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:

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


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

    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

Author: Dylan Holmes

Created: 2019-04-15 Mon 13:39

Emacs 26.1 (Org mode 8.3beta)