3 ms·

> 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

> 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.
        eesmith 5y ago
        A photon has no mass. Newton's equation specified massive bodies. Ergo, Newton's equation does not apply to photons. Einstein's equation does apply to photons. Ergo, Einstein's equation is not Newton's equation. For velocities much less than the speed of light, Einstein's equation is well approximated by Newton's equation.
eesmith 5y ago
Not every value is divided by 2. Only the K.E. term is divided by 2. The P.E. term is not. If you ever find where Young worked out the equivalency to P.E. then you'll find he had m v² = 2 m g h. But I suspect he did not. The underlying topic is vis viva. As https://en.wikipedia.org/wiki/Vis_viva https://en.wikipedia.org/wiki/Vis_viva points out, Leibniz was the first to point out that m v² was conserved, Bernoulli used 1/2 m v² in 1741, and at about the same time du Châtelet derived the notion of conservation of energy using Newtonian mechanics, Young called the concept "energy" in 1807 (though without the 1/2), and the 1/2 'recalibration' was due to Gaspard-Gustave Coriolis and Jean-Victor Poncelet during 1819–1839. Quoting https://en.wikipedia.org/wiki/Gaspard-Gustave_de_Coriolis https://en.wikipedia.org/wiki/Gaspard-Gustave_de_Coriolis : > In 1829, Coriolis published a textbook, Calcul de l'Effet des Machines ("Calculation of the Effect of Machines"), which presented mechanics in a way that could readily be applied by industry. In this period, the correct expression for kinetic energy, ½ mv2, and its relation to mechanical work, became established. Coriolis gives the specific reason at https://archive.org/details/ducalculdeleffe00corigoog/page/n41/mode/2up?q=vives https://archive.org/details/ducalculdeleffe00corigoog/page/n... : > .. nous appliquerons cette dénomination à la moitié de ce produit, en sorte que la force vive sera le produit de la masse par la moitié du carré de la vitesse. Cette légère modification à l'usage ancien introduira plus de simplicité dans les énoncés des principes que nous avons à donner. From Google Translate: > we will apply this denomination to half of this product, so that the living force will be the product of the mass by half of the square of the speed. This slight modification to the old usage will introduce more simplicity in the statements of the principles that we have to give. In other words, it's not "a useless convention" as you write, but a convention which simplifies the resulting mathematics.
- SirIsaac 5y ago
  I understand French. I looked through the document. I could not find any place where Coriolis derived kinetic energy. He was apparently using someone else's derivation. By simple substitution, anyone can easily derive E = mv² from Newtonian equations. One starts with W = Fd = mad, and go from there. There is no need to integrate the difference in initial and final velocities (as is currently being done) because this is already assumed in the definition of acceleration. The 1/2 is a mistake and I stand by it.

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.
      - eesmith 5y ago
        A photon has no mass. Newton's equation specified massive bodies. Ergo, Newton's equation does not apply to photons. Einstein's equation does apply to photons. Ergo, Einstein's equation is not Newton's equation. For velocities much less than the speed of light, Einstein's equation is well approximated by Newton's equation.

eesmith 5y ago

Not every value is divided by 2. Only the K.E. term is divided by 2. The P.E. term is not. If you ever find where Young worked out the equivalency to P.E. then you'll find he had m v² = 2 m g h. But I suspect he did not. The underlying topic is vis viva. As https://en.wikipedia.org/wiki/Vis_viva https://en.wikipedia.org/wiki/Vis_viva points out, Leibniz was the first to point out that m v² was conserved, Bernoulli used 1/2 m v² in 1741, and at about the same time du Châtelet derived the notion of conservation of energy using Newtonian mechanics, Young called the concept "energy" in 1807 (though without the 1/2), and the 1/2 'recalibration' was due to Gaspard-Gustave Coriolis and Jean-Victor Poncelet during 1819–1839. Quoting https://en.wikipedia.org/wiki/Gaspard-Gustave_de_Coriolis https://en.wikipedia.org/wiki/Gaspard-Gustave_de_Coriolis : > In 1829, Coriolis published a textbook, Calcul de l'Effet des Machines ("Calculation of the Effect of Machines"), which presented mechanics in a way that could readily be applied by industry. In this period, the correct expression for kinetic energy, ½ mv2, and its relation to mechanical work, became established. Coriolis gives the specific reason at https://archive.org/details/ducalculdeleffe00corigoog/page/n41/mode/2up?q=vives https://archive.org/details/ducalculdeleffe00corigoog/page/n... : > .. nous appliquerons cette dénomination à la moitié de ce produit, en sorte que la force vive sera le produit de la masse par la moitié du carré de la vitesse. Cette légère modification à l'usage ancien introduira plus de simplicité dans les énoncés des principes que nous avons à donner. From Google Translate: > we will apply this denomination to half of this product, so that the living force will be the product of the mass by half of the square of the speed. This slight modification to the old usage will introduce more simplicity in the statements of the principles that we have to give. In other words, it's not "a useless convention" as you write, but a convention which simplifies the resulting mathematics.

SirIsaac 5y ago
I understand French. I looked through the document. I could not find any place where Coriolis derived kinetic energy. He was apparently using someone else's derivation. By simple substitution, anyone can easily derive E = mv² from Newtonian equations. One starts with W = Fd = mad, and go from there. There is no need to integrate the difference in initial and final velocities (as is currently being done) because this is already assumed in the definition of acceleration. The 1/2 is a mistake and I stand by it.