joigus

Fingers crossed.

In a nutshell, it is the contention that you can understand a system by analysing its parts and the relationships between them. Assuming that it makes sense to consider it as made up of distinct parts, that is. https://en.wikipedia.org/wiki/Reductionism

I'd say that given that both \( \omega_{\textrm{Earth}}R_{\textrm{Earth}}\ll c \) --slow-rotating Earth-- and \( \frac{GM_{\textrm{Earth}}}{R_{\textrm{Earth}}}\ll c^{2} \) --not very intense gravitational field--, you're quite safe using Lorentz transformations that factor out into a boost --jump to a constant-velocity frame-- and a slow rotation. For finer effects you would want to consider GR --Lens-Thirring effect, and such. Does that answer you question?

Now that I think about it, it's also dangerous when you're in a Euclidean context too.

Yes, that was exactly my point. Swapping indices cavalierly is dangerous when you're in a non-Euclidean context. If you want to correlate observers and you need boost + rotation to do that, transpose Lorentz transformations do not coincide with given transformation. BT not equal to B. @Genady, what's your code for that? I use, \left.B_{a}\right.^{b} which produces, \[ \left.B_{a}\right.^{b} \] Never mind. I've just realised while I was making myself a sandwich! LOL

You bet. You can't imagine how many people I've met who don't know this. As well as the reason why one set of coordinates is co-variant, and the other contra-variant. It's all in the language of differential forms, but that's not gonna be a problem for the likes of you.

Exactly. Wrong notation! Ambiguous, as you rightly said.

As to index gymnastics, I'm very fond of Anomalies in Quantum Field Theory, by Reinhold A. Bertlmann --the famous mathematician of John Bell's article. The first third of the book has a lot of it, because he deals with gravitation a lot. Not in detail, but you can check the calculations page by page as a good gymnastics.

In other words, one should never write the tensors with vertical alignment. That could lead to errors if you have rotations mixing in in you brew. Some people do it, I know.

Vertical alignment is exactly what I'm talking about. If you have a 2-index tensor, the first index is the column index, and the second one is the row index. Unless I'm missing something, that's what the vertical alignment is all about.

This is because the first and the second indices generally act on different indices irrespective of whether they are covariant --they transform with the same matrix as the basis members-- or contravariant --the transform with the inverse matrix. For the Euclidean case, this is of no importance, but as you well know, for Minkowski, it matters. Consider the Lorentz transformations, 1) Boost in the t-x plane: \[ B=\left(\begin{array}{cccc} \gamma & \beta\gamma & 0 & 0\\ \beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right) \] \[ \left.B_{\mu}\right.^{\nu}=\left.B_{\nu}\right.^{\mu} \] 2) Rotation in the x-y plane \[ R=\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & \cos\theta & -\sin\theta & 0\\ 0 & \sin\theta & \cos\theta & 0\\ 0 & 0 & 0 & 1 \end{array}\right) \] But, \[ \left.R_{\mu}\right.^{\nu}\neq\left.R_{\nu}\right.^{\mu} \] The way I do it in LaTeX is, \left.B_{\mu}\right.^{\nu} The Schaum series perhaps? PS: Sorry, Markus. I think I swapped co- and contravariant indices in my answer. Let me fix it. I meant, \[ \left.B_{\mu}\right.^{\nu}=\left.B^{\nu}\right._{\mu} \] for the boost, and, \[ \left.R_{\mu}\right.^{\nu}\neq\left.R^{\nu}\right._{\mu} \] for the rotation. I hope that helps.

Both DM and DE are consistent with GR. Sigh.

AAMOF, relativity makes your intent to supersede F=dp/dt with the 'bold new physics' F=ma even more implausible, as p has an extra dependence on v which makes the connection between F (in relativity the 4-force) one step more logically --and calculationally-- removed from acceleration. It has terms that do not involve second-order time derivatives. One term is proportional to velocity, and the other is proportional to acceleration. So no.

Agreed. I was thinking more about a rocket orbiting a planet. So external forces would be conservative.

My answer would be a resounding 'no' based on reasons pretty much pointed out by other members. I have little much significant to add to what, eg, Markus has said. If anything, it strikes me as a sample of our most primitive instincts hijacking our reason under the guise of being a 'rational' solution. It's essentially what our traditional approach to garbage has been throughout centuries: Round it up and put it away, I don't wanna see it anywhere near me.

Ok. In the notes that I'm taking to in order to understand the problem better, I'm distinguishing Fe,ext, Fr,ext, and Fext=Fe,ext+ Fr,ext, which might seem a little bit overthinking it, but is not too bad an approach if one wants to make sure we're not missing anything.

Nice pic. I do remember having watched cousins of these guys in some documentary. Different colour perhaps, but general appearance very similar.

Absolutely. I'm looking at it from every perspective I know, and I can't see any reason why this shouldn't hold true. I can't see any way in which dp/dt is not valid here either. I think we agree on that. Am I right?

I'm checking up on everything I can. I get the same as you. If motion of respective CoM of both rocket and exhaust are collinear and in the absence of external fields. Of course, forces must be identified with dp/dt. What's tricky is the momentum of what and what force on what 'object.' I'm considering the exhaust as one big indistinct thing, although its CoM must move in a predictable way.

OK. I'm analising everything in terms of Newtonian mechanics. I need some time to react to @studiot and @sethoflagos, and see if we agree on reference systems and everything else. You can either ignore external fields --or consider the system in free space-- or not. I'll think about a situation that's as general as possible without the whole thing being a mess. Real rockets, of course, require many other considerations.

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.