When we have vibration excitation then the radius of nucleus is define like:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]R=R_0[1+\sum^{\infty}_{\lambda=0}\sum^{\lambda}_{\mu=-\lambda}\alpha_{\lambda\mu}Y^{\lambda}_{\mu}(\theta,\phi)][/tex]

where [tex]\alpha_{\lambda,\mu}=\alpha_{\lambda,-\mu}[/tex] and [tex]\alpha_{\lambda,\mu}=\alpha_{\lambda,\mu}(t)[/tex]

How you measure this [tex]\alpha[/tex] parametar?

[tex]Y^{\mu}_{\lambda}=\frac{(-1)^{\mu+\lambda}}{2^{\lambda}\lambda!}\sqrt{\frac{2\lambda+1}{4\pi}\frac{(\lambda-\mu)!}{(\lambda+\mu)!}}e^{i\mu\varphi}(sin\Theta)^{\frac{\mu}{2}}\frac{d^{\mu+\lambda}}{d(cos\Theta)^{\mu+\lambda}}sin^{2\lambda}(\Theta)[/tex]

And more:

Kinetic energy of system is define like:

[tex]T=\frac{1}{2}\sum_{\lambda,\mu}B_{\lambda}|\frac{d \alpha_{\lambda,\mu}}{d t}|^2[/tex]

Rayleight use [tex]\rho=\frac{3M}{4R^3_0\pi}[/tex], and get [tex]B_{\lambda}=\frac{3MR^2_0}{4\pi\lambda}[/tex]. How?

Thanks for answers

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Vibration excitation

**Physics Forums | Science Articles, Homework Help, Discussion**