This post basically further develops my speculative thoughts about the real meaning of the E = m·*c*^{2} formula. However, I’ll use the relativistically correct formulas for the calculations this time, so it may *look* somewhat more complicated. However, I think you should be able to digest it relatively easily, as none of the math is exceedingly difficult.

My previous post explored the similarity between the formula for the energy of a harmonic oscillator and the E = m·*c*^{2} formula. Now, there is another formula that sort of resembles it: the E = m·*v*^{2}/2 formula for the kinetic energy. Could we relate them somehow and – in the process – gain a better understanding of Einstein’s famous formula? I think we can, and I want to show you how. In fact, in this post, I will try to relate all three.

We should first note that the E = m·*v*^{2}/2 is a non-relativistic formula. It is only correct if we assume the mass – defined as a measure of *inertia*, remember? – to be constant, which we know isn’t true. As an object accelerates and gains kinetic energy, its *effective *mass will increase. In fact, the *relativistically correct *formula for the kinetic energy just calculates it as the difference between (1) the total energy (which is given by the E = m·*c*^{2} formula, *always*) and (2) its rest energy, so we write:

K.E. = E − E_{0} = m* _{v}*·

*c*

^{2}− m

_{0}·

*c*

^{2}= m

_{0}·γ·

*c*

^{2}− m

_{0}·

*c*

^{2}= m

_{0}·

*c*

^{2}·(γ − 1)

The γ in this formula is, of course, the ubiquitous Lorentz factor. Hence, the correct formula for the kinetic energy is m_{0}·*c*^{2}·(γ − 1). We shouldn’t use that m·*v*^{2}/2 formula. Still, the two formulas are remarkably similar: there is a squared velocity (*v*^{2} and *c*^{2}) and some factor (1/2 versus γ − 1). Why the squared velocity? That’s child play, right? Yep, I effectively wrote a post on that for my kids. We have a force that acts on some object over some time and over some distance, and so that force is going to do some *work*. While it’s child play, we’re calculating a path or line *integral *here:

Child play? Perhaps, but many kids don’t know what a vector dot product is (the **F**·*d***x**), and they also don’t realize we can only solve this because we assume the mass *m* to be constant (i.e. *not *a function of the velocity *v*). So… Well… In our flywheel model of an electron, we’ve been using a non-relativistic formula, but we’ve calculated the tangential speed as being equal to *c*. A recipe for disaster, right? 🙂 Can we re-do the calculations? We can. You can *google* a zillion publications on relativistic harmonic oscillators but I took the derivation below from a fairly simple one I’d recommend. The only correction we’ll do here is to use the relativistically correct expression of Newton’s force law: the force equals the time rate of change of the (relativistic) momentum *p* = m* _{v}v* = γm

_{0}

*v*. So we write:

F* = dp*/*dt* = F = –*kx *with *p* = m* _{v}v* = γm

_{0}

*v*

Multiplying both sides with *v *= *dx/dt* yields the following expression: Now, when we combine *two *oscillators – think of the metaphor of a frictionless V-twin engine, as illustrated below 🙂 – then we know that – because of the 90° angle between the two cylinders, the motion of one piston will be equal to *x* = *a*∙cos(ω∙t), while the

motion of the other is given by *y* = *a*∙cos(ω∙t–π/2) = *a*∙sin(ω∙t).Now how do we calculate the *total* energy in this system? Should we distinguish the *x*– and *y*– components of the total momentum ** p**? We can easily do that. Look at the animation below, and you’ll immediately understand that we can easily calculate the velocities in the

*x*– and a

*y*-direction:

*v*/

_{x}= dx*dt*= −

*a*·ω·sin(ω∙t) and

*v*/

_{y}= dy*dt*=

*a*·ω·cos(ω∙t). The sum of the square of both then gives us the tangential velocity

*v*:

*v*

^{2}

*=*

*a*

^{2}∙ω

^{2}∙sin

^{2}(ω∙t) +

*a*

^{2}∙ω

^{2}∙cos

^{2}(ω∙t) =

*a*

^{2}∙ω

^{2}⇔

*v*=

*a*∙ω. But how do we add energies here? It’s a tricky question: we have potential energy in one oscillator, and then in the other, and these energies are being transferred from one to another through the flywheel, so to speak. So there is kinetic energy there. Can we just add it all? Let us think about our perpetuum mobile once more, noting that the equilibrium position for the piston is at the center of each cylinder. When it goes past that position, extra pressure will build up and eventually push the piston back. When it is below that position, pressure is below equilibrium and will, therefore, also want to pull the piston back. The dynamics are as follows:

- When θ is zero, the air in cylinder 1 is fully compressed, and the piston will return to the equilibrium position (
*x*= 0) as θ goes to 90°. The flywheel will transfer energy to cylinder 2, where the piston goes from the equilibrium position to full compression. Cylinder 2 borrows energy, and will want to return to its equilibrium position. - When θ is 90°, the air in cylinder 2 is fully compressed, and the piston will return to the equilibrium position (
*y*= 0) as θ goes to 180°. The flywheel will transfer energy back to cylinder 1, where the piston goes past the equilibrium position to create a vacuum. The piston in cylinder 1 borrows energy, and will want to return to its equilibrium position. - When θ is 180°, the piston in cylinder 1 is fully extended, and will want to return to equilibrium because the pressure is
*lower*than when in equilibrium. As θ goes from 180° to 270°, the piston in cylinder 1 does effectively return to equilibrium and, through the flywheel, pushes the piston in cylinder 2 past the equilibrium to create vacuum. The piston in cylinder 2 borrows energy, and will want to return to equilibrium. - Finally, between 270° and 360°, the piston in cylinder 2 returns to equilibrium and, through the flywheel, causes the piston in cylinder 1 to compress air. The piston in cylinder 2 borrows energy, and will want to return to equilibrium.

It is a funny thing. Where is the energy in this system? Energy is *not *supposed to be thought as being directional but, here, direction clearly matters! We need to think about averages here (kinetic energy is a non-directional (scalar) quantity but it’s a function of velocity, and velocity is directional. If we have two directions only (*x* and *y*), then we can write: 〈*v*_{x}^{2}〉 = 〈*v*_{y}^{2}〉 = [〈*v*_{x}^{2}〉 + 〈*v*_{y}^{2}〉]/2 = 〈*v*^{2}〉/2. So this gives us a clue, but we won’t make things to complicated here. Think of it like this. While transferring energy from one piston to the other, the crankshaft will rotate with a constant angular velocity: linear motion becomes circular motion, and vice versa. So what is the total energy in the system? What if we would want to use it? What can we take from it? You’d agree we would have to take it from the flywheel, right? The *usable *energy is in the flywheel. Let’s have a look at that energy conservation law we derived above: The usable energy in the flywheel is the E = m·*c*^{2} term. This, and my previous post, suggests we may interpret the mass of an electron as a two-dimensional oscillation. In fact, I think my previous post is an easier read because I use the classical (non-relativistic) formulas there. This post, hopefully, demonstrates that a relativistically correct mathematical treatment doesn’t alter the picture that I’ve been trying to offer. 🙂

Of course, the more difficult thing is to go beyond this metaphor and explain *how exactly *this motion from borrowing and returning energy to space would actually work. 🙂 So that would be a proper ‘ring theory’ of matter. 🙂