Subject: |
Re: Pioneer 10 looks like red shift, not blue |
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From: |
Richard Saam |

Date: |
Tue, 06 Dec 2005 17:02:25 GMT |

Newsgroups: |
sci.astro, sci.physics.relativity, sci.physics |

George Dishman wrote: George Here is something to contemplate and perhaps you have some insight. When you indicate "direction" perhaps you are indicating the direction of longitudinal and transverse forces as indicated as possibly archaic but still valid concepts presented in my physics 101 text?I am talking of the direction is which the craft are moving and in which they are accelerated. There is nothing "archaic" about the concept of direction. In fact making sure that a spacecraft is moving in the correct direction is quite important!Now for general relativistic newtonian force within the context of work energy theorem: F = dp/dtNoting that F and p are vectors.Constant mass m Work Energy theorem F = m v/dtNope, again you have lost the dm/dt term and it is the Newtonian formula while you start using the relativistic version later.... = m v dv/dxYou are again losing the y and z components. Richard, I've pointed out all these errors before. Can't you correct them before posting again?de Broglie cell: x*y*zYou have omitted any definitions but presumably x, y and z are the dimensions of the cell.defined by de Broglie condition: m * v_x * x = h m * v_y * y = h m * v_z * z = hYou have not defined m or v. They are not related to the symbols in your previous equations.Work - Energy on object moving through de Broglie space cell in x direction: General form: F dx = m v dv = (m*v*(1-v^2/c^2)^(n)) dv integration: F x = work = energy = - mc^2 *(1-v^2/c^2)^(n+1) /(2*(n+1)) for n = -1/2 (transverse condition)Total energy is E = m c^2 (1-v^2/c^2) ^ (-1/2) but note that v this time is the magnitude of the vector so your "transverse condition" in meaningless.F x = work = energy = - 0 * mc^2 for 0<v<<cE ~ m c^2 for v << c. Your "0<v" is nonsense since for v=0, E = m c^2 exactly. "0 =< v" would be correct but redundant since v is the magnitude of the velocity (speed) hence v >=0 by definition.Transverse forces in y & z direction cancel with zero force on object moving in x direction through space de Broglie cell.Prove that for the general case where the cell is moving with some arbitrary velocity (e.g. the 620km/s of the CMBR dipole) in your chosen rest frame (e.g. heliocentric). George George Clarification: For general relativistic newtonian force within the context of work energy theorem: F = dp/dt de Broglie cell volume: x*y*z = volume defined by de Broglie condition: m * v_x * x = h m * v_y * y = h m * v_z * z = h mass(m) density of de Broglie cell (rho) = m/volume Define cells to completely fill space in the form of a lattice i*x,j*y,k*z where i,j&k = 1 to infinity Heisenberg Uncertainty in each de Broglie cell m * Dv_x * Dx = h/(4pi) m * Dv_y * Dy = h/(4pi) m * Dv_z * Dz = h/(4pi) Work = Energy on object moving through de Broglie space cell in x direction: General Newton Second Law relativistic form: F dx = (m*v_x*(1-v_x^2/c^2)^(n)) dv_x F dy = (m*v_y*(1-v_y^2/c^2)^(n)) dv_y F dz = (m*v_z*(1-v_z^2/c^2)^(n)) dv_z integration: F x = work = energy = - mc^2 *(1-v_x^2/c^2)^(n+1) /(2*(n+1)) F y = work = energy = - mc^2 *(1-v_y^2/c^2)^(n+1) /(2*(n+1)) F z = work = energy = - mc^2 *(1-v_z^2/c^2)^(n+1) /(2*(n+1)) Given that i = number of de Broglie cells the object crosses with the quantity i*v_x equal to any arbitrary velocity <<c such as CMBR dipole of 620 km/sec or sun referenced Pioneer velocity of 12km/s or earth (around sun) orbital velocity of 29.8 km/sec or sun (around milky way center) rotation velocity of 400 km/sec (noting that c ~ 300,000 km/sec). Define [L1,L2] as integration limits. for n = -1/2 (transverse condition) F y [0-j*y,0+j*y] = work = energy [0-Dv_y,(0+Dv_y)] = - mc^2 *(1-v_y^2/c^2)^(-1/2+1) /(2*(-1/2+1)) [0-Dv_y,(0+Dv_y)] = - mc^2 *(1-v_y^2/c^2)^(1/2) / 1 [0-Dv_y,(0+Dv_y)] = - mc^2 *(1-v_y^2/c^2)^(1/2) [0-Dv_y,(0+Dv_y)] ~ 0 Similar calculation for z direction Transverse forces in y & z direction cancel with resultant y & z zero force on object moving in x direction through space de Broglie cell. for n = -3/2 (longitudinal condition) defining integration limits from (i*v_x) to (0+Dv_x) (total momentum transfered to object over multicell distance i*x F x [0,i*x] = work = energy [(i*v_x),(0+Dv_x)] = - mc^2 *(1-v_x^2/c^2)^(-3/2+1)/(2*(-3/2+1)) [(i*v_x),(0+Dv_x)] = - mc^2 *(1-v_x^2/c^2)^(-1/2) [(i*v_x),(0+Dv_x)] ~ - mc^2 [(i*v_x),(0+Dv_x)] F = Ma = M dv_x / dt ~ -area/volume m c^2 ~ -area rho c^2 area = object area M = object mass a = object acceleration a ~ -area rho c^2 / M relative to de Broglie cell lattice. An object observer at any arbitrary point i*x,j*y,k*z relative to object at x,y,z will observe object negative acceleration 'a' or object decceleration 'a' Richard |

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