DimaMazin

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.

I made correct math. Genady don't understand it because he is ortodoxal. You don't understand it because you understant no math at all.

Concrete mathematics was created by smoked brain. We cannot apply it anywhere.

We consider physics and the mathematics has no relation to the physics.

What experiment did Einstein present to the formula?

We should consider million experiments simultaneously?

What experiment proves the mathematics?

My definition is observing therefore the formula of scientists is disinformation.

Correct formula should be u'=v-u/gamma We can use it for velocities of non-simultaneouses ,wich are defined by MD65536.

I incorrectly started to define. There is simple definition. Instant pure acseleration has double disareement in simultaneity. Therefore neutral simultaneity is negative relative to it. t=t0-(gamma-1)x/(gamma×v) Time between events in the non-simultaneity is the same in 2 frames.

Thanks. That looks like triangle with sides gamma*v , gammau*u , and gammau'*u' . If we have many travelers with velocities between 0 and v and we add their gamma*velocity between them then can we get circular arc?Is then gamma*v a diameter or chord of the arc?

Theoretically I should get the same module of the velosity in S', but my brain is not working.

When body A is moving with velocity v relative to frame S then it is observer in S' and has velocity o in S'. Body O is observer in S. What is velocity of body B in frame S' when it travels with velocity u=v(4-3v2/c2)1/2 in S frame?

For example body travels at v and has momentum m*gamma*v. Then let's define at what velocity the body has 1/2 the momentum. gamma*v/2=v/2(1-v2/c2)1/2 v/2(1-v2/c2)1/2=gamma'*u=u/(1-u2/c2)1/2 u=v/(4-3v2/c2)1/2 I don't understand how relativity works, but do you think the velosity is another in S'(the frame is frame of the body at v)? And if non-simultaneities of u(in S) and u'(in S') are simultaneous then why we cannot use them as neutral simultaneity?