# SI units and prefixed points

Question. A liter is smaller than a cubic meter. But a milli-liter is larger than a cubic milli-meter — the prefix changes which one is larger. Is there a prefix which makes them equal? In other words, can you find a prefix "pre-" such that one pre-liter is equal to one cubic pre-meter?

I found this puzzle difficult to even set up at first, because I'm still learning how to think about unit conversions. I solved it eventually, though, and found some interesting algebraic insights. The general solution follows.

General solution.

1. In the general setup, you have units A (e.g. liters) and B (e.g. meters).

2. The units may have different dimensions (e.g. volume and length), so they need dimensional exponents to make them commensurate. Call the dimensional exponents r and s; they ensure that Ar is commensurate with Bs. (e.g. liters L1 are commensurate with cubic meters m3).

3. With commensurate units, there's now some constant conversion factor. Say that $$a$$ units of Ar equal $$b$$ units of Bs . (For example, there are 1000 liters per one cubic meter.) The conversion equation is:

$$a \text{A}^r = b \text{B}^s$$

4. We're looking for a prefix which makes the two quantities equal. We can think of prefix in terms of powers of 10, as follows:

exponent $$k$$ number $$10^k$$ named prefix
-9 0.000000001 nano
-6 0.000001 micro
-3 0.001 milli
-2 0.01 centi
-1 0.1 deci
0 1 (prefix-less)
1 10 deca
2 100 hecto
3 1000 kilo
6 1000000 mega
9 1000000000 giga

For some problems, the prefix we're looking for might not exist in this table. For example, it might be a fractional number instead of a whole number.

When we don't have a name for a prefix $$k$$, we can just use $$10^k$$ as a placeholder name. For example, a 103-meter is a kilometer, and a 104-meter is ten kilometers.

5. The proper setup for the prefixed-point problem is to find the prefix (exponent) $$k$$ which makes the following two quantities equal:

$$(10^k \text{A})^r = (10^k \text{B})^s$$

Note that the dimensional exponents are placed on the outside so that, for example, we have "cubic micrometers" rather than "micro cubic-meters"

6. Shuffling around terms, we have:

$$10^{k(r-s)} = \text{B}^s / \text{A}^r$$

And based on the conversion equation, we know that the right-hand ratio of one Bs to one Ar is $$a/b$$:

$$10^{k(r-s)} = a/b$$

Taking the logarithm of both sides:

$$k(r-s) = \log_{10}(a) - \log_{10}(b)$$

or

$$k = \frac{\log_{10}(a) - \log_{10}(b)}{r-s}.$$

In slogan form, if you like, the formula for finding $$k$$ is "The difference between the logs of the conversion coefficients, divided by the difference between the dimensional exponents."

7. For example, in the original problem we had A (liters), B (meters), dimensional exponents r=1 and s=3, conversion coefficients a=1000, b=1. Hence the prefixed point is:

$$k = \frac{\log_{10}(1000) - \log_{10}(1)}{1 - 3} = \frac{3-0}{-2} = \;-\frac{3}{2}$$

This is a prefix of -1.5; situated halfway between the "deci" (-1) and "centi" (-2) prefixes. You can confirm that one 10-1.5-liter is equal to one cubic 10-1.5-meter. As a prefix name, I suggest sesqui, as in sesquimeter and sesquiliter. (Or a variant of sesqui- to indicate the negative sign.)

8. When you apply a prefix to the units A and B, the difference in dimensionality causes them to scale at different rates. This property is what allows for there to be a prefixed point at all: because of that difference, the prefix k multiplies both sides by a different amount. As k becomes large, the unit with larger dimensionality gets much larger. As k becomes small, the unit with larger dimensionality gets much smaller. And they meet somewhere in between.

To see this disparate scaling, note that a liter is only 10001 (one thousand) times larger than a milliliter, but a cubic meter is 10003 (one billion) times larger than a cubic millimeter.

If the units A and B have different dimensional exponents, you can always find a unique solution for $$k$$. On the other hand, if the units A and B have the same dimensional exponents (for example, inches and meters), then they will remain at a fixed ratio regardless of what prefix is added. If they were different without prefixes, they will be different after scaling by prefix. There will be no solution for $$k$$, and correspondingly the formula contains division by zero.

9. Actually even if two different units have the same dimensional exponents, we can find a solution for $$k$$ as long as we allow infinities. We can define the omega prefix (±∞) which makes the two units equal. The omega prefix corresponds to 10+∞ (or 10-∞, or both, log(0/0), whatever). For example, one omega-mL is equal to one omega-L. This is because any two infinite piles of commensurate stuff are equal. Similarly any two empty piles of commensurate stuff are equal.

10. If we want to have a different prefix for positive and negative infinities, we can borrow Greek letters and refer to them as, respectively, omega- and omicro- (for example omega-meter and omicro-meter).

Date: 2019/Apr/14

Created: 2019-04-15 Mon 13:39

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