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November 24th, 2016, 01:45 AM   #1
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From: belgrade

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POTENTIAL: sum ∑ to integral

Hi, I'm new to this forum.
I have a task to do: positive charged particle (proton in this case) is traversing throught channel in crystal lattice ((100) direction in FCC crystal for example). But in my case thing is little more simple: positive particle i passing near row (string) of atoms that are all on same spacing "d".
Total potential is sum of all potentials from every single atom in row:
U(total) = ∑ Ui(Ri),
where Ri is distance from positive particle to i-th atom in row. So sum goes from first particle in row to last particle in row, and "i" is number of particles in row.
How come that this sum can be changed with integral?
Integral is:
What is procedure, step by step for this equation?
Thank You,

Last edited by NikolaS; November 24th, 2016 at 01:49 AM.
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November 24th, 2016, 07:24 AM   #2
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Think of this problem as a little bit like centre of mass.

In a centre of mass problem you have

$\displaystyle \bar{x} = \frac{1}{M} \sum_{i=1}^n m_i x_i$

The centre of mass of a system of particles is equal to the sum of the products of mass and relative distance for each individual particle, divided by the total mass.

But what are you actually working out using this formula? You are performing a weighted-average. A weighted-average is like a mean, except that each value contributes more or less to the centre of mass based on the weight. In the above formula, the weight is the ratio of the individual mass to the total mass. You can show this by moving the factor 1/M into the summation:

$\displaystyle \bar{x} = \frac{1}{M} \sum_{i=1}^n m_i x_i$
$\displaystyle = \sum_{i=1}^n \frac{m_i}{M} x_i$
$\displaystyle = \sum_{i=1}^n R_i x_i$

where $\displaystyle R_i$ is the mass ratio for particle $\displaystyle i$. A regular mean uses equal weights for every item in the sum.

If we have a continuous medium, then the we can use the integral instead (for example, over a length L):

$\displaystyle \bar{x} = \frac{1}{M} \int_{0}^{M} x dm = \frac{1}{M} \int_{0}^{L} \Gamma(x) x dx$

Notice that the mass in the latter formula is now a density ($\displaystyle \Gamma$) which has units of kg/m. In the problem you are given you have a somewhat similar situation, except that the weighting is now performed by distance rather than by mass, you are not dividing by a quantity because you want the total potential (not average potential) and you'll have a potential density rather than a mass density (more on this later).

Summations terms:

$\displaystyle U_{total} = \sum_{i=1}^n U_i R_i$
$\displaystyle = \sum_{i=1}^n U_i \sqrt{\rho^2 + x_i^2}$

This formula can work for any distribution of particles in any configuration. However, in your example problem you have a particular configuration of particles interacting with the proton, is a single row of particles extending across an line of infinite length (in both directions). Let's define particle i = 0 to be the one directly above the proton. Therefore you have

$\displaystyle U_{total} = ... U_{-2} R_{-2} + U_{-1} R_{-1} + U_0 R_0 + U_1 R_1 + U_2 R_2 + ...$
$\displaystyle = ... U_{-2} \sqrt{\rho^2 + x_{-2}^2} + U_{-1} \sqrt{\rho^2 + x_{-1}^2} + U_0 \sqrt{\rho^2 + x_0^2} + U_1 \sqrt{\rho^2 + x_1^2} + U_2 \sqrt{\rho^2 + x_2^2} + ...$

In your configuration the particles have equal spacing d, therefore $\displaystyle x_i = id$

$\displaystyle U_{total} = ... U_{-2} \sqrt{\rho^2 + (-2d)^2} + U_{-1} \sqrt{\rho^2 + (-d)^2} + U_0 \rho + U_1 \sqrt{\rho^2 + (d)^2} + U_2 \sqrt{\rho^2 + (2d)^2} + ...$

Now for the integral... if we just copy the way it is achieved for the centre of mass problem, we have

$\displaystyle U_{total} = \int_{-\infty}^{\infty} \Gamma(x) \sqrt{\rho^2 + x^2} dx$

We don't divide the result by anything because we are working out the total, not an average.

Just like the centre of mass problem we have a density term. In our case, the density is the potential for each particle, $\displaystyle U(x)$, multiplied by a number density, (with units of per-metre) which tells us how many particles contribute to the potential per unit length. The lattice has a constant number density which is 1/d (since there is 1 particle every d metres). Therefore, the integral becomes

$\displaystyle U_{total} = \frac{1}{d} \int_{-\infty}^{\infty} U_i \sqrt{\rho^2 + x_i^2} dx$

Last edited by Benit13; November 24th, 2016 at 07:30 AM.
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