Four-vector algebra
The notation for four-vectors is different than it is for three-vectors.                         In the case of three-vectors, if we were to talk about the ordinary                         three-vector momentum we would write it $\FLPp$. If we wanted to be more                         specific, we could say it has three components which are, for the axes                         in question, $p_x$, $p_y$, and $p_z$, or we could simply refer to a                         general component as $p_i$, and say that $i$ could either be $x$, $y$,                         or $z$, and that these are the three components; that is, imagine                         that $i$ is any one of three directions, $x$, $y$, or $z$. The notation                         that we use for four-vectors is analogous to this: we write $p_\mu$                         for the four-vector, and $\mu$ stands for the four possible                         directions $t$, $x$, $y$, or $z$. We could, of course, use any notation we want; do not laugh at                         notations; invent them, they are powerful. In fact, mathematics is, to a                         large extent, invention of better notations. The whole idea of a                         four-vector, in fact, is an improvement in notation so that the                         transformations can be remembered easily. $A_\mu$, then, is a general                         four-vector, but for the special case of momentum, the $p_t$ is                         identified as the energy, $p_x$ is the momentum in the $x$-direction,                         $p_y$ is that in the $y$-direction, and $p_z$ is that in the                         $z$-direction. To add four-vectors, we add the corresponding components.                          If there is an equation among four-vectors, then the equation is true                         for each component. For instance, if the law of conservation of                         three-vector momentum is to be true in particle collisions, i.e., if the                         sum of the momenta for a large number of interacting or colliding                         particles is to be a constant, that must mean that the sums of all                         momenta in the $x$-direction, in the $y$-direction, and in the                         $z$-direction, for all the particles, must each be constant. This law                         alone would be impossible in relativity because it is incomplete;                         it is like talking about only two of the components of a three-vector.                         It is incomplete because if we rotate the axes, we mix the various                         components, so we must include all three components in our law. Thus, in                         relativity, we must complete the law of conservation of momentum by                         extending it to include the time component. This is                         absolutely necessary to go with the other three, or there cannot                         be relativistic invariance. The conservation of energy is the                         fourth equation which goes with the conservation of momentum to make a                         valid four-vector relationship in the geometry of space and time. Thus                         the law of conservation of energy and momentum in four-dimensional                         notation is                         \begin{equation}                         \label{Eq:I:17:13}                         \sum_{\substack{\text{particles}\\\text{in}}}p_\mu=                         \sum_{\substack{\text{particles}\\\text{out}}}p_\mu                         \end{equation}                         or, in a slightly different notation                         \begin{equation}                         \label{Eq:I:17:14}                         \sum_ip_{i\mu}=\sum_jp_{j\mu},                         \end{equation}                         where $i = 1$, $2$, … refers to the particles going into the                         collision, $j= 1$, $2$, … refers to the particles coming out of                         the collision, and $\mu = x$, $y$, $z$, or $t$. You say, “In which                         axes?”  It makes no difference. The law is true for each component,                         using any axes. In vector analysis we discussed one other thing, the dot product of two                         vectors. Let us now consider the corresponding thing in space-time. In                         ordinary rotation we discovered there was an unchanged quantity $x^2 +                         y^2 + z^2$. In four dimensions, we find that the corresponding quantity                         is $t^2 - x^2 - y^2 - z^2$ (Eq. 17.3). How can we write that?                         One way would be to write some kind of four-dimensional thing with a                         square dot between, like $A_\mu \boxdot B_\mu$; one of the notations                         which is actually used is                         \begin{equation}                         \label{Eq:I:17:15}                         \sideset{}{'}\sum_\mu A_\mu A_\mu=A_t^2-A_x^2-A_y^2-A_z^2.                         \end{equation}                         The prime on $\sum$ means that the first term, the “time” term, is                         positive, but the other three terms have minus signs. This quantity,                         then, will be the same in any coordinate system, and we may call it the                         square of the length of the four-vector. For instance, what is the                         square of the length of the four-vector momentum of a single particle?                         This will be equal to $p_t^2 - p_x^2 - p_y^2 - p_z^2$ or, in other                         words, $E^2 - p^2$, because we know that $p_t$ is $E$. What is $E^2 -                         p^2$? It must be something which is the same in every coordinate system.                         In particular, it must be the same for a coordinate system which is                         moving right along with the particle, in which the particle is standing                         still. If the particle is standing still, it would have no momentum. So                         in that coordinate system, it is purely its energy, which is the same as                         its rest mass. Thus $E^2 - p^2 = m_0^2$. So we see that the square of                         the length of this vector, the four-vector momentum, is equal                         to $m_0^2$. From the square of a vector, we can go on to invent the “dot product,”                         or the product which is a scalar: if $a_\mu$ is one four-vector                         and $b_\mu$ is another four-vector, then the scalar product is                         \begin{equation}                         \label{Eq:I:17:16}                         \sideset{}{'}\sum a_\mu b_\mu=a_tb_t-a_xb_x-a_yb_y-a_zb_z.                         \end{equation}                         It is the same in all coordinate systems. Finally, we shall mention certain things whose rest mass $m_0$ is zero.                         A photon of light, for example. A photon is                         like a particle, in that it carries an energy and a momentum. The energy                         of a photon is a certain constant, called Planck’s                         constant, times the frequency of the                         photon: $E = h\nu$. Such a photon also carries a momentum,                         and the momentum of a photon (or of any other particle, in                         fact) is $h$ divided by the wavelength: $p = h/\lambda$. But, for a                         photon, there is a definite relationship between the frequency and the                         wavelength: $\nu= c/\lambda$. (The number of waves per second, times the                         wavelength of each, is the distance that the light goes in one second,                         which, of course, is $c$.)  Thus we see immediately that the energy of a                         photon must be the momentum times $c$, or if $c = 1$, the energy                             and momentum are equal. That is to say, the rest mass is zero. Let us                         look at that again; that is quite curious. If it is a particle of zero                         rest mass, what happens when it stops? It never stops! It always                         goes at the speed $c$. The usual formula for energy is $m_0/\sqrt{1 -                         v^2}$. Now can we say that $m_0 = 0$ and $v = 1$, so the energy is $0$?                         We cannot say that it is zero; the photon really can (and does)                         have energy even though it has no rest mass, but this it possesses by                         perpetually going at the speed of light! We also know that the momentum of any particle is equal to its total                         energy times its velocity: if $c = 1$, $p = vE$ or, in ordinary units,                         $p = vE/c^2$. For any particle moving at the speed of light, $p = E$ if                         $c = 1$. The formulas for the energy of a photon as seen from a moving                         system are, of course, given by Eq. (17.12), but for the                         momentum we must substitute the energy divided by $c$ (or by $1$ in this                         case). The different energies after transformation means that there are                         different frequencies. This is called the Doppler                         effect, and one can calculate it easily                         from Eq. (17.12), using also $E = p$ and $E = h\nu$.                          As Minkowski said, “Space of                         itself, and time of itself will sink into mere shadows, and only a kind                         of union between them shall survive.”