Challenge on angular momentum

Challenge on angular momentum PhL 2.15.17 The L section of frames is very complicated and has no external verification. But I was able to make it work in the dumbbell satellite situation. As I now resume Goldstein rotational dynamics, I realize that there is a basic equation which I never used and which should be investigated. It is this: (dL/dt)S = (dL/dt)S' + ω x L (1) This simple equation yields a simple analysis of things like tops. It appears on Goldstein page 158. In frames doc I deal a bit with L before I have stated the G rule which the above is for L. I call this doc a "challenge" because I need to constantly challenge the validity and clarity of frames doc in order to get it right. I have no one else who is available to do this task. I don't mind it at all, because each such challenge increases my understanding of the rotating frame topic and erases my misconceptions. Notation for equation (1) When dealing with other vectors, for each concept there are two different vectors. For example, (12.1.2) gives us r = b + r' r, r' v = v' + ω x r' + S v, v' a = a' + x r' + 2 ω x v' + ω x (ω x r') + S a, a' Here the primed variables are always "natural" ones such as v' = (dr'/dt)S' a' = (dv'/dt)S' Not only is there ()S' surrounding, but the inside vector is primed as well. So equation (1) above can be written = (dL/dt)S' + ω x L (2) where (dL/dt)S' is NOT a natural vector. I have encountered such "mixed" vectors before, such as vS' = v' + S – ω x b = v' + S' (6.3b,a) v'S = v' + ω x r' (6.8a) (6.9) Here vS' and v'S are both mixed vectors. I guess I could write (2) above as = S'+ ω x L and that is my standard notation for a mixed vector S' . My frames doc has no "section" which relates different forms of the way it does with v and a [wrong]. In such a section, presumably the object S' would appear. I guess one such equation would be S' = - ω x L Question: Generally when doing dynamics, I don't use mixed variables. But Goldstein has this mixed S' sitting there on page 158 [correct]. Now it is true that = N so I can say S' = N - ω x L which is G (5-35). or (dL/dt)S' = N - ω x L I have always said that the rotating observer in Frame S' can see both L and L' and can work with either one. He can compute (dL/dt)S' for example. So I guess it is OK to have a mixed vector in a solution. [yes] Let's pause to ask how L and L' are related, and also to consider the "reference point". In this section I am encountering MANY notational issues, so this is a good place to try and resolve them. I will summarize the conclusions later. G does not deal with reference points and I think he avoids that because he co-sites the two origins (why not do that simplifying action! ) which causes b = c = c' = 0. For a spinning top, perhaps the contact point is the common origin, it is the rigid body "fixed point", I am not sure of that. Even in my general case, c and c' represent "the same point" in space as my figure (1.9.1) shows. So how are L and L' related ? From (1..9.3) I can say L(c)S ≡ (r-c) x mvS = (r-c) x mv = (r-c) x p ≡ L(c) = L L'(c')S' ≡ (r'-c') x mv'S' = (r'-c') x mv' = (r'-c') x p' ≡ L'(c') . = L' (1.9.3) I don't attempt to show how they are related at this point in Section 1.9. But eventually I have L(c) = L'(c') + m(r'-c') x [ (ω x r') + S] (11.2.14) If I co-locate both reference points at a common origin this becomes L(0) = L'(0') + m r' x (ω x r') (11.2.14) or L = L' + m r' x (ω x r') ** So yes, these really are different vectors. Now before continuing, suppose m is a particle at rest in a rigid body in Frame S'. In fact all particles of the rigid body are at rest, so v' = 0 and this means p' = 0 and then L' = 0. In this case the above becomes L = m r' x (ω x r') Question: How do you write (dL/dt)S' = N - ω x L in components? Write first as S' = N - ω x L Then in Frame S' I would write (S')'i = N'i - εijk(ω)'j(L)'k Now the true torque I think is the same in both frames, just as with true force F. For example, the force on a particle would be mg in either frame, but in Frame S you have m(g)i and in Frame S' have m(g)'i but as a vector equation you say F = F'. So N'i in the above equation is the true torque whatever it might be. But now we come to the inertial business. I will now go brush up on that section of G. Chapter 5.1. So holding here for the moment. // Got it, very simple. On page 144 G shows that with everything in the body or Frame S' frame, we can write in my Frame S' notation. ( I just wrote this above so repeating) L = m r' x (ω x r') where the particle is at rest in Frame S' (as are all points in the body). This agrees with ** above because in the body, L' = 0 since L' = r' x p' and p' = 0 since v' = 0 etc etc. We then write out the cross product r' x (ω x r') = (r'r')ω - (r'ω)r' = r'2ω - (r'ω)r' so L = m [ r'2ω - (r'ω)r'] Now we take the Frame S' components to get (L )'i = m [ r'2(ω)'i - (r'ω)(r')i] = m [ r'2(ω)'i -Σj(r')j(ω)'j(r')i] = m [Σj r'2(ω)'jδij -Σj(r')j(ω)'j(r')i] = Σj [m r'2δij - m(r')i(r')j](ω)'j = Σj I'ij (ω)'j I'ij ≡ m r'2δij - m(r')i(r')j and thus appears the "inertia tensor I' ". For example (L )'1 = Σj I'1j (ω)'j = I'11 (ω)'1 + I'12 (ω)'2 + I'13 (ω)'3 = m( r'2 - m(r')12) (ω)'1 + ( - m(r')1(r')2)(ω)'2 + ( - m(r')1(r')3)(ω)'3 = m( r'2 - mx'2)(ω)'x + ( - mx'y')(ω)'y + ( - mx'z')(ω)'z and this is Gold p 144 (5-45) if you sum over all the particles in the rigid body. So then here is what we have (L )'i = Σj I'ij (ω)'j (S')'i = N'i - εijk(ω)'j(L)'k Note that (S')'i = (∂S'L)'i = [(∂L/∂t)S']'i so restate the above as (L)'i = Σj I'ij (ω)'j [(∂L/∂t)S']'i = N'i - εijk(ω)'j(L)'k Notice that (∂L/∂t)S' is indeed a mixed object! L is the Frame S vector, but we take time derivative in Frame S'. In vector notation the above read L = I' ω (∂L/∂t)S' = N - ω x L N = N' **** Question: Why do I refer to I' instead of I? Well, its definition was this I'ij ≡ m r'2δij - m(r')i(r')j Note that this thing is symmetric in ij ! and this seems to be an object in Frame S' . I claim this really is a tensor in S'-space not S-space. So I continue to use I' . Later this is vindicated when I say (I' ω )'k = I 'kn (ω)'n . This is another topic I should write up. You can only do this is I' is a tensor in Frame S'. Question: Is there a vector ω' and if so, what is it? In my Fig 1 or other figures you see ω and b as vectors. The drawing shows r, r' and v,v' because these are things related to the two Frames. But b and ω are vectors in Fig 1 which don't relate to either Frame. In some sense we could argue b' = -b and ω' = -ω if you make Frame S' fixed to the paper, as I have shown in a few places. I might want to say something about that. These two vectors are very different from the other vectors r,v,a,L . Now write **** in components [(∂L/∂t)S']'i = Ni - εijk (ω)'j (L)'k for example [(∂L/∂t)S']'1 = N1 - ε1jk (ω)'j (L)'k = N1 - (ω)'2 (L)'3 + (ω)'3 (L)'2 or [(∂L/∂t)S']'x = Nx - (ω)'y (L)'z + (ω)'z (L)'z or [(∂L/∂t)S']'x + (ω)'y (L)'z - (ω)'z (L)'z = Nx matches G (5-35) G's notation is MUCH more compact. In our swap notation the above would read [(∂L'/∂t)S]x + (ω)y (L')z - (ω)z (L')z = Nx [(∂L'/∂t)S]x + ωy L'z - ωz L'z = Nx One must keep in mind that the object [(∂L'/∂t)S]x is "unnatural" in either notation. So G is using neither our non-swap nor our swap notation! He is in between. He uses L for the fixed frame L and he uses just (∂L/∂t) for the "cross derivative" thing. Then he just has (∂L/∂t) + ωy Lz - ωz Lz = Nx (5-35) Again, there is nothing wrong with having an object (∂L/∂t)S' sitting in an equation and measuring it and working with it as he does. Now go his next step : we had L = I' ω (∂L/∂t)S' = N - ω x L N = N' **** so insert L from the first to get (∂L/∂t)S' = N - ω x (I' ω ) Now write this out in components in Frame S' [(∂L/∂t)S']'i = Ni - εijk (ω)'j (I' ω )'k (I' ω )'k = I 'kn (ω)'n = Ni - εijk (ω)'j I'kn (ω)'n For example [(∂L/∂t)S']'1 = N1 - ε1jk (ω)'j I'kn (ω)'n = N1 - ε123 (ω)'2 I'3n (ω)'n - ε132 (ω)'3 I'2n (ω)'n = N1 - (ω)'2 I'3n (ω)'n + (ω)'3 I'2n (ω)'n = N1 - (ω)'2 [I'31 (ω)'1 + I'32 (ω)'2 + I'33 (ω)'3] + (ω)'3 [I'21 (ω)'1 + I'22 (ω)'2 + I'23 (ω)'3] = N1 - (ω)'2 [I'31 (ω)'1 + I'32 (ω)'2 + I'33 (ω)'3] + (ω)'3 [I'21 (ω)'1 + I'22 (ω)'2 + I'23 (ω)'3] This certainly is a give mess. Now G adds a new wrinkle. Since I'ij is symmetric, it can be diagonalized, so from now on let's assume that we select the body frame to align with the principle axes which make I be diagonal. Then the above becomes [(∂L/∂t)S']'i = Ni - εijk (ω)'j I'kn (ω)'n = Ni - εijk (ω)'j I'nδkn (ω)'n = Ni - εijk (ω)'j I'k(ω)'k ***** Then for example [(∂L/∂t)S']'1 = N1 - (ω)'2 I'3(ω)'3 + (ω)'3 I'2(ω)'2 = N1 - (ω)'2 (ω)'3 [ I'3 - I'2] Now meanwhile we have L = I' ω (∂L/∂t)S' = I' // recall is the same in both frames Then [(∂L/∂t)S']'i = I'ii ()'i = I'i ()'i // no sum on i Then we have from ***** I'i ()'i = Ni - εijk (ω)'j I'k(ω)'k or I'i ()'i + εijk (ω)'j I'k(ω)'k = Ni For example I'1 ()'1 + (ω)'2 I'3(ω)'3 - (ω)'3 I'2(ω)'2 = N1 or I'1 ()'1 + (ω)'2(ω)'3 [ I'3 - I'2] = N1 or I'1 ()'x + (ω)'y(ω)'z [ I'3 - I'2] = Nx or I'1 ()'x - (ω)'y(ω)'z [ I'2 - I'3] = Nx // matches G page 158 A and then G is off and running. We no longer have any ( )S' mixed objects in the equation. So I think I have figured out what G's compact notation means in that section. Question: Have I overlooked a simple way to express fictional torques? I have this result in frames doc N(c) = (c) N'(c')eff = '(c') (11.3.4) N'(c')eff = N(c) + N'(c')fict // defines N'(c')fict (11.3.5) N'(c')fict = '(c') - (c) . (11.3.7) It is admittedly messy when written out. Then above I have these equations, N = (dL/dt)S (dL/dt)S = (dL/dt)S' + ω x L (1) (0) = (0)S' + ω x L(0) I can write the last as (0)S' - (0) = - ω x L(0) which compare with '(c') - (c) = N'(c')fict The left hand sides are not the same !!! (0)S' is a cross -object, '(c') is a natural.