joigus

mu and nu representing coordinates and i representing the extension of your body. The "parts" it's made of.

But when describing rigid bodies, points like the one you describe play a role as, say, kinematical references. An example is the instantaneous centre of rotation for a cilinder rolling over a plane. But it's not like the mass can be considered lying at that point. You must add the, \[\boldsymbol{\omega}\wedge\boldsymbol{r}_{i}\] to the velocity. And you get for the kinetic energy something like this: \[K.E.=\frac{1}{2}M\boldsymbol{V}^{2}+\boldsymbol{\omega}\wedge\boldsymbol{P}_{{\scriptscriptstyle \textrm{COM}}}\cdot\boldsymbol{V}+\frac{1}{2}\omega_{\mu}\sum_{i}m_{i}\left[\boldsymbol{r}_{i}^{2}\delta_{\mu\nu}-\left(x_{i}\right)_{\mu}\left(x_{i}\right)_{\nu}\right]\omega_{\nu}\] If V (the velocity of the reference point tracking the motion of the body as a whole) coincides with the COM of the body (nut), then you have a further simplification: \[K.E.=\frac{1}{2}M\boldsymbol{V}_{{\scriptscriptstyle \textrm{COM}}}^{2}+\frac{1}{2}\omega_{\mu}\sum_{i}m_{i}\left[\boldsymbol{r}_{i}^{2}\delta_{\mu\nu}-\left(x_{i}\right)_{\mu}\left(x_{i}\right)_{\nu}\right]\omega_{\nu}\] so that the energy can be expanded as translational plus rotational. The quantity, \[\frac{1}{2}\sum_{i}m_{i}\left[\boldsymbol{r}_{i}^{2}\delta_{\mu\nu}-\left(x_{i}\right)_{\mu}\left(x_{i}\right)_{\nu}\right]\] is the matrix of inertia. It's not a point-particle problem. Are you familiar with this?

But isn't this what I told you, @John2020? Yes, but the nut's thread is not a dynamical element of any relevance in the problem. It's rather its COM and moment of inertia.

https://en.wikipedia.org/wiki/Linear_actuator Quite different from helical trajectories.

Probably not. I'm of the same mind as Markus here --and I don't mean in genius, and probably he didn't mean that either. In my case it's more of an "on the shoulders of giants" kind of idea. Though to me, that doesn't mean I don't value contributions from people analyzing complicated scenarios. Quite the contrary. Some explanations I've found here are nothing short of a masterpiece. But I for one need the shortcuts that the big picture gives you (these words "big picture" have appeared on another thread recently.) If only some people saw what many of us can see thanks to these great minds. It's as if someone had given you night-vision goggles to see in the dark. Why won't Michel put on the goggles and see the vistas? That's what I ask myself. The big picture gives you power, even if you're not a powerful thinker. Mathematical tools take you farther and farther afield, and uphill. It's as if someone took you with a helicopter to the top of the mountain and you said, "oh, I see." Then you can follow the terrain downstream. It's such a pleasure! I think that this self-indulgence is both the greatest advantage and the first deadly sin of the theorist. You need to see the landscape, and you must cover a lot of ground.

Wait a minute. Has the claim changed again? If there is gravity, the nature of the claim changes completely. That's not what I understood.

Yes.

You're not the one who gets to say when the discussion is over, may I point out.

That's easy: No.

I'm starting to suspect you don't understand how a nut works. It does not follow an helical trajectory. A point at the contact surface does, not the nut as a whole. For every point that moves in that helical orbit, a symmetrical point moves opposite to it, and the nut advances along the axis. No "overall helical motion" results. Is that what you don't see?

I'm always open to surprises. Your new concept of space distorted by a nut and screw I can only eagerly await. Get ready for some heavy-duty vector algebra and some real analysis too. And by real I mean... Get ready to get real. Pun intended.

That is implied in h. Don't you see it? Bigger angle, bigger step h. As Guideon noted with his image. I just hope the step is not changing with time! Rotation translates into relative linear displacement of both pieces. I hope you agree with that. You further have, \[dx_{s}+dx_{n}=0\] That's why your system stays in place as a whole. Sorry.

Rotational motion is converted into linear motion, as Swansont and Ghideon say. I'm working on more additional explanations. In terms of: (1) Systems of particles (dynamics of rigid solids) (2) Lagrangian constraints (parametrizing the positions of these rigid solids) \[d\theta_{n}=k_{n}dx_{n}\] \[d\theta_{s}=k_{s}dx_{s}\] \[2\pi=k_{n}h\] \[2\pi=k_{s}h\] theta_n is the nut's angular variable theta_s is the screw's angular variable x_n is the nut's COM position x_s is the screw's COM position k_n = k_s because nut and screw fit together with the same h (the linear "step" of the thread) Later.

It could be a value not accessible by elementary operations. It could be pi, or e. In that sense, it's at the core of space, rather than at the edge. The "camera lens" is made of deltas and epsilons that I defined for you before. Calculus allows you to forget about the lens and operate with the points.

Exactly, you're getting closer. Literally. Calculus is about formalising the operation of "getting closer." You can use it to calculate the gentlest slope on a mountain, where the summit is, etc. It's not necessarily about "orbits." It's no coincidence that KE=(1/2)mv2 (Newtonian kinetic energy) reminds you of E=mc2 (Einstein's kinetic energy). One is obtained from the other by means of... guess what. Calculus!

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