rbwinn
2008-03-29 02:24:26 UTC
If there are two frames of reference, S and S', with S' moving with a
velocity of v relative to S, we can showthis by the Galilean
transformation equations.
� � � � � � � � � � � � � � � � � �x'=x-vt
� � � � � � � � � � � � � � � � � � y'=y
� � � � � � � � � � � � � � � � � � z'=z
� � � � � � � � � � � � � � � � � � �t'=t
� � But scientists say that they have proven that a clock in S' is
running slower than a clock in S, therefore, the Galilean
transformation equations cannot be used. �However, using the equation
n'=t(1-v/c) for the time of a clock in S', we can see that an observer
in S' will calculate a faster speed for S' than an observer in S
because less time transpires on a clock in S', and the Galilean
transformation equations show that the distance is the same as
measured from either frame of reference, (there is no distance
contraction).
� � �Scientists have informed me that the velocity calculated by the
observer in S', using the S' clock, cannot be used with equations for
momentum. �If v2 is the velocity calculated by the observer in S',
then it is true that � � p=m(v2) � �is incorrect. �
That is not the problem. The problem is that m*v is not a good formulavelocity of v relative to S, we can showthis by the Galilean
transformation equations.
� � � � � � � � � � � � � � � � � �x'=x-vt
� � � � � � � � � � � � � � � � � � y'=y
� � � � � � � � � � � � � � � � � � z'=z
� � � � � � � � � � � � � � � � � � �t'=t
� � But scientists say that they have proven that a clock in S' is
running slower than a clock in S, therefore, the Galilean
transformation equations cannot be used. �However, using the equation
n'=t(1-v/c) for the time of a clock in S', we can see that an observer
in S' will calculate a faster speed for S' than an observer in S
because less time transpires on a clock in S', and the Galilean
transformation equations show that the distance is the same as
measured from either frame of reference, (there is no distance
contraction).
� � �Scientists have informed me that the velocity calculated by the
observer in S', using the S' clock, cannot be used with equations for
momentum. �If v2 is the velocity calculated by the observer in S',
then it is true that � � p=m(v2) � �is incorrect. �
for momentum, unless you're willing to ignore how off it is. When
speeds are small, then that's not too bad. When speeds are larger,
it's simply the wrong formula, period.
what Isaac Newton said.
good formula except as an approximation. If you want an education on
what the law of conservation of momentum says and what the proper
expressions for momentum are, I'd be happy to give you a reading
reference, complete with chapter numbers. I see no reason to
transcribe those pages here to save you the effort of removing your
ass from your chair.
PD
orbit and radius of a satellite are less when measured by a clock in
the satellite. I think the orbit and radius of orbit stay the same.
You said that my equations would not work for momentum in this
particular problem, and I showed that they did. If you want to talk
about particle accellerators, why don't you talk to some of the other
scientists? I do not know a thing about particle accellerators.
Robert B. Winn
equation in sci.physics.relativity? Why don't you take a few minutes
and go read the charter of the newsgroup?
Robert B. Winn
subject you don't understand then point to the charter to defend your
stupidity?- Hide quoted text -
you find wrong with the equations I posted since you do not like
them. You mentioned that you disagreed about S and S', and then I
explained what frames of reference were. What else do you see wrong?
Robert B. Winn
- Show quoted text -
besides reading my posts. Maybe some of these nice scientists would
talk to you.
Robert B. Winn