Chapter 4 tutorial

Problem 6

The pressure is given by equation 3.26:





Note The partition function would be obtained from its definition:


if we were at equilibrium.  Because we are not at thermal equilibrium, the partition function is not useful.

By definition, we get:


[Graphics:Images/ch4_gr_7.gif] is zero, but I am not entirely sure why unless we are at equilibrium (and therefore [Graphics:Images/ch4_gr_8.gif] does not change).  In general, I am not sure why it would be zero.

We know from Eq. 4.15 that:


therefore, using chain rules, we can calculate:


Note: if the volume is increased, the energy (or frequency) of a mode decreases as we can see from the sign of the derivative.

We substitute and obtain:


Now, we are asked to compare the  kinetic pressure with the thermal radiation pressure.  Because we now deal with thermal radiation pressure,(i. e.  pressure of radiation at thermal equilibirum), we can use the results of p. 94 in the page, which are valid for thermal equilibrium.  We know that at thermal equilibrium, the energy density of radiation is:


And therefore, using the derived result p=U/3V, we can say that the thermal radiation pressure is:


The pressure of an ideal gas at that temperature is given by Eq. (4.74):


where C is the concentration of molecules per volume. With C= [Graphics:Images/ch4_gr_17.gif], the pressures are equal when:


and therefore


In the  sun, the kinetic pressure is roughly:


and in the center it is a hundred times more:



Problem 7



We substitute for clarity:


to obtain:






We can actually find the limit of the series:


Therefore we obtain the final result:


Converted by Mathematica      September 26, 2002