# Why electromagnetic fields are defined as 4-D Function?

If electromagnetic fields propagate along 3-D spaces why are they defined as function of space (X,Y,Z) and time t.

For me would be enough to use two functions E(x,y,z) and H(x,y,z) to model the electric and magnetic field propagation, respectively. But in the literature I always found that they are defined as E(x,y,z,t) and H(x,y,z,t).

• They have to propagate in 3D space. How can you say it would be enough for you to have $E(x,y,z)$ and $B(x,y,z)$? How would you distinguish between the field at time $t_1$ and the field at time $t_2$? – ACuriousMind Oct 8 '15 at 12:45
• ACuriousMind, why would I distinguish the field at different times?. I think that the fields has the same "shape" all the time. d1u1p2xjjiahg3.cloudfront.net/… – zeellos Oct 9 '15 at 18:33

You must include the $t$ because the field configuration can change with time. Perhaps this seems counterintuitive because you are used to thinking about stationary charges and constant currents. However, imagine the following situation: you start with a charge at the origin. Then $E$ will have a peak in magnitude at the origin. Now at some later time you move your charge somehwere else; then $E$ will have a peak in magnitude at the new location of the charge. Thus $E$ depends not just on position in space but also on time.
This is really no different than how any other quantity such as velocity or position can depend on time. I could just write the position of a particle as $\vec{r}$, but if I want to emphasize that it depends on time, I will write $\vec{r}(t)$.
Variables of a function are anything the function depends upon, namely all those characteristics such that changing any of them will result in change of the overall value of the function. To directly answer your question, electromagnetic fields depend on time as well because if you sit on the exact place $(x,y,z)$ in the universe you still measure different things according to when you measure them. That means that the field must depend on time too.
Maxwell's equations define (provided boundary conditions and differentiability) the electromagnetic field anywhere as solution of $$\textrm{div }\mathbf{D} = \rho,\qquad\textrm{div }\mathbf{B}=0$$ $$\textrm{curl }\mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t},\qquad\textrm{curl } \mathbf{H} =\mathbf{j}+\frac{\partial \mathbf{D}}{\partial t}.$$ As you can see, time does appear in there despite everything; consequently the solutions will depend on it.
Equivalently you may want to consider that if you place a charge at $t=0$ somewhere in the universe, it takes a finite amount of time for the propagation to reach an observer who is located elsewhere; this is because the velocity of propagation is not infinite, hence the same observer measures different quantities at different times, since different parts of the field propagate accordingly.