Re: Inverse Galilean transformation equations.

On Apr 21, 3:15 pm, PD <TheDraperFam...@xxxxxxxxx> wrote:
On Apr 21, 4:38 pm, rbwinn <rbwi...@xxxxxxxx> wrote:





On Apr 20, 6:08 pm, Eric Gisse <jowr...@xxxxxxxxx> wrote:

On Apr 20, 4:49 pm, rbwinn <rbwi...@xxxxxxxx> wrote:

On Apr 20, 5:22 pm, Eric Gisse <jowr...@xxxxxxxxx> wrote:

On Apr 20, 4:11 pm, rbwinn <rbwi...@xxxxxxxx> wrote:

On Apr 20, 4:48 pm, Eric Gisse <jowr...@xxxxxxxxx> wrote:

On Apr 20, 3:16 pm, rbwinn <rbwi...@xxxxxxxx> wrote:

On Apr 20, 3:58�pm, Eric Gisse <jowr...@xxxxxxxxx> wrote:

On Apr 20, 2:49�pm, rbwinn <rbwi...@xxxxxxxx> wrote:

Several scientists have expressed concern because they could not
invert the Galilean transformation equations with the variables that I
use. �I am therefore showing how it is done in order to address the
misconceptions they have been posting concerning these equations. �The
equations are:

� � � � � � � � � � � �x'=x-vt
� � � � � � � � � � � �y'=y
� � � � � � � � � � � �z'=z
� � � � � � � � � � � �t'=t

The inverse Galilean transformation equations are:

� � � � � � � � � � � � x=x'-v't'
� � � � � � � � � � � � y=y'
� � � � � � � � � � � � z=z'
� � � � � � � � � � � � t=t'

Since these are the Galilean transformations and the inverse, I fail
to see why you had to write anything else.

Well, if you had read the rest of the post, Eric, you would have seen
that the equations had to work with two clocks running at different
rates.
Robert B. Winn

But you said Galilean transformations. You wrote t' = t. Why would you
say t' = t and then say t' isn't equal to t?- Hide quoted text -

- Show quoted text -

Where did I say t' was not equal to t?
Robert B. Winn

"From either frame of reference, the clock in S' says .1 sec and the
clock in S says 1 sec. "- Hide quoted text -

- Show quoted text -

That is correct.  If the clock in S' says .1 sec, then t  is .1 sec.
as seen from S'.

If the clock in S says 1 sec, then t' is 1 sec as seen from S.
Robert B. Winn

And there we go - t' is not equal to t. You aren't discussing the
Galilean transformations - you are discussing some bull*** you just
made up.

How many thousands of posts over the years could have been stopped if
you'd just admit the obvious?- Hide quoted text -

- Show quoted text -

I admit that you seem to be unable to grasp the concept that if two
clocks are running at different rates, you cannot represent them by
the equation t'=t.

And I admit that you seem to be unable to grasp the concept that one
clock cannot be represented by the equation t'=t.- Hide quoted text -


Any number of clocks can be represented by the equation t'=t.
Including one clock. Or no clocks at all. It depends on how many
clocks you have. In this problem, scientists said they had two
clocks, a cesium clock in S and a cesium clock in S' running at a
slower rate.
Robert B. Winn
.

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