DimaMazin

I consider instant acceleration to v. And my equation is created for it. All parts of object are instantly accelerated to v , but they can't be simultaneously accelerated to have length contraction. Why do you think they are obligated be simultaneously accelerated? Events can be non-simultaneous. You even don't know where is length contracting to.

Agen you make math of break, it is when connection force is weaker than force of acceleration. The forces initially pull tail. If tail is very massive then the forces some pull nose to tail. New math is always wrong for simple folk.

Even when engines are on rocket nose their force with connection force reduce tail length and move the tail earlier than nose. Still I think I do make math of length contraction. You don't make.

Tail, losing own length,approaches to middle. whole length=tail / gamma + length of most of the rocket

The equation is derived from law of length contraction, all data are there . Only moving part of rocket is contracted by gamma factor to motionless part of the rocket. Still middle of the rocket isn't moving therefore it isn't contracted but it is moving to nose as mathematical middle.

With two simultaneously accelerated rockets there is no distance contraction between them when they have weak connection. Your question was about derivation of the math, I showed. What is your next question? Why do you think a tactic of mistakes correction isn't helpful?

Part of distance disappears with any speed and at the speed. In my math is no different between 1 rocket or 2 rockets, because theoretical laws correctly work independently and each with other.The different acceleration is only in time , moving tail is losing length and approaching to the nose, some instant without acting force to the nose(due to losing length) when we consider 1 rocket. I think you don't understand that contracted length of traveler can be created only by two way: with unsimultaneous accelerations or with force of connection when the accelerations try to be simultaneous.

Length of string between the rockets in frameS is r Length of the string in frameS' relative to frameS is r/gamma Contracted distance is r-r/gamma or (gamma-1)r/gamma t=contracted distance/v t=(gamma-1)r/(gamma*v) Sorry. I have mistaken about simultaneity of stops in frameS'. Of course the forward rocket stops later than back rocket in S'. This is single case when the accelerated rockets don't create force to break off or bend the string.

Length of spaceship before motion = r Length of spaceship at v = r/gamma Then time of motion of tail before nose motion=(gamma-1) r / (gamma * v) I have understood this is math of non-simultaneity of instant starts of motions of rockets at v relative to frameS and this is math of simultaneity of stops of the rockets in frameS'.

Events of starts of motions are not simultaneous relative to one of two frames. Why do you think non-simultaneity of starts of motions doesn't cause force between the rockets?

Moving object is moving and contracted in frame S, it is motionless in own frame S'.

I don't see confused me. Our frame is frame S,agen you don't understand that contracted moving objects are in frame S. They don't make necessary math.They didn't make even math of non-simultaneity of start of object motion. Tstart of tail motion=Tstart of head motion -(gamma-1)*length/(gamma*v)

Study this: https://en.wikipedia.org/wiki/Frame_of_reference Moving frame is contracted in our frame and it is in our frame. And at all: show math of contraction of moving object in our frame please. If you don't make the math then you can use my math.

Then all this means that spaceship tail can't be simultaneously accelerated with spaceship nose because it breaks off the spaceship. It additionally works in gravitational fall.

It means dualism of simultaneity in motionless frame.

When people can't make changes then they want to get them.

It is when motionless observers trace an immovability and motion of objects in motionless coordinates of own structure.

Length of spaceship before motion = r Length of spaceship at v = r/gamma Then time of motion of tail before nose motion=(gamma-1) r / (gamma * v) then T' = T + (gamma-1)r/(gamma*v)

How does that relate to attraction of laser beams?

I don't think an approaching of photons increases their number. I think any increase of energy concentration increases energy of approaching( to each other) particles,though the increasing energy is tiny.

Medium doesn't increase energy of photons. Sticking together of laser beams increases energy of their photons.

T'=T-r/c is equation of relativity of simultaneity in motionless frame, this doesn't relate to Lorentz transformation. This defines that seeing events are simultaneous. I think we need to create another non-simultaneities also for explanation of relativity in one frame. Second equation explains reason of indication of moving or moved clock for the non-simultaneity.Here you should use gamma if it considerably more than one. Sorry,I am confused in derivation.

Don't confuse T with t . T is clock indication or age t is time or dT Any motionless observer has relative top of age. The equation defines reason why brought clock from remote place has such defined indication.

T'=T - r/c T' is simultaneous time of motionless remote object T is time of observer r is distance between observer and remote object c is speed of light Moving object to observer creates additional rate of time. Object motion from observer creates additional slowing of time by factor of motion. t=gamma * t'+ dr/(c*gamma)

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