Ghideon Posted September 29, 2020 Share Posted September 29, 2020 (edited) 18 minutes ago, John2020 said: I don't understand where I get contradictions. According to what I presented so far the findings are the following: a) Newton's F=ma addresses external forces in order a body (m) to accelerate Another way to describe the contradiction is to attach the box, with the device inside, to a dynamometer. The dynamometer should show a measurement of zero at all times since no force is acting on the box; you state that F=0. At the same time the dynamometer has to apply force to keep the box from escaping with means the dynamometer shows a measurement other than zero. Again, your claims result in a contradiction. Edited September 29, 2020 by Ghideon Link to comment Share on other sites More sharing options...
John2020 Posted September 29, 2020 Author Share Posted September 29, 2020 (edited) Your mistake is that you try to justify the motion of the system using the F=ma that applies for external forces. Moreover, you still ignore the possibility of change of CoM by means of induced internal forces as I presented above. As also collimear forces will not do the trick (see Fig.1 - Lower). I cañ show all these with the momentum conservation too. See you tomorrow. I have to sleep now. Edited September 29, 2020 by John2020 Link to comment Share on other sites More sharing options...
Ghideon Posted September 29, 2020 Share Posted September 29, 2020 (edited) 16 minutes ago, John2020 said: Your mistake is that you try to justify the motion of the system using the F=ma that applies for external forces. Not in my last post. I did a thought experiment regarding measurements using your descriptions and claims as input. 42 minutes ago, John2020 said: electron cannot be propelled by means of internal forces We are moving into @joigus* territories so I'll just offer a side note: So far we have discussed Newton, if you wish to introduce particle physics the Hamiltonian will be important and your claims, if correct, would have profound impact. Have you studied Hamiltonian mechanics and connection to Newton and other more recent work? 16 minutes ago, John2020 said: I have to sleep now. Damn! I guess we are in the same time zone, but I seem to have been too active on this forum today and postponed some things. Back to work... *) and other experts active here Edited September 29, 2020 by Ghideon Link to comment Share on other sites More sharing options...
John2020 Posted September 29, 2020 Author Share Posted September 29, 2020 No I have not study Hamiltonian mechanics but what I have done in my paper is that I have applied these findings to special relativity and Lorentz transformations that lead to the discovery of a more general framework that is relevant for quasiparticles that may reach and even surpass the light speed without contradicting Einstein or Lorentz. The problem is addressed with very simple straightforward maths (no tensor, no Hamiltonians and no Langragians). It will be a huge challenge (since now comes Einstein and Lorentz in the picture) but I don't believe you want to see this, after the classical mechanics challenge I just presented. Otherwise, tell me to send you the link to my paper in private (through mail). Link to comment Share on other sites More sharing options...
Ghideon Posted September 30, 2020 Share Posted September 30, 2020 (edited) 7 hours ago, John2020 said: No I have not study Hamiltonian mechanics but what I have done in my paper is that I have applied these findings to special relativity and Lorentz transformations that lead to the discovery of a more general framework that is relevant for quasiparticles that may reach and even surpass the light speed without contradicting Einstein or Lorentz. The problem is addressed with very simple straightforward maths (no tensor, no Hamiltonians and no Langragians). Ok, we can address that later and keep focus on the more immediate issues: 10 hours ago, John2020 said: In order to find the equivalent external force that is required to achieve the same acceleration then: The above is incorrect. Either you have an external force or you don't have an external force. You claim that F=ma is not applicable so I modified the setup slightly. Please describe what would be measured** by a dynamometer connected to the box* when the drive inside is trying to accelerate it. By "trying" I mean that the drive is enabled an operating according to your claims but the box is still at rest since, held back by the dynamometer attached to the wall. 7 hours ago, John2020 said: It will be a huge challenge (since now comes Einstein and Lorentz in the picture) but I don't believe you want to see this, after the classical mechanics challenge I just presented. Otherwise, tell me to send you the link to my paper in private (through mail). It depends on how this thread evolves and the how the above issues are resolved. *) I continue to use a box with a working reactionless drive inside while moving away from F=ma for now. **) Exact numbers not required, "zero" or "more than zero" will do Edited September 30, 2020 by Ghideon unnecessary sentence Link to comment Share on other sites More sharing options...
John2020 Posted September 30, 2020 Author Share Posted September 30, 2020 (edited) Good morning Ghideon! I am a little bit late. I will try to answer this while I am in the bus to work and we will continue this conversation at the evening. On Friday evening and the coming weekend I will have more time available to clear out all these issues. Stay tuned! For the other stuff in the paper we need to open a new thread in case there is interest and after the completion of this thread. 58 minutes ago, Ghideon said: The above is incorrect. Either you have an external force or you don't have an external force. You claim that F=ma is not applicable so I modified the setup slightly. Please describe what would be measured** by a dynamometer connected to the box* when the drive inside is trying to accelerate it. By "trying" I mean that the drive is enabled an operating according to your claims but the box is still at rest since, held back by the dynamometer attached to the wall. a) There is no external force acting on the box b) The box will start to accelerate by means of induced internal forces c) The dynamometer will show a reading e.g. 1N d) Let us say the mass of the box is 1 Kgr e) The reading of dynamometer is not aware the type of force being acted on the box (internal or external) f) Due to (e) we may use the F=ma-> a=1 m/sec^2 without knowing the propulsion mechanism of the box What you fail to understand is in case we have a box that moves by means of internal forces, we cannot describe the cause of motion through F=ma but just through measurements. I arrived at my workplace. See you later in the evening. Edited September 30, 2020 by John2020 Link to comment Share on other sites More sharing options...
joigus Posted September 30, 2020 Share Posted September 30, 2020 (edited) 12 hours ago, John2020 said: d) Something that none noticed so far is that I speak about systems with inner structure. I think everybody has noticed that. You can argue that most everything has some kind of inner structure. 12 hours ago, John2020 said: e) A direct consequence of (d) is that bare particles e.g. electron cannot be propelled by means of internal forces. They can't and they never do. They can decelerate, though, by means of radiation reaction. But it's the radiation field that pulls the electron almost to a standstill. Electrons, of course, are decelerated by their own radiation field, not by means of internal forces. You can argue that the electron "ejects" something (radiates photons). Nothing can either brake or accelerate itself without ejecting or absorbing something or having an external field producing these effects. It's a property of space-time (symmetries) not a property of any particular system you come up with. Your system looks very much to me like a frictionless nut turning around in a bolt. No new physics there that I can intuit. Transfer of mass doesn't apply to the system that you seem to be representing in your drawing --as noted by Swansont. Rockets obtain momentum by liberating exhaust mass to space. That's what makes the dm/dt term relevant. Your equations for force and torque do not apply to a system where there is mass transfer. Mass transfer does not apply to solid systems. The mass "stays there." Please, consider this possibility: What if you're about to embark on an ill-conceived project that will make you waste a number of years in something that can't work just because you won't listen to criticism? Your enthusiasm is praiseworthy, but you seem to be applying physical principles incorrectly. Edited September 30, 2020 by joigus Link to comment Share on other sites More sharing options...
swansont Posted September 30, 2020 Share Posted September 30, 2020 14 hours ago, John2020 said: Obviously, the F=ma applies just for external forces. In order to find the equivalent external force that is required to achieve the same acceleration then: Finternal→=nrmTu⃗ reldt=ma⃗ =Fexternal→ @Ghideon @swansont and @joigus Induced Force Meaning: In Fig.1 - Upper the pair F_A (input force) and F_A' applies a torque about the axis of the translation screw having a clockwise direction. The resulting force (called output force, see wikipedia: https://en.wikipedia.org/wiki/Screw_(simple_machine) that pushes the mass m_T to the right is perpendicular to the action force and opposes (counterclockwise) to the cause that created it (clockwise direction) due to the conservation of angular momentum. It is similar to the electromagnetic induction and Lenz Law. Unfortunately, the definition "induced" was reserved just for the electromagnetism. Regarding the r_A and r_R, I created the following drawing: F_A is exerted tangentially on the translation screw that is always perpendicular to the m_T motion and F_R is the reaction force being exerted on the mass m_T or vice versa. The r_A is the position vector of the τ_A having as start point the center of the translation screw and end point the surface of it. The r_R is the position vector of the τ_R having as start point the center of the translation screw and end point the internal surface of the mass m_T. If FA is up, what force is there keeping the screw from accelerating upward? What force causes MT to move? 13 hours ago, John2020 said: d) Something that none noticed so far is that I speak about systems with inner structure. The corresponding known body problems related to classical Newtonian motion, are always processed as point like. Considering bodies or systems as point like, they can never be self-propelled (they must have an inner structure) even if it was possible. Everybody has noticed. But here's the thing: you could simply define your system to include fewer components "inside the box" i.e. internal vs external is an artificial constraint you put on the system. Where does the violation of the third law occur? Analyze just that interaction, and get rid of the extraneous discussion. Make that the system. But here's the conundrum: that means Newton's 3rd law would apply. IOW, Newton's law doesn't cease to apply simply because you put something in a box. If you think that's the case, it's most likely because you aren't accounting for some force introduced when you make the system more complicated. (this is nothing new; it happens all the time in relativity discussions with people who have decided that relativity isn't correct.) 13 hours ago, John2020 said: e) A direct consequence of (d) is that bare particles e.g. electron cannot be propelled by means of internal forces. Quasiparticles e.g. particle trapped within a standing wave can be propelled by means of internal forces (by applying a standing wave phase shift) No, not then either. As above, if you think that's the case then you just haven't accounted for some interaction. 13 hours ago, John2020 said: Electrodynamics Hint: Replace the translation screw mechanism and the rest of the periphery of the system with a standing wave. A standing wave of what? Is this an EM standing wave? 13 hours ago, John2020 said: Replace m_T with the mass of a particle (trapped within the standing wave). Applying a phase shift, results in the phase shift of the nodes and the redeployment of the entrapped particle mass. This is not my discovery but of someone else and proves the idea of this thread with very simple trigonometric identities (just for the standing wave part). "Apply a phase shift" isn't magic. There has to be an interaction between the particle and the EM wave. That interaction is what moves the particle. 13 hours ago, John2020 said: m_T is not refer to dm/dt but just to dm. The dm/dt is the mass transfer ratio. In our case (see Fig.1-Upper) the amount of mass being transferred is dm=m_T. The equation having the dm/dt addresses the general case (variable mass transfer ratio). That's not what dm refers to with Newton's laws. dm is the change in mass of the system, but you've define the mass of the system as a constant. Link to comment Share on other sites More sharing options...
Ghideon Posted September 30, 2020 Share Posted September 30, 2020 (edited) 8 hours ago, John2020 said: a) There is no external force acting on the box b) The box will start to accelerate by means of induced internal forces c) The dynamometer will show a reading e.g. 1N d) Let us say the mass of the box is 1 Kgr e) The reading of dynamometer is not aware the type of force being acted on the box (internal or external) f) Due to (e) we may use the F=ma-> a=1 m/sec^2 without knowing the propulsion mechanism of the box What you fail to understand is in case we have a box that moves by means of internal forces, we cannot describe the cause of motion through F=ma but just through measurements. Your attempts at explanation does not match how physics work; maybe due to misunderstanding Newtonian mechanics? But that can be addressed if you wish to learn and understand. Are you aware how action and reaction pairs of forces work? Maybe we could look at the issues using Newtons third law*? *) As Joigus said, you must have either all three Newton's laws or none. I'm just testing my approach as an attempt at highlighting the consequences and the contradictions in a way that may drive the discussion forward. Edited September 30, 2020 by Ghideon Link to comment Share on other sites More sharing options...
John2020 Posted September 30, 2020 Author Share Posted September 30, 2020 I am driving back home and while reading the responses on this thread, I put myself in everybody position for a moment and I made the following thought: Why you cannot see what I see? Since I am not a physicist I see things from a different perspective and I use a slightly different language than you expect. When I looked to all requests to me to justify aspects of the concept I realized that we were actually not in tune (from joigus view this sounds as “not hearing criticism”. In text messaging platforms sometimes the written words sound differently, I can identify this with myself, too). Today, I conducted a quick search on the internet while I was working regarding the Action-Reaction law together with something else and I finally got it! In order to make it more interesting, I am not going to tell you why the construction will work (although I showed it a little bit in a different way), therefore I have the following challenge for you: When or what are the conditions where Newton’s 3rd law or Newton’s laws in general stop to hold as they are originally formulated? If you find the answer to this question, look at Fig.1 – Upper. From that moment you are 4 steps (four words) away from confirming the construction will definitely work! Link to comment Share on other sites More sharing options...
swansont Posted September 30, 2020 Share Posted September 30, 2020 5 minutes ago, John2020 said: Since I am not a physicist I see things from a different perspective and I use a slightly different language than you expect. When I looked to all requests to me to justify aspects of the concept I realized that we were actually not in tune (from joigus view this sounds as “not hearing criticism”. In text messaging platforms sometimes the written words sound differently, I can identify this with myself, too). Today, I conducted a quick search on the internet while I was working regarding the Action-Reaction law together with something else and I finally got it! Be warned that there are some engineering treatments where the term "reaction force" is used incorrectly. I noticed this in a recent thread about action/reaction force pairs. Quote When or what are the conditions where Newton’s 3rd law or Newton’s laws in general stop to hold as they are originally formulated? They won't work where Newton tells you they won't. Newton explains that an object in uniform motion maintains that state unless acted upon by a net force. If you are in a situation where that is not the case (e.g. a rotating frame) then you can't expect the other laws to apply. Link to comment Share on other sites More sharing options...
John2020 Posted September 30, 2020 Author Share Posted September 30, 2020 (edited) 20 minutes ago, swansont said: If you are in a situation where that is not the case (e.g. a rotating frame) then you can't expect the other laws to apply Congratulations swansont, you made it! Rotating frame is the correct answer. Now look at Fig.1 - Upper. Does it fill the bill? Edited September 30, 2020 by John2020 Link to comment Share on other sites More sharing options...
Ghideon Posted September 30, 2020 Share Posted September 30, 2020 29 minutes ago, John2020 said: When or what are the conditions where Newton’s 3rd law or Newton’s laws in general stop to hold as they are originally formulated? In addition to @swansont: Behaviour of massless particles (invariant mass = 0) and massive objects at relativistic speeds (velocity approaching speed of light in vacuum) are outside the applicability of Newtons laws. (I stated such conditions in my first post.) Link to comment Share on other sites More sharing options...
John2020 Posted September 30, 2020 Author Share Posted September 30, 2020 1 minute ago, Ghideon said: In addition to @swansont: Behaviour of massless particles (invariant mass = 0) and massive objects at relativistic speeds (velocity approaching speed of light in vacuum) are outside the applicability of Newtons laws. (I stated such conditions in my first post.) Well done! Since we are still in classical mechanics the key that leads to the solution of this puzzle is the rotating frame. Now check Fig.1-Upper and think on your feet. I am going to eat something and I will be back in a few minutes. Link to comment Share on other sites More sharing options...
Ghideon Posted September 30, 2020 Share Posted September 30, 2020 (edited) 6 minutes ago, John2020 said: Well done! Since we are still in classical mechanics the key that leads to the solution of this puzzle is the rotating frame. Now check Fig.1-Upper and think on your feet. You have a device with rotating parts in it. Not a rotating frame of reference with a device in it. Edited September 30, 2020 by Ghideon spelling Link to comment Share on other sites More sharing options...
John2020 Posted September 30, 2020 Author Share Posted September 30, 2020 5 minutes ago, Ghideon said: You have a device with rotating parts in it. Not a rotating frame of reference with a device in it. You are 50% correct. Try to think on your feet. Link to comment Share on other sites More sharing options...
joigus Posted September 30, 2020 Share Posted September 30, 2020 (edited) 1 hour ago, John2020 said: When or what are the conditions where Newton’s 3rd law or Newton’s laws in general stop to hold as they are originally formulated? I know what you're trying to get at, and I'm trying to tell you it won't work and why. You can't transform internal angular momentum into COM momentum. You're trying to obtain translation from rotation. I know that from the very beginning. It won't work. I can invest a certain amount of time in telling you why in more detail, but I must assess carefully how much it's going to be worth it* in terms of a useful (hopefully for both of us and other interested users) dialogue in terms of elucidating meaningful physical concepts. The patronizing treatment ("well done!," "congratulations!," as if they were the advanced students that "understand better" what you're trying to do won't sell your idea any better. *Edit: Also, other users are handling this very well, so my help here is not much needed. Edited September 30, 2020 by joigus Addition Link to comment Share on other sites More sharing options...
swansont Posted September 30, 2020 Share Posted September 30, 2020 47 minutes ago, John2020 said: Congratulations swansont, you made it! Rotating frame is the correct answer. Now look at Fig.1 - Upper. Does it fill the bill? No. You're standing outside the device. Your coordinate system is inertial. A rotating object in an inertial frame follows Newton's laws. If you are analyzing something from a rotating frame, you have to analyze everything from that rotating frame. (i.e. you need to add in all of the fictitious forces if you want to apply Newton's laws). If you mix frames, your conclusions are invalid. Link to comment Share on other sites More sharing options...
John2020 Posted September 30, 2020 Author Share Posted September 30, 2020 12 minutes ago, joigus said: The patronizing treatment ("well done!," "congratulations!," as if they were the advanced students that "understand better" what you're trying to do won't sell your idea any better. Please don't be so hard. We are just making a discussion and I am not trying to sell anything. Does it sound selling or support the words "congratulations" and "well done"? Link to comment Share on other sites More sharing options...
swansont Posted September 30, 2020 Share Posted September 30, 2020 1 minute ago, John2020 said: Please don't be so hard. We are just making a discussion and I am not trying to sell anything. Does it sound selling or support the words "congratulations" and "well done"? You're the one who hasn't studied physics, by your own admission. You aren't in a position to tell someone they're 50% right and they need to think on their feet. 1 Link to comment Share on other sites More sharing options...
John2020 Posted September 30, 2020 Author Share Posted September 30, 2020 Just now, swansont said: You're the one who hasn't studied physics, by your own admission. You aren't in a position to tell someone they're 50% right and they need to think on their feet. You are overreacting. Don't behave like those in NSF forum. Let people breath. Do you want to close this thread? Please do it. -2 Link to comment Share on other sites More sharing options...
Ghideon Posted September 30, 2020 Share Posted September 30, 2020 (edited) 41 minutes ago, John2020 said: You are 50% correct. Thanks. Thats at least better than being 100% wrong. 19 hours ago, John2020 said: I have not study Hamiltonian mechanics If you had it would probably be easier for you to understand how your device is impossible according to known laws of physics. May I suggest that you get back to the topic; address the objections to the presented claims? Edited September 30, 2020 by Ghideon missing sentence Link to comment Share on other sites More sharing options...
joigus Posted September 30, 2020 Share Posted September 30, 2020 (edited) 1 hour ago, John2020 said: Why you cannot see what I see? It's not that we can't see what you see. It's rather that you're the only one here who can't see what everybody else can. Anybody who knows anything about physics has spent some time in their youth trying to make a toy model like the one you're proposing in their mind. It's a rite of passage. You must do all these checks in your mind if you want to understand better why you can't violate the momentum and angular momentum conservation laws. And then you must learn to unlearn them when you're studying general relativity. Because they don't always apply there. Do you know that there is a weak and a strong formulation of Newton's 3rd law concerning mutual directions? You seem not to be aware of it. It is the isotropy of space that forbids that action-reaction forces be in any other direction than the relative position vector for any two parts of the system. That's the basis for the strong statement of Newton's third law: Mutual forces are equal in magnitude and opposite in direction, and their direction coincides with the relative position vector. So that, \[\left(\boldsymbol{r}_{i}-\boldsymbol{r}_{j}\right)\wedge\boldsymbol{F}_{ij}=0\] for every pair, i, j. Otherwise, empty space would be anisotropic. Edited September 30, 2020 by joigus Link to comment Share on other sites More sharing options...
swansont Posted September 30, 2020 Share Posted September 30, 2020 5 minutes ago, John2020 said: You are overreacting. Don't behave like those in NSF forum. Let people breath. Do you want to close this thread? Please do it. There have been no rules violations that would suggest that closure is the right course of action. Plus I'm involved, so I am far less inclined to take any such action (there would be no inclination if this weren't in speculations. I would defer to my fellow moderators to make the call) But perhaps we could return to discussing the physics, rather than this sideshow? Link to comment Share on other sites More sharing options...
John2020 Posted September 30, 2020 Author Share Posted September 30, 2020 1 minute ago, joigus said: Do you know that there is a weak and a strong formulation of Newton's 3rd law concerning mutual directions? You seem not to be aware of it. No I don't. I am aware in terms of maths some classical mechanics, special relativity and elwctromagnetism. Math knowledge: 1st order differential equations, some vector calculus, something about integrals and derivatives. Aren't those enough one to describe the device in Fig.1? Or should I abandon this thread? Thinking on your feet is more important than knowledge. Eibstein used the word"imagination". Link to comment Share on other sites More sharing options...
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