4 ms·

Textbook Kinetic Energy Equation (E = 1/2 mv²) is Wrong

3 points by SirIsaac 5y ago

eesmith 5y ago
This is silly. Energy and momentum are preserved. m g h = 1/2 m v² is tested over and over again in ballistics experiments in intro physics labs. https://www.youtube.com/watch?v=QkzbLMQMFck https://www.youtube.com/watch?v=QkzbLMQMFck . A wrong factor of 2 would stand out. > it hides the ease with which we can derive E = mc² from classical physics without using Einstein’s relativity Einstein's full equation is: E² = (p c)² + (mc²)² The (p c) term is the kinetic energy term comparable to Young's "product of the momentum and velocity" (quoting Young), not the mc² rest-mass term. > E = mc² is the maximum kinetic energy a particle can have. Thus setting an upper limit to the speed any non-massless particle might have (and incidentally disproving special relativity). The Large Hadron Collider runs at 6.5 TeV per proton - https://en.wikipedia.org/wiki/Large_Hadron_Collider https://en.wikipedia.org/wiki/Large_Hadron_Collider . The proton has a rest mass of 938.3 MeV/c² - https://en.wikipedia.org/wiki/Proton https://en.wikipedia.org/wiki/Proton . The 6.5 TeV of kinetic energy is some 7,000 times larger than the supposed maximum of 938.3 MeV. Ergo, this quoted 'maximum kinetic energy' claim is wrong. > The original derivation of kinetic energy in 1807 by British physicist Thomas Young was correct and should not have been changed Young didn't derive kinetic energy in "A course of lectures on natural philosophy and the mechanical arts", he defined energy as "the product of the mass or weight of a body, into the square of the number expressing its velocity." Page 78 of "A course of lectures on natural philosophy and the mechanical arts" at https://archive.org/details/lecturescourseof01younrich/page/78/mode/2up https://archive.org/details/lecturescourseof01younrich/page/... . While he points out that doubling the speed of a dropped object requires four times the height, I wasn't able to find where he calculates potential energy and makes the equivalence between K.E. and P.E.
- SirIsaac 5y ago
  > Energy and momentum are preserved. m g h = 1/2 m v² is tested over and over again in ballistics experiments in intro physics labs. If every result is divided by 2, that is not a test. It's a useless convention. It is silly to solve W = fd by integrating over velocities. The equation is already given. Going from fd to mv² is a simple matter of substitutions. It is easy to derive Ek = mv² from Newtonian equations. Young was highly educated in the math and physics of his time. You're right, I can't find his derivation anywhere but this does not mean he did not derive it mathematically. > Ergo, this quoted 'maximum kinetic energy' claim is wrong. Not true. The values you quoted are measured values. They are as much determined by the observed phenomena as by the instruments used to measure them. When you measure, include the measurer. This is something that relativists need to learn. For example, the constancy of the speed of light is false. Physicists should always make sure they are referring to the constancy of the measured speed of light. They are not the same thing.
  - SirIsaac 5y ago
    By the way, using the correct kinetic energy formula, one can arrive at this striking analogy: E/Ek = c/v This tells us that the rest energy (E) of a body is to its kinetic energy (Ek) what the speed of light (c) is to its velocity (v).
    - eesmith 5y ago
      Again, Einstein's full equation is E² = (p c)² + (mc²)². You left out the momentum term, which means your equation assumes v=0, in which case K.E. = 0 is trivially valid. The momentum term is also important as massless photons still have kinetic energy, which cannot be derived from classical Newtonian mechanics.
      - SirIsaac 5y ago
        I don't need Einstein's equation. Momentum is already assumed in kinetic energy and kinetic energy assumes massive bodies. This is the reason for using mass in the formula. I have no idea why you want the kinetic energy to be 0. The equation assumes a moving body. Thus v cannot be equal to zero.
        eesmith 5y ago
        You need Einstein's equation to have a "striking analogy". You used Einstein's equation for v=0; the rest-mass/energy equivalence. If you use Einstein's full equation, there is no "striking analogy." Your definition of kinetic energy excludes photons, which have kinetic energy and momentum but no mass.
        SirIsaac 5y ago
        In my opinion, it is not Einstein's equation. It's Newton's equation applied to the speed of light. It's not v=0 but v=c. It just so happens that mc² is also the potential energy of a body at rest. Yes, a photon's energy is all kinetic but so what? Newton's equation specified massive bodies. Again, the 1/2 is BS but the winner of this debate is still to be determined.

3 points by SirIsaac 5y ago

This is silly. Energy and momentum are preserved. m g h = 1/2 m v² is tested over and over again in ballistics experiments in intro physics labs. https://www.youtube.com/watch?v=QkzbLMQMFck https://www.youtube.com/watch?v=QkzbLMQMFck . A wrong factor of 2 would stand out. > it hides the ease with which we can derive E = mc² from classical physics without using Einstein’s relativity Einstein's full equation is: E² = (p c)² + (mc²)² The (p c) term is the kinetic energy term comparable to Young's "product of the momentum and velocity" (quoting Young), not the mc² rest-mass term. > E = mc² is the maximum kinetic energy a particle can have. Thus setting an upper limit to the speed any non-massless particle might have (and incidentally disproving special relativity). The Large Hadron Collider runs at 6.5 TeV per proton - https://en.wikipedia.org/wiki/Large_Hadron_Collider https://en.wikipedia.org/wiki/Large_Hadron_Collider . The proton has a rest mass of 938.3 MeV/c² - https://en.wikipedia.org/wiki/Proton https://en.wikipedia.org/wiki/Proton . The 6.5 TeV of kinetic energy is some 7,000 times larger than the supposed maximum of 938.3 MeV. Ergo, this quoted 'maximum kinetic energy' claim is wrong. > The original derivation of kinetic energy in 1807 by British physicist Thomas Young was correct and should not have been changed Young didn't derive kinetic energy in "A course of lectures on natural philosophy and the mechanical arts", he defined energy as "the product of the mass or weight of a body, into the square of the number expressing its velocity." Page 78 of "A course of lectures on natural philosophy and the mechanical arts" at https://archive.org/details/lecturescourseof01younrich/page/78/mode/2up https://archive.org/details/lecturescourseof01younrich/page/... . While he points out that doubling the speed of a dropped object requires four times the height, I wasn't able to find where he calculates potential energy and makes the equivalence between K.E. and P.E.

SirIsaac 5y ago
> Energy and momentum are preserved. m g h = 1/2 m v² is tested over and over again in ballistics experiments in intro physics labs. If every result is divided by 2, that is not a test. It's a useless convention. It is silly to solve W = fd by integrating over velocities. The equation is already given. Going from fd to mv² is a simple matter of substitutions. It is easy to derive Ek = mv² from Newtonian equations. Young was highly educated in the math and physics of his time. You're right, I can't find his derivation anywhere but this does not mean he did not derive it mathematically. > Ergo, this quoted 'maximum kinetic energy' claim is wrong. Not true. The values you quoted are measured values. They are as much determined by the observed phenomena as by the instruments used to measure them. When you measure, include the measurer. This is something that relativists need to learn. For example, the constancy of the speed of light is false. Physicists should always make sure they are referring to the constancy of the measured speed of light. They are not the same thing.
- SirIsaac 5y ago
  By the way, using the correct kinetic energy formula, one can arrive at this striking analogy: E/Ek = c/v This tells us that the rest energy (E) of a body is to its kinetic energy (Ek) what the speed of light (c) is to its velocity (v).
  - eesmith 5y ago
    Again, Einstein's full equation is E² = (p c)² + (mc²)². You left out the momentum term, which means your equation assumes v=0, in which case K.E. = 0 is trivially valid. The momentum term is also important as massless photons still have kinetic energy, which cannot be derived from classical Newtonian mechanics.
    - SirIsaac 5y ago
      I don't need Einstein's equation. Momentum is already assumed in kinetic energy and kinetic energy assumes massive bodies. This is the reason for using mass in the formula. I have no idea why you want the kinetic energy to be 0. The equation assumes a moving body. Thus v cannot be equal to zero.
      - eesmith 5y ago
        You need Einstein's equation to have a "striking analogy". You used Einstein's equation for v=0; the rest-mass/energy equivalence. If you use Einstein's full equation, there is no "striking analogy." Your definition of kinetic energy excludes photons, which have kinetic energy and momentum but no mass.
        SirIsaac 5y ago
        In my opinion, it is not Einstein's equation. It's Newton's equation applied to the speed of light. It's not v=0 but v=c. It just so happens that mc² is also the potential energy of a body at rest. Yes, a photon's energy is all kinetic but so what? Newton's equation specified massive bodies. Again, the 1/2 is BS but the winner of this debate is still to be determined.