Sunday, November 4, 2018

Selected Questions- Answers From All Experts Astronomy Forum (The Zeeman Effect And Sunspot Magnetic Fields)

Question: I've just been reading a book on the Sun ('Our Sun' by Donald Menzel )  and found references to sunspots having magnetic fields of up to 4000 gauss. How can astronomers or solar physicists obtain such values? How do they know this?

The Zeeman effect is  a broadening, i.e. of a spectral line from the Sun,  due to strong magnetic fields such as in sunspots. An example is depicted below: The left image shows the photo of the line-centered sunspot, i.e. the sunspot for which a spectral line has been obtained at line center, and in classic "triplet" form.  That is there exists a normal (unaffected)  line of wavelength  l on either side of which are lesser and greater wavelength lines, hence "splits".  Thus the triplet is presented in terms of the normal wavelength as follows:

lo   +   D  l H

lo

l-   D  l H

George Ellery Hale was the first to apply the Zeeman splitting of a solar spectral line to the problem of quantifying the strength of the magnetic field associated with a sunspot.  He thereby arrived at the following cgs version of the equation:

D  l H  =     (lo)e H / 4 π  me c2

Here:  H is the intensity of the sunspot magnetic field to be found,  e is the electron charge in electrostatic  units (e.s.u.),  me    is the electron mass in grams and c is the speed of light in cm/sec.  To obtain the intensity in Gauss then, we first need to use basic algebra to solve for H:

H =   4 π  me c 2   D  l H  /    (lo)2  e

We then must pay attention to the units, so that we have:

e =    4.8 x  10  -10  esu

me     =   9.1   x 10-28    gram

c  =  3   x 10  10 m/s

To  illustrate the application we will let the  undisturbed solar line  ( lo ) be the H-alpha line which  has wavelength:  6.62  x 10 - 5  cm.   We then let the line displacement (shift owing to H) on either side be:  D  l H    = 0.05 A = 5.0 x 10 -10 cm
The equation with units substituted in for computation, then becomes:

H  =

4 π (9.1   x 10-28  g) (x 10  10 cm/s )2  (5.0 x 10 -10 cm) / (6.62  x 10 - 5  cm)2 (4.8 x  10  -10  esu)

The calculated sunspot magnetic field intensity is:  H = 2440 G  approximately.

An interesting further exercise is to compute the field strength in Tesla (T) instead of Gauss. Tesla is the S.I. unit of measurement for the magnetic field intensity.  To do this basically requires changing all the units used above to consistent S.I. units.   Thus, cm now becomes meters (m), and the e.s.u. becomes coulombs (C). The electron mass is now in kg instead of grams and so on.