Electromagnetic Wave and its EField and BField
Formula
What do the formula symbols mean?
Electric field
$$ \class{purple}{\boldsymbol E} $$ Unit $$ \frac{\mathrm{V}}{\mathrm{m}} = \frac{\mathrm{N}}{\mathrm{C}} = \frac{\mathrm{kg} \, \mathrm{m}}{\mathrm{A} \, \mathrm{s}^3} $$Solving the vectorial wave equation with the respective boundary conditions yields the electric field. For example, a simple solution of the wave equation yields the Efield in the form of plane waves.
Vacuum Permittivity
$$ \varepsilon_0 $$ Unit $$ \frac{\mathrm{As}}{\mathrm{Vm}} $$Magnetic field constant
$$ \mu_0 $$ Unit $$ \frac{\mathrm{Vs}}{\mathrm{Am}} = \frac{ \mathrm{kg} \, \mathrm{m} }{ \mathrm{A}^2 \, \mathrm{s}^2 } $$Nabla operator
$$ \nabla $$ Unit $$ \frac{1}{\mathrm m} $$Applying \(\nabla^2\) to the Efield yields a vector quantity. The first component of this vector quantity is: \[ \frac{\partial^2 E_x}{\partial x^2} + \frac{\partial^2 E_x}{\partial y^2} + \frac{\partial^2 E_x}{\partial z^2} ~=~ \mu_0 \, \varepsilon_0 \, \frac{\partial^2 E_x}{\partial t^2} \]
Table of contents
 Formula
 Video
 E and B fields are orthogonal to each other Here you will learn why the electric field component must always be perpendicular to the magnetic field component.
 Special case: plane electromagnetic waves Here you will learn what plane waves are and what the wave equation for it looks like.
Video
An electromagnetic wave (short: EMwave) has an electric field \( \boldsymbol{E}(x,y,z,t) \) and a magnetic field \( \boldsymbol{B}(x,y,z,t) \). The two fields are vectorial quantities, each having three components in threedimensional space. So they are threedimensional vector fields:
Both fields generally depend on the space coordinates \(x,y,z\). The amplitude (magnitude of the corresponding vector) is therefore different from location to location. Furthermore, the amplitude at a certain location does not always remain the same, but changes with time \( t \). So the fields also depend on time.
How the electromagnetic wave changes exactly in space and time, that is how it moves and propagates in space, is described by the following two wave equations:
The two wave equations, are partial differential equations of second order and can be derived from Maxwell's equations in chargefree space. They were derived for chargefree and currentfree space and therefore hold only under these conditions.
 'Chargefree' means that the electric charge density is zero at any location: \(\rho = 0\).
 'Currentfree' means that the electric current density is zero at any location: \(\boldsymbol{j}\).
The electric field constant \(\varepsilon_0\) as well as the magnetic field constant \(\mu_0\) ensure that both sides of the wave equation have the same unit.
The general form of a wave equation looks like this:
Here \( \boldsymbol{F} \) is any vector field satisfying the wave equation and \( v \) is the phase velocity of the wave. It indicates how fast the wave travels in space.
If you compare the EM wave equation 2
or 3
with the general form of a wave equation 4
, you find out how phase velocity \( v \) of an electromagnetic wave is related to the two field constants: \( \varepsilon_0 \) and \( \mu_0 \):
Rearrange for the phase velocity \( v \):
&~=~ 3 \cdot 10^8 \, \frac{\mathrm m}{ \mathrm s} \\\\
&~=~ c \end{align} $$
So you can therefore express the two wave equations 2
and 3
by using the speed of light \( c \):
\nabla^2 \, \boldsymbol{B} &~=~ \frac{1}{c^2} \, \frac{\partial^2 \boldsymbol{B}}{\partial t^2} \end{align} $$
For example, let's write out the wave equation for the Efield component to better understand its structure:
On the lefthand side of the wave equation, each component of \(\boldsymbol{E}\) is differentiated twice with respect to \(x\), \(y\) and \(z\). On the righthand side, each Efield component is differentiated twice with respect to time \( t \). The wave equation thus relates spatial derivatives of the Efield to the time derivatives and thus represents a system of differential equations.
The wave equation has three components, each of which is a partial differential equation of second order:

Formula anchor $$ \begin{align} \frac{\partial^2 E_{\text x}}{\partial x^2} + \frac{\partial^2 E_{\text x}}{\partial y^2} + \frac{\partial^2 E_{\text x}}{\partial z^2} ~=~ \frac{1}{c^2} \, \frac{\partial^2 E_{\text x}}{\partial t^2} \end{align} $$
If you solve this DEQ for \( E_{\text x}(x,y,z,t) \), then you will know how the Efield changes spatially and temporally on the \(x\) axis.

Formula anchor $$ \begin{align} \frac{\partial^2 E_{\text y}}{\partial x^2} + \frac{\partial^2 E_{\text y}}{\partial y^2} + \frac{\partial^2 E_{\text y}}{\partial z^2} ~=~ \frac{1}{c^2} \, \frac{\partial^2 E_{\text y}}{\partial t^2} \end{align} $$
If you solve this DEQ for \( E_{\text y}(x,y,z,t) \), then you will know how the Efield changes spatially and temporally on the \(y\) axis.

Formula anchor $$ \begin{align} \frac{\partial^2 E_{\text z}}{\partial x^2} + \frac{\partial^2 E_{\text z}}{\partial y^2} + \frac{\partial^2 E_{\text z}}{\partial z^2} ~=~ \frac{1}{c^2} \, \frac{\partial^2 E_{\text z}}{\partial t^2} \end{align} $$
If you solve this DEQ for \( E_{\text z}(x,y,z,t) \), then you will know how the Efield changes spatially and temporally on the \(z\) axis.
The three DEQ's are not coupled with each other and thus can be solved independently. To state it physically: What happens on the \(x\)axis with the Efield does not influence what happens with the Efield on the \(y\) or \(z\)axis. The same is true for \(y\) and \(z\) axes.
E and B fields are orthogonal to each other
The solution to the wave equation 5
is indeed a wave, but not necessarily an electromagnetic wave! An electromagnetic wave in vacuum exists only if the solution to the wave equation also satisfies all Maxwell's equations in vacuum.
If the solution to the wave equation is a plane wave, then both the third and fourth Maxwell's equations demand it from the solution:
The Efield component and Bfield component of an electromagnetic wave in vacuum must always be orthogonal to each other. To see this, let's consider the fourth Maxwell's equation concerning the rotation of the Bfield:
We know from mathematics that the result vector \(\nabla \times \boldsymbol{B}\) of the cross product is always orthogonal to the vectors between which the cross product is formed. In this case, the \(\boldsymbol{B}\)field vector is perpendicular to the derivative of the \(\boldsymbol{E}\)field: \(\frac{\partial \boldsymbol{E}}{\partial t}\). However, the time derivative does not change the direction of a vector! So, the \(\boldsymbol{E}\)field vector and its derivative point in the same direction. Thus, the Bfield is not only perpendicular to the derivative of the Efield but also to the Efield vector itself.
Special case: plane electromagnetic waves
A possible solution of the wave equation 9
are plane waves. These are characterized by the fact that their Efield (and Bfield), besides the time dependence \(t\), depend only on ONE spatial coordinate. For example only on the space coordinate \(z\):
Since the Efield does not depend on \(x\) and \(y\), in the wave equation 9
the derivatives with respect to \(x\) and \(y\) disappear. Thus 9
simplifies to:
Graphically, this independence of \(x\) and \(y\) means that the electric field \( \boldsymbol{E}(z,t) \) has a constant value at a fixed time \(t = t_0\) and at \(z=z_0\) in the \(x\)\(y\) plane: \( \boldsymbol{E}(z_0, t_0) = \text{const} \).
Since the wave equations hold only in chargefree space, the first Maxwell equation \(\nabla \cdot \boldsymbol{E} = 0 \) (with \(\rho = 0\)) can be used to further simplify 14
. Written out, the first chargefree Maxwell equation is:
The first and second summands disappear, because the Efield does not depend on \(x\) and \(y\). Only the second derivative with respect to \(z\) remains:
Eq. 16
is an ordinary first order DEQ and can be solved very easily. The derivative of a function (here \(E_{\text z}\)) is zero if the function is constant. The third component of the Efield does not depend on \(z\), so it is a constant which we define as \(E_0\):
With the chosen boundary condition that the Efield at the location \( z \) is zero: \( E_{\text z}(z) = 0 \), \(E_0\) can be eliminated: \( E_0 = 0\). Thus, the Efield of a plane wave has only two variable components:
So the Efield (it is also valid for the Bfield) of a plane wave has no changing \(z\) component at all. Only two of the three components of \(\boldsymbol{E}\) can change with \(z\) and \(t\).
The wave equation 14
simplifies to:
The solution of the first or second component of the wave equation for plane waves has always the form:
E_{\text y}(z,t) &~=~ f_{\text y}(zc\,t) + g_{\text y}(z+c\,t) \end{align} $$
Here \(f\) and \(g\) are twice continuously differentiable functions. One function depends on \(zc\,t\) and the other on \(z+c\,t\). These (\(zc\,t\)) and (\(z+c\,t\))dependencies characterize the wave behavior. \(f_{\text x}(zc\,t)\) is shifted to the right (in the positive \(c\) direction) and \(g_{\text x}(z+c\,t)\) is shifted to the left (in the negative \(z\) direction). With increasing time \(t\), this shift along the \(z\) axis becomes larger. So one field part \(f_{\text x}(zc\,t)\) of \(E_{\text x}\) propagates to the right and the other field part \(g_{\text x}(z+c\,t)\) propagates to the left.
Since the electromagnetic wave (here specifically the Efield component) propagates along the \(z\) axis but has no \(E_{\text z}\) component, the electromagnetic wave is a transverse wave (i.e. oscillation of the Efield is orthogonal to the direction of propagation).