joigus

Yes. The rocket --never mind it being an open system-- has its own Lagrange equation. The beauty of Lagrange is you don't even have to think about forces. It's all in describing so-called configurations of the system. That is: How many variables do I need to know where it's at in its evolution? You don't even need to consider whether you are in an inertial system or not. Coordinates could be curvilinear for all you care. Lagrange's method takes care of everything else, including inertial forces.

This is not even dimensionally correct. Besides, m in both sides of the equation would simplify. Please, review your maths before you commit another badly wrong comment.

No. Fext stands for 'total external force' for both the rocket --with unburnt fuel-- and the exhaust. You didn't understand me. I will need some more time, as comments are piling up. I see no end to this...

That's fairly clear to me. You don't understand... Here (from page 8, where you apparently answered 'several times': And indeed, \[ \frac{d}{dt}\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x}=0\Rightarrow\frac{d}{dt}\left(m\dot{x}\right)=-\frac{dV}{dx}\Rightarrow m\ddot{x}+\dot{m}\dot{x}=F_{\textrm{ext}}=-V'\left(x\right) \] Take \( F_{\textrm{ext}}=0 \), and there's your super-special instance. I told you, didn't I? More from unbearable page 8... I told you conservation of momentum applies for overall rocket + exhaust in free space, and I told you how. I've highlighted it. Maybe I'm lucky enough that you happen to read it this time. Etc. PS: By fuel + rocket I mean 'exhaust + rocket.' Good day, sir.

Dear @martillo, there's only one person here who's just been adhering to his own opinion, gritting his teeth against all evidence and reasoning. The evidence being that these are the concepts that engineers apply on a regular basis. All of this is established science. Nothing more to say on my part, as you totally ignored my arguments. Other members seem to have come to similar conclusions.

Lee Smolin: "The trouble with physics..." Martillo: "...is F=dp/dt" "is F=ma"

I know. This is very close to coming full circle, or is already there. Sigh.

What system? There's your vagueness again. Which is included as a particular case in my analysis, of which you said nothing.

Absolutely non-sensical piece of WAG pseudo-reasoning. You obviously don't understand what gravity and mass --inertial or gravitational-- are. You obviously don't understand what electric charge is. You obviously don't understand what zero-point energy is. And, BTW, axioms never justify your logic. Axioms are initial assumptions, and logic is a given.

No. It (momentum of either rocket, fuel or both) only has to be conserved if there are no external forces on the system. In the more general case with an external force that I presented to you --in the presence of this external force acting on both the rocket and the exhaust-- neither momentum nor (mechanical) energy have to be conserved. You say "momentum" and "energy" as if they must mean something independent of what system you apply it to, or which level of detail you want to describe it. They don't. Rocket: Energy is not conserved Momentum is not conserved Exhaust: Energy is not conserved Momentum is not conserved Rocket + exhaust: Mechanical energy is not conserved Mechanical energy + chemical energy stored in fuel is conserved Momentum is not conserved if rocket + exhaust are in a gravitational field Momentum is conserved if fuel + rocket are in free space You see, when you burn fuel, some kind of potential energy that's stored in the molecular bonds gets converted into heat + kinetic energy for both rocket and exhaust. We call this energy Gibbs free energy (of fuel). So no, mechanical energy is not conserved in this case. If the rocket is stationary in free space at initial time, K.E = P.E. = 0 initially. Then it starts burning fuel and starts moving. Now P.E. is not cero, but K.E. is K.E. of exhaust + K.E. of rocket. So where did this energy come from?

I wasn't showing lack of concern. I forgot to write a sentence that I had in my mind. Namely: "which was every bit as horrible as a WW-nth, never mind the number of casualties." The Cold War was pretty horrific, if you ask me, or people who suffered it in Angola, Chile, Korea, Vietnam, and on, and on. You can add to that many indirect consequences, like Chernobyl, which IMO was a consequence of the Soviets beeing obsessed with getting ahead of the US.

LOL ...

I agree. I'm no historian or political analyst, but this looks very much like a CWII rather than a WWIII, because the nukes are still there. Technology has completely changed the rules, and it keeps doing so. This has led to a whole newfangled model of warfare and intelligence. You only have to look at the influence of such things as the Internet, drones, and what's in store with the possibilities that AI offers.

Not necessarily. Here's a sketch of how it would be done. 1) We don't care about the exhaust at all. Our system is the rocket; but it's an open system. To simplify, let's make it 1-degree of freedom, with coordinate of rocket = \( x \). \[ L=\frac{1}{2}m\dot{x}^{2}-V\left(x\right) \] Momentum of rocket is partial derivative of \( L \) with respect to generalised velocity. Just following our noses, and blindly believing in variational calculus and our guessed-at Lagrangian being good: \[ p_{x}=\frac{\partial L}{\partial\dot{x}}=m\dot{x} \] Euler-Lagrange equation: \[ \frac{d}{dt}\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x}=0\Rightarrow\frac{d}{dt}\left(m\dot{x}\right)=-\frac{dV}{dx}\Rightarrow m\ddot{x}+\dot{m}\dot{x}=F_{\textrm{ext}}=-V'\left(x\right) \] which is, of course, the rocket equation. 2) We care about the exhaust. Take all this with a pinch of salt, because I'm not 100% sure this is correct. It produces consistent equation for the rocket though. It's possible that the law of change of \( m \) with time had better be treated as a time-dependent constraint, rather than a coordinate. The exhaust is complicated to describe. But, as we don't care about it very much, we will only describe it with one generalised coordinate, \( X \) wich accounts for the CoM of the whole thing at time \( t \). Very important: The system configuration needs specifying three coordinates: \( x \), \( X \) and \( m\left( t \right) \). We will have 3 Euler-Lagrange equations. The Lagrangian should be, \[ L=\frac{1}{2}\left(M-m\right)\dot{X}^{2}+\frac{1}{2}m\dot{x}^{2}-V_{r}\left(x\right)-V_{e}\left(X\right) \] and \( V_{r}\left(x\right) \), \( V_{e}\left(X\right) \) the respective potential energies of rocket and exhaust CoM. We will also treat \( m \) as a generalised coordinate. The Euler-Lagrange equations are, \[ \frac{d}{dt}\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x}=0\Rightarrow m\ddot{x}+\dot{m}\dot{x}=-V_{r}'\left(x\right) \] \[ \frac{d}{dt}\frac{\partial L}{\partial\dot{X}}-\frac{\partial L}{\partial X}=0\Rightarrow\frac{d}{dt}\left[\left(M-m\right)\dot{X}\right]=-V_{e}'\left(X\right) \] \[ \frac{d}{dt}\frac{\partial L}{\partial\dot{m}}-\frac{\partial L}{\partial m}=0\Rightarrow V_{r}\left(x\right)=V_{e}\left(X\right) \] And those are your evolution equations, at least within this very simple model, and as you can see you get a generalised expression for the momentum of the rocket that accounts for the mass it's losing. My conclusion thus is that @Genady --and everybody else-- is right. As I told you, your trusty momentum can be generalised. This version is called 'canonical momentum associated to coordinate \( x \).' In the case of a particle in a magnetic field, it has a 'tail' that contains the vector potential. It doesn't have to conform to any previous ideas that you have about what it is. The only thing that spoils your Lagrangian methods is dissipation in those cases where you can't follow where the energy has gone. If you don't care about details, and are happy to describe only the CoM of the bulk of exhaust, you can get a good understanding of what's going on. Forgot to mention: \( M \) is a certain total initial mass of rocket when it's fully charged of fuel.

I've never interviewed a muon. And nobody is saying that. Please, read @studiot's answer and ponder what he' saying.

Oh, come on. As Leonard Susskind put it, you can make a pretty good living today out of criticising every new idea that comes up. Beating string theory is beating a dead horse. It sells books too. There are many other ideas besides string theory. What's his problem with Witten in particular? Do you know anything about string theory? Your technique for renormalising a divergent logarithmic integral is making up its numerical value. Do I really need to say more?

Now I realise I meant to emphasise "indirect", not "evidence."

It's perfectly understood. It's called proper time, and @Markus Hanke gave you the recipe to calculate it. The observer on the ground, who is aware of the theory of relativity, can use it to infer how much time has ellapsed in the frame associated to the muon. She, OTOH, can use her own coordinate time and observe the difference. You've been told: It's been measured. It's common knowledge. Misunderstood only by people who haven't studied physics. So maybe you're right about the 'widely misunderstood.' For the wrong reasons, of course.

I would say something like, virtual particles actually contribute to the mass, charge, and other quantum numbers of real particles. So, in a way, we do have indirect evidence of them playing a role. It's just that they never show up. They are modes of the quantum field. It's just they generally keep a low profile, so to speak. I would try to stay away from any metaphysical language here.

You don't, for all I can see. Euler forces* are transversal (centrifugal) or tangential (thrust experienced from inside the rocket, when measured in a non-inertial system), or a combination of the two. They're kinematical in nature. I agree it's not the same case. My comment was about maths and definitions, and what you decide to call 'the force on the system.' This requires to define what you mean by 'the system.' If 'the system' is both the rocket, the fuel and the exhaust, then all forces implied are internal and there is no net force. If it's just the rocket with the fuel that's left, well... You see? Thrusts are not kinematical, but sure enough passengers on the rocket will experience these forces as very much real. That's why astronauts must buckle themselves pretty tightly. * Sorry I may have used the term 'Euler forces' in a sense that's a bit more general. I meant ficticious forces in general, centrifugal and Coriolis included.

And I totally understand your dissatisfaction, believe me. It's part of the conceptual issues of QM. Is it there when I'm not measuring it? What does it mean that now it is? How does the formalism incorporate that fact? It's not that one virtual particle contributes to the mass of the measured particle, it is rather that all the wildest things you can imagine the particle as possibly doing within the strictures of HUP are somehow contributing to the measurable properties of the particle. And all of these ghostly presences interfere in every which way to produce the observed behaviour. That't what's difficult to swallow. But that's what the quantum formalism tells us. How to wrap your head around it in a pictorial way that helps you understand it better is another matter.

Can I interest you in, \[ f=ma \] \[ f=F-v\frac{dm}{dt} \] ? with \( f \) being the rocket-generalised force? That's similar to what you're doing when you go to a non-inertial frame with acceleration \( A \) and re-write, \[ F=ma+mA \] as, \[ F-mA=ma \] \( f=F-mA \) being the force and \( -mA \) the Euler force. To be honest, I'm not nearly as much concerned on how we call it as I am in preventing the possibility of rockets falling on my head because people don't apply the equation correctly.

Well, I don't know. A measurement is not just a particle hitting a screen. You can separate isotopes by their mass, for example. There's your nucleus. You certainly cannot centrifuge virtual particles. They do contribute to the particle's mass, but you can't separate them and analyse them as independent things. That's why people sometimes talk about a 'cloud of virtual particles' around the real, detectable particle. Again, a pictorial handle for the concept, not quite what it 'is' in the theory.

It's actually needed if what you mean is to argue that 'physics is in trouble.' You see, for some physical systems the concepts of force, or acceleration for that matter, or even position, are ambiguous or cannot be defined at all. It's also possible for them to have a surprising and totally non-intuitive definition that nevertheless makes mathematical sense. Take the case of a particle in a constant magnetic field. These systems are better treated in terms of a Lagrangian. And when you do that, lo and behold, the momentum of the system is not what you would expect, and includes the vector potential, turning out to be mv-qA, instead of just mv. In the case that you guys are discussing, it's because the definition of 'the system' is what's ambiguous. The rocket is losing fuel by the second, so part of 'the system' is going away. What do you do? You drop the definition that happens to be useless for this particular case --of F=ma--, and stick to the more secure, more general one of F=dp/dt, as knowledgeable members are telling you. Then you have two terms mdv/dt, and vdm/dt. The same equation that's been applied for decades to investigate how rockets move.

Actually, no. He's a well-known hackler of serious scientists in public talks. I know enough about him to know he's a crackpot magnet. He claims that all of physics since Planck is wrong --I was patient enough to watch one of his videos or two. That's some time down the drain I'm not getting back. He certainly doesn't understand the ideas behind renormalisation. I'm not saying quantum physics is problem-free, and there are no consistency issues. There are. But I see nothing of value in trying to substitute renormalisation strategies with WAG numerical games and numerical analysis. And all hand-waving. That's what he does. Very similar to what you did by copying and pasting some formulae from a 1968 paper and pulling some numbers from a part of your anatomy, and substituting a logarithmically divergent integral by your wild guess. Zero value from a scientific POV IMO.