Question 13.5: The surface of a neutron star (sphere) vibrates slowly, so t...
The surface of a neutron star (sphere) vibrates slowly, so that the principal moments of inertia are harmonic functions of time:
I_{zz} = \frac{2}{5} mr^{2}(1+ ε cos ωt),I_{xx} = I_{yy} = \frac{2}{5}mr^{2} \left(1 −ε \frac{cos ωt}{2}\right), ε\ll1.
The star simultaneously rotates with the angular velocity Ω(t).
(a) Show that the z-component of Ω remains nearly constant!
(b) Show that Ω(t) nutates about the z-axis and determine the nutation frequency for Ω_{z} \ll ω.
(a) If the total angular momentum is given in an inertial system, then
\left(\frac{dL}{dt}\right)_{inertial}= 0.The principal moments of inertia are, however, given in a body-fixed system that rotates itself with the angular velocity Ω with respect to the inertial system. Then
\left(\frac{dL}{dt}\right)_{inertial}=\left(\frac{dL}{dt}\right)_{k}+Ω×L = 0.One therefore gets in the body-fixed system (Euler equations)
\frac{d}{dt}(I_{zz}Ω_{z}) = 0, (13.8)
\frac{d}{dt} (I_{xx}Ω_{x} )+ \frac{3}{2} I_{0}Ω_{y}Ω_{z}ε cos ωt = 0, (13.9)
\frac{d}{dt} (I_{yy}Ω_{y} )− \frac{3}{2} I_{0}Ω_{x}Ω_{z}ε cos ωt = 0, (13.10)
where I_{0} = (2/5)mr^{2} is the moment of inertia of the sphere. (13.8) has the solution
Ω_{z} = \frac{Ω_{0z}}{1 +ε cos ωt},where Ω_{0z} follows from the initial conditions; this means that Ω_{z} is only very weakly time dependent.
(b) We suppose that ω \ll Ω_{z} , i.e.,
\frac{dI_{xx}}{dt}≈ 0 and \frac{ dI_{yy}}{dt}≈ 0.From this, we find
I_{xx} \dot{Ω}_{x} + \frac{3}{2} I_{0}Ω_{z} ε cos ωtΩ_{y} = 0, I_{yy}\dot{Ω}_{y} − \frac{3}{2} I_{0}Ω_{z} ε cos ωt Ω_{x} = 0. (13.11)
Differentiating again and inserting (13.8), (13.9), and (13.10) yield
I_{xx} \ddot{Ω}_{x} + \frac{1}{I_{yy}} \left(\frac{3}{2} I_{0} Ω_{z} ε cos ωt\right)^{2} Ω_{x} = 0,I_{yy} \ddot{Ω}_{y} + \frac{1}{I_{xx}} \left(\frac{3}{2} I_{0} Ω_{z} ε cos ωt\right)^{2} Ω_{y} = 0,
If I_{xx} = I_{yy} ≈ I_{0}, then
\ddot{Ω}_{x} +\left(\frac{3}{2} ε Ω_{z} cos ωt\right)^{2} Ω_{x} = 0, \ddot{Ω}_{y} +\left(\frac{3}{2} ε Ω_{z} cos ωt\right)^{2} Ω_{y} = 0.
Since ω \ll Ω_{z} (we further assume that ω \ll εΩ_{z}), we find
ω_{n} = \frac{3}{2}ε Ω_{z} cos ωt (nutation frequency),
i.e., Ω_{x} and Ω_{y} perform a nutation motion with ω_{n}.