Re: Pioneer Anomoly


"John C. Polasek" <jpolasek@xxxxxxxxxx> wrote in message
news:5ufpb19vf278o0v01cfj6qrgb01r2g1gu2@xxxxxxxxxx
> On Fri, 24 Jun 2005 22:30:50 +0100, "George Dishman"
> <george@xxxxxxxxxxxxxxxxx> wrote:
>>"John C. Polasek" <jpolasek@xxxxxxxxxx> wrote in message
>>news:c2mmb118be041dv2cno85m8sv1kon05bqe@xxxxxxxxxx
>>> On Thu, 23 Jun 2005 23:55:24 +0100, "George Dishman"
>>> <george@xxxxxxxxxxxxxxxxx> wrote:
....
>>>>The relativistic formula approximates the usual
>>>>1/2 m V^2 for small V and remains accurate at
>>>>energies of 300 GeV/c and above.
>>> You might think you're getting 300 GeV into your target but looking at
>>> my Fig. 1b #2 you know that the higher the force the more ineffectual
>>> it is.
>>
>>The 300GeV is what is measured, the force
>>would be calculated.
>>
>>My point was that since your equations mean
>>the total energy is independent of speed,
>>and the energy when the particle is brought
>>to rest must be the same for both particles
>>(0.1c and 0.9c in my toy example), then
>>there would _equal_ amounts of energy given
>>to the target for both speeds. Clearly
>>that is wrong.
>>
> Put .1c or .9c as V in Fig. 1 #2. That is the real velocity axis. Then
> the energies are .5mV^2.
> The hypotenuse is where the .5mc^2 of creation lies and it is the new
> rotated time-arrow of the mass. The horizontal axis is the "cosmic
> velocity" of the object and it has a shortened c.
> You have to make an effort to see the 4th dimension as a legitimate
> entity and our speed through time a vital part of the energy equation.

The 4th (time) dimension is equally real in
relativity so there is no change in that.
The point is that as the 1/2 m v^2 factor
increases, the other factor decreases so
there is no change in the total hence no
energy delivered to the target.

Regardless of that, starting with the measured
energy of 300GeV, what speed does your theory
predict for the proton?

George


.

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