Re: ABOUT THE GALILEI PRINCIPLE OF RELATIVITY

On Nov 17, 7:47�am, PD <TheDraperFam...@xxxxxxxxx> wrote:
On Nov 17, 8:26�am, rbwinn <rbwi...@xxxxxxxx> wrote:





On Nov 17, 6:14 am, PD <TheDraperFam...@xxxxxxxxx> wrote:

On Nov 15, 8:54 pm, rbwinn <rbwi...@xxxxxxxx> wrote:

Psychological issue? I walked out of the front door of that place.
They did not want me there any more. What do you think is an issue?
Robert B. Winn

I thought I just told you that. It's directly above.
Being unable to read for comprehension is another issue, of a
different sort.

PD

Well, everyone has problems, PD. �I know that you think people should
spend more time helping you with your problems. �Here in
sci.physics.relativity, the idea is to discuss relativity, not to
solve the problems that scientists are having in their personal
lives. �In any event, the difference we see between the Galilean
transformation equations and the Lorentz equations is that the
Galilean transformation equations show space to be undistorted.

No, I'm sorry, I don't see that as a difference. Lorentz
transformations do not show a distortion of space at all. Space is
still flat.

The difference between Galilean transformations and Lorentz
transformations is that Galilean transformations say that space
behaves much differently than time. That is, by changing reference
frames, Galilean transformations say that the spatial components for
points all change, but that the time component does not.

Well, let's see.
x'=x-vt
t'=t
According to a clock in S, the frame of reference at rest, if you
subtract the distance traveled by S',the frame of reference in motion,
from a coordinate x in S, you have x', the coordinate in S'.
According to a clock in S, this happens in a time of t'=t in both
frames of reference.
However, since we know from experiment that light is traveling at a
speed of c in both frames of reference according to clocks in both
frames of reference, then the clock in S' has to be going slower, so
time on the clock in S' is n'=t(1-v/w), where w is the velocity of
light. I hope this will clear it up for you.

Lorentz
transformations say that, by changing reference frames, time is not
special and space and time components all change.

Well, a clock running slower than another clock is not going to show
the same time as the other clock, if that is what you mean.

This is not a distortion, any more than the changes in spatial
components that a Galilean transformation invokes is a distortion.

Well, what is a distortion in The Lorentz equations is the length
contraction. A length contraction is a distortion by definition.
Robert B. Winn


Consequently, if light is traveling at c= 186,000 miles per second in
two different frames of reference, then, as YBM has so kindly shown,
n', the time on a clock in the moving frame of reference, is equal to
n'=t(1-v/w), where w is velocity of light. �This is true anywhere in
the moving frame of reference.

� � � � � x^2 + y^2 + z^2 = w^2t^2
� � � � �(x')^2 + (y')^2 + (')^2 = w^2(n')^2

since y'=y and z' =z,

� � � � � x^2 - (x')^2 = w^2t^2 - w^2(n')^2
� � � � � � � (x')^2 = w^2(n')^2
� � � � � � � �x'=wn'
� � � � � � � wn'=wt-vt
� � � � � � � �n'=t(1-v/w)

We therefore see that if all clocks in S and S' have the same rate if
S' is at rest, then if S' is moving with a velocity of v relative to
S, all clocks in S will show a time of t, and all clocks in S' will
show a time of n'.
Robert B. Winn- Hide quoted text -

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