Law of Conservation of Energy

Theorem. Let \(\vec{c} (t)\) as a position of the particle of mass \(m\) at time \(t\). Let \(V : \mathbb {R}^3 \rightarrow \mathbb{R}\) as a potential function, having \( - \nabla V \) as a force field. Then

\[\frac 1 2 m \| \vec{c}'(t) \|^2 + V( \vec{c}(t))\] 

is constant.

proof. Let above equation as \( E(t)\), and set our aim as showing \(E'(t) = 0 \). Thereby approaching \(V(\vec{c}(t) )\) first.

By chain rule, we get following equation. Note that \(\cdot\) is an inner product.

\[ \frac {d}{dt} V( \vec{c} (t) ) ) = \nabla V (\vec{c} (t)) \cdot \vec{c} ' (t)\] 

Since force field is \(-\nabla V\), we can use Newton's law.

\[ - \nabla V ( \vec{c}(t)) = m \vec{c}'' (t) \]

Using this fact,

\[ \frac {d}{dt} V(\vec{c} (t) ) =\nabla V (\vec{c} (t)) \vec{c} ' (t) = - m \vec{c} '(t) \cdot \vec{c}''(t) \]

Now consider the other part.

\[ \frac {1}{2} m \left( \frac {d}{dt} \| \vec{c}'(t) \|^2 \right) = \frac 1 2 m \frac {d}{dt} ( \vec{c}'(t)  \vec{c}'(t) ) = \frac 1 2 m ( 2 \vec{c}'(t) \cdot \vec{c}''(t) )= m \vec{c}' (t) \cdot \vec{c}'' (t) \]

Therefore \(E'(t)=0\) and \(E(t)\) is constant among \(t\).