Genady

No, space does not accelerate. The matter on the Earth surface accelerates downwards.

This is a dishonest response. He doesn't tell you to discuss the problem in terms of Variational Calculus. He tells you that F=dP/dT follows from more general principles rather than is just made up, and thus it makes no sense to claim that it is wrong, as you do.

Sure, he is.

This is what I think, too.

Does it?

Why not?

If you are talking about the Earth spinning around its axis, then the points on the surface everywhere accelerate downwards rather than upwards. (except the poles)

Yes, they appear in fewer cases than nuclei, but still the same virtual particles appear in all Feynman diagrams. Regarding separating and analyzing them individually, one word: quarks. I understand what you mean and do not disagree. It is just that the distinction is not good enough for me to decide that they are profoundly different. Like we say here often, it's a model, just like everything else.

True. But not distinct enough for me. For example, atom nuclei don't appear in measurements either; they appear in the calculation of a spread of recoiled alpha particles. The latter also don't appear in the measurements; they appear in calculations of trajectories from the gold screen to the detectors. Etc.

There is no time dilation (split from The twin Paradox revisited) - Speculations - Science Forums The same applies here.

All this has been already said and debunked. Nevertheless, the OP repeats themselves without any progress. The OP is going in circles ignoring input from various members and not supporting their claims with any evidence. This thread is now a troll. The OP does not discuss in good faith. * the OP above refers to @martillo

I understand that in evaluating == the LS gets evaluated first, then the RS, then they get compared. Let's say in the beginning x is 1 and y is 2. So, in x == (x=y): 1) the LS evaluates to 1 2) x=y gets executed; x is now 2 3) the RS evaluates to 2 4) 1 == 2 returns false. In (x=y) == x: 1) the LS gets executed; x is 2 2) the LS evaluates to 2 3) the RS evaluates to 2 4) 2 == 2 returns true

OK, one more time. They start from dP/dt=0. Then they rewrite this equation in such a way that the characteristics of the rocket are separate from the characteristics of the ejection. This allows them to express force on the rocket separately from the force on the ejection. The end. They don't need and don't apply F=ma anywhere, and they don't need dP/dt separately for the rocket. If you don't get this, I can't help you anymore.

That is how it should be applied.

Saying that the units match the units of force is not the same as applying F=ma. It is only your misinterpretation of their derivation, that they apply F=ma.

No, this is not. They don't apply anything else to get the thrust. They already got it from dP/dt=0.

No, they don't apply F=ma to say it. They take the result of previous calculation which they got from F=dP/dt, and call a component of that result thrust. It is a force, but they did not derive it from F=ma, but from F=dP/dt.

Right. And then they say, "Because there are no external forces, dP/dt=0." That is, dP/dt = external forces.

Yes, they do. They start the derivation with calculating dP/dt by expressing delta-P/delta-t and then taking delta-t to dt; see 1.3 and after it. Then they separate two parts of the dP/dt to get the equation 1.5.

From your link:

Is it an opinion poll? My answer: the rest of the universe.

It is not my original idea. It took me some time to find where I got it from, but here it is: Quantum Field Theory for the Gifted Amateur, Tom Lancaster and Stephen J. Blundell, 2014, Oxford University Press

It is perfectly correct, but I have a caveat to add. I think that all photons are virtual. We never observe photons directly, only their interactions with some detectors. The so-called real photons are just virtual photons that last so long that their deviation from the shell is immeasurably small.

They will not match in the case of Earth, too, if length is in cm, or length in feet and mass in pounds, etc. (in case the OP does not realize what mismatched units mean.)

Right, and that was my simple comment above: