Re: symmetric twins paradox
- From: "Paul B. Andersen" <paul.b.andersen@xxxxxxxxxxxxxxxx>
- Date: Tue, 21 Nov 2006 22:57:48 +0100
Sue... wrote:
dlzc wrote:Dear Sue:
Sue... wrote:N:dlzc D:aol T:com (dlzc) wrote:Three clocks, two different elapsed times. Argument? Beads, hoses?Dear lkoluk2003:In response to your complaints about the elastic properties
<lkoluk2003@xxxxxxxxx> wrote in message
news:1164106119.311251.148050@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
If you did the math, observer A sees observer B's clock
go very slowly on A's outbound, continue slowly for some
time with A inbound (since B's start-of-return doesn't reach
A until later) then sees B's clock rate increase until it
meets A at Earth, with net 0 difference at Earth.
So three observers in a symmetric situation yields two
results, because only two unique paths are described.
It is the words that trick your common sense, not the facts.
Use the math.
of today's garden hose material and the possibility of
"bead bunching", we are proud to announce the all new and
improved
I'm not looking to buy Manhattan...
:o)
Uhmm Kimosabe, Manhattan cost you heapum wampum.
I'll let the web publishing of Unnikrishnan, Fitzpatrick and MIT
serve as my argument for *this* thread because the OP doesn't
mention resolution of the postulates and that is where an SR
discussion needs to go to be productive for the OP.
We can discuss equal length path techniques in a thread
where electromagnetism is not the focus. I was just injecting
some humor, not intending to do anything to interfere with
the remote possibility that the discussion might wander off
into the physics arena. ;-)
Regards,
Sue...
There is no risk that you ever will wander into the physics arena.
You are babbling a lot about hoses and marbles, but you are
utterly unable to explain how you would use them to prove
your claims.
You flee like hell every time someone asks you to do so.
Like you did now.
Paul
.
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