Speed of propagation of energy

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An electric field which has the structure of a harmonic progressive plane wave, of amplitude \(E_0\), of angular frequency \(\omega\), of the wave vector \(\vec k\), propagating along Oz and polarized along Ox.

Question

Write the expression of the electric field \(\vec E\).

Solution

In complex notation :

\(\underline {\vec E} = E_0 e^{i(\omega t - k z)} \vec u_x\)

Question

Calculating the mean value of the volume density of electromagnetic energy at a point in space.

Solution

The magnetic field is calculated using the relationship of structure :

\(\underline{\vec B} = \frac {\vec k}{\omega}\wedge \underline {\vec E} = \frac {\vec u_z}{c}\wedge \underline {\vec E}\)

So :

\(\underline {\vec B} = \frac {E_0}{c} e^{i(\omega t - k z)} \vec u_y\)

The average value of the volume density of electromagnetic energy is then :

\(<u_{em}>=\frac {1}{2}\frac {1}{2}Re(\varepsilon_0 \underline {\vec E} \;\underline {\vec E^*})+\frac {1}{2}\frac {1}{2}Re(\frac{1}{\mu_0 }\underline {\vec B} \;\underline {\vec B^*})\)

Either, with \(\varepsilon_0 \mu_0 c^2 = 1\) :

\(<u_{em}>=\frac {1}{2} \varepsilon_0 E_0^2\)

Question

Calculate the average value of Poynting vector.

Solution

The average value of the Poynting vector is :

\(\left\langle {\vec \Pi } \right\rangle = \frac{1}{2}{\mathop{\rm Re}\nolimits} \left( {\frac{{\underline {\vec E} \wedge {{\underline {\vec B} }^*}}}{{{\mu _0}}}} \right) = \frac{1}{2}{\varepsilon _0}cE_0^2\;{\vec u_z}\)

Question

Calculate the average value of energy \(<dU>\) on a volume of surface \(dS = dxdy\) and thickness \(dz\).

Solution

The average value of the energy in the volume is :

\(<dU>=\frac {1}{2} \varepsilon_0 E_0^2 \;dxdydz\)

Question

Calculate the flux \(<d\phi>\) of energy through the surface \(dS\) during a time interval \(dt\).

Solution

The flux of energy \(<d\phi>\) through the surface \(dS\) during a time interval \(dt\) is connected to the flux of the Poynting vector, which is a power :

\(<d\phi>=\frac{1}{2}{\varepsilon _0}cE_0^2 \;dxdydt\)

Question

Derive the value of the speed of propagation of electromagnetic energy. Comment.

Solution

If \(v_e\) is the velocity of propagation of energy, then :

\(dz=v_e dt\)

Equating the two energies calculated to the previous questions :

\(<dU>=\frac {1}{2} \varepsilon_0 E_0^2 \;dxdyv_edt=<d\phi>=\frac{1}{2}{\varepsilon _0}cE_0^2 \;dxdydt\)

It remains :

\(v_e=c\)

Thus, in the vacuum (non-dispersive medium), the velocity of propagation is that of the propagation of the wave, or \(c\).

Phase and group velocities (often corresponding to the speed of propagation of energy) are equal.