There is a derivation of relativistic action that treats space and time symetricaly which is just playing arround with the square of kinetic energy in the non-relativistic action and plugging in speed of light. Nothing more! Just by rewriting this we get relativistic action, and mathematics of it are clear. But how did relativity actually enter this? Through the speed of light?

Here you can see what I mean

Reference: A. Zee, QFT in a Nutshell.


1 Answer 1

  1. For a clean derivation, it is easier to go the other way: Start with the relativistic action $$S_0~=~-E_0\Delta\tau,\qquad E_0~:=~m_0c^2.\tag{1}$$ Eq. (1) itself is severely restricted by

    • (i) dimensional arguments,

    • (ii) relativistic invariance, i.e. it must be a function of proper time $\Delta\tau$, rest/invariant mass $m_0$, and the speed of light $c$; and

    • (iii) it must reproduce the geodesic equation.

  2. We can not take the non-relativistic limit $c\to\infty$ directly because of the rest energy (1) becomes infinite. However we can add a constant to the Lagrangian without affecting the EOM. So consider instead the equivalent action (without manifest relativistic symmetry) $$\begin{align} \underbrace{S}_{\text{non-rel.}}~=~&\int_{t_i}^{t_f}\! dt~L~=~ \underbrace{S_0}_{\text{rel.}} + \underbrace{E_0\Delta t}_{\text{const.}}\cr ~\stackrel{(1)}{=}~& E_0(\Delta t-\Delta\tau) ,\end{align} \tag{2}$$ up to boundary terms.

  3. After this shift, we can now consistently take the non-relativistic limit $c\to\infty$ and derive the well-known formula for the non-relativistic kinetic energy $$\begin{align} L~\stackrel{(2)}{=}~E_0(1-\gamma^{-1})~=~&\frac{1}{2}m_0{\bf v}^2+{\cal O}(c^{-2}), \cr \gamma ~:=~&\frac{1}{\sqrt{1-\frac{{\bf v}^2}{c^2}}}. \end{align}\tag{3}$$


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