In a non-relativistic regime, time 't' is considered to be a parameter by changing which we change \(q(t), \dot{q}(t)\) and therefore \(L(q, \dot{q}, t)\) as a whole. Now we consider 't' (time) to be an independent co-ordinate itself, and \(\theta\) is the parameter varying which we vary \(q(\theta), q'(\theta) = \frac{dq}{d\theta}, t(\theta)\) and therefore \(L_{\theta}(q(\theta), q'(\theta), t(\theta),\) \(t'(\theta) = \frac{dt}{d\theta}, \theta)\). If the action (I) remains invariant under this re-parameterisation, then :
MH SET - 2025, SET Exam - 2025,
Answer: A
The correct relation is \(L_{\theta} = t'L\left(q, \frac{q'}{t'}, t\right)\).
The physical velocity is \(\dot{q} = dq/dt = q'/t'\), and the time element is \(dt = t'\,d\theta\). For the action to be the same number in both descriptions:
\[ \begin{aligned} I &= \int L(q, \dot{q}, t)\,dt \\ &= \int L\left(q, \frac{q'}{t'}, t\right) t'\,d\theta \\ &\equiv \int L_{\theta}\,d\theta \end{aligned} \]
So \(L_{\theta}\) must carry the factor \(t'\) from the change of integration variable. This is the standard parameterisation-invariant form used when time is promoted to a coordinate. Without the factor (\(L_{\theta} = L\)) the action would change with the choice of \(\theta\), and the factors \(t'/q'\) or \(q'/t'\) do not correspond to \(dt/d\theta\).
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