tide potential calculationv4

Tide potential Calculation PhL 1.18.17 We seek a potential V(x,y) which solves this equation (no minus sign) V(x,y) = Ftid = - [(GM2m)/r2] + (mM1G/d03) ( 2x - y ) // r = (8.8.63) The solution is the following (by inspection) V(x,y) = (GM2m/r) + (GM1m/d03)( x2 - y2/2) . // (1/r) = -(1/r2), r > 0 (8.8.64) Of interest are surfaces on which V is a constant, so (M2/r) + (M1/d03)( x2 - y2/2) = k (8.8.65) M22 = r2[ k -(M1/d03)( x2 - y2/2) ]2 = k2r2 [1 - (M1/kd03)( x2 - y2/2)]2 ≈ k2r2 [1 - (M1/kd03)( 2x2 - y2)] (M2/k)2 = r2 [1 -ε(2x2 - y2)] ε ≡ (M1/kd03) = (x2+y2) (1 - 2εx2 + εy2) dim(M1/k) = L = r2 (1 - 2εr2cos2θ + εr2sin2θ) dim(ε) = L-2 (8.8.66) Maple file "tides by potential method" has code and graphic: (8.8.67) This is a quadratic equation in r2 which has this solution (after some work), to lowest order in ε r(θ) ≈ (M2/k)[ 1 - (1/8)ε(sin2θ - 2cos2θ)(M2/k)2] // see v3 of current doc (8.8.68) As was done in **, we require that the average value of r(θ) over θ be R2 . We find <r(θ)> = c[ 1 -(1/8)ε{(1/2)- 2(1/2)}c2 ] = c[ 1 - (1/16)c2 ε {1-2} ] = (M2/k)[ 1+ (1/16)(M2/k)2ε ] = R2 (8.8.69) Solving for (M2/k) gives (M2/k) = R2(1-αεR22) (M2/k) ≈ R2 (8.8.70) so the locus of the cubic is this R22(1-αεR22)2 = (x2+y2) [1 -ε(2x2 - y2)] (8.8.71) ε = (M1/kd03) ≈ (M1/[M2/R2]d03) = (M1/M2)R2/d03 (8.8.71) At the right edge of the quartic we set y = 0 and find that x ≈ R2 [ 1 + (15/16)(M1/M2)(R2/d0)3] (8.8.72) which says that the high tide is |Δ|high = (15/16)(M1/M2)R24/d03 At the top edge of the quartic we set x = 0 and find that |Δ|low = (9/16)(M1/M2)R24/d03 (8.8.72) The height difference is then H = |Δ|high + |Δ|high = (M1/M2)R24/d03 { (15/16)+ (9/16) } = (3/2)(M1/M2)R24/d03 (8.8.72) In (8.8.33) we have H = 2a so the result is then a = (3/4)(M1/M2)R24/d03 (8.8.73) in agreement with (8.8.41) . What happens now at the right edge of the quadric where y = 0: R22(1-αεR22)2 = (x2) [1 -ε(2x2)] = x2(1 - 2εx2) ≈ x2(1 - 2εR22) Then it seems that x = R2(1-αεR22) * (1 - 2εR22)-1/2 ≈ R2(1-αεR22) * (1+ εR22) ≈ R2 [ 1 + εR22(1-α)] ≈ R2 [ 1 + (M1/M2)R2/d03R22(15/16)] ≈ R2 [ 1 + (M1/M2)(R2/d0)3(15/16)] (8.8.72) So the high tide is then hhigh = (M1/M2)(R2/d0)3R2(15/16) Now what happens at the top edge of the quadric where x = 0 R22(1-αεR22)2 = (x2) [1 -ε(2x2)] = y2(1+εy2) ≈ y2(1 + εR22) It seems then that y = R2(1-αεR22) * (1 + εR22)-1/2 ≈ R2(1-αεR22) * (1-[1/2]εR22) ≈ R2(1 -αεR22 -[1/2]εR22) ≈ R2(1 -εR22(1/16 + 8/16)) ≈ R2(1 -εR22(9/16)) ≈ R2(1 -(M1/M2)R2/d03R22(9/16)) ≈ R2(1 -(M1/M2)(R2/d0)3(9/16)) (8.8.72) So the low tide is then (abs val) hlow = (M1/M2)(R2/d0)3R2(9/16) We then add these to get "2a" = (M1/M2)(R2/d0)3R2 [ (15/16)+ (9/16) ] = (M1/M2)(R2/d0)3R2 [ (24/16) ] = (M1/M2)(R2/d0)3R2 [ (3/2) ] (8.8.72) Then "a" = (3/4) (M1/M2)(R2/d0)3R2 // it works!!!! (8.8.73) This locus is close to the locus of a circle of some radius R R2 = (x2+y2) or R2 = (x2+y2) (1 - 2εx2+εy2) ε ≡ (M1/kd03) R = M2/k = constant Since d0 is large, ε is small, and R2 ≈ (x2+y2) so 1/k ≈ R/M2. Then (M2/k)2 ≈ (x2+y2) (1 - 2εx2+εy2) ε = (M1/M2)/d03 This is a quartic curve which resembles an ellipse. For example, for ε = .05 and M2/k = 1: where the blue curve is a circle of radius 1. Gm(M1/d03) (x2/2) - Gm (M1/d03) (y2/2) } Start with Ftid = mM1G ( 0/d02 – /d2) = mM1G ( d0/d03 – d/d3) and here is our picture In Cartesian coordinates we can write dx = d0 + R2 cosθ dy = R2sinθ d = (d0 + R2 cosθ) + R2sinθ d0 = d0 Then, Ftid = mM1G ( d0/d03 – [ (d0 + R2 cosθ) + R2sinθ ]/d3) now use d2 = R22 + d02 - 2d0R0cos(π-θ) = R22 + d02 +2d0R0coθ = d02[ 1 + (R2/d0)2 + 2 (R2/d0)cosθ ] ≈ d02[ 1 + 2 (R2/d0)cosθ ] to get d-3 ≈ d0-3 [ 1 + 2(R2/d0)cosθ]-3/2 ≈ d0-3 [ 1 - 3(R2/d0)cosθ ] . Then, Ftid = mM1G ( d0/d03 – d/d3) ≈ mM1G ( d0/d03 – [ (d0 + R2 cosθ) + R2sinθ ] ) d0-3 [ 1 - 3(R2/d0)cosθ ] ) = (mM1G/d02) ( – [ (1 + (R2/d0) cosθ) + (R2/d0)sinθ ] [ 1 - 3(R2/d0)cosθ ] ) = (mM1G/d02) ( – - (R2/d0) cosθ) - (R2/d0)sinθ + 3(R2/d0)cosθ ) + O((R2/d0)2) ≈ (mM1G/d02) ( - (R2/d0) cosθ) - (R2/d0)sinθ + 3(R2/d0)cosθ ) = (mM1G/d02)(R2/d0) ( - cosθ - sinθ + 3cosθ ) = (mM1G/d02)(R2/d0) ( 2cosθ - sinθ ) Now check this at the compass points Ftid(0) = (mM1G/d02)(R2/d0) 2 point B to the right Ftid(π) = - (mM1G/d02)(R2/d0) 2 point A to the left Ftid(π/2) = - (mM1G/d02)(R2/d0) down at top Ftid(-π/2) = +(mM1G/d02)(R2/d0) up at bottom Compass agrees. I like the general formula: Ftid(θ) ≈ (mM1G/d02)(R2/d0) ( 2cosθ - sinθ ) How might I plot this? x = rcosθ y = rsinθ Ftid(r,θ) ≈ (mM1G/d02)(1/d0) ( 2rcosθ - rsinθ ) = (mM1G/d03) ( 2x- y ) Very good. You want to think of R2 = r as a variable. What is it in polar coordinates? ( r = R2) Ftid(θ) ≈ (mM1G/d02)(r/d0) ( 2cosθ - sinθ ) = (mM1G/d02)(r/d0)( 2cosθ [cosθ – sinθ ] - sinθ [sinθ + cosθ ]) = (mM1G/d02)(r/d0)[ (2cos2θ-sin2θ) + (-2sinθcosθ - sinθcosθ)] = (mM1G/d02)(r/d0)[ (3cos2θ-1) + (-3sinθcosθ) ] = (mM1G/d02)(r/d0)[ (3{1/2)(1+cos2θ)}-1) + (-(3/2)sin2θ) ] = (1/2)(mM1G/d02)(r/d0)[ (3{(1+cos2θ)}-2) + (-3sin2θ) ] = (1/2)(mM1G/d02)(r/d0)[ (1 + 3cos2θ) - 3sin2θ ] = (3/2)(mM1G/d02)(r/d0)[ ([1/3] + cos2θ) - sin2θ ] This says piece = - (3/2)(mM1G/d02)(r/d0) sin2θ horizontal (5) Butikov piece = (3/2)(mM1G/d02)(r/d0) ( cos2θ +1/3) vertical (6) Butikov So far so good! Can this force be represented by a potential? In electrostatics we have E = -φ div E = -2φ = 0 region of no charge curlE = - x (φ) = 0 So E must have zero divergence and zero curl in order to be rep by a potential. So consider div Ftid = div( 2x- y ) = ∂2x(2x) +∂2y(-y) = 0 + 0 = 0 curl Ftid = [∂xFy- ∂yFx] + cyclic [ 0 - 0] + 0 + 0 = 0 Therefore you NOT represent Ftid by a potential. WRONG!!! First of all, div E = -2φ = "ρ" and ≠ 0. Second of all, conservativity only involves the curl! So the tidal force is a conservative force. Here then is the force for which we seek a potential Fg2 = -[(GM2m)/r3] r Earth gravity ok Ftid = (mM1G/d03) ( 2x - y ) dim(F) = M L/T2 ok dim(GMm/r2) = M L/T2 dim(GMm) = M L3/T2 Now consider dim(V) = M L2/T2 V(x,y) = -[(GM2m)/r3] r + (mM1G/d03) ( 2x - y ) ok = -[(GM2m)/r3]( x + y ) + (mM1G/d03) ( 2x - y ) = [ -(GM2m)/r3) + (mM1G/d03)2 ] x + [ -(GM2m)/r3) - (mM1G/d03) ] y = Gm { [ -M2/r3 + 2(M1/d03) ] x + [ -M2/r3 - (M1/d03) ] y } ok Now suppose you try as ansatz, WRONG!!!! V(x,y) = k[A2/x2 + B2/y2 - 1] ok // an arbitrary ellipse centered at Earth center V(x,y) = -2kA2/x3 -2kB2/y3 ok // dim(k) = M L2/T2 This does not seem to work, why is that? You would need now to have -2kA2/x3 = Gm [ -M2/r3 + 2(M1/d03) ] x -2kB2/y3 = Gm [ -M2/r3 - (M1/d03) ] y or -2kA2 = Gm [ -M2/r3 + 2 (M1/d03) ] x4 -2kB2 = Gm [ -M2/r3 - (M1/d03) ] y4 r2 = x2+y2 Just check dimensions please. dim(kA2) = M L4/T2 dim(GmMx4/r3) = M L3/T2 * L = M L4/T2 so dimensions are OK. I do know that r << d0 x << d0 y << d0 but this does not seem to help at all. I am stumped, this will take another day or two of burned time to figure out. It is so simple, what am I doing wrong here? Paradox #1. Suppose I let M1→ 0 so the moon goes away slowly. The last two equations then become -2kA2 = Gm [ -M2/r3 ] x4 -2kB2 = Gm [ -M2/r3 ] y4 r2 = x2+y2 I expect the potential surface to be a circle rather than an ellipse since there are now no tidal forces, and that means I expect to have A = B. But instead I have A ≠B. This should give A = B = R2 somehow. Maybe I have to do this in 3D space and not fake 2D space? Start Over! Use spherical coordinates r,θ,φ centered on the Earth with z to the right. Ftid = mM1G { d0/d03 – d/d3 } d0 = d0 C = r = rsinθcosφ + rsinθsinφ + rcosθ = x Moon center = cm = -d0 d = C - cm = rsinθcosφ + rsinθsinφ + rcosθ + d0 = rsinθcosφ + rsinθsinφ + (rcosθ+d0) d2 = ( rsinθcosφ)2 + ( rsinθsinφ)2 + (rcosθ+d0)2 = r2sin2θ + (d0 + rcosθ)2 = r2sin2θ + d02(1 + [r/d0]cosθ)2 ≈ r2sin2θ + d02(1 + 2 [r/d0]cosθ) = d02 [ [r/d0]2sin2θ + d02(1 + 2 [r/d0]cosθ) ≈ d02(1 + 2 [r/d0]cosθ) // ignore quadratic term This is same as in Section 8.8, no surprise. So still get d-3 ≈ d0-3 [ 1 + 2(R2/d0)cosθ]-3/2 ≈ d0-3 [ 1 - 3(R2/d0)cosθ ] . (8.8.22) Now: Ftid = mM1G { d0/d03 – d/d3 } = mM1G { [d0]/d03 – [ rsinθcosφ + rsinθsinφ + (rcosθ+d0)]/d3 } = mM1G { d02 – [ rsinθcosφ d-3 + rsinθsinφd-3 + (rcosθ+d0)d-3]} = mM1G { – [ rsinθcosφ d-3 + rsinθsinφd-3 + {d02+(rcosθ+d0)d-3} ]} STOP. Why not do this all in Cartesians. Start over Ftid = mM1G { d0/d03 – d/d3 } d0 = d0 C = r = x + y + z Moon center = cm = -d0 d = C - cm = x + y + z + d0 = x + y + (d0+z) d2 = x2 + y2 + (d0+z)2 = d02{ [ 1 + (z/d0) ]2 + (x/d0)2 + (y/d0)2 } ≈ d02{ [ 1 +2 (z/d0) + (z/d0) 2 ] + (x/d0)2 + (y/d0)2 } ≈ d02[ 1 +2 (z/d0)] dropping all quadratic terms d-3 ≈ d0 -3[ 1 +2 (z/d0)] -3/2 ≈ d0 -3( 1- 3(z/d0) ) Now have Ftid = mM1G { d0/d03 – d/d3 } = mM1G { d0/d03 – [ x + y + (d0+z) ]/d3 } = mM1G { /d02 – [ (x/d3) + (y/d3) + ((d0+z)/d3) ] } = mM1G {– (x/d3) – (y/d3) – ((d0+z)/d3 - 1/d02) ] } = – mM1G { (x/d3) + (y/d3) + ((d0+z)/d3 - 1/d02) ] } Now look at each component separately (x/d3) ≈ d0 -3( 1- 3(z/d0) ) x ≈ x/d03 (y/d3) ≈ d0 -3( 1- 3(z/d0) ) y ≈ y/d03 (d0+z)/d3 - 1/d02 = [(d0+z)/d03] ( 1- 3(z/d0) ) - 1/d02 = [(1+(z/d0))/d02] ( 1- 3(z/d0) ) - 1/d02 = { [(1+(z/d0))] ( 1- 3(z/d0) ) - 1 } /d02 = { 1 + (z/d0) - 3(z/d0) - 3(z/d0)2 - 1} /d02 ≈ { - 2(z/d0) } /d02 = -2z/d03 We then end up with Ftid = – mM1G{ (x/d03) + (y/d03) + ( -2z/d03) } = – (mM1G/d03){ x + y - 2z } and this then is my 3D generalization of previous result where I had y = 0. Could make 3D field plot. I also have Fg2 = -[(GM2m)/r3] r The total force acting on a particle m of water is then F = Ftid + Fg2 = – (mM1G/d03){ x + y - 2z }-[(GM2m)/r3] r = mG { (M1/d03 - M2/r3)x + (M1/d03 - M2/r3)y + (-2M1/d03 - M2/r3)z } Now suppose I drop the 1/d03 terms since d0 >> r and this is OK for M1→ 0 as well. then F = Ftid + Fg2 = – (mM1G/d03){ x + y - 2z }-[(GM2m)/r3] r = mG { - M2/r3)x + (- M2/r3)y + ( - M2/r3)z } and only gravity is left. NOW: Here is an ansatz ellipsoid for the potential surface : V(x,y,z) = k[A2/x2 + A2/y2 + C2/z2 ] V = -2k[ (A2/x3) + (A2/y3) + (C2/z3) ] Now I have the same problem as before I think. I have to identify: V = Ftid and then the following must be true: 2k[ (A2/x3) + (A2/y3) + (C2/z3) ] = (mM1G/d03){ x + y - 2z } - [(GM2m)/r3] r = (mM1G/d03){ x + y - 2z } - [(GM2m)/r3] {x + y + z } = mG { (M1/d03 - M2/r3)x + (M1/d03 - M2/r3)y + (-2M1/d03 - M2/r3) } which then requires that 2k(A2/x3) = mG (M1/d03 - M2/r3)x 2k(A2/y3) = mG (M1/d03 - M2/r3)y 2k(C2/z3) = mG (-2M1/d03 - M2/r3) So converting to 3D coordinates has done NOTHING to explain my problem. Paradox 1 Reappears: Suppose I turn off the moon. Then I get 2k(A2/x3) = mG (- M2/r3)x 2k(A2/y3) = mG (- M2/r3)y 2k(C2/z3) = mG (- M2/r3) and this is supposed to give a sphere!!! So I just wasted 2 hours maybe seeing if going from 2D to 3D would fix this paradox. A bad hunch. Let's try a simpler problem! There is no moon, only earth gravity so F = mG { - M2/r3)x + (- M2/r3)y + ( - M2/r3)z } This is supposed to be related to a sphere. x2+ y2+z2 = A2 How is this sphere connected? It is not hard to see in sphericals F = -(mG/r2) V(r) = (mG/r2) So V is constant on a sphere of radius r. But how does that work in Cartesians? V = (mG) The sphere is "on the bottom". So you don't say V(r) = k (x2+y2+z2) You just need to have "V be constant on a sphere". In this case I can integrate the force to get a potential: F = -(mG/r2) V(r) = !Syntax Error, IF dr = -!Syntax Error, I(mG/r2) = mG/r If I try to do that in Cartesians, how does it go? F = mG { (-M2/r3)x + (- M2/r3)y + ( - M2/r3)z } V(r) = ∫path F dr = ∫path { { (-M2/r3)xdx + (-M2/r3)ydy + (- M2/r3)zdz } You can pick any path you want to make the integral be simple. Suppose I try a path from the divergent origin and do it in Cartesian steps. Then V(r) = !Syntax Error, I (-M2/r3)xdx + !Syntax Error, I (-M2/r3)ydy + !Syntax Error, I (-M2/r3)zdz I will need: !Syntax Error, I = // says Maple Somehow it integral is path independent it seems that you get + { - } + { - } = 1/r and there is your hard-won result. Int (∞,0,0) to (x,0,0)] x integrand + Int (x,0,0) to (x,y,0)] y integrand + Int (x,y,0) to (x,y,z)] z integrand giving what I quote just above. So in the sphere case, you just want to show that V = V(r) so it is constant on a sphere. Another way: V(r) = ∫∞00x00 (-M2/r3)xdx + ∫x00xy0 (-M2/r3)ydy + ∫xy0xyz (-M2/r3)zdz The three indefinite integrals are all the same!! So could write V(r) = I1(x,0,0) - I1(∞,0,0) + I1(x,y,0) - I1(x,0,0) + I1(x,y,z) - I1(x,y,0) = I1(x,y,z) - I1(∞,0,0) = I1(x,y,z) = What do you do for an ellipse? What is constant on an ellipse? The string distance. I think I have stated the problem correctly, going back to 2D, V = + F V(x,y) = Gm { [ -M2/r3 + 2(M1/d03) ] x + [ -M2/r3 - (M1/d03) ] y } Compute V(x,y) and show it is constant on the desired ellipse. How about just guessing the V(x,y) ! For the two 1/r3 terms I know that V13(x,y) = GmM2/r r = For term #2 my guess is V2(x,y) = Gm 2(M1/d03) (x2/2) For term #4 my guess is V2(x,y) = - Gm (M1/d03) (y2/2) Then my candidate potential is this V(x,y) = GmM2/r + Gm 2(M1/d03) (x2/2) - Gm (M1/d03) (y2/2) } = Gm { M2/r + 2(M1/d03) (x2/2) - (M1/d03) (y2/2) } So this is constant not on a circle but on a distorted circle. We then have this equation M2/r + 2(M1/d03) (x2/2) - (M1/d03) (y2/2) = k // some constant M2/r = k - 2(M1/d03) (x2/2) + (M1/d03) (y2/2) M22 = (x2+y2)[ k - 2(M1/d03) (x2/2) + (M1/d03) (y2/2)]2 = (x2+y2)k2 [1 - 2(M1/kd03) (x2/2) + (M1/kd03) (y2/2)]2 ≈ k2(x2+y2) [ 1 + 2*{-2(M1/kd03) (x2/2) + (M1/kd03) (y2/2)} ] ≈ k2(x2+y2) [ 1 - 4 (M1/kd03) (x2/2) + 2(M1/kd03) (y2/2) ] ≈ k2(x2+y2) [ 1 - 2 (M1/kd03)x2 + (M1/kd03)y2 ] M22 ≈ k2(x2+y2) [ 1 - 2εx2 + εy2 ] ε = (M1/kd03) (M2/k)2 ≈ (x2+y2) [ 1 - 2εx2 + εy2 ] Since ε << 1. we see that R2 ≈ M2/k = radius of the basic circle (R2 = radius of Earth). My quartic is now R22 ≈ (x2+y2) [ 1 - 2εx2 + εy2 ] k = (M2/R2) ε = (M1/kd03) = (M1/M2) R2/d03 Imagine plotting this thing. So there is your ellipse-like curve. How far to the right of 1 does it go? At that point we have R22 ≈ x2 [ 1 - 2εx2 ] which you have to solve for x. Write as a ≈ z [ 1 - 2εz ] So the solution is that z = a + 2ea2 which I translate to x2 = R22 + 2εR24 = R22(1 + 2εR22) x = R2 ≈ R2(1 + εR22) = R2 + εR23 Then the little tide height is h = εR23 = (M1/M2) R2/d03 * R23 = R2 (M1/M2) (R2/d0)3 My answer to this question from Section 8,8 is this: a = (3/4) R2 (M1/M2) (R2/d0)3 which is pretty close. So why don't I have a 3/4 sitting there? I think I could redo this doing a better job. You see the distortion, but this is a quartic which I have to approximate with a quadratic somehow. General problem: f(x,y) = x2+y2 + ε1(x2+y2)x2 + ε2(x2+y2)y2 = x2+y2 + ε1x4 + (ε1+ε2)x2y2 + ε2y4 The equation of an ellipse is this A2y2+B2x2 = x2y2 Could do a least squares fit to estimate A and B. Go back to M12 ≈ k2(x2+y2) [ 1 + 2εx2 - εy2 ] ` ε = 2(M1/kd03) Accept this as a quartic, not an ellipse. Have (x2+y2) [ 1 + 2εx2 - εy2 ] = (M1/k)2 At the left edge how far is it displaced from the circle : xcircle = (M1/k) xq2 [ 1 + 2εxq2 ] = (M1/k)2 2εxq4 + xq2 - (M1/k)2 = 0 xq2 = [ -1 ±]/(4ε) ≈ [ -1 ±(1 + 4ε(M1/k)2 -(1/8) 82ε2(M1/k)4 ) ] /(4ε) ≈ [ -1 +(1 + 4ε(M1/k)2 -(1/8) 82ε2(M1/k)4 ) ] /(4ε) ≈ [ 4ε(M1/k)2 -8ε2(M1/k)4 ) ] /(4ε) = [ (M1/k)2 -2ε(M1/k)4 = (M1/k)2 -2ε(M1/k)4 = (M1/k)2 ( 1 -2ε(M1/k)2) =xcircle2 ( 1 -2ε(M1/k)2) xq = xcircle Sign is wrong, but there is a plan here where you evaluate the thing "a". I am not sure why this method is so horrible! It gives a quartic to which you have to fit an ellipse, and then you can get A and B for the ellipse, incredibly ugly! Taylor never mentions "ellipse" but just does what I am doing above to get the tide height! Go back to the surface which I think is exact M12 ≈ k2(x2+y2) [ 1 - 2εx2 - εy2 ] ` ε = 2(M1/kd03)