DaemonicSigil comments on Intuitive Lagrangian Mechanics

DaemonicSigil 31 Oct 2020 18:00 UTC
5 points
Nice. Some people in the comments were asking about the actual Taylor expansion for kinetic energy. So here it is (assuming $v << c = 1$ ): $S = \int_{τ_{1}}^{τ_{2}} m d τ = \int_{t_{1}}^{t_{2}} m \sqrt{1 - v^{2}} d t \approx \int_{t_{1}}^{t_{2}} m (1 - \frac{v^{2}}{2} + \dots) d t$ (Since the derivative of $\sqrt{x}$ is $\frac{1}{2 \sqrt{x}}$ , we have $\sqrt{1 + x} \approx (1 + x / 2)$ for small $x$ .) So, up to an additive constant, we have $L = - \frac{m v^{2}}{2}$ . Also, if we want to account for non-gravitational potential energy as well, we can note that in special relativistic units, $E = m$ in free space, and $E = m + V$ in a potential. ( $E$ being the energy of the particle measured in that particle’s frame.) So, assuming $V << m$ : $L = E τ = (m + V) \sqrt{1 - v^{2}} \approx (m + V) (1 - \frac{v^{2}}{2}) \approx m + V - \frac{m v^{2}}{2}$
- habryka 31 Oct 2020 18:10 UTC
  3 points
  Parent
  Note that you can use actual LaTeX in your comments by pressing CMD/CTRL+4 in the editor and typing LaTeX in there (or using the toolbar option for inserting math).
  - DaemonicSigil 31 Oct 2020 18:21 UTC
    5 points
    Parent
    Ah thanks, I’ve got it now. My browser seems to not like CMD-4, but putting dollar signs in markdown worked.