Cartesian coordinate transformations

[steve waterman]

I found a big typo in my final proof. I assume that this post will be "locked", and none will be able to comment. I did not want the final proof to be riddled with typos, and to represent the results of our conversations, on the Pressures" thread.
Hopefully, this post will not be deleted, though I do not know what powers you have xkcd to delete a math premise, on your own, likely you do.

"mathematically given cannot be changed by changing the given, mathematically*

example; let A = 4. Do some math. Therefore saying that A = 3, would be mathematically incorrect.

mathematically given by the Galilean Coordinate transformation equations, S (x,y,z) = S' (x',y',z') at t = t' = 0,

therefore S (1,2,3) = S' (1,2,3') at t = t' = 0,

x' = x -vt OR x = x' + vt, [ Galilean transformation and corresponding Galilean inversion transformation equations ]

S' (x',y',z') = S (x,y,z) - vt OR S (x,y,z) = S' (x',y',z') + vt

To be invariant transformations, then the Galilean must also be invariant in three dimensions. What if vt = (1,1,1)?

let vt to S' = +(1,1,1). [ keeping S stationary ] then S (1,2,3) = S' (1,2,3) + (1,1,1) = S' (2,3,4) OR
let vt to S = -(1,1,1). [ keeping S' stationary ] then S' (1,2,3) = S (1,2,3) - (1,1,1) = S (0,1,2)

S (1,2,3) = S' (1,2,3') at t = t' = 0 AND* S (1,2,3) = S' (1,2,3') at t > 0

therefore, S' (1,2,3) = S(1,2,3) = S' (2,3,4) OR S (1,2,3) = S' (1,2,3) = S' (0,1,2)

+vt to S does not equal -vt to S' transformation results

[Meteoric]

mathematically given by the Galilean Coordinate transformation equations, S (x,y,z) = S' (x',y',z') at t = t' = 0,

therefore S (1,2,3) = S' (1,2,3') at t = t' = 0,

What if vt = (1,1,1)?

If vt = (1,1,1), then t =/= 0, and neither of the equalities in the first quote are applicable to the situation.

[gmalivuk]

If vt = (1,1,1), then t =/= 0, and neither of the equalities in the first quote are applicable to the situation.

Welp, there you have it.

/thread