Collection of Solved Problems in Physics

Complex Representation and the Harmonic Plane Wave

Task number: 1967

• 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:

  \[ (e^{ix})^\prime = i e^{ix}. \]

• 1. Hint solution — Time derivative

  We differentiate the complex exponential with respect to time as follows:

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

• 2. 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:

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

• 2. Hint solution — Divergence

  The three partial derivatives are independent. We calculate below the derivative \(\frac{\partial E_x}{\partial x}\) explicitly:

  \[ \color{MidnightBlue}{ \begin{align} \frac{\partial E_x}{\partial x} &= \frac{\partial}{\partial x} E_{0x} e^{i(\omega t - \vec{k}\cdot \vec{r} )} = \\ &= E_{0x} e^{i\omega t} \frac{\partial }{\partial x} e^{i(- k_x x - k_y y - k_z z)} = \\ &= E_{0x} e^{i\omega t} e^{i(- k_y y - k_z z)} \frac{\partial }{\partial x} e^{- i k_x x} = \\ &= E_{0x} e^{i\omega t} e^{i(- k_y y - k_z z)} k_x e^{- i k_x x} = \\ &= -i k_x E_{0x} e^{i\omega t} e^{-i\vec{k}\cdot\vec{r}} = \\ &= -i k_x E_{0x} e^{i(\omega t - \vec{k}\cdot \vec{r} )}. \end{align} } \]

  The derivatives of the other components are analogous.

  We sum the partial derivatives to get the divergence:

  \[\mathrm{div\,} \vec{E} = \vec\nabla \cdot \vec E = \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} = \\ = -i k_x E_{0x} e^{i(\omega t - \vec{k}\cdot \vec{r} )}-i k_y E_{0y} e^{i(\omega t - \vec{k}\cdot \vec{r} )} -i k_z E_{0z} e^{i(\omega t - \vec{k}\cdot \vec{r} )} = \]

  which is nothing less than the scalar product of the plane wave electric field vector and its wave vector:

  \[ = -i(\underbrace{k_x E_{0x} + k_y E_{0y} + k_z E_{0z}}_{\vec{k}\cdot\vec{E_0}}) e^{i(\omega t - \vec{k}\cdot \vec{r} )} = -i \vec{k}\cdot \underbrace{\vec{E_0} e^{i(\omega t - \vec{k}\cdot \vec{r} )}}_{\vec{E}} = -i\vec{k}\cdot \vec{E}. \]

• 3. 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:

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

• 3. Hint solution — Curl

  The symbolic vector product tells us what partial derivatives should be subtracted in the components of the resulting vector 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 explicitly 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 previous hint solution:

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

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

• 4. Hint solution — Magnetic field

  The curl of the electric field was calculated in part 3 to be

  \[ \mathrm{curl\,} \vec E = -i\vec{k}\times\vec{E} \]

  and we substitute into the Maxwell–Faraday equation as indicated in the hint

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

  We calculate the magnetic field from this equation by integrating with respect to time:

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

  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.

• 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 \(\vec{B} = \frac{1}{\omega} \vec{k}\times\vec{E}\).