Genady

(Speed of light) * (1 ns) = (1 ft) (Avogadro number) = (number of stars in the observable universe) g = (Human population in 2050)

And, as I and, evidently, you know, it was Russians who defeated Germany, while allies jumped in when it got clear to them that Russian are winning.

Yes, it's as much significance as you can have without committing major acts of sedition or heroism. But it only works once. After you've become part of a new community or nation, you are expected to contribute to its welfare, help shape its future. You no longer matter to the old one, but you still can matter. Only once? I think it works every day. I keep taking my money from a country A and spend it in the country B (B for Bonaire) for many years. Every time I do it, I transfer my economic activity, don't I? (Of course, I do other things also in the country where I live, for example, volunteer for the Animal Shelter, recycle, etc.)

idk. I've never heard of such a parallel. AFAIK, conservation of quantum information is a consequence of unitarity of quantum system evolution. But this unitarity gets broken by an act of measurement.

I don't think it is strictly so. There is always a (theoretical) observer for whom a specific piece of quantum information is still there, but for any given observer the quantum information disappears as soon as they interact with it. This is my understanding of it.

Some hostile act added to the picture almost daily. Today (yesterday?) they suspended new nuclear talks...

Disagree. Not slow.

No Walmart here, but the situation in our stores similar. About a dime difference. (It could be double price if it were a dime vs two dimes, but it is not. About 5% difference in price.)

All actions have consequences... do they? We don't know ahead if our vote will or will not have consequences, but we know afterwards. If our candidate won, our vote had consequences, otherwise it did not, I think.

It has been already explained to you in various ways by various contributors in various replies in this thread. Enjoy!

Exactly. +1

You say that you are considering only momentum of a rocket, but when you write P = mv, you work with the momentum of the entire system, and it does not matter what you say about it. You can insist on making the same mistake, and it will not lead you anywhere. But if it makes you happy, enjoy!

This is incorrect. Both F = ma and F = dP/dt are correct. Your mistake is that you apply them incorrectly. In F = ma you consider force on the rocket only, while in F = dP/dt, F and P are the force and the momentum of the entire system. So, the F in the first equation is not the same as the F in the second. Of course, dP/dt should be ≠ a, because dP/dt is rate of change of the momentum of the entire system, while a is an acceleration of its part only, the rocket. The whole is not equal to its part. This is a nice back-of-envelope derivation! Thank you.

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.