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

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.

• If you're already familiar with relativity, then you'll know that $\frac{1}{2}mv^2$ is only an approximation to the kinetic energy. The "true" expression for the total energy of a body is $E = mc^2(1-\frac{v^2}{c^2})^{-\frac{1}{2}} = mc^2 + \frac{1}{2}mv^2 + \frac{3}{8}m\frac{v^4}{c^2} + ...$. When $v$ is much less than $c$ then this is very close to $mc^2$ (the rest energy) plus $\frac{1}{2}mv^2$. Sep 16 at 11:09
• I do know which is why I mentioned the binomial expansion. However, your reply didn't address the mechanism I described and asked an opinion for. Sep 16 at 11:24
• There's no need for any other "mechanism", and it wouldn't be correct anyway if it only addressed one coefficient (the 1/2) and not all of the other infinitely many coefficients. BTW, the expansion is a Taylor (or Maclaurin) series, not a binomial one. Sep 16 at 11:55
• A math equation may represent what's happening in physics but it cannot replace physics and the physical mechanism. So what's physically happening to an object in motion to increase its energy? It's not just motion as that's just momentum. Also, according to this a binomial expansion and a Taylor series are the same function - math.stackexchange.com/questions/905361/… Sep 16 at 12:02
• Motion is relative, as is energy, so trying to seek some "physical" mechanism to explain an "increase" in energy is hopeless. Relative to the chair you are sitting in you are motionless and have 0 kinetic energy. Relative to the sun you have a considerable amount of kinetic energy. Sep 16 at 14:28

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.

• Thank you for the mathematical interpretation. However I'm looking for the physical interpretation. If the moving mass is performing work to accelerate the stationary mass, and for every action there is an equal and opposite reaction, then where is the energy coming from to oppose that reaction? The moving object pushes on the stationary object, the stationary object pushes back. Therefore the moving object must be pushed back, with the same amount of force. Where is the energy coming to resist that force? Sep 18 at 4:40
• Also isn't force x distance = mv^2. As force = (ma) = (m(v/t)) & distance = (vt). The t's cancel leaving (mvv) or mv^2. So where did you get that 1/2 or how are you dividing by 2? Sep 18 at 10:14
• I can not understand what you are asking me.
– hft
Sep 20 at 16:16
• Regarding your second comment: No. What you have written is wrong. It seems like you are trying to work out the constant acceleration version of the general problem. In that case, just look at the other answer from R. W. Bird. His answer applies for constant acceleration.
– hft
Sep 21 at 0:44
• Force = (ma) = (m(v/t)) & distance = (vt). Where exactly is the error? Sep 21 at 4:53

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.)

• My formula refers to a single mass which starts from rest and is accelerated through a distance (1/2)a$t^2$ by by an applied force. Sep 18 at 14:54