Four-dimensional space? It is very simple! Or "the Theory of Spaces".
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Chapter 9.
Paradox of Twins.
The simplified flight on about at horizon speeds.
For the best understanding of my example described in 5 the chapter (flight on planet "Chobruchi"), I bring here more simple example of flight on about at horizon speeds and as all this will look from a position of the Earth and the cosmonaut. One of two brothers - twins remains on the Earth, another flies up to a conditional planet, and will return back to the Earth.
Distance from the Earth up to a conditional planet (let it will be a planet "Pliug") 30 light day. The rocket, with the cosmonaut onboard is smoothly dispersed within one hour till the speed by 299 thousand in km/s. Then deenergizing the engine and free flight with this constant speed follows. On as much not reaching up to a planet, she as is smoothly braked within 1 hour (on terrestrial time), and landing to a planet. On a planet the cosmonaut will be 1 day, and in the same way will come back to the Earth. Onboard there is a television camera which will be included all flight from start before returning to the Earth, and the TV showing terrestrial translation.
Start
So, the rocket has started in recently past New year's eve:
On January, 1, 2005 at 0 o'clock 00 minutes.
Approximately in 28 minutes the cosmonaut has seen deenergizing the engine, after an awful overload has come [1] weightlessness.
The board showed speed - to 299 thousand in km/s. Kd = Ö ((C + V) / (C V))
Where Kd Dopler`s factor.
On the screen of the TV showing the Earth (the screen of the Earth), has passed approximately 8 mines, and time for it was slowed down in 24,474 times. The cosmonaut "has settled" to sleep.
On January, 1 9 o'clock in the morning. On the screen of the Earth: on January, 1 0 hours of 28 minutes.
Have passed even day, on hours of the cosmonaut:
On January, 2 9 o'clock in the morning.
On the screen of the Earth: on January, 1 1 more hour of a new year's eve of 26 minutes. The image as in a slowed-up way in 24 with superfluous time.
On hours of the cosmonaut: on January, 3 11 hour. 13 minutes,
on the screen of the Earth: on January, 1 2 hour. 32 minutes.
In the staff on the Earth the image gets also his (cosmonaut) a body, on it the cosmonaut sees itself in 598,9999 times in a slowed-up way, and sees the date: on January, 1 0 hour of 29 mines (i.e. a minute ago the cosmonaut has switched off the engine after start and dispersal).
Engines on braking have joined. Time, on the screen of the terrestrial TV, began to be accelerated (from slowed down to come to normal), and at the moment of landing became normal.
Landing to a planet:
- on hours of the cosmonaut: on January, 3 11 hour. 41 minutes.
- on the screen of the Earth: on January, 1 2 hour. 40 minutes.
Time now on the screen of the Earth went equally with time of the cosmonaut: 1:1, but on the screen on the Earth, it sees itself in 24,474 times in a slowed-up way.
On the Earth now: on January, 31 2 hours of 40 mines, but the TV show events which were one month ago because distance 1 light month (30 light day). It means, that light (radio waves) go up to this planet of 30 day, therefore and the television image, and the Sun, and all events there, will be visible from this planet to monthly prescription.
The difference in the age of twins has made: 27 day of 14 hours and 59 minutes.
The cosmonaut on a planet 1 day has stayed, terrestrial telecasts looked during this period in time 1:1 i.e. as it is usual, as well as on the Earth.
Rise home
- on hours of the cosmonaut: on January, 4 11 hour of 41 minutes.
- on the screen of the Earth: on January, 2 2 hour of 40 minutes.
Through 28 mines the cosmonaut has seen deenergizing the engine, after an awful overload there has come weightlessness.
The board showed speed - to 299 thousand in km/s.
On the screen of the TV showing the Earth (the screen of the Earth), has passed about 1 hour of 52 mines, and time for it was sped up in 24,474 times.
- on hours of the cosmonaut: on January, 4 12 hour. 09 minutes.
- on the screen of the Earth: on January, 2 4 hour. 32 minutes.
Have passed day of flight:
- on hours of the cosmonaut: on January, 5 12 hours of 09 minutes.
- on the screen of the Earth: on January, 26 15 hours of 56 minutes.
Braking
- on hours of the cosmonaut: on January, 6 22 hour of 54 minutes.
- on the screen of the Earth: on March, 3 3 hour of 28 minutes.
On the screen on the Earth, it sees itself in 598,9999 times accelerated, and sees the date on the screen: on January, 5 4 hours of 30 minutes.
In engines on braking Have joined.
- on hours of the cosmonaut: on January, 6 2005 23 hours of 22 minutes.
- on the screen of the Earth, and date terrestrial: on March, 3 2005 5 hours of 20 minutes.
Both TVs show now one and too.
The difference in the age of twins has made: 55 day of 5 hours and 58 minutes.
That earthmen saw, observing on the TV for the cosmonaut:
On January, 1, 2005 at 0 o'clock the rocket has started 00 mines.
Approximately at 1 o'clock 52 mines, earthmen have seen deenergizing the engine. On the telescreen hours showed: 0 hour of 28 mines (onboard time), time on the telescreen was gradually slowed down, and at the moment of deenergizing the engine became slowed down in 24,474 time. The cosmonaut "has settled" to sleep.
Have passed day:
On January, 2 1 hour of 52 minutes.
On the telescreen (time of the cosmonaut): on January, 1 1 hour of 26 mines, the cosmonaut sleeps.
It has overslept up to 9 o'clock in the morning. His rise, earthmen observed:
On January, 10 at 4 o'clock 20 mines
In the staff a body of the image there is an onboard TV which shows terrestrial TV, and earthmen now observe on it themselves and the date: on January, 1 0 hour of 28 mines, also see the image slowed down almost in 600 time (24,4742 = 598,9999).
On February, 3 15 hours of 45 minutes .
On the screen: on January, 2 9 o'clock in the morning, rise of the cosmonaut next day. On his TV, showing the Earth, just: on January, 1 1 hours of 26 mines of a new year's eve, the terrestrial image all as almost in 600 times in a slowed-up way.
On braking earthmen have seen inclusion of the engine: on March, 2 at 0 o'clock 48 minutes.
And the end of flight (landing to a planet):
On March, 2 2 hours of 40 minutes, on the telescreen (at the cosmonaut): on January, 3 11 hour of 41 minutes.
On the onboard screen saw the date: on January, 1 2 hours of 40 minutes.
Now time for both TVs went 1:1.
The cosmonaut has stayed on a planet 1 day and rise back, home.
On March, 3 2 hours of 40 minutes, observe Start:
- on hours of the cosmonaut: on January, 4 11 hour of 41 minutes.
- on the onboard screen of the Earth: on January, 2 2 hours of 40 minutes.
On March, 3 2 hours of 48 minutes. The Earth has seen deenergizing the engine.
Hours on the screen (cosmonaut) showed: on January, 4 12 hour of 09 minutes.
Hours on the screen of the onboard TV showed: on January, 2 4 hour of 32 minutes.
Time now on the screen has run in 24,474 times accelerated, and on the onboard TV on which earthmen see themselves, time has run almost in 600 times accelerated (24,4742 = 598,9999).
Has not passed also hour: on March, 3 3 hours of 47 mines as earthmen have seen, that at the cosmonaut day have already run.
The telescreen showed time (at the cosmonaut): on January, 5 12 hours of 09 minutes.
And the onboard TV : on January, 26 15 hours of 56 minutes.
On March, 3 5 hour of 12 minutes the Earth will see inclusion of the engine on braking, and in 8 minutes a landing:
On March, 3, 2005 5 hours of 20 minutes of morning.
- on hours of the cosmonaut: on January, 6, 2005 23 hours of 22 minutes.
Both TVs show now one and too.
Let's assume now that beside and in parallel with a rate of a rocket (on a line of free flight), kilometer columns (anchor buoy) are spread out. In a rocket the counter (frequency meter) which counts columns (anchor buoys) is established.
Question:
How many columns (anchor buoys) the counter for a second will count?
The answer:
Speed of a rocket: - 299 thousand in km/s (in terrestrial second).
Kt = (Kd + 1/Kd)/2 where Kt Factor of a parity of time.
And in a rocket 1 second = (24,474476 + 1/24,474476)/2 = (24,474476 + 0,040858)/2 = 12,257667 terrestrial seconds.
It means, that when in a rocket 1 second will pass, on the Earth will pass - 12,257667 seconds, and quantity of the lanterns, the counted gauge for a second, will be:
299000*12,257667 =3042
pieces.
To you three and more one million in km a second really ("terrestrial" kilometers in rocket second), at imaginary Lorenc`s reductions of space the Earth a planet which was in flight under "corner" to rocket space in the fourth measurement (in time).
Time at the cosmonaut did not go in a slowed-up way,
it goes everywhere equally. Simply at him has passed it
less, it
flied on a planet through other space where the way up to a planet is
shorter, both on distance, and on time.
But all spaces are equal in rights. If the rocket, with the kilometer lanterns will fly by by the Earth also the terrestrial observer (photo gauge) will register 3042 flown by lanterns for terrestrial second. And in rocket space the Earth flies with the same mutual speed 299 thousand in km/s (rocket kilometers to rocket second).
Time goes equally, but separately, it does not go one in one, it goes separately in other coordinates. The beginning and the end readout, occurs in different systems and is not simultaneous. All paradoxes are resolved only by set of three-dimensional spaces in four-dimensional space (in time).
See, how the small chest
simply opens if to solve problems!
Four-dimensional space - time - for his understanding is required other level of thinking!
To whom the kitchen of calculation of all flight is interesting, let's consider this flight in more detail with instant dispersal and an instant stop in one party.
Detailed analysis of flight.
So, the Earth and planet Pliug, distance between them 30 over day or 720 over hours.
Flight will pass in an automatic mode, in one party, with an onboard computer, a television transmitter and two television receivers.
On Pliug`s people and they the same as also the Earth, will accept signals of the television image of onboard hours from a rocket too live.
In a rocket the onboard computer will count time (hours, minutes, seconds, the tenth shares of second) to write down these indications to itself on the Winchester in the attitude 1:1, and simultaneously to conduct translation of the image of these hours to the Earth and on Pliug, as 1:1.
Onboard television receivers will accept images of the same hours from the Earth and with Pliug`s, and they as will enter the name computers on Winchesters.
Now we yet do not know, what is the time will pass onboard, and what is the time will write onboard computers, signals of the Earth and Pliug `s. But we then can see these records on Pliug `s.
So, have gone!
The rocket has started in recently past New year's eve:
On January, 1, 2005 at 0 o'clock 00 minutes.
Dispersal from the Earth and braking on Pliug`s instant. The same speed - 299 thousand in km/s
Now let's count:
What is the time flight last?
(C/V) *S (300/299) *720 hour = 722,4 hours (722 hours of 24 minutes)
We have received Terrestrial time of flight tz = 722,4 hours flight last.
Now we be asked a question: and, what is the time Pliug-pipl looked this flight on the TV?
Radio waves moved there 720 hours (distance S = 720 over hours).
Start means (from a board) they will see the beginning of translation in 720 hours, and the end of flight (and the end of translation) will see in 722,4 hours. Leaves, that translation of all flight at them last only 722,4 720 = 2,4 hours. We shall designate it as t.
And as much time for a rocket the telereporting from the Earth, from the moment of start up to an arrival last, i.e. as though there was no time in a rocket, and on the Winchester hours, a cat will be continuously written down. Have counted 2 hours of 24 minutes of what will be convinced Pliug-pipl, having seen record on an arrival of a rocket.
Means:
tz = S + t
where tz time for the Earth own, time on Pliug`s own.
S - light distance.
t - Pliug`s time of the telereporting from a rocket, television time of the Earth in a rocket.
Now we be asked a question: and, the telereporting from a board will last how many for the Earth, and how many Pliug`s time will write down an onboard computer?
From a board the Earth will observe the telereporting from the moment of start. When the rocket will arrive on Pliug, will pass time - tz and further while the signal about the moment Pliug will not come to the Earth.
Means, from the moment of start S + t and still + S
Total we have: T = 2S + t
And how many television time Pliug`s will write down an onboard computer?
From the moment of start the rocket will cross waves which have already left from Pliug`s 720 hours back, i.e. S yes while flies, will pass still + S + t
Total we have: T = 2S + t
where T Terrestrial time of the telereporting from a rocket, television time Pliug`s in a rocket.
You see, that time one and too! Yes differently also cannot be, whether there are at someone doubts?
Now it is necessary to find out, how many time will pass in a rocket?
That is, if now on Pliug`s (after an arrival) to look record of Terrestrial hours they will count: t = 2,4 hours.
And if to see (overlook) record Pliug`s of hours they will count: T = 2S + t
2*720 + 2,4 = 1442,4 hours.
Leaves, that one hours will accelerated go, others in a slowed-up way.
We admit, that in a rocket of time has passed as much, how many and on the Earth (or on Pliug) - tz = S + t = 722,4 hours. (tr)
Then Pljug`s hours will go on the screen accelerated in 1442,4/722,4 = 1,9966 times.
And Terrestrial hours will go in a slowed-up way in 722,4/2,4 = 301 times.
In as much, in 301 times accelerated and Pliug-pipl will observe the reporting from a board, and Earthmen the same reporting will see all in 1,9966 times slowed down.
It is impossible!
In practice the Dopler's-effect is completely symmetric!
To not be proofless, we shall take, for example, a militian radar "Spark - 1".
Direct it to a forehead moving on you with constant speed to the automobile, then when the automobile will pass by you, direct the device to it following.
The device will show the same speed!
Then you sit with the device in the automobile, and direct it on standing at roads the automobile, on approach (approximation), and then removal.
And again the device will show the same speed!
To you symmetry Dopler`s of effect!
Let's return to our calculations.
Full symmetry turns out at such parity:
where
tr
time of a rocket own.
tr = Ö (1442,4 * 2,4) = 58 hours per a rocket flight last. [2]
We check symmetry:
where
Kd
dopler`s
factor.
(1442,4/58) = (58/2,4) = Ö (1442,4/2,4) = Kd
24 = 24 = 24 = Kd
Now we see full symmetry of Dopler's-effect:
- The Earth saw (on the TV) hours per a rocket, going in 24 times more slowly
- In a rocket were visible (on the TV) the Terrestrial hours going to 24 times more slowly
- On Pliug`s observed (on the TV) hours per a rocket, going in 24 times are faster
- In a rocket were visible (on the TV) Pliug `s the hours going to 24 times are faster.
 |
On figure 9: - The Earth is in a point "O", Pliug in a point "X" - The big circle the beginning of translation (start of a rocket) - Small circle too the beginning of translation (the same circle, but for a rocket) - The radius of the big circle, it is - tz (time of flight, terrestrial) - The radius of a small circle, it is tr time of flight of a rocket own. - t - Pliug `s time of the telereporting from a rocket, television time of the Earth in a rocket. |
| Fig. 9 | So, television images will be in @ 24 times accelerated or in a slowed-up way, and time in a rocket will pass in @ 12 times are less, than on the Earth (on Pliug). |
I take out formula Dopler`s of factor:
-
though on approach, though on
removal:
FOR OBSERVERS BEHIND:
Where: Kd Dopler`s factor; F - frequency;
L - it is long waves; LET - new value of argument.
Kd (apparently from the formula) always> 1. At V = 0 Kd = 1.
Range d: from 1 up to ¥ (infinity).
In our example: Kd = Ö ((300 + 299) / (300 299)) @ 24 (1442,4/58 @ 24)
And as I take out the formula of factor of a parity (ratio) of time:
Kt = (Kd + 1/Kd)/2
In our example Kt = (24 + 1/24)/2 @ 12 (722,4/58 @ 12)
If after flight on the Earth or on Pliug`s to scroll these video recordings we shall see the same picture, as the cosmonaut in flight: records of a course of Terrestrial hours will be visible during 58 hours per 24 times slowed down, and records of course Pliug`s of hours as during the same 58 hours per 24 times accelerated.
Conclusion: 58 hours of flight are equal 58 to hours on the Earth!
Time goes equally!
It in flight has simply passed less because it does not go one in one, it goes separately! To you and the same speed of light in any measurement!
There is no uniform Global time in the universe, in different places and on different relative speeds the it!
Here it is necessary to change psychology of thinking, it is necessary to accept set of equal in rights three-dimensional spaces in the fourth measurement time!
Otherwise IMPASSE, and other way is NOT PRESENT!
On April, 17, 2005 Ljubar V. J.
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Extreme editing on October, 13, 2005.
[1] The reader should understand, that such overload who will not sustain, even a cockroach. It only a theoretical example about paradox of twins where it is possible to suppose both instant dispersal, and an instant stop.
[2] If speed of light to take precisely 299792,458 certainly there will be other figures, but for simplification of calculations I have taken in round figures With = 300 thousand in km/s.