Is the potential energy of an ion in a lattice ##U_0## or ##U_0/2##?

[yucheng]

Homework Statement

Purcell Electricity and Magnetism Problem 1.7

Potential energy in a two-dimensional crystal Use a computer to calculate numerically the potential energy, per ion, for an infinite two-dimensional square ionic crystal with separation $a$; that is, a plane of equally spaced charges of magnitude $e$ and alternating sign (as with a checkerboard).

Relevant Equations

N/A

Potential energy in a two-dimensional crystal

Consider the potential energy of a given ion due to the full infinite plane. Call it$U_{0}$. If we sum over all ions (or a very large number$N$) to find the total$U$of these ions, we obtain$N U_{0}$. However, we have counted each pair twice, so we must divide by 2 to obtain the actual total energy. Dividing by$N$then gives the energy per ion as$\left(N U_{0} / 2\right) / N=U_{0} / 2$.

(part of the answer given)

My question is, why is the potential energy of a given ion due to to the full infinite plane is $U_0$, but then the author then moves on to show that the energy per ion is $U_0/2$. Somehow, this seems scandalous to me!

 

[haruspex]

Homework Statement:: Purcell Electricity and Magnetism Problem 1.7

Potential energy in a two-dimensional crystal Use a computer to calculate numerically the potential energy, per ion, for an infinite two-dimensional square ionic crystal with separation $a$; that is, a plane of equally spaced charges of magnitude $e$ and alternating sign (as with a checkerboard).
Relevant Equations:: N/A

Potential energy in a two-dimensional crystal

Consider the potential energy of a given ion due to the full infinite plane. Call it$U_{0}$. If we sum over all ions (or a very large number$N$) to find the total$U$of these ions, we obtain$N U_{0}$. However, we have counted each pair twice, so we must divide by 2 to obtain the actual total energy. Dividing by$N$then gives the energy per ion as$\left(N U_{0} / 2\right) / N=U_{0} / 2$.

(part of the answer given)

My question is, why is the potential energy of a given ion due to to the full infinite plane is $U_0$, but then the author then moves on to show that the energy per ion is $U_0/2$. Somehow, this seems scandalous to me!

It depends what you mean by the PE of a single ion within the lattice. The author is taking it to be the energy associated with plucking out that single ion and removing it to infinity.
Consider what that means in another context, a cluster of N ions of the same charge. As you move each in turn to infinity, the energy associated with each move reduces linearly. So the average energy per ion is half that associated with the first move.

 

[Orodruin]

Put in a different way:

The potential energy of a system consisting of two charges $q$ and $Q$ is $kqQ/r$. This is the same as the energy required to bring one of the charges from infinity. However, it is the potential of the pair of charges.

If you consider a system of charges and sum the Coulomb potentials of each charge you will therefore get a factor of two too much as you will add the potential of each pair twice (once for each charge). You therefore need to divide by two to obtain the actual potential - which is the sum of the Coulomb potential over the pairs of charges.

 

[yucheng]

It depends what you mean by the PE of a single ion within the lattice. The author is taking it to be the energy associated with plucking out that single ion and removing it to infinity.
Consider what that means in another context, a cluster of N ions of the same charge. As you move each in turn to infinity, the energy associated with each move reduces linearly. So the average energy per ion is half that associated with the first move.

Umm, what then is the difference between

the potential energy of a given ion due to the full infinite plane, $U_0$.

and energy per ion, $U_0/2$? I understand why it is divided by 2 (summed over the same bond twice, but I don't understand which should I use the first or the second)

 

[yucheng]

Put in a different way:

The potential energy of a system consisting of two charges $q$ and $Q$ is $kqQ/r$. This is the same as the energy required to bring one of the charges from infinity. However, it is the potential of the pair of charges.

If you consider a system of charges and sum the Coulomb potentials of each charge you will therefore get a factor of two too much as you will add the potential of each pair twice (once for each charge). You therefore need to divide by two to obtain the actual potential - which is the sum of the Coulomb potential over the pairs of charges.

How is summing over the whole plane counting twice?

 

[haruspex]

Umm, what then is the difference between

the potential energy of a given ion due to the full infinite plane, $U_0$.

and energy per ion, $U_0/2$? I understand why it is divided by 2 (summed over the same bond twice, but I don't understand which should I use the first or the second)

The wording "the potential energy of a given ion" is misleading. The ion does not "have" the energy. The energy resides in the relationship between ions. If a single ion is removed to infinity, the energy change relates to every pair of ions in which one of the pair is the removed ion and the other is an ion that remains.

 

[yucheng]

The wording "the potential energy of a given ion" is misleading. The ion does not "have" the energy. The energy resides in the relationship between ions. If a single ion is removed to infinity, the energy change relates to every pair of ions in which one of the pair is the removed ion and the other is an ion that remains.

I think it would be better to scale it down. Suppose we have two ions, one at the origin (A) and one a distance $d$ from it (B). So if I move a single ion to infinity (B), and the potential energy $V_0 =$ work done (let's not worry about signs), so... the energy change of one ion is, $V_0/2$?

 

[yucheng]

It depends what you mean by the PE of a single ion within the lattice. The author is taking it to be the energy associated with plucking out that single ion and removing it to infinity.
Consider what that means in another context, a cluster of N ions of the same charge. As you move each in turn to infinity, the energy associated with each move reduces linearly. So the average energy per ion is half that associated with the first move.

Actually, if it's the energy associated with plucking out a single ion, shouldn't it be $V_0 \, \text{not} \, V_0/2$? Assuming the crystal is held at rest.

 

[yucheng]

The wording "the potential energy of a given ion" is misleading. The ion does not "have" the energy. The energy resides in the relationship between ions. If a single ion is removed to infinity, the energy change relates to every pair of ions in which one of the pair is the removed ion and the other is an ion that remains.

Oh wait, I remember learning that the gravitation potential energy function represents the potential energy not just of the object (like meteor) but also of the Earth!

So what you mean is that when we have a pair of ions, the P.E. function tells us about the energy of both ions (in this case it is shared equally)!

And now Purcell wants us to calculate the energy of one of them, no wonder the factor $1/2$!

 

[haruspex]

Oh wait, I remember learning that the gravitation potential energy function represents the potential energy not just of the object (like meteor) but also of the Earth!

So what you mean is that when we have a pair of ions, the P.E. function tells us about the energy of both ions (in this case it is shared equally)!

And now Purcell wants us to calculate the energy of one of them, no wonder the factor $1/2$!

Right.