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Electromagnetic Wave and its E-Field and B-Field

Important Formula

Formula: Wave Equation for E-Field

$$\nabla^2 \, \class{purple}{\boldsymbol{E}} ~=~ \mu_0 \, \varepsilon_0 \, \frac{\partial^2 \class{purple}{\boldsymbol{E}}}{\partial t^2}$$

What do the formula symbols mean?

Electric field

$$ \class{purple}{\boldsymbol E} $$ Unit $$ \frac{\mathrm{V}}{\mathrm{m}} = \frac{\mathrm{N}}{\mathrm{C}} = \frac{\mathrm{kg} \, \mathrm{m}}{\mathrm{A} \, \mathrm{s}^3} $$

Electric field specifies the force that would act on an electric charge if it were placed in a location where the electric field exists.

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 E-field in the form of plane waves.

Vacuum Permittivity

$$ \varepsilon_0 $$ Unit $$ \frac{\mathrm{As}}{\mathrm{Vm}} $$

The vacuum permittivity is a physical constant that appears in equations involving electromagnetic fields. It has the following experimentally determined value: $$ \varepsilon_0 ~\approx~ 8.854 \, 187 \, 8128 ~\cdot~ 10^{-12} \, \frac{\mathrm{As}}{\mathrm{Vm}} $$

Magnetic field constant

$$ \mu_0 $$ Unit $$ \frac{\mathrm{Vs}}{\mathrm{Am}} = \frac{ \mathrm{kg} \, \mathrm{m} }{ \mathrm{A}^2 \, \mathrm{s}^2 } $$

It is a natural constant and occurs whenever electromagnetic fields are involved. It has the value $ \mu_0 = 4\pi \cdot 10^{-7} \, \frac{ \text{N} }{ \text{A}^2 } $.

Nabla operator

$$ \nabla $$ Unit $$ \frac{1}{\mathrm m} $$

The operator $\nabla^2$ is applied to the electric field to differentiate the components of the E-field with respect to the spatial coordinates.

Applying $\nabla^2$ to the E-field 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

Electromagnetic wave with E-field and B-field component.

An electromagnetic wave (short: EM-wave) 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 three-dimensional space. So they are three-dimensional vector fields:

$$ \begin{align} \boldsymbol{E} ~=~ \begin{bmatrix}E_{\text x}(x,y,z,t)\\ E_{\text y}(x,y,z,t) \\ E_{\text z}(x,y,z,t) \end{bmatrix}; ~~~ \boldsymbol{B} ~=~ \begin{bmatrix}B_{\text x}(x,y,z,t)\\ B_{\text y}(x,y,z,t) \\ B_{\text z} (x,y,z,t)\end{bmatrix} \end{align} $$

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:

$$ \begin{align} \nabla^2 \, \class{purple}{\boldsymbol{E}} ~=~ \mu_0 \, \varepsilon_0 \, \frac{\partial^2 \class{purple}{\boldsymbol{E}}}{\partial t^2} \end{align} $$

$$ \begin{align} \nabla^2 \, \class{violet}{\boldsymbol{B}} ~=~ \mu_0 \, \varepsilon_0 \, \frac{\partial^2 \class{violet}{\boldsymbol{B}}}{\partial t^2} \end{align} $$

The two wave equations, are partial differential equations of second order and can be derived from Maxwell's equations in charge-free space. They were derived for charge-free and current-free space and therefore hold only under these conditions.

'Charge-free' means that the electric charge density is zero at any location: $\rho = 0$.
'Current-free' means that the electric current density is zero at any location: $\boldsymbol{j}$.

$$ \begin{align} \nabla^2 ~=~ \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \end{align} $$

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:

$$ \begin{align} \nabla^2 \, \boldsymbol{F} ~=~ \frac{1}{{v_{\text p}}^2} \, \frac{\partial^2 \boldsymbol{F}}{\partial t^2} \end{align} $$

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.

Phase velocity here indicates how fast a wave crest moves from A to B.

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 $:

$$ \begin{align} \mu_0 \, \varepsilon_0 ~=~ \frac{1}{{v_{\text p}}^2} \end{align} $$

Rearrange for the phase velocity $ v $:

$$ \begin{align} v_{\text p} &~=~ \frac{1}{\sqrt{ \mu_0 \, \varepsilon_0 }} \\\\
&~=~ 3 \cdot 10^8 \, \frac{\mathrm m}{ \mathrm s} \\\\
&~=~ c \end{align} $$

What is the phase velocity of an electromagnetic wave?

The phase velocity of an EM wave corresponds to the speed of light $ c $. Electromagnetic waves propagate at the speed of light.

So you can therefore express the two wave equations 2 and 3 by using the speed of light $ c $:

$$ \begin{align} \nabla^2 \, \boldsymbol{E} &~=~ \frac{1}{c^2} \, \frac{\partial^2 \boldsymbol{E}}{\partial t^2} \\\\
\nabla^2 \, \boldsymbol{B} &~=~ \frac{1}{c^2} \, \frac{\partial^2 \boldsymbol{B}}{\partial t^2} \end{align} $$

For example, let's write out the wave equation for the E-field component to better understand its structure:

$$ \begin{align} \begin{bmatrix} \frac{\partial^2 E_{\text x}}{\partial x^2} + \frac{\partial^2 E_{\text x}}{\partial y^2} + \frac{\partial^2 E_{\text x}}{\partial z^2} \\ \frac{\partial^2 E_{\text y}}{\partial x^2} + \frac{\partial^2 E_{\text y}}{\partial y^2} + \frac{\partial^2 E_{\text y}}{\partial z^2} \\ \frac{\partial^2 E_{\text z}}{\partial x^2} + \frac{\partial^2 E_{\text z}}{\partial y^2} + \frac{\partial^2 E_{\text z}}{\partial z^2} \end{bmatrix} ~=~ \frac{1}{c^2} \, \begin{bmatrix} \frac{\partial^2 E_{\text x}}{\partial t^2} \\ \frac{\partial^2 E_{\text y}}{\partial t^2} \\ \frac{\partial^2 E_{\text z}}{\partial t^2} \end{bmatrix} \end{align} $$

On the left-hand side of the wave equation, each component of $\boldsymbol{E}$ is differentiated twice with respect to $x$, $y$ and $z$. On the right-hand side, each E-field component is differentiated twice with respect to time $ t $. The wave equation thus relates spatial derivatives of the E-field 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:

$$ \begin{align} \frac{\partial^2 E_{\text x}}{\partial x^2} + \frac{\partial^2 E_{\text x}}{\partial y^2} + \frac{\partial^2 E_{\text x}}{\partial z^2} ~=~ \frac{1}{c^2} \, \frac{\partial^2 E_{\text x}}{\partial t^2} \end{align} $$

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

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

If you solve this DEQ for $ E_{\text z}(x,y,z,t) $, then you will know how the E-field 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 E-field does not influence what happens with the E-field 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 E-field component and B-field 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 B-field:

$$ \begin{align} \nabla ~\times~ \boldsymbol{B} ~=~ \mu_0 \, \varepsilon_0 \frac{\partial \boldsymbol{E}}{\partial t} \end{align} $$

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 B-field is not only perpendicular to the derivative of the E-field but also to the E-field 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 E-field (and B-field), besides the time dependence $t$, depend only on ONE spatial coordinate. For example only on the space coordinate $z$:

$$ \begin{align} \boldsymbol{E}(z,t) ~=~ \begin{bmatrix} E_{\text x}(z,t) \\ E_{\text y}(z,t) \\ E_{\text z}(z,t) \end{bmatrix} \end{align} $$

Since the E-field 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:

$$ \begin{align} \begin{bmatrix} \frac{\partial^2 E_{\text x}}{\partial z^2} \\ \frac{\partial^2 E_{\text y}}{\partial z^2} \\ \frac{\partial^2 E_{\text z}}{\partial z^2} \end{bmatrix} ~=~ \frac{1}{c^2} \, \begin{bmatrix} \frac{\partial^2 E_{\text x}}{\partial t^2} \\ \frac{\partial^2 E_{\text y}}{\partial t^2} \\ \frac{\partial^2 E_{\text z}}{\partial t^2} \end{bmatrix} \end{align} $$

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 charge-free space, the first Maxwell equation $\nabla \cdot \boldsymbol{E} = 0 $ (with $\rho = 0$) can be used to further simplify 14. Written out, the first charge-free Maxwell equation is:

$$ \begin{align} \frac{\partial E_{\text x}}{\partial x} + \frac{\partial E_{\text y}}{\partial y} + \frac{\partial E_{\text z}}{\partial z} ~=~ 0 \end{align} $$

The first and second summands disappear, because the E-field does not depend on $x$ and $y$. Only the second derivative with respect to $z$ remains:

$$ \begin{align} \frac{\partial E_{\text z}}{\partial z} ~=~ 0 \end{align} $$

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 E-field does not depend on $z$, so it is a constant which we define as $E_0$:

$$ \begin{align} E_{\text z} ~:=~ E_0 \end{align} $$

With the chosen boundary condition that the E-field at the location $ z $ is zero: $ E_{\text z}(z) = 0 $, $E_0$ can be eliminated: $ E_0 = 0$. Thus, the E-field of a plane wave has only two variable components:

$$ \begin{align} \boldsymbol{E}(z, t) ~=~ \begin{bmatrix} E_{\text x}(z, t) \\ E_{\text y}(z, t) \\ 0 \end{bmatrix} \end{align} $$

So the E-field (it is also valid for the B-field) 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 $t$.

What characterizes a plane wave?

A plane wave oscillates in only one plane, such as the $x$-$y$ plane.

The wave equation 14 simplifies to:

$$ \begin{align} \begin{bmatrix} \frac{\partial^2 E_{\text x}}{\partial z^2} \\ \frac{\partial^2 E_{\text y}}{\partial z^2} \\ 0 \end{bmatrix} ~=~ \frac{1}{c^2} \, \begin{bmatrix} \frac{\partial^2 E_{\text x}}{\partial t^2} \\ \frac{\partial^2 E_{\text y}}{\partial t^2} \\ 0 \end{bmatrix} \end{align} $$

The solution of the first or second component of the wave equation for plane waves has always the form:

$$ \begin{align} E_{\text x}(z,t) &~=~ f_{\text x}(z-c\,t) + g_{\text x}(z+c\,t) \\\\
E_{\text y}(z,t) &~=~ f_{\text y}(z-c\,t) + g_{\text y}(z+c\,t) \end{align} $$

Here $f$ and $g$ are twice continuously differentiable functions. One function depends on $z-c\,t$ and the other on $z+c\,t$. These ($z-c\,t$)- and ($z+c\,t$)-dependencies characterize the wave behavior. $f_{\text x}(z-c\,t)$ 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 $f_{\text x}(z-c\,t)$ 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 E-field 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 E-field is orthogonal to the direction of propagation).