rotation rule debug v1

Rotation rule debug PhL 7.5.12 Status: In temp3 v4 I finally got the "right angle" involved and my dΩ complicated angle went away. But the L paradox still remains, so I will ponder them here a bit. I am concerned that even my fancy [V]S notation is insufficient for what is needed here. Example 1. Go into the Marion prime context for a while, page 341. Space is S' and body is S in this system. The vector shown in the picture as r is the one of interest in dr = dθ x r. It is a vector with its tail on the body origin. All three objects in this equation can be expressed in terms of S or S' coordinates. So I could write dr = dθ x r [dr]S = [dθ]S x [r]S [dr]S' = [dθ]S' x [r]S' I think I understand what these last two equations mean. And this reinforces the idea that you can take a vector rotation like the first above and write it out in any coordinate system you want. Notice that the vector r' appearing in the picture becomes r if we take out the origin shift, so don't worry now about r'. Example 2. How let's try this not for position r, but for velocity v. I want to write dv = dθ x v [dv]S = [dθ]S x [v]S [dv]S' = [dθ]S' x [v]S' But somehow things seem less clear in this case. I could draw the vector v with tail on the origin if I liked, so it looks like r in the Marion picture. It is the orientation of v that counts, not so much where I put its tail. Except I need the tail at the origin to understand dv = dθ x v with a cone picture. But I guess the tip of the cone could be at any location, not just the origin, and the picture looks the same. We are just translating the cone around, the equation dv = dθ x v stays the same. So again, we can take each of the three vectors and write them in either frame S or S' and then the two lower equations become meaningful. But there is a problem, Houston. I intend [v]S' to mean the vector v just expressed in frame S' coordinates. But if the frame S is rotating, you might mean by [v]S' the velocity in frame S' that arises from the two usual sources. Thus, the notation [v]S' is ambiguous, and that spells trouble. This ambiguity did no arise with the vector r, but it does with the vector v. Example 3. Same velocity example, but I want to get the primes back the way I have been using them in all my temp3 vn documents. So space = S body = S' dv = dθ x v [dv]S = [dθ]S x [v]S [dv]S' = [dθ]S' x [v]S' Now our ambiguity is with [v]S . I might denote the simple coordinate-change version as [v]Sω=0 Then the above equations become dv = dθ x v [dv]S = [dθ]S x [v]Sω=0 [dv]S' = [dθ]S' x [v]S' But let's first revert back to r with this notation and see what it says dr = dθ x r [dr]S = [dθ]S x [r]Sω=0 [dr]S' = [dθ]S' x [r]S' Here we don't need ω=0 on [r]Sω=0 because the fact that the body frame is rotating or even translating does not instantaneously affect r. Of course we still have that [r]S ≠ [r]S' . Now back to velocity v again, where I add yet more qualifiers, dvjustω = dθ x v [dv]Sjustω = [dθ]S x [v]Sω=0 [dv]S' = [dθ]S' x [v]S' Now I need some equations for the "other part". dvother [dvother]S [dvother]S' This represents a change in v because v is moving in the body frame S'. The last two objects are as usual not equal because a vector in different systems has different components. Then we have the additive situation dv = dvother + dvjustω = dvother + dθ x v [dv]S = [dvother]S + [dvjustω]S = [dvother]S + [dθ]S x [v]Sω=0 [dv]S' = [dvother]S' + [dvjustω]S' = [dvother]S' + [dθ]S' x [v]S' where no qualifier on something means it is the total thing, such as with [dv]S. I suspect the most important of these equation is this one [dv]S = [dvother]S + [dθ]S x [v]Sω=0 Goldstein might have written body instead of other. But OK. Now divide by dt [dv]S/dt = [dvother]S/dt + [dθ]S/dt x [v]Sω=0 [dv]S/dt = [dvother]S/dt + [ω]S x [v]Sω=0 Question: Why does "divide by dt" give a different result from "differentiate wrt t" ? Above you see the divide by dt result. Here I will do the differentiation result. Repeat the three equations above but for r dr = drother + drjustω = drother + dθ x r [dr]S = [drother]S + [drjustω]S = [drother]S + [dθ]S x [r]Sω=0 [dr]S' = [drother]S' + [drjustω]S' = [drother]S' + [dθ]S' x [r]S' Then do the divide by dt thing, dr/dt = drother/dt + drjustω/dt = drother/dt + dθ/dt x r [dr]S/dt = [drother]S/dt + [drjustω]S/dt = [drother]S/dt + [dθ]S/dt x [r]Sω=0 [dr]S'/dt = [drother]S'/dt + [drjustω]S'/dt = [drother]S'/dt + [dθ]S'/dt x [r]S' which says v = drother/dt + drjustω/dt = drother/dt + ω x r [v]S = [drother]S/dt + [drjustω]S/dt = [drother]S/dt + [ω]S x [r]Sω=0 [v]S' = [drother]S'/dt + [drjustω]S'/dt = [drother]S'/dt +[ω]S' x [r]S' and the key result here is [v]S = [drother]S/dt + [ω]S x [r]Sω=0 or [v]S = [vbody]S + [ω]S x [r]S Now if we differentiate this wrt time, we get a term d[v]S/dt = d[vbody]S/dt + [ω]S x (d[r]S/dt) + []S x [r]Sω=0 or d[v]S/dt = d[vbody]S/dt + [ω]S x [v]S + []S x [r]Sω=0 which compare to the divide by dt result [dv]S/dt = [dvbody]S/dt + [ω]S x [v]Sω=0 If these are both valid, then it must be true that [ω]S x [v]S + []S x [r]Sω=0 = [ω]S x [v]Sω=0 or [ω]S x {[v]S –[v]Sω=0 } = – []S x [r]S Everything has components in frame S. Rewrite as [dv]S/dt = [dvbody]S/dt + [ω]S x [v]Sω=0 Maybe go back to the uncommitted version, dv = dvbody + dθ x v dv/dt = dvbody/dt + dθ/dt x v a = (dv/dt) = (dvbody/dt ) + ω x v Is this true? I guess so. Now can I generate the acceleration paradox? Well write as [a]S = [dvbody]S/dt + [ω]S x [v]Sω=0 Suppose the velocity is at rest in the body frame. Then the first term should be 0 and then [a]S = [dv]S/dt = [ω]S x [v]Sω=0 But I am expecting some sort of term and I don't see it. How do I express the acceleration in the body frame? Maybe it is this [abody]S' = [dvbody]S'/dt Now maybe I can write the following vector equation, dv = dvjustω + dvω=0 [dv]S = [dvjustω]S + [dvω=0]S [dv]S' = [dvjustω]S' + [dvω=0]S'