- #1
Wavefunction
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- Homework Statement
- Consider a symmetric top ##I_1=I_2\neq I_3## with constant charge to mass ratio ##\frac{q}{m}## and hence constant gyro magnetic ratio ##\gamma=\frac{q}{2m}##. As discussed in lecture, such an object will have Lagrangian:
$$L=T-V=T-\boldsymbol{\omega}_l\cdot\mathbf{L}$$
a) Show that the kinetic energy ##T## is constant and find the other constants of motion.
b) Under the assumption that ##\omega_l## is much less than the initial component of the angular velocity along the symmetry axis, obtain expressions for the frequency and amplitude of nutation and the average precession frequency.
- Relevant Equations
- $$T=\frac{I_1}{2}\Big(\dot{\theta}^2+\dot{\phi}^2\sin^2(\theta)\Big)+\frac{I_3}{2}\Big(\dot{\psi}+\dot{\phi}\cos(\theta)\Big)^2$$
$$\boldsymbol{\omega}_l=-\gamma\mathbf{B}$$
$$V=-\gamma\mathbf{B}\cdot\mathbf{L}$$
$$\boldsymbol{\omega}=\dot{\phi}\hat{\mathbf{e}}_3+\dot{\theta}\hat{\boldsymbol{\rho}}_1+\dot{\psi}\hat{\mathbf{e}}'_3$$
Setup: Let ##\hat{\mathbf{e}}_1,\hat{\mathbf{e}}_2,\hat{\mathbf{e}}_3## be the basis of the fixed frame and ##\hat{\mathbf{e}}'_1,\hat{\mathbf{e}}'_2,\hat{\mathbf{e}}'_3## be the basis of the body frame. Furthermore, let ##\phi## be the angle of rotation about the ##\hat{\mathbf{e}}_3## axis, ##\theta## be the rotation about the ##\boldsymbol{\rho}_1## axis, and ##\psi## be the angle of rotation about the ##\hat{\mathbf{e}}_3'## axis (##\phi##, ##\theta##, and ##\psi## are euler angles)
Assumption 1: ##\mathbf{B}## can be oriented along the fixed ##\hat{\mathbf{e}}_3## axis; thus,
\begin{equation}
\mathbf{B}=B\hat{\mathbf{e}}_3
\end{equation}
which expressed in the basis of the body frame is
\begin{equation}
\mathbf{B}=B\sin(\theta)\sin(\psi)\hat{\mathbf{e}}'_1+B\sin(\theta)\cos(\psi)\hat{\mathbf{e}}'_1+B\cos(\theta)\hat{\mathbf{e}}'_3
\end{equation}
Assumption 2: I can express ##\mathbf{L}## in the body frame using the fact that in the body frame, the inertia tensor ##\mathbf{I}## becomes diagonal.
\begin{equation}
I=\begin{bmatrix}
I_1 & 0 & 0\\
0 & I_2 & 0\\
0 & 0 & I_3
\end{bmatrix}
\end{equation}
However, since ##I_1=I_2##, then this becomes
\begin{equation}
I=\begin{bmatrix}
I_1 & 0 & 0\\
0 & I_1 & 0\\
0 & 0 & I_3
\end{bmatrix}
\end{equation}
Additionally, ##\boldsymbol{\omega}## in the body frame is:
\begin{equation}
\boldsymbol{\omega}=\Big(\dot{\phi}\sin(\theta)\sin(\psi)+\dot{\theta}\cos(\psi)\Big)\hat{\mathbf{e}}'_1
+\Big(\dot{\phi}\sin(\theta)\cos(\psi)-\dot{\theta}\sin(\psi)\Big)\hat{\mathbf{e}}'_2
+\Big(\dot{\psi}+\dot{\phi}\cos(\theta)\Big)
\end{equation}
Which leads to ##\mathbf{L}## in the body frame being:
\begin{equation}
\begin{split}
\mathbf{L}=I_1\Big(\dot{\phi}\sin(\theta)\sin(\psi)+\dot{\theta}\cos(\psi)\Big)\hat{\mathbf{e}}'_1\\
+I_1\Big(\dot{\phi}\sin(\theta)\cos(\psi)-\dot{\theta}\sin(\psi)\Big)\hat{\mathbf{e}}'_2\\
+I_3\Big(\dot{\psi}+\dot{\phi}\cos(\theta)\Big)\hat{\mathbf{e}}'_3
\end{split}
\end{equation}
Furthermore, since ##V=-\gamma\mathbf{B}\cdot\mathbf{L}##
\begin{equation}
V=-\gamma B\Big(I_1\dot{\phi}\sin^2(\theta)+I_3\cos(\theta)(\dot{\psi}+\dot{\phi}\cos(\theta))\Big)
\end{equation}
Thus, the lagrangian is simply
\begin{equation}
\begin{split}
L=\frac{I_1}{2}\Big(\dot{\theta}^2+\dot{\phi}^2\sin^2(\theta)\Big)+\frac{I_3}{2}\Big(\dot{\psi}+\dot{\phi}\cos(\theta)\Big)^2\\
+\gamma B\Big(I_1\dot{\phi}\sin^2(\theta)+I_3\cos(\theta)(\dot{\psi}+\dot{\phi}\cos(\theta))\Big)
\end{split}
\end{equation}
Clearly, from the lagrangian, ##\psi## and ##\phi## are cyclic; thus,
For ##\phi##
\begin{equation}
\begin{split}
I_1\dot{\phi}\sin^2(\theta)+I_3(\dot{\psi}+\dot{\phi}\cos(\theta))\cos(\theta)\\
+\gamma B\Big(I_1\sin^2(\theta)+I_3\cos^2(\theta)\Big)=I_1b
\end{split}
\end{equation}
where ##b## is a constant
For ##\psi##
\begin{equation}
I_3(\dot{\psi}+\dot{\phi}\cos(\theta))+\gamma B I_3\cos(\theta)=I_1a
\end{equation}
where ##a## is a constant.
These equations can be used to solve for ##\dot{\phi}## and ##\dot{\psi}## respectively.
\begin{equation}
\dot{\phi}=\frac{b-a\cos(\theta)-\gamma B\sin^2(\theta)}{\sin^2(\theta)}
\end{equation}
and
\begin{equation}
\dot{\psi}=\frac{I_1a}{I_3}-\Big[\frac{b-a\cos(\theta)-\gamma B\sin^2(\theta)}{\sin^2(\theta)}+\gamma B\Big]\cos(\theta)
\end{equation}
We can substitute these expressions into both the kinetic energy and potential energy to get
\begin{equation}
T=\frac{I_1}{2}\Big(\dot{\theta}^2+\Big[\frac{b-a\cos(\theta)-\gamma B\sin^2(\theta)}{\sin^2(\theta)}\Big]^2\sin^2(\theta)\Big)+\frac{I_3}{2}\Big(\frac{I_1a}{I_3}-\gamma B\cos(\theta)\Big)^2
\end{equation}
\begin{equation}
V=-\gamma B\Big(I_1\Big[b-a\cos(\theta)-\gamma B\sin^2(\theta)\Big]+I_1a\cos(\theta)-\gamma BI_3\cos^2(\theta)\Big)
\end{equation}
Now, ##T## can be put into the following form
\begin{equation}
T=\frac{1}{2}\frac{I_1^2a^2}{I_3}+\frac{1}{2}I_1\gamma^2B^2-\frac{1}{2}I_1b\gamma B+\frac{1}{2}I_1\dot{\theta}^2+\frac{I_1\gamma B(b-a\cos(\theta)^2}{2\sin^2(\theta)}+\frac{1}{2}\gamma^2B^2(I_3-I_1)\cos^2(\theta)
\end{equation}
and ##V## can be put into this form
\begin{equation}
V=-\gamma BI_1b+\gamma^2B^2I_1+\gamma^2B^2(I_3-I_1)\cos^2(\theta)
\end{equation}
Both of these terms contain constants which can be removed such that
\begin{equation}
T'=\frac{1}{2}I_1\dot{\theta}^2+\frac{I_1\gamma B(b-a\cos(\theta))^2}{2\sin^2(\theta)}+\frac{1}{2}\gamma^2B^2(I_3-I_1)\cos^2(\theta)
\end{equation}
And
\begin{equation}
V'=\gamma^2B^2(I_3-I_1)\cos^2(\theta)
\end{equation}
Now this is the part where I am stuck: It seems counter-intuitive that the kinetic energy would be constant. Assuming total energy of the system is conserved, and ##V'## varies with ##\theta##, then ##T'## can't be constant; however, if total energy is not conserved and ##V## varies with the total energy in such a way that makes ##E'-V'## a constant, then it would follow that ##T'## is constant. It seems like a pretty large leap of faith to simply assume the total energy varies in such a manner so as to make ##E'-V'## constant though. Am I on the right track here or have I made an egregious error that I simply can't see?
Note: I will try part b) on my own, but I am including it in this post just in case I get stuck on that part too (So, I don't make a duplicate post). So to be clear, I am not seeking help on part b) right now.
For reference, I will also attach the latex document that I working with. The part of interest is Homework 3>Problem 3>Part a). Thank you for all of your help in advance.
Assumption 1: ##\mathbf{B}## can be oriented along the fixed ##\hat{\mathbf{e}}_3## axis; thus,
\begin{equation}
\mathbf{B}=B\hat{\mathbf{e}}_3
\end{equation}
which expressed in the basis of the body frame is
\begin{equation}
\mathbf{B}=B\sin(\theta)\sin(\psi)\hat{\mathbf{e}}'_1+B\sin(\theta)\cos(\psi)\hat{\mathbf{e}}'_1+B\cos(\theta)\hat{\mathbf{e}}'_3
\end{equation}
Assumption 2: I can express ##\mathbf{L}## in the body frame using the fact that in the body frame, the inertia tensor ##\mathbf{I}## becomes diagonal.
\begin{equation}
I=\begin{bmatrix}
I_1 & 0 & 0\\
0 & I_2 & 0\\
0 & 0 & I_3
\end{bmatrix}
\end{equation}
However, since ##I_1=I_2##, then this becomes
\begin{equation}
I=\begin{bmatrix}
I_1 & 0 & 0\\
0 & I_1 & 0\\
0 & 0 & I_3
\end{bmatrix}
\end{equation}
Additionally, ##\boldsymbol{\omega}## in the body frame is:
\begin{equation}
\boldsymbol{\omega}=\Big(\dot{\phi}\sin(\theta)\sin(\psi)+\dot{\theta}\cos(\psi)\Big)\hat{\mathbf{e}}'_1
+\Big(\dot{\phi}\sin(\theta)\cos(\psi)-\dot{\theta}\sin(\psi)\Big)\hat{\mathbf{e}}'_2
+\Big(\dot{\psi}+\dot{\phi}\cos(\theta)\Big)
\end{equation}
Which leads to ##\mathbf{L}## in the body frame being:
\begin{equation}
\begin{split}
\mathbf{L}=I_1\Big(\dot{\phi}\sin(\theta)\sin(\psi)+\dot{\theta}\cos(\psi)\Big)\hat{\mathbf{e}}'_1\\
+I_1\Big(\dot{\phi}\sin(\theta)\cos(\psi)-\dot{\theta}\sin(\psi)\Big)\hat{\mathbf{e}}'_2\\
+I_3\Big(\dot{\psi}+\dot{\phi}\cos(\theta)\Big)\hat{\mathbf{e}}'_3
\end{split}
\end{equation}
Furthermore, since ##V=-\gamma\mathbf{B}\cdot\mathbf{L}##
\begin{equation}
V=-\gamma B\Big(I_1\dot{\phi}\sin^2(\theta)+I_3\cos(\theta)(\dot{\psi}+\dot{\phi}\cos(\theta))\Big)
\end{equation}
Thus, the lagrangian is simply
\begin{equation}
\begin{split}
L=\frac{I_1}{2}\Big(\dot{\theta}^2+\dot{\phi}^2\sin^2(\theta)\Big)+\frac{I_3}{2}\Big(\dot{\psi}+\dot{\phi}\cos(\theta)\Big)^2\\
+\gamma B\Big(I_1\dot{\phi}\sin^2(\theta)+I_3\cos(\theta)(\dot{\psi}+\dot{\phi}\cos(\theta))\Big)
\end{split}
\end{equation}
Clearly, from the lagrangian, ##\psi## and ##\phi## are cyclic; thus,
For ##\phi##
\begin{equation}
\begin{split}
I_1\dot{\phi}\sin^2(\theta)+I_3(\dot{\psi}+\dot{\phi}\cos(\theta))\cos(\theta)\\
+\gamma B\Big(I_1\sin^2(\theta)+I_3\cos^2(\theta)\Big)=I_1b
\end{split}
\end{equation}
where ##b## is a constant
For ##\psi##
\begin{equation}
I_3(\dot{\psi}+\dot{\phi}\cos(\theta))+\gamma B I_3\cos(\theta)=I_1a
\end{equation}
where ##a## is a constant.
These equations can be used to solve for ##\dot{\phi}## and ##\dot{\psi}## respectively.
\begin{equation}
\dot{\phi}=\frac{b-a\cos(\theta)-\gamma B\sin^2(\theta)}{\sin^2(\theta)}
\end{equation}
and
\begin{equation}
\dot{\psi}=\frac{I_1a}{I_3}-\Big[\frac{b-a\cos(\theta)-\gamma B\sin^2(\theta)}{\sin^2(\theta)}+\gamma B\Big]\cos(\theta)
\end{equation}
We can substitute these expressions into both the kinetic energy and potential energy to get
\begin{equation}
T=\frac{I_1}{2}\Big(\dot{\theta}^2+\Big[\frac{b-a\cos(\theta)-\gamma B\sin^2(\theta)}{\sin^2(\theta)}\Big]^2\sin^2(\theta)\Big)+\frac{I_3}{2}\Big(\frac{I_1a}{I_3}-\gamma B\cos(\theta)\Big)^2
\end{equation}
\begin{equation}
V=-\gamma B\Big(I_1\Big[b-a\cos(\theta)-\gamma B\sin^2(\theta)\Big]+I_1a\cos(\theta)-\gamma BI_3\cos^2(\theta)\Big)
\end{equation}
Now, ##T## can be put into the following form
\begin{equation}
T=\frac{1}{2}\frac{I_1^2a^2}{I_3}+\frac{1}{2}I_1\gamma^2B^2-\frac{1}{2}I_1b\gamma B+\frac{1}{2}I_1\dot{\theta}^2+\frac{I_1\gamma B(b-a\cos(\theta)^2}{2\sin^2(\theta)}+\frac{1}{2}\gamma^2B^2(I_3-I_1)\cos^2(\theta)
\end{equation}
and ##V## can be put into this form
\begin{equation}
V=-\gamma BI_1b+\gamma^2B^2I_1+\gamma^2B^2(I_3-I_1)\cos^2(\theta)
\end{equation}
Both of these terms contain constants which can be removed such that
\begin{equation}
T'=\frac{1}{2}I_1\dot{\theta}^2+\frac{I_1\gamma B(b-a\cos(\theta))^2}{2\sin^2(\theta)}+\frac{1}{2}\gamma^2B^2(I_3-I_1)\cos^2(\theta)
\end{equation}
And
\begin{equation}
V'=\gamma^2B^2(I_3-I_1)\cos^2(\theta)
\end{equation}
Now this is the part where I am stuck: It seems counter-intuitive that the kinetic energy would be constant. Assuming total energy of the system is conserved, and ##V'## varies with ##\theta##, then ##T'## can't be constant; however, if total energy is not conserved and ##V## varies with the total energy in such a way that makes ##E'-V'## a constant, then it would follow that ##T'## is constant. It seems like a pretty large leap of faith to simply assume the total energy varies in such a manner so as to make ##E'-V'## constant though. Am I on the right track here or have I made an egregious error that I simply can't see?
Note: I will try part b) on my own, but I am including it in this post just in case I get stuck on that part too (So, I don't make a duplicate post). So to be clear, I am not seeking help on part b) right now.
For reference, I will also attach the latex document that I working with. The part of interest is Homework 3>Problem 3>Part a). Thank you for all of your help in advance.
Last edited: