8 ms·

The Controversial Origin of the Kinetic Energy Equation Ek = ½mv²

5 points by SirIsaac 5y ago

_Microft 5y ago
How is this definition compatible with the derivation from special relativity? Express total energy as function of the momentum E(p) = (m^2*c^4 + p^2*c^2)^0.5 and do a Taylor expansion for momentum p_0. You'll arrive at E(p_0) = mc^2 + p_0^2/(2m /* <-- here is the factor 1/2 coming in */) + p_0^4/(8m^3*c^2) + ... The first term is the rest energy, the second term is the kinetic energy, higher ones are relativistic corrections to that. Setting kinetic energy arbitrarily to p^2/m (= mv^2 i.e. without the factor 1/2) breaks that.
- eesmith 5y ago
  It's not. My exchange with the author is at https://news.ycombinator.com/item?id=27974055 https://news.ycombinator.com/item?id=27974055 .
  - SirIsaac 5y ago
    Thanks. I admit that my previous article was in error but this new article uses a different approach to the problem. It shows that the current equation was never meant to replace the previous one. Coriolis clearly wrote that he used vis viva = ½mv² for purposes of convenience. Edit: I'm grateful for our previous exchange. You pointed out new places and ideas that were useful to my research.
- SirIsaac 5y ago
  In my opinion, Ek = mv² contradicts special relativity. The maximum energy is E = mc². There can be no doubt about it. And it is not rest energy. It is obviously kinetic energy. But that's just me.
  - adrian_b 5y ago
    There is no contradiction, because the 2 "energies" from the 2 formulas are 2 different quantities. The energy is not a primitive physical quantity. It is a quantity that is defined. The definition of energy is not unique, it can contain an arbitrary multiplicative factor without changing anything in physics, as long as the same definition is used everywhere. If the kinetic energy is defined as mv^2, everything is fine, but then the mechanical work is 2Fd and the total energy from the special relativity is 2mc^2. If the kinetic energy is defined in the usual way, all the values of energy or work are halved. There is nothing mysterious about this. It is just a matter of convention. In the beginning, the kinetic energy was defined as mv^2, because it was the simplest formula. After more complex mechanical problems began to be studied and solved, requiring differential and integral calculus, the definition was modified to be that from today, because there are more formulas that become simpler than formulas that become more complex, like the direct computation of the kinetic energy from mass and velocity. The same has happened with the Coulomb law. Initially the electric force was defined as the product of charges divided by the square of the radius, because that was the simplest possible formula. Later, it was realized that making the Coulomb formula more complex, by defining the force as the product of charges divided by the area of the sphere centered on one charge, many other formulas become simpler, so the old definition was changed. The change of the kinetic energy formula was a good change, because computing a half of mv^2 is a negligible complication, while a large number of formulas containing integrals or derivatives become simpler.
    - SirIsaac 5y ago
      Sorry. Ek = mv² is the correct formula, not the simplest. And it does not express rest energy but kinetic energy. This is obvious, no?
woofie11 5y ago
I dislike this. The ½ comes from an integration. Whenever you see a square term in an equation, there's often a corresponding ½. Yes, we could "simplify" equations by getting rid of the ½ and squashing it into the units, but at that point, all of the stuff makes less sense. I do physics with kids, and often show area (usually of a triangle) as the distance traveled with constant acceleration, the energy of something, the accumulated debt or what-not. The source of the ½ is very obvious -- we're looking at a triangle and not a rectangle.
- SirIsaac 5y ago
  The ½ comes from integrating W = Fd. No one is arguing against that. The equation should be W = mv² = 2Fd. Why? because v² = 2 a*d.
  - woofie11 5y ago
    Yes, the algebra works, but you lose the intuition for what's going on. There's a pattern of factorial terms you see with successive integrations: 1/1, 1/2, 1/6, 1/24, 1/n!, Multiplying by 2 changes this to 2, 1, 1/3, 1/12, 2/n!. I view that as a clear loser, not a winner.
    - SirIsaac 5y ago
      Ok. Thanks for the exchange.
scotty79 5y ago
If you define kinetic energy as mv^2 it basically means that you redefine the word "energy" to mean double energy (as we currently definine it). It could work just as well, but we would have to have the factor 2 in all of the other equations that define any kind of energy or work so the "double energy" is conserved. The author alread gives such doubled equation for potential energy Epd=2mgh. You'd need to do the same for every other equation that involves any kind of energy in physics. I wonder how author feels about 1/2 factor in equation that gives distance traveled by moving with constant acceleration. Should it be just at^2 ? This would redefine distance to be double distance and we would have to consider what it means for acceleration and velocity. All these 1/2 come straight from integration of linear functions. To get rid of them we'd have to patch math results and since math holds this would mean factors of 2 pop up around everywhere else.
- SirIsaac 5y ago
  Thanks for the comment. I see your point but I don't think it's a matter of definition. There is no doubt that the correct equation is Ek = mv². Since velocity v² = 2ad, it follows that mv² = 2mad = 2Fd. The work done is the acceleration of the body which is perfectly expressed in mv². The distance traveled is a result of the acceleration not the cause of it. Only the change in velocity matters. At least, that is my take on it.
  - scotty79 5y ago
    But the work done is Fd so 2Fd is double the work, and hence mv^2 is double energy. Why do you think work is given by 2Fd? Do you want it to be that? Do you want to define work as two times the force times the distance the force acted over? You can, if you like, but you'll be plagued by a factor of 2 in every place where you will talk about any work. It's way easier to keep this factor of 1/2 just in definition of kinetic energy instead. You can name mv^2 kinetic energy, and 2Fd work. But then you can name (mv^2)/2 kinetic half-energy and Fd half-work and physicist will still prefer to do calculations for half-work and half-energy because there will be less places where this factor of 2 will crop up in calculations.

5 points by SirIsaac 5y ago

How is this definition compatible with the derivation from special relativity? Express total energy as function of the momentum E(p) = (m^2*c^4 + p^2*c^2)^0.5 and do a Taylor expansion for momentum p_0. You'll arrive at E(p_0) = mc^2 + p_0^2/(2m /* <-- here is the factor 1/2 coming in */) + p_0^4/(8m^3*c^2) + ... The first term is the rest energy, the second term is the kinetic energy, higher ones are relativistic corrections to that. Setting kinetic energy arbitrarily to p^2/m (= mv^2 i.e. without the factor 1/2) breaks that.

eesmith 5y ago
It's not. My exchange with the author is at https://news.ycombinator.com/item?id=27974055 https://news.ycombinator.com/item?id=27974055 .
- SirIsaac 5y ago
  Thanks. I admit that my previous article was in error but this new article uses a different approach to the problem. It shows that the current equation was never meant to replace the previous one. Coriolis clearly wrote that he used vis viva = ½mv² for purposes of convenience. Edit: I'm grateful for our previous exchange. You pointed out new places and ideas that were useful to my research.
SirIsaac 5y ago
In my opinion, Ek = mv² contradicts special relativity. The maximum energy is E = mc². There can be no doubt about it. And it is not rest energy. It is obviously kinetic energy. But that's just me.
- adrian_b 5y ago
  There is no contradiction, because the 2 "energies" from the 2 formulas are 2 different quantities. The energy is not a primitive physical quantity. It is a quantity that is defined. The definition of energy is not unique, it can contain an arbitrary multiplicative factor without changing anything in physics, as long as the same definition is used everywhere. If the kinetic energy is defined as mv^2, everything is fine, but then the mechanical work is 2Fd and the total energy from the special relativity is 2mc^2. If the kinetic energy is defined in the usual way, all the values of energy or work are halved. There is nothing mysterious about this. It is just a matter of convention. In the beginning, the kinetic energy was defined as mv^2, because it was the simplest formula. After more complex mechanical problems began to be studied and solved, requiring differential and integral calculus, the definition was modified to be that from today, because there are more formulas that become simpler than formulas that become more complex, like the direct computation of the kinetic energy from mass and velocity. The same has happened with the Coulomb law. Initially the electric force was defined as the product of charges divided by the square of the radius, because that was the simplest possible formula. Later, it was realized that making the Coulomb formula more complex, by defining the force as the product of charges divided by the area of the sphere centered on one charge, many other formulas become simpler, so the old definition was changed. The change of the kinetic energy formula was a good change, because computing a half of mv^2 is a negligible complication, while a large number of formulas containing integrals or derivatives become simpler.
  - SirIsaac 5y ago
    Sorry. Ek = mv² is the correct formula, not the simplest. And it does not express rest energy but kinetic energy. This is obvious, no?

woofie11 5y ago

I dislike this. The ½ comes from an integration. Whenever you see a square term in an equation, there's often a corresponding ½. Yes, we could "simplify" equations by getting rid of the ½ and squashing it into the units, but at that point, all of the stuff makes less sense. I do physics with kids, and often show area (usually of a triangle) as the distance traveled with constant acceleration, the energy of something, the accumulated debt or what-not. The source of the ½ is very obvious -- we're looking at a triangle and not a rectangle.

SirIsaac 5y ago
The ½ comes from integrating W = Fd. No one is arguing against that. The equation should be W = mv² = 2Fd. Why? because v² = 2 a*d.
- woofie11 5y ago
  Yes, the algebra works, but you lose the intuition for what's going on. There's a pattern of factorial terms you see with successive integrations: 1/1, 1/2, 1/6, 1/24, 1/n!, Multiplying by 2 changes this to 2, 1, 1/3, 1/12, 2/n!. I view that as a clear loser, not a winner.
  - SirIsaac 5y ago
    Ok. Thanks for the exchange.

scotty79 5y ago

If you define kinetic energy as mv^2 it basically means that you redefine the word "energy" to mean double energy (as we currently definine it). It could work just as well, but we would have to have the factor 2 in all of the other equations that define any kind of energy or work so the "double energy" is conserved. The author alread gives such doubled equation for potential energy Epd=2mgh. You'd need to do the same for every other equation that involves any kind of energy in physics. I wonder how author feels about 1/2 factor in equation that gives distance traveled by moving with constant acceleration. Should it be just at^2 ? This would redefine distance to be double distance and we would have to consider what it means for acceleration and velocity. All these 1/2 come straight from integration of linear functions. To get rid of them we'd have to patch math results and since math holds this would mean factors of 2 pop up around everywhere else.

SirIsaac 5y ago
Thanks for the comment. I see your point but I don't think it's a matter of definition. There is no doubt that the correct equation is Ek = mv². Since velocity v² = 2ad, it follows that mv² = 2mad = 2Fd. The work done is the acceleration of the body which is perfectly expressed in mv². The distance traveled is a result of the acceleration not the cause of it. Only the change in velocity matters. At least, that is my take on it.
- scotty79 5y ago
  But the work done is Fd so 2Fd is double the work, and hence mv^2 is double energy. Why do you think work is given by 2Fd? Do you want it to be that? Do you want to define work as two times the force times the distance the force acted over? You can, if you like, but you'll be plagued by a factor of 2 in every place where you will talk about any work. It's way easier to keep this factor of 1/2 just in definition of kinetic energy instead. You can name mv^2 kinetic energy, and 2Fd work. But then you can name (mv^2)/2 kinetic half-energy and Fd half-work and physicist will still prefer to do calculations for half-work and half-energy because there will be less places where this factor of 2 will crop up in calculations.