Re: Galilean transformation equations

On Feb 6, 6:34�am, PD <TheDraperFam...@xxxxxxxxx> wrote:
On Feb 6, 5:33�am, rbwinn <rbwi...@xxxxxxxx> wrote:





On Feb 5, 6:49 pm, PD <TheDraperFam...@xxxxxxxxx> wrote:

On Feb 5, 7:01 pm, rbwinn <rbwi...@xxxxxxxx> wrote:

On Feb 5, 6:02 am, PD <TheDraperFam...@xxxxxxxxx> wrote:

On Feb 4, 8:42 pm, rbwinn <rbwi...@xxxxxxxx> wrote:

On Feb 4, 6:20 am, PD <TheDraperFam...@xxxxxxxxx> wrote:

On Feb 4, 5:44 am, rbwinn <rbwi...@xxxxxxxx> wrote:

Well, it is similar to scientists when they are asked about two
different clocks running at different rates which scientists say show
the same velocity. They disappear for a while.
Robert B. Winn

I know you take the fact that, after a while, scientists start
ignoring you as some kind of victory on your part.

You have an interesting way of viewing your own influence.

PD

If I had any influence other than the ability to make scientists run,
PD, there would be a scientist somewhere capable of talking something
other than garbage.
Robert B. Winn

I don't even think you have the ability to make scientists run, Bobby.
I think you do have the ability to bore scientists until they walk
away. I imagine that, from your perspective, those two are
indistinguishable.

PD- Hide quoted text -

- Show quoted text -

Well, you personally have not said a word about relativity for more
than a year. How do two clocks running at different rates show the
same velocity, PD?

They show the same *relative* velocity, Robert. Why would you think
they don't.- Hide quoted text -

- Show quoted text -

I think they do not because the two clocks show that something travels
a certain distance in two different times.

That's in error, Bobby, because there is no "certain distance" that is
common between the two frames, just as there is no "certain time" that
is common between the two frames. Different distances, different
times, same relative velocities.

This is what I meant when I said that 1.00/3.00 = 1.01/3.03. Those
ratios are the same, even though the numerators are different AND the
denominators are different.

Is this really that hard for you? It appears that it is taking an
*extraordinarily* long time for you to grasp this, even for a welder.

I can grasp it perfectly.

c= x/t =x'/n' =(x-vt)/(t-v/c)=(x-vt)/(t-vx/c^2)

= [(x-vt)/sqrt(1-v^2/c^2)]/[(t-vx/c^2)/sqrt(1-v^2/c^2)

= x'Lorentz/t'Lorentz

I know everything there is to know about how the Lorentz equations
relate to relativity.
Robert B. Winn
.

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