joigus

The ancient Egyptians used a primitive version of calculus to make the pyramids, rather. (Small incremental sums.) Maybe you mean "similar to KE=(1/2)mv2 Or perhaps F=ma? F=mv2 is no standard physics.

It's hard for me to understand what your point about the point of calculus is. But as I can't sleep very well tonight, I've thought I might as well tell you a little bit about what your problem may be. Your problem may be that you don't understand real numbers. Real numbers go beyond what intuitive numbers (numbers you may be used to) are. They're not like the number of people in a room, nor like the reading of a ruler, nor like the money in an account or exchange rates between currencies. These numbers can be classed into a list: Counting numbers: 1, 2, 3,... (natural numbers) Whole numbers: ..., -3, -2, -1, 0, 1, 2, 3,... Ratios: 2/3, -1/5, 10132/11, (plus all the whole numbers), etc. (rational numbers) In order to define real numbers, besides all the usual algebraic assumptions, you need this axiom: It is impossible to approach a number arbitrarily closely without that number being part of my system of numbers. This is called "completeness." You can rephrase it as "limits of numbers must be numbers." Calculus was being used very fruitfully by many mathematicians and natural philosophers for almost 200 years before mathematicians like Cauchy and Weierstrass defined it rigorously. <ignore if you don't understand> Another way to understand an idea could be to understand when it fails and how. This link may be interesting in that regard: https://amsi.org.au/ESA_Senior_Years/SeniorTopic3/3a/3a_4history_4.html There you can find a function (Weierstrass' function) which cannot be differentiated meaningfully: \[f(x)=\sum_{n=1}^{\infty}\dfrac{1}{2^{n}}\cos(4^{n}x)\] And a graphic representation to an approximation by taking only 50 terms of the sum: \[y=\sum\limits _{n=1}^{50}\frac{1}{2^{n}}\cos(4^{n}x)\] </ignore if you don't understand> Now let's go with your example. First, you've copied the formula wrongly. It should read: \[\lim_{x\rightarrow0}\frac{x^{2}-25}{x-5}=5\] What does that mean? (what you wrote has no meaning.) It means you can get as close as you want to number 5 by substituting in the expression for f(x), \[f\left(x\right)=\frac{x^{2}-25}{x-5}\] a number x as close to 0 as you want. The key that may be confusing you is that "as close as you want." Now, when x is not 5, the expression, \[\frac{x^{2}-25}{x-5}\] of course simplifies to, \[x+5\] And it is obvious that you can get as close as you want to 5 by substituting x in x+5 for a number as close as you want to zero. That's the key to the "delta" that seems to ring a bell to you, but not the right bell. Here's the rigorous definition of limit: A function f(x) of one variable x as limit at x=a, and the limit is L if, \[\forall\varepsilon>0\;\exists\delta>0/\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-L\right|<\varepsilon\] Literally read: for all positive epsilon there is at least one positive delta such that, when x is closer to a than delta, then f(x) is closer to L than epsilon. In other words: You can get as close as you want to L in the expression of f(x) by substituting x for a value as close as you want to a. I hope that helps. Derivatives come later.

The key to the Lagrangian formalism is not to consider the composite system as a point. The key is to think about variables in an analytic manifold rather than vectors and directions. Positions of all parts of the system become points in an n-dimensional surface. They are called "configurations." Generally there are two ways to proceed. One is to identify all the independent variables to specify a configuration and write down the evolution equations, which involve so-called "generalised momenta" and "generalised forces." The other is to ignore dependences (constraints) and solve the constraints later by a method called Lagrange multipliers. In your case you would need 3 variables to specify where your system as a whole is. Then, internal variables to specify where every internal part is with respect to this centre (typically the COM.) Because no internal interaction energy depends on where the system is, no energy exchange depends on where you place it, there can be no "generalised force" on the COM, and thereby no second derivative of it involved in the dynamics. I don't have time to be more explicit now. Maybe tonight, or tomorrow.

This is another possible and interesting way to interpret your question I had missed. Composition of motion from local references to more cosmic-based ones.

Sorry. Let's say some additional angles to account for articulations.

If there's no lab experiment, how do you know it's not spinning faster enough? There are good reasons why it's not spinning faster enough, but they probably have to do with fluids being pumped farthest away from the rotation axis, and thus making a higher-than-expected-by-geometry moment of inertia. Friction further complicates things. \[ I\omega = \textrm{constant} \] \[ I_{\textrm{expected}}<I_{\textrm{real}} \] \[ I \sim mr^2 \] Discrepancies are expected, rather than revealing the theory being wrong. \[ \omega_{\textrm{real}} < \omega_{\textrm{expected}} \]

Depends on the level of approximation. If you consider the human as a point (centre of mass): 3 If you consider the human as a solid (COM & orientation): 6 If you want to account for motion of articulations: 6 plus (number of relative angles needed to specify articulations) If you want to account for every cell's position: somewhere between 1012-1016 If you want to account for every atom's position... You get the idea? It depends on how precisely you want to be able to tell the dynamical state of the human. Up-down motion is considered the same "way" (so-called degree of freedom) no matter whether it comes from an earthquake or an elevator.

Force doesn't make much sense when dealing with quantum mechanical systems, but if you want to get an idea of the magnitude of the interaction in Newtons, there's a trick you can do. You take the Coulomb force law and substitute the distance r by the Bohr radius. Somebody's made the calculation for us here: https://www.toppr.com/en-es/ask/question/calculate-the-electrostatic-force-of-attraction-between-a-proton-and-an-electron-in-a-hydrogen/ I've roughly checked the calculation and it is correct. About a tenth of a millionth of a Newton. It must be interpreted as an average. The strong force doesn't play any significant role in this.

Keep it simple? How more simply could I have put it? You cannot have a force producing an acceleration on x_com because you don't have any dependence on x_com in the potential energy. It cannot come from fictitious forces because ficticious forces are obtained by dependence on corresponding coordinates in the kinetic energy. The Lagrangian formalism also allows you to obtain fictitious forces quite simply: \[ \frac{\partial}{\partial r}\frac{1}{2}mr^2\dot{\theta}^2=mr\dot{\theta}^2 \] The kinetic energy acting to these effects pretty much as an effective potential energy. As you have no dependence on the COM coordinates anywhere (neither kinetic, nor purely potential terms), you can have no force on them, and thereby no acceleration. Simpler it can't be, but you need to understand it, which you don't. You are mixing everything. Inertial, non-inertial frames, particles, rigid bodies and deformable mechanical elements, changing mass, cosmology, the speed of light. There is practically no field of physics whence you haven't taken some magical word to help you reason the unreasonable. Including "bare particle", which is a concept from quantum field theory and is playing no role there. Rigid solids have 6 degrees of freedom, so they are not particles in any sense. That's why they have moments of inertia characterized by certain integrals of their mass density. "Variable inertia", a magnetic dipole "Carrying excitation"? You don't make any sense. Do you mean a magnetic-dipole-carrying excitation? I'm completely sure the laws of electromagnetism don't allow you to obtain COM momentum either. You shouldn't play that game, if you don't know how to play it. You're playing chess, you announce check mate in three moves, and you're doing it by having the rook move along a diagonal. Don't you understand nobody who knows anything about chess can be cheated with that? It doesn't help if you say: "keep it simple," or "you overlook something." You overlook something: physics.

Nice. Zhuang Zhou

What about this for exotic assumptions?: I must confess I'm in awe as to the amount of detailed analysis that wiser folks than me are willing to offer you to explain why your idea cannot work. I'm far lazier. That's why I go for the arguments why your idea is a non-starter because it is ill-conceived from the get go, instead of going into the details. Energy can be hidden inside a system and suddenly be released. But momentum is a vector quantity, so it doesn't work that way. Momentum cannot be hidden "under the rug" of internal variables. It also must be conserved for very robust, absolutely fundamental reasons that have been explained to everybody's satisfaction but yours. It is not only a mathematical theorem about symmetries and conservations laws. It has been checked ad nauseam experimentally. It is valid across different theories. It is quite simply how Nature works. The errors have been pointed out to you from every which direction. You are mixing forces that are only valid in a non-inertial frame with forces that are only valid in an inertial frame so that, in your mind, your system can do what you want it to do, never mind it's known to be impossible. When pressed about a fundamental point, you claim that maybe general relativity is wrong, that maybe there is a special place in the universe where the speed of light is null. You use concepts like "quasiparticle" or "bare particle" (that make no sense in the context you're using them) so that your nut-and-bolt system can produce momentum. Nothing in your system can be sensically considered a quasiparticle. Also, you use a mathematical term like dm/dt (a varying mass term implying m(t), only justifyable if there is a continuous mass jettison of some kind) for a mechanical element which actually stays in your system and thus keeps contributing inertia, internal forces, and torques; so it cannot be represented by such term. Neither your system's mass, nor its parts' can be represented by a mass that is a function of time, as all your masses are constants. Now suddenly magnetic fields make an appearance. This looks more and more like one of those open-ended problems, constantly morphing, of which only the conclusions remain constant. What's next? The Dirac equation must be corrected?

Sorry, I meant "vacuum energy". It is a part of Einstein's general theory of relativity. It is a term responsible for expansion. The farther away galaxies are from each other, the faster the recede from each other. So the "repulsion" (it's not really a repusive force, it's rather space itself expanding) is proportional to the distance. It also depends on time, increasing exponentially with it, so you would never be able to compensate for it by gathering mass locally. Expansion of the universe will always win eventually, no matter how much mass you cluster together to compensate for it.

Why do you say all space-time to be curved is inside the mass? (My emphasis.) I'm interested in your question, but I need clarification of the previous point.

I agree that there are important differences. I said that in the spirit of an analogy, which was --I think-- Swansont's original intention. Many physical systems look inert, but they have certain "tension" stored in them. A rubber band, a compressed spring, chemical energy --free energy is the useful concept in the acid's case, rather than potential energy,* as you know very well. In an analogous way, magnets have this tension stored in them. Whether this tension is released to produce work (where there's work, there's force), depends on the presence of other magnets. Once you put a magnetic dipole in front of a magnet, there is a potential energy of interaction that depends both on the distance and the relative orientation of both. The most elementary version of this potential energy is that between a magnetic dipole and a magnetic field: https://en.wikipedia.org/wiki/Potential_energy#Magnetic_potential_energy The OP, and then Charles, seemed to be confused with how something "inert" can produce work. * Although they can be related from a fundamental POV

Not enough in the visible universe. You would have to go beyond the cosmic horizon (3-odd billion light years away) and pick up a lot that was lost when the universe was inflating more than a 3-odd billion years ago. And that's impossible. We suspect lots and lots of matter to have been lost forever beyond the cosmic horizon. Other users are suggesting different ways of understanding your question, though. That's interesting. Black holes are usually associated with very intense gravitational fields, but that's because they are behind their horizons; they are extremely compressed. But as pointed out by MigL, the outer field at a distance far enough away is indistinguishable from the field of any other source of the same mass. Be aware though, that the farther away you go, the more intense is the effect of energy vacuum, and the feebler is the "source effect" one. So you would never be able to compensate by accreting local mass. I hope the argument is clear. They pull, not only in different directions, but with opposite varying tendencies depending on the distance. I think that's key to say; no, it would not be possible. Still IOW, you're trying to compensate for k'r with a -k/r2 by playing with k. There will always be a distance r such that your "compensation" breaks down.

You can picture magnets as pieces of metal that package circulating currents inside. In the case of ferromagnetism, it's somewhat more involved, because you need quantum mechanics for ferromagnetism to be possible. But you can start by seeing from Maxwell's equations that small magnetic dipoles exert forces on each other, because they have an energy depending on their relative position. Then you should go on to picture the chunck of metal as having many of these little magnetic dipoles aligned with each other. I don't know if that helps. Trying to understand these things with vague concepts like "inert" is perhaps not the best way. As iNow says, magnets are not "inert." Something is "moving" inside so as to keep the fields in place. An as Swanson says too, the magnet, left to itself, looks quiet, but there is some "tension" inside, if you will. The acid example is a very good one. You look at sulfuric acid in a vessel and it looks pretty much inert. But there's a lot of potential energy in it that can cause quite some damage.

The point of calculus is studying change and balances of change. The point of Cartesian coordinates is to parametrize space.

Magnets do have energy density proportional to the field squared. Maybe the bell that's ringing in your head is that a magnetic field doesn't do work on a moving charged particle. If that's the case, it's true that a magnetic field doesn't do work on a charged particle with velocity v, because the force is perpendicular to the displacement. Magnets acting on each other also have energy.

As @MigL says, those are periodic trajectories in phase space, not CTCs. CTCs involve time as a dimension too.

I don't understand what you mean. Can you rephrase? On a humorous note, the point of calculus is x.

Black holes are very small in relation to their mass. The whole Milky Way is somewhere between 100-200 thousand light years across. Yet its Schwarzschild radius is about 0.31 light years only. On the contrary, the accelerated expansion is only noticeable beyond the range of billions of light years. So black holes would never overcome expansion. The mass would have to be ridiculously big. You can do the comparison yourself. G=6.674×10−11N⋅m2/kg2 c=3×108 ms-1 1 light year = 3×108 ms-1 × 365×24×3600 s M = (1 billion light years)× c2 /2×G Somewhere around 6 × 1042 kg I think. 1030 Milky Ways... Ridiculous A million million million million million Milky Ways. No way.

That's what I believe.

Nice historical account. Thank you.

=> Newton pairs => No contribution to overall motion => Not significant to COM evolution equations => Again. Rotation of one part not relevant to COM coordinates motion => No. Fictitious forces only relevant to objects moving with respect to non inertial frames (nut, bolt, etc.) subject to acceleration. Not relevant to COM motion. => Already addressed. Non-sequitur => Already addressed. Non-sequitur That's exactly what I think. There will also be a push and pull effect along the axial direction, to be studied in terms of action and reaction, if you will. Again: no overall momentum. Even though the detailed analysis would be quite involved. Assuming there's no friction, any angular displacement would result in a linear displacement, depending on the pitch of the screw. As said, and so far unanswered, the Lagrangian formalism makes it quite transparent that there can be no thrust for the COM system. It is true that the Lagrangian formalism can only be applied to conservative systems. But friction would only make it worse, not better, for the OP's claims.

You cannot dismiss essential points by saying they are "just assumptions" (to be addressed later) and then rush to mention in passing Planck's length (or a tiny fraction of it?), alleged axial anysotropies in the CMB (unobserved), etc. and then expect everybody to believe these bizarre phenomena that would require a complete re-work of everything physics is based on, all coming from a simple diagram. Your machine cannot turn the world as we know it upside down in one fell swoop. The physical principle you're up against is even valid for systems of quantum fields and is a version of Ehrenfest's theorem. I would (and will) demand nothing short of arguments of such dazzling clarity as to make me think it's worth giving up everything I've learnt during a lifetime. 1) Momentum conservation is tied to space symmetries 2) Fictitious forces only appear in non-inertial systems 3) Internal constrictions can always be resolved into action-reaction pairs in the inertial frame (I told you why this is necessary to be able to apply Newton's laws either to whole systems or to their constituent parts) 4) Mass transfer only results in thrust when mass is permanently ejected, not when it's kept inside the system May I also remind you of the adage, "Extraordinary claims require extraordinary evidence." Carl Sagan