# When “unphysical” solutions are not actually unphysical

When solving problems in physics, one often finds, and ignores, "unphysical" solutions. For example, when solving for the velocity and time taken to fall a distance h (from rest) under earth gravity:

$\Delta t = \pm \sqrt{2h/g}$

$\Delta v = \pm \sqrt{2gh}$

One ignores the "unphysical" negative-time and positive-velocity solutions (taking x-axis as directed upwards normal to the earth's surface). However, this solution is not actually unphysical; it is a reflection of the fact that the equation being solved is invariant with respect to time-translation and time-reversal. The same equation describes dropping an object with boundary conditions ($t_i$ = 0, $x_i$ = h, $v_i$ = 0) and ($t_f$ = $|\Delta t|$, $x_f$ = 0, $v_f$ = $-\sqrt{2gh}$), or throwing an object backward in time with boundary conditions ($t_i$ = $-|\Delta t|$, $x_i$ = 0, $v_i$ = $+\sqrt{2gh}$) and ($t_f$ = 0, $x_f$ = h, $v_f$ = 0). In other words, both solutions are physical, but they are solutions to superficially different problems (though one implies the other), and this fact is an expression of the underlying physical time-translation and time-reversal invariance.

My question is: is there a more general expression of this concept? Is there a rule for knowing when or if an "unphysical" solution is or is not truly unphysical, in the sense that it may be a valid physical solution corresponding to alternate boundary conditions?

• Of course, the most famous example of this is a certain theory of the strong interaction that was thrown out because it contained all of those pesky massless spin-2 particles... – Jerry Schirmer Apr 4 '12 at 13:23
• ...aka string theory. – Qmechanic Mar 19 '18 at 13:58