🚀 go-pugleaf

"Max Keon" <maxkeon@optusnet.com.au> wrote in message 
news:47c15d5c$0$25392$afc38c87@news.optusnet.com.au...
|
| "Paul B. Andersen" <paul.b.andersen@hiadeletethis.no> wrote in message
| news:47BC2905.9080708@hiadeletethis.no...
| > Max Keon wrote:
| >> Paul B. Andersen wrote:
|
| >>> Please read the following carefully.
| >>> We have a tower.
| >>> If em-radiation is emitted from one end of the tower, and received at
| >>> the other end, all frequencies must change by the exactly the same
| factor.
| >>> That means that if the em-radiation is gamma, the frequency shift
| >>> delta_f/f must be the same as when the em-radiation is RF.
| >>> If the em-radiation is a series of pulses (or flashes),
| >>> the repetition frequency of these flashes must also change by
| >>> the same delta_f/f as above.
| >>> Now, imagine that we have a clock at each end of the tower,
| >>> emitting one flash per second (let the clocks be Cs clocks,
| >>> emitting one flash per SI-second).
| >>> If the em-radiation going up is shifted to lower frequencies,
| >>> then the observer on the top will count fewer flashes from
| >>> clock at the bottom, than from his own clock. While he
| >>> is counting N pulses from his own clock, he will have counted
| >>> only N-n pulses from the bottom clock. This process is cumulative,
| >>> so he will see that the bottom clock is lagging more and more
| >>> behind his own clock. The bottom clock is running slower than
| >>> the top clock.
| > [..]
| >> Firstly, you state that the process is cumulative, which suggest
| >> to me that the frequency transmitted from a low altitude clock
| >> to a more elevated clock was actually higher than is observed
| >> when it arrives there. Meaning that more ticks have been
| >> transmitted than are actually received. It's not possible for
| >> that to continue on indefinitely unless the space between the
| >> two clocks is expanding. Even if the big bang theory allowed for
| >> local expansion, it would be 1/44 of that required to store the
| >> missing cycles.
|
| > You seem to have a serious reading comprehension problem.
| >
| > The receiver doesn't work.
| > Communication impossible.
|
| You clearly stated that the em-radiation going up is shifted to
| lower frequency. But it's only the clock rate stationed with the
| observer at the top that is blueshifted. There is no "photon
| frequency" change at all.
|
| I've updated the web page at
| http://members.optusnet.com.au/maxkeon/redshft2.html
| Your error should become apparent to you.
|
| -----
|
| Max Keon

The way to tell the time on the Moon (which is always facing the Earth)
is to look at the terminator (the line between night and day).
If the terminator is across London it is 6 o'clock +/- 4hours GMT, UTC,
Zulu time, American military time or whatever else you want to call it,
depending on Summer or Winter, and 1 o'clock EST in New York.
If it's across New York it is 6:00 o'clock EST in New York, 11 o'clock
GMT in London and 3 o'clock PST in Los Angeles. That's the clock.

According to the crank Andersen,
|" While an observer on the Earth measure 1,000,000 years,
| an observer on the Moon (if we ignore the mass of the Moon)
| would measure 1000000.0006797 years.
| That is approximately 6 hours more than the observer on the Earth."

So after a million years it takes light approximately six hours to reach
the Moon (or is it 18 hours, I'm confused).

I can assure you that a crank's errors never become apparent to the
crank. He'll go on arguing for the rest of eternity, certain of his own
infallible faith in the crank Einstein.
Quite. Blatantly obviously. Hilarious, no. :-)