goldstein v equation attempt 2 on 7_12_12

This is the Title PhL 3.26.05 Note that page numbering is turned on in this template. [ This entire document is an example of : garbage in, garbage out ] (a) A Pre-calculation. Recall from above that vbody(t) = R(t) v'body(t) [ what is vbody ? ] [ OK, let's imagine that v'body is the velocity of our ant as measured in the rotating frame S'. Later I will call this vector v'S' = v' = (dr'/dt)S' with components (v')'i in the rotating frame. What then is vbody in the above equation? It cannot be vS= v because we know the equation is not right for that. And v'S = (dr'/dt)S which from the G Rule we know is v' – ω x r' and that doesn't seem to match our equation either. The fourth is vS' = (dr/dt)S' and that seems pretty distant. So I think vbody is none of the actual four velocities in the problem. It is just an actively rotated version of v' . So why do a calculation on something that you don't know what is ? I think I am just continuing my erroneous ways. Recall how we keep getting that "extra factor of R" in the G Rule which I though you had to then set to R = 1] Then dt vbody(t) = (dtR) v'body(t) + R(t) dtv'body(t) or abody(t) = (dtR) v'body(t) + R(t) a'body(t) . And we must carefully take note that we don't just have abody(t) = R(t) a'body(t). In fact abody is not a "vector" due to the above transformation rule with its extra piece. Using our usual method, [(dtR) v'body(t)]i = [(dtexp(-iφ(t)Jz)) v'body(t)]i = {[-iφ'(t) Jz] [(exp(-iφ(t)Jz) ) v'body(t)] }i // i (Jk)ij = kij = {[-iφ'(t) Jz] [Rv'body(t)] }i = {[-iφ'(t) Jz] [vbody(t)] }i = -iφ'(t) (Jz)ij [vbody(t)]j = - φ'(t) ε3ij [vbody(t)]j = - ω3 ε3ij [vbody(t)]j = εi3j ω3[vbody(t)]j = εikj ωk[vbody(t)]j So we have shown that [(dtR) v'body(t)]i = εikj ωk[vbody(t)]j [(dtR) v'body(t)] = ω x vbody(t) Therefore, going way back we then have abody(t) = (dtR) v'body(t) + R(t) a'body(t) = R(t) a'body(t) + ω x vbody(t) [ and here is that fully factor of R mentioned above ] and so we have restarted our "extra term". (b) The Goldstein equation for v . That equation says this dv = dφ x v + dvbody dv/dt = ω x v + dvbody/dt [ OK, if correctly interpreted: dv/dt = ∂Sv and dvbody/dt = ∂S'v ] a = ω x v + abody Installing our pre-calculation result, this says a = ω x v + R(t) a'body(t) + ω x vbody(t) We may regard our first equation as a first order PDE in variable v(t) dv/dt – ω x v = R(t) a'body(t) + ω x vbody(t) We think we know everything on the RHS, but let's make sure. For the first term we get, a'body = dv'body /dt = dt[' ' + r' ' '] = five terms = ' ' + ' ' ' + ' ' ' + r'' ' + r' ' [- ' '] = (' - r''2) ' + (r'' + 2 ' ' ) ' and then (this removes the primes from the unit vectors) R(t) a'body(t) = (' - r''2) + (r'' + 2 ' ' ) For the second term we have, vbody = ' + r' ' // from Section 3 ω x vbody(t) = [ω] x [' + r' ' ] = ω' - ω r' ' The combined RHS is then (' - r''2) + (r'' + 2 ' ' ) + ω ' - ω r' ' = (' - r''2- ω r' ') + (r'' + 2 ' ' + ω ') Therefore, our Goldstein equation is this dv/dt – ω x v = (' - r''2- ω r' ') + (r'' + 2 ' ' + ω ') Now we want to get the components for the terms on the LHS: v = vr + vθ ω = ω => ω x v = [ω] x [vr + vθ] = ω vr – ω vθ dv/dt = r + vr + θ + vθ[- ] = (r - vθ) + (vr + θ) So here is our grand equation (r - vθ) + (vr + θ) – ω vr + ω vθ = (' - r''2- ω r' ') + (r'' + 2 ' ' + ω ') Separating by components, we get these two equations (r - vθ) + ω vθ = (' - r''2- ω r' ') (vr + θ) – ω vr = (r'' + 2 ' ' + ω ') or (r - vθ)+ ω vθ = (' - r''2- ω r' ') (vr + θ)– ω vr = (r'' + 2 ' ' + ω ') or r - vθ + ω vθ = ' - r''2- ω r' ' vr + θ – ω vr = r'' + 2 ' ' + ω ' or r - (-ω) vθ = ' - r''2 – ω r' ' θ + ( – ω)vr = r'' + 2 ' ' + ω ' And now the two equations are not decoupled! Things are much more complicated than before. No doubt we could solve these equations somehow, but I am not going to try. My main interest is to verify that these equations are true!! From the previous section we found that v(t) = vr + vθ where vr = vθ = r So we can say vr = r = vθ = r θ = r + Then our two coupled equations above become r - (-ω) vθ = ' - r''2 – ω r' ' θ + ( – ω)vr = r'' + 2 ' ' + ω ' or - (-ω) r = ' - r''2 – ω r' ' r + + ( – ω) = r'' + 2 ' ' + ω ' Let's whittle away on the first equation: - (-ω) r = ' - r''2 – ω r' ' - (-ω) r = - r'2 – ω r ' // since r = r' - (-ω) r = - r'2 – ω r ' - ' r = - r'2 – ω r ' - ' = - '2 – ω ' - = - ' – ω = ' + ω and this is true, so we have verified our first equation ! Now let's whittle away at the second equation r + + ( – ω) = r'' + 2 ' ' + ω ' r + + ( – ω) = r' + 2 ' + ω // since r = r' r + + – ω = r' + 2 ' + ω r + 2 – ω = r' + 2 ' + ω r ( –' ) + 2 ( - ') = 2ω r + 2 ω = 2ω r = 0 Close but something is still wrong.