Four-dimensional space? It is very simple! Or "the Theory of Spaces".

© Любарь Владимир Яковлевич

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The appendix 2

  Analysis  The chapter - 4:   « Experimental flight »                  

 So, we with you were dispersed stage by stage till speeds of 150 thousand km/s, leaving at each stage a shuttle (a label in space). Concerning the shuttle left once again, speed at us always reached 150 thousand km/s.

The new system of readout differs nothing from previous and consequently dispersal concerning varying system, always looks linear. Acceleration does not fall, the weight does not increase. I about it already wrote in The appendix - 1. The previous shuttles will lag behind me not so linearly, and seen speeds between them will be less. Define their speeds we can only on their radio beacons which frequency, to us is known. We compare frequency of the accepted signal with what it radiates (known to us), the greater we divide on smaller, and we receive Doplers`s factor (for light) - K_d, and from him we calculate speed:

V = ((K_d² – 1)/(K_d² + 1)) *C

To learn speed between shuttles, it is necessary to divide the greater factor on smaller, it and will be K_d between them on which we learn speed between them (subtraction of speeds). At addition of their speeds K_d are multiplied. It is possible to put speeds on Einstein:

 V₃ = C² * (V₁ + V₂)/(C² + V₁ * V₂),    that is completely identical.

And to not trouble itself with bulky calculations, speed of light C – I took equally 300 thousand km/s.

Have made 7 stages, 6 shuttles are left. After speed of 150 thousand km/s, has been once again achieved., concerning last 6-th shuttle, I have switched off the engine in a rocket.

Now we shall look, as all this looks, from a position of eight observers:

I in a rocket (a Fig. 1):		I stand on a place. All from me leave to the left [1]:
	Fig. 1	6 shuttle with speed of 150 thousand km/s 5 shuttle with speed of 240 thousand km/s 4 shuttle with speed of 278,57 thousand km/s 3 shuttle with speed of 292,68 thousand km/s 2 shuttle with speed of 297,54 thousand km/s 1 shuttle with speed of 299,17 thousand km/s The Earth with speed of 299,72 thousand km/s

I in 6-th shuttle (a Fig. 2):		I stand on a place, the rocket from me leaves to the right with speed of 150 thousand km/s,
	Fig. 2	Other all to the left: 5 shuttle with speed of 150 thousand km/s 4 shuttle with speed of 240 thousand km/s 3 shuttle with speed of 278,57 thousand km/s 2 shuttle with speed of 292,68 thousand km/s 1 shuttle with speed of 297,54 thousand km/s The Earth with speed of 299,17 thousand km/s

I in 5-th shuttle (a Fig. 3):		I stand on a place, to the right from me leave:
	Fig. 3	6 shuttle with speed of 150 thousand km/s, rocket with speed of 240 thousand km/s To the left from me leave: 4 shuttle with speed of 150 thousand km/s 3 shuttle with speed of 240 thousand km/s 2 shuttle with speed of 278,57 thousand km/s 1 shuttle with speed of 292,68 thousand km/s The Earth with speed of 297,54 thousand km/s

I in 4-th shuttle (a Fig. 4):		I stand on a place, to the right from me leave:
	Fig. 4	5 shuttle with speed of 150 thousand km/s 6 shuttle with speed of 240 thousand km/s rocket with speed of 278,57 thousand km/s To the left from me leave: 3 shuttle with speed of 150 thousand km/s 2 shuttle with speed of 240 thousand km/s 1 shuttle with speed of 278,57 thousand km/s The Earth with speed of 292,68 thousand km/s

I in 3-rd shuttle (a Fig. 5):		I stand on a place, to the right from me leave:
	Fig. 5	4 shuttle with speed of 150 thousand km/s 5 shuttle with speed of 240 thousand km/s 6 shuttle with speed of 278,57 thousand km/s rocket with speed of 292,68 thousand km/s To the left from me leave: 2 shuttle with speed of 150 thousand km/s 1 shuttle with speed of 240 thousand km/s The Earth with speed of 278,57 thousand km/s

I in 2-nd shuttle (a Fig. 6):		I stand on a place, to the right from me leave:
	Fig. 6	3 shuttle with speed of 150 thousand km/s 4 shuttle with speed of 240 thousand km/s 5 shuttle with speed of 278,57 thousand km/s 6 shuttle with speed of 292,68 thousand km/s rocket with speed of 297,54 thousand km/s To the left from me leave: 1 shuttle with speed of 150 thousand km/s The Earth with speed of 240 thousand km/s

I in 1-st shuttle (a Fig. 7):		I stand on a place, to the right from me leave:
	Fig. 7	2 shuttle with speed of 150 thousand km/s 3 shuttle with speed of 240 thousand km/s 4 shuttle with speed of 278,57 thousand km/s 5 shuttle with speed of 292,68 thousand km/s 6 shuttle with speed of 297,54 thousand km/s rocket with speed of 299,17 thousand km/s To the left from me the Earth with speed of 150 thousand km/s leaves.

I on the Earth (a Fig. 8):		We cost(stand) on a place, all from us fly to the right with speeds:
	Fig. 8	1 shuttle with speed of 150 thousand km/s 2 shuttle with speed of 240 thousand km/s 3 shuttle with speed of 278,57 thousand km/s 4 shuttle with speed of 292,68 thousand km/s 5 shuttle with speed of 297,54 thousand km/s 6 shuttle with speed of 299,17 thousand km/s rocket with speed of 299,72 thousand km/s

As you can see, any speed can be accepted for a zero. From the point of view of any observer, any speed will not exceed a rod – 300 thousand km/s, not because it is a limit of speed, that is why that, it at all speed, and observable "speed", speed of current of time in own system of readout.  

Addition of speeds cannot be accepted in literal sense of this word, addition here only conditional. It is necessary to understand four-dimensional space, time and distance – concepts relative, in the present sense of this word. The person speaks about a relativity, and in subconscious ness this all the same sits absolute. Take you, for example the globe, and photograph it. Distances to measure in his photo that is vertically under an objective is possible only, instead of that is near edge of the globe in a photo. There it will be simple a projection of one plane to other (photo) which are under a corner to each other.

Let's consider my additions of speeds in  The appendix - 1

  If a particle moving with speed:

299,602239389441111963018984171545 thousand in km/s to brake so that she has lost speed on 200 thousand in km/s what will have it residual speed?

It will have a speed:

298,840992451745453694519903346968 thousand in km/s  

Still we shall brake on 200 thousand in km/s

It will have a speed:

295,057333754764957000678687844567 thousand in km/s  

Still we shall brake on 200 thousand in km/s

It will have a speed:

276,805111063643564796742822907365 thousand in km/s  

Still we shall brake on 200 thousand in km/s

It will have a speed:

200 thousand in km/s

And from 200 thousand in km/s it is possible to take away gradually, subtraction will be not linear, as well as additions. These of no linearity begin with zero speed, is simple on small speeds they will be so small, that they can be neglected. And energy at dispersal (or braking) will be distributed the same as also the speeds above mentioned.

To you and seeming increases in weight!    

 Einstein has deduced the formula of addition of speeds, and it was on a correct way!

But for some reason did not understand physical essence of this addition, and has gone not to that party, representing it as reduction of acceleration, at the same power expenses. And what to explain it, it has presented it as increase in weight. Also there was all on a false way.

The weight of rest, weight for the speed, one mistake (misunderstanding), gives an avalanche of the next errors. Inert weight, heavy weight.

Even who be, has given precise definition what is that?

Speak: the inert weight is equal to heavy weight???      All this is equal what to tell: I am equal to myself!  

And are such which speak: no, I am not equal to myself a little!  Why?       That is why that!    

And that it is intelligible, in this occasion neither who nor that cannot tell.

He/she is the Person has equated weight with weight on the Earth, and then is surprised: how it they are equal?

And for the nature all the same, than we shall measure her, in arching or bats shoes, we can measure though by parrots.

 Analysis  The chapter - 3: « And if speed will be big … »

 Let's disassemble cases, as well as in 3 chapter:

The rocket flies on a straight line And – with constant speed, and flying by a point About, short light flash is radiated. Then during any moment when the rocket will be in an any point – X, we stop an instant (a Fig. 9).

Fig. 9

So, in a point About the light pulse has been given. We stop an instant when the rocket is in a point of X terrestrial space. The iridescent ring from a position of the terrestrial observer is a big ring on figure. Light has overtaken a rocket on size – t. But light, as is known, has constant speed in any system of readout so the iridescent ring, from a position of the observer from rocket space will be another, and the epicenters will be at a rocket (a small ring in figure).

The way gone by a rocket from About up to X is equal – S.

The radius of the big ring is there will be time past on the Earth (or terrestrial time - t_z).     t_z = S + t

The radius of a small ring is there will be time past in a rocket (or rocket time - t_r).

 What physical essence – t?

It is a terrestrial television signal which has overtaken a rocket, with the same speed – With, this television time of the Earth in a rocket so it will be stretched(will be dragged out) on size of rocket time – t_r. The parity  t_r/t will be Doplers`s factor – K_d.

But back and forward signals are symmetric, so also the attitude of back front to rocket time will be same, and will be equal - K_d.

Back front – from a point And up to a point About (S + t) + from a point About up to a point X (S),

Total we have:    T = 2*S + t

The attitude of back front to rocket time, will be same, as well as the attitude of rocket time to forward front, and will be equal - Doplers`s to factor – K_d.

          and    

 It means, that the cosmonaut will look terrestrial telecasts in K_d time in a slowed-up way (t_r/t), and the Earth a body the reporting from a board of the ship, will as see in K_d time in a slowed-up way (T/t_r).

We admit, that speed of a rocket such, at which K_d = 3 (see. A Fig. 9).

  On the screen of terrestrial TV in a rocket has passed two day (t = 2), and in a rocket will pass:

t_r = t * K_d        t_r = 2 * 3 = 6 day, i.e. the cosmonaut will see terrestrial telecasts in 3 times in a slowed-up way.

T = t_r * K_d           T = 6 * 3 = 18

Precisely as also the Earth will see the cosmonaut:      T/t_r         18/6 = 3,    i.e. in 3 times in a slowed-up way (on TV).

Now we shall calculate, gone by a rocket, a way – S             T = 2*S + t

From here:        S = (T – t)/2 S = (18 – 2)/2 = 8 light day

Time past on the Earth:       t_z = S + t        t_z = 8 + 2 = 10 day.

Parity of terrestrial time by time past in a rocket (K_t):       10/6 =1,666666   or:     K_t = (K_d + 1/K_d)/2 K_t = (3 + 0,333333)/2 = 1,666666

Now we shall take speed, at which: K_d = 10 (a Fig. 10):

     Fig. 10

As t - we shall take 2 day, it means, that by the moment X, on the screen of terrestrial TV in a rocket, has passed 2 day.

Own time in a rocket will pass:

t_r = t * K_d         t_r = 2 * 10 = 20 day.

T = t_r * K_d        T = 20 * 10 = 200

T = 2*S + t       from here:       S = (T – t)/2 S = (200 – 2)/2 = 99 over day.

 Own time of the Earth:

t_z = S + t            from here:            t_z = 99 + 2 = 101 day.

Parity of terrestrial time by time past in a rocket  (K_t):

101/20 =5,05    or:      K_t = (K_d + 1/K_d)/2           K_t = (10 + 0,1)/2 = 5,05

 So, the picture will be following:

From the moment of flight of a point About up to a point X on the Earth will pass 101 day. The distance gone by a rocket – 99 light day. Speed of a rocket will be: (99/101)*300 = 294,0594 thousand km/s.

          we check:   

As was to be shown!                    (formula K_d as look in 5 chapter)

In a rocket will pass – 20 day.

The cosmonaut will see terrestrial telecasts in 10 times slowed down (20/2 = 10).

Telecasts from a rocket, the Earth will see too in 10 slowed down (200/20 = 10).

Analysis  The chapter - 5: « Travel on planet Chobruchi »

 Let's take a site of a way: from a point where I have switched off the engine after dispersal, up to a point In where I have included the engine on braking (under the schedule). It was free flight, with constant speed, without acceleration (a Fig. 11). 

    Fig. 11

 K_d – (Doplers`s factor) was equal this free flight 19.

At desire it is possible to find and speed:       (and it is possible and to do without it)

V = ((K_d² – 1)/(K_d² + 1)) *C      

V = ((361 – 1)/(361 + 1)) *300 = 298,34253 thousand in km/s.

 If to accept t = 1 that an allocation will be such: 

Time in a rocket:      t_r = t * K_d           t_r = 1 * 19 = 19

 Terrestrial time of the telereporting from a rocket, television time Chobruchi`s in a rocket:

T = t_r * K_d          T = 19 * 19 = 361                  

Chobruchi`s  time of the telereporting from a rocket, television time of the Earth in a rocket:

t = t_r/K_d              t = 19/19 = 1

 Light distance:      S = (T – t)/2      S = (361 – 1)/2 = 180

 Factor of a parity of time:    K_t = (K_d + 1/K_d)/2         K_t = (19 + 1/19)/2 = 9,5263155

  Time for the Earth own, time on Chobruchi`s own:      t_z = S + t             t_z = 180 + 1 = 181

         or so:     t_z = t_r * K_t         t_z = 19 * 9,5263155 = 181

   Now we shall substitute our real values which were in flight:

 t = 3,27 day

 Time in a rocket:     t_r = t * K_d         t_r = 3,27 * 19 = 62,13 day.

 Terrestrial time of the telereporting from a rocket, television time Chobruchej in a rocket:

T = t_r * K_d        T = 62,13 * 19 = 1180,4 day.              

Chobruchej time of the telereporting from a rocket, television time of the Earth in a rocket:

t = t_r/K_d         t = 62,13/19 = 3,27 day.

 Light distance:        S = (T – t)/2      S = (1180,4 – 3,27)/2 = 588,56 over day.

 Factor of a parity of time:     K_t = (K_d + 1/K_d)/2                 K_t = (19 + 1/19)/2 = 9,5263155

  Time for the Earth own, time on Chobruchej own:

t_z = S + t            t_z = 588,56 + 3,27 = 591,83 day.

         or so:

t_z = t_r * K_t                t_z = 62,13 * 9,5263155 = 591,83 day.

Extreme changes: on October 13, 2005                                                                                                     Ljubar  V. J.