# Susskind's treatment of a rotor in a magnetic field

• I

## Main Question or Discussion Point

My question is related to the chapter 10 of L.Sussknd's book on analytical mechanics named "The theoretical minimum". There he considers dynamics of a charged rotor in magnetic field using Hamiltonian formalism and poisson brackets.

He also introduces a treatment of the same kind for a gyroscope starting at 1:11:35 at his online lecture at http://theoreticalminimum.com/courses/classical-mechanics/2011/fall/lecture-8

In both cases, first relations for poison brackets of angualar momentum for a freely rotating body are introduced, then the very same relations are used for a hamiltonian in which coupling energy of magnetic field (first case) or potential energy in gravitational field (second case) is added. Everything looks nice and simple...

Yet as the hamiltonian formalism is originally defined, a (conjugate) momentum is a derivative of lagrangian by the velocity, which means that as soon as we introduce any potential energy (change the lagrangian), the angular momentum is no more the same one which we defined for the freely rotating body, so that we cannot just blindly reuse the formulas which we got for a case of freely rotating body. We cannot just add the coupling energy to the hamiltonian when we add the field, we need rather to reevaluate this hamilton with adjusted conjugate momentum.

So my questiin is how can the approach of Susskind be justified (it provably can). I.e. the approach in which after adding a potential field, we do not change the momentum definition at all, but are using hamilton formalism.

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if the Hamiltonian is changed by adding a potential energy V which is velocity-independent then the conjugate momentum will not change. For the gravitational field case, the potential energy V = mgz which has no velocity dependence so the derivative by velocity of any lagrangian with any v-independent V will be the same either with or without this V. This is true for any V which has no velocity dependence not just the gravitational case. Are you worried about cases where V is explicitly a function of v only or will the gravitational case satisfy your actual issue?

• MichPod
I see. Thank you. I do not have good intuition in this field, so I may be lost easily.

Actually, in the gravitational case Susskind (in the video) says that the potential energy is proportional to the... vertical component of the angular momentum(!) so he effectively gets a formula of potential energy the same for a gyroscope in gravitational field and for a charged rotor in magnetic field. Because the energy is proportional to the angular momentum, I somehow believed that the potential energy in lagrangian will depend on the angular velocity even for a gyroscooe in potential field, as well as for a rotor in magnetic field (doesn't it?). In fact, in both cases it is not even clear for me which coordinates are used (Susskind says nothing of that, just using angular momentums formally). I suppose, this may be a the root of misunderstanding.

Not that I now understand everything with this lecture, but thank you, this helped.

Yes I saw that in the Susskind lecture which was nice for you to provide a link for (as I would never have gone searching for something like that but like it now that I have seen it -- that is a cool trick to prove those angular momentum properties with Poisson Brackets)

Potential energy is actually $V=mgh$ where h is height and height is usually taken as the $z$ axis so just the value $z$ goes in for $h$. Susskind was saying that in the absence of gravity the angular momentum of the spinning disc does not move so that its $z$ component $L_z$ would not change and also be proportional to just the $z$ coordinate. (As the $z$ coordinate goes up and down, so does $L_z$.) "Proportional to" means there is some constant of proportionality which he calls $c$ so that $z=cL_z$ and hence that $V=mgz=cmgL_z$ and then he wrote $cmg$ as actually a new $c'=cmg$ since usually constants can be collected together and just called a "new" $c'$ but which he still calls $c$. So he changed his definitions for the constants a few times along the way there, but you should get all the points by now.