# Symmetric top with constant charge to mass ratio in a magnetic field

Wavefunction
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,

\mathbf{B}=B\hat{\mathbf{e}}_3

which expressed in the basis of the body frame is

\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

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.

I=\begin{bmatrix}
I_1 & 0 & 0\\
0 & I_2 & 0\\
0 & 0 & I_3
\end{bmatrix}

However, since ##I_1=I_2##, then this becomes

I=\begin{bmatrix}
I_1 & 0 & 0\\
0 & I_1 & 0\\
0 & 0 & I_3
\end{bmatrix}

Additionally, ##\boldsymbol{\omega}## in the body frame is:

\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)

Which leads to ##\mathbf{L}## in the body frame being:

\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}

Furthermore, since ##V=-\gamma\mathbf{B}\cdot\mathbf{L}##

V=-\gamma B\Big(I_1\dot{\phi}\sin^2(\theta)+I_3\cos(\theta)(\dot{\psi}+\dot{\phi}\cos(\theta))\Big)

Thus, the lagrangian is simply

\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}

Clearly, from the lagrangian, ##\psi## and ##\phi## are cyclic; thus,

For ##\phi##

\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}

where ##b## is a constant

For ##\psi##

I_3(\dot{\psi}+\dot{\phi}\cos(\theta))+\gamma B I_3\cos(\theta)=I_1a

where ##a## is a constant.

These equations can be used to solve for ##\dot{\phi}## and ##\dot{\psi}## respectively.

\dot{\phi}=\frac{b-a\cos(\theta)-\gamma B\sin^2(\theta)}{\sin^2(\theta)}

and

\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)

We can substitute these expressions into both the kinetic energy and potential energy to get

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

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)

Now, ##T## can be put into the following form

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)

and ##V## can be put into this form

V=-\gamma BI_1b+\gamma^2B^2I_1+\gamma^2B^2(I_3-I_1)\cos^2(\theta)

Both of these terms contain constants which can be removed such that

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)

And

V'=\gamma^2B^2(I_3-I_1)\cos^2(\theta)

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.

#### Attachments

• Homework Problems.pdf
1.2 MB · Views: 196
Last edited:
Delta2 and PhDeezNutz

## Answers and Replies

Wavefunction
Update: I have figured out the solution. Basically, you need to think about the torque on the magnetic moment of the top due to the magnetic field; then relate this torque to the change in potential energy. Once you figure out this relationship you can show that the angle of nutation is constant which allows you to make the desired conclusion: Additionally you get conservation of the systems energy which is reassuring in this case. Hope this helps anyone who gets stuck where I did. Cheers everyone.

PhDeezNutz
Homework Helper
Gold Member
In general, there will be some nutation. That is, ##\theta## will vary with time. The precession rate will also vary. This goes along with what you are asked to do in part (b).

The function ##V## of the lagrangian depends on the angular velocities ##\dot \psi## and ##\dot \phi## as well as the angle ##\theta##. So, ##V## is not a true potential energy function.

If you form the energy ##E =\dot \theta p_{\theta} + \dot \phi p_{\phi} + \dot \psi p_{\psi} - L##, you will find that ##E## is just the kinetic energy ##T##. Therefore, ##E## and ##T## are the same constant of the motion. A direct way to show that ##T## is constant is to calculate ##\frac {dT}{dt}## using your expression (17) for ##T## and the equation of motion for ##\theta## derived from the Lagrangian.

Last edited:
Wavefunction
Wavefunction
Hi Tsny and thank you for the reply.

Eventually, I was able to get that ##\frac{dE}{dt}=\frac{dT}{dt}## since the change in angular momentum is perpendicular to the ##B## field; thus, ##\frac{dV}{dt}=0##. Once I got there, then I was able use ##\frac{dE}{dt}=\frac{dT}{dt}## to show that ##\dot{\theta}=0\Rightarrow \theta=\theta_0## which means that the kinetic energy is constant. I originally did not elect to use the energy function since it was unclear to me if the energy function was indeed the total energy due to the fact that ##V## depends on generalized velocities. I suppose the conditions to conclude that the energy function is the total energy is an implication rather than a bi-conditional, but I would have to re-read that section of Goldstein.

Homework Helper
Gold Member
I'm not sure how you get ##\frac{dV}{dt} = 0##. This would be true only for the very special case where the initial conditions are chosen specifically to make the top precess without any nutation.

A typical initial setup would be to have the top initially spinning at some rate ##\dot \psi_0## about the ##I_3## axis while this axis is held at an angle ##\theta_0## to the B-field direction. The top is released without any initial precession or nutation: ##\dot \theta_0 = \dot \phi_0 = 0##. The top will then start to precess and nutate.

For a top in a gravitational field, the kinetic energy ##T## will vary as the top nutates.

But in the case of the charged top in the B field, ##T## will remain constant as the top nutates and the function ##V = \boldsymbol {\omega}_l \cdot \mathbf L## will vary with time.

Goldstein wrote a paper on the charged top in a B field which gives a thorough analysis. The reference is
American Journal of Physics, Vol 19, page 100 (1951) by Herbert Goldstein.

---------------------------------------------

In looking at your Latex document, it appears to me that your method of obtaining the equation of motion for ##\theta## (equation 3.65) is not correct. I believe you need to use the initial form of the Lagrangian as given in equation 3.49. Your Lagrangian ##L'## as express in equation 3.64 will not give the correct equation of motion.

Here's a rather trivial example that illustrates this. Consider a free particle of unit mass. The Lagrangian expressed in polar coordinates is

(1) ##L = T-V = T = \frac{1}{2}\left(\dot r^2 + r^2 \dot \theta^2 \right)##

The cyclic coordinate ##\theta## yields the constant of motion ##l = r^2 \dot \theta## (the angular momentum). Thus, ##\dot \theta = \frac{l}{r^2}##. Substituting this into (1) gives

(2) ##L = \frac{1}{2} \left(\dot r^2 + \frac{l^2}{r^2} \right)##

But the two forms of ##L## given in (1) and (2) do not yield the same equation of motion for ##r##. (1) gives

##\ddot r = r^2 \dot \theta = +\frac{l^2}{r^3}## (correct)

If you use (2) and treat ##l## as a constant, then you get

##\ddot r = -\frac{l^2}{r^3}## (not correct)

So, you should use your Lagrangian as expressed in 3.49 to get the correct equation of motion for ##\theta##.

Wavefunction
Wavefunction
So the logic goes as follows (Assuming a constant magnetic field oriented along the z-axis)

##\frac{dV}{dt}=\gamma \mathbf{B}\cdot \frac{d\mathbf{L}}{dt}##

Now the torque on a magnetic moment ##\mathbf{M}## is given by ##\mathbf{N}=\frac{d\mathbf{L}}{dt}=\mathbf{M}\times\mathbf{B}##

Thus,

##\frac{dV}{dt}=\gamma \mathbf{B}\cdot (\mathbf{M}\times\mathbf{B})##

However, ##\mathbf{B}\cdot (\mathbf{M}\times\mathbf{B})=0##

Therefore, ##\frac{dV}{dt}=0##

(Upon thinking about it, yes I did implicitly assume special initial conditions; however, these happen to be the initial condition discussed in lecture: start spinning a symmetric top very fast about it's axis of symmetry. I should also note here that in this example, there is no gravity in the problem)

Ah, I see you have my old document: I decided to go a different route than what I did in the attached document in my final answer: I decided against getting the E.L for theta and instead used ##\frac{dE'}{dt}=\frac{dV'}{dt}## to get that ##\theta=\theta_0## and ##\dot{\theta}=0##. I do see what you mean about plugging in the constants of motion before getting the E.L. equations: I think at that point I was just trying to get something nice to fall out for ##\theta##.

Homework Helper
Gold Member
Now the torque on a magnetic moment ##\mathbf{M}## is given by ##\mathbf{N}=\frac{d\mathbf{L}}{dt}=\mathbf{M}\times\mathbf{B}##
It is an interesting fact that this equation for the torque is not completely valid for the charged top in a uniform B field. This is because this equation for the torque assumes that the magnetic moment ##\mathbf M## has constant magnitude as the magnetic moment changes its orientation. But for the top, ##|\mathbf M|## changes as ##\theta## changes due to the fact that the magnitude of the angular momentum ##\mathbf L## changes as ##\theta## changes (i.e., as the top nutates). This is discussed in Goldstein's paper where he derives an expression for ##L^2## as a function of ##\theta##. He also derives the correct equation for the torque which contains the term ##\mathbf{M}\times\mathbf{B}##, but also contains additional terms.

(Upon thinking about it, yes I did implicitly assume special initial conditions; however, these happen to be the initial condition discussed in lecture: start spinning a symmetric top very fast about it's axis of symmetry.
So, I think you are taking ##\dot \theta_0 = 0##. If you also take ##\dot \phi_0 = 0##, then the top will begin to both precess and nutate. If you are assuming that ##\dot \phi_0## is chosen specifically to prevent any nutation, then what is the purpose of part (b) of the problem statement?

I decided against getting the E.L for theta and instead used ##\frac{dE'}{dt}=\frac{dV'}{dt}## to get that ##\theta=\theta_0## and ##\dot{\theta}=0##.
The equation ##\frac{dE'}{dt}=\frac{dV'}{dt}## is not valid except for the special case where the initial conditions are chosen to produce no nutation. In general, ##\frac{dV'}{dt} \neq 0## even though ##\frac{dE'}{dt}## is always zero.

Wavefunction
Wavefunction
So as for the expression on the torque, I think we were meant to just use ##\mathbf{N}=\mathbf{M}\times\mathbf{B}##, just for the sake of making the problem tractable: I'm starting to see the argument I made in my final answer a little bit more clearly.

##\mathbf{N}=\mathbf{M}\times\mathbf{B}## is the major assumption of my argument as everything else follows from there. If the antecedent of the implication isn't true, then that puts doubt on the truth value of the consequent. I think if I were to do the problem in full generality (probably much beyond my ability at this point, but I might come back to it when I have some time), then ##T## would not in general be constant which is what I think you're getting at here which is something that was also initially confusing me to a great extent.

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.

Basically, this problem would be one long proof by contradiction that for the full general system where you have additional terms in the torque, that ##T## is not in general constant. This would lead me to believe that there is a component of the torque which depends on ##\theta## along ##\mathbf{B}##, then the dot product would have a non-zero value.

However, if ##\mathbf{N}=\mathbf{M}\times\mathbf{B}##, then I'm fairly confident it follows that ##\theta## is constant. Which was the end goal of the problem to show.

As a final note: I'd have to read more about when you can use this approximation because to be completely honest with you this is beyond my current knowledge.

Homework Helper
Gold Member
I think it's important to see that there is no need to use the concept of torque in order to answer the questions in the problem statement. Everything follows from the Lagrangian. Using the Euler-Lagrange equations of motion derived from the Lagrangian, you can show that ##T## is constant. And you can work out the precession rate as well as the nutation frequency and amplitude.

Wavefunction
I think it's important to see that there is no need to use the concept of torque in order to answer the questions in the problem statement. Everything follows from the Lagrangian. Using the Euler-Lagrange equations of motion derived from the Lagrangian, you can show that ##T## is constant. And you can work out the precession rate as well as the nutation frequency and amplitude.

I'll have to retry it at a later date, my when I tried to do it the last time it got quite messy and it didn't work out so nice. I see your point that the lagrangian is a more general method and you don't get stuck with the assumption that ##\mathbf{N}=\mathbf{M}\times\mathbf{B}##, but I'm going to have to do it on my own later to see it when I have the time. Currently, I'm working on the next problem set so I have to move on for right now, lest I fall behind in the course. I do thank you for taking the time out of your schedule to talk to me about this though! It was definitely enlightening.

TSny