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WHY m1*v1=M1*v.



 
 
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Old April 30th 06, 02:00 PM posted to sci.space.policy,sci.philosophy.tech,sci.astro,rec.org.mensa
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Default WHY m1*v1=M1*v.

$$ Bill Hobba wrote -=-
Physics is an experimental science.
Every assumption - it does not matter how 'obvious' -
requires experimental support.

-=-
And in its domain of applicability (inertial frames)
SR has yet to find experimental refutation.

-=-
Thanks Bill nss *************


$$ Note "iNERTiAL" means (..is a synonym for), "REST";
$$ And, in a REST FRAME (as SR has distinguished contrary to Newton)
$$ ..of M1, M1 has no THEORETiCAL acceleration towards m1,
$$
$$ ..where (the OPPOSiTE radial vectors of ) m1*v1 = M1*v.
$$
$$ [EVEN THOUGH the THEORETiCAL acceleration M1*v is SMALL].
$$ [EVEN if the EXPERiMENTAL acceleration M1*v, negligible].

$$ The GR-"equations" ARiTHMETiCALLY eliminate Newton's M1*v.
$$ The GR-"equations" *SYSTEMiCALLY* eliminate Newton's M1*v.
$$ snicker
$$ This is WHY the *SR* (synonym) "iNERTiAL" replaced "REST".
$$ This is WHY the *SR* (synonym) "iNTRiNSiC" ..means "REST".

$$ Also, OPPOSiTE central RADiAL VECTORs have the SAME sign.
$$ [An OUT-going vector is (+) ..iN-coming vectors are (-)].
$$ Note EQUAL radial vectors areN'T PARALLEL, anyway.

$$ Go-go NETSCAPE news alt.sci.nanotech WHY m1*v1=M1*v .


 




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