# Time-reversibility symmetry in classical mechanics

Newton's laws are invariant under time reversal transformation $$t \longrightarrow -t$$ for time-independent potentials.

But Hamilton-Jacobi equation is too an equivalent description of classical mechanics which is not invariant under this transformation because of being 1st order in time.

Where is the fallacy?

The Hamilton -Jacobi equation:

$$\frac{\partial S}{\partial t}=-H$$

with:

$$H=\frac{\partial \mathscr L}{\partial \dot{q}}\,\frac{d q}{dt}- \mathscr L$$

you get:

$$\frac{\partial S}{\partial t}=-\frac{\partial \mathscr L}{\partial \dot{q}}\,\frac{d q}{dt}+ \mathscr L\tag 1$$

reverse time $$t\mapsto -t$$ we obtain for eq. (1)

$$-\frac{\partial S}{\partial t}=+\frac{\partial \mathscr L(-t)}{\partial \dot{q}}\,\frac{d q}{dt}+ \mathscr L (-t)\tag 2$$

so equation (2) is equal equation (1), if $$\mathscr L (-t)=\mathscr L (+t)$$ this is the case in your for your question.

you get the same results for Euler- Lagrange equations.

Assuming the Hamiltonian is invariant under $$T$$-transformation, the Hamilton-Jacobi (HJ) equation is also invariant, since Hamilton's principal and characteristic functions change sign.