Re: 7 problems with usenet aether theorists

From: John Kennaugh (JKNG_at_kennaugh2435hex.freeserve.co.uk)
Date: 02/27/05 

Date: Sun, 27 Feb 2005 15:40:28 +0000

In message <cvmq6v$r9t$1@beech.fernuni-hagen.de>, Ilja Schmelzer
<q6867901@mailstore.fernuni-hagen.de> writes
>
>"John Kennaugh" <JKNG@kennaugh2435hex.freeserve.co.uk> schrieb
>> Ballistic theory says that c is the natural
>> speed relating photons and matter. Light therefore leaves a source at c
>> but from then on nothing is controlling it. i.e. the source does not
>> maintain any influence upon it. If it is travelling in empty space it
>> will continue at c relative to the source but if it encounters matter
>> and if its speed is not c relative to that matter this represents an
>> unnatural state and the speed will have a tendency to be restored to be
>> its natural speed c relative to the matter. - called extinction. Now
>> when we say 'encounters matter' we really mean 'encounters the field
>> within and near to matter'. One can therefore assume that if light is
>> travelling at c+v or c-v relative to a field it will end up as c
>> relative to the field. A rotating star will have a field surrounding it
>> and by the time light leaves the fields influence, it will be travelling
>> at c relative to the star as a whole and have lost its c+v and c-v
>> components.
>
>Yep. And now imagine a double star, where one visible star is rotating,
>like Earth around the Sun, around the other star.

Talk to Henri about that. I don't know enough about it. I think the
problem with double stars is that that you have no direct confirmation
of what it is you are looking at, only data. You only infer what
configuration you are looking at by interpreting the data and in order
to interpret the data you have to assume a theory. You cannot then test
that theory assuming the configuration is correct. All you can do is say
whether you can get a good fit. Henri claims double stars prove source
dependence - he certainly obtains very convincing fits with data in some
cases. I know he isn't the only one working along the same lines.
Certainly there is no clear cut evidence one way or the other from
double stars. De Sitter's evidence based upon double star observation
was discredited by Fox in 1965.

[snip]

>> >> If an oscillator orbits a central point its frequency is lower by a
>> >> factor Sqr(1-vv/cc) as predicted by relativity. Relativity calls it
>time
>> >> dilation. The alternative ballistic theory says there is no such thing
>> >> as time dilation but while it does not predict time dilation it does
>> >> predict that the frequency will be lower by a factor Sqr(1-vv/cc). The
>> >> maths of the ballistic theory shows that it is the result of nothing
>> >> more sophisticated than a velocity triangle and conventional Doppler
>> >> shift.
>> >
>> >How? The simplest case, the observer is in the center, he observes
>> >no Doppler shift because the distance to the source remains constant.
>>
>> You are wrong but I accept it isn't obvious.
>
>> You are wrong but I accept it isn't obvious.
>> [some argumentation snipped]
>> Light leaves source at time = 0 and a sphere of light will expand
>> outwards, the centre of which continues to move at v -->
>
>I see your point - there is, indeed, some sort of "dilation" in ballistic
>theory: The light does not arrive at the center at l/c but later.
>
>But you incorrectly apply the Doppler shift formula to this situation.

No I am not. Take the following:

  S-->v----------A------B------C------------------D--------

      FIG 1

                        O
A Source is moving along the line at v and O is receiving a radio signal
transmitted. You will I think agree that with the source at S it has a
component of velocity towards O so there will be an increased frequency
due to Doppler. Again we know that at D the source is moving away and
the frequency will become lower.

Q - At which point on the line will there be zero Doppler shift?

The answer is that the signal transmitted at A such that A-B = vt and
B-O = ct. Light emitted at A will move out in a sphere and reach O when
the source *reaches* B. The direction of the light received will be B-O.

Light emitted when the source is at B will reach O when the source is at
C which is the case as the source going around the observer previously
dealt with. The direction of the light received will be C-O which has a
component in the -x direction.

>
>To explain why, it seems useful to consider the Doppler shift not for
>light, but for a signal send by some experimenter. The experimenter
>sits in the spaceship and once in a second presses a button which
>sends some signal (a beep). The other experimenter sits in O,
>receives the beeps and measures their arrival time.
>
>First, lets consider the standard case where the spaceship moves away
>with constant speed u and sends signals back with another constant speed c.
>The distance increases with each beep by l=u*1sec, therefore the next beep
>arrives not after one second but later, at
>1sec+l/c = 1 sec + 1 sec*u/c. = 1 sec(1+u/c). The
>frequency of receiving signals at O is 1/(1+u/c). This is essentially your
>formula
>
>> f'/f = ( 1 - u/c)
>
>You see (I hope) that the description with the experimenter sending beeps
>once a second gives, indeed, the classical Doppler shift formulas.

It is a valid way of describing what Doppler is.

>
>Now look at such an experiment in the spaceship orbiting O. He sends
>a signal once a second. But his distance to O remains constant. That the
>signal comes late doesn't matter, the effect remains the same for all beeps.
>The difference between the arrival times of the beeps remains unchanged,
>1 sec. Thus, the frequency of receiving signals remains unchanged. No
>Doppler effect for the beeps.

It is hard to visualise and I understand what you are saying. I will
think it through and ask you to do the same. 

-- 
John Kennaugh
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