Just a few words about chapter 7:

OK, I've started worrying about you guys, reading this stuff on your own. Unfortunately I can't type in equations on the web. Nevertheless, let me stress a few things:

(1) You've been working with one four-vector for over a week now. That's the "displacement four-vector". We've usually ignored two of its components (y-position and z-position), and worked with only two of its components, x-position and time. You might want to think of the time component as ct instead of t - that way, it's clear that all components of this vector have the same dimensions.

(2) Think back to the beginning of the course. We talked about good old-fashioned 3-dimensional space, and imagined people working with two different ideas about which way is north. We agreed that, if we asked both groups of people how to get from here to Denver, say, the two groups would give us different stories about how far north and how far east or west we need to go, but both groups would tell us the same total distance we'd have to travel to get there. In our current vocabulary, we'd say that the components of the two different (three-dimensional) displacement vectors are different, but their magnitudes are the same. Our great leap into 20th Century relativity was to understand that position and time are components of the same vector. According to whom you ask, the components of the displacement between two events (delta-x and delta-t) are different, but the magnitude of the interval between them (c-squared delta-t-squared minus delta-x squared) is the same. Get the similarity to the situation with magnetic north vs geographic north? Hope so.

(3) The point of chapter 7 is to introduce ANOTHER vector that, like the displacement vector, (a) is important and (b) has different components in different reference frames and (c) has a magnitude that is the same in different reference frames. In Newtonian physics, things just don't get more important than momentum and energy.

(4) Personally, I don't think the authors give a convincing argument for why you have to divide the displacement components (delta-x, etc.) by proper time instead of time as measured in the laboratory frame. After all, we're measuring the displacements in the laboratory frame, aren't we? Really, the reason is that dividing by proper time gives us a four-vector that satisfies the three criteria listed in the preceding paragraph.

(5) The units and dimensions thing arises again: The x-component of the momentum is m times gamma times v-sub-x. The y and z components are defined similarly, and the fourth component ought to be m times gamma times c. This gives everything the same dimensions and units (mass times length divided by time). Of course, in the book c=1.

(6) If you multiply that fourth component of the momentum by c, you get gamma times m times c-squared. That's the particle's total energy. It's equal to the sum of its rest energy (which is m times c-squared) plus its kinetic energy. At low speeds, that kinetic energy is equal to one-half m v-squared, which is the Newtonian kinetic energy.

(7) The magnitude of the momentum-energy four-vector is the square of the fourth component minus the squares of the first three components; this magnitude is ALWAYS equal to m-squared times c-squared, which is just the square of the rest energy, divided by c-squared. You can multiply through by c-squared, and find that

the total energy squared - (p-sub-x squared + p-sub-y squared + p-sub-z squared)c-squared = the rest energy squared.

This is a hugely important fact.

(8) Think about the reference frame in which a particle is at rest. In that frame, p-sub-x, and the other two components, are zero, and the kinetic energy is zero, and so the total energy is just the rest energy. Notice how that's consistent with the above equation. Then, try to appreciate how cool it is that, in any other reference frame, this magnitude (of the momentum-energy vector) is still the same.

So, the bottom line on chapter 7 is, try to appreciate how the magnitude of a particle's momentum-energy vector is the same in all reference frames, just as the squared interval between two events is the same in all reference frames.