On Feb 4, 12:50*am, "Jeckyl" wrote:
"snapdragon31" wrote in message
"snapdragon31" wrote in message
...
On Feb 1, 10:49 pm, "Jeckyl" wrote:
Let's look at this more clearly...
Let us assume our coordinates are such that at x=0,t=0 we also have
x'=0,
t'=0
Let us assume that we have a moving rod of length L travelling at
speed v
Let us assume the light is shining along the rod in the direction of
travel
Let us assume the light is emitted at t=t'=0, when the rear end of the
rod
is at x=x'=0

In the stationary frame of reference, the rod is moving while the
light
is
travelling at c
So the light gets to the other end of the rod, as seen by the
stationary
observer, at x, t where
t = L / (c - v)
x = c . t ... because x / t = c

That corresponds in the rods frame of reference to a point (x',t')

x' = gamma . ( x - v.t ) ... Lorentz
x' = gamma . ( c.t - v.t ) ... subs for x
x' = gamma . ( c - v ) . t ... factorise
x' = gamma . ( c - v ) . L / (c - v) ... subs for t
x' = gamma . L ... cancel

t' = gamma . ( t - v.x / c^2 ) ... Lorentz
t' = gamma . ( t - v.t / c ) ... subs for x
t' = gamma . (1 - v/c ) . t ... factorise
t' = gamma . (c - v) / c . L / (c - v) ... subs for t
t' = gamma . L / c ... cancel

Please try and come up something better. I believe you can because
you are smarter than the other guy.

Thanks for your patience .. you picked up my error in the above at the
line:

Let us assume that we have a moving rod of length L travelling at
speed v

It should have said that the L is the length of the moving rod measured
in
the stationary frame (ie the contracted length)

Your speed is incredible.

It helps when you know what you're talkin about .. even if one can stuff up
occasionaly on expressing it . *Indeed .. the equations would look better if
I used L' instead of L throughout .. not that it changes anything

I admit that I cannot think that fast.
The equations look good. *But I still need to plug in some numbers to
have a better feeling of what they are.

L' = 100m, Gamma = 1.34164
== L = 100/1.34164 = 74.54m

OK



Of course, one can do the analysis the other way .. by seeing that light
moving along the stationary rod at c transforms to light moving along the
moving rod at c

Lets assume the length of the stationary rod in the stationary frame is L
When light is emitted, we have at x=x'=0, t=t'=0

When light gets to the other end of the rod, we have x=L, t=L/c.
That point corresponds to (x',t') where

x = L = 74.54 m

No .. in this case, L = 100m

t = L/c = 2.48E-07 = 0.000000248 sec
Should t be L/(c-v)?

No .. but your L is wrong .. L is the length of the stationary rod in the
stationary frame

x' = gamma . ( x - v.t ) ... Lorentz
x' = 1.34 * (74.54 - 2E8 * 2.48E-07) = 33.33 m
(Sorry, it is not equal to 100m)

Because you stuffed up your L
Hi Jeckyl,

All calculations were based on the equations you provided. It showed
that your equations failed to work properly.


x' = gamma . ( L - v.L/c ) ... subs x=L, t=L/c
x' = gamma . (c - v) . L / c ... factorise

t' = gamma . ( t - v.x / c^2 ) ... Lorentz
t' = gamma . ( L/c - v.L / c^2 ) ... subs x=L, t=L/c
t' = gamma . ( c - v) L / c^2 ... factorise

x'/t' = ( gamma . (c - v) . L / c ) / ( gamma . ( c - v) L / c^2 ) = c

Again, the speed of light is c

Another way to prove x' / t' = c, given x = c * t is
x' = gamma . ( x - v.t ) ... Lorentz
t' = gamma . ( t - v.x / c^2 ) ... Lorentz
x'/t' = [gamma . (x - v.t)]/[gamma . (t - v.x / c^2)]
= (x - v.t)/(t - v.x / c^2)
= (c.t - v.t)/(t - v.c.t / c^2) * * *substitute x = c.t
= [t.(c-v)]/[t.(1-v/c)]
= (c-v)/[(c-v)/c]
= c

Indeed

In this case, there is no way to check what x, t, x' and t' are.

But if they are related by a Lorentz transformation, then the speed of light
is c.

But
one thing is sure that
if t t' due to time dilation then

You don't understand time dilation and simulatneity .. t is less that t'
because we are looking at different positions in space
Unfortunately, your equations were not correct. I could not show you
that the time dilation was not given by the formula t = gamma * t'
based on your formula.

Given v = 2E8 m/s, c = 3E8 m/s and x' = 100m
t' = x'/c = 3.33E-7 sec
Using Lorentz transformation to work backward, we get
x = gamma * (x' + v * t') = 223.61 m and
t = gamma * (t' + v * x' / c^2) = 7.45E-7 sec

t = 7.45E-7 sec t' = 3.33E-7 sec sorry your statement of t is less
than t' is not correct.

t/t' = 2.236 which is not equal to gamma (= 1.342)
2.236 = sqrt((c + v)/(c - v))
1.342 = gamma = 1 / sqrt(1 - v^2/c^2)

Interesting? The ratio between t and t' is not equal to gamma? It is
not a surprise if you understand Lorentz transformation thoroughly.


x must be greater than x', right?

At least you're thinking .. but you need to do some more learning first.
The idea of length elongation seems a bit difficult to you. Just
leave it until you fully understand the Lorentz transformation and
time dilation. One recommendation is to plug numbers into equations
to get a better feeling of what they are.