DimaMazin

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

Well. 0/0=unknown Then E1=unknown2×0=0 E2=unknown2×0=0

Teachets teached us 0/0=1 What is in Israel?

Do you want to dispruve the law? Or do you remake it?

Where?

I don't understand you. 0/0=1 Ask mathexperts.

Then p1 and p2=0. E=0 0/0=1 E1=12×0=0 E2=12×0=0 E is kinetic energy E1 and E2 are kinetic energies.

In S .

It is exactly incorrect because it violates energy momentum law. S' is traveling with velocity v1 in S. We acselerate body to S' then the body has momentum p1 and energy E1. Then we acselerate the body to v2. Then the body has momentum p and energy E . In the same time the body has momentum p2 and energy E2 in S'. The enegy momentum law is E1=(p1/p)2E E2=(p2/p)2E

Thanks.

My idea is so. Velocity v is between S and S' . gamma=1/(1-v2/c2)1/2. Object travels with velocity u in rame S. Norm of relation between distances and times is defined by gamma. u is rate of change of distance at time in S and it is the same when S travels in S'. We should only transform it relative to distance and time of S'. Then u transforms into u/gamma. And we should add velocity v. u'=v-u/gamma Einstein's velocity u' is another and therefore it changes norm of contracted and not contracted distances and times. Therefore it is incorrect.