0
$\begingroup$

potential energy vs intermolecular distance r
(source: a-levelphysicstutor.com)

I'm trying to understand this curve better, but I can't quite figure out what "negative potential energy" means.

The graph should describe a molecule oscillating between $A$ and $B$, however where I'm stuck in reasoning this is that the PE is equal in $A$ and $B$, but then why does this mean $r$ will increase in $A$ (repel) and decrease in $B$?

$\endgroup$
2
$\begingroup$

Suppose that two molecules are at distance $B$ and have zero kinetic energy. There's a lower potential energy position in $C$ and therefore the molecules will attract.

They will convert $\epsilon$ potential energy into kinetic energy and reach $C$.

Now, the law of inertia states, and the fact that they have positive kinetic energy indicates, that they will maintain their state of motion at $C$ towards $A$.

Going towards $A$ they will gain potential energy by converting kinetic energy into it (in other words, slowing down).

Once at the distance $A$, they will have gained exactly $\epsilon$ potential energy and will have therefore zero kinetic energy.

At this point the cycle will repeat inverted, they will move towards a distance of $C$ because it's lower energy, surpass it and reach $B$ with zero kinetic energy, which will make the cycle repeat from the start.

$\endgroup$
1
$\begingroup$

This post answers your question in mathematical terms. But I suspect you are looking for a more intuitive answer.

I suspect the confusion about negative energy comes from thinking about energy as some kind of "stuff" that lives in objects like flying balls. Energy can change from one form to another (kinetic to potential) or move from one ball to another, but can never be created or destroyed. This would naturally lead to think that you run out of energy when $E = 0$.

This concept doesn't work when there is more than one frame of reference. Think of two ants on two balls. Ant$_a$ on ball$_a$ sees ant$_b$ on ball$_b$ approaching at $1 m/s$. Suppose $m_a = 1 kg$ and $m_b = 2 kg$.

Then ant$_a$ calculates the total kinetic energy as $E_a + E_b = 0 Joule + 1 Joule$.

Ant$_b$ gets $E_a + E_b = 1/2 Joule + 0 Joule$.

Both answers are right. Just like both ants are right to say the velocity of their ball is 0.

It doesn't work to think of velocity as some kind of "stuff" that is gone when a ball is at rest. It doesn't work for kinetic energy either.


Energy is more like an accounting system.

Two balls fly through space. Ant$_a$, Ant$_b$, and anybody else calculate the energy. Something happens - the balls bounce off each other. Everybody calculates the energy again. Some balls have gained energy, and some have lost. But everybody gets the same total energy as they got before.

So what is the accounting system keeping track of? For balls, it is the effect of forces.

A ball rolls up a hill, against gravity, and slows down. There is a relationship between the distance the ball travels in the against-gravity direction and the change in velocity. $mg(h{_2}-h{_1}) = \tfrac{1}{2} mv{_1}^2 - \tfrac{1}{2} mv{_2}^2$.

This can be expressed as changes in potential and kinetic energy, where one increases as much as the other decreases.

$U = mgh$ and $T = \tfrac{1}{2} mv^2$

$U{_2} - U{_1} = T{_1} - T{_2}$
or
$U{_2} + T{_2} = U{_1} + T{_1}$

You can roll balls up the hill at the beach, where $h = 0$ at the bottom. You can do it in Death Valley, where $h = 0$ at the top.

It doesn't matter where $h = 0$ is. It doesn't matter that $U < 0$ in Death Valley. You get the same $U{_2} - U{_1}$ either way.

$\endgroup$
  • $\begingroup$ You've pointed out to me the importance of frame of reference for PE, thanks! $\endgroup$ – andy Jan 19 '14 at 15:03
0
$\begingroup$

It seems you are equating value of potential energy and repulsion/attraction. Identical values of potential energy do not mean identical behavior of the particle/system in question. Recall that force $F_r=-dU/dr$. If you want to find out whether there is attraction ($F_r<0$) or repulsion ($F_r>0$), you should look at the negative derivative of potential energy, not the value of potential energy.

You also claimed that you were unsure about "negative potential energy." In classical mechanics, there isn't really in physical meaning to it, since only changes in potential energy matter. For example, it's the derivative that determines force, not the value. As another example, recall $U_\text{grav}=mgy$, where one can set $y=0$ anywhere; here, again, the particular value doesn't matter.

$\endgroup$
  • $\begingroup$ Thanks for making it clear and link it to the work-formula! I accepted the other answer because it's more intuitive. $\endgroup$ – andy Jan 19 '14 at 14:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.