on relative motion

[Sam Wormley]

How can v the same for both A and B when a second to define v in A
and B have different duration?

Let x be the distance between A and B at some point in time.
Then the velocity between A and B at that point in time is v = dx/dt .

Remember the Lorentz transformation that applies to A and B

http://en.wikipedia.org/wiki/Relativity_of_simultaneity#Lorentz_transformations
The relativity of simultaneity can be calculated using Lorentz
transformations, which relate the coordinates used by one observer to
coordinates used by another in uniform relative motion with respect
to the first.
Assume that A uses coordinates labeled t, x, y, and z, while B uses
coordinates labeled t', x', y', and z'. Now suppose that A sees B
moving in the x-direction at a velocity v. And suppose that the
observer's coordinate axes are parallel and that they have the same
origin.
Then, the Lorentz transformations show that the coordinates are
t' = (t-vx/c^2)/(√(1-v^2/c^2))
x' = ((x-vt)/(√(1-v^2/c^2))
y' = y
z' = z
where c is the speed of light. If two events happen at the same time
in the frame of the first observer, they will have identical values
of the t-coordinate. However, if they have different values of the
x-coordinate (different positions in the x-direction), we see that
they will have different values of the t' coordinate; they will
happen at different times in that frame. The term that accounts for
the failure of absolute simultaneity is that vx/c^2.

[Sam Wormley]

Length Contraction Problem for Seto

Problem: At what speed does a meter stick move if its length is
observed to shrink to 0.5 m?

Solution: We assume that the meter stick is at rest in S'. As
observed by stationary observers in S, the meter stick moves
in the positive x-direction with speed v.

x' = γ(x−vt)

which relates the position x' measured in S' with the position
x measured in S.

Let ∆x' be the length of the meter stick measured by an observer
at rest in S'. (∆x' is the proper length of the meter stick.)

The meter stick is moving with speed v along the x axis in S. To
determine its length in S, the positions of the front and back of
the meter stick are observed by two stationary observers in S at
the same time. The length of the meter stick as measured in S is
the distance ∆x between the two stationary observers at ∆t = 0.

Then

∆x' = γ ∆x

where γ = 1/√(1-v^2/c^2)

It follows that

β = √(1 - (∆x/∆x')^2)

And that β = 0.866 when ∆x = ∆x'/2.

∆x' = γ ∆x indicates that the length ∆x of an object measured by
observers at rest in S is smaller than the length ∆x' measured by
an observer at rest with respect to the meter stick. That is,
“moving rods contract”.

It is important to note that ∆x' = γ ∆x compares an actual length
measurement in S' (a proper length) with a length measurement
determined at equal times on two separate clocks in S.

[benj]

Post by Sam Wormley
Length Contraction Problem for Seto
Problem: At what speed does a meter stick move if its length is
observed to shrink to 0.5 m?
Solution: We assume that the meter stick is at rest in S'. As
observed by stationary observers in S, the meter stick moves
in the positive x-direction with speed v.
x' = γ(x−vt)
which relates the position x' measured in S' with the position
x measured in S.
Let ∆x' be the length of the meter stick measured by an observer
at rest in S'. (∆x' is the proper length of the meter stick.)
The meter stick is moving with speed v along the x axis in S. To
determine its length in S, the positions of the front and back of
the meter stick are observed by two stationary observers in S at
the same time. The length of the meter stick as measured in S is
the distance ∆x between the two stationary observers at ∆t = 0.
Then
∆x' = γ ∆x
where γ = 1/√(1-v^2/c^2)
It follows that
β = √(1 - (∆x/∆x')^2)
And that β = 0.866 when ∆x = ∆x'/2.
∆x' = γ ∆x indicates that the length ∆x of an object measured by
observers at rest in S is smaller than the length ∆x' measured by
an observer at rest with respect to the meter stick. That is,
“moving rods contract”.
It is important to note that ∆x' = γ ∆x compares an actual length
measurement in S' (a proper length) with a length measurement
determined at equal times on two separate clocks in S.

Sammy! Moving rods do not contract! They only APPEAR shorter to the
observer in the other frame. Appearances are not the same thing as
actually shrinking! You are FAR too gullible to do science, Worm.

[Lloyd Buckner]

Muon are just a clock!
And it slows down about 10 times, because we measure the time of fly
22us and the muon localy gets 2.2us only.

Nope. The muon's clock does not slow down at all. But an OBSERVER on
earth MEASURES the moving muon's clock to be slower than the clock in a
muon at rest.

This essentially an impossible task to do. Since you need two clocks, one
fixed at muon creation, then the other at earth sea level.

However, that would be another muon. Not that one created. Since once
measured, even considered as invasive (another impossibility), you would
never catch that muon again at the level of the sea.

As you can see, there are many levels of impossibilities in Relativity.

[nntp]

Obviously more than nntp.

How is that obvious. Are we sure?

I've a lot of your name shifting alias in my plonkers cellar-- Keeps
down the screaming noise of trolls. It takes way less time to plonk
you (one click) than it does you to create a new alias.

What are you talking about? First time I see. If you have nothing to say,
go to sleep.

Hmmm, "name shifting". What is it, is "Sam Wormley" your real name? Funny
name, no doubts about that.

I thought coming here we would talk physics and relativity. Too bad.

[nntp]

Post by nntp

Obviously more than nntp.

How is that obvious. Are we sure?

I've a lot of your name shifting alias in my plonkers cellar-- Keeps
down the screaming noise of trolls. It takes way less time to plonk
you (one click) than it does you to create a new alias.

What are you talking about? First time I see. If you have nothing to
say,sci.physics.relativity
go to sleep.
Hmmm, "name shifting". What is it, is "Sam Wormley" your real name?
Funny name, no doubts about that.
I thought coming here we would talk physics and relativity. Too bad.

In fact in stead of Relativity, we can use Classical Physics with Gravity
as friction. I bet it is first time your read this theory.

[Walt Montgomery]

No it means that a clock second is not a universal interval of time.
They redefined the GPS second to have 4.46 more periods of Cs 133
radiation than the ground clock second to make the redefined GPS second
contain the same amount of absolute time.

But this is not true for the clocks and personnel up there. For them as
second is EXACTLY as large in duration as it is for you here on Earth.

I bet you disagree.

[Walt Montgomery]

Post by Walt Montgomery
For them as
second is EXACTLY as large in duration as it is for you here on Earth.

Only if they use the redefined GPS second to make the comparison.

Post by Walt Montgomery
I bet you disagree.

I bet you are right.

Lets assume you are able to boil an egg up there in space. If it takes you
2 minutes to boil an egg here on Earth, how many minutes takes to boil
that egg in space?

Or maybe this would made the point. You measure a muon lifetime here on
earth 2.2us. You take another muon in space and repeat the experiment. How
long would that muon live in space?