DimaMazin

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)

Friendship with Putin causes military development of Russia. Military development of Russia causes expansion of Russia in Europe and war in Europe. War in Europe causes war of USA with Russia. Trump's words contradict themselves.

That is my ununderstanding of relativity in one frame.

Greeks have chosen problems therefore they should live (some time) with them.

Putin's friends are nasty people.

I consider all these events in one frame of motionless observers.Therefore all the events are simultaneous.Why the Lorentz transformation doesn't happen in one frame of observation for the moving objects?The elapsed time is the same in our frame therefore speed is the same.The observers don't use t' for observation of the events.

Is the math wrong? Are you not sure that the same acceleration creates the same speed in one frame?

Let's consider the math : first object starts acceleration on x1, second object simultaneously starts the same acceleration on x2, third object simultaneously starts the same acceleration on x3. First object ends the acceleration on point x1+dx, second object ends the acceleration on point x2+dx, third object ends the acceleration on point x3+dx. They all have the same speed v. The distance between centers of first and second objects was=/x2-x1/ At v the distance between centers of first and second objects =/x2+dx-x1-dx/ The distances are equal before acceleration and after acceleration in the frame.Tail of train can be the first object, center of train can be the second object, head of train can be the third object.How do observers see the contraction?

Can ends simultaneously be in ends and near to the center of the contraction for all co-located observers?

I am afraid of that we have relativity of simultaneity in one frame.

I want to know the simple. We should consider only in S. They all are equal before motion. And red objects with their gaps are contracted after acceleration . Are they contracted (by the speed) to tail, head or center?