Genady

OK. This answers the first question. The second question is still open:

Yes, I do. The expression, dP/dt = mdv/dt + wdm/dt, represents the total force on the combined system of the two bodies. One component of this total force is mdv/dt, the force on the rocket.

How do we know what is better? Better for whom or for what?

I get a different result. I assume that P = mv applies to variable masses, but I apply dP/dt carefully, keeping track of what P refers to. Before the change of mass, it refers to a body with mass m and velocity v. At this time, P1 = mv After losing a mass dm, it does not refer to a body with mass (m-dm) but rather to two bodies: one with mass (m-dm) and velocity (v+dv), and another one with mass dm and some velocity u. At this time, P2 = (m-dm)(v+dv) + udm So, in time dt the momentum change is dP = P2 - P1 = (m-dm)(v+dv) + udm - mv = (mv - vdm + mdv - dmdv + udm) - mv = mdv + (u - v - dv)dm = mdv + wdm - dvdm where w = u - v is a relative velocity between the two bodies. Now, dividing by dt I get the result, dP/dt = mdv/dt +wdm/dt - dvdm/dt The last term, dvdm/dt, is infinitesimally small, and the final result is, dP/dt = mdv/dt + wdm/dt rather than dP/dt = mdv/dt + vdm/dt

Yes, you are.

It depends on the specifics of the system. No general formula. You need to construct a system Lagrangian and then apply some formulas to this Lagrangian.

Sorry, I don't understand your question. As I said, momentum is mass times velocity only for a system with constant mass. What is p in your question? Momentum or something else?

I can't, because but that is where p is defined for general systems.

Oh, I see. Got it.

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?