evaluating vector equations in different frames

Question 7: Rotating Vectors Questions PhL 3.1.17 Consider equation (1.5.8) which says, da = dφ x a (1.5.8) 1. Is this a vector equation which can be evaluated in different Frames? I am pretty sure the answer is YES. In my graphical derivation, you could view my picture from any direction, or from any frame of reference. 2. What about this equation? a = (dv/dt) = ω x v where ω ≡ dφ/dt . (1.6.1) Again I draw a picture in (1.6.2) showing cone motion. I draw this say in Frame S. Then it says a = [(dv/dt)S] = ω x v It still seems you could compute ai = [(dv/dt)S]i = [ω x v]i a'i = [(dv/dt)S]'i = [ω x v]'i The vector a and the vector ωxv certainly have components in any frame you want. So I guess again the answer is YES. Notice that we have all kinematic vectors here. Another way to answer: taking components in Frame S" just means dotting with e"n . You cannot stop someone from dotting a vector equation with a vector! 3. What about this equation? (de'n/dt)S = ω x e'n . (1.7.1) or better c ≡ (de'n)S = ω x e'n dt Now we do not have all Kinematic Vectors in this equation, we have a mix. So more suspicious. In Frame S there is some vector c and there exists some vector quantity ω x e'n dt . You can dot both sides into either ei or e'i and you don't get zero. So ci = [(de'n)S]i = [ω x e'n]i dt c'i = [(de'n)S]'i = [ω x e'n]'i dt In particular, I have no reason to say that [(de'n)S]'i = c e'i = 0 . Conclusion: You can evaluate (de'n/dt)S = ω x e'n in either Frame S or in Frame S'. 4. What about these equations? en(φ) = R(φ)e'n dc ≡ (den(φ))S' ≡ en(φ+dφ) - en(φ) = R(φ+dφ)e'n - R(φ)e'n = [ R(φ+dφ)e - R(φ)]e'n Maybe use time t as a parameter en(t) = R(t)e'n dc ≡ (den(t))S' ≡ en(t+dt) - en(t) = R(t+dt)e'n - R(t)e'n = [ R(t+dt) - R(t)]e'n // assumes Frame S' If you wanted to be general, you should have written en(t) = R(t)e'n(t) dc ≡ (den(t))S' ≡ en(t+dt) - en(t) = R(t+dt)e'n(t+dt) - R(t)e'n Then e'n(t+dt) = e'n(t) + [∂te'n(t)]dt R(t+dt) = R(t) + ∂tR(t)dt and then dc = (den(t))S' ≡ en(t+dt) - en(t) = [R(t+dt)] [e'n(t+dt)] - R(t)e'n(t) = [R(t) + ∂tR(t)dt] [e'n(t) + [∂te'n(t)]dt] - R(t)e'n(t) = [ R(t) e'n(t) - R(t)e'n(t)] + [R(t)[∂te'n(t)] + [∂tR(t)]e'n(t)] dt + O(t2) ≈ [ R(t)(∂te'n(t)) + (∂tR(t))e'n(t) ] dt and there are two terms! In general you do NOT get dc = [ R(t+dt) - R(t)]e'n . Let's evaluate the above equation in both frames dc = (den(t))S' = [ R(t)∂te'n(t)) + (∂tR(t))e'n(t) ] dt [dc] =[(den(t))S'] = R(t)[∂te'n(t)]dt + [∂tR(t)] e'n(t) dt [dc]i =[(den(t))S']i = R(t)ij[∂te'n(t)]jdt + [∂tR(t)]ij (e'n(t))j dt [dc]'i =[(den(t))S']'i = R'(t)ij[∂te'n(t)]'jdt + [∂tR(t)]'ij (e'n(t))'j dt Is it OK to evaluate the above equations in the two frames as I have done? I think the answer is YES, because the equation is valid in both frames. Now back to this: Conclusion: dc ≡ (den(t))S' ≡ en(t+dt) - en(t) valid in both Frame S and Frame S' dc ≡ (den(t))S' = [ R(t+dt) - R(t)]e'n valid ONLY in Frame S' 4a. Can you "evaluate" the second equation above in both Frames? NO! Yes it is a vector equation, but a vector equation is just a shorthand for an equation in components. If you say that a certain vector equation is not valid in Frame S, what you mean is that when that vector equation is written in Frame S components, it is invalid. So this is certainly something to ponder. Correction to the above. The equation (den(t))S' = [ R(t+dt) - R(t)]e'n has a Frame S' descriptor on the derivative. The equation says vector = vector. You can dot this equation with any vector you like. Thus, you can evaluate the equation in Frame S or in Frame S'. It is wrong to say this equation is valid ONLY in Frame S' . It is true that the equation (den(t))X = [ R(t+dt) - R(t)]e'n is valid only when X = S', but once you X = S', you can evaluate the equation in any frame you like.