Counting Dots Kwok-Wai Ng Feb 1, 2007

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1 Counting Dots Kwok-Wai Ng Feb 1, 007 This sounds so easy (indeed it is not difficult), yet so simple that we never think about it carefully. When we are asked to do it, suddenly we do not know what to do! In this document, I provide three examples, from simple two dimensional case to the third example of three dimensional k-space, which is just the stuff we are now learning in class. Through these examples, we will pick up two skills. The first skill is simple arithmetic, even grade school students can do it. The problem is they do not understand what the problem is asking and why it is important. The second skill involved calculus, which is very common in physics (like eating bread every day). Example 1. et us look at a graph paper (in the computer age, I hope you know what a graph paper is). In the graph paper, you can find little squares of 1mm x 1mm in size (see figure 1). The intersections of these lines form a grid. These are the dots we want to count. If I cut a BIG piece of graph paper and all I know is the area (say,.4m ). Do I know how many dots are there? You may ask will the answer depends on the shape of the cut out. The exact answer probably depends on the shape, but answers for different shapes should not be too far apart. That s what we want to do here. We want to estimate how many dots are there. If the area is BIG, the error will be extremely small. (I leave a question for you to think about: How big is BIG? Compare to what?) 1mm 1mm Figure 1 What I am suggesting here is to first calculate how much space (area) belong to one dot. For example, the red region in figure belongs to dot A because it is closest to A than other dots. This region ends up to be a 1mm x 1mm square again. Actually at this point you should have many questions to think about. For example, why the red region have the same size of the original grid? Can I cut up the space differently? Will I get the A 1mm same answer by different cutting methods? If we know how much area belongs to a single dot, we can just divide the total area of the graph paper by the area belongs to a single dot, we will get the total 1mm Figure

2 number of dots. In this example, the total area is.4 m and the area belongs to one dot is 1mmx1mm 10 - x 10 - m 10-6 m, so the number of dots.4 / x 10 6 dots (.4 million dots!). Now it is your turn to try different graph paper and grid sizes. Skill In the last example, I gave you the area of the graph paper. However, it occurs more often that we only know the shape of the graph paper and its size and we have to estimate the area by ourselves. For example, it may be told that the graph paper is a circle of 1.5 m in radius, we will then have to calculate the area of that circle. In two dimension case, it ends up that the most important shape is a ring (because it is an infinitesimal element of a circle), as shown in figure. Now the graph paper is becoming the ring in figure (with the same grid size). How many dots are there? We need to first calculate the area of the ring, and then we can apply skill #1 to get the answer. To calculate the area of the ring, you can first calculate the area of the outer circle, and then use it to subtract the area of the inner circle. Your answer will be near perfect (except not knowing the exact value of π). However, this is the type of calculate I want to discourage you here. First, it is too slow. Second, more importantly, this type of thinking is not calculus compatible. r1.5m t0.1m Figure So the proper way to get a not so proper value of the area of the ring is like this. If the ring is very thin (again, I leave a question for you to think about: How thin is thin? Compare to what?), we can cut it open and stretch it straight. We now have a very long trapezoid, like the one shown in figure 4. It is so long that we can even ignore how the two ends look and treat it like a rectangle! The length of this rectangle is πr and the width is t, so the area must be πrt (so is the area of the ring). Plug in the number, the area of the ring is xπx1.5x m. You can compare this answer with the exact answer, it is probably not off too much. So the number of dots in the ring is 0.94 / x 10 5 dots t0.1m πr with r1.5m Figure 4

3 Example. Now another example (actually a more useful example) to expand our skill to three dimensional space. We consider an array of atoms as shown in figure 5. To make thing slightly more complicated, distance between neighbors is different along the three directions. The problem is the same as before, given a volume of sample (say, 0.4 m ), how many atoms are there? Just like before, we first calculate how much space (volume) belong to one atom. For example, the red region in figure 6 belongs to atom A because it is closest to A than other atoms. This region ends up to have the same size and shape as the cell (figure 6). In the particular example, the volume belongs to one atom is x10-10 x 1x10-10 x 1.5x10-10 x10-0 m, so the number of atoms 0.4 / x x 10 9 atoms. Now it is your turn to try different volume and cell sizes. Skill A x10-10 m 1x10-10 m 1.5x10-10 m Figure 5 Figure 6 Similar to example 1, it occurs more often that we only know the shape of the sample and its size and we have to estimate the volume by ourselves. For example, it may be told that the sample is sphere of 0.5 m in radius, we will then have to calculate the volume of that sphere. In three dimensional case, it ends up that the most important shape is a spherical shell (because it is an infinitesimal element of a sphere), as shown in figure 7. In here, let us consider a spherical shell of 0.01m thickness, and the inner radius is 0.5m. The proper way to get a not so proper value of volume of the shell is like this. If the shell is very thin (again, I leave a question for you to think about: How thin is thin? Compare to what?), we can cur it open and hammer it into a flat thin sheet. The area of this flat sheet is the same as the surface area of the sphere, 4πr and the thickness of this sheet is t. So the volume r 0.5m t 0.01m Figure 7

4 must be 4πr t (so is the volume of the spherical shell). Plug in the number, the volume of the spherical shell is 4πx(0.5) x x 10 - m. You can compare this answer with the exact answer (by subtracting the volume of the inner sphere from the volume of the outer sphere), it is probably not off too much. So the number of atoms in the shell is.14 x 10 - / x x 10 8 atoms. Example. Now we apply what we learnt from the last example to the problem we are studying in class. It is essentially the same as example, except that we are now working in the k- space, instead of the more comfortable real space. Note that the unit length in the k-space is not even in meter. Since kπ/λ, so the unit length in k-space is actually m -. Except a change in length, the way we calculate volume is exactly the same as in real space. Of course, unit of volume in k-space is m -, instead of m π/ x. The dots in figure 8 are not atoms. Each dot represents a state corresponds to a possible (actually two because of polarization) standing wave in the cavity. As demonstrated in class, because of the standing wave requirement, separations between these dots in k-space is (π/ x, π/ y, π/ z ). We want to count how many dots are there between k and k+dk. Is this the same problem as examples 1 and? π/ y π/ z Figure 8 I don t need to say more here now. The volume of space belongs to a dot is π π π π π x y z x y z V The calculation is exactly the same as that in example, except we do not use number this time. Pay attention to V in the above equation. V x y z, so V is a volume in real space (the volume of the cavity). Do not confuse it with the volume in k-space. k z k dk Skill k y k x Figure 9

5 We want to count how many dots are there between k and k+dk. The region in k-space satisfying this condition is just the spherical shell with inner radius k and outer diameter d+dk. So the thickness of the shell is dk (figure 9). Applying what we learnt from skill in example, the volume of this spherical shell is 4πk dk. How many dots are there within the shell? Simple now, # of dots (or states) Volume of the spherical shell Volume belongs to one dot (or state) 4π k dk π / V 4 V k dk π Since in this particular problem, we consider only positive k x, k y and k z, we have only 1/8 of the complete shell, so # of states between k 1 4 V k dk and k + dk 8 π V k dk π