Expert: Philip A. Stahl - 2/18/2010

Question
I have been trying to work out a practice test problem on the Stromgren sphere with no success. Can you help? The problem reads:

“The Rosette Nebula has a brightness temperature of 200K, measured at 242 MHz. assuming an electron temperature of 10,000K   Find:

a)the optical depth (assuming optically thin conditions)
b)the emission measure
c)the average electron density assuming the radius of the Stromgren sphere is 37 pc

Thanks for any help!

Answer
Hello,

This is not too difficult a problem, but does require some attention to the units used. Let's look at each part in turn:

First, bear in mind the definition of the Stromgren sphere, which is the effective range or radius of radiative influence of the exciting source- which depends upon the type of star (or other emission object). The radiative intensity of the source then determines the dimensions of the Stromgren sphere.


a)For an electron temperature T_e and a brightness temperature, T_b , the optical depth (t_d) can be obtained from:

T_b = T_e (1 – exp(-t_d))

Where t denotes the optical thickness. For the optically thin approximation, t_d << 1 so we can write:

T_b = T_e(t_d)

So: t_d =  T_b/ T_e   =  200K / 10,000 K   =  0.02


b) The optical depth as a function of the emission measure (EM) can be written:

t_d =   0.4/f^2  INT _0 to L  {N^2 ds } = 0.4/f^2 (EM)

where f is the frequency in megahertz (Mhz),  INT denotes integral (from 0 to L), N is the electron number density and L is the path length in parsecs.

Writing the emission measure as a function of optical depth, we get (noting the units are customarily expressed in electrons per cm^3 for N and parsecs for r, with f in MHz)

EM = t_d f^2/ 0.4  =  2.5 (t_d)(f^2)


= (0.02) (242)^2 (2.5) = 2.9 x 10^3


c) The number density is easily obtained from EM using the integral from (b), since:

 INT _0 to L  {N^2 ds } = (EM)

Integrating from 0 to L over path ds:   N^2 L  = EM


And N =  SQRT[EM/L]  =   SQRT [2.9 x 10^3/ 37]  = 8.89 or about 9 electrons per cubic centimeter.

Note that we use 37 pc for the path length, i.e. effectively the *diameter* of the Stromgren sphere for the Rosette Nebula (check your text again for an error)

Hope this helps!