 If this is your first visit, be sure to check out the FAQ by clicking the link above. You may have to register before you can post: click the register link above to proceed. To start viewing messages, select the forum that you want to visit from the selection below.  What is or is not a paradox?
 Site Map Home Authors List Search Today's Posts Mark Forums Read Web Partners

## What is or is not a paradox?

#1 December 31st 12, 07:04 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
 Koobee Wublee external usenet poster Posts: 815 What is or is not a paradox?

On Dec 30, 4:17 pm, Sylvia Else wrote:

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).

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

** c^2 dt1^2  ds1^2 = c^2 dt2^2  ds2^2 = c^2 dt3^2  ds3^2

Where

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

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

The above spacetime equation can also be written as follows.

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

Where

** B^2 = (ds/dt)^2 / c^2

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

** dt1^2 (1  B1^2) = dt2^2 . . . (1)

Where

** B2^2 = 0, #2 is observing itself

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

** dt1^2 = dt2^2 (1  B2^2) . . . (2)

Where

** B1^2 = 0, #1 is observing itself

According to relativity, the following must be true.

** B1^2 = B2^2

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

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

Where

** B^2 = B1^2 = B2^2

The only time the equations (3) and (4) can co-exist is when B^2 = 0.
Thus, the twins paradox is very real under the Lorentz transform.
shrug

Within the Lorentz transform, the little professor from Norway, paul
andersen, was able to play a mathemagic trick, and he is not alone.
In doing so, the Minkowski spacetime was not recognized in his little
applet. He is out in the very left field chasing chickens again.
shrug

Tom and other self-styled physicists have recognized that fault and
moved on to claim a mythical proper time flow where all local time
flow is a projection of this absolute time flow. Oh, excuse Koobee
Wublee. Not absolute time but proper time whatever \$hit it is.
However, these guys cannot explain why the projection did not cancel
out on the traveling twins return trip. So, equations (3) and (4)
are still indicating the paradox regardless if projection or not.
shrug

Well, sooner or later, these bozos are going to wake up someday and
ask themselves what the fvck was I thinking?. Guess what? The time
projection crap is the last piece of float the self-styled physicists
are clinging on to. Take that away. They will sink. That is why the
self-styled physicists are very reluctant to give the time projection
crap a serious thought. shrug

#2 December 31st 12, 08:31 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
 Sylvia Else external usenet poster Posts: 1,063 What is or is not a paradox?

On 31/12/2012 5:04 PM, Koobee Wublee wrote:
On Dec 30, 4:17 pm, Sylvia Else wrote:

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).

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

** c^2 dt1^2  ds1^2 = c^2 dt2^2  ds2^2 = c^2 dt3^2  ds3^2

Where

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

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

The above spacetime equation can also be written as follows.

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

Where

** B^2 = (ds/dt)^2 / c^2

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

** dt1^2 (1  B1^2) = dt2^2 . . . (1)

Where

** B2^2 = 0, #2 is observing itself

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

** dt1^2 = dt2^2 (1  B2^2) . . . (2)

Where

** B1^2 = 0, #1 is observing itself

According to relativity, the following must be true.

** B1^2 = B2^2

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

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

I assume you meant to write

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

Where

** 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

the two twins back together, which necessitates accelerating at least
one of them, making their frame non-inertial./blink

Sylvia.
#3 December 31st 12, 09:49 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
 Koobee Wublee external usenet poster Posts: 815 What is or is not a paradox?

On Dec 30, 11:31 pm, Sylvia Else wrote:
On 31/12/2012 5:04 PM, Koobee Wublee wrote:

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

** c^2 dt1^2  ds1^2 = c^2 dt2^2  ds2^2 = c^2 dt3^2  ds3^2

Where

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

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

The above spacetime equation can also be written as follows.

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

Where

** B^2 = (ds/dt)^2 / c^2

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

** dt1^2 (1  B1^2) = dt2^2 . . . (1)

Where

** B2^2 = 0, #2 is observing itself

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

** dt1^2 = dt2^2 (1  B2^2) . . . (2)

Where

** B1^2 = 0, #1 is observing itself

According to relativity, the following must be true.

** B1^2 = B2^2

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

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

I assume you meant to write

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

No, Koobee Wublee meant every letter in the equations (3) and (4).
shrug

Where

** 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?

Have you not been reading Koobee Wublee? Did Koobee Wublee not say
the Lorentz transform? shrug

the two twins back together, which necessitates accelerating at least
one of them, making their frame non-inertial./blink

So, you believe in the nonsense of Born? He was the first one to
propose acceleration thing breaking the symmetry. Can you show any
mathematics that support your/Borns claim? No self-styled physicists
have now believed in such nonsense. shrug

#4 December 31st 12, 09:52 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
 Lord Androcles, Zeroth Earl of Medway[_5_] external usenet poster Posts: 74 What is or is not a paradox?

"Sylvia Else" wrote in message ...

the two twins back together, which necessitates accelerating at least
one of them, making their frame non-inertial./blink

Sylvia.

==================================================
"If one of two synchronous clocks at A is moved in a closed curve with
constant velocity until it returns to A, the journey lasting t seconds, then
by the clock which has remained at rest the travelled clock on its arrival
at A will be 1/2 tv^2/c^2 second slow." -- Einstein.
blink/ Non-inertial? Where did that come from?
The twin "paradox" involves bringing the two twins back together, which
necessitates keeping one at absolute rest, but the phenomena of
electrodynamics as well as of mechanics possess no properties corresponding
to the idea of absolute rest.
Oh wait, I get it. You are discussing Phuckwit Duck's special relativity,

-- This message is brought to you from the keyboard of
Lord Androcles, Zeroth Earl of Medway.
When I get my O.B.E. I'll be an earlobe.

#5 December 31st 12, 10:48 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
 Sylvia Else external usenet poster Posts: 1,063 What is or is not a paradox?

On 31/12/2012 7:49 PM, Koobee Wublee wrote:
On Dec 30, 11:31 pm, Sylvia Else wrote:
On 31/12/2012 5:04 PM, Koobee Wublee wrote:

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

** c^2 dt1^2  ds1^2 = c^2 dt2^2  ds2^2 = c^2 dt3^2  ds3^2

Where

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

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

The above spacetime equation can also be written as follows.

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

Where

** B^2 = (ds/dt)^2 / c^2

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

** dt1^2 (1  B1^2) = dt2^2 . . . (1)

Where

** B2^2 = 0, #2 is observing itself

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

** dt1^2 = dt2^2 (1  B2^2) . . . (2)

Where

** B1^2 = 0, #1 is observing itself

According to relativity, the following must be true.

** B1^2 = B2^2

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

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

I assume you meant to write

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

No, Koobee Wublee meant every letter in the equations (3) and (4).

(2) doesn't become (4) just be writing B for B2.

shrug

Where

** 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).

In the classical twins paradox, there is no symmetry. The travelling
twin has to change velocities in order to be able to get back to the
stay at home twin.

To get symmetry, both twins have to travel, and if the travel is really
symmetrical, their ages will match when they return.

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

blink Where did that come from?

Have you not been reading Koobee Wublee? Did Koobee Wublee not say
the Lorentz transform? shrug

the two twins back together, which necessitates accelerating at least
one of them, making their frame non-inertial./blink

So, you believe in the nonsense of Born? He was the first one to
propose acceleration thing breaking the symmetry. Can you show any
mathematics that support your/Borns claim? No self-styled physicists
have now believed in such nonsense. shrug

The symmetry can be broken without acceleration though to bring an
actual person back then involves cloning. It's simpler to forget the
twin, and just take a clock whose time is copied onto another clock
going in the opposite direction halfway through the travel.

But the symmetry is still broken, and once that happens, you have no

Sylvia.

#6 December 31st 12, 05:58 PM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
 kenseto[_2_] external usenet poster Posts: 17 What is or is not a paradox?

On Dec 31, 2:31*am, Sylvia Else wrote:
On 31/12/2012 5:04 PM, Koobee Wublee wrote:

On Dec 30, 4:17 pm, Sylvia Else wrote:

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).

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

** *c^2 dt1^2  ds1^2 = c^2 dt2^2  ds2^2 = c^2 dt3^2  ds3^2

Where

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

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

The above spacetime equation can also be written as follows.

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

Where

** *B^2 = (ds/dt)^2 / c^2

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

** *dt1^2 (1  B1^2) = dt2^2 . . . (1)

Where

** *B2^2 = 0, #2 is observing itself

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

** *dt1^2 = dt2^2 (1  B2^2) . . . (2)

Where

** *B1^2 = 0, #1 is observing itself

According to relativity, the following must be true.

** *B1^2 = B2^2

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

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

I assume you meant to write

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

Where

** *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

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?
#7 January 1st 13, 10:59 PM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
 Koobee Wublee external usenet poster Posts: 815 What is or is not a paradox?

On Dec 31 2012, 1:48 am, Sylvia Else wrote:
On 31/12/2012 5:04 PM, Koobee Wublee wrote:

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

** c^2 dt1^2  ds1^2 = c^2 dt2^2  ds2^2 = c^2 dt3^2  ds3^2

Where

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

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

The above spacetime equation can also be written as follows.

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

Where

** B^2 = (ds/dt)^2 / c^2

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

** dt1^2 (1  B1^2) = dt2^2 . . . (1)

Where

** B2^2 = 0, #2 is observing itself

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

** dt1^2 = dt2^2 (1  B2^2) . . . (2)

Where

** B1^2 = 0, #1 is observing itself

According to relativity, the following must be true.

** B1^2 = B2^2

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

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

Where

** B^2 = B1^2 = B2^2

(2) doesn't become (4) just be writing B for B2.

Are you complaining about the typo? It is corrected above. shrug

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

In the classical twins paradox, there is no symmetry. The travelling
twin has to change velocities in order to be able to get back to the
stay at home twin.

[snip more nonsense]

This is the second time, you are asked to show the math that shows
this acceleration breaking the symmetry nonsense. There is no way you
can, and that is because you are totally wrong just like Born.
shrug

So, you believe in the nonsense of Born? He was the first one to
propose acceleration thing breaking the symmetry. Can you show any
mathematics that support your/Borns claim? No self-styled physicists
have now believed in such nonsense. shrug

The symmetry can be broken without acceleration though to bring an
actual person back then involves cloning. It's simpler to forget the
twin, and just take a clock whose time is copied onto another clock
going in the opposite direction halfway through the travel.

But the symmetry is still broken, and once that happens, you have no

You have no idea what you are talking about, and there is no need to
discuss any further. shrug

#8 January 1st 13, 11:35 PM posted to sci.physics.relativity,sci.physics,sci.astro,sci.math
 Koobee Wublee external usenet poster Posts: 815 What is or is not a paradox?

On Jan 1, 11:46 am, Tom Roberts wrote:

OK, consider the case in which "home" is at rest in an inertial frame; one twin
stays at home in a centrifuge, spinning at a few tens of meters per second with
a proper acceleration of 1g; the other twin blasts off in a rocket accelerating
at 1g for a year on his clock, turns around with a maneuver that maintains the
1g acceleration, and returns home executing another maneuver that beings him to
rest at home (the trip takes 4 years on the rocket twin's clock: 1y speed up
headed out, 1y slow down, 1y speed up headed back, 1y slow down, stop). The
rocket twin is unequivocally younger when they meet again, because his average
speed relative to the inertial frame of home is much greater than that of the
centrifuge twin, even though their proper accelerations are equal.

Just as Koobee Wublee has predicted:

** Divine visions of showing no paradox to
** Euphoria of rejoicing no paradox to
** Careful examination of these divine visions to
** what the fvck was I thinking? rude awakening to
** Desperation once again and the cycle repeats.

shrug

The divine visions a

1) Acceleration breaking the symmetry by Born
2) Spacetime time diagram of a few lines
3) Mathemagic trick by paul andersen and others
4) Coordinate time as projection of proper time

The last time, Tom was in 4) so happy with the projection thing.
Apparently, he had gone through the rude awakening stage and more
desperation. Now, he is back to the first one. That is the life of a
self-styled physicist. Koobee Wublee thinks that is a fvcked up life
style, but what does Koobee Wublee know? Perhaps, Tom enjoys stuck in
these cycles chasing his own tail. shrug

Anyway, one can simply redesign the scenario to have both twins
traveling where each twin will experience the same acceleration
profile and thus nullifying the effect of acceleration. Given an
arbitrary amount of time to allow both twins coasting with null
acceleration but with a significant speed between them, the mutual
time dilation building up will obviously never be rectified. Thus,
showing the paradox is very real. shrug

Bottom line: here it is speed relative to the inertial frame of home that
matters, not acceleration. The actual computation involves integrating the
metric over the path followed by each twin between meetings, and comparing the
results; when written in terms of home's inertial coordinates, this integral
involves only speed, not acceleration, not position, and not direction.

Remember that in SR (no gravity) it simply is not possible for
two twins that both move inertially to separate and rejoin.

Tom, care to show some math to back up your babbling? shrug
#9 January 2nd 13, 04:17 PM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
 G=EMC^2[_2_] external usenet poster Posts: 2,655 What is or is not a paradox?

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

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

On Dec 30, 4:17 pm, Sylvia Else wrote:

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).

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

** *c^2 dt1^2  ds1^2 = c^2 dt2^2  ds2^2 = c^2 dt3^2  ds3^2

Where

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

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

The above spacetime equation can also be written as follows.

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

Where

** *B^2 = (ds/dt)^2 / c^2

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

** *dt1^2 (1  B1^2) = dt2^2 . . . (1)

Where

** *B2^2 = 0, #2 is observing itself

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

** *dt1^2 = dt2^2 (1  B2^2) . . . (2)

Where

** *B1^2 = 0, #1 is observing itself

According to relativity, the following must be true.

** *B1^2 = B2^2

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

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

I assume you meant to write

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

Where

** *B^2 = B1^2 = B2^2

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

#10 January 2nd 13, 05:41 PM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
 benj external usenet poster Posts: 23 What is or is not a paradox?

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:
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 â ds1^2 = c^2 dt2^2 â ds2^2 = c^2 dt3^2 â 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 â B1^2) = dt2^2 (1 â B2^2) = dt3^2 (1 â 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 â 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 â 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 â B^2) = dt2^2 . . . (3)
** Â*dt2^2 = dt2^2 (1 â 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 twinsâ paradox is very real under the Lorentz transform.
shrug

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?

 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts vB code is On Smilies are On [IMG] code is On HTML code is Off
 Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Space Science     Space Science Misc     News     Space Shuttle     Space Station     Science     Technology     Policy     History Astronomy and Astrophysics     Astronomy Misc     Amateur Astronomy     CCD Imaging     Research     FITS     Satellites     Hubble     SETI Others     Astro Pictures     Solar     UK Astronomy     Misc About SpaceBanter     About this forum Similar Threads Thread Thread Starter Forum Replies Last Post Fermi Paradox mike3 SETI 7 September 23rd 08 09:11 AM The Cow Paradox Keith Wood SETI 5 December 30th 06 01:10 AM what if paradox kjakja Misc 130 December 12th 04 05:09 AM Ashmore's Paradox Lyndon Ashmore Misc 3 November 22nd 03 01:04 PM Ashmore's Paradox Lyndon Ashmore UK Astronomy 0 November 14th 03 10:32 AM

All times are GMT +1. The time now is 02:09 AM. - Contact Us - SpaceBanter Home - FAQ - Links - Privacy Statement - Top Copyright ©2004-2019 SpaceBanter.com.