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Re: Pioneer 10 looks like red shift, not blue

Subject: Re: Pioneer 10 looks like red shift, not blue
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/dt


Noting that F and p are vectors.


Constant mass m
Work Energy theorem

F = m v/dt


Nope, 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/dx


You 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*z


You 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 = h


You 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<<c


  E ~ 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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