Question 15.16: Write down the Lagrangian of the heavy asymmetric top. Use t...
Write down the Lagrangian of the heavy asymmetric top. Use the Euler angles as generalized coordinates and determine the related generalized momenta. Which coordinate is cyclic? Which further cyclic coordinate appears for a symmetric top?
In the system of principal axes, the kinetic energy of motion is given by
T = \frac{1}{2} Θ_{1}ω^{2}_{1}+ \frac{1}{2} Θ_{2}ω^{2}_{2}+ \frac{1}{2} Θ_{3}ω^{2}_{3}.The potential energy is
V = mgh = mgl cos β.We take the Euler angles (α,β,γ ) as generalized coordinates. The angular velocities expressed by these coordinates read (see (13.43)),
T = \frac{1}{2}(Θ_{1}ω^{2}_{1}+Θ_{2}ω^{2}_{2}+Θ_{3}ω^{2}_{3})= \frac{1}{2}Θ_{1}(\dot{α} sin β sin γ + \dot{β} sin γ )^{2}
+ \frac{1}{2}Θ_{2}(\dot{α} sin β cos γ − \dot{β} sin γ )^{2}
+ \frac{1}{2}Θ_{3}(\dot{α} cos β + \dot{γ} )^{2}. (13.43)
ω_{1} = \dot{α} sin β sin γ + \dot{β} cos γ,
ω_{2} = \dot{α} sin β cos γ − \dot{β} sin γ,
ω_{3} = \dot{α} cos β + \dot{γ} .
By inserting this into the Lagrangian L = T − V , we get
L = \frac{1}{2} Θ_{1}(\dot{α}^{2} sin^{2} β sin^{2} γ + \dot{β}^{2} cos^{2} γ + 2\dot{α} \dot{β} sin β sin γ cos β)+ \frac{1}{2} Θ_{2}(\dot{α}^{2} sin^{2} β cos^{2} γ + \dot{β}^{2} sin^{2} γ − 2\dot{α} \dot{β} sin β sin γ cos β)
+ \frac{1}{2}Θ_{3}(\dot{α} cos β + \dot{γ} )^{2} − Mgl cos β.
The Euler angles as generalized coordinates obey the Euler–Lagrange equations
\frac{d}{dt} \frac{∂L}{∂\dot{α}}= \frac{∂L}{∂α}and the analogous equations for β and γ . The Lagrangian does not depend on the angle α; hence, this coordinate is cyclic, and the related generalized momentum is conserved.
We determine the generalized momenta:
p_{α} = \frac{∂L}{∂\dot{α}}= Θ_{1}ω_{1} \frac{∂ω_{1}}{∂\dot{α}}+Θ_{2}ω_{2} \frac{∂ω_{2}}{∂\dot{α}}+ Θ_{3}ω_{3} \frac{∂ω_{3}}{∂\dot{α}}=Θ_{1}(\dot{α} sin β sin γ + \dot{β} cos γ ) sin β sin γ
+Θ_{2}(\dot{α} sin β cos γ − \dot{β} sin γ ) sin β cos γ
+Θ_{3}(\dot{α} cos β + \dot{γ} ) cos β
= \dot{α} sin^{2} β(Θ_{1} sin^{2} γ +Θ_{2} cos^{2} γ )+ (Θ_{1} −Θ_{2}) \dot{β} cos γ sin β sin γ
+Θ_{3}(\dot{α} cos β + \dot{γ} )cos β.
In an analogous way, we obtain
p_{β} = \frac{∂L}{∂ \dot{β}}=Θ_{1} \dot{β} cos^{2} γ +Θ_{2} \dot{β} sin^{2} γ+ (Θ_{1} −Θ_{2})\dot{α} cos γ sin β sin γ +Θ_{3} (\dot{α} cos β + \dot{γ} ),
p_{γ} = \frac{∂L}{∂\dot{γ}}= Θ_{3}(\dot{α} cos β + \dot{γ} ).
For the symmetric top, Θ_{1}=Θ_{2}, and thus, the Lagrangian simplifies considerably:
L = \frac{1}{2} Θ_{1}(\dot{α} sin^{2} β + \dot{β}^{2})+ \frac{1}{2} Θ_{3}(\dot{α} cos β + \dot{γ} )^{2} − Mgl cos β.The Lagrangian of the symmetric top no longer depends on the angle γ ; therefore, the angle γ becomes cyclic too. Hence, the momentum p_{γ} is also conserved.
The generalized momenta then read
p_{β} = \dot{β} Θ_{1} +Θ_{3}(\dot{α} cos β + \dot{γ} ),
p_{γ} =Θ_{3}(\dot{α} cos β + \dot{γ} ) = constant.
The generalized momenta, being the projection of the total angular momentum onto the rotational axis related to the particular Euler angle, have a direct physical meaning.
p_{α} is the projection of the total angular momentum onto the space-fixed z-axis (see Exercise 13.12):
This projection is a conserved quantity for the asymmetric and the symmetric top. Since the gravitational force acts only along the z-direction, the angular momentum about this axis remains unchanged.
p_{β} is the projection of the total angular momentum onto the nodal line, i.e., the axis about which the second Euler rotation is being performed:
p_{β} = L · e_{β} = L · e_{x}.This momentum is not conserved.
p_{γ} can be interpreted as the projection of the total angular momentum onto the body-fixed e_{z^{′}} -axis:
p_{γ} = L · e_{γ} = L · e_{z^{′}} .
For a symmetric top, the body-fixed z^{′} -axis is a symmetry axis, and the angular momentum projection L · e_{z^{′}} is conserved.