Parker's solar wind model

Parker (1958) suggested that the corona could not remain in static equilibrium but must be continually expanding since the interstellar pressure cannot contain a static corona. The continual expansion is called the solar wind. The existence of a solar wind had been known from comet observations but the properties predicted by Parker were confirmed by the satellites Lunik III and Venus I in 1959 and by Mariner II in the early 1960s.

The main assumptions of Parker's model are that the outflow is steady, spherically symmetric and isothermal. It is straightforward to relax the isothermal assumption and consider an adiabatic or polytropic atmosphere. The basic steady equations are

$$\nabla \cdot \left (\rho {\bf v}\right ) = 0,$$ (6.7)

$$\rho \left ({\bf v}\cdot \nabla \right ){\bf v} = - \nabla p + \rho {\bf g},$$ (6.8)

$$p = \rho R T,$$ (6.9)

and

$$T = T_{0}.$$ (6.10)

The velocity is taken as purely radial so that ${\bf v} = v \hat{\bf r}$ and gravitational acceleration obeys the inverse square law, ${\bf g} = - G M_{\circ}/r^{2}$. In spherical coordinates and assuming a steady flow, (6.7) gives

$${d\over dr}\left ( r^{2}\rho v \right ) = 0 \qquad \Rightarrow \qquad r^{2}\rho v = \hbox{const.}$$ (6.11)

and the raidial component of (6.8) becomes

$$\rho c {dv\over dr} = - {dp\over dr} - {GM_{\circ}\rho \over r^{2}}.$$ (6.12)

Defining the isothermal sound speed as $(p/\rho )^{1/2} = c_{s}$, the gas law gives

$$p = c_{s}^{2}\rho,$$ (6.13)

and substituting (6.11) and (6.13) into (6.12) we get

$$\rho v {dv\over dr} = - c_{s}^{2}{d\rho \over dr} - {GM_{\circ}\rho \over r^{2}}.$$

Then dividing by $\rho$ we obtain

$$v{dv\over dr} = - {c_{s}^{2}\over \rho}{d\rho \over dr} - {GM_{\circ} \over r^{2}}.$$

Now we can use (6.11) to express $d\rho /dr$ and $\rho$ in terms of $r^{2}v$ as

$$v{dv\over dr} = - c_{s}^{2}r^{2}v{d\over dr}\left ( {1\over r^{2}v}\right ) - {GM_{\circ}\over r^{2}}.$$

Expanding the pressure gradient term and rearranging gives the final radial equation of motion as

$$\left (v - {c_{s}^{2}\over v}\right ){dv\over dr} = 2{c_{s}^{2}\over r^{2}}\left (r - r_{c}\right ),$$ (6.14)

where $r_{c}= {GM_{\circ}/c_{s}^{2}}$. $r_{c}$ is important since $v = c_{s}$ and $r = r_{c}$ is a critical point of (6.14). If the velocity of the plasma reaches the sound speed then the radius must either equal $r_{c}$ or else the velocity gradient becomes infinite. Another way of saying this is to say that if $r = r_{c}$ then either $dv/dr = 0$ or $v = c_{s}$ and if $v = c_{s}$ then either $dv/dr = \infty$ or $r = r_{c}$.

(6.14) can be integrated to give a transendental equation for the velocity in terms of the radius as

$$\left ({v\over c_{s}}\right )^{2} - \log \left ({v\over c... ...4 \log \left ({r\over r_{c}}\right ) + 4 {r_{c}\over r} + C,$$ (6.15)

where $C$ is a constant. The solutions for different values of $C$ are shown in Figure 6.1.

Figure 6.1: The solar wind velocity, $v$, as a function of the radius, $r$ for various values of the constant $C$. The five different classes of solution are indicated.

In Figure 6.1 five distinct types of solution are indicated. Solution I is doubled valued and so is unphysical. It is not possible for the plasma to leave the solar surface with a velocity below the sound speed, reach a maximum radius below $r_{c}$ and then turn round and return to the Sun with a super-sonic speed. Solution II is also double valued but it never even starts from the solar surface and it is also unphysical. Solution III starts with a velocity greater than the sound speed but such a fast steady outflow is not observed. Hence, this solution must also be neglected. The only two types of solution that are physically reasonable at this stage are IV and V. Solution V is the particular case where the plasma leaves the solar surface with a particular speed and passes through the critical point (also called the sonic point) at $r = r_{c}$ and $v = c_{s}$. Solution IV always remains below the sound speed and is called the solar breeze solution. For solution V we choose the constant $C$ so that $r = r_{c}$ and $v = c_{s}$ and this requires $C = -3$.Is it possible to decide between solutions IV and V?

Example 6.2.1   The behaviour of solution V for large $r$ can be obtained quite easily. From Figure 6.1, we may assume that $v \gg c_{s}$ so that (6.15) is approximated by

$$\left ({v\over c_{s}}\right )^{2} \approx 4 \log \left ({r\... ...ox 2 \left (\log \left ({r\over r_{c}}\right)\right )^{1/2}.$$

Hence, from (6.11) the density is given by

$$\rho = {\hbox{const.}\over r^{2}v} \approx {\hbox{const.}\over r^{2}\sqrt{\log(r/r_{c})}}.$$

Thus, $\rho$ tends to zero as $r$ tends to infinity. Since the plasma is isothermal the pressure also tends to zero. This means that the solution can eventually match onto the interstellar plasma at large distances from the Sun. Thus, solution IV is a physically realistic model of the solar wind. It predicts that the plasma will be super-sonic beyond the critical point.

Example 6.2.2   The behaviour of solution IV for large $r$ is quite different. Again using Figure 6.1, we see that $v$ tends to zero as $r$ tends to infinity. Thus, (6.15) may now be approximated by

$$-\log\left ({v\over c_{s}}\right )^{2} \approx 4 \log \left... ...ad {v\over c_{s}} \approx \left ({r_{c}\over r}\right )^{2}.$$

The mass continuity equation gives the density as

$$\rho = {\hbox{const.}\over r_{c}^{2}c_{s}}.$$

Since the density tends to a constant value so will the pressure. Thus, the solar breeze solution is unphysical since it cannot be contained by the extremely small interstellar pressure.

Thus, the Parker solar wind model is given by solution IV. The plasma starts at the solar surface with a small velocity that increases towards the critical point. At the critical point the speed reaches the sound speed. Then the flow becomes (and remains) super-sonic while the gas pressure decreases.

Example 6.2.3   We may calculate the critical radius, $r_{c}$. The radius of the Sun is $R_{\circ} = 6.96 \times 10^{8}$m. Assuming a typical coronal temperature of $10^{6}$K the sound speed is

$$c_{s} = \left (RT\right )^{1/2} = \left (8.3\times 10^{3}\t... ...}\right )^{1/2} = 9\times 10^{4} \approx 10^{5}\hbox{ms}^{-1}.$$

The critial radius is

$$r_{c} = {G M_{\circ}\over 2 c_{s}^{2}} = 8 \times 10^{9}\hbox{m} \approx 10 R_{\circ}.$$

To put this into context, the radius of the Earth's orbit is $R_{E}\approx 214 R_{\circ}$. Thus, the solar wind is highly super-sonic by the time it reaches the Earth. To calculate the actual wind speed from Parker's model we set $r = R_{E}$ and solve for $v$. Hence,

$$\left ({v\over c_{s}}\right )^{2} - \log \left ({v\over c... ...g \left ({214\over 10}\right ) + 4 {10\over 214} - 3 = 9.44.$$

This may be solved using the Newton-Raphson method to give

$$v = 3.45 c_{s} = 310 kms^{-1}.$$

Observations at $1AU$ give the quiet solar wind as

$$v \approx 320 kms^{-1}.$$

Thus, Parker's solar wind model gives quite a good estimation of the velocity.

There are three points to note

1. As an alternative energy equation it is possible to try a polytrope of the form
   
   $$p = K \rho^{\gamma}.$$
   
   

2. The real solar wind does not come from the whole of the solar surface but only from the coronal hole regions where the magnetic field is open. This is only about 20 percent of the whole surface. The strong magnetic field of the other 80 percent is closed and effectively holds in the hot corona.

3. The real solar wind probably ends in a shock called the heliopause at about $100 AU$.