Twin paradox (Pete and MacM)

[Pete]

Are you just showing that what I did was reverse the cards on SR?

I'm not sure what you mean by "reverse the cards on SR"...

4 Seconds? To start with, light has a speed of 299,792,458 meters per second.
How did the travel distance become 1199169832 meters? The spaceship and the laser were separated by 299,792,458 meters when the light reflected off the prism on the spaceship.

The distance between the laser and the spaceship at the moment of reflection isn't directly relevant.

The travel distance is:
1) The distance between the laser at the moment of emission and the spaceship at the moment of reflection, plus
2) The distance between the laser at the moment of reception and the spaceship at the moment of reflection

In the spaceship frame, the light follows a diagonal path, which is more than 1 light second each way.

Look at this labelled diagram:
<img src=/attachment.php?attachmentid=3401&stc=1>

To get from the laser at emission to the ship at reflection in two seconds, the light beam must travel 1 light-second at right angles to the Earth's path, and 1.732 light-seconds (2 x 0.866) parallel to Earth's path.

Using pythagorus's theorem, you can check the length of the path: &radic;(1.732² + 1²) = 2 light-seconds

As I keep repeating, a moving inertial frame is NOT the same as a rest frame.

I'm not convinced we're attaching the same concept to some of these phrases...
In your mind, is it meaningful to talk about the rest frame of the spaceship? Or is that something you would call a "moving inertial frame"?

Yes, the spaceship clock would record MORE time than the Earth clock. Thanks, Pete.

What do you conclude from that fact?

 

[geistkiesel]

i think he got my message.

2. MacM believes in absolute time.
He believes that causing clocks to stop simultaneously in the frame of reference of one observer must stop them in every simultaneously in every frame of reference. He provides no reasoning as to why this should be true, except that he feels it makes more sense to him.

MacM did not say " . . .that causing clocks to stop simultaneously in the frame of reference of one observer must stop them in every simultaneously in every frame of reference." This is your editing, conscious editing of course.

You are perverting this thread by your post. MacM has suggested only that in the moving frames that a separate clock be installed and programed to tick at the earth frame rate,the A frame, using whatever SR formulae predict, programmed by the movingf observers. Then when the adjusted clock reaches 36000 all clocks in the moving frame stop and their output is recorded. Putting such clocks in the B and C frames to run simultaneously with the clocks that are not adjusted, the dilating clocks, then in all three frames, A , B and C all clocks should terminate ticking simultaneously without intruding into SR theory in the slightest. The adjustments are made using SR formulae.

In any event the experiment need not be conducted with both the B and C frame moving simultaneously. Do the B frame test on Monday and the C test on Friday and compare the results. The test is only a test of time dilation and is not a test of simultaneity.

If you want to accuse me of "believing in absolute time", making it read like an indictment, like such a person is some how immoral, I will save you the trouble: Absolute time is more than a belief, absolute time is a physical reality, a fact.

Just to make sure of the accuracy of the calculations used in adjusting the clocks an SR theorist should make the calulations in the A ftame and in the moving frames to assure himself, just in case, that SR time dilation implications of extensive velocity do not slow down the moving observer brain such that a different tick rate would result.

Like I stated before chroot, crawl back into your own forum hole.

 

[2inquisitive]

quote:
"I'm not sure what you mean by "reverse the cards on SR"..."
========================================================

I mean the only way to solve this paradox is to have the spaceship stopwatch record
MORE time than the Earth's stopwatch, reverse of what SR states. That is unless you
use Newtonian reality, in which both stopwatches record the SAME time, two seconds.
Reality would have Earth ALWAYS in the rest frame and the spaceship ALWAYS in the
moving frame. The laser beam would travel straight up and down in BOTH frames, as
it should. If you try to solve the example by Special Relativity, you end up with the
spaceship stopwatch recording more time than the Earth stopwatch.
================================================================

by Pete:
"I'm not convinced we're attaching the same concept to some of these phrases...
In your mind, is it meaningful to talk about the rest frame of the spaceship? Or is that something you would call a "moving inertial frame"?"
================================================================

I admit that sounded a little confusing. In a true rest frame, an event can be viewed
simaltaneously by two observers of equal distance from the said event. An observer
in relative motion to the two stationary observers/event will not disagree that the
event can happen simaltaneously between them. However, do what SR does and
let the 'moving' observer consider himself at rest and place the 'stationary observers/event' in the moving frame and the physics ARE different. Events cannot
happen simaltaneously between the 'now moving' observers and event. Straight up
and down paths now become 'V's and lengthen distances and travel times. By always
specifying the observer at rest in his own frame and the other moving, physics are
changed. SR manipulates time and distance this way, based on faulty physics.
===========================================================

by Pete:
"What do you conclude from that fact?" (In reference to "the spaceship clock would
record MORE time than the Earth clock)
==============================================================

That what comes around, goes around. I simply found a way to reverse the conclusions of Special Relativity and show the spaceship clock as running FASTER
than the Earth clock. It does not in reality, of course, but is due to the faulty physics
of Special Relativity. I am sure a lot of physicists have discovered the same thing I
have, but the faults are not exposed. Fear of embarrassing the mainstream physics
community or what? I, of course, may have it all wrong but that will have to be shown.

 

[MacM]

Pete, you are one of the physicists on these forums that I believe has integrity.

I just want to step in and second that notion. I believe Pete has shown an integrity that should he become convienced a sound arguement had been made he wouuld so state. Other I do not believe would.

 

[Pete]

quote:
"I'm not sure what you mean by "reverse the cards on SR"..."
========================================================

I mean the only way to solve this paradox is to have the spaceship stopwatch record MORE time than the Earth's stopwatch, reverse of what SR states.

It is not the reverse of what SR states, although it can be difficult to see why. I think that you're thinking that the spaceship has to be the moving frame, and the Earth has to be the stationary frame...

When working out time dilation, it helps to keep in mind that the time between a particular pair of events is shortest in the frame in which those events occur in the same place*, and longer in all other frames.

(* Assuming the events have timelike separation. If they don't, then there is no frame in which they occur in the same place)

Reality would have Earth ALWAYS in the rest frame and the spaceship ALWAYS in the moving frame. The laser beam would travel straight up and down in BOTH frames, as it should.

Is the Earth's surface the only stationary frame?
Is the Earth's surface a stationary frame at all?
How do you distinguish the one true stationary frame? How do you determine the real path of the laser?

In a true rest frame, an event can be viewed simaltaneously by two observers of equal distance from the said event. An observer in relative motion to the two stationary observers/event will not disagree that the event can happen simaltaneously between them. However, do what SR does and let the 'moving' observer consider himself at rest and place the 'stationary observers/event' in the moving frame and the physics ARE different. Events cannot happen simaltaneously between the 'now moving' observers and event. Straight up and down paths now become 'V's and lengthen distances and travel times. By always specifying the observer at rest in his own frame and the other moving, physics are changed. SR manipulates time and distance this way, based on faulty physics.

I'm not sure exactly what you're disagreeing with here.
What exactly do you mean by "the physics are different", and how do you conclude that "faulty physics" is the basis of SR?

 

[2inquisitive]

by Pete:
"It is not the reverse of what SR states, although it can be difficult to see why. I think that you're thinking that the spaceship has to be the moving frame, and the Earth has to be the stationary frame..."
==========================================================

That was why I stated the laser was fired upward at a 90 degree angle to intercept
the spaceship. I was thinking of how SR always puts the observer at rest in his own
frame. I am very well aware the paradox can be solved by REVERSING what SR specifically states and having the Earth-based observer with the laser consider himself in MOVING frame of reference in the opening sequence. Then the path of the
laser is altered when firing at the 'stationary' spaceship. As I said before, I know how
Special Relativity does its tricks.

by Pete:
How do you distinguish the one true stationary frame?"
========================================================

It's the one in which the whole universe is not whizzing past you at .866c. hehe!
It's the one in which objects are not floating around you while in an inertial frame
aboard your spaceship. It's the one in which (according to SR) planets and stars are
not contracted in one direction only, appearing flattened and distorted. It's the one
in which two observers can witness an explosion happen exactly halfway between
them and agree on the time that the explosion occured. It's the one in which physics
do not have to change to suit someone else's distorted view. Need I go on?

 

[Pete]

by Pete:
"It is not the reverse of what SR states, although it can be difficult to see why. I think that you're thinking that the spaceship has to be the moving frame, and the Earth has to be the stationary frame..."
==========================================================

That was why I stated the laser was fired upward at a 90 degree angle to intercept the spaceship. I was thinking of how SR always puts the observer at rest in his ownframe. I am very well aware the paradox can be solved by REVERSING what SR specifically states and having the Earth-based observer with the laser consider himself in MOVING frame of reference in the opening sequence. Then the path of the laser is altered when firing at the 'stationary' spaceship. As I said before, I know how Special Relativity does its tricks.

I'm not convinced that you do...

How does the resolution mean that the Earth based observer considers himself moving?
The diagram shown is for the spaceship frame... the spaceship observer considers the Earth based observer to be moving.

Earth frame:
The beam travels 299792458m each way (perpendicular to the spaceship's path), total time 2 seconds.

Spaceship frame:
The beam travels 599584916m each way (diagonally to the Earth's path), total time 4 seconds.

I'm not sure where you think the paradox lies?

It's the one in which the whole universe is not whizzing past you at .866c. hehe!

The averaged CMBR frame? In which (almost) the whole universe is whizzing away from you at (very!) high speeds, but averaging to zero? That's a good fram,e to be sure, but it's not Earth's frame.

It's the one in which objects are not floating around you while in an inertial frame aboard your spaceship.

?? I don't understand

It's the one in which (according to SR) planets and stars are not contracted in one direction only, appearing flattened and distorted.

Some planet and stars are flattened in all frames... do you want to choose a particular star or planet to define your absolute frame? What's special about that star/planet?

It's the one in which two observers can witness an explosion happen exactly halfway between them and agree on the time that the explosion occured.

That applies to any two observers that share the same frame, regardless of where the explosion occurs.

It's the one in which physics do not have to change to suit someone else's distorted view. Need I go on?

That would be the frame defined by the ether... which seems to be undetectable. Ether based models force different physics in moving frames.

The whole foundation of SR, the fundamental principle of relativity, is that the laws of physics should not change in moving reference frame - and they do not, in the SR model.

 

[2inquisitive]

I based my paradox on certain flaws in Special Relativity.
(1) This is a large flaw. Lorentz contractions occur only along direction of travel,
but time dilation is not directional. This is a paradox within itself as the speed of
light is not constant in all directions. Again, the speed of light defines the length
of a meter. Contract the meter in one direction only, and clock would have to run
slow in that direction only. Contract the meter in one direction only, but have the
clock run slow in all and the measured speed of light will change to the sides.
I knew I had to design my paradox to take advantage of this breech of physics, so
it had to have a measured distance at a 90 degree angle to direction of motion. It
had to use the speed of light as the measuring stick and the timer.
(2) I was fully aware of how Special Relativity swaps frames to dilate time and distance, so I had to have a firm starting frame with that frame at rest. The Earth-
based laser firing straight up to intercept the moving spaceship. I knew SR states
'moving clocks run slow' but I knew the spaceship could only record an increased
travel time for the 'V' path of the laser when the Earth-based laser was put into
motion via SR's keeping the spacecraft at rest and the laser in motion. SR's trick.
I also knew I could have NO non-inertial frames and no turn-arounds for Relativists
to use as a crutch to explain the paradox. I also had to use stopwatches to time
the flight of the laser so they could be compared later WITHOUT changing during
a non-inertial decceleration phase. The paradox looks deceptively simple, but was
complex to actually piece together. I believe it is simple and bulletproof, without
resorting to obvious complete changes in the way it is presented. And, by the way,
relativity of simultaneity will not work in solving it either because of the clocks timed
by the constant speed of light. I am not a physicist and have trouble making a coherent argument in accordance with physics, but I did figure out where Special
Relativity's faults were and used a paradox to highlight them. I do not expect any
acknowledgement of the paradox's success or any changes in the use of Special
Relativity in physics, however. Life goes on and this thread will just fade into oblivion.
But, SOMEDAY, JamesR, the house of cards will fall.

 

[Pete]

I don't understand what the apparent paradox is...

Are you maintaining that SR predicts different times for the stopwatches when calculated in different frames or in different ways?

For example, it would be a paradox if SR predicted both that:

1) The Earth stopwatch will read 2 seconds for the round trip, and
2) The Earth stopwatch will read 1 second for the round trip

Is this the sort of thing you have in mind?
Can you spell it out for me?

 

[2inquisitive]

Pet, the title of this thread "Twin Paradox." Special Relativity states 'moving
clocks run slow.' The reason the traveling twin returns younger than the
Earth twin who does not travel. The laser does not travel, exactly the same
as the stay-at-home twin. The spaceship is traveling overhead at .866c.
Special Relativity states it is the frame that will record less time on its clock.
When the clocks are brought together and compared, the traveling clock
will record more seconds on its stopwatch than the Earth-based stopwatch
for a common event, the laser travel time to the moving spaceship and back
to Earth. You can say the traveling clock is 'really' running slower than the
Earth clock, but the traveling clock still has to accumilate displayed time
defined by the speed of light in its frame. If the traveling clock is running
'half-as-fast' as the Earth clock, then the distance between the Earth and
the spaceship ALSO has to contract by one half. The result is the same,
the ships stopwatch will record more than two seconds for the event (laser
travel time), while the Earth's clock will record two seconds. The result is
a paradox within SR, which specifically predicts the traveling clock will
record LESS time on its stopwatch if the spaceship is traveling at .866c.

 

[James R]

2inquisitive:

I based my paradox on certain flaws in Special Relativity.
(1) This is a large flaw. Lorentz contractions occur only along direction of travel, but time dilation is not directional.

Where did you get that idea from? Please show us how this comes about from the theory.

This is a paradox within itself as the speed of light is not constant in all directions.

That is contrary to the basic postulates of relativity.

 

[2inquisitive]

by JamesR:
"Where did you get that idea from? Please show us how this comes about from the theory."
================================================================

The first step in the twin paradox. As the traveling twin receeds from Earth, the
Earth-based observer sees the ship contracted along the direction of travel only
and HER CLOCK RUNNING SLOW. Is the clock supposed to run slow along the direction
of travel and speed up as speed up as the second hand nears the perpendicular?

 

[Pete]

The spaceship is traveling overhead at .866c. Special Relativity states it is the frame that will record less time on its clock.

No it doesn't.
You need to read up on exactly what SR does and doesn't say for time dilation.

Relativity demands that in flat space, if spaceship clocks run slow in Earth's frame, then Earth clocks run slow in the spaceship's frame.

In your scenario, the laser is effectively an Earth clock - it must run slowly in the spaceship frame.

The spaceship observation of the laser events on Earth are not a spaceship clock - they are a spaceship observation of the Earth clock.


Nothing in SR says that the time measured between the two events should be shorter in the spaceship frame.

 

[2inquisitive]

by Pete:
"In your scenario, the laser is effectively an Earth clock - it must run slowly in the spaceship frame."
================================================================

No it isn't an 'Earth clock'. It is precisely defined by the speed of light. You cannot make the spaceship clock record less time for a defined two light second trip than two
seconds or you change the speed of light in that frame. Remember? A meter is defined
as 1/299,792,458seconds. The 'laser events' is the travel of light. Are you saying the
laser light travels slower in the spaceship frame?

 

[Paul T]

2inquisitive and Pete,

With respect to 2inquisitive's exercise, I think the following should be considered.

1) Laser beam, as we know, does not spread out as normal light does
2) If in the test, the laser beam is directed vertically perpendicular to the trajectory of the spaceship, then the only position where the spaceship's observer "see" the laser is when it passes the point directly above the laser source point.

To evaluate 2inquisitive's original proposed test with laser (not normal light), I propose to inspect the spaceship clock reading at three points, they are when the ship is at z=1 light second and respectively:

a) at x=-0.866c second
b) at x=0
c) at x=0.866c second

Let's evaluate spaceship clock reading at those three points using Lorentz transformation [t'=g(t-(g/c)x)].

• Laser leaves earth at t=0, when the spaceship is at x=-0.866c sec. Spaceship clock reading is t' = 2*[0-(0.866/c)(-0.866c)] = 1.5 sec.

• Laser reaches the spaceship at t=1 sec, when the spaceship is at x=0. Spaceship clock reading is t' = 2*[1-0] = 2.0 sec.

• Laser back to earth at t=2 sec, when the spaceship is at x=0.866c sec. Spaceship clock reading is t' = 2*[2-(0.866/c)(0.866c)] = 2.5 sec.

There is no doubt that the time interval for 1 cycle according to earth observer is 2 seconds. Based on the above evaluation, the time interval for the same cycle according the spaceship observer is 1 second (2.5 sec - 1.5 sec). A simple computation using time dilation formula will certainly give the same result, that is total t' = t/g = 1 second.

 

[2inquisitive]

You see Pete, the laser is the kicker in my paradox. You can't just look at a clock and say it is running slower. You HAVE to consider the laser as traveling the speed of light
in BOTH frames. You can increase the travel distance by lengthening the path the laser beam takes, but you CANNOT make it travel faster than the speed of light by
any clock as the second, meter and speed of light are interconnected. Distance is NOT
contracted at the 90 degree angle of the ships flight path.

 

[2inquisitive]

So, Paul T, are you saying the spaceship stopwatch will record 1 second for light
to travel a distance of over 2 light seconds? Remember the spaceship observer has
to record the speed of light as 'c' also.

 

[Pete]

by Pete:
"In your scenario, the laser is effectively an Earth clock - it must run slowly in the spaceship frame."
================================================================

No it isn't an 'Earth clock'. It is precisely defined by the speed of light. You cannot make the spaceship clock record less time for a defined two light second trip than two seconds or you change the speed of light in that frame.

Of course... that's what makes it a clock.
We know how far it goes in the Earth frame.
It returns to the same point in the Earth frame.
This means we can predict the travel time in the Earth frame.
We know (because of the fixed speed of light) that one "tick" of this laser appartus is equal to two seconds, and that the apparatus is stationary in the Earth frame.
That makes it a clock.

Are you saying the laser light travels slower in the spaceship frame?

No, I'm saying it travels further in the spaceship frame. Twice as far.
This means that the "clock" runs slower in the spaceship frame; the laser appartus ticks once every four seconds.

This is as it should be... in the spaceship frame, the laser apparatus (the clock) is moving, so its time is dilated.

You HAVE to consider the laser as traveling the speed of light
in BOTH frames.

Aren't you following?

Earth frame:
Beam goes 2 light-seconds in 2 seconds.

Spaceship frame:
Beam goes 4 light-seconds in 4 seconds.


As predicted by relativity.

 

[2inquisitive]

And as recorded on each stopwatch: 2 seconds on the Earth stopwatch and 4 seconds
on the spaceship stopwatch. The Earth stopwatch ages less than the traveling stopwatch.

 

[Pete]

1) Laser beam, as we know, does not spread out as normal light does
2) If in the test, the laser beam is directed vertically perpendicular to the trajectory of the spaceship, then the only position where the spaceship's observer "see" the laser is when it passes the point directly above the laser source point.

In Earth's frame, yes... but in the spaceship frame, the laser beam is projected at an angle.

• Laser leaves earth at t=0, when the spaceship is at x=-0.866c sec. Spaceship clock reading is t' = 2*[0-(0.866/c)(-0.866c)] = 1.5 sec.

• Laser reaches the spaceship at t=1 sec, when the spaceship is at x=0. Spaceship clock reading is t' = 2*[1-0] = 2.0 sec.

• Laser back to earth at t=2 sec, when the spaceship is at x=0.866c sec. Spaceship clock reading is t' = 2*[2-(0.866/c)(0.866c)] = 2.5 sec.

Note that these are the times shown on the spaceship clock simultaneously with the laser light sent and received in the Earth frame.

The times shown on the spaceship clock simultaneously with the light sent and received in the spaceship frame will be different. To find the times of those events, you need to transform the coordinates of the send and receive events (not the spaceship position at the time of those events):
Laser beam leaves Earth at t=0, x=0
Laser beam back to Earth at t=2, x=0

Transformed to spaceship frame:
Laser beam leaves Earth at t'=0, x'=0
Laser beam back to Earth at t'=4, x'=-3.464

There is no doubt that the time interval for 1 cycle according to earth observer is 2 seconds. Based on the above evaluation, the time interval for the same cycle according the spaceship observer is 1 second (2.5 sec - 1.5 sec).

No - the time interval you calculated is the time passing on spaceship clocks moving in Earth's frame during the laser's transit.
The time interval during the laser's transit on spaceship clocks in their rest frame requires a different calculation (above), which results in 4 seconds.

A simple computation using time dilation formula will certainly give the same result, that is total t' = t/g = 1 second.

This illustrates how easy it is to misuse equations... and how tricky it can be to pin down which way the relationships work.