John2020 Posted September 27, 2020 Share Posted September 27, 2020 (edited) Hi everybody! First of all I have to make clear that I am not a physicist, however as many other science fiction enthusiasts, I have an interest in alternative propulsion methods (reactionless drive etc). My theory is published in a non-arxiv.org repository and essentially explores the feasibility of constructing a reactionless drive (an initial mechanical prototype). As we know Newton's 3rd law forbids the construction of a device that may move by means of internal forces, something that all propellantless propulsion/reactionless drive theories tend to ignore. The conclusion of my research can be briefly stated as follow: A reactionless drive may become feasible if and only if Newton's 3rd law is not the whole story that means it may probably be incomplete. Fig.1 - Proof of concept. Action--reaction forces in isolated systems. Upper: Induced internal forces. Redeployment of the center of mass and acceleration of the system. Lower: Collinear internal forces. No change in the center of mass and no acceleration of the system. Attention! The induced internal reaction force was not deliberately inverted (same direction with the induced internal action force) but as a consequence of the mathematical proof. The mathematical description assumes the reactionless drive (see Fig.1.Upper) is internally powered (motor and power supply on board (appears hidden in Fig.1 - Upper)) as also there is no dissipation of energy due to friction (ideal machine). Starting from the conservation of angular momentum, the net external and internal torques in FIG.1 - Upper, are \[\sum \vec{\tau}_{ext} = \vec{0} \] \[\overbrace{\vec{F}_{A}}^{\longrightarrow} + \overbrace{\vec{F}_{R}}^{\longleftarrow} = \vec{0} \] \[\vec{r}_{A} \neq \vec{r}_{R} \Rightarrow \vec{\tau}_{A} \neq \vec{0} \text{ and } \vec{\tau}_{R} \neq \vec{0} \] \[\sum \vec{\tau}_{int} + \vec{\tau}_{A} + \vec{\tau}_{R} =\vec{0} \] \[\sum \vec{\tau}_{int} + \left(\vec{r}_{A} \times \vec{F}_{A} \right) + \left(\vec{r}_{R} \times \vec{F}_{R} \right) = \vec{0} \] \[\sum \vec{\tau}_{int} + \left(\vec{r}_{A} \times \vec{F}_{A} \right) + \left(\vec{r}_{R} \times \left(-\vec{F}_{A} \right)\right) = \vec{0} \] \[\overbrace{\sum \vec{\tau}_{int}}^{\curvearrowleft} + \overbrace{\left(\left(\vec{r}_{A} - \vec{r}_{R} \right) \times \vec{F}_{A} \right)}^{\curvearrowright} = \vec{0} \] An ideal translation mechanism (translation screw in FIG.1 - Upper) can maintain, amplify or reduce the magnitude of the input force by delivering the same amount of energy (no energy dissipation through friction) entering the system (energy conservation). At this point, developing a general expression for the net induced force requires the introduction of the dimensionless factor \(n_{T}\) (ideal mechanical advantage) along with a definition of the net induced torque. Hence, \[n_{T} = \frac{|\vec{\omega} \times \left(\vec{r}_{A} - \vec{r}_{R}\right)|}{|\vec{u}_{T}|} = \frac{\left(2 \pi |\vec{r}_{A} - \vec{r}_{R}|\right)}{|\vec{l}_{T}|} \] \[\overbrace{n_{T} \sum \vec{\tau}_{int}}^{\curvearrowleft} + \overbrace{n_{T} \left(\left(\vec{r}_{A} - \vec{r}_{R} \right) \times \vec{F}_{A} \right)}^{\curvearrowright} = \vec{0}\] \[\sum \vec{\tau}_{T} = n_{T} \sum \vec{\tau}_{int}\] \[\overbrace{\sum \vec{\tau}_{T}}^{\curvearrowleft} + \overbrace{n_{T}\left(\left(\vec{r}_{A} - \vec{r}_{R} \right) \times \vec{F}_{A} \right)}^{\curvearrowright} = \vec{0}\] Dividing the above by the position-vector magnitude \(|\vec{r}_\mathrm{A} - \vec{r}_\mathrm{R}|\) yields \[\overbrace{\frac{\sum \vec{\tau}_{T}}{|\left(\vec{r}_{A} - \vec{r}_{R} \right)|}}^{\longleftarrow} + \overbrace{\frac{n_{T} \left(\left(\vec{r}_{A} - \vec{r}_{R} \right) \times \vec{F}_{A} \right)}{|\left(\vec{r}_{A} - \vec{r}_{R} \right)|}}^{\longrightarrow} = \vec{0} \] \[\sum \vec{F}_{T} = \frac{\sum \vec{\tau}_{T}}{|\left(\vec{r}_{A} - \vec{r}_{R} \right)|}\] Due to conservation of energy, in an ideal machine (FIG.1- Upper), the power output equals to the power input: \[P_{T} = P_{A} \Rightarrow \frac{|\sum \vec{F}_{T}|}{|\vec{F}_{A}|} = \frac{|\vec{\omega} \times \left(\vec{r}_{A} - \vec{r}_{R}\right)|}{|\vec{u}_{T}|} = n_{T} \] The above shows when the \(\vec{F}_\mathrm{A}\) is constant then, the angular velocity of the translation screw \(\vec{\omega}\), the induced force \(\vec{F}_{T}\) and the translational velocity \(\vec{u}_{T}\) of mass \(m_{T}\) are also constant. The force \(\vec{F}_T \) can be also written as \[\sum \vec{F}_{T} = - \frac{n_{T} \left(\left(\vec{r}_{A} \times \vec{F}_{A} \right) + \left(\vec{r}_{R} \times \vec{F}_{R} \right) \right)}{|\left(\vec{r}_{A} - \vec{r}_{R} \right)|} \] \[\sum \vec{F}_{T} = - n_{T} \left(\vec{F}_{A_T} + \vec{F}_{R_T} \right) = \text{const.}\] Nevertheless, the above equations address just the motion of mass \(m_{T}\). By expanding the ideal mechanical advantage to include varying angular and translational velocities, a new net force expression is derived that may apply for the motion of the system as a whole. Thus, \[d \vec{\omega}\begin{cases} = \vec{0} &:\quad \vec{F}_{A} = \text{const.} \text{ and } \vec{F}_{R} = \text{const.} \\ \neq \vec{0} &:\quad \vec{F}_{A} \neq \text{const.} \text{ and } \vec{F}_{R} \neq \text{const.} \end{cases} \\ n_{r} = \frac{|\mathrm{d} {\omega} \times \left(\vec{r}_{A} - \vec{r}_{R}\right)|}{|\mathrm{d} \vec{u}_{T}|}, \\ \sum \vec{F}_{ind} = - \frac{n_{r} \left(\left(\vec{r}_{A} \times \vec{F}_{A} \right) + \left(\vec{r}_{R} \times \vec{F}_{R} \right) \right)}{|\left(\vec{r}_{A} - \vec{r}_{R} \right)|}, \\ \sum \vec{F}_{ind} \propto d \vec{\omega}, \\ \sum \vec{F}_{ind} = - n_r \left(\vec{F}_{A_I} + \vec{F}_{R_I} \right),\\ d \vec{\omega} \begin{cases} = \vec{0} \Rightarrow \vec{d} \vec{u}_{T} = \vec{0} &:\quad - n_{r} \left( \vec{F}_{A_I} + \vec{F}_{{R_I}} \right) =\vec{0}, \\ \neq \vec{0} \Rightarrow \vec{d} \vec{u}_{T} \neq \vec{0} &:\quad - n_{r} \left( \vec{F}_{A_I} + \vec{F}_{{R_I}} \right) \neq \vec{0}. \end{cases}\] Applying the mass transfer mechanism (FIG.1 - Upper), yields \[\mathrm{d} \vec{\omega} \neq \vec{0} \Rightarrow {d} \vec{u}_{T} = \vec{u}_{rel} \neq \vec{0}, \\ \frac{\mathrm{d} \vec{p}}{\mathrm{d}t} = \sum \vec{F}_{ind} = - n_{r}\left(\vec{F}_{A_I} + \vec{F}_{R_I} \right) \neq \vec{0},\\ \frac{\mathrm{d} \vec{p}}{\mathrm{d}t} = -n_{r} \vec{u}_{rel} \frac{\mathrm{d} m}{\mathrm{d}t} \neq \vec{0} \Rightarrow \vec{a} \neq \vec{0}, \\ \] I will be very happy if you can help me evaluate the above idea! Edited September 27, 2020 by John2020 Link to comment Share on other sites More sharing options...

swansont Posted September 27, 2020 Share Posted September 27, 2020 You have not adequately labeled you diagrams and described your variables. r⃗ A≠r⃗ R⇒τ⃗ A≠0⃗ and τ⃗ R≠0⃗ To start: what are r_{A} and r_{R} ? Link to comment Share on other sites More sharing options...

John2020 Posted September 27, 2020 Author Share Posted September 27, 2020 r_A and r_R are the position vectors of the action and reaction torques and the τ_A and τ_R are the corresponding action and reaction torques. Link to comment Share on other sites More sharing options...

John2020 Posted September 28, 2020 Author Share Posted September 28, 2020 @swansont May I share the link to my paper over a post and through my profile? It will help the readers of this thread to understand better the whole idea. Link to comment Share on other sites More sharing options...

Ghideon Posted September 28, 2020 Share Posted September 28, 2020 (edited) Hello! The laws of Newton is widely accepted* On 9/27/2020 at 6:08 PM, John2020 said: The conclusion of my research can be briefly stated as follow: A reactionless drive may become feasible if and only if Newton's 3rd law is not the whole story that means it may probably be incomplete. Is the rig you have provided the only way to break the laws and build a reactionless drive? Maybe you can describe in more general terms how Newton's laws are incorrect*, if that is what you claim. I would like to understand the concept before looking into the math. *) Within their range of applicability so that for instance velocities << c, invariant mass > 0 Edited September 28, 2020 by Ghideon Link to comment Share on other sites More sharing options...

John2020 Posted September 28, 2020 Author Share Posted September 28, 2020 19 minutes ago, Ghideon said: Hello! The laws of Newton is widely accepted* Is the rig you have provided the only way to break the laws and build a reactionless drive? Maybe you can describe in more general terms how Newton's laws are incorrect*, if that is what you claim. I would like to understand the concept before looking into the math. *) Within their range of applicability so that for instance velocities << c, invariant mass > 0 Hi Ghideon! Maybe I was a little bit misunderstood. My work as also what I briefly stated above is that Newton's laws of motion always hold, however the construction above and the maths that backed it show that there is a particular case where Newton's 3rd law for isolated systems appears to be incomplete. In other words, according to my view Newton didn't actually consider the case of the induced internal forces that are direction of translation screw rotation depended (not collinear) that means whatever the induced forces (action and reaction) will evolve in the same direction (since the translation screw rotates only in one or the other direction and never both at the same moment). In case the moderators of this forum allow me to share the link to my work there you may find more details (along with the implications of these findings in special relativity and Lorentz transformations) about the concept which is based on two basic findings: a) The induced internal reaction force is mathematically proven to have the same direction with the induced internal action force b) From a purely mathematical perspective, a second interpretation of the rocket equation is also possible. An internal force causes a mass transfer between two points inside the system, resulting in a change of system’s effective inertia (reduction) after a time interval. In order this to work, you cannot use collinear forces (as seen in Fig.1 - Lower) but induced. The paper develops this argument as follow: "The rate of mass ejection corresponds to a mass transfer from the system’s interior to a point away from it (rocket) after a time interval, without affecting its center of mass. From a purely mathematical perspective, a second interpretation is also possible. An internal force causes a mass transfer between two points inside the system, resulting in a change of system’s effective inertia (reduction) after a time interval.". These two views can be described with the same rocket equation. Seeing the reactionless drive from the point of an external observer and have no knowledge of what is going on inside, its motion can be justified just by a change of its inertia (caused by the internal mass transfer) or better reduction of its inertia (gain of inertia would never have as result, motion). And about your question:"Is the rig you have provided the only way to break the laws and build a reactionless drive?". The answer is: The construction (linear actuator) in Fig.1 - Upper is the simplest Reactionless Drive ever as also serves as the basic mechanical equivalent of all Reactionless Drives. If this may work, it is very easy (it is a little bit tricky actually) to develop a pure electromagnetic device (no need of superconductors and exotic material. Just known applied electromagnetism) without moving parts. Any reactionless drive should have the following ingredients: 1.Translation mechanism (e.g. translation screw) 2.Induced internal forces (without translation mechanism there are no induced forces) 3.A part of the system or all the parts of the system should be coupled on the translation mechanism Hint: The inner functionality of Fig.1-Upper resembles the electromagnetic induction and Lenz law (check the equations above). Link to comment Share on other sites More sharing options...

swansont Posted September 28, 2020 Share Posted September 28, 2020 On 9/27/2020 at 12:36 PM, John2020 said: r_A and r_R are the position vectors of the action and reaction torques and the τ_A and τ_R are the corresponding action and reaction torques. Not helpful if they aren’t identified in a diagram. 4 hours ago, John2020 said: @swansont May I share the link to my paper over a post and through my profile? It will help the readers of this thread to understand better the whole idea. That’s not in compliance with rule 2.7. Discussion needs to take place here, without requiring people to click on links. 2 hours ago, John2020 said: Hi Ghideon! Maybe I was a little bit misunderstood. My work as also what I briefly stated above is that Newton's laws of motion always hold, however the construction above and the maths that backed it show that there is a particular case where Newton's 3rd law for isolated systems appears to be incomplete. In other words, according to my view Newton didn't actually consider the case of the induced internal forces that are direction of translation screw rotation depended (not collinear) that means whatever the induced forces (action and reaction) will evolve in the same direction (since the translation screw rotates only in one or the other direction and never both at the same moment). Internal forces don’t cause motion of the CoM. Action and reaction forces act on different objects What is an “induced” force? 2 hours ago, John2020 said: In case the moderators of this forum allow me to share the link to my work there you may find more details (along with the implications of these findings in special relativity and Lorentz transformations) about the concept which is based on two basic findings: a) The induced internal reaction force is mathematically proven to have the same direction with the induced internal action force That’s what you should be deriving here Link to comment Share on other sites More sharing options...

Ghideon Posted September 29, 2020 Share Posted September 29, 2020 (edited) Thanks for your reply. I have some trouble following your description of the internals of the device, maybe it gets clear once @swansont's questions are answered. In the meantime let's test your device with a thought experiment that can be performed without a clear understanding of the internals: 10 hours ago, John2020 said: Seeing the reactionless drive from the point of an external observer and have no knowledge of what is going on inside Let's assume that your device performs as you expect. We put the device in closed box with low* mass in the vacuum of space, far from ay gravitational sources. There are no forces acting on the box and the box is at rest in our frame of reference. The momentum of the box is zero. Picture 1: At time t=0 a remote control is used to start the device in the box. According to your description the result is that the box with the device inside accelerates**. At some time t>0 the box have moved some distance x>0 and it keeps moving. The momentum of the box is not zero; p>0. There is no mass ejected from the box and still no external forces acting on the box. Picture 2: Is the above a correct description of your expected outcome when testing your device? If so, we can analyse the consequences it would have on the laws of Newton and from there move on to the rig and math presented in opening post. *) low enough so we can neglect any effects on performance of the rig. **) lets assume it is to the right Edited September 29, 2020 by Ghideon grammar Link to comment Share on other sites More sharing options...

John2020 Posted September 29, 2020 Author Share Posted September 29, 2020 (edited) Yes, what you presented above is what is expected. I have a daily job and at the moment I am having a meal pause (Local time UTC+1). Don't worry I will answer all of yours questions. Swansont is right. The position vectors and the torque vectors must be visible in the drawing, otherwise the concept is unclear. I hope I can find some time this evening to answers all those questions. In the meantime, the nut (m_T) evolves around the screw that means the Newtonian Action-Reaction is perpendicular to the direction of motion of the nut itself. Having the nut a constant translation speed appears no change in momentum that leads to no change in momentum of the system as a whole. IF the nut changes its momentum then, this will be reflected to the entire system where due to momentum conservation the whole system will follow the direction of the nut (since there is no force opposing the momentum of the nut). Note: Action-reaction Newtonian forces are developed perpedicular to nut momentum. 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) 1 hour ago, John2020 said: Yes, what you presented above is what is expected. Thanks. That means that the law F=ma does not apply to the system*. The force F applied to the system is always zero but the acceleration a is not always zero. I also note that momentum conservation does not apply; initially momentum is zero and at a later time the momentum is not zero but there is no interaction** with anything external to the system. So when we looking at the accelerating system from the outside, what laws can we use to describe the whole system's behaviour? *) box with a working reactonless device inside. **) The remote control signal in the thought experiment is neglected except for its ability to start the device. Edited September 29, 2020 by Ghideon clarification about interaction Link to comment Share on other sites More sharing options...

John2020 Posted September 29, 2020 Author Share Posted September 29, 2020 3 hours ago, Ghideon said: So when we looking at the accelerating system from the outside, what laws can we use to describe the whole system's behaviour? *) box with a working reactonless device inside. **) The remote control signal in the thought experiment is neglected except for its ability to start the device. The answer to your question is given by the last six expressions of my first post where center stage has the rocket equation that may apply for the change in the CoM (see my posts above). As I mentioned above an external observer may justify the motion of the isolated system (no external forces) because of the change in CoM that implies reduction of the effective inertia, otherwise motion cannot take place. Swansont and generally in established physics supports no change in CoM by internal forces and more accurately by collinear internal forces action - reaction). This is what I also support with the construction in Fig.1-Lower (rod with no threads. I am now driving back home. In a couple of hours I will post more stuff (questions and answers). Link to comment Share on other sites More sharing options...

joigus Posted September 29, 2020 Share Posted September 29, 2020 What you've got at the end is a rocket equation, in contradiction with your opening statement that it's reactionless. The force, \[\frac{d\boldsymbol{p}}{dt}=-n_{r}\boldsymbol{u}_{\textrm{rel}}\frac{dm}{dt}\] Is the reaction force (with mass transfer) of the fuel against a rocket. How did this reaction term come about from an allegedly reactionless mechanism? As pointed out above, also, you can't have self-propulsion from internal forces in outer space, because that would imply momentum is not conserved for a system of particles. This, in turn, would imply that empty space is inhomogeneous at small scales. There can be no exception. I could explain in further detail, but you must have either all three Newton's laws or none. They are not really independent. Otherwise you may have a system satisfying Newton's laws but it would be completely impossible to consider it made up of smaller parts that also satisfy Newton's laws. You can't just cross out the third law without giving up Newtonian mechanics altogether. Newton's third law, \[\boldsymbol{F}_{ij}=-\boldsymbol{F}_{ji}\] Is really a statement already implied by Newton's 1st and 2nd laws for binary partitions of a system into two subsystems exerting mutual forces (internal). These must cancel in pairs, \[\boldsymbol{F}_{ij}+\boldsymbol{F}_{ji}=0\] Otherwise the composite system in the absence of external forces (free) would be subject to accelerations. It would contradict Newton's laws for the COM (all of them, in particular F=ma for the COM and the corolary, the 1st law too). Link to comment Share on other sites More sharing options...

Ghideon Posted September 29, 2020 Share Posted September 29, 2020 (edited) 39 minutes ago, John2020 said: The answer to your question is given by the last six expressions of my first post where center stage has the rocket equation that may apply for the change in the CoM (see my posts above). As I mentioned above an external observer may justify the motion of the isolated system (no external forces) because of the change in CoM that implies reduction of the effective inertia, otherwise motion cannot take place. Can you express that using basic mathematics? I'll formulate the question differently: A box of mass m with unknown content* is at rest in our frame of reference in space far, from gravitational sources. I am asked to write an equation of motion for the box when: no external forces act on the box. mass m is constant; nothin is ejected Since I know that Newton's laws of mechanics are applicable I write: [math]F=m*a[/math] and [math]F=0, m>0 [/math] which means that [math]a=\frac{0}{m}=0[/math] The box will remain stationary, it can't possible accelerate. I double check the impulse with [math] \Delta p= \Delta t*F [/math] and again [math]F=0[/math] so [math] \Delta p= 0[/math]. There is no movement. Now you start the device and the box moves! Note that the conditions above still holds according to your description; F=0 all the time. Yet my application of Newton's laws now fails to predict the box movement, I should have used some other formulas. Can you show which formulas that I should have used? I would prefer to sort out the basic application of your new physics as seen from the outside of the system first. Then we could move on to the internals of the box. Edit: I notice that @joigus will handle the internal parts in great detail. 6 minutes ago, joigus said: made up of smaller parts *) I do not know that it may contain a reactionless drive performing according to your description. Edited September 29, 2020 by Ghideon x-post with joigus Link to comment Share on other sites More sharing options...

John2020 Posted September 29, 2020 Author Share Posted September 29, 2020 (edited) 1 hour ago, joigus said: What you've got at the end is a rocket equation, in contradiction with your opening statement that it's reactionless. Is the reaction force (with mass transfer) of the fuel against a rocket. How did this reaction term come about from an allegedly reactionless mechanism? @joigus Hi joigus! Let's start first with the above. As I mentioned previously my paper justifies the use of the rocket equation as follows: "The rate of mass ejection corresponds to a mass transfer from the system’s interior to a point away from it (rocket) after a time interval, without affecting its center of mass. From a purely mathematical perspective, a second interpretation is also possible. An internal force causes a mass transfer between two points inside the system, resulting in a change of system’s effective inertia (reduction) after a time interval.". These two views can be described with the same rocket equation. Seeing the reactionless drive from the point of an external observer and have no knowledge of what is going on inside, its motion can be justified just by a change of its inertia (caused by the internal mass transfer) or better reduction of its inertia (gain of inertia would never have as result, motion). The only thing that needs to be clarified regarding the rocket equation is the (-) sign in front of the mass transfer. Since the mass transfer itself is not a negative quantity (see ejection), the (-) is justified (see my first post) by the induced torque. As you see the translation screw rotates clockwise and the mass evolves counterclockwise (to the right) which is something known from the mechanism of the linear actuator. Mathematically is justified as follow: \[ \overbrace{\sum \vec{\tau}_{T}}^{\curvearrowleft} + \overbrace{n_{T}\left(\left(\vec{r}_{A} - \vec{r}_{R} \right) \times \vec{F}_{A} \right)}^{\curvearrowright} = \vec{0} \] What requires further justification is what Swansont correctly noted and this is the r_A and r_R along with the corresponding torques. Edited September 29, 2020 by John2020 Link to comment Share on other sites More sharing options...

joigus Posted September 29, 2020 Share Posted September 29, 2020 39 minutes ago, John2020 said: What requires further justification is what Swansont correctly noted and this is the r_A and r_R along with the corresponding torques. Swansont's point is well taken. And it is, of course, pertinent. If you don't specify your variables on a diagram it's impossible to point out where the mistake is. What is possible to declare beyond any doubt is that there must be a mistake in your analysis, (Ghideon has pointed that out too) unless empty space be inhomogeneous or anysotropic at small scales. And the reason is the part of my analysis that you seem to have chosen not to address. Namely, 1 hour ago, joigus said: As pointed out above, also, you can't have self-propulsion from internal forces in outer space, because that would imply momentum is not conserved for a system of particles. This, in turn, would imply that empty space is inhomogeneous at small scales. There can be no exception. I could explain in further detail, but you must have either all three Newton's laws or none. They are not really independent. Otherwise you may have a system satisfying Newton's laws but it would be completely impossible to consider it made up of smaller parts that also satisfy Newton's laws. You can't just cross out the third law without giving up Newtonian mechanics altogether. Newton's third law, \[\boldsymbol{F}_{ij}=-\boldsymbol{F}_{ji}\] Is really a statement already implied by Newton's 1st and 2nd laws for binary partitions of a system into two subsystems exerting mutual forces (internal). These must cancel in pairs, \[\boldsymbol{F}_{ij}+\boldsymbol{F}_{ji}=0\] Otherwise the composite system in the absence of external forces (free) would be subject to accelerations. It would contradict Newton's laws for the COM (all of them, in particular F=ma for the COM and the corollary, the 1st law too). Linear momentum appearing from nowhere requires space not to be symmetric. That's why I know it cannot be correct. If you were as kind as to provide more detail about the position of what exactly is it that r_A and r_R represent, maybe someone could, upon reflection, tell you were the mistake is. Link to comment Share on other sites More sharing options...

John2020 Posted September 29, 2020 Author Share Posted September 29, 2020 (edited) 2 hours ago, Ghideon said: Can you express that using basic mathematics? I'll formulate the question differently: A box of mass m with unknown content* is at rest in our frame of reference in space far, from gravitational sources. I am asked to write an equation of motion for the box when: no external forces act on the box. mass m is constant; nothin is ejected ...... Now you start the device and the box moves! Note that the conditions above still holds according to your description; F=0 all the time. Yet my application of Newton's laws now fails to predict the box movement, I should have used some other formulas. Can you show which formulas that I should have used? @Ghideon In case there is no external force and there is no knowledge how the box moves then, it is required to measure the change in speed using a doppler like radar in order to calculate its acceleration: \[\vec{a} = \frac{\mathrm{d} \vec{u}}{\mathrm{d}t} \] Let's say now you are aware about the propulsion mechanism then (see last expression on my first post): \[\frac{\mathrm{d} \vec{p}}{\mathrm{d}t} = \sum \vec{F}_{ind} = -n_{r} \vec{u}_{rel} \frac{\mathrm{d} m}{\mathrm{d}t} \] The \(\mathrm{d} m \) on the mechanical construction (Fig.1 - Upper) corresponds to the quantity \(m_{\mathrm{T} } \), thus: \[\frac{\mathrm{d} \vec{p}}{\mathrm{d}t} = \sum \vec{F}_{ind} = -n_{r} \vec{u}_{rel} \frac{m_{\mathrm{T}} }{\mathrm{d}t} \]. Finally the acceleration of the system is: \[\vec{a} = n_{r} \frac{\vec{u}_{rel}}{\mathrm{d}t} \] 16 minutes ago, joigus said: Swansont's point is well taken. And it is, of course, pertinent. If you don't specify your variables on a diagram it's impossible to point out where the mistake is. What is possible to declare beyond any doubt is that there must be a mistake in your analysis, (Ghideon has pointed that out too) unless empty space be inhomogeneous or anysotropic at small scales. And the reason is the part of my analysis that you seem to have chosen not to address. ..... Linear momentum appearing from nowhere requires space not to be symmetric. That's why I know it cannot be correct. If you were as kind as to provide more detail about the position of what exactly is it that r_A and r_R represent, maybe someone could, upon reflection, tell you were the mistake is. @joigus As I mentioned above I have some time in the evening (due to a daily job) to respond, therefore please be patient. I am working on this. I cannot draw r_A and r_R on the image I shared (it is a 2D without a third dimension perspective). I am working on a new drawing to show all these. It will take some time. Stay tuned! 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) Thanks for your reply. 1 hour ago, John2020 said: In case there is no external force and there is no knowledge how the box moves then, it is required to measure the change in speed using a doppler like radar in order to calculate its acceleration: Ok. I measure the acceleration with your method and then apply Newton's law F=ma. But a measured acceleration a>0 means F>0. That is a contradiction. Note that the total mass is constant, the sealed box we agreed on does not allow any ejection of mass. 1 hour ago, John2020 said: Let's say now you are aware about the propulsion mechanism That is not what I asked, I asked about the external view. Once the contradictions regarding Newton's laws are solved we may have a detailed look inside the box. Also note; as you have not defined what an induced force is I can't comment on the internals yet. Edited September 29, 2020 by Ghideon missing word Link to comment Share on other sites More sharing options...

John2020 Posted September 29, 2020 Author Share Posted September 29, 2020 (edited) 1 hour ago, Ghideon said: Ok. I measure the acceleration with your method and then apply Newton's law F=ma. But a measured acceleration a>0 means F>0. That is a contradiction. Note that the total mass is constant, the sealed box we agreed on does not allow any ejection of mass. 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: \[\vec{F_{internal}} = n_{r} m_{\mathrm{T}} \frac{\vec{u}_{rel}}{\mathrm{d}t} = m \vec{a} = \vec{F_{external}} \] 1 hour ago, Ghideon said: That is not what I asked, I asked about the external view. Once the contradictions regarding Newton's laws are solved we may have a detailed look inside the box. Also note; as you have not defined what an induced force is I can't comment on the internals yet. @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. 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 17 minutes 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: The external force is either zero or not zero, you can't have both at the same time. Would you like a drawing and some more mathematics to highlight the issues? Link to comment Share on other sites More sharing options...

John2020 Posted September 29, 2020 Author Share Posted September 29, 2020 1 minute ago, Ghideon said: The external force is either zero or not zero, you can't have both at the same time. Would you like a drawing and some more mathematics to highlight the issues? There are two issues here: a) Newton discovered the F=ma since all of his observation and experiments were related to external forces. b) Obviously, (a) cannot apply to what I presented here because in case the construction in Fig.1 - Upper works, the resulting motion is attributed to internal forces. This leads to what I mentioned on my first post which is Newton's laws of motion are probably incomplete. Link to comment Share on other sites More sharing options...

Ghideon Posted September 29, 2020 Share Posted September 29, 2020 25 minutes ago, John2020 said: a) Newton discovered the F=ma since all of his observation and experiments were related to external forces. Physics have progressed a lot since Newton did the initial work. See for instance @joigus posts, there are more fundamental issues than what Newton did or did not confirm in his work. 25 minutes ago, John2020 said: b) Obviously, (a) cannot apply to what I presented here because in case the construction in Fig.1 - Upper works, the resulting motion is attributed to internal forces. This leads to what I mentioned on my first post which is Newton's laws of motion are probably incomplete. You claim that Newton's laws are incomplete; Newton's formulas does not predict its behaviour. I showed that above, you get contradictions that Newton can't explain. You need to present a completely new set of formulas*. You can't use Newton to explain a behaviour and at the same time claim that Newton does not describe the behaviour. Internal forces can't move the center of mass. Idea: Maybe we should write down the Lagrangian for your box and look at the result? *) Which has happened in the past, for instance to predict and describe the motion of massless photons (that is, invariant mass = 0) or massive bodies moving at speeds approaching the speed of light. Link to comment Share on other sites More sharing options...

swansont Posted September 29, 2020 Share Posted September 29, 2020 Not seeing how m_{T} can correspond to dm/dt Is the mass being ejected? Link to comment Share on other sites More sharing options...

John2020 Posted September 29, 2020 Author Share Posted September 29, 2020 4 minutes ago, Ghideon said: You claim that Newton's laws are incomplete; Newton's formulas does not predict its behaviour. I showed that above, you get contradictions that Newton can't explain. You need to present a completely new set of formulas*. You can't use Newton to explain a behaviour and at the same time claim that Newton does not describe the behaviour. Internal forces can't move the center of mass. 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 b) Newton seems to have overlooked the nature of the induced mechanical force that is direction of rotation of the translation mechanism (screw) depended and processes all problems as being collinear forces. This works perfectly as long as these forces are external. c) We agree that collinear forces cannot move the center of mass, however due to the nature of the induced mechanical force it might be possible to move the center of mass through mass transfer. 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. 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) Electrodynamics Hint: Replace the translation screw mechanism and the rest of the periphery of the system with a standing wave. 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). 14 minutes ago, swansont said: Not seeing how mT can refer to dm/dt 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). Link to comment Share on other sites More sharing options...

studiot Posted September 29, 2020 Share Posted September 29, 2020 This whole thread reminds me of Edward De Bono's Black Cylinder Experiment. Why did it topple over ? Link to comment Share on other sites More sharing options...

John2020 Posted September 29, 2020 Author Share Posted September 29, 2020 Good night guys, I have to go to sleep and to wake up early tomorrow. Local Time is about 23.00. See you tomorrow! Link to comment Share on other sites More sharing options...

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