drop something from a moving horse and its gravity
fall path continuous to move with the horse...
at the original speed.

Two motions are joined as there is a moving path.

Galileo had also a moving ship with throwing
something having a gravity parabola fall path
and at the same time shares the motion
of the ship at the impetus where it was
thrown...

Gravity as a motion curve will keep other motions.

Mitchell Raemsch

On Thursday, February 23, 2023 at 6:19:24 PM UTC-7, mitchr...@gmail..com wrote:
> drop something from a moving horse and its gravity
> fall path continuous to move with the horse...
> at the original speed.
>
> Two motions are joined as there is a moving path.
>
> Galileo had also a moving ship with throwing
> something having a gravity parabola fall path
> and at the same time shares the motion
> of the ship at the impetus where it was
> thrown...
>
> Gravity as a motion curve will keep other motions.
>
> Mitchell Raemsch
Yes, I know. But here is the puzzle. If you drop a ball in a flying airplane, it will follow a parabolic path relative to the ground. The question for Einstein disciples is, If the clock in the airplane is slower, explain how the laws of physics are the same in both frames of reference, since the ball is falling faster in the airplane than it is according to the time of a clock on the ground?

On Friday, February 24, 2023 at 7:03:16 AM UTC+1, Robert Winn wrote:
> On Thursday, February 23, 2023 at 6:19:24 PM UTC-7, mitchr...@gmail.com wrote:
> > drop something from a moving horse and its gravity
> > fall path continuous to move with the horse...
> > at the original speed.
> >
> > Two motions are joined as there is a moving path.
> >
> > Galileo had also a moving ship with throwing
> > something having a gravity parabola fall path
> > and at the same time shares the motion
> > of the ship at the impetus where it was
> > thrown...
> >
> > Gravity as a motion curve will keep other motions.
> >
> > Mitchell Raemsch
> Yes, I know. But here is the puzzle. If you drop a ball in a flying airplane, it will follow a parabolic path relative to the ground. The question for Einstein disciples is, If the clock in the airplane is slower,

It's not intrinsically slower (meaning, not as measured inside that airplane).

One would think after so many years you'd get at least the basics of special
relativity. No?

--
Jan

On Thursday, February 23, 2023 at 11:21:13 PM UTC-7, JanPB wrote:
> On Friday, February 24, 2023 at 7:03:16 AM UTC+1, Robert Winn wrote:
> > On Thursday, February 23, 2023 at 6:19:24 PM UTC-7, mitchr...@gmail.com wrote:
> > > drop something from a moving horse and its gravity
> > > fall path continuous to move with the horse...
> > > at the original speed.
> > >
> > > Two motions are joined as there is a moving path.
> > >
> > > Galileo had also a moving ship with throwing
> > > something having a gravity parabola fall path
> > > and at the same time shares the motion
> > > of the ship at the impetus where it was
> > > thrown...
> > >
> > > Gravity as a motion curve will keep other motions.
> > >
> > > Mitchell Raemsch
> > Yes, I know. But here is the puzzle. If you drop a ball in a flying airplane, it will follow a parabolic path relative to the ground. The question for Einstein disciples is, If the clock in the airplane is slower,
> It's not intrinsically slower (meaning, not as measured inside that airplane).
>
> One would think after so many years you'd get at least the basics of special
> relativity. No?
>
> --
> Jan
I do not need Special Relativity. I use the Galilean transformation equations. Here is the problem with Special Relativity. In his book, Relativity, the Special and General Theories, Einstein says he has extracted two little equations from the Lorentz equations, x=ct and x'=ct', that show that light is traveling at c in both frames of reference. The reason these two little equations exist in conjunction with the Lorentz equations is because of a false interpretation of the Galilean transformation equations which results in the numerator of Lorentz's equation for t'.
t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
If you take the two little equations Einstein said he extracted from the Lorentz equations and substitute them into the Galilean transformation equations, you get

x'=x-vt
ct' = ct - vt
t' = t-vt/c = t-vct/c^2 = t-vx/c^2

Which is obviously how Lorentz came up with the numerator for his equation for t'.

The problem with this expression t-vx/c^2 is that it is an incorrect representation of time and only works for one value of x, the distance light travels in a time of t. t is the time it takes for frame of reference S' to travel a distance of vt relative to frame of reference S. It is not the time it takes light to travel a distance of x unless x is the distance light travels in time t, meaning there is only one value for x that satisfies the equation. If t is one second, x is 186,000 miles. But x can be any coordinate on the x axis. Suppose x is one mile. Then x does not equal ct if t is one second. x equals c(1/186,000) sec. If x is two miles, then x = c(2/186,000)sec., and so on. So if light is traveling at c in both frames of reference, then x does not equal ct and x' does not equal ct' except in the one special case. So the correct way to represent this is with different variables for the times that light takes to travel x and x'.

x=cn
x=cn'

cn' = cn-vt
n' = n -vt/c

Then if there is some reason to do what Lorentz and Einstein were trying to do, you would have

n' = n - v(?)/c^2

and (?) would be ct, which is not equal to x except in the one special case..
I hope this will help scientists with their errors in algebra. If you have any more questions, just let me know.