DimaMazin

How to define gravitational slowing of clock in center of mass?

Relativity of simaltaneity is t'=[t-vx/c2]/(1-v2/c2)1/2 or t'=gamma*t - gamma*vx/c2 In the my case t=0 because events of the accelerations of the spaceships are simultaneous in frame S x= - l The my question was because I thought that I have opened another relativity of simultaneity( like you think). But it is the same thing. Our level of knowledge of math is less than professional . Therefore we don't understand some things.

Is d a quantity of contracted meters? Is vesc a velosity wich is unlimited by ultimate speed?

If speed of escape is c , then do you think that light has less speed than c ?

Are clocks slowed by energy-momentum tensor there where momentum of falling photon is increased by the energy -momentum tensor?

We can do the formula without c . gamma=1/(1-c2/v2)1/2 then c2=gamma2v2/(gamma2-1) then kinetic energy=m0(gamma-1)gamma2v2/(gamma2-1) kinetic energy =m0(gamma -1)gamma2v2/[(gamma-1)(gamma+1) kinetic energy=m0*gamma2v2/(gamma+1) In newtonian physics gamma=1 therefore kinetic energy=m0*12v2/(1+1)

Excuse me. I didn't simplify the equation. dt'=(gamma2-1)l/(gamma*v)=gamma*vl/c2 It corresponds to Einstein's relativity of simultaneity.

Let's consider Bell's spaceship experiment. Forward and backward spaceships are simultaneously and instantly accelerated in frame S. l is distance between the spaceships.As soon as forward spaceship is accelerated it sends light signal for measurement of distance between the ships.As soon as backward spaceship is accelerated it sends light signal for measurement of distance between the ships. tf is time of measurement of the distance by forward spaceship tb is time of measurement of the distance by backward spaceship tf = l /(gamma(c+v))+l/(gamma(c-v))=2lc/(gamma(c2-v2))=2 l /(c2-v2)1/2 tb = l/(gamma(c-v)+l/(gamma(c+v))=2 l/(c2-v2)!/2 l'f is distance measured by forward spaceship l'b is distance measured by backward spaceship l'f=[2 l/(c2-v2)1/2]*c/2=gamma*l l'b=[2 l/(c2-v2)1/2]*c/2=gamma*l Firstly distance between the unstoped spaceships in frame S' was l /gamma l'-l/gamma is a change of distance between the spaceships l' - l/gamma=(gamma2l-l)/gamma=(gamma2- 1)*l/gamma then time between stops of the spaceships in frame S' is dt' dt'=(gamma2-1)l/(gamma*v)

Can energy-momentum tensor of rotating mass increase photon fraquency without slowing of time in place where the photon is?

If big mass is rotating and has energy-momentum tensor, then the tensor creates spacetime? Does energy-momentum tensor of big mass attract photons or it can attract only objects with mass?

Change of photon momentum in gravitational field is an action. What is counteraction? Does photon energy-momentum attract the big mass?

Photons are not mass therefore they don't create space/time. But when they are falling to big mass their energy is increasing. Does the increse of their energy create space/time field?

I have confused escape velocity and falling velocity because my question was about falling velocity,

You have said about zero acceleration.

My question is about change of momentum. Is Einstein's change of momentum more than Newton's change of momentum near black hole?

Is Einstein's acceleration bigger than Newton's acceleration near to black hole?

Is it stronger nearer to mass than Newton's force of attraction?

Expense of general time is dt+dt' (dt+dt')/dx=(dx/v+dx/(gamma*v))/dx=(gamma+1)/(gamma*v) p/KE=gamma*v*m/((gamma-1)mc2) = (gamma+1)/(gamma*v) (dt+dt')/dx=p/KE

Expense of general time for crossing of one meter= p /KE = (gamma+1 ) /(gamma*v) It is true for both frames. Do you think that physics law can be useless?

GT= p * dx / KE GT is general time=observer time+traveler time p is momentum KE is kinetic energy

I am bad in math therefore I can't correctly define the force. But physicists can experimentaly define the force. Two weights are connected by two long bars and a force gauge between them. We should measure force when the back weight is accelerating and pushing the forward weight.And we should measure when the forward weight is accelerating and pulling the back weight. In the first case cosmological force doesn't increase a pushing force . In the second case cosmological force increases a pulling force.

Let's define cosmological force of rupture of thread between accelerating rockets. uncontracted distance=(gamma-1)x/gamma increased distance between accelerating rockets=uncontracted distance * gamma=(gamma-1)x dv=(gamma-1)x/dt a'=dv/dt=(gamma-1)x/dt2 a' is cosmological acceleration between accelerating rockets a is acceleration of the rockets relative to us gamma=1/(1-a2dt2/c2)1/2 Fc=a' * (m1+m2)=(gamma-1)x *(m1+m2)/dt2 Fc={1-(1-a2dt2/c2)1/2}x*(m1+m2)/[(1-a2dt2/c2)1/2*dt2 Fc is force of rupture of filament between accelerating rockets

It is clearer. Thanks.

Because photon delay has some time. For example gravitational delay of photon . Also electrons are moving, the moving could make change during some time.

It doesn't in one medium. Why entrance in atom and exit are on one straight line which coincides with the direction?