More about four-vectors
Let us now return to our consideration of the analogy of the Lorentz                         transformation and rotations of the space axes. We have learned the                         utility of collecting together other quantities which have the same                         transformation properties as the coordinates, to form what we call                         vectors, directed lines. In the case of ordinary rotations, there                         are many quantities that transform the same way as $x$, $y$, and $z$                         under rotation: for example, the velocity has three components, an                         $x$, $y$, and $z$-component; when seen in a different coordinate system,                         none of the components is the same, instead they are all transformed to                         new values. But, somehow or other, the velocity “itself” has a greater                         reality than do any of its particular components, and we represent it by                         a directed line.                          We therefore ask: Is it or is it not true that there are quantities                         which transform, or which are related, in a moving system and in a                         nonmoving system, in the same way as $x$, $y$, $z$, and $t$? From our                         experience with vectors, we know that three of the quantities, like $x$,                         $y$, $z$, would constitute the three components of an ordinary                         space-vector, but the fourth quantity would look like an ordinary scalar                         under space rotation, because it does not change so long as we do not go                         into a moving coordinate system. Is it possible, then, to associate with                         some of our known “three-vectors” a fourth object, that we could call                         the “time component,” in such a manner that the four objects together                         would “rotate” the same way as position and time in space-time? We                         shall now show that there is, indeed, at least one such thing (there are                         many of them, in fact): the three components of momentum, and the                             energy as the time component, transform together to make what we call a                         “four-vector.” In demonstrating this, since it is quite inconvenient                         to have to write $c$’s everywhere, we shall use the same trick                         concerning units of the energy, the mass, and the momentum, that we used                         in Eq. (17.4). Energy and mass, for example, differ only by a                         factor $c^2$ which is merely a question of units, so we can say energy                         is the mass. Instead of having to write the $c^2$, we put $E = m$, and                         then, of course, if there were any trouble we would put in the right                         amounts of $c$ so that the units would straighten out in the last                         equation, but not in the intermediate ones. Thus our equations for energy and momentum are                         \begin{equation}                         \begin{alignedat}{2}                         &E&&=m=m_0/\sqrt{1-v^2},\\[1ex]                         &\FLPp&&=m\FLPv=m_0\FLPv/\sqrt{1-v^2}.                         \end{alignedat}                         \label{Eq:I:17:6}                         \end{equation}                         Also in these units, we have                         \begin{equation}                         \label{Eq:I:17:7}                         E^2-p^2=m_0^2.                         \end{equation}                         For example, if we measure energy in electron volts, what does a mass of                         $1$ electron volt mean? It means the mass whose rest energy is                         $1$ electron volt, that is, $m_0c^2$ is one electron volt. For example,                         the rest mass of an electron is $0.511\times10^6$ eV. Now what would the momentum and energy look like in a new coordinate                         system? To find out, we shall have to transform Eq. (17.6),                         which we can do because we know how the velocity transforms. Suppose                         that, as we measure it, an object has a velocity $v$, but we look upon                         the same object from the point of view of a space ship which itself is                         moving with a velocity $u$, and in that system we use a prime to                         designate the corresponding thing. In order to simplify things at first,                         we shall take the case that the velocity $v$ is in the direction of $u$.                         (Later, we can do the more general case.)  What is $v'$, the velocity as                         seen from the space ship? It is the composite velocity, the                         “difference” between $v$ and $u$. By the law which we worked out                         before,                         \begin{equation}                         \label{Eq:I:17:8}                         v'=\frac{v-u}{1-uv}.                         \end{equation}                         Now let us calculate the new energy $E'$, the energy as the fellow in                         the space ship would see it. He would use the same rest mass, of course,                         but he would use $v'$ for the velocity. What we have to do is                         square $v'$, subtract it from one, take the square root, and take the                         reciprocal:                         \begin{equation*}                         \begin{aligned}                         v'^2&=\frac{v^2-2uv+u^2}{1-2uv+u^2v^2},\\[1.5ex]                         1-v'^2&=\frac{1-2uv+u^2v^2-v^2+2uv-u^2}{1-2uv+u^2v^2},\\[1ex]                         &=\frac{1-v^2-u^2+u^2v^2}{1-2uv+u^2v^2},\\[1.5ex]                         &=\frac{(1-v^2)(1-u^2)}{(1-uv)^2}.                         \end{aligned}                         \end{equation*}                         Therefore                         \begin{equation}                         \label{Eq:I:17:9}                         \frac{1}{\sqrt{1-v'^2}}=\frac{1-uv}{\sqrt{1-v^2}\sqrt{1-u^2}}.                         \end{equation} The energy $E'$ is then simply $m_0$ times the above expression. But we                         want to express the energy in terms of the unprimed energy and momentum,                         and we note that                                                  \begin{equation*}                         E'=\frac{m_0-m_0uv}{\sqrt{1-v^2}\sqrt{1-u^2}}=                         \frac{(m_0/\sqrt{1-v^2})-(m_0v/\sqrt{1-v^2})u}{\sqrt{1-u^2}},                         \end{equation*}                        
                         \begin{align*}                         E'&=\frac{m_0-m_0uv}{\sqrt{1-v^2}\sqrt{1-u^2}}\\[2ex]                         &=\frac{(m_0/\sqrt{1-v^2})-(m_0v/\sqrt{1-v^2})u}{\sqrt{1-u^2}},                         \end{align*}                                                 or                         \begin{equation}                         \label{Eq:I:17:10}                         E'=\frac{E-up_x}{\sqrt{1-u^2}},                         \end{equation}                         which we recognize as being exactly of the same form as                         \begin{equation*}                         t'=\frac{t-ux}{\sqrt{1-u^2}}.                         \end{equation*}                         Next we must find the new momentum $p_x'$. This is just the energy $E'$                         times $v'$, and is also simply expressed in terms of $E$ and $p$:                                                  \begin{equation*}                         p_x'=E'v'=\frac{m_0(1-uv)}{\sqrt{1-v^2}\sqrt{1-u^2}}\cdot                         \frac{v-u}{(1-uv)}=                         \frac{m_0v-m_0u}{\sqrt{1-v^2}\sqrt{1-u^2}}.                         \end{equation*}                         
                         \begin{align*}                         p_x'=E'v'&=\frac{m_0(1-uv)}{\sqrt{1-v^2}\sqrt{1-u^2}}\cdot                         \frac{v-u}{(1-uv)}\\[2ex]                         &=\frac{m_0v-m_0u}{\sqrt{1-v^2}\sqrt{1-u^2}}.                         \end{align*}                                                   Thus                         \begin{equation}                         \label{Eq:I:17:11}                         p_x'=\frac{p_x-uE}{\sqrt{1-u^2}},                         \end{equation}                         which we recognize as being of precisely the same form as                         \begin{equation*}                         x'=\frac{x-ut}{\sqrt{1-u^2}}.                         \end{equation*} Thus the transformations for the new energy and momentum in terms of the                         old energy and momentum are exactly the same as the transformations                         for $t'$ in terms of $t$ and $x$, and $x'$ in terms of $x$ and $t$: all                         we have to do is, every time we see $t$ in (17.4)                         substitute $E$, and every time we see $x$ substitute $p_x$, and then the                         equations (17.4) will become the same as Eqs.                         (17.10) and (17.11). This would imply, if                         everything works right, an additional rule that $p_y' = p_y$ and                         that $p_z' = p_z$. To prove this would require our going back and                         studying the case of motion up and down. Actually, we did study the case                         of motion up and down in the last chapter. We analyzed a complicated                         collision and we noticed that, in fact, the transverse momentum is                         not changed when viewed from a moving system; so we have already                         verified that $p_y' = p_y$ and $p_z' = p_z$. The complete                         transformation, then, is                         \begin{equation}                         \begin{aligned}                         p_x'&=\frac{p_x-uE}{\sqrt{1-u^2}},\\                         p_y'&=p_y,\\[1ex]                         p_z'&=p_z,\\                         E'&=\frac{E-up_x}{\sqrt{1-u^2}}.                         \end{aligned}                         \label{Eq:I:17:12}                         \end{equation} In these transformations, therefore, we have discovered four quantities                         which transform like $x$, $y$, $z$, and $t$, and which we call the                         four-vector momentum. Since the momentum is a four-vector, it can                         be represented on a space-time diagram of a moving particle as an                         “arrow” tangent to the path, as shown in Fig. 17–4. This                         arrow has a time component equal to the energy, and its space components                         represent its three-vector momentum; this arrow is more “real” than                         either the energy or the momentum, because those just depend on how we                         look at the diagram.