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## Re: Pioneer 10 looks like red shift, not blue

 Subject: Re: Pioneer 10 looks like red shift, not blue "George Dishman" Sat, 3 Dec 2005 12:43:23 -0000 sci.astro, sci.physics.relativity, sci.physics
 ```"Richard Saam" wrote in message news:bzZjf.114994\$qk4.69663@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx > George Dishman wrote: >> "Richard Saam" wrote in message >> news:ENEhf.90886\$qk4.60323@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx >> >>>Basic Physics 101 >>> >>>Newton's Second Law >>> >>>F = m dv/dt >> >> >> Sorry Richard, that's wrong. If your going to do >> "Basic Physics 101" you need to get it right. Force >> is defined by: >> >> F = dp/dt >> >> where p is the momentum. For v << c we can approximate >> the momentum by the non-relativistic p = m v hence >> >> F = m dv/dt + v dm/dt >> >> Note that the second term may not be negligible where >> you have a moving object being slowed by drag. However, >> let's gloss over that and see where this goes. >> >> Note importantly that F and v are vectors. >> >> >>>differential identity >>> >>>dv/dt = v dv/dx >>> >>>Therefore: >>> >>>F = m v dv/dx >> >> >> Noting again that F, v and dv/dx are all vectors. >> What about dv/dx and dv/dz ? For Physics 101 you >> need to write the complete formula and then for >> example you could say the trajectory is in the >> x,y plane and set z to zero but you can only >> ignore one since the path is hyperbolic. >> >> >>>F dx = m v dv >> >> >> Noting again that F, v and dv/dx are all vectors >> and you are again neglecting y and z components. >> >> >>>integrate and then >>> >>>Work = Ek2 - Ek1 >> >> >> Hold on there Richard, force * distance is the >> work done only when the distance is measured in >> the direction of force. You have forgotten that >> most of your quantities are vectors. >> >> >>>Statement of Work Energy Principle or Theorem >>> >>>"The work of the resultant force >>>exerted on a particle >>>equals the change in kinetic energy >>>of the particle" >>> >>>In terms of relativistic mass >>> >>>F dx = m (1 - v^/c^2)^(-1) v dv >> >> >> And since you assumed above that v << c, we can >> neglect the gamma term. If you don't like that, >> start again but use the relativistic formula for >> momentum from the beginning. >> >> >>>where m is constant >> >> >> You started by neglecting the dm/dt term. >> >> >>>and m v x = h / (4 pi) (Heisenberg's Uncertainty) >> >> >> Oh dear, no. The Uncertainty Principle applies to >> the _uncertainties_ in the measurements, not the >> measurements themselves. > > but for constant m, the uncertainties exist within the de Broglie > condition see below. > > Here mv is the momentum >> and x is one coordinate of the location (previously >> you used x as the distance moved in the x direction) >> but it is the product of the uncertainties in the >> measurements that is constrained. Also note that >> it is an inequality: >> >> dx * dp > h_bar / 2 > > do not forget the equal sign possibility > > dx * dp >= h_bar / 2 True, my mistake. > dx * dp >= h / (4pi) > >> >> That says we cannot measure both the location and >> momentum to unlimited accuracy. >> >> http://en.wikipedia.org/wiki/Uncertainty_principle >> >> In the case of Pioneer, practical limitations mean >> we are nowhere near that limit, the uncertainty in >> the location alone is thousands of km. > > No No, see below >> >> At the end of the day Richard, all you are showing >> is that applying a force to a mass changes its >> momentum and kinetic energy which as you say is >> basic physics, but what you haven't attempted to do >> is state the mechanism that is applying the force >> in the case of the Pioneer craft. As I have said >> several times, the kinetic energy in that case is >> tiny compared to the RAG waste heat so energy is >> not of interest, it is the _mechanism_ by which the >> force is applied that is being sought. If you are >> claiming it is drag then the dm/dt factor is >> fundamental and you have simply ignored it. > > No it is not ignored, see clarification below. > > George > > Clarification: > > Apparently you forgot that "m" > is not the mass of the spacecraft > but is 110 * electron mass No Richard, apparently you forgot we were discussing "Basic Physics 101". The above equations are entirely general and you have yet to apply them to anything. On that basis, they _should_ include the v dm/dt term to be correct. > and represents the mass in the space density of 6E-30 g/cc. > and available as work energy 6E-30*c^2 erg/cc or 5.4E-9 erg/cc. > > Above all "m" is constant > or more specifically rest mass. M _might_ be constant once you apply the equations to a specific case. Later you say of your hypothesis: > Conceptually, it is like a disc sweeping out a cylinder > through a quantum mechanical medium of virtual particles > in equilibrium with CMBR > having the characteristics of a Bose Einstein Condensate In that case, m is the total mass of the craft plus the "virtual particles" which have been swept up. Of course the latter is negligible in comparison so we can treat m as constant but for "Basic Physics 101", you cannot neglect the term. > Newton's Second Law > > F = dp/dt > > Newton's Relativistic Second Law > > F = (m v (1 - v^2/c^2)^(-1) / dt > > m = constant > > One dimensional assumption No Richard, F, v and p are all vectors and in the case of the Pioneer craft they are not aligned. > therefore > > F = m (v (1 - v^2/c^2)^(-1) / dt > > differential identity > > dt = dx/dv Where "x" is the distance moved by m in time dt. > work - energy theorem > > F dx = m (1 - v^2/c^2)^(-1) v dv > > m = constant (reminder) > > Heisenberg Uncertainty > > Dp * Dx >= h/(4pi) > > m * Dv * Dx >= h/(4pi) > > m >= h/(4pi) * 1/(Dv * Dx) > > de Broglie relation > > p * x = h p * lambda = h where lambda is the wavelength of the particle. > m * v * x = h Wrong, you have confused the wavelength of the particle with the distance it moves in some time. The next part rested on this and is therefore invalid so I'll snip it to save space. However, For the sake of argument, I'll assume your conclusion in the rest of the reply, that each cell produces an amount of slowing which is independent of the speed of the craft through it. > and specifically for: > 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 > or the CMBR dipole of 620 km/s > or any other velocity 'v' magnitude (0 do not contribute to the observed Pioneer deceleration > (as observed) > (noting that c ~ 300,000 km/sec). > > if deceleration or slowing down is observed in one space lattice cell > the same will be observed in a series of space lattice cells > and independent of v or n vd Each cell would contribute the same amount of slowing so the net effect will depend on the rate at which cells are traversed therefore it should be proportional to v. That isn't necessarily in conflict with observation since the speed of the Pioneers was fairly constant over the period of the analysis even relative to the sun, and relative to the CMBR, the variation is even smaller. > The observed deceleration will be > in the line of observer to object > in accordance with a = F/M > where M is the Mass of the object. No, the deceleration should be in the same direction as the _vector_ F, that's "Physics 101" Richard > This deceleration or slowing rate > will be numerically the same whether object > is moving away from the observer or towards the observer. > > Conceptually, it is like a disc sweeping out a cylinder > through a quantum mechanical medium of virtual particles > in equilibrium with CMBR > having the characteristics of a Bose Einstein Condensate This is the same basic mistake I pointed out some time ago, if the particles are in equilibrium with CMBR, the deceleration shold be aligned with and opposite to the direction of motion of the craft in a local frame in which there is no CMBR dipole. The milky way is moving towards Hydra at ~ 600km/s so the deceleration of both craft should be opposite to that direction. > If the observer and pioneer/object > where in line with CMBR 640 km/sec dipole > and having + or - relative velocity 'v' > the same pioneer/object (deceleration or slowing down) > would be observed as if > observer and pioneer line was orthogonal > to CMBR 640 km/sec dipole. No but what you say next the correct: > The Force and (deceleration or slowing rate) > is always the same for a given F/M. > (M = Mass of Pioneer or object) Exactly, F is the same for both craft but you are forgetting that F is a vector. The acceleration should be in the same direction as F and should therefore be in the same direction (away from Hydra) for both craft. The observed acceleration is roughly towards the Sun for both craft and they are on opposite sides of the Sun. > At the end of the day, > the above hypothesis falls within available experimental data. Not for the direction, which is what I pointed out right at the beginning. George ```
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