jeffrey de rycke

[jefdryck]

In the example of the nucleus and the electron we defined the energy at infinite distance = 0, and the energy was more negative when the electron came closer, due to it being an attractive force. The gravitational force is also a negative force so we can say that the closer the masses, the more negative (lower) the energy of the system will be. This is logical as you will need to expend energy if you want to separate 2 masses as you work against the force.

Take the same L (distance between 2 rings) for the three systems (3 alpha cases). For alpha>0 the masses of the rings will be much closer to the dumbbell than for the alpha<0 case.

Closer means lower energy and therefore alpha>0 is the lowest-energy orientation (we do not need to look at alpha=0 as this is just an in between sate).

We do not need to look at the attraction between the two rings as this will always be the same for a given L.

We can do a quick check using the given formula. Choose alpha>0 = 1. We now have the lowest energy for theta = 0 (+k*pi) (looking at the goniometric functions) as we want the cosine term to dominate, this gives an energy of -2. Choose alhpa<0 = -1. We now have the lowest energy for theta = pi/2 (+k*pi) as we want the sine term to dominate, this gives an energy of -1.

[jefdryck]

Looking at the other answers I now realize I read the question wrong.

So
alpha>0 just as in the clip (and last part of my answer), theta=0 (+k*pi) or dumbell parallel to z-axis
alpha0= -> doesn’t matter (formula is always 0)
alpha<0 theta = pi/2 (+k*pi) or dumbell in xy-plane.

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