The four-dimensional gradient
The next thing that we have to discuss is the four-dimensional analog of the gradient. We recall (Chapter 14, Vol. I) that the three differential operators $\ddpl{}{x}$, $\ddpl{}{y}$, $\ddpl{}{z}$ transform like a three-vector and are called the gradient. The same scheme ought to work in four dimensions; that is, we might guess that the four-dimensional gradient should be $(\ddpl{}{t},\ddpl{}{x},\ddpl{}{y},\ddpl{}{z})$. This is wrong.  To see the error, consider a scalar function $\phi$ which depends only on $x$ and $t$. The change in $\phi$, if we make a small change $\Delta t$ in $t$ while holding $x$ constant, is \begin{equation} \label{Eq:II:25:13} \Delta\phi=\ddp{\phi}{t}\,\Delta t. \end{equation} On the other hand, according to a moving observer, \begin{equation*} \Delta\phi=\ddp{\phi}{x'}\,\Delta x'+\ddp{\phi}{t'}\,\Delta t'. \end{equation*} We can express $\Delta x'$ and $\Delta t'$ in terms of $\Delta t$ by using Eq. (25.1). Remembering that we are holding $x$ constant, so that $\Delta x=0$, we write \begin{equation*} \Delta x'=-\frac{v}{\sqrt{1-v^2}}\,\Delta t;\quad \Delta t'=\frac{\Delta t}{\sqrt{1-v^2}}. \end{equation*} Thus, \begin{align*} \Delta\phi&=\ddp{\phi}{x'}\Biggl( -\frac{v}{\sqrt{1-v^2}}\,\Delta t \Biggr)+\ddp{\phi}{t'}\Biggl( \frac{\Delta t}{\sqrt{1-v^2}} \Biggr)\\[1.5ex] &=\biggl( \ddp{\phi}{t'}-v\,\ddp{\phi}{x'} \biggr)\frac{\Delta t}{\sqrt{1-v^2}}. \end{align*} Comparing this result with Eq. (25.13), we learn that \begin{equation} \label{Eq:II:25:14} \ddp{\phi}{t}=\frac{1}{\sqrt{1-v^2}}\biggl( \ddp{\phi}{t'}-v\,\ddp{\phi}{x'} \biggr). \end{equation} A similar calculation gives \begin{equation} \label{Eq:II:25:15} \ddp{\phi}{x}=\frac{1}{\sqrt{1-v^2}}\biggl( \ddp{\phi}{x'}-v\,\ddp{\phi}{t'} \biggr). \end{equation}  Now we can see that the gradient is rather strange. The formulas for $x$ and $t$ in terms of $x'$ and $t'$ [obtained by solving Eq. (25.1)] are: \begin{equation*} t=\frac{t'+vx'}{\sqrt{1-v^2}},\quad x=\frac{x'+vt'}{\sqrt{1-v^2}}. \end{equation*} This is the way a four-vector must transform. But Eqs. (25.14) and (25.15) have a couple of signs wrong!  The answer is that instead of the incorrect $(\ddpl{}{t},\FLPnabla)$, we must define the four-dimensional gradient operator, which we will call $\fournabla$, by  \begin{equation} \label{Eq:II:25:16} \fournabla=\biggl(\ddp{}{t},-\FLPnabla\biggr)= \biggl( \ddp{}{t},-\ddp{}{x},-\ddp{}{y},-\ddp{}{z} \biggr). \end{equation} 
 \begin{align} \label{Eq:II:25:16} \fournabla&=\biggl(\ddp{}{t},-\FLPnabla\biggr)\\[1ex] &=\biggl(\ddp{}{t},-\ddp{}{x},-\ddp{}{y},-\ddp{}{z}\biggr).\notag \end{align}  With this definition, the sign difficulties encountered above go away, and $\fournabla$ behaves as a four-vector should. (It’s rather awkward to have those minus signs, but that’s the way the world is.) Of course, what it means to say that $\fournabla$ “behaves like a four-vector” is simply that the four-gradient of a scalar is a four-vector. If $\phi$ is a true scalar invariant field (Lorentz invariant) then $\fournabla\phi$ is a four-vector field.  All right, now that we have vectors, gradients, and dot products, the next thing is to look for an invariant which is analogous to the divergence of three-dimensional vector analysis. Clearly, the analog is to form the expression $\fournabla b_\mu$, where $b_\mu$ is a four-vector field whose components are functions of space and time. We define the divergence of the four-vector $b_\mu=(b_t,\FLPb)$ as the dot product of $\fournabla$ and $b_\mu$:  \begin{equation} \begin{aligned} \fournabla b_\mu&=\ddp{}{t}\,b_t- \biggl(-\ddp{}{x}\biggr)b_x- \biggl(-\ddp{}{y}\biggr)b_y- \biggl(-\ddp{}{z}\biggr)b_z\\[1ex] &=\ddp{}{t}\,b_t+\FLPdiv{\FLPb}, \end{aligned} \label{Eq:II:25:17} \end{equation} 
 \begin{gather} \fournabla b_\mu=\notag\\[1.5ex] \ddp{}{t}b_t\!-\! \biggl(\!-\ddp{}{x}\!\biggr)b_x\!-\! \biggl(\!-\ddp{}{y}\!\biggr)b_y\!-\! \biggl(\!-\ddp{}{z}\!\biggr)b_z=\notag\\[1.25ex] \label{Eq:II:25:17} \ddp{}{t}\,b_t+\FLPdiv{\FLPb}, \end{gather}  where $\FLPdiv{\FLPb}$ is the ordinary three-divergence of the three-vector $\FLPb$. Note that one has to be careful with the signs. Some of the minus signs come from the definition of the scalar product, Eq. (25.7); the others are required because the space components of $\fournabla$ are $-\ddpl{}{x}$, etc., as in Eq. (25.16). The divergence as defined by (25.17) is an invariant and gives the same answer in all coordinate systems which differ by a Lorentz transformation.  Let’s look at a physical example in which the four-divergence shows up. We can use it to solve the problem of the fields around a moving wire. We have already seen (Section 13-7) that the electric charge density $\rho$ and the current density $\FLPj$ form a four-vector $j_\mu=(\rho,\FLPj)$. If an uncharged wire carries the current $j_x$, then in a frame moving past it with velocity $v$ (along $x$), the wire will have the charge and current density [obtained from the Lorentz transformation Eqs. (25.1)] as follows: \begin{equation*} \rho'=\frac{-vj_x}{\sqrt{1-v^2}},\quad j_x'=\frac{j_x}{\sqrt{1-v^2}}. \end{equation*}  These are just what we found in Chapter 13. We can then use these sources in Maxwell’s equations in the moving system to find the fields.  The charge conservation law, Section 13-2, also takes on a simple form in the four-vector notation. Consider the four divergence of $j_\mu$: \begin{equation} \label{Eq:II:25:18} \fournabla j_\mu=\ddp{\rho}{t}+\FLPdiv{\FLPj}. \end{equation} The law of the conservation of charge says that the outflow of current per unit volume must equal the negative rate of increase of charge density. In other words, that \begin{equation*} \FLPdiv{\FLPj}=-\ddp{\rho}{t}. \end{equation*} Putting this into Eq. (25.18), the law of conservation of charge takes on the simple form \begin{equation} \label{Eq:II:25:19} \fournabla j_\mu=0. \end{equation} Since $\fournabla j_\mu$ is an invariant scalar, if it is zero in one frame it is zero in all frames. We have the result that if charge is conserved in one coordinate system, it is conserved in all coordinate systems moving with uniform velocity.  As our last example we want to consider the scalar product of the gradient operator $\fournabla$ with itself. In three dimensions, such a product gives the Laplacian \begin{equation*} \nabla^2=\FLPdiv{\FLPnabla}= \frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2}+ \frac{\partial^2}{\partial z^2}. \end{equation*} What do we get in four dimensions? That’s easy. Following our rules for dot products and gradients, we get  \begin{align*} \fournabla\fournabla&=\ddp{}{t}\,\ddp{}{t}- \biggl(-\ddp{}{x}\biggr)\biggl(-\ddp{}{x}\biggr)- \biggl(-\ddp{}{y}\biggr)\biggl(-\ddp{}{y}\biggr)- \biggl(-\ddp{}{z}\biggr)\biggl(-\ddp{}{z}\biggr)\\[1ex] &=\frac{\partial^2}{\partial t^2}-\nabla^2. \end{align*} 
 \begin{align*} \fournabla\fournabla=\ddp{}{t}\,\ddp{}{t}&-\biggl(-\ddp{}{x}\biggr)\biggl(-\ddp{}{x}\biggr)\\ &-\biggl(-\ddp{}{y}\biggr)\biggl(-\ddp{}{y}\biggr)\\ &-\biggl(-\ddp{}{z}\biggr)\biggl(-\ddp{}{z}\biggr)\\ =\,\frac{\partial^2}{\partial t^2}\;\;&-\;\;\nabla^2. \end{align*}  This operator, which is the analog of the three-dimensional Laplacian, is called the d’Alembertian and has a special notation: \begin{equation} \label{Eq:II:25:20} \Box^2=\fournabla\fournabla=\frac{\partial^2}{\partial t^2}-\nabla^2. \end{equation} From its definition it is an invariant scalar operator; if it operates on a four-vector field, it produces a new four-vector field. (Some people define the d’Alembertian with the opposite sign to Eq. (25.20), so you will have to be careful when reading the literature.)  We have now found four-dimensional equivalents of most of the three-dimensional quantities we had listed in Table 25–1. (We do not yet have the equivalents of the cross product and the curl operation; we won’t get to them until the next chapter.) It may help you remember how they go if we put all the important definitions and results together in one place, so we have made such a summary in Table 25–2. 