About the Sun
Fundamental properties of the Sun include its size (radius R), its mass M, and the power
it puts out, L. We also want to know its composition.
How do we know R?
A
B
the angle is ~ 30 arcmin = ½ degree.
How long to transit from A to B?
Time its transit: ∆t/24 hours = angle/360°; 24x3600 sec/720 = 120 sec = 2 min
How do we know L
Inventory all forms of energy coming from the Sun.
Light
all EM radiation
gamma-X-UV-Vis-IR-Radio
all mass with its kinetic and gravitational potential energy.
(why include GPE? It takes energy to lift matter to infinity)
(total energy for taking m to v at infinity: ½ m (v2 + vesc2) )
all wave energy including acoustic, shock, and alfven wave energies.
Fortunately, nearly all is in EM radiation for most stars, including our Sun. Also,
fortunately, the peak of solar power comes in the visual where most of the energy can
reach the ground (except on cloudy days).
We capture some small fraction, = area of telescope aperture / total area at 1AU =
= (telescope diameter in m /2)2/4*1AU2 (Where’d the πs go? they canceled)
How dow we know M?
Kepler’s 3rd law says (M+m) P2 = a3 with P in years, a in AU, and M+m in solar masses.
This is trivially true and does not yield MSun.
Kepler’s 3rd law in Newtonian form says M = ((2π)2/G) a3/P2 = M where now M is in kg, r
in m, and P in seconds. Look up G, the number of seconds in a year, and m in an AU and
find M:
1 year ≈ π e7 sec (actually = 3.1557 e7); a = 1.5e11 m; G = 6.6743 e-11
=> M = 4 π2 (1.5e11)3/(πe7)2 (6.7e-11) = 2e30 kg
So: How do we know G? How do we know the length of an AU?
G: Lab experiments… Hard! because gravity is so weak unless the gravitating bodies are
quite massive; and, for massive bodies, what we can measure is GM, not M and G
separately. The usual method to get G: Torsion balance (Cavendish experiment).
G = 6.67428±0.0007 x10-11 m3 kg-1 s-1 or N m2 kg-2.
Better known: GMEarth – 3.986004418 km3s-2 or the k in (Mtot ) k2 P2 = 4π2 a3 for the Sun:
k = 0.01720209895.
How do we know the number of meters in an AU: Near-Earth asteroids whose orbits are
known in AU; when they get close to Earth we can measure their distance by
triangulation from the ground; then we have that distance in AU and km both. This gave
a value for the AU that was pretty good, before modern instrumentation allowed other
methods.
Radar ranging to a planet or a moon -> again, distance to X in both km and AU. Note
that you need a good mirror at the other end – such as the surface of Mercury or one of
Jupiter’s moons. Of course, for this one you first need a good measure of the speed of
light – and the first measurement of that, by Ole Rømer in Copenhagen, used the time
delay of the motions of Jupiter’s moons together with an estimate for the length of the
AU. He estimated that it would take 11 minutes for light to travel one AU; in fact it is
closer to 8 minutes.
Space missions: If the distance is wrong, you miss the target; small corrections to the
velocity in transit provide => small corrections to a combination of the length of the AU
and the value of GMSun.
How do we know the surface temperature and composition for the Sun?
If the Sun were a solid (or an opaque gas) body with uniform temperature near the
surface) then the spectrum of the Sun would be that of a black body. What we observe
depends on both the absorption of the atmosphere (first figure) and absorption in the
Sun’s atmosphere.
Bν = (2hc2/ν3)/(exp(hν/kT) – 1) which has the limiting forms of an exponential in the UV,
and 1/ν2 in the IR and radio parts of the spectrum. We can fit the observed bits of the
solar spectrum to this form and get a measure of T that way. It will be pretty good for the
Sun but not great for some other kinds of stars.
The total power radiated by an opaque solid “blackbody” is sigma T4, so we usually
define the “effective temperature” of a star as that temperature for which
L = 4πR2 σT4
It turns out to be quite hard to measure R for most stars. The distances to stars are
measured in parsecs, and 1 pc = 206265 AU. If the Sun were 200,000 times farther
away, how big an angle would it subtend? It’s angular diameter now is 30 arc minutes,
so that would be 30x60 arcsec / 200,000 or about 1 milli-arc-sec. A dime (1cm) would
need to be 2063 km away to be the same size. (1 cm/2πd) = 1 mas / 360 degrees
= > d = 2.06e8 cm.
The spectrum of the Sun is not a pure blackbody, because the temperature in the
atmosphere first decreases then increases with height, and the atmosphere is more opaque
at some wavelengths than at others. There is a lot of information in the spectrum, and a
model fit to it lets us deduce the temperature structure and also the surface composition
within limits – unfortunately, however, He/H is one of the hardest, and 98% of the
surface material is H and He.
The temperature at the center of the Sun is approximately 14 Million Kelvins.
How do we know the temperature in the middle of the Sun!?
Models.
How to construct a model: The 4-4-4 recipe.
Mass conservation: dM(r)/dr = 4πr2ρ
Hydrostatic equilibrium: dP/dr = - GM(r) ρ / r2
(Note: momentum conservation is a special case of ∑F = ma with ∑F = 0 (bold =
vector). Hydrostatic equilibrium is also a special case of ∑F = ma with a = 0. So I
sometimes refer to this as the force-momentum equation; it is more commonly called
Newton’s second law.)
Energy conservation:
(a) Generation dL(r)/dr = 4πr2 (ε - εneutrino-TdS/dt)
Here S = entropy and this takes into account that compression -> heating. This last term
is usually small when nuclear reactions are present, but it is very important for stars that
are contracting or expanding before, between, or after nuclear burning stages.
NOTE: This is energy conservation or continuity in a way that is parallel to the mass
conservation equation. However, the energy generation rate (ε - εneutrino) is built from
nuclear reactions and E=mc2.
(b) Transport (radiation) dT/dr = – (3κρ/4acT3) L(r)
Transport (convection) dT/dr = - ((Γ2 - 1)/ Γ2) (T/P) dP/dr
where Γ2 = dlnP/dlnT = T/P dP/dT
Note: T must decrease with increasing r in order for energy to be transported outward.
We also need at least 4 constitutive relations (=information on how the particular material
of the star behaves as a function of density and temperature or pressure and temperature):
P = P(ρ, T, composition)
ε - εneutrino = function of (ρ, T, composition)
κ = κ(ρ, T, composition)
mean molecular mass = function of (ρ, T, composition)
and sometimes
Γ2 = Γ2(ρ, T, composition) (related to the first one)
and possibly (if dS/dt is not small enough)
S = S(ρ, T, composition)
Finally, we need 4 boundary conditions:
L(0) = 0 and M(0) = 0
P(R) = 0 and M(R) = M at the surface *
*These BC are expressed in different ways in different books. In particular, M(R) = M is
not very useful in practice, because one can’t easily construct a model from the center
out. Nor can one easily construct one from the surface down. What is done in practice is
to start from both ends and adjust parameters until the two solutions meet at some
intermediate M(r) or r. To do this, the surface T is sometimes approximated as = 0; this
produces a model almost indistinguishable from one with finite, realistic, T(R).
The internal structure of the Sun, based on models:
Nuclear reactions in the core => a composition profile that has low H, high He at the
center; energy is transported by radiation from the core out until the convection zone is
reached. The source of energy is 4H becoming 4He by several paths, such as H+H -> 2H
+ neutrino; 2H + H -> 3He; 3He+3He -> 4He + H + H. Because 2H is very unstable at 14
million K, the process converts H to He very slowly, and thus the Sun has a modest L and
a long lifetime.
Exercise: One H atom (or proton – the electron mass and energy are very small) has a
mass of 1.67354x10-27kg; one 4He atom or nucleus has a mass of 6.6463x10-27 kg. How
much energy is produced when 1 kg H is converted into 0.99… kg He? So how much H
is converted into He every second inside the Sun? Hint: From the data given, find
∆M/MH.
Ans: 4H = 6.69416 and the difference is 0.0479 x10-27 kg. Thus for every kg of H that is
converted, 0.0479/6.69416 = 0.007kg becomes energy. E = mc2 = 0.007kg x c2 =
6.43x1014 Joules for each kg. The Sun’s luminosity is about 4x1026 J/s so this can be
provided by conversion of 4e26/6.43e14 = 6.2x1011 kg per second.
Models did not do well at predicting the velocity fields in the convection zone, but the
Sun is rotating and it is not rotating as a solid body – this is evident from the surface
features we will discuss next time. These velocity fields have now been studied via
helioseismology.
The velocity pattern in the solar convection zone is very important because the solar
magnetic dynamo is driven by velocities / velocity gradients in the base of the CZ, we
think. This is very hard to model convincingly, however. The magnetic field in the Sun
is very important because it governs the surface and atmospheric phenomena including
flares and Coronal Mass Ejections (CMEs) that affect us, disrupting radio
communications and endangering astronauts.
A direct check on the nuclear reaction rates in the Sun, and thus on Tcentral, is to measure
the neutrino emission, because the various ways that 4H ends up in He all give rise to
neutrinos, with different energies and different ratios depending on Tc. In the 1960s, an
experiment was set up in a mine, to measure neutrinos. It captured a few neutrinos each
month in a giant tank of drycleaning fluid, giving rise to a few atoms that could be
captured and counted. It consistently captured about 1/3 of the predicted rate, and after
decades of trying to figure out what could be wrong with the experiment or with the
models, new detectors came on line that could measure more of the neutrino spectrum.
The final result was the problem wasn’t with solar models or with the experiment, but
with the neutrinos – the detectable kind could become a different, undetectable by the
mine experiment, kind. This result can be true even if neutrinos have mass, but their
mass is very tiny and has not yet been measured.
L. C. An electron comes into a region of magnetic field. The field is perpendicular to the
motion of the electron, and the field is uniform. There is no electric field and gravity
may be neglected. Describe the motion of the electron, giving formulae if you can but if
not, then give a qualitative description.
Answer: The force is proportional to qvxB so perpendicular to v and to B. Acceleration
perpendicular to the motion => circular motion with the force = mv2/r. If we know v and
B we can solve for the radius of the circle, r, and the frequency of the oscillation, v/2πr.