Genady

Well, you are wrong about this and you are wrong about E = mc2 .

Why do you call it absolute? Isn't it specific for two observers and two events (start and finish of the duration)?

Yes. No absolute time. Each observer has its own proper time.

You are wrong. This is not a general definition. A general definition is based on Lagrangian (remember Variational Calculus?). It becomes mv in this specific case. Did you ever see formula for momentum of photon, for example? It is not mv, because for photon m=0.

It makes sense to compare proper times of different observers. That was what we did in the twins exercise. The proper time of one observer advanced 10 years while the proper time of another observer advanced 1 week. It is very physical result.

No, this formula holds only for momentum of a system with constant, non-zero mass. Also in your book, they substituted P by mv after they assumed that the mass is constant. In other cases, the momentum formula is different and depends on specifics of the case.

There is nothing to defend. You are mistaken when you say that textbooks say otherwise. Also in your book. It says, if mass is constant then from F=dP/dt we get F=ma. But it does not say that the mass must be constant.

A passage of the proper time of an observer does not change. This is so in SR and in GR. (Proper time is time in a reference frame in which observer is at rest.)

Even with my poor Spanish - it is not one of the four languages I've mentioned earlier - I understand that it is not what the book says. It does not say that F=ma is valid for constant value of m only. It says, that if a momentum changes only due to a change of velocity while the mass is constant, then from F=dP/dt we can derive a familiar form of the Newton second law, F=ma. It does not mean, that this form, F=ma, is not valid otherwise. And it is valid. An instant value of force equals an instant value of mass multiplied by an instant value of acceleration. F(t) = m(t) x a(t). Nowhere in textbooks on Newtonian mechanics I ever saw that this is not so.

It means to me that F=ma applies to momentary value of a if a changes in time.

Yes, what you say is GR. What I replied to above, is not.

I can't believe you don't get this simple question, but I try again. You have acceleration changing in time, you observe it, you get the values as they change. Mass is known. You need to find force. One equation, one unknown.

But you need to put some number for a to find F. Which number will you put if a is changing?

Let's say you observe a body moving with a variable acceleration and you want to know a force.

I see. Sorry, but it is not how GR works.

My question is, what value of a should be used in F=ma if a is changing in time?

OK. How it works if the acceleration changes, for example, continuously increases?

I doubt it says that the law is not applicable if mass changes. Does it apply for constant a only?

Could you show me what exactly that physics text says?

It applies to a momentary mass, a mass at a moment in time.

I mean misunderstanding. I have four in total. But everyday language is English.

I see. When you said, 'considered for' you meant, 'relative to'. OK. Lost in translation.

Yes.

The rocket is not an inertial frame because it accelerates. Consider relative to the ground.

Constant velocity relative to the rocket, but it changes relative to an inertial frame together with a changing velocity v of the rocket. Check in your source, how that ve is defined.