Find each of the following:

a) The energy going into the half sphere each second

b) The change in (a) over each absorption and re-emission over vertical optical depth.

c) The total absorption and total emission and the relationship between then over all space.

for the star Eta Ophiuchi which has a (B – V) color index of -0.30.

**:**

*Solution*We need to use the Table in the blog link at the end of the article to identify the (B – V) index with the log of effective temperature, log Teff. On doing so we find: log T_eff = 4.65 because we see:

**(B – V) = -0.30**corresponds to

**log T_eff = 4.65**.

antilog (log T_eff) = antilog (4.65 ) = 44,700

So: T_eff = 44,700 K

Then: π( F) = σ (Teff)^4

(5.67 x 10-8 W m^-2 K^-4) (4.47 x 10^ 4 K)^4 =

(5.67 x 10-8 W m^-2 K^-4)(4 x 10 ^18 K^4) = 2.2 x 10^11 W m-2

Whence:

S/τ = 3π F/ 4π = 3(2.2 x 10^11 W m^-2 ) / 4π

Or: S/τ = 5.4 x 10^10 W m^-2

Next:

2π S = πF or:

S = π F / (2π) = (2.2 x 10^11 W m^-2)/ 2π

Or:

S = J = 3.5 x 10^10 W m^-2

Another problem:

a) Estimate the specific intensity I (Θ=π/4) if the surface flux from the Sun is 6.3 x 10^ 7 Jm^-2 s-1.

(b) Find the mean intensity if I (Θ=π/4) is taken over all space.

(c) Find the effective temperature of the Sun and the boundary temperature (T_o) and account for any difference.

(d) Estimate the net flux, H, passing through the Sun’s surface.

**Solutions**:

The specific intensity is defined from:

π( F_o ) = 2 π (I(cos (Θ))

And for Θ = π/4, then cos (π/4) = [2]^½/2 and:

I = π( F_o ) / 2 π ([2]^½/2)= π( F_o ) / π([2]^½)

Therefore:

I = (6.3 x 10^ 7 Jm^-2 s^-1) / π([2]^½)

I » 1.4 x 10^ 7 Jm^-2 s^-1

(b) If I is taken over all space (i.e. 4 π steradians), the mean intensity is:

J = 1/4 π INT (I) dw = 1/4 π (1.4 x 10 ^7 Jm^-2 s^-1) 4 π

So: J = 1.4 x 10^ 7 Jm^-2 s^-1

(c) The effective temperature is obtained using:

π( F_o ) = σ (Teff)^4

So: Teff = [π( F_o ) / σ] ^¼

Where σ = 5.67 x 10^-8 W m^-2 K^-4

Is the Stefan-Boltzmann constant. Then:

T_eff = [6.3 x 10 ^7 Jm^-2 s^-1/ 5.67 x 10^-8 W m^-2 K^-4] ^¼

Teff = 5800 K

The boundary temperature is found from:

T_eff = (2)^¼T_o = 1.189 T_o

Or: T_o = T_eff /1.189 = 5800K/ 1.189= 4800 K

The boundary temperature differs because of being referenced to a

*different optical depth*. The boundary temperature (T_o)

*approaches the value of the effective (or surface) temperature*when τ = 0, but this still exhibits a difference in layers so will not be exactly the same!

(d) The net flux (H) passing through the Sun’s surface is estimated using: σT_o 4/ π= 2H.

Therefore: H = σT^o ^4/2 π

Where: σT^o ^4= (5.67 x 10^-8 W m^-2 K^-4)(4800K)

σT^o ^4 = 2.7 x 10^-4 W m^-2

H = (2.7 x 10^-4 W m^-2) /2 π = 4.3 x 10^-5 W m^-2

**:**

*Other Problems*1) A star has a gray atmosphere for which the Eddington approximation:

T^4 = ¾ T_e^4(τ + 2/3)

is valid, where T_e denotes the effective temperature. Use this approximation to obtain the fraction of outward intensity escaping from the star’s surface.

2) The star Suhail has a (B – V) color index of +1.7. Use this to obtain the net flux (H) passing through the Suhail’s surface. How might you estimate the intensity I from this and the mean intensity J?

3) For many stars, the solar constant S can be computed if its angular diameter is known. If the angular radius of a star is: a= (R/r) with r the distance to Earth and R the star’s linear radius then:

π F= S (r/R)^2

(Note: a is measured in radians where 1 rad = 57.3 deg)

If the Sun’s angular radius is 959.63 arcsec then find the solar constant S.

For more ambitious readers:

Find the solar constant S for alpha Lyrae (Vega) if we know (from Hanbury and Brown’s measurements) that its angular diameter is 0.0032 arcsec, and it has a (B- V) color index of 0.00. Show all working and state any assumptions.

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