Complex Representation and the Harmonic Plane Wave

Task number: 1967

A plane wave propagates though vacuum. Its electric field is described with the complex function:

\[\vec{E} = \vec{E}_0 e^{i(\omega t - \vec{k}\cdot \vec{r} )}.\]

Assess

  1. the time derivative \(\frac{\partial \vec{E}}{\partial t}\),
  2. the divergence \(\mathrm{div\,} \vec{E}\),
  3. the curl \(\mathrm{curl\,} \vec{E}\),
  4. the magnetic field \(\vec{B}\) (use Maxwell’s equations).

The assignment gives the complex representation of the electric field of the wave. We keep the complex formalism in the solution below also for the vector of the magnetic field \(\vec{B}\). (We do not emphasize the complex character of the vectors by any special notation.)

  • Theory — complex formalism

    The theory on the description of the plane wave in the complex representation is comprehensively handled in another task within Wave optics here.

  • 1. Hint — Time derivative

    Assess the time derivative \(\frac{\partial \vec{E}}{\partial t}\).

    Let us remind that the complex exponential is differentiated in the same manner as the real one:

    \[ (e^{ix})^\prime = i e^{ix}. \]
  • 2. Hint — Divergence

    The divergence of a vector in cartesian coordinates is obtained as a symbolic scalar product of the vector differential operator “del” (also called “nabla”) \(\vec\nabla = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right)\) with the assigned vector. To calculate the divergence of the vector field \(\vec{E}\), calculate the following sum of the partial derivatives:

    \[\mathrm{div\,} \vec{E} = \vec\nabla \cdot \vec E = \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z}. \]
  • 3. Hint — Curl

    The curl of the assigned vector field is found through the symbolic vector product of the vector differential operator “del” (also called “nabla”) \(\vec\nabla = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right)\) with the vector of the field. Calculate the curl of the assigned vector field \(\vec{E}\) as the product:

    \[\mathrm{curl\,} \vec{E} = \vec\nabla \times \vec E = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) \times \left(E_x,E_y,E_z\right).\]
  • 4. Hint — Magnetic field

    From the four Maxwell’s equations, we use Faraday’s law of induction (the Maxwell–Faraday equation)

    \[ \mathrm{curl\,} \vec E = -\frac{\partial \vec B}{\partial t} \]

    and the result of part 3.

  • Solution

    1. We differentiate the complex exponential with respect to time:

    \[ \frac{\partial \vec{E}}{\partial t} = \frac{\partial }{\partial t} \left( \vec{E}_0 e^{i(\omega t - \vec{k}\cdot \vec{r} )}\right) = \frac{\partial }{\partial t} \left( \vec{E}_0 e^{i\omega t }e^{-i\vec{k}\cdot \vec{r}}\right) = \\ = \vec{E}_0 e^{-i\vec{k}\cdot \vec{r}} \frac{\partial }{\partial t} e^{i\omega t } = i \vec{E}_0 e^{-i\vec{k}\cdot \vec{r}} \omega e^{i\omega t } = i\omega \underbrace{\vec{E}_0 e^{i(\omega t - \vec{k}\cdot \vec{r} )}}_{\vec{E}} = i\omega \vec{E}. \]

    2. We calculate the divergence of the vector of the magnetic field:

    \[\mathrm{div\,} \vec{E} = \vec\nabla \cdot \vec E = \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} = \]
    We calculate explicitly the partial derivative \(\frac{\partial E_x}{\partial x}\) \[ \color{MidnightBlue}{ \frac{\partial E_x}{\partial x} = \frac{\partial}{\partial x} E_{0x} e^{i(\omega t - \vec{k}\cdot \vec{r} )} = E_{0x} e^{i\omega t} \frac{\partial }{\partial x} e^{i(\overbrace{- k_x x - k_y y - k_z z}^{-\vec{k}\cdot\vec{r}})} = \\ = -i E_{0x} e^{i\omega t} k_x e^{-i\vec{k}\cdot\vec{r}} = -i k_x E_{0x} e^{i(\omega t - \vec{k}\cdot \vec{r} )}. } \] We get the derivatives \(\frac{\partial E_y}{\partial y}, \frac{\partial E_z}{\partial z}\) analogously.
    \[ = -i k_x E_{0x} e^{i(\omega t - \vec{k}\cdot \vec{r} )}-i k_y E_{0y} e^{i(\omega t - \vec{k}\cdot \vec{r} )} -i k_z E_{0z} e^{i(\omega t - \vec{k}\cdot \vec{r} )} = \\ = -i(\underbrace{k_x E_{0x} + k_y E_{0y} + k_z E_{0z}}_{\vec{k}\cdot\vec{E_0}}) e^{i(\omega t - \vec{k}\cdot \vec{r} )} = -i \vec{k}\cdot \underbrace{\vec{E_0} e^{i(\omega t - \vec{k}\cdot \vec{r} )}}_{\vec{E}} = -i\vec{k}\cdot \vec{E}. \]

    3. We calculate the curl of the vector of the electric field

    \[\mathrm{curl\,} \vec{E} = \vec\nabla \times \vec E = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) \times \left(\vec{E}_0 e^{i(\omega t - \vec{k}\cdot \vec{r} )}\right) = \\ = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) \times \left(E_x,E_y,E_z\right) = \\ = \left( \frac{\partial E_z}{\partial y} - \frac{\partial E_y}{\partial z}, \frac{\partial E_x}{\partial z} - \frac{\partial E_z}{\partial x}, \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y}, \right) \]

    We calculate separately the derivatives in the first component:

    \[\color{MidnightBlue}{\small \frac{\partial E_z}{\partial y} - \frac{\partial E_y}{\partial z} = \frac{\partial}{\partial y} E_{0z} e^{i(\omega t - \vec{k}\cdot \vec{r} )} - \frac{\partial}{\partial z} E_{0y} e^{i(\omega t - \vec{k}\cdot \vec{r} ) } =\\ = E_{0z} \frac{\partial}{\partial y} e^{i(\omega t - \vec{k}\cdot \vec{r} )} - E_{0y} \frac{\partial}{\partial z} e^{i(\omega t - \vec{k}\cdot \vec{r} ) } = }\]

    The partial derivative of \(e^{- i \vec{k} \cdot \vec{r} }\) with respect to the \(j\)th coordinate puts a factor \(-ik_j\) in front of the exponential as shown in more detail in the solution of Hint 3.:

    \[\color{MidnightBlue}{\small = -i \left(E_{0z} k_y - E_{0y} k_z\right) e^{i(\omega t - \vec{k}\cdot \vec{r} )} .} \]

    We get the other components of the curl by cyclic permutation of the indices.

    And thus we get the vector:

    \[ = -i\underbrace{\left( E_{0z} k_y - E_{0y} k_z, E_{0x} k_z - E_{0z} k_x, E_{0y} k_x - E_{0x} k_y \right)}_{\vec{k}\times\vec{E}_0 }e^{i(\omega t - \vec{k}\cdot \vec{r} )} = \]

    where we have identified the vector product \((k_x,k_y,k_z)\times(E_{0x},E_{0y},E_{0z})\)

    \[ = -i\vec{k}\times\underbrace{\vec{E}_0\,e^{i(\omega t - \vec{k}\cdot \vec{r} )}}_{\vec{E}} = -i\vec{k}\times\vec{E}. \]

    4. We use the Maxwell–Faraday equation

    \[ \mathrm{curl\,} \vec E = -\frac{\partial \vec B}{\partial t} \]

    together with the curl of the vector of the electric field calculated in the third section

    \[ -i\vec{k}\times\vec{E}_0\,e^{i(\omega t - \vec{k}\cdot \vec{r} )} = -\frac{\partial \vec B}{\partial t}. \]

    We calculate the magnetic field by integrating of the equation above:

    \[ \vec B = \int i\vec{k}\times\vec{E}_0\,e^{i(\omega t - \vec{k}\cdot \vec{r} )}\,\mathrm{d}t = i \vec{k}\times\vec{E}_0 \int e^{i(\omega t - \vec{k}\cdot \vec{r} )}\,\mathrm{d}t= \] \[ = i \vec{k}\times\vec{E}_0 \frac{^{i(\omega t - \vec{k}\cdot \vec{r} )}}{i\omega}= \frac{\vec{k}\times\vec{E}}{\omega}. \]

    We can see that the vector of the magnetic field \(\vec{B}\) is perpendicular to the vector of the electric field \(\vec{E}\) and to the wave vector \(\vec{k}\) (the direction of propagation of the wave). The set of vectors \(\vec{k},\vec{E},\vec{B}\) make a right-handed basis.

  • Answer

    Using the complex representation, we have easily found the following results:

    1. the time derivative \(\frac{\partial \vec{E}}{\partial t} = i\omega \vec{E},\)
    2. the divergence \(\mathrm{div\,} \vec{E} = -i\vec{k}\cdot \vec{E}\),
    3. the curl \(\mathrm{rot\,} \vec{E} = -i\vec{k}\times\vec{E}\),
    4. the magnetic field \(\vec{B} = \frac{1}{\omega} \vec{k}\times\vec{E}\).
Difficulty level: Level 4 – Undergraduate level
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