On Wed, 02 Jan 2013 07:17:18 -0800, G=EMC^2 wrote:

On Dec 31 2012, 11:58Â*am, kenseto wrote:
On Dec 31, 2:31Â*am, Sylvia Else wrote:









On 31/12/2012 5:04 PM, Koobee Wublee wrote:

On Dec 30, 4:17 pm, Sylvia Else wrote:
I started writing a post about this yesterday, then scrubbed it -
too

What is a paradox in special relativity (hereinafter SR)?

I've expressed the view that to contain a paradox, SR has to
predict,
from different frames, outcomes that are mutually incompatible. An
example that comes to mind (though not directly arising) from a
recent discussion is that in one frame, there is massive
destruction on a citywide scale, and in another other frame,
nothing much happens.

Clearly, if SR were to make such predictions for two frames, it
would have to be regarded as seriously wanting. Of course, it does
no such thing.

But people seem to want to regard measurements in two frames as
mutually incompatible if they give different results. I am at a
loss to understand why people would seek to regard those different
results as constituting a paradox that invalidates SR (well,
leaving intellectual dishonesty aside).

Â*From the Lorentz transformations, you can write down the
Â*following
equation per Minkowski spacetime. Â*Points #1, #2, and #3 are
observers. Â*They are observing the same target.

** Â*c^2 dt1^2 – ds1^2 = c^2 dt2^2 – ds2^2 = c^2 dt3^2 – ds3^2

Where

** Â*dt1 = Time flow at Point #1 ** Â*dt2 = Time flow at Point #2 **
Â*dt3 = Time flow at Point #3

** Â*ds1 = Observed target displacement segment by #1 ** Â*ds2 =
Observed target displacement segment by #2 ** Â*ds3 = Observed
target displacement segment by #3

The above spacetime equation can also be written as follows.

** Â*dt1^2 (1 – B1^2) = dt2^2 (1 – B2^2) = dt3^2 (1 – B3^2)

Where

** Â*B^2 = (ds/dt)^2 / c^2

When #1 is observing #2, the following equation can be deduced from
the equation above.

** Â*dt1^2 (1 – B1^2) = dt2^2 . . . (1)

Where

** Â*B2^2 = 0, #2 is observing itself

Similarly, when #2 is observing #1, the following equation can be
deduced.

** Â*dt1^2 = dt2^2 (1 – B2^2) . . . (2)

Where

** Â*B1^2 = 0, #1 is observing itself

According to relativity, the following must be true.

** Â*B1^2 = B2^2

Thus, equations (1) and (2) become the following equations.

** Â*dt1^2 (1 – B^2) = dt2^2 . . . (3)
** Â*dt2^2 = dt2^2 (1 – B^2) . . . (4)

I assume you meant to write

dt1^2 = dt2^2 (1 - B^2) . . . (4)

Where

** Â*B^2 = B1^2 = B2^2

The only time the equations (3) and (4) can co-exist is when B^2 =
0.

Which tells us nothing more than that when two observers observe each
other, the situation is symmetrical. Each will measure the same time
for equivalent displacements of the other. Or more simply, they share
a common relative velocity (save for sign).

Thus, the twins’ paradox is very real under the Lorentz transform.
shrug

blink Where did that come from? The twin "paradox" involves
bringing the two twins back together, which necessitates accelerating
at least one of them, making their frame non-inertial./blink

There is no inertial frame exists on earth ....does that mean that SR
is not valid on earth?

Well think of this. "Time in a plane flying east is less than that for
those flying west". The Earth speed of rotation sees to it. Get the
picture TreBert

Treeb is right. Every schoolkid knows that if you fly east, it's a time
machine. Every time you go around the earth you go back in time a day!
Get the Picture?