DESTROYING Energy

[gravityblock]

The momentum does not have the second unit of time
Because momentum is a relative term.
We require the other relation to determine energy.
It is not energy, it holds no energy, and cannot
Until you interact with it from another relative reference.
If I move at the same velocity as your Kg
It has no momentum.
In fact it has no velocity
We are both standing still

When your Kg impacts something, Then there is energy
Because you do not have one velocity, but Two
m/s * m/s = m^2/s=2
The velocity of your Kg, and the velocity of what it hit.
This is a relative term.

you did not "impart energy into your Kg"
but rather you increased the Kg's relative velocity with
respect to your inertial reference.
Which equates to a Kinetic Energy potential, to your reference.
The energy you put into it to change the velocity
is also respective of your reference.

Change the reference, change the energy
Because the relative velocity has changed.
You can do this without making any changes to the Kg.

If your Kg is standing still
and the Earth moves towards it at 1000m/s
it will have the same momentum as if the Kg were moving
1000m/s and the Earth stood still.

without a stationary inertial frame, we don't know the true
velocity of anything.
only it's movement relative to our own.

Momentum by itself, cannot be related to Energy.
you need the other factor.

"The momentum does not have the second unit of time, because momentum is a relative term".  That is total B.S.!  Energy, like velocity, is also a relative measurement!

Why is the velocity squared in the kinetic energy equation, E = ½mv².  Why should the energy depend on the square of the velocity?  We have the same question with the equation E = mc².  Why square the speed of light?  Why should the energy depend on c²?  Or, to extend the question, why should the energy of any moving particle, moving with a constant velocity, depend on the square of that velocity?

The measured wavelength and the real wavelength of the photon differ by a factor of c². This is because the linear motion of the photon stretches the spin wavelength. The linear velocity is c, of course, and the circular velocity approaches 1/c. The difference between the two is c².  Energy, like velocity, is a relative measurement.  A quantum with a certain energy has that energy only relative to us, since it has its velocity only relative to us.  If the wavelength has to be multiplied by c² in order to match it to our measurements, then the mass or mass equivalence will also. Hence the equation E = mc².  In this way, c² is not a velocity or a velocity squared, it is a velocity transform. It tells us how much the wavelength is stretched, and therefore how much the mass and energy are stretched, due to the motion of the object.

The short answer is that the kinetic energy equation, like the equation E = mc², always included the spin energy.  Just as with the photon, all matter has a wavelength (see de Broglie), and the wavelength is determined by spin. The spin has a radius, and this radius is the local wavelength.  Any linear velocity of the spinning particle will stretch our measurement of this wavelength, in a simple mechanical manner. As the linear velocity increases, the spin velocity relative to the linear velocity decreases, by a factor of 1/v. This makes the difference between the linear velocity and the spin velocity v². The term v² transforms the local wavelength into the measured wavelength. This is why we find the term in the energy equation.

The only question remaining is why we have the term ½ in the kinetic energy equation. The reason is simple. We are basically multiplying a wavelength transform by a mass, in order to calculate an energy. So we have to look at how the mass and the wavelength interact.  What we have is a material particle spinning end-over-end. If we look at this spin over any extended time interval, we find that half the time the material particle is moving in the reverse direction of the linear motion. Circular motion cannot follow linear motion, of course, and if we average the circular motion over time, only half the circular motion will match the linear vector. This means that half the effective mass will be lost, hence the equation we have.

Gravock

[magneat]

@EHT


Have you built the testbed?


respectfully

[Reiyuki]

Thanks EHT, you've started a really interesting thought experiment (and maybe soon some actual experiments).