Energetics

In this section we use the general MHD equations to obtain an energy equation that contains all the different types of energy, including kinetic energy due to plasma motions, gravitational energy, the internal energy due to the gas pressure and the magnetic energy, $B^{2}/2 \mu$ per unit volume. We repeat the MHD equations for convenience.

$${\partial \rho \over \partial t} + \nabla \cdot (\rho {\bf v}) = 0,$$ (2.68)

$$\rho {D{\bf v}\over Dt} = - \nabla p + {\bf j} \times {\bf B} - \rho \nabla \Phi + \rho \nu \nabla^{2}{\bf v}.$$ (2.69)

$${\partial p \over \partial t} + {\bf v} \cdot \nabla p = - \gamma p \nabla \cdot {\bf v} - (\gamma - 1){\textsc{L}}.$$ (2.70)

$${\partial {\bf B}\over \partial t} = \nabla \times ({\bf V} \times {\bf B}) + \eta \nabla ^{2} {\bf B},$$ (2.71)

$${{\bf j}\over \sigma } = {\bf E} + {\bf v} \times {\bf B},$$ (2.72)

Here we have defined a gravitational potential such that $-\nabla \Phi = {\bf g}$. To obtain an energy equation containing all the sources of energy, we begin by constructing the kinetic energy of the plasma. This is obtained by taking $v^{2}/2$ times (2.68) and the dot product of ${\bf v}$ and (2.69). The left hand side is then

$${1\over 2} v^{2} {\partial \rho \over \partial t} + \rho {\... ...) + \nabla \cdot \left ({1\over 2} \rho v^{2} {\bf v}\right ).$$ (2.73)

The right hand side is

$$-{\bf v} \cdot \nabla p + {\bf v}\cdot ({\bf j} \times {\bf... ...f v}\cdot \nabla \Phi + {\bf v} \cdot \nu \nabla ^{2} {\bf v}.$$ (2.74)

The terms in (2.74) are taken term by term, rearranged and manipulated using the other MHD equations. Consider the gravitational term

$$-{\bf v}\cdot \rho \nabla \Phi = - \nabla \cdot (\rho \Phi {\bf v}) + \Phi \nabla \cdot (\rho {\bf v}).$$ (2.75)

Now me may use the continuity equation (2.68) and the fact that $\partial \Phi /\partial t = 0$ to obtain

$${\bf v}\cdot \rho \nabla \Phi = \nabla \cdot (\rho \Phi {\bf v}) + {\partial \over \partial t}(\rho \Phi ).$$ (2.76)

The first term on the right hand side of (2.76) is the flux of gravitational potential energy and the second term is the rate of change of gravitational potential energy in time.

The next term we consider is the Lorentz force. Using Ohm's law, (2.72), we have

$${\bf v}\cdot ({\bf j} \times {\bf B}) = - {\bf j} \cdot ({\... ...\times {\bf B}) = - {j^{2}\over \sigma} + {\bf j}\cdot {\bf E}$$ (2.77)

Now

$${\bf j}\cdot {\bf E} = {\nabla \times {\bf B} \over \mu } \... ...\right ) + {1\over \mu }{\bf B} \cdot {\nabla \times {\bf E}},$$ (2.78)

and on using Faraday's law (2.3) we obtain

$${\bf v}\cdot ({\bf j} \times {\bf B}) = - {j^{2}\over \sigm... ...rtial \over \partial t} \left ( {B^{2} \over 2 \mu } \right ).$$ (2.79)

Next the pressure gradient term gives

$$-{\bf v} \cdot \nabla p = - \nabla \cdot (p {\bf v}) + p \nabla \cdot {\bf v}.$$ (2.80)

The last term, $p \nabla \cdot {\bf v}$, can be obtained from the energy equation (2.70). Firstly, we rewrite (2.70) in the form

$${\partial p \over \partial t} + \nabla \cdot (p{\bf v}) = -... ...amma -1 ) p \nabla \cdot {\bf v} - (\gamma - 1){\textsc{L}}.$$ (2.81)

so that

$$p \nabla \cdot {\bf v} = - {\partial \over \partial t} \lef... ... \left ( {p\over \gamma - 1}{\bf v} \right ) - {\textsc{L}}.$$ (2.82)

Finally, we can combine all the expressions to express the full energy equation as

$${\partial \over \partial t}\left [ {1\over 2}\rho v^{2} + \... ...}\over \mu }\right \} =- {j^{2}\over \sigma } - {\textsc{L}}$$ (2.83)