tide potential calculationv2

Tide potential Calculation PhL 1.18.17 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 ) Here I have again derived the hybrid Cartesian/polar coordinates version of Fttid 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 ) // and now Cartesian! 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! (8.8.31) 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. yes 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, 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 Comment: Here I am assuming that V(x,y) is a multiple of the desired ellipse equation so I guess the surface V(x,y) = 0 would be an ellipse. Should have said V(x,y) = k[A2/x2 + B2/y2] and then any surface V(x,y) = constant would be an ellipse. What I did not know is that the actual surface is a quartic and not an ellipse, so this was all hopeless. (Hence Paradox below) 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 ] // repeating the above hoping for ellipsoid 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. Comment: Sometimes to fix a problem you take a wrong tack in hopes of a solution. Going from 2D to 3D was such a wrong tack and it did not resolve the paradox! 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) But still V = constant would give a sphere either way.