Question Statement

Consider the following Lagrangian for a classical system: $$\mathcal{L}(x,\dot{x})=\frac{1}{2}m\dot{x}^2-\frac{\alpha}{x^2}$$ Show that the action is invariant under the following symmetry transformations: $$\begin{cases}t'=\frac{at+b}{ct+d}\\x'=\frac{1}{ct+d}x\end{cases}$$ With $\text{Det}\begin{pmatrix}a&b\\c&d\end{pmatrix}=1.$

Attempt at a solution

$$t'=\frac{at+b}{ct+d}\implies \text{d}t'=\frac{1}{(ct+d)^2}\text{d}t$$ $$x'=\frac{1}{ct+d}x\implies \text{d}x'=\frac{1}{ct+d}\text{d}x$$ Using both of these relations gives: $$\implies \frac{\text{d}x'}{\text{d}t'}=\dot{x}'=(ct+d)\dot{x}$$ So,

$$\begin{align}S'(x',\dot{x}')&=\int^{t_2'}_{t_1'}\text{d}t'\mathcal{L}(x',\dot{x}')\\\\ &=\int^{t_2'}_{t_1'}\text{d}t'\bigg\{\frac{1}{2}m(\dot{x}')^2-\frac{\alpha}{x'^2}\bigg\}\\\\&=\int^{t_2'}_{t_1'}\frac{1}{(ct+d)^2}\text{d}t\bigg\{\frac{1}{2}m(ct+d)^2\dot{x}^2-(ct+d)^2\frac{\alpha}{x^2}\bigg\}\\\\&=\int^{t_2'}_{t_1'}\text{d}t\bigg\{\frac{1}{2}m\dot{x}^2-\frac{\alpha}{x^2}\bigg\}\\\\&=\int^{t_2'}_{t_1'}\text{d}t~\mathcal{L}(x,\dot{x})\end{align}$$

This is almost in the right form except for the time bounds in the integral. I'm probably missing something quite obvious but I just can't think of it right now, hopefully someone can point it out for me.

Is it simply that when I changed the integrating variable from $t'\rightarrow t$ the bounds also correspondingly change?


1 Answer 1


You aren't missing anything, your solution is perfect.

There is no need for the boundaries to have the same numerical values. Whenever $t$ transforms, so do the boundaries.

The important thing is: the value of the action doen't change (provided you didn't forget to adjust the boundaries, of course). Also note that this doesn't have an impact on the equations of motion.

  • $\begingroup$ Ah, I knew it was something obvious like that. So technically, from the third line and downwards I should have unprimed times in the bounds of the integral? $\endgroup$ Commented Jun 13, 2017 at 4:00
  • $\begingroup$ @NormalsNotFar yup $\endgroup$ Commented Jun 13, 2017 at 6:32

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