C01-TOPIC- A quick look at the sun

A quick look at the sun
Treating the sun as a self-gravitating mass M of radius R, its selfforce is just
[Eq. 1]
F  GMM/R2.
Since pressure P = F/A and since A  R2, from Eq. 1 we get
[Eq. 2]
P  GM2/R4.
From the perfect gas law we have
[Eq. 3]
PV  NkT,
where k is Boltzmann's constant, N is the number of particles in the
volume V, and T is the absolute temperature. Since the sun is mostly
[Eq. 4]
N = M/mp,
where mp is the mass of a single proton. Since V  R3, Eq. 3 and Eq. 4
give us
[Eq. 5]
PR3  MkT/mp.
Substitution of Eq. 2 into Eq. 5 gives us
(GM2/R4)R3  MkT/mp.
which cleans up to
[Eq. 6]
T  GMmp/Rk.
Eq. 6 predicts (fairly well) the internal temperature of the sun. We
find that due to self-gravitation, the internal temperature of the sun
[Eq. 7]
T  2.3107 K.
Note the large value V(r) takes on as r
approaches the value ro = 110-15 m, one
femtometer (the width of a nucleus), at
which point the strong nuclear force
can cause fusion.
From the figure, we
see that if the total energy E of the
protons just equals V(ro), the protons
can just make it into the correct range
for fusion. From Eq. 8 we have
In order to get two protons to fuse, the Coulomb potential wall must be
A picture of this wall is easily gotten from the Coulomb
potential energy function:
[Eq. 8]
V(r) = e2/4or.
E = V(ro) = e2/4oro = 2.310-13 J.
[Eq. 9]
Since the average kinetic energy of a typical particle in a gas is
given by E = 3kT/2, from Eq. 9 we see that the temperature necessary
for fusion is
[Eq. 10]
T  11010 K.
Comparing Eq. 10 to Eq. 7, it is clear that something is wrong with our
analysis--the self-gravitational temperature does not appear to be high
enough for fusion to occur, by a factor of 1000.
Early work on sun
models came up against the same problem. Studies showed, however, that
the temperature of Eq. 7 was about correct, and fusion was taking
So, how close do the protons actually come to one another at the
temperature provided by self-gravity? Working backward, the distance r
can be found from
[Eq. 11]
E = 3kT/2 = e2/4or
to be
[Eq. 12]
r = 4.810-13 m,
short by a factor of 10 to 100.
How do we resolve this difficulty?
The solution comes about through quantum mechanics and Heisenberg's
uncertainty principle,
pr  h.
[Eq. 13]
Eq. 13 tells us the more precise one measurement is, the less precise
the other is--at least to the extent that the product is  h. To use
Eq. 13 we need to determine p: To this end, at a temperature of T, the
average kinetic energy of a proton is given by mpv2/2 = 2kT/3, so that
[Eq. 14]
p2  3mpkT.
From Eq. 14 we see that at the temperature predicted
gravitation that the average momentum of a proton is
p = 1.210-21 kgm/s.
[Eq. 15]
r  5.510-13 m.
[Eq. 16]
From Eq. 12 we see that at the
selfgravitation, the protons can be pushed
to within r = 4.810-13 m of each other.
principle, that distance can vary by r
= 5.510-13.
Thus protons can "breech"
the Coulomb wall and come under the
influence of the strong force, allowing
fusion to take place.
We call this
phenomenon tunneling.
From Eq. 13 we have