The Calculus of Momentum

In physics, the amount of force an object has is described as \[ F = dp/dt \], so one could take the derivative of relativistic momentum to find the relativistic force by using the product rule. Given the equation for relativistic momentum, \[ p = mv \gamma \], we could assume that the mass is constant, and it is the only variable that doesn't change due to a relative velocity. Then we can find the derivative using the product rule on the two remaining variables.

\[ F = \frac{dp}{dt} = m \frac{d}{dt} (\gamma v) \] -> \[ F = m (\frac{d \gamma}{dt} v + \gamma a) \] from \[ a = \frac{dv}{dt} \]

Then \[ F = ma \gamma + mv \frac{d \gamma}{dt} \] by distribution and rearrangement of the variables. The first term is what failed to achieve experimental verification, but it is followed by a second term that describes the classical momentum as being a factor of the rate of change of the Lorentz Factor \[ \gamma \]. This calculus suggest that the Lorentz Factor changes over time during acceleration, and it affects the momentum of the object as it approaches a high relative speed. I have found that the rate of change of the Lorentz Factor actually starts out at zero when there is no relative velocity, so the second term would completely diminish. The first term would reduce to Newtonian Physics from the Lorentz factor becoming equal to one.

The calculus would suggest that there is a natural resistance to the amount of change in the relative velocity as though there is a Lorentzian Force. That force is scaled by the objects rest momentum. An object would have to overcome the rate of change of the Lorentz factor with its momentum in an accelerated frame of reference. This would have to be a key ingredient to deal with equations of motion in an accelerating relativistic framework, if calculus was to hold true to relativity theory. Basically, one would have to always consider the rate of change of the Lorentz factor in an accelerating frame of reference.

Edited November 5Nov 5 by Conjurer