Second Virial Coefficient, numerically

[Selveste]

Since the integrand is spherically symmetric I use spherical coordinates

[tex] \int d\vec{r} = \Omega_d\int_{0}^{\infty}dr r^{d-1} [/tex]

where [itex] \Omega_d [/itex] is the solid angle i [itex] d [/itex] dimensions. Since I am to plot as a function of [itex] 1/\beta U_0 (= k_bT/U_0) [/itex], I introduce a new varible [itex] \Theta = 1/\beta U_0 = k_bT/U_0 [/itex]. This gives

[tex] B_2(T) = \frac{1}{2} \Omega_d \Bigg( \int_0^R dr r^{d-1} + \int_R^{\infty} dr r^{d-1}(1-e^{\frac{1}{\Theta}(\frac{R}{r})^{\alpha}}) \Bigg)[/tex]

Now I wonder if this is correct. And how to procede to compute the integrals numerically. Here [itex] R [/itex] is not given. Should I just set it to some value? I thought about integrating with respect to some varible [itex] Rr [/itex] or [itex] r/R [/itex], but I can't seem to get rid of [itex] R [/itex].

 

[Nate810]

Yes, this is correct. All that remains is to calculate the two integrals numerically. To do that, you can use a numerical integration technique such as the trapezoidal or Simpson's rule, or you can use a computer algebra system such as Mathematica or Maple. The first integral will be a simple one-dimensional integral, while the second integral will require more work since it involves an exponential function.When calculating the integrals, you should set R to some value in order to obtain a result. It is best to choose a value of R that is appropriate for the given problem. In this case, it might be reasonable to set R equal to the average radius of the particles.