Re: More Trouble for Relativity

From: Henri Wilson (H_at_..(Henri)
Date: 09/11/04 

Date: Sat, 11 Sep 2004 04:26:22 GMT

On Fri, 10 Sep 2004 21:16:50 +0200, "Paul B. Andersen" <paul.b.andersen@hia.no>
wrote:

>
>"Henri Wilson" <H@..> skrev i melding news:hlp1k05ra88ee3m0lb91q7lrbetqdsqpou@4ax.com...
>> On Wed, 08 Sep 2004 23:05:36 GMT, H@..(Henri Wilson) wrote:
>>
>> >On Tue, 7 Sep 2004 22:13:40 +0200, "Paul B. Andersen" <paul.b.andersen@hia.no>
>> >wrote:
>> >
>> >>
>> >>"Henri Wilson" <H@..> skrev i melding news:paqpj0h49g7nn5olsifjo985s9ggj9thpk@4ax.com...
>> >>> On Mon, 6 Sep 2004 16:30:38 +0200, "Paul B. Andersen" <paul.b.andersen@hia.no>
>> >>> wrote:
>> >>> Thus there is no way clock A can run FAST in B's frame, during any part of its
>> >>> orbit.
>> >>
>> >>This is the crux of the matter.
>> >>Clock A WILL run FAST in B's frame part of the orbit.
>> >>This IS what GR predicts.
>> >>(SR would predict the same for two contra moving clocks
>> >>forced to move in a circle with the same speed as the
>> >>orbiting clocks.)
>> >>Why do you say this is impossible?
>> >
>> >Do you know what is A's path in B's frame, Paul? Try it both with B not
>> >rotating or in 'tidal lock'.
>> >
>> >Have a look at it, then tell me how A can run fast (according to your sacred
>> >relativity)
>> >
>> >>
>> >>> According to SR, there is no way A can have the same reading as B when they
>> >>> meet.
>> >>
>> >>You are simply plain wrong, Henry.
>> >>
>> >>> But we know they do, don't we Paul. You agreed on that.
>> >>
>> >>Indeed.
>> >>Exactly as predicted by GR.
>> >
>> >Forget GR, the clocks are always experiencing the same gravity potential and
>> >radial accelerations.
>> >This is an SR problem.
>
>Not really, but we can make it an SR problem.
>See below.
>
>> >>(Or SR if we remove the gravitating Earth and let the clocks
>> >> be forced to move in the same circle.)
>> >
>> >Just tell me in which part of A's path does A run fast in B's frame.
>
>I have done so a several days ago, Henri.
>Why didn't you respond to the posting where I did?
>
>Here are the most signifacant parts of that posting repeated:
>
>You are assuming that the measured "rate of the other clock"
>is constant. It isn't.
>That is, you are insisting that:
> d(tau(t))/dt < 1 when t = nT, n integer, T period
>implies:
> tau < t when t = nT
>
>This is the same logical blunder as Androcles made.
>
> d(tau(t))/dt < 1 when t = nT, n integer, T period
>and
> tau = t when t = nT
>are NOT contradictory
>
>---
>
>If you insist on using SR (as opposed to GR), you must change your scenario
>somewhat. Remove the gravitating planet, and let the clocks be forced
>to move in a circle (rocket) with the same radius and with the same speed
>as the orbiting clocks. SR will then predict exactly the same for this scenario
>as GR predicts for the satellites orbiting the planet.
>
>The important point you obviously miss, is than in such a scenario,
>the clock frames are accelerated frames.
>SR does indeed predict that distant clocks in an accelerated frame
>may run fast.
>This IS the case in this scenario. The other clock will run slow when
>they are adjacent, and it will run fast when is on the other side of
>the circle.

Cut the crap Paul. It doesn't impress me at all.

In my experiment, the clocks are together when they are on the other side of
the circle. What is more they must both read the same (because the GO has a
detector there)

SR claims clock A will run slow in B's frame.
You claim iit will run both slow and fast and end up running the same average
rate.

All I want you to do is tell me in which part of A's path in B's frame does A
run fast and in which part does it run slow?

Also tell me how this can happen when gamma enters into the picture.

>
>----
>
>For example, if two clocks A and B are instantly at rest wrt each other
>at a distance d, and A is accelerating at a with direction towards B,
>then the rate of B measured in A's frame is dt_B/dt_A = (1 + ad/c^2).
>And no, you do not have to invoke the equivalence principle, it is
>easy to show that this is what SR predicts. It follows from the LT.
>Calculus.
>
>It is a little more complicated than that because the two clocks
>are never instantly at rest, but it is the above effect that is at play.
>The observing clock has an acceleration component in the direction
>of the other clock, and when they are at the opposite side of
>the circle, this effect will make the other clock run fast.

What a load of old codswallop!

The GO can put 360 synched clocks around the planet and observe the orbiting
clocks' readings as they pass. It will surely be found that no matter which
detector they pass, they will always be in synch with both themselves and the
GC.

The two clocks remain exactly in synch and running at the same rate no matter
where they are or what they are doing. The same argument applies to orbiting
rods. Their lengths will not vary.

>
>Paul
>

HW.

www.users.bigpond.com

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