Astrophysics PS11

Author

Will St. John + Theo Darci-Maher + Caedan Miller

Problem 1

  1. Using the equation that gives the FWHM at a given wavelength thermal and turbulent motions (given below. Note in class, we just saw the factor by which the line is broaden), estimate full width at half-maximum of the hydrogen H\(\alpha\) absorption line due to random thermal motions in the Sun’s photosphere. Assume the temperature is the Sun’s effective temperature.
  2. Using H\(\alpha\) redshift data for solar granulation (discussed in Ch. 11 of the textbook), estimate the FWHM when convective turbulent motions are included with thermal motions.
  3. What is the ratio of \(v^2_{turb}\) to \(2kT/m\)?
  4. Determine the relative change in the FWHM due to Doppler broadening when turbulence is included. Does turbulence make a significant contribution to FWHM in the solar photosphere?

Answer: The following code chunk calculates the answers to the questions above. Given the relative change in part d), turbulence does not make a significant contribution to FWHM in the solar photosphere.

import astropy.units as u
import astropy.constants as c
import numpy as np

T = 5777 * u.K
halpha = 656 * u.nm
v_turb = 0.4 * u.km / u.s
m = c.m_p
k = c.k_B

# a
fwhm_no_turb = 2 * halpha / c.c * np.sqrt(2 * k * T / m * np.log(2))
print(f"fwhm without turbulence: {fwhm_no_turb.to(u.angstrom):.4f}")

# b
fwhm = 2 * halpha / c.c * np.sqrt((2 * k * T / m + v_turb ** 2) * np.log(2))
print(f"fwhm with turbulence: {fwhm.to(u.angstrom):.4f}")

# c
ratio = v_turb ** 2 / (2 * k * T / m)
print(f"v_turb^2 is {100 * ratio.to(u.dimensionless_unscaled):.4f}% of 2kT/m")

# d
relative_change = (fwhm - fwhm_no_turb) / fwhm_no_turb
print(f"relative change in FWHM with turbulence: {100 * relative_change:.4f}%")
fwhm without turbulence: 0.3558 Angstrom
fwhm with turbulence: 0.3561 Angstrom
v_turb^2 is 0.1678% of 2kT/m
relative change in FWHM with turbulence: 0.0838%

Problem 2

Suppose you are attempting to make observations through an optically thick gas that has a constant density and temperature. Assume that the density and temperature of the gas are \(2.2\times10^{-4}\) kg m\(^{-3}\) and 5777 K, respectively, typical of the values found in the Sun’s photosphere.
a) If the opacity of the gas at one wavelength (\(\lambda_1\)) is \(\kappa_{\lambda 1} = 0.026\) m\(^2\) kg\(^{-1}\) and the opacity at another wavelength (\(\lambda_2\)) is $_{} = 0.030 $ m\(^2\) kg\(^{-1}\), calculate the distance into the gas where the optical depth equals 2/3 for each wavelength.
b) At which wavelength can you see farther into the gas? c) How much farther?

Answer: The code chunk below calculates the distances for \(\lambda_1\) and \(\lambda_2\) from the given the opacities, density, and optical depths. From this result, we find that \(\lambda_1\) has the larger distance, thus we can see further into the gas at \(\lambda_1\) by about 15 km.

tau_1, tau_2 = 2/3, 2/3
rho = 2.2E-4 * u.kg / u.m ** 3
kappa_1 = 0.026 * u.m ** 2 / u.kg
kappa_2 = 0.030 * u.m ** 2 / u.kg

s_1 = tau_1 / (kappa_1 * rho)  # distance for lambda 1
s_2 = tau_2 / (kappa_2 * rho)  # distance for lambda 2
print(f"s1: {s_1:.2e}")
print(f"s2: {s_2:.2e}")
print(f"difference: {s_1 - s_2:.2f}")
s1: 1.17e+05 m
s2: 1.01e+05 m
difference: 15540.02 m

Problem 3

Assume that a large solar flare erupts in a region where the magnetic field strength is 0.03 T and that it releases 1025 J in one hour. a) What was the magnetic energy density in that region before the eruption began? b) What minimum volume would be required to supply the magnetic energy necessary to fuel the flare? c) Assuming for simplicity, that the volume involved in supplying the energy for the flare eruption was a cube, compare the length of the one side of the cube with the typical size of a large flare. d) How long would it take an Alfven wave to travel the length of the flare?

Answer: The code chunk below calculates the answers the questions above. The energy density was found using \[u_m = \frac{B^2}{2\mu_0},\]

which allowed us to find the volume by dividing the given energy by our answer for energy density. Assuming the volume was a cube, then a side length is the cube root of the total volume, which results in a length \(1.42\cdot^{-8}\) times the size of a large flare (100,000 km).

The alfven speed of a wave is given by \[v_a = \frac{B}{\sqrt{\mu_0 \rho}},\]

where \(\rho\) is the density of the photosphere, which is given in the textbook to be \(4.9\cdot10^{-6}\) kg / m\(^3\). Dividing the length of our cube volume by the alfven speed gives the total time for an alfven wave to traverse the flare, which we estimate to be around \(1.17\cdot10^{-4}\) s.

B = 0.03 * u.T
energy = 1025 * u.J
density = 4.9E-6 * u.kg / u.m ** 3

# a
u_m = B ** 2 / (2 * c.mu0)
print(f"energy density of flare: {u_m.to(u.J / u.m ** 3):.2e}")

# b
volume = energy / u_m
print(f"minimum volume of flare: {volume.to(u.m ** 3):.2e}")

# c
length = volume ** (1/3)
large_flare_length = 1E5 * u.km
ratio = (length / large_flare_length).to(u.dimensionless_unscaled)
print(f"length of cube with volume above: {length.to(u.m):.2e}")
print(f"the cube length is {ratio:.2e} times the size of a large flare")

# d
v_alfven = B / np.sqrt(c.mu0 * density)
time = length / v_alfven
print(f"alfven speed of wave: {v_alfven.to(u.m / u.s):.2e}")
print(f"time for alfven wave to traverse length of flare: {time.to(u.s):.2e}")
energy density of flare: 3.58e+02 J / m3
minimum volume of flare: 2.86e+00 m3
length of cube with volume above: 1.42e+00 m
the cube length is 1.42e-08 times the size of a large flare
alfven speed of wave: 1.21e+04 m / s
time for alfven wave to traverse length of flare: 1.17e-04 s