Mazulu
I want to get a better understanding of the mathematics of reference frames. Galillean transformations is a good place to start. http://en.wikipedia.org/wiki/Galilean_transformation

A Galillean transformation is used to transform coordinates from one reference frame to another. I elect to create two reference frames, S and S'. It makes it easier to think of two geometric boxes that are initially located in the same place at the same time. They each have a center at, (x,x'=0, y,y'=0, z,z'=0, t,t'=0). Let's make them square boxes of length "a" along each edge.
Mazulu
QUOTE (Mazulu+May 23 2012, 11:31 PM)
I want to get a better understanding of the mathematics of reference frames.  Galillean transformations is a good place to start.  http://en.wikipedia.org/wiki/Galilean_transformation

A Galillean transformation is used to transform coordinates from one reference frame to another.  I elect to create two reference frames, S and S'.  It makes it easier to think of two geometric boxes that are initially located in the same place at the same time.  They each have a center at, (x,x'=0, y,y'=0, z,z'=0, t,t'=0).  Let's make them square boxes of length "a" along each edge.

Since box S and box S' are just coordinate systems (not physical boxes), I can have box S' travel past (and through) box S, while traveling at velocity v along the x-coordinate. The clocks for each box starts at t = 0, t' = 0, when the center of box S (x=0,y=0,z=0) aligns with the center of box S' (x'=0,y'=0,z'=0). Box S just sits at the origin (x,y,z=0). Box S' can track the movement (or lack of movement) of the center of box S. Box S' will denote the position of box S as,
X'=-vT'

If S' was traveling past S at 50meters/sec, what will be the position of box S according to the S' coordinate system after 10 seconds?
Answer: X' = -vT', v=50m/s, T'=10s, X' = -(50m/s)(10s) = -500meters.

The Gallilean transformation, from S to S', is,

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

If we ignore the y,z coordinates, we can write the Galillean transformation from S to S' as,
|x'| = |1 -v||x|
|t' | |0 1||t |

Which is x' = x - vt, and t' = t.

What is the coordinate of box S' in the S coordinate system?
X = X' + vT = 0 + 50m/s*10sec = X' = 500meters
T' = T = 10 seconds

waitedavid137
QUOTE (Mazulu+May 23 2012, 06:46 PM)
Since box S and box S' are just coordinate systems (not physical boxes), I can have box S' travel past (and through) box S, while traveling at velocity v along the x-coordinate. The clocks for each box starts at t = 0, t' = 0, when the center of box S (x=0,y=0,z=0) aligns with the center of box S' (x'=0,y'=0,z'=0). Box S just sits at the origin (x,y,z=0). Box S' can track the movement (or lack of movement) of the center of box S. Box S' will denote the position of box S as,
X'=-vT'

If S' was traveling past S at 50meters/sec, what will be the position of box S according to the S' coordinate system after 10 seconds?
Answer: X' = -vT', v=50m/s, T'=10s, X' = -(50m/s)(10s) = -500meters.

The Galilean transformation, from S to S', is,

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

If we ignore the y,z coordinates, we can write the Galilean transformation from S to S' as,
|x'| = |1 -v||x|
|t' | |0 1||t |

Which is x' = x - vt, and t' = t.

What is the coordinate of box S' in the S coordinate system?
X = X' + vT = 0 + 50m/s*10sec = X' = 500meters
T' = T = 10 seconds

You are calculating the location of the spatial Origin of S' with respect to S at that time, not the location of "the box" at that time as "the box" is actually an imagined infinitely extent boxed grid extent everywhere throughout all space at all times.
Confused1
@Mazulu, This looks interesting - where is it heading? (I have some suggestions if you don't). -C2.
Mazulu
QUOTE (Confused1+May 24 2012, 11:26 AM)
@Mazulu, This looks interesting - where is it heading? (I have some suggestions if you don't). -C2.

I'd like to move into space-time geometry.
Mazulu
From general relativity, the square of the interval between two events is invariant for all reference frames. The interval between two events in reference frame S is (in natural units,
(Delta s)^2 = -(Delta t)^2 + (Delta x)^2 + (Delta y)^2 + (Delta z)^2.

The interval between the same two events in reference frame S' is written as,
(Delta s')^2 = -(Delta t')^2 + (Delta x')^2 + (Delta y')^2 + (Delta z')^2.

Since the square of the interval is invariant for all reference frames,
(Delta s')^2 = (Delta s)^2.

QUOTE (waitedavid+)
You are calculating the location of the spatial Origin of S' with respect to S at that time, not the location of "the box" at that time as "the box" is actually an imagined infinitely extent boxed grid extent everywhere throughout all space at all times.

I will define box A, which represents coordinate system S, with vertices located at the following points.
(-a/2,-a/2,-a/2,t),
(-a/2,-a/2, a/2,t),
(-a/2, a/2,-a/2,t),
(-a/2, a/2, a/2,t),
( a/2,-a/2,-a/2,t),
( a/2,-a/2, a/2,t),
( a/2, a/2,-a/2,t),
( a/2, a/2, a/2,t).
waitedavid137
QUOTE (Mazulu+May 24 2012, 04:31 PM)
...I will define box A, which represents coordinate system S...

Thats nonsensicle.
Mazulu
QUOTE (waitedavid137+May 25 2012, 03:59 AM)
Thats nonsensicle.

Would you prefer that I define a coordinate system S that represents box A?
AlexG
QUOTE (waitedavid137+May 24 2012, 10:59 PM)
Thats nonsensicle.

That's Mazulu.
Confused1
@Mazulu,
It seems possible that we maybe ought to start with something a bit simpler.
Suggestion..
1/ Draw the x and t coordinates of a clock in it's own frame
2/ Draw the x and t coordinates of a clock seen from a frame moving at velocity v relative to the clock.
-C2.
Confused1

After drawing the graphs we see
1/ The clock has moved through time by an amount t and has stayed in the same place so distance moved (x) is zero.
2/ The clock has moved through time by an amount t' and has moved a distance x' where x' = vt'

In SR we know the invariant distance is constant (we'll call it s)
s² = -c²t² + x²
and
s² = -c²t'² + x'²
so
-c²t² + x² = -c²t² + x'²
in the clock frame x was zero so
-c²t² = -c²t'² + x'²

Can we do more?

we know x'=vt' so
-c²t² = -c²t'² + v²t'²
dividing by t'²
-c²t²/t'² = -c² + v²
dividing by -c²
t²/t'² = 1-v²/c²
or
t = t' √(1-v²/c²) which, all being well, is the Lorentz transform for time.
if instead/also we solved for x we'd have the result that mik needs to know.
-C2.
Mazulu
QUOTE (Confused1+May 26 2012, 10:16 AM)

After drawing the graphs we see
1/ The clock has moved through time by an amount t and has stayed in the same place so distance moved (x) is zero.
2/ The clock has moved through time by an amount t' and has moved a distance x' where x' = vt'

In SR we know the invariant distance is constant (we'll call it s)
s² = -c²t² + x²
and
s² = -c²t'² + x'²
so
-c²t² + x² = -c²t² + x'²
in the clock frame x was zero so
-c²t² = -c²t'² + x'²

Can we do more?

we know x'=vt' so
-c²t² = -c²t'² + v²t'²
dividing by t'²
-c²t²/t'² = -c² + v²
dividing by -c²
t²/t'² = 1-v²/c²
or
t = t' √(1-v²/c²) which, all being well, is the Lorentz transform for time.
if instead/also we solved for x we'd have the result that mik needs to know.
-C2.

Nice.
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