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Suppose you want to estimate the number of atoms in a rectangular sheet of graphene. You might estimate the sheet to have $10^{7}$ atoms along one edge and $2*10^{7}$ atoms along the other edge. Multiplying while keeping track of units, we get

$$10^{7}\text{atoms} * 2*10^{7} \text{atoms} = 2*10^{14} \text{atoms}^2$$

But obviously, there are $2*10^{14}$ atoms, not $2*10^{14} \text{atoms}^2$. What is wrong with this calculation?

Qmechanic
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Mark Eichenlaub
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    As a general rule in physics, "counting" stuff (atoms, molecules, pears, bananas and whatnot) means no unit. Eventually any physical quantity can be broken into 7 primal ones. You can also solve the contradiction like this: "I have 2 * 10⁷ atoms per row and 10⁷ rows". Likewise when you have 10 baskets each containing 10 apples, you do not say that you own 100 apples². – Valentin B. Sep 07 '17 at 08:28
  • No reason to make the numbers so big or to involve atoms in the mess: the situation is left basically unchanged if we consider a $2 \times 2$ grid of apples, say. – Qiaochu Yuan Sep 08 '17 at 06:20
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    @QiaochuYuan This community is not particularly open-minded about what inquiries are relevant. As it is, the question has 6 downvotes. If I ask about a 2x2 grid of apples, it would be even worse. – Mark Eichenlaub Sep 08 '17 at 06:26
  • @MarkEichenlaub Some in this community might not be particularly open-minded about this specific question, but it did get 22 upvotes, and it is one of the most viewed in the last few days. – ZeroTheHero Sep 08 '17 at 19:18

5 Answers5

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Your "dimensions" are not quite "correct". The calculation should be something like $10^{7} \frac{\hbox{atoms}}{\hbox{row}}\times 2\times 10^7 \hbox{row}=2\times 10^{14}$ atoms. In fact, atoms are objects to be counted and added, like cars or pears.

I believe I remember from a math course that the Greeks could not (apparently) abstract numbers and so would always think of "$5$" as associated to objects: $5$ apples, $5$ pebbles, etc. Thus you could add apples: $5$ apples + $5$ apples + $5$ apples is $15$ apples.

Multiplication was different and considered as a geometrical operation. A rectangle of sides $3$ m and $4$ m had an area of $3\times 4 =12\hbox{m}^2$.

As a result, they (apparently) never "discovered" the general abstract result that $a\times b=b+b+b\ldots$ ($a$ times) since the two operations were in some sense "incompatible". Moreover, since we live in $\mathbb{R}^3$, it didn't make sense to them to multiply more than $3$ numbers together.

The OP likewise would like to equate two "incompatible" operations (in the sense of the Greeks), the outcome of which numerically agree because one needs to sum all atoms of all rows rather than "multiply" atoms together.

Unfortunately, I cannot find a source to confirm this.

Community
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ZeroTheHero
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    On the other hand, the old Babylonians, several centuries (over a thousand years) before the Greeks, had a more algorithmic approach to mathematics, and in their toy problems (exercises for students etc.) were happy adding areas to volumes, and so on. (Source: see p. 673 of Knuth's 1972 CACM article Ancient Babylonian Algorithms.) – ShreevatsaR Sep 06 '17 at 18:14
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    @ShreevatsaR Thanks for the reference, which I am very much looking forward to read! – ZeroTheHero Sep 06 '17 at 18:19
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There is no such unit as an atom$^2$.

You are counting the number of objects (atoms) and so adding

$10^{7} \,\rm atoms\, + 10^{7} \,\rm atoms\,+10^{7} \,\rm atoms\,+10^{7} \,\rm atoms\,+ . . . . . + 10^{7} \,\rm atoms\,$

with $2*10^{7}$ terms in the summation and you write this in "shorthand" as

$(10^{7} * 2*10^{7})\, \rm atoms = 2*10^{14} \,atoms$.

Farcher
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Atoms are not really a unit. They don't combine in the right way when you multiply. If you are careful, you can make them work in some situations, as both other answers show (+1 to both).

But as you saw, they don't work everywhere. Not like a length would. If the distance from atom to atom is 1 Angstrom, there is no problem with an area of $10^{14}$ Angstrom$^2$.

So what is the unit? A count is dimensionless.

Even so, people often will use it as a unit where it works. You just have to be careful not to use it where it doesn't.

Physicists are sometimes sloppy in ways like this where mathematicians are much more careful. For example, a function can have a value of $0$ most places, but have a tall thin spike near $0$, so the area under the spike is 1. This is useful in some situations. Physicists find they need the spike to be infinitely narrow. So they created the Dirac delta function, which is $0$ everywhere except at $0$. The value at $0$ is infinite. The area under the spike is $1$.

A mathematician would find problems with such a "function" and say it doesn't exist. A physicist is careful to use it where it works.

mmesser314
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    Could you please explain why you would say it doesn't exist. Or point to some relevant literature to help me understand that statement? – Griffin Sep 06 '17 at 14:39
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    @Griffin - mmesser314 is just giving a short overview of the mathematical problems with the Dirac delta function. And as is always the case with short overviews, it simplifies the matter so much that it is technically incorrect. However, it still illustrates the issue without getting bogged down in details. Mathematicians do define the function, but it requires either a more complicated concept of "function" or of "infinity" (two different approaches). In either case, a more complicated concept of integration is also needed. – Paul Sinclair Sep 06 '17 at 16:51
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    @Griffin The contradiction happens when we talk about the measure of a set, if a set has a measure of zero it does not contribute to an integral. A single point is the canonical set of zero measure. This is where the Delta function causes issues, because it is only non-zero at a single point the integral must be zero. As Paul mentions there are ways around this, but they all require us to alter the definitions at bit. – Ukko Sep 06 '17 at 17:41
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    Mentioning distributions would have helped make things technically correct, but would have obscured the point. Thanks to Paul Sinclair and Ukko for their clarifications. They have pointed out examples of mathematicians being careful. – mmesser314 Sep 06 '17 at 21:15
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Your units aren't atoms, but atom-bounding-widths; the square of atom-bounding-widths naturally comes in units of atom-bounding-areas.

Russell Borogove
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What is wrong with your calculation, is that you substituted a "count" for a dimension.
The correct way to do it, would be to determine the length of $10^7$ atoms. Assuming they take 1cm, then $1cm^2$ would have $1x10^{14}$ atoms, and the sheet (1 cm by 2 cm, or 2 $cm^2$) would contain ($1x10^{14}$ atoms/$cm^2$ x 2 $cm^2$ =) $2x10^{14}$ atoms.

Guill
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