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3.3.7 Kinetic Energy

Calculating the kinetic energy of an object using Newton's laws of motion is complicated because, as kinetic energy increase so does mass. Newton considered mass to be constant and offered no explanation concerning the location of kinetic energy. Eventually, Einstein used Relativity to show that it was stored as extra mass. A more direct proof is given below.

A fixed force acting on a particle will incur increasing inertial resistance to acceleration as kinetic energy is incorporated as mass. Hence, the rate of increase in kinetic energy will decrease. In the classical expression for kinetic energy, \(mv^2/2\), the definition of mass is ambiguous. It cannot be rest mass, \(m_o\), since it increases with kinetic energy. This situation is analogous to that of simple and compound interest. A more rigorous derivation focusing on the energy content of mass, is as follows.

For a particle moving within the continuum, its Newtonian momentum, \(P=mv\). By the second law of motion, a force, \(F_v\), acting to accelerate the particle is equal to its rate of change of its momentum. Hence \begin{align} \frac{\partial P}{\partial t}= F_v \nonumber \end{align}.
Change in the velocity of the particle results in a change in its kinetic energy and hence a change in its total energy content. The reactive (inertial) force, \(F_E\), associated with this change in total energy, \(E\), is \begin{align} \frac{\partial E}{\partial x} = F_E \nonumber
\end{align}
In terms of differential algebra, \(\partial E = F_E\ \partial x \ \ and\ \partial P = F_v\ \partial t \)
\begin{align} dividing\ \partial E\ by\ \partial P \ yields\ \ \ \
\frac{\partial E}{\partial P} =&\ \frac{F_E\ \partial x}{F_v\ \partial t}\ = \frac{F_E}{F_v}\ v \nonumber
\end{align}
If the two forces acting on the particle are equal, then, \(F_v = F_E\), and the equation simplifies to \begin{align}
\frac{\partial E}{\partial P} = \frac{F_E}{F_v} v =\ v \ \ hence\ \partial E = v\ \partial P \nonumber \\ where\ \ \partial P = (\partial m\ v + m\ \partial v) \nonumber
\end{align}

It is convenient at this stage to convert mass to energy content, and since the mass is moving we may write particle energy \( E=m_v\ c_v^2\), where \(m_v\) is some unknown function of velocity v. The total velocity, \(c_v\), is the vector addition of expansion velocity, \(c_o\) and local velocity, \(v\). That is \(c_v^2 = (c_o^2 + v^2)\). The problem of the unknown function, \(m_v\) can be resolved if it is written as \(m_v \propto E/c_v^2\), and \(\partial m_v \propto \partial E /c_v^2\), the equation then becomes
\begin{align}
\partial E = v \partial P = v( \partial m_v\ v + m_v\ \partial v) =&\ v (\frac{\partial E}{c_v^2}\ v + \frac{E}{c_v^2}\ \partial v) \nonumber \\
So\ that\ \partial E\ c_v^2 =&\ \partial E\ v^2 + E\ v\ \partial v \nonumber \\
and\ \partial E\ (c_o^2 + v^2) - \partial E\ v^2 =&\ E\ v\ \partial v \nonumber \\
\partial E\ ( c_0^2) =&\ E\ v\ \partial v \nonumber \\
\frac{\partial E}{E} =&\ \frac{v\ \partial v}{c_0^2} \nonumber
\end{align}

This is a partial differential equation. The integral of the left hand side is \begin{align}
\int_{E_0}^{E} \frac{\partial E}{E} = ln(\frac{E}{E_0}) \nonumber \end{align} where the constant of integration, \(E_0\) is rest energy - the equivalent of rest mass, which represents the stationary situation, \(v=0\).
The integral on the right hand side is \begin{align} \int_{0}^{v} \ \frac{v}{c_0^2}\ \partial v = \frac{1}{2} \frac{v^2}{c_0^2}\ \nonumber \end{align}

\begin{align} \ hence\ \ ln(\frac{E}{E_0}) = \frac{1}{2} \frac{v^2}{c_0^2}\ \ and\ E = E_0 \exp({\frac{1}{2} \frac{v^2}{c_0^2}}) \nonumber \end{align}

If Kinetic energy, \(E_k\), is defined as \(E_k=E-E_0\), then \begin{align}
E_k = E_0\ exp(\frac{v^2}{2c_0^2}) -E_0 =E_0\ (exp(\frac{v^2}{2c_0^2}) -1) \nonumber \end{align}

This result is not the traditional expression for kinetic energy, which is \(m_0v^2/2\). Its clear that the accepted formula is incorrect since it depends on rest mass, \(m_0\), and doesnt take into account the fact that the mass is continuously increasing. This analysis shows that in calculations involving motion, it makes more sense to use energy content rather than mass, which is inherently ambiguous.

However, the traditional expression may be recovered as an approximation to the correct expression given above by noting that the exponential function can be replaced by its power series, \(\exp (x) = 1 + x + x^2/2 + x^3/6 + .... x^n/n! \). Hence,
\begin{align}
E_k =&\ E_0\ ( \exp(\frac{1}{2} \frac{v^2}{c_0^2}) -1) \nonumber \\ E_k=&\ E_0 (\ (1 + \frac{1}{2} \frac{v^2}{c_0^2} + higher\ order\ terms) -1) \nonumber \\ so\ approximately,\ E_k =&\ E_0 ( \frac{1}{2} \frac{v^2}{c_0^2})\ \ \ \ \ and\ since\ \frac{E_o}{c_o^2} = m_o, \nonumber \\ \ E_k =&\ \frac{m_o v^2}{2} \ \ \ where\ m_o\ is\ rest\ mass \nonumber \end{align}

In the real world, terms of higher order than \(v^2/c_0^2\) are very small. For example, at the very high speed of 30,000 meters per second, the third term is of order \(10^{-16}\). Hence, this final expression is an approximation valid for most calculations of kinetic energy. The exception is in particle accelerators where physicists tend to use total energy rather than mass. In these situations it is necessary to use \( E = E_0 \exp({\frac{1}{2} \frac{v^2}{c_0^2}})\), as the correct relationship between Energy content and velocity.

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