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Using a naive view of Einstein's Energy Mass Equivalence $E=mc^2$ (where m is mass and c is the speed of light), it seems tempting to interpret this as a quantum mechanical version of the inherent kinetic energy for a collection of relativistic particles.

If we assume that all quantum particles travel with constant velocity c through 4-dimensional spacetime, this means that if a particle has attained geometrical lightspeed, it's velocity in the time dimension must be zero. If we try to freeze a particle at some location, it will still move randomly in spacetime with velocity c, and consequently it becomes increasingly blurred, as we try to fix it's position by cooling it down.

In this naive view, quantum uncertainty is related to the inherent kinetic energy of quantum particles. The expectation would then be that the kinetic energy equivalence of mass would be $\frac{1}{2} m c^2$.

Is there any actual relation between the equations for mass-energy and kinetic energy, and if so what happens with the $\frac{1}{2}$ factor?

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2 Answers 2

The non-relativistic expression

$$ E_{kin} = \frac{p^2}{2m} $$

is just a low-energy approximation.

Using the relation between intrisic mass, energy and momentum

$$ E^2 - p^2c^2 = m^2c^4 $$

and restricting ourselves to positive energy solutions, we get

$$ \begin{align*} E_{kin} &= E - E_0 \\&= \sqrt{m^2c^4 + p^2c^2} - mc^2 \\&= mc^2\left[\sqrt{1 + \frac{p^2}{m^2c^2}} - 1\right] \\&\approx mc^2\left[\left(1 + \frac12 \frac{p^2}{m^2c^2}\right) - 1\right] \\&= \frac{p^2}{2m} \end{align*} $$

where our approximation is just the first-order Taylor expansion of $\sqrt{1+x}$ at $0$ as $p^2 \ll m^2c^2$ in the non-relativistic case.

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What you are describing is the discredited LeSage ether theory popular in the 19th century. It does not explain Heisenberg uncertainty, nor does it explain relativity. But Olinto dePretto and others used it to predict $E=kmc^2$, where k is a numerical constant of order unity, before Einstein. Their arguments are bogus, because neither relativity nor quantum mechanics works like the LeSage ether.

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Don't quite understand how you reach this conclusion. LeSage's theory was an attempt at describing gravity by a postulated particle field, which according to the theory would explain gravity. According to Wikipedia, LeSage did at some point suggest these particles would travel at lightspeed until later on adjusting that to $10^5$ times lightspeed. I was merely suggesting that in a naive view the energy of mass could be seen as composed of the kinetic energy of elementary particles on the assumption that they travel at constant speed of light (+ potential energy from particle bindings). –  Halfdan Faber Dec 13 '11 at 5:34
    
@Grigory: I was describing the dePretto version of LeSage ether, where all matter is described from jiggling LeSagian particles. This is a modification of LeSage's ideas, but the speed of the particles is c. –  Ron Maimon Dec 13 '11 at 6:02
    
Ron, Thanks for the explanation. Agree, that it basically is the same idea, though an ether doesn't seem essential. References can be seen here: en.wikipedia.org/wiki/Mass%E2%8%93energy_equivalence#Others. –  Halfdan Faber Dec 13 '11 at 6:53
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