Question 4.2.3: a) Calculate the energy per unit volume for shear and compre......

Consider first a shear Alfven wave, for which the magnetic field and fluid velocity perturbations can be written as (see Problem 4.2.1):

b_{y} = b_{1} \cos(\vec{k} · \vec{r} − ωt), v_{y} = − \frac{k_{z}B}{4πρω} b_{1} \cos(\vec{k}  · \vec{r} − ωt)

Such a wave possesses the additional kinetic and magnetic energies per unit volume

\Delta W_{k} = \frac{ρ < v^{2}_{y} >}{2} = \frac{ρk^{2}_{z}B^{2}}{4(4πρ)^{2}ω^{2}} b^{2}_{1} = \frac{b^{2}_{1}}{16π},

\Delta W_{M} = \frac{< (\vec{B} + \vec{b})^{2} >}{8π} − \frac{B^{2}}{8π} = \frac{b^{2}_{1}}{16π},

where the symbol < > means an average over spatial coordinates, and the dispersion law ω^{2} = k^{2}_{z}V^{2}_{A} has been used. Hence, the energy per unit volume of this wave is

\Delta W = \Delta W_{k} + \Delta W_{M} = \frac{b^{2}_{1}}{8π} ,

which is evenly divided between the kinetic and magnetic contributions. This is not a coincidence, but comes from the general rule, that for a small amplitude oscillation the associated potential energy (in this case the magnetic one) and kinetic energy are, on average, equal to each other. Thus, it is also the case for a compressional Alfven wave (see below).

The flux of energy in the wave is given by the averaged Poynting vector < \vec{P} >= \frac{c}{4π} < (\vec{E} × \vec{B}) >. Since in the linear approximation \vec{E} = −\frac{1}{c} (\vec{v} × \vec{B}), in this case its only non-zero component is E_{x} = −Bv_{y}/c. Hence

< P_{z} >= \frac{c}{4π} < E_{x}b_{y} >= \frac{k_{z}B^{2}}{32π^{2}ρω} b^{2}_{1} = \frac{b^{2}_{1}}{8π} V_{A},

which, when represented as < \vec{P} >= (\Delta W)\vec{V}_{A}, confirms the group velocity of this wave derived in Problem 4.2.1.

In the case of a compressional Alfven wave it is convenient to express the wave energy and its flux in terms of the velocity amplitude, so that

v_{x} = v_{1} \cos(\vec{k}  · \vec{r} − ωt), b_{x} = − \frac{k_{z}B}{ω} v_{1} \cos(\vec{k}  · \vec{r} − ωt),

b_{z} = \frac{k_{x}B}{ω} v_{1} \cos(\vec{k}  · \vec{r} − ωt)

Then, the energies are given by \Delta W_{k} = ρ < v^{2}_{x} > /2 = ρv^{2}_{1}/4, \Delta W_{M} = (< b^{2}_{x} > + < b^{2}_{z} >)/8π = ρv^{2}_{1}/4 = \Delta W_{k} (the dispersion equation ω^{2} = k^{2}V^{2}_{A} is used here), so that the total energy

\Delta W = \Delta W_{k} + \Delta W_{M} = \frac{1}{2} ρv^{2}_{1}

The only non-zero component of the electric field is E_{y} = Bv_{x}/c, therefore

< P_{x} >= \frac{B}{4π} < v_{x}b_{z} >= \frac{B^{2}k_{x}}{8πω} v^{2}_{1}, < P_{z} >= − \frac{B}{4π} < v_{x}b_{x} >= \frac{B^{2}k_{z}}{8πω} v^{2}_{1}

This yields

< \vec{P} >= \frac{ρv^{2}_{1}}{2} V_{A} \frac{\vec{k}}{k} = (\Delta W)V_{A} \frac{\vec{k}}{k} ,

again in agreement with the group velocity derived from the dispersion relation in Problem 4.2.1.