Re: Pioneer 10 looks like red shift, not blue

[Richard Saam]

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