DimaMazin

If you don't understand relativity then you should consider usual analogue. For example you push or pull very long and hot metal rod. Cold wind cools the rod. Then force of length contraction of the rod increases force of acceleration of tail of the rod and reduces force of acceleration of head of the rod. Do you deny even this?

Net force can't simultaneously accelerate and contract all parts of object even mathematically. Object at v has contracted distance of own forces balance. Therefore preservation of distance causes force of reaction against the distance.We can call the force by force of length contraction.

net force=2F for tail +F for middle + 0 for head at start of motion =F for tail+F for midde +F for head at end of acceleration

Do you think that object forces aren't in balance on defined distance?And the distance isn't changed relative frameS? https://en.wikipedia.org/wiki/Reaction_%28physics%29 Force of length contraction is caused by reaction of forces connection to past distance which try to break off the object with force of acceleration.

Let's consider math of length contraction of ideal material. Forces of object at v are equal on contracted distance. Therefore acceleration causes forces of length contraction to mathematical middle of the object. For example masses of tail,middle and head are equal. Therefore: Force of head acceleration = F Force of length contraction for head at start of motion= -F Force of tail acceleration=F Force of length contraction of object for tail=F Therefore force of tail motion at start of motion=2F Force of head motion at start of motion=0

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Inequality is danger for development of some people already now.

When culture creates rights inequality then our protection should destroy the culture.

Let's consider example when you try to accelerate head first. Head,middle and tail are connected by strong interaction. You,trying to accelerate head, increase distance between the head and the middle. Increase of gap increases strong interaction.Also acceleration increases speed and length contraction, which increase gap too. It creates more strong interaction even between the middle and the head.When the strong interaction force is bigger than acceleration force for the head then the strong interaction force accelerates the middle faster than the head acceleration. The tail accepts sufficient force for acceleration and reducing of distance to the head from bigger force of strong interaction.

I think necessary distance for forces balance is reducing therefore forces of counteraction have different shifts. Counteraction force of tail has blueshift, counteraction force of head has redshift to motion direction.

Even when they are accelerated at the same time then they are initially accelerated to different speeds. Physics should be idealized only by math.

.Agreed. If you have no such material then what proves that all parts of accelerated and contracted object are accelerated at the same time?

Really my math is considering acceleration and speed after one push. Thank you for idea about speed of force propagation. And so speed of force propagation is faster than c . And reducing object reduces speed of force propagation. Maybe therefore my math contradicts GR, but does it contradict QM ?

I confused some things. Speed of force propagation = r/t = r/[(gamma-1)r/(gamma*v)] = gamma*v/(gamma-1)

I didn't say that. Sorry. Seems I have made mistake about propagation at v, I will reconsider it. Do you think all railroad cars in train are simultaneously accelerated after a push?

They are not simultaneously accelerated because forces don't instantly transfer momentum especially when losing length of parts creates delay of subsequent collisions. Therefore speed of force propagation is v.

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'.