Re: SR cannot determine Contraction
- From: "kenseto" <kenseto@xxxxxxxxxx>
- Date: Mon, 25 Feb 2008 10:04:09 -0500
"wugi" <brol@xxxxxxxxxx> wrote in message
news:47bda2bf$0$831$5f6aeac3@xxxxxxxxxxxxxxxxxx
"kenseto" :expansion???
"wugi" :Lorentz
BTW, IRT agrees that there is no physical length contraction of a
SR does mean a physical Lorentz length contraction.
No it doesn't....if there is Lorentz contraction then there must be
expansion!!!!! So where is the SR equation that shows Lorentz
Of course it is there.
Suppose an object OA' is moving past your "rest" object OA when both left
sides coincide in O :
OxxxxxxxxxxxxxA'xxxAxxxxxA"xxxxxxxxA"'
Length contraction tells you
OA' = OA / gamma
The point is:
why can't OA' = OA*gamma???
IOW, why can't OA' be longer than OA? Is it because the observer is in a
state of absolute rest? Why do I say that? The reason is that only the
observer who is in a state of absolute rest can see all the rods in the
universe are contracted.
Ken Seto
The moving system tells you that a later position A" (and not A') is
simultaneous with O, so that
OA = OA" / gamma, or
OA" = OA * gamma.
But he will also tell you that actually your event A to him coincides
spatially with an event A"' so that
OA" = OA"' / gamma, or
OA"' = OA * gamma
So, either contraction (in both systems) is paired with an expansion when
taking in account the other party's simultaneity.
rest
Minkowsky makes it clear for you:
( http://home.scarlet.be/~pin12499/MySRT/LorentzObjects.Detail.png )
As you see, a matter of homothecy in triangles ;-)
Besides if Lorentz contraction is real then why is the observer's rod isthe
longest rod in the universe?? Does that mean that the observer is at
inmoving
Each observer's rod is the longest to himself...
a preferred frame? OTOH, if the SR contraction is a geometric projection
... because each observer experiences a different, relativistic,
simultaneity. No preferred frames involved.
effect there is no problem with the above problems. In addition every SR
observer can claim that his rod is longer than any projection of a
merod......much like I see you to be shorter from a distance and you see
to
be shorter from a distance.
Right, homothecy in vanishing triangles ;-)
Same.Regards,
guido
http://home.scarlet.be/~pin12499/paratwin.htm
.
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