electric dipole in a sphere

Electric dipole in a sphere PhL 3.29.17 Step 1. If the dipole is embedded in a uniform density sphere, I know at once that all three inertia components are the same no matter how you choose the principal axes, so the Frame S' equations of motion are : (N)'1 = I'1 ()'1 - (ω)'2(ω)'3 (I'2 - I'3) (N)'2 = I'2 ()'2 - (ω)'3(ω)'1 (I'3 - I'1) (N)'3 = I'3 ()'3 - (ω)'1(ω)'2 (I'1 - I'2) . (I.4.3) and then (N)'1 = I'1 ()'1 (N)'2 = I'1 ()'2 (N)'3 = I'1 ()'3 I'1 = (2/5)Mr2 looking it up on the web (could calc it here) Let I'1 = (2/5)Mr2 ≡ J. then the above three equations say (N)'i = J ()'i Since this is true for all Frame S' components, it must be true as a vector equation, so N = J and therefore in Frame S we could write Ni = J ()i Step 2. We know that the torque on the dipole is this N = p x E based on our I.10 problem work. Thus, in Frame S we have εijkpjEk = J ()i But only E3 = E exists in Frame S, so set k = 3 to get εij3pjE = J ()i Write out these equations i=1 ε123p2E = J ()1 p2E = J ()1 i=2 ε213p1E = J ()2 -p1E = J ()2 i = 3 ε3j3pjE = J ()3 0 = J ()3 We then end up with : ω3 = constant ()1 = [p2E/J] ()2 = - [p1E/J] Now since p = p by fiat (I declare the embedded p to be in the direction) , we have p = p [ sinθcosφ, sinθsinφ,cosθ] = [p1,p2,p3] This our equations can be written ω3 = constant ()1 = [pE/J] sinθsinφ ()2 = - [pE/J] sinθcosφ At this point I know nothing about θ(t) and φ(t) so cannot simply integrate. So what do I do next?? What is L ? L = Iω = Jω J = scalar = (2/5)MR2 = J What does Newton say in Frame S? N = Thus we know that p x E = J εijkpjEδk3 = J ()i but I already know this, so nothing new here! Maybe look for some conserved quantities! Maybe I should be starting by writing force and torque as I did in the electric dipole dumbbell problem. Plan A: Imagine that the dumbbell is embedded in the sphere so that there are charges q and -q as before, and the stick length is 2r where r is the radius of the mass sphere. . Then p = 2rq. So this is just the dumbbell problem with a different inertia tensor. In this case I have F = 0 N = N1+ N2 = 2N1 = -2rqEsinθ Thus the equation of motion can be written p x E = J N = J = -2rqEsinθ = J = - (2rqE/J)sinθ or = - (pE/J) sinθ So here then is the equations of motion in terms of spherical unit vectors. I have to be careful now not to make my previous mistake in trying to solve this equation! I think I can still claim that ()r = 0 ()r = 0 Nr = 0 ()θ = 0 ()θ = 0 Nθ = 0 ()φ = -pEsinθ . ()φ = -(pE/J) sinθ Nφ = -2rqEsinθ (I.10.9) But now I no longer have (F.2.2) for L, maybe I have to complete L somehow!! Go back to ω3 = constant ()1 = [pE/J] sinθsinφ ()2 = - [pE/J] sinθcosφ I have four unknowns ω1(t), ω2(t), θ(t) and φ(t), but only two equations! I need more equations! How about these equations, ω1 = sinθcosφ - sinφ ω2 = sinθsinφ + cosφ ω3 = cosθ + . // Frame S (H.6.18) Can I use them somehow? Can I assume ψ = constant? Is there torque about the axis? Well above I have written Nr = 0 so do I know somehow that = Nr ? Look at ()r = 0 = 0 (dL/dt)S = 0 [(dL/dt)S]r = 0 I used this kind of fact in (H.4.1). ωφ = ωθ = ' ωψ = ' . (H.4.1) How did I get these three equations? How do I know that does not contribute to ωφ ? // OK, I pondered this and added a new paragraph to frames doc below (H.4.1), I think this was worth pondering and getting stated. So OK, what does the information ()r = 0 tell me? Here is what I want to say Nr = 0 ψ = constant How do I justify this? N = Nr = ()r = 0 Can I say this (dL/dt)r = dLr/dt ? If I could say this, then I would have Nr = dLr/dt = 0 Lr = constant Then I have to argue that Lr = Jωr and then ωr = 0 and so then ψ = constant But this is a commutation theorem issue. I really have L = Lr + Lθ + Lφ Then = r + Lr ∂t + θ + Lθ ∂t + φ + Lφ ∂t = r + Lr [ + sinθ ] + θ + Lθ [ – + cosθ ] + φ + Lφ [ – sinθ – cosθ ] = {r - Lθ - sinθLφ } + {Lr + θ - cosθ Lφ} + { Lr sinθ + Lθ cosθ + φ } This tells me that ()r = r - Lθ - sinθLφ ()θ = Lr + θ - cosθ Lφ ()φ = Lr sinθ + Lθ cosθ + φ I then combine these with results from above ()r = 0 ()r = 0 Nr = 0 ()θ = 0 ()θ = 0 Nθ = 0 ()φ = -pEsinθ . ()φ = -(pE/J) sinθ Nφ = -2rqEsinθ (I.10.9) and I end up with these equations of motion r - Lθ - sinθLφ = 0 Lr + θ - cosθ Lφ = 0 Lr sinθ + Lθ cosθ + φ = -pEsinθ This is very ugly stuff. I suppose I can say = J and rewrite this mess in terms of ω components. I hate it! But recall my conjecture from above (dL/dt)r = dLr/dt ? Well in fact (dL/dt)r = r - Lθ - sinθLφ so forget that pathway! I really need to solve things within Frame S'! Status: Today 3.29.17 I have spent perhaps 4 hours on this problem, and I have made zero progress. Typical for the course! Is this perhaps the same as the top problem? There the torque is N = r x [-mg ] about the center of mass of the top, whereas here the torque is N = p x E = 2q r x E , sure looks the same. Both torques are about the CMS. Then connection ought then to be 2qE = -mg or 2qE = -mg. So I think it really is the top problem, but with a simpler inertia tensor.