Infinite chain of alternating charges (+/-)

[starstruck_]

Homework Statement

A crystal is a periodic lattice of positively and negatively charged ions.

(a) Consider an infinite one-dimensional crystal of alternating charges +q and −q, separated by distance d:
(+q)--_d--(-q)--_d--(+q)--_d--(-q)--_-d--(+q)--_d--(-q)--_d--(+q)--_d--(-q)--_d--(+q)--_d--(-q)--_d--(+q)--_d--(-q)--_d-- [...]

What is the potential energy per ion?

Homework Equations

This is a second year electricity and magnetism problem set that I'm attempting ( found my professor's previous problem sets online, I haven't started the actual course - it starts in the Fall)

From first year, I know that the potential energy between two charges is V= kq₁q₂/r

The Attempt at a Solution

I know I definitely don't have the answer (might need someone to "talk" me through it ) but this is what I was thinking.

If there is an infinite number of (+q) and (-q) sets , I could take an infinite sum of the potential of one set (so between one +q and -q )and then divide out by the number of ions. If n is the number of (+q) and (-q) sets then the number of ions would be 2n?This seems wrong though, because there is potential between q1 and q2 and then q1 and q3 and q3 and q2, so on and so forth.

 

[jbriggs444]

What if you imagine assembling this crystal one charge at a time? How much work is done adding a charge to the chain?

 

[haruspex]

What if you imagine assembling this crystal one charge at a time? How much work is done adding a charge to the chain?

Not sure that helps. Seems more straightforward just to pick one charge and add up all the potentials at that point from all the others.
One ambiguity though: it says "infinite" but the diagram shows semi-infinite.

 

[starstruck_]

Not sure that helps. Seems more straightforward just to pick one charge and add up all the potentials at that point from all the others.
One ambiguity though: it says "infinite" but the diagram shows semi-infinite.

I'm almost certain we need to work with it being infinite, that's just how the diagram was drawn on the paper- a few ions and then a [...].

If we take one charge and then add up the potentials from that charge to the others, would it be something like
∑ (-1)^(n+1) kq₁(q_n+1)/(d(n-1))
and then divide that by the number of ions (n) ?

 

[haruspex]

would it be something like
∑ (-1)^(n+1) kq₁(q_n+1)/(d(n-1))

Not quite. What is the lower bound on n in your sum? What is the least distance?
All the charges are equal.

then divide that by the number of ions (n) ?

Why?

 

[starstruck_]

Not quite. What is the lower bound on n in your sum? What is the least distance?
All the charges are equal.

Why?

The lower bound on n in my sum is 1

I was using n to represent the number of ions from 1 to infinity.

It asked for the potential for each ion so I assumed I could divide out the total potential by the total number of ions.

 

[haruspex]

The lower bound on n in my sum is 1

What does the n=1 term look like in your sum?

divide out the total potential

But your sum is not the total PE of the chain. It is only that of one ion.

 

[starstruck_]

What does the n=1 term look like in your sum?

But your sum is not the total PE of the chain. It is only that of one ion.

Ahh because i’ve written my sum relative to ONE ion?

The n=1 term is undefined

 

[haruspex]

Ahh because i’ve written my sum relative to ONE ion?

Yes.

The n=1 term is undefined

I don't understand why you have d(n-1). The closest any two can be is d.

Can you perform the sum?

 

[starstruck_]

Yes.

I don't understand why you have d(n-1). The closest any two can be is d.

Can you perform the sum?

The distance of any charge from the first one would be d(n-1)

If n=1 is the first charge, the potential is undefined since d(1-1) = 0

when we add another charge, so n= 2, we know that the distance between them is d, d(n-1) = d(2-1) = d

When we add a third charge, so n= 3, we know that the distance between the 1st and 3rd charge should be 2d.

d(n-1)= d(3-1) = 2d

And so on.

If I preform the sum I’ve written

When n= 1, the sum = undefined

When n= 2, the sum = (-1)^3 (e)(e) /d
= -e^2/d

When n= 3, the sum = (-1)^4 (e)(e)/2d
= e^2/2d

The signs keep changing as the charges are alternating between +/-

When n=4, the sum = -e^2/3d
When n= 5, the sum = e^2/4d

 

[haruspex]

If n=1 is the first charge

... then the sum should start at n=2. No work is done positioning a charge at itself.

When n=4, the sum = -e^2/3d
When n= 5, the sum = e^2/4d

Right, but can you perform the sum? It is one you should recognise.
If not, try using generating functions. Multiply the n^th term by sⁿ and see whether integrating or differentiating wrt s simplifies matters.

 

[jbriggs444]

Not sure that helps. Seems more straightforward just to pick one charge and add up all the potentials at that point from all the others.

It is the same result as long as you do not over-count. The hazard is that by adding up the potentials at that point from all the others you have potentially double-counted. You have to remember to divide by two because you've already implicitly counted the potential at all the other points from the one.

 

[J Hann]

What if you take one charge out of the infinite chain and then consider the work done
in adding the charge back into the chain?
For the first two adjacent charges : W1 = -2kq / d
and then for the next 2 charges W2 = 2 k q / (2 d)
then you get an infinite series of terms like (2 k q / d) * (-1)^n / n
Then the problem would be to find the sum of that series.

 

[haruspex]

What if you take one charge out of the infinite chain and then consider the work done
in adding the charge back into the chain?
For the first two adjacent charges : W1 = -2kq / d
and then for the next 2 charges W2 = 2 k q / (2 d)
then you get an infinite series of terms like (2 k q / d) * (-1)^n / n
Then the problem would be to find the sum of that series.

That is where starstruck_ got to in post #10; except that, correctly, the sum there only counts the charges to one side of the given charge. If you count both sides you are effectively double counting, as warned against in post #12.

 

[J Hann]

That is where starstruck_ got to in post #10; except that, correctly, the sum there only counts the charges to one side of the given charge. If you count both sides you are effectively double counting, as warned against in post #12.

But it seems if you place one end of the chain at the origin then you have to divide by n (infinity) to
get the potential energy per ion. But if you place the middle of the chain at the origin then you have the
difference of two infinite series that proceed as:
1/2 + 1/4 + 1/6 ... 1/2n and 1 + 1/3 + 1/5 ... 1/ (2n -1)

 

[jbriggs444]

But it seems if you place one end of the chain at the origin then you have to divide by n (infinity) to
get the potential energy per ion. But if you place the middle of the chain at the origin then you have the
difference of two infinite series that proceed as:
1/2 + 1/4 + 1/6 ... 1/2n and 1 + 1/3 + 1/5 ... 1/ (2n -1)

As you know, subtracting one infinity from another or dividing one infinite quantity by another is a mathematically undefined operation. Such things are to be avoided.

Interleaving the two series termwise as gives you a convergent series. It can be googled.

Edit: removed some of the more explicit hints.

 

[J Hann]

As you know, subtracting one infinity from another or dividing one infinite quantity by another is a mathematically undefined operation. Such things are to be avoided.

Interleaving the two series termwise as 1/2 - 1/3 + 1/4 - 1/5... gives you a convergent series. It can be googled "alternating harmonic series".

Thanks!

 

[jbriggs444]

Coming from a mathematical background, I find concepts like "potential energy per charge" in an infinite array to be poorly defined. One would like to have a more formal definition to use. Something along the lines of:

"Suppose that we have finite array of n alternating unit charges with unit separation in one dimension. This finite array has some total potential energy P(n). It has potential energy per charge given by P(n)/n. Evaluate the limit (if it exists) of P(n)/n as n increases without bound."

This probably does not help much with physical intuition. Or even with calculation. But it gives the mathematical approach a well defined starting point that avoids infinities.

 

[starstruck_]

... then the sum should start at n=2. No work is done positioning a charge at itself.

Right, but can you perform the sum? It is one you should recognise.
If not, try using generating functions. Multiply the n^th term by sⁿ and see whether integrating or differentiating wrt s simplifies matters.

right so my sum would just be
Σ(-1)^(n+1)(e)^n/(d(n-1)) from n=2 to infinity

as for recognizing the sum, I'm don't believe this was covered in my calculus 2 course ( a lot of the math is covered by our physics professors instead)

would I just be required to find the limit of the sum?

 

[jbriggs444]

right so my sum would just be
Σ(-1)^(n+1)(e)^n/(d(n-1)) from n=2 to infinity

What is that $e^n$ term supposed to represent? I think that you should have written it as $q^2$.
You can factor d out of the sum, simplifying it that way.
You dropped the q's out of the sum. That's somewhat OK since those could have been factored out.
Having factored the constants out of the sum, what is left? Can you write down the first four or five terms?

 

[starstruck_]

What is that $e^n$ term supposed to represent? I think that you should have written it as $q^2$.
You can factor d out of the sum, simplifying it that way.
You dropped the q's out of the sum. That's somewhat OK since those could have been factored out.
Having factored the constants out of the sum, what is left? Can you write down the first four or five terms?

ohhhh, I think I see where to go with this. Right, factoring out the constants from the sum, you would factor out that (-1) ^(n-1),k (forgot to put that into my sum,oops), q^2 and d. You would be left with 1/n-1. This looks like " a/(r-1) "

 

[haruspex]

What is that $e^n$ term supposed to represent? I think that you should have written it as $q^2$.

In post #10 @starstruck_ wrote e², so I think the eⁿ is just a typo.
@starstruck_, did you try the trick I described in post #11?
What do you get if you differentiate $G(s)=\Sigma_1\frac{(-s)^n}n$?

 

[haruspex]

ohhhh, I think I see where to go with this. Right, factoring out the constants from the sum, you would factor out that (-1) ^(n-1),k (forgot to put that into my sum,oops), q^2 and d. You would be left with 1/n-1. This looks like " a/(r-1) "

You cannot factor out (-1)ⁿ since it contains an n.

 

[starstruck_]

You cannot factor out (-1)ⁿ since it contains an n.

And yes! Sorry I meant e^2, getting confused with Ns everywhere.

Oh right the n term, I can still factor out (-1) ^ 1, right? So that’ll leave me with (-1)^n/ (n-1)

 

[haruspex]

And yes! Sorry I meant e^2, getting confused with Ns everywhere.

Oh right the n term, I can still factor out (-1) ^ 1, right? So that’ll leave me with (-1)^n/ (n-1)

Yes.

 

[starstruck_]

Yes.

Just got back to this again!

Okay so (-1)^n/(n-1)

I searched up alternating harmonic series as someone mentioned before ( haven't been taught them yet) but looking at what I've read, the limit of this series should be ln(2) ?

 

[jbriggs444]

Okay so (-1)^n/(n-1)

You should be able to rewrite that summing from 1 to infinity, making it even prettier.

I searched up alternating harmonic series as someone mentioned before ( haven't been taught them yet) but looking at what I've read, the limit of this series should be ln(2) ?

I believe that @haruspex was attempting to get you to derive this result for yourself.

If you multiply each term by $s^n$ and differentiate with respect to s, you get a result that can be further manipulated. Try it and see.

 

[starstruck_]

You should be able to rewrite that summing from 1 to infinity, making it even prettier.I believe that @haruspex was attempting to get you to derive this result for yourself.

If you multiply each term by $s^n$ and differentiate with respect to s, you get a result that can be further manipulated. Try it and see.

Sorry, why do we multiply by s^n?

 

[jbriggs444]

Sorry, why do we multiply by s^n?

Because of the nifty way things cancel when you differentiate.

[It is not obvious until you actually perform the manipulation what magic can then be invoked to provide the result you seek. @haruspex probably knows it inside and out. Me, you have to beat over the head with it]

 

[haruspex]

Sorry, why do we multiply by s^n?

I was trying to get you to work with the form $-\Sigma_1\frac{(-1)^n}{n}$, in which you would multiply by sⁿ. But if you are using the form $\Sigma_2\frac{(-1)^n}{n-1}$ then it will be s^n-1. You'll see why.
It's a technique worth acquiring.