraph, and let us write the replacement for Δt, i.e., Δs/c, in full. Δs/c is equal to. This is a bit of a mouthful, but a little mathematical manipulation allows us to write it in a simpler form, i.e., it can also be written as Δt/γ where. To obtain that, we have used the fact that υ = Δx/Δt is the speed of the object. Now γ is none other than the quantity we met in Chapter 3 that quantifies the amount by which time slows down from the point of view of someone observing a clock fly past at speed.
We are actually nearly where we want to be. The whole point of that piece of mathematics is that it allows us to figure out by exactly how much the momentum vector points off in the space and time directions separately. First let’s recap how we dealt with the momentum vector in three-dimensional space. Figure 11 helped us picture this. The three-dimensional momentum vector points off in exactly the same direction as the arrow in Figure 11, because it points in the same direction that the ball is moving in. The only difference is that its length is changed because we need to multiply it by the mass of the ball and divide by the time interval. The situation is entirely analogous in the four-dimensional case. Now the momentum vector points off in the direction in spacetime in which the ball is moving, which is the direction of the arrow in Figure 12. Again, to get the momentum, we need to rescale the length of the arrow, but this time we are to multiply by the mass and divide by the invariant quantity Δs/c (which we showed in the last paragraph is equal to Δt/γ). If you look carefully at the arrow in Figure 12, you should be able to see that if we want to change the length by some amount while keeping