In this answer, it is said that the invariance of the action under the transformation $$ x \rightarrow (1+\epsilon)x\tag{0}$$ gives, up to some boundary terms the virial theorem.
I tried to interpret this but I'm very unsure:
Given the previous transformation, the action varies as follow: $$\delta S(x) = \epsilon\int_{0}^{T} \left(m\dot{x}^2 -x\frac{\partial V}{\partial x}\right)\mathrm{d}t\tag 1$$
but since $\delta(0)\neq0$ and $\delta(T)\neq0$, the boundary terms in the usual action variation doesn't vanish:
$$\delta S(x)=\epsilon\left[ x\frac{\partial L}{\partial \dot x}\right]_0^T+\int_0^T\left(\frac{\partial L}{\partial x}-\frac{\mathrm d}{\mathrm d t}\frac{\partial L}{\partial \dot x}\right)\mathrm{d} t\tag 2$$
If $x$ satisfies the equations of motion, then (2) is equal to: $$\delta S(T,0) = \epsilon\left[ x\frac{\partial L}{\partial \dot x}\right]_0^T\tag 3$$
Now, saying that (1) = (3) (inserting the $x$ which satisfy the equation of motion in (1)) we have:
$$\frac{1}{T}\left[ x\frac{\partial L}{\partial \dot x}\right]_0^T=\frac{1}{T}\int_{0}^{T} \left(m\dot{x}^2 -x\frac{\partial V}{\partial x}\right)\mathrm{d}t\tag 4$$
The LHS vanishes for bounded trajectories at $T\rightarrow\infty$ and we obtain the virial theorem.
I have three questions:
- Is this right?
- If so, why can't we just say that $\delta S = 0$ in (1) and thus get the wrong equation: $m\dot{x}^2 -x\frac{\partial V}{\partial x}=0$? Is this because, we have to ask $\delta S = 0$ for every $\delta x(t)$ with $\delta x(0) = \delta x(T) =0$?
- Then, if this is right, we are not really asking that (1) $= 0$ to find the virial theorem, we are asking: $\delta S = \textrm{boundary term}$, where the boundary terms are given by a variation (here the scaling) of the on-shell action (if i'm not misusing the term, by-on shell I'm trying to say that to go from (2) to (3) we used the E-L equation)? If this view is correct, I find it somewhat reminiscent of the Noether's theorem...
