Roadmap for Formal Mathematical Physics Content


Explanation: often steps are skipped in conveying a derivation. To facilitate formal validation, the implicit steps need to be documented.

Who does this work: human content creator

Motive for doing this work: clearer explanation of mathematical steps. Facilitates student learning and reproducibility in a research context.

  1. lecture video
  2. hand-written notes
  3. Latex document
  4. Content tags
  5. tags for sections and words and expressions
  6. concepts to variables
  7. all steps
  8. derivation graph
  9. replace symbols with numeric ID
  10. validation of steps
  11. validation of dimension
  12. proof of inference rule
Back to layers overview

source: "Derivation of Gravitational Potential Energy" by Rhett Allain

Suppose an object starts an infinite distance from a moon and is dropped, falling towards the moon due to gravitational acceleration. What is the speed of the object when it is distance \(r\) from the moon?

Figure 1: small mass falling towards a moon from initial position at infinity.

added step There are two ways to characterize work in the context of a system. The two ways are equivalent \begin{equation} W_{\rm to\ system} = W_{\rm by\ system} \end{equation}
The initial conditions are \begin{equation} v(x=\infty) =0 \label{eq:initial_velocity} \end{equation}
added step and \begin{equation} U_g(r=\infty) = 0 \end{equation}
The force acting on the object is \begin{equation} \vec{F} = \frac{-G m_1 m_2}{x^2} \hat{x} \label{eq:gravitational force} \end{equation} The work is calculated using W_{\rm to\ system} = \(\Delta E\) since the force changes. To find the cumulative work done on the object, integrate over all positions between \(\infty\) and \(r\) \begin{equation} W_{\rm to\ system} = \int_{\infty}^r \vec{F}\cdot d\vec{r} \label{eq:work as function of force} \end{equation} Substituting the gravitational force into Eq. \ref{eq:work as function of force}, \begin{equation} W_{\rm to\ system} = \int_{\infty}^r \frac{-G m_1 m_2}{x^2} dx \end{equation} Factor out the constants, \begin{equation} W_{\rm to\ system} = -G m_1 m_2\int_{\infty}^r \frac{1}{x^2} dx \end{equation}
added step Integrate to get \begin{equation} W_{\rm to\ system} = -G m_1 m_2 \left(\left.\frac{-1}{x}\right|^r_{\infty}\right) \end{equation} Then expand \begin{equation} W_{\rm to\ system} = -G m_1 m_2 \left(\frac{-1}{r} - \frac{-1}{\infty}\right) \end{equation} Simplify to get
\begin{equation} W = \frac{G m_1 m_2}{r} \end{equation} Another definition of work is that it is the change in energy for a system: \(W_{\rm by\ system} = \Delta KE \).
added step In this characterization \(\Delta PE = 0\).
Because the initial velocity was zero, the work here is \begin{equation} W_{\rm by\ system} = \frac{1}{2} m_1 v^2 \end{equation} Thus we can combine the two definitions of work to get \begin{equation} \frac{1}{2} m_1 v^2 = \frac{G m_1 m_2}{r} \end{equation} The \(m_1\) cancels, leaving \begin{equation} v(r) = \sqrt{\frac{2Gm_2}{r}} \end{equation}