Expert: Philip A. Stahl - 4/6/2007

Question
I have read that the assumption of a static corona (no gas expanding into space as solar wind) can be shown to be so from basic plasma physics. How can this be shown?

Answer
Hello,

Of course, a static corona seems on first blush to be reasonable but that is why it is necessary to test that this is so. The first one to do this was Sydney Chapman. He began by first assuming the condition for hydrostatic equilibrium applied:

dp/ dr = - rho {GM_s/ r^2}

where G is the usual Newtonian gravitational constant, and rho defines the plasma density for the corona, M_s is the mass of the Sun, and r the distance from the solar center:

rho = n(m_p)

with n the number density for protons

The coronal pressure (p) is given by:

P = 2 n T

Provided both protons and electrons are assumed to have the same temperature.

The thermal conductivity of the corona is dominated by electron thermal conductivity and takes the form:

k =  k_o T^5/2

for typical coronal conditions the value of k is about 20 times the value of copper at room temperature.

Now, the coronal heat flux density is:

q = -k grad T

A static corona means heat inputs cancel heat outputs so that the divergence:

div q = 0

Assuming a spherical symmetry for the corona one can write:

1/r^2 [d/dr (r^2 k_o T^5/2 dT/dr) = 0

Obviously the preceding assumptions mean there must be some distance where the coronal temperature becomes zero.

From the above eqn. you should be able to show:

d(T^7/2) =  7/2 (FT_o^5/2)/ 4 pi k_o  d(1/r) = C d(1/r)

where C is a constant.

The integral is:

T_o ^7/2  - T^ 7/2 = C[ 1/R_o  - 1/r]

Now, set the temperature at infinity (T) to zero and obtain:

C = R_o T_o ^7/2

which fixes the total flux at:

F = 2/7  4 pi R_o k_o T_o

After another step, one finds:

T(r) = T_o (R_o/ r) ^2/7

this gives the temperature T at a distance from the Sun r . This is based on using a defined value (say T_o = 2 x 10^6 K) at a defined distance, say R_o = 7 x 10^8 m.

For example, at the Earth’s distance (r = 1.5 x 10^11 m) one would find: T = 4.3 x 10^5 K

This seems fine, until one examines the pressure.

Analogous to the temperature formalism, we have, the pressure p(r) at some distance defined by:

p(r) = p(R_o) exp [7/5 GM_sm(p)/ 2 T(R_o) R_o {(R_o/ r)^5/7 – 1}]

Now, if one allows r to approach infinity, e.g. r -> oo an interesting thing occurs in the equation, as you can see. That is, the denominator of the first term in the end brackets becomes so large (R_o/ oo) that the first term vanishes.

Then you are left with the expression for the pressure:

p(oo) = p(R_o) [exp – 7k/5  * 1/ T(R_o) R_o]

where ‘k’ denotes a constant composed of all the constant quantities in the previous eqn. (G, M, m(p) etc)

Substituting the given values into the above, one finds p(R_o) multiplied by a factor

exp[0] = 1

The reason is that the exponential of a very small and negative valued magnitude -> 0

Then:  

p(oo) ~  p(R_o)

But this can’t be since the pressure of the coronal base would then be the same as the value at infinity!

This led astrophysicists to conclude an unphysical result, and that the static coronal model couldn’t be accurate.

If the static model were accurate, the pressure at infinity should be zero, p(oo) = 0, not a small finite pressure that’s effectively equal to the coronal base pressure. This finding led to the further investigations that disclosed a solar “wind” had to flow outwards from the corona.