Does the 1/2 coefficient in kinetic energy stem from Newton, the De Broglie wavelength and length contraction?

Question

This is my thinking. Everything has a De Broglie wavelength. From this the energy of an object can be calculated.

Now due to Special Relativity anything with a velocity experiences length contraction. This length contraction reduces an object's De Broglie wavelength thereby increasing that object's energy - relative to a stationary object/observer.

The original object now acts like a “compressed spring” .

However, when that “uncompressed” object, makes contact with a second stationary object, it begins to uncompress as the length contraction of Special Relativity stops.

Now in order for that uncompressing object to move the second object forward, 1/2 of its energy/momentum moves forward and via (work = force x distance) begins to accelerate the second object.

This is kinetic energy, but we can only measure it by the energy imparted into the second object.

But Newton's third law must now kick in order to conserve momentum. As such the other 1/2 of the uncompressing object's energy/momentum (the spring) must move in the opposite direction, in order to give the stationary object something to push off from.

As for every action there is an equal and opposite reaction.

This is basically the way rocket propulsion works. A propellant explodes, 1/2 pushes the rocket forwards giving the rocket kinetic energy but the other 1/2 must exit in the opposite direction to conserve momentum.

This is where I think the 1/2 originates from in kinetic energy and what its physical interpretation means.

The question is...

If this is the case, in the binomial expansion for relativistic kinetic energy - the 1/2 term encoded within it. How does the binomial expansion know about the conservation of momentum. Or is it just some coincidence?

Of course all of this could just be a fanciful delusion. But I'd like an opinion.

Answer 1

Does the 1/2 coefficient in kinetic energy stem from Newton, the De Broglie wavelength and length contraction?

No. It stems entirely from Newtonian dynamics and definitions.

The differential work (dW) is defined as below ($\vec F$ is the force and $\vec x$ is the distance):

$$ dW = \vec F \cdot d\vec x = m \frac{d\vec v}{dt} \cdot d\vec x = m \vec v \cdot d\vec v\;, $$ where the second equation follows from Newton's second law and the third equation follows from the definition of velocity (symbolically just move the $dt$ over to the denominator of the $d\vec x$ term (and recall that these numbers all commute)).

Now, recognize that $$ d(v^2) = 2 \vec v \cdot d\vec v\;, $$ where the $2$ on the RHS comes from the usual power rule of differentiation.

Or, dividing both sides by $2$: $$ \frac{1}{2}d(v^2) = \vec v \cdot d\vec v\;, $$

Plugging the bottom equation into the top equation gives: $$ dW = m \frac{1}{2}d(v^2) = d(\frac{1}{2}mv^2) $$

In other words, the Work is equal to the change in kinetic energy, where kinetic energy is defined as:

$$ \frac{1}{2}mv^2 $$

This is why kinetic energy is a useful concept. The $\frac{1}{2}$ comes from moving the factor of two from the power rule of differentiation to the other side of the equation.

Answer 2

The factor of $\frac12$ in the formula for kinetic energy originates in basic kinematics. If you apply a constant force $F$ to a mass $m$ which starts from rest, the work you do is $Fx = (ma)\left(\frac12at^2\right) = m\frac vt\left(\frac12\frac vt t^2\right) = \frac12mv^2$. (You can run this calculation in reverse to find the work which the moving mass can do on you.)