Re: coordinates and equations.
- From: rbwinn <rbwinn3@xxxxxxxx>
- Date: Sat, 29 Nov 2008 19:00:14 -0800 (PST)
On Nov 28, 9:32�am, YBM <ybm...@xxxxxxxx> wrote:
rbwinn a �crit :
This is a simple way to think of coordinates for some of our
scientific friends. �See, there is an x axis and a y axis and a
z axis. �So if we have an equation that says
� � � � � � � � x^2 + y^2 + z^2 - c^2t^2 = 0
This is not what the equation *says*, this is what the equation *is*.
Couldn't you realize that in algebra, as well as in physics, writting
what the equation means is as important as writing them down ?
So what this is equation represents is a relation between coordinates
of specific events: (x,y,z) are coordinates of points on the spherical
light front at time t in S, light having been emitted from (0,0,0) at
t=0 in S.
I hope you agree.
where x, y, and z are coordinates, and t is time on a clock in S, the
frame of reference, and c is the speed of light, then if you
substitute numerical values into the equation, what you end up with is
a sphere with a radius of ct with its center at the origin of S.
Now we will do the same thing with another set of coordinates S'.
� � � � � �(x')^2 + (y')^2 + (z')^2 - c^2(n')^2 = 0
where n' is the time on a clock in S'. �Scientists have told us that
according to scientific experiments, light is traveling at c=300,000
km/sec according to a clock in S'. �So what we end up with here is a
sphere with a radius of c(n') with its center at the origin of S'.
� �I hope this will help some of our scientific friends to understand
how light propagates. �I know it is confusing to them because there is
no length contraction in these equations.
You only did the first step, now try to find out how you could express
a transformation giving x',y',z',n' as function of x,y,z,t depending on
v such as :
IF x^2 + y^2 + z^2 - c^2t^2 = 0
THEN (x')^2 + (y')^2 + (z')^2 - c^2(n')^2 = 0
Don't forget the condition that S' is moving with respect to S at speed
v.
If you solve this problem correctly you'll end up with Lorentz
Transformation.
Then you could try to understand what people refer to when talking about
"lenght contraction" with LT, something you don't understand yet which
is not very surprising given that you don't understand what a
transformation is.- Hide quoted text -
- Show quoted text -
Well, no, the transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
so you have the equation
(x')^2 + (y')^2 + (z')^2 - (c-v)^2(t'^2) = 0
and the equation
(x')^2 +(y')^2 +(z')^2 -c^2(n')^2 = 0
c^(n')^2 = (c-v)^2(t')^2
n'=t(1-v/c)
Funny how mathematics works, isn't it?
Robert B. Winn
.
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