Comment by eesmith | Hacker News Reader

Comment by eesmith | Hacker News Reader

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 . 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 .

The proton has a rest mass of 938.3 MeV/c² - 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/... .

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.

> 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.

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).

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.

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.

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.

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.

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.

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 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 :

> 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/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.

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.

I pointed out you likely need to look to du Châtelet to discover the derivation of what Coriolis refers to as the "old usage".

As to the "1/2 is a mistake", one question in physics is, is gravitational mass exactly equivalent to inertial mass?

You can omit the 1/2 and continue with the old definition. If you do so, then your system of physics will have an inertial mass which is half the gravitational mass.

As such, it's a difference only of philosophy. The main advantage to the 1/2 is the math is cleaner if we assume inertial mass = gravitational mass.

I don't yet see how inertial mass would differ from gravitational mass if the K.E. equation is changed. I'll need to think about it some more.

We are currently preparing an experiment to test the prediction of our theory that will leave no doubt. If we are right, textbooks everywhere will have to change, at the very least.

> Not every value is divided by 2. Only the K.E. term is divided by 2. The P.E. term is not.

I'm not sure what that means.

> 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.

I don't get it. mv² is not equal to 2mgh. That makes no sense.

> mv² = 2mgh

I think I now know what you mean. You mean that a mass will gain kinetic equal to 1/2 mv² by the time it hits the ground from a height of h. This doesn't prove your claim. Using my own derivation of kinetic energy, I get mgh = mv². Here's why:

PE = mgh is the same thing as Newton's Work equation, W = Fd = mad. Why? Because g is acceleration (a) and h is distance (d).

mgh = mv² is experimentally disproven in the ballistics lab experiments I mentioned.

I don't believe it. They're probably measuring mass and velocity and applying the formula afterwards. That doesn't prove anything. Back in Leibniz' days, they would use soft clay and measure the depths of the indentations made by falling weights. This was accurate enough to tell them that the kinetic energy was proportional to mv².

Your "probably" means you've refused to look at the relevant experimental evidence?

"proportional to mv²" also means "proportional to 1/2 mv²" and "proportional to 123.45 mv²".

Assuming your gravitational and inertial masses are supposed to be the same, then 1/2 m v² is, by Noether's theorem, a consequence of conservation of energy. https://en.wikipedia.org/wiki/Noether%27s_theorem#Example_1:... .

>"proportional to mv²" also means "proportional to 1/2 mv²" and "proportional to 123.45 mv²".

Yes. This is my point. The error has never been found experimentally because everything eventually gets resolved to Newton's E = Fd. Dividing every result by 2 makes no difference because the proportions do not change.

I still don't see why gravitational mass is a problem but I'll keep looking into it.

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