awaterpon Posted June 3, 2021 Share Posted June 3, 2021 (edited) Suppose I have a rest mass m0 of 10 kg, part of this mass 1 kg converts to energy, E=mc² , E=e joules. In this case I will have a new rest mass m0 which is 10-1 kg or 9 kg . pushed by the 1 kg energy This system of rest mass 9 kg will move by the 1 kg energy with speed v1 in which the relativistic mass for the system will be : m=m0/√(1-v²/c²) or m=9 / √(1-v1²/c²) Because no energy comes from out side, then m will be 10 kg and this for speed v1 and rest mass 9 kg * Now let's calculate the kinetic energy of the system : The relativistic kinetic energy of the system is energy of the rest mass or m0c^2 plus energy due to motion: K.E= m0c²/√(1-v1²/c²) + m0c² K.E= 9c²/√(1-v1²/c²)+9c² According to * and because m0=9 kg : 9 / √(1-v1²/c²)=10 kg K.E= 10c²+9c² or 19c² joules : However the whole energy of the system must be the converted mass energy 1 kg plus the remained rest mass 9 kg which is 10 kg or E=10c² joules but K.E is 19c² joules which also must be the whole energy of the moving system This means the system generates 9c^2 joules free energy while it moves with some speed v1 Edited June 3, 2021 by awaterpon Link to comment Share on other sites More sharing options...

Ghideon Posted June 3, 2021 Share Posted June 3, 2021 (edited) 37 minutes ago, awaterpon said: The relativistic kinetic energy of the system is energy of the rest mass or m0c^2 plus energy due to motion: K.E= m0c²/√(1-v1²/c²) + m0c² K.E= 9c²/√(1-v1²/c²)+9c² What kinetic energy does the formulas give you for v_{1}=0? Maybe you need to correct the calculation? Hint: at low speeds the relativistic kinetic energy should be approximately be the non-relativistic version mv^{2}/2 (There are other issues with the reasoning; may have more time to read more carefully later) Edited June 3, 2021 by Ghideon Misinterpreted the formulas Link to comment Share on other sites More sharing options...

swansont Posted June 3, 2021 Share Posted June 3, 2021 53 minutes ago, awaterpon said: Suppose I have a rest mass m0 of 10 kg, part of this mass 1 kg converts to energy, E=mc² , E=e joules. In this case I will have a new rest mass m0 which is 10-1 kg or 9 kg . pushed by the 1 kg energy Energy doesn’t push things. What is happening that results in this motion, and conserves both energy and momentum? Link to comment Share on other sites More sharing options...

awaterpon Posted June 3, 2021 Author Share Posted June 3, 2021 (edited) 5 hours ago, swansont said: Energy doesn’t push things. What is happening that results in this motion, and conserves both energy and momentum? Nothing, it is a matter of equations and that why I refer to the energy as pushing the system.The energy of 1 kg is a kinetic energy that the system can have. And it is true considering the atomic bomb energy. Edited June 4, 2021 by awaterpon Link to comment Share on other sites More sharing options...

Ghideon Posted June 4, 2021 Share Posted June 4, 2021 9 hours ago, awaterpon said: Nothing, it is a matter of equations and that why I refer to the energy as pushing the system.The energy of 1 kg is a kinetic energy that the system can have. And it is true considering the atomic bomb energy. If you use the equations to an unphysical situation, for instance a system that breaks conservation of momentum is not conserved*, the equations does not predict the behaviour of the system. The equations loose their predictive power; conclusions you draw from using the equations outside of their scope of applicability, will be incorrect. 15 hours ago, awaterpon said: K.E= m0c²/√(1-v1²/c²) + m0c² More detail regarding the above equation. The consequence is that you claim kinetic energy (K.E) to be 2m_{0}c^{2} for an object at rest (v_{1}=0). That is false, kinetic energy at rest is 0. Maybe you got wrong sign in the formula? Please clarify and adjust your predictions and conclusions, also taking @swansont's comments into account. *) Conservation of momentum is a mathematical consequence of the homogeneity of space. That is, conservation of momentum is a consequence of the fact that the laws of physics do not depend on position. (adopted from wikipedia.org, Momentum) Link to comment Share on other sites More sharing options...

exchemist Posted June 4, 2021 Share Posted June 4, 2021 15 hours ago, awaterpon said: Suppose I have a rest mass m0 of 10 kg, part of this mass 1 kg converts to energy, E=mc² , E=e joules. In this case I will have a new rest mass m0 which is 10-1 kg or 9 kg . pushed by the 1 kg energy This system of rest mass 9 kg will move by the 1 kg energy with speed v1 in which the relativistic mass for the system will be : m=m0/√(1-v²/c²) or m=9 / √(1-v1²/c²) Because no energy comes from out side, then m will be 10 kg and this for speed v1 and rest mass 9 kg * Now let's calculate the kinetic energy of the system : The relativistic kinetic energy of the system is energy of the rest mass or m0c^2 plus energy due to motion: K.E= m0c²/√(1-v1²/c²) + m0c² K.E= 9c²/√(1-v1²/c²)+9c² According to * and because m0=9 kg : 9 / √(1-v1²/c²)=10 kg K.E= 10c²+9c² or 19c² joules : However the whole energy of the system must be the converted mass energy 1 kg plus the remained rest mass 9 kg which is 10 kg or E=10c² joules but K.E is 19c² joules which also must be the whole energy of the moving system This means the system generates 9c^2 joules free energy while it moves with some speed v1 If you use a wrong formula, you can expect to get a nonsensical outcome. As @Ghideon is gently hinting, that is what you have done here. Link to comment Share on other sites More sharing options...

awaterpon Posted June 4, 2021 Author Share Posted June 4, 2021 35 minutes ago, Ghideon said: If you use the equations to an unphysical situation, for instance a system that breaks conservation of momentum is not conserved*, the equations does not predict the behaviour of the system. The equations loose their predictive power; conclusions you draw from using the equations outside of their scope of applicability, will be incorrect. It is a physical situation with few details. consider this: A rest mass of 6 kg, 2 kg of it exploded,it pushed the 4 kg with 1 kg energy and the other 1 kg went away. Now I have a new rest mass of 4kg and energy of 1 kg: relativistic mass m is= m=m0/sqr (1-v^2/c^2) I will have a velocity v1 m=4/sqr(1-v1^2/c^2) But at the same time m is 1+4 or5 kg Then 4/sqr(1-v1^2/c^2)=5 kg relativistic kinetic energy: K.E=m0c^2/sqr(1-v^2/c^2)+m0c^2 =4c^2/sqr(1-(v1^2/c^2))+4c^2 But: =4/sqr(1-(v1^2/c^2))=5 kg so K.E=5c^2+4c^2= 9c^2 joules The actual mass is 1+4 kg, and the energy of this system must be: E=mc^2 = 5c^2 joules instead of 9c^2 joules Link to comment Share on other sites More sharing options...

joigus Posted June 4, 2021 Share Posted June 4, 2021 Just to clarify --although Swansont and Ghideon are doing a very good job of it--. For a particle of mass m --mass is just rest energy: Total energy: E=mc21−v2/c2−−−−−−−−√ Rest energy: E0=mc2 Kinetic energy: K.E.=mc21−v2/c2−−−−−−−−√−mc2 Rest energy is akin to positive potential energy. These concepts were clarified in Taylor & Wheeler Space-Time Physics a long time ago. Mass is better understood as rest energy. Link to comment Share on other sites More sharing options...

exchemist Posted June 4, 2021 Share Posted June 4, 2021 12 minutes ago, awaterpon said: It is a physical situation with few details. consider this: A rest mass of 6 kg, 2 kg of it exploded,it pushed the 4 kg with 1 kg energy and the other 1 kg went away. Now I have a new rest mass of 4kg and energy of 1 kg: relativistic mass m is= m=m0/sqr (1-v^2/c^2) I will have a velocity v1 m=4/sqr(1-v1^2/c^2) But at the same time m is 1+4 or5 kg Then 4/sqr(1-v1^2/c^2)=5 kg relativistic kinetic energy: K.E=m0c^2/sqr(1-v^2/c^2)+m0c^2 =4c^2/sqr(1-(v1^2/c^2))+4c^2 But: =4/sqr(1-(v1^2/c^2))=5 kg so K.E=5c^2+4c^2= 9c^2 joules The actual mass is 1+4 kg, and the energy of this system must be: E=mc^2 = 5c^2 joules instead of 9c^2 joules You are still using a wrong sign in the formula for relativistic k.e. In your formula, if you set v to zero, you get a kinetic energy of 2mc², which is an obvious nonsense. This has already been pointed out to you. If you fix that, you have a chance of making sense, at least. Link to comment Share on other sites More sharing options...

swansont Posted June 4, 2021 Share Posted June 4, 2021 11 hours ago, awaterpon said: Nothing, it is a matter of equations and that why I refer to the energy as pushing the system.The energy of 1 kg is a kinetic energy that the system can have. And it is true considering the atomic bomb energy. This is unphysical. Momentum is a conserved quantity, which means that you can’t assume all the energy is converted from mass into KE of one particle. That can’t happen. In atomic bombs, you would have multiple massive parts and radiation carrying away energy. If it’s isotropic, momentum is zero, so momentum is conserved. But this is not what you are describing. 1 hour ago, awaterpon said: It is a physical situation with few details. No, it’s unphysical with few details 1 hour ago, awaterpon said: consider this: A rest mass of 6 kg, 2 kg of it exploded,it pushed the 4 kg with 1 kg energy and the other 1 kg went away. Now I have a new rest mass of 4kg and energy of 1 kg: What posseses this 1 kg’s worth of energy that “went away”? Energy doesn’t go away. It is conserved. Energy is a property, not a substance. (also, don’t introduce new examples. Stick with one until you acknowledge it’s wrong and you correct it) Link to comment Share on other sites More sharing options...

MigL Posted June 4, 2021 Share Posted June 4, 2021 1 hour ago, awaterpon said: A rest mass of 6 kg, 2 kg of it exploded,it pushed the 4 kg with 1 kg energy and the other 1 kg went away. What do you mean '1 kg pushed the 4 kg' and the other '1 kg went away' ??? That is not conserving momentum. The 1 kg that goes along with the 4 kg is not pushing it ( that makes no sense ). Momentum conservation demands that, as a consequence of the 1 kg 'going away', the 4 kg +1 kg has to go in the opposite direction at about 1/5 velocity ( modified by the Lorentz factor at relativistic speeds ). that is how a rocket works. X-posted with Swansont Link to comment Share on other sites More sharing options...

awaterpon Posted June 4, 2021 Author Share Posted June 4, 2021 (edited) 2 hours ago, exchemist said: You are still using a wrong sign in the formula for relativistic k.e. In your formula, if you set v to zero, you get a kinetic energy of 2mc², which is an obvious nonsense. This has already been pointed out to you. If you fix that, you have a chance of making sense, at least. I wrote the kinetic energy of the system at rest 5c^2 instead of S.R 2*5C^2 but this still give similar results I will rewrite it giving K.E at rest 2*5c^2 instead of 5c^2: I also would like to add this question: How a system contains only m0c^2 joules at rest have K.E of 2m0c^2? A rest mass of 6 kg, 2 kg of it exploded,it pushed the 4 kg with 1 kg energy and the other 1 kg went away. Now I have a new rest mass of 4kg and energy of 1 kg: relativistic mass m is= m=m0/sqr (1-v^2/c^2) I will have a velocity v1 m=4/sqr(1-v1^2/c^2) But at the same time m is 1+4 or5 kg Then 4/sqr(1-v1^2/c^2)=5 kg relativistic kinetic energy: K.E=m0c^2/sqr(1-v^2/c^2)+m0c^2 =4c^2/sqr(1-(v1^2/c^2))+4c^2 But: =4/sqr(1-(v1^2/c^2))=5 kg so K.E=5c^2+4c^2= 9c^2 joules Because the whole energy of the system with speed v1 does not change " energy from inside mass" then I can consider the 5 kg as rest mass 1 kg energy and 4 kg mass and both as rest mass because energy and mass are equivalent in which K.E=2mc^2 or 2*5c^2 = 10c^2 joules instead of 9c^2 joules S.R K.E Edited June 4, 2021 by awaterpon Link to comment Share on other sites More sharing options...

swansont Posted June 4, 2021 Share Posted June 4, 2021 1 hour ago, awaterpon said: I also would like to add this question: How a system contains only m0c^2 joules at rest have K.E of 2m0c^2? It doesn’t. Your analysis is of an unphysical situation (no conservation of momentum) and your KE equation is wrong, therefore no valid conclusion can be drawn from it. Link to comment Share on other sites More sharing options...

awaterpon Posted June 4, 2021 Author Share Posted June 4, 2021 (edited) 1 hour ago, swansont said: What posseses this 1 kg’s worth of energy that “went away”? Energy doesn’t go away. It is conserved. Energy is a property, not a substance. What I mean is only part of the 2 kg energy pushed the mass and this can be a physical situation 27 minutes ago, awaterpon said: Edited June 4, 2021 by awaterpon Link to comment Share on other sites More sharing options...

swansont Posted June 4, 2021 Share Posted June 4, 2021 27 minutes ago, awaterpon said: What I mean is only part of the 2 kg energy pushed the mass and this can be a physical situation Well, yes, but the rest of the information is needed to determine what’s going on. The mass of any other particles, photon energy, direction of emission, etc. You can’t solve for anything without this. Link to comment Share on other sites More sharing options...

exchemist Posted June 4, 2021 Share Posted June 4, 2021 (edited) 53 minutes ago, awaterpon said: I wrote the kinetic energy of the system at rest 5c^2 instead of S.R 2*5C^2 but this still give similar results I will rewrite it giving K.E at rest 2*5c^2 instead of 5c^2: I also would like to add this question: How a system contains only m0c^2 joules at rest have K.E of 2m0c^2? A rest mass of 6 kg, 2 kg of it exploded,it pushed the 4 kg with 1 kg energy and the other 1 kg went away. Now I have a new rest mass of 4kg and energy of 1 kg: relativistic mass m is= m=m0/sqr (1-v^2/c^2) I will have a velocity v1 m=4/sqr(1-v1^2/c^2) But at the same time m is 1+4 or5 kg Then 4/sqr(1-v1^2/c^2)=5 kg relativistic kinetic energy: K.E=m0c^2/sqr(1-v^2/c^2)+m0c^2 <- =4c^2/sqr(1-(v1^2/c^2))+4c^2 But: =4/sqr(1-(v1^2/c^2))=5 kg so K.E=5c^2+4c^2= 9c^2 joules Because the whole energy of the system with speed v1 does not change " energy from inside mass" then I can consider the 5 kg as rest mass 1 kg energy and 4 kg mass and both as rest mass because energy and mass are equivalent in which K.E=2mc^2 or 2*5c^2 = 10c^2 joules instead of 9c^2 joules S.R K.E You are still using the wrong formula for k.e. Do you bother to read the replies to your threads? Why do you not check it? The second term should not be +mc², but -mc². That's MINUS. I've highlighted it for you in red. That way, when v = 0, k.e. = 0 which makes sense. Rework your maths using the correct formula and see what you get. Edited June 4, 2021 by exchemist Link to comment Share on other sites More sharing options...

Ghideon Posted June 4, 2021 Share Posted June 4, 2021 (edited) 3 hours ago, awaterpon said: consider this: No. You have not addressed important issues: (color added by me) 1 hour ago, awaterpon said: K.E=m0c^2/sqr(1-v^2/c^2)+m0c^2 As pointed out by several members the plus marked with red is incorrect. Added below* is a link that may help, It shows the correct formula and also how it mathematically relates to the low velocity approximation known from Newtonian physics. (edit: @exchemist nicely shows the error in the full context) (Side note: If, for some reason, you believe your version of the equation is actually correct, then you need to provide some backing evidence) *) https://courses.lumenlearning.com/boundless-physics/chapter/relativistic-quantities/ Edited June 4, 2021 by Ghideon Just noted exchemist's answer 1 Link to comment Share on other sites More sharing options...

awaterpon Posted June 4, 2021 Author Share Posted June 4, 2021 (edited) I will use the equations wrong and they will be invalid for sure and I will use the equations right and they turn out to be wrong as I say Two kg of a 6 kg mass exploded and the rest 4 kg will be moved with part of this exploded energy: The rest mass is 4 kg the energy pushing it is the energy of 1 kg: relativistic mass m is= m=m_{0}/sqr (1-v^{2}/c^{2}) I will have a velocity v_{1} m=4/sqr(1-v_{1}^{2}/c^{2}) But m is 1+4 or 5 kg Then 4/sqr(1-v_{1}^{2}/c^{2})=5 kg relativistic kinetic energy: K.E=m_{0}c^{2}/sqr(1-v^{2}/c^{2})-m_{0}c^{2} =4c^{2}/sqr(1-(v_{1}^{2}/c^{2}))-4c^{2} But: =4/sqr(1-(v_{1}^{2}/ c^{2}))=5 kg so K.E=5c^{2}-4c^{2}= c^{2} joules The total mass 5 kg did not change when it moved with the v_{1} velocity it is internal Mass/Energy conversion then I can consider the 5 kg as rest mass . If I can consider the 5 kg as rest mass then the kinetic energy of the 5 kg is also zero , the above equation gives the 5 kg kinetic energy of c^{2 }and that means a according to the S.R equations free energy of c^{2 }joules is generated. Edited June 4, 2021 by awaterpon Link to comment Share on other sites More sharing options...

swansont Posted June 4, 2021 Share Posted June 4, 2021 5 minutes ago, awaterpon said: I will use the equations wrong and they will be invalid for sure and I will use the equations right and they turn out to be wrong as I say When are you going to start using them right? Link to comment Share on other sites More sharing options...

exchemist Posted June 4, 2021 Share Posted June 4, 2021 28 minutes ago, awaterpon said: I will use the equations wrong and they will be invalid for sure and I will use the equations right and they turn out to be wrong as I say Two kg of a 6 kg mass exploded and the rest 4 kg will be moved with part of this exploded energy: The rest mass is 4 kg the energy pushing it is the energy of 1 kg: relativistic mass m is= m=m_{0}/sqr (1-v^{2}/c^{2}) I will have a velocity v_{1} m=4/sqr(1-v_{1}^{2}/c^{2}) But m is 1+4 or 5 kg Then 4/sqr(1-v_{1}^{2}/c^{2})=5 kg relativistic kinetic energy: K.E=m_{0}c^{2}/sqr(1-v^{2}/c^{2})-m_{0}c^{2} =4c^{2}/sqr(1-(v_{1}^{2}/c^{2}))-4c^{2} But: =4/sqr(1-(v_{1}^{2}/ c^{2}))=5 kg so K.E=5c^{2}-4c^{2}= c^{2} joules The total mass 5 kg did not change when it moved with the v_{1} velocity it is internal Mass/Energy conversion then I can consider the 5 kg as rest mass . If I can consider the 5 kg as rest mass then the kinetic energy of the 5 kg is also zero , the above equation gives the 5 kg kinetic energy of c^{2 }and that means a according to the S.R equations free energy of c^{2 }joules is generated. Nope, all it means is you are double counting the k.e. Relativistic mass has fallen into disfavour nowadays, as it is not seen as very helpful. You are inadvertently illustrating why. The full version of the mass-energy relation, for bodies in motion relative to the observer measuring their energy, is E² = (mc²)² +p²c², in which p is the momentum. For objects at rest, p=0 and this reduces to the well-known E=mc². (For a photon, which has zero rest mass, it reduces to E=pc, thus accounting for why photons still have momentum in spite of no rest mass.) As I understand it - others better qualified can jump in - relativistic mass is the "fudge" you have to apply if you want to carry on using E=mc² even though the object is moving (which, if you are being rigorous, you should not really be doing). In effect it is just reconverting the k.e. back into its mass equivalent. In other words you either have a 4kg rest mass object with an additional k.e. of c², or you have an object with a relativistic mass of 5kg. 1 Link to comment Share on other sites More sharing options...

MigL Posted June 4, 2021 Share Posted June 4, 2021 16 minutes ago, exchemist said: As I understand it - others better qualified can jump in - Normally that wouldn't be requred, as you seem to have an excellent understanding of the Physics involved. However, Awaterpon might not even be convinced by the 'brute force' method of everyone else on the Forum presenting valid evidence contrary to his calculations. Link to comment Share on other sites More sharing options...

Ghideon Posted June 4, 2021 Share Posted June 4, 2021 3 hours ago, awaterpon said: The total mass 5 kg did not change when it moved with the v_{1} velocity it is internal Mass/Energy conversion then I can consider the 5 kg as rest mass . One more issue: the statement above is not consistent with the setup: 3 hours ago, awaterpon said: Two kg of a 6 kg mass exploded and the rest 4 kg will be moved with part of this exploded energy: The rest mass is 4 kg (bold by me in the quotes above) A friendly reminder: You are posting on a forum where plenty of members are experts; scientists, teachers, engineers and more. Imagine what you would be able to learn if you posted questions instead of incorrect claims. You would get advanced answers, good references, reading advices, links to free online courses etc. You seem interested in physics and capable of attempting basic mathematics. The predictive power of the current models of mainstream physics is quite good, why not try to find out how those models work before attempting to change them? 2 Link to comment Share on other sites More sharing options...

StringJunky Posted June 4, 2021 Share Posted June 4, 2021 2 hours ago, MigL said: Normally that wouldn't be requred, as you seem to have an excellent understanding of the Physics involved. However, Awaterpon might not even be convinced by the 'brute force' method of everyone else on the Forum presenting valid evidence contrary to his calculations. He's probably read about Galileo... Link to comment Share on other sites More sharing options...

awaterpon Posted June 4, 2021 Author Share Posted June 4, 2021 (edited) 2 hours ago, Ghideon said: A friendly reminder: You are posting on a forum where plenty of members are experts; scientists, teachers, engineers and more. Imagine what you would be able to learn if you posted questions instead of incorrect claims. You would get advanced answers, good references, reading advices, links to free online courses etc. You seem interested in physics and capable of attempting basic mathematics. The predictive power of the current models of mainstream physics is quite good, why not try to find out how those models work before attempting to change them? I did what all these parrot professors could not do. I have my own discovery with observations and experiments . Even though my discovery is fundamental no previous scientist was able to discover it. What evidence do you need than a clear experiment you did it yourself ? I am a great discoverer , greater than Newton or Einstein " both were not be able to discover this fundamental discovery" So why disrespect? 2 hours ago, StringJunky said: He's probably read about Galileo... Yes. And I completed what Newton started. Edited June 4, 2021 by awaterpon -4 Link to comment Share on other sites More sharing options...

swansont Posted June 4, 2021 Share Posted June 4, 2021 38 minutes ago, awaterpon said: I did what all these parrot professors could not do. I have my own discovery with observations and experiments . Even though my discovery is fundamental no previous scientist was able to discover it. I see no reference to experiment, and your point seems to be that incorrect equations give unphysical results. Nothing original about that. You’ve basically said 2 + 2 = 7 and concluded math is wrong. Quote What evidence do you need than a clear experiment you did it yourself ? or i is just rejection? I am a great discoverer , greater than Newton or Einstein " both were not be able to discover this fundamental discovery" So why disrespect? You’re wrong, and demonstrably so. Pointing this out is not disrespect. Declaring yourself the be greater than people who were right is disrespect. Link to comment Share on other sites More sharing options...

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