DimaMazin

Let's consider hypotheses. When velocity U is bigger than c, then we use velocity of MD65536 V=v/(1-1/gamma)) U=u/(1-1/gamma)) Then u=2Uc2/(U2+c2) Then we should define u' u'=(v+u) / (1+vu/c2) Then we can define U' using again formula of MD65536 U'=u'/(1-1/gamma) U'=u'/(1-(1-u'2/c2)1/2)😛

t'=gamma(t-vx/c2) Let's use velocity u as distance t'=gamma(1-vu/c2) u'=u/(gamma×t')=u/(gamma2×(1-vu/c2)) I have got equation for addition of any velocity, even when it is as nonsimultaneity bigger than c .

Then t=t'=1second on distance 2v(1-1/gamma) between point of S and point of S'. Why so?

Then Jews were obliged to serch for consensus with Hittler. Due to your wrong policy you do no success in science. Likes of backward Americans led you in delusion.

America created, uses and supports nucklear blackmail policy like all members of nucklear klub . Ukraine destroys the policy . Therefore Trump and all Biden's team are against victory of Ukraine.

Why it cannot be so?

Let's consider Minkowski diagram with general coordinates of two frames

I cannot correctly make complex exploring therefore that is incorrect. Let's consider what is quantum velocity. Quantum velosity is velocity which is created by constant and instant acceleration from frame S in frame S' and back. For example velocity between S and S' is v then quantum velocity is v/2. Time of such observer is t/2+t/(2gamma) because the observer has equal time of travel in both frames.

Seems I have got absolutely neutral velocity V = 2×51/2×c/5 gamma=51/2 Then non-simultaneity t =t0 + (51/2 - 1)x/(2c) is absolutely neutral simultaneity.

Rether Biden needs to destroy Iranian nuclear program.

My experiment does experimentally define the three velocities. You don't understand my experiment because you don't understand relativity of observers and travelers. Wy the experiment can be bad for Relativity? When the experiment can prove Relativity,if it correct.

The equation has three velocities. How do you experimentally define them?

It can be checked experimentaly. 3 travelers is traveling relative to each other. They send light with the same fraquency to each other and resieve the light with other fraquency. Then they can define their velocities and check the laws. Energy factor has square power of velocity c, therefore delta energy can be variable in different frames. But velocity and gamma factor have unit power in formula of momentum. Because square root annihilate square power. Therefore delta momentum should be constant in any frame.

On distance 0 time is 0. On distance 1 million km time is 1 second in S frame. What is velocity of non-simultaneity? Non-simultaneity is not velocity but it can have imaging velocity. In frame S' on distanse 0 time is 0 but on what distance and what time was or will be in co-located point relative to second point of frame S?

I have confused. We should create law of addition of non-smultaneities. We can image any velocity(and bigger than c) as non-simultaneity. Then we can check any law of addition of velocities.

The Einstein's law is incorrect because we cannot use the same module of reverse velocities in two frames. 😝

Yes, i mistaken.

A very simple refutation of Einstein's law of addition of velocities.

When you have got change of momentum dp then any observer,in any frame, defines the your change of momentum as the same dp.

When they have equal masses.

When travelers ,relative to each other , have d momentum then it is the same in any frame.

Let's use conservation of momentums for correct addition of velocities V3×gamma3=v1×gamma1+v2×gamma2

Do you think that undefined opinios of mathematicians are more useful in physics than laws of physics? For y=x/x When we approach x to 0 there is no changing of the result.

You don't know what is 0/0. Then what do you try to explain?

Well. any2×0=0