# Mechanical energy $E<0 \implies$ the system is in a bound state?

In a gravitational field if a body has negative mechanical energy, then the system body-planet (or star) is in a gravitationally bound state. The explanation that I have in mind for that is: since mechanical energy is negative and constant, and potential energy increases with distance (even if always negative), there is a position in which the mechanical energy equals the potential energy. The body cannot go any further since, for the conservation of mechanical energy, negative kinetic energy would be necessary (which is absurd).

Is this a general fact valid for any system in Physics? Does the following implication hold?

mechanical energy $E<0 \implies$ the system is in a bound state

Moreover how to justify so? Is it correct to think about it (at least in the gravitational case) as a consequence of the conservation of mechanical energy?

• @AccidentalFourierTransform Don't you mean $E<$ max$V(r)$? Consider the gravitational case. The minimum $V(r)\to -\infty$ and $E$ is larger than that. The maximum $V(r)$ is zero. Another case is the infinite square well. min $V(r)=0$ and $E>0$. Commented Mar 30, 2016 at 20:34
• @AccidentalFourierTransform That isn't ever possible (for then kinetic energy is negative). The potential energy is undetermined by a constant shift, through which the energy is determined. Commented Mar 30, 2016 at 20:35
• You need to clarify exactly what you mean by 'bound state', especially if you're looking for a general answer. For instance, by some definitions, a gas is 'bound' entropically, but not enthalpically, in a particular region of phase space... Commented Mar 30, 2016 at 20:47

Your understanding is actually perfectly correct. In the general case we have indeed something like $E_m = K(r) + V(r)$ for a central force say.

Now, as you rightly point out, the valid positions are those that satisfy $K(r) \geq 0$. This implies that

$E_m - V(r) \geq 0$

If $V(r)$ is an ever increasing function of $r$ that has an asymptote at $V_0$, then we also know that $V(r) \leq V_0 \: \implies E_m - V(r) \geq E_m -V_0$. The criterion on the existence of a bound state or not depends on the ordering of these inequalities:

• If $E_m - V(r) \geq E_m - V_0 \geq 0$, then all possible values of $r$ are valid from $0_+$ to $+\infty$

• On the contrary, if $E_m - V(r) \geq 0 \geq E_m - V_0$, then it means that there exists a limiting distance $r_l$ beyond which the kinetic energy would be negative and thus the corresponding distances are not valid.

Of course, one could calculate directly $r_l$ by simply asking whether there is any solution to the equation $V(r) = E_m$ but such an equation could be hard to solve analytically whereas, as long as the existence of a bound state is concerned, it is enough to know that $E_m - V_0 \leq 0$ to answer the question.

This of course translates as $E_m \leq V_0$ for a bound state to exist, which transforms into $E_m \leq 0$ in the case where $V_0$ is chosen to be zero (remember that the potential energy is defined up to an additional constant).