Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am having trouble with notations like 1 cm$^{-3}$, especially since I am converting them between compound units.

Is there a way to express 1 cmcm$^{-3}$ without writing the negative exponent?

The particular queston I was asked was:

The EC standard for lead in the atmosphere is 2 µg/m$^{−3}$. Express this value in scientific notation in g/cm$^{−3}$, to an appropriate number of significant figures.

I have never run into these before, and I have serious problems coming to grips with the notation.

$1\cdot10 ^{ -3}$ would be fine, but 1 * cm$^{−3}$ just seems weird.

But, I simply have a hard time visualising these quantities. Can someone suggest a way to imagine them, or motivate the need for a negative exponent attached to a number of units?

What does the negative exponent actually mean?

Also, is there a term for these?

Typing into google gives.

1 (centimetre^(-3)) = 1 000 000 m^-3

Which is not helpful at all.

share|improve this question
    
Express two micrograms per cubic metre as a number of grams per cubic centimetre. –  Mark Bennet May 29 '12 at 12:38
    
So in particular what google is telling you is if you have $1$ unit (e.g. gram) per cubic centimetre, you have $1,000,000$ units per cubic metre, which is what you expect. –  Matt Pressland May 29 '12 at 12:46
6  
The notation $\mathrm{g}/\mathrm{cm}^{-3}$ does not mean 'gram per cubic centimeter' but 'gram times cubic centimiter'. Gram per cubic centimeter would be either $\mathrm{g}/\mathrm{cm}^3$ or $\mathrm{g}\,\mathrm{cm}^{-3}$. –  Egbert May 29 '12 at 13:04
3  
Egbert's right - either you've miscopied the question, or the question is nonsense. –  Gerry Myerson May 29 '12 at 13:20
add comment

2 Answers

up vote 7 down vote accepted

I believe the quoted passage contains an error.

The EC standard for lead in the atmosphere is 2 µg/m$^{−3}$. Express this value in scientific notation in g/cm$^{−3}$, to an appropriate number of significant figures.

It should either be 2 µg m$^{−3}$ or 2 µg/m$^3$, both of which have the same meaning, but not 2 µg/m$^{−3}$. Likewise, g/cm$^{−3}$ should either be g cm$^{−3}$ or g/cm$^3$. The negative exponent is just an alternative notation for division.

Think of µg/m$^3$ as "micrograms per cubic meter". This is a density. So a higher number means a greater density of lead. Visualize 2 µg of substance spread out over the volume of a cube with dimensions 1m $\times$ 1m $\times$ 1m.

Now as a warmup to doing unit conversions, ask yourself whether 2 µg/m$^3$ is a higher or lower density than 2 µg/cm$^3$. If you think about it, you will realize that the former represents 2 µg of substance spread out over a large cube with dimensions 1m $\times$ 1m $\times$ 1m, while the latter represents the same amount of substance spread out over a tiny cube with dimensions 1cm $\times$ 1cm $\times$ 1cm. So 2 µg/cm$^3$ is the higher density since the same amount of substance is crammed into a much smaller volume.

Now lets go back to the original density, 2 µg/m$^3$, and try to write it, first in units of µg/cm$^3$, and then in units of g/cm$^3$. A cube with dimensions 1m $\times$ 1m $\times$ 1m is the same as a cube with dimensions 100cm $\times$ 100cm $\times$ 100cm. Thinking about that cube as composed of many tiny 1cm $\times$ 1cm $\times$ 1cm cubes, how many of those tiny cubes are there? The answer is $100\times100\times100=100^3=10^6=1,000,000$. So if the big cube contains 2 µg of substance, which we imagine to be uniformly distributed, how much substance does each tiny cube contain? Since there are 1,000,000 tiny cubes, each contains $1/1000,000$ of the total amount of substance, or $2\times10^{-6}$ µg. Therefore, a density of 2 µg/m$^3$ is the same as a density of $2\times10^{-6}$ µg/cm$^3$. This insight can be encoded in the following algebraic manipulation: $$ 2\frac{{\mu}\text{g}}{\text{m}^3}=2\frac{{\mu}\text{g}}{\text{m}^3}\left(\frac{1\,\text{m}}{100\,\text{cm}}\right)^3=2\frac{{\mu}\text{g}}{\text{m}^3}\frac{\text{m}^3}{1,000,000\,\text{cm}^3}=\frac{2}{1,000,000}\frac{{\mu}\text{g}}{\text{cm}^3}. $$ In the first step we are free to multiply by (1 m/100 cm) since that's just a way of writing 1.

Now let's deal with the conversion from µg to g. One µg is a tiny amount of material - just 1 millionth of a gram. If you have 1 µg of material in a certain volume, that's the same as having 1/1,000,000 g of material in the same volume. This insight allows us to complete the calculation: $$ \frac{2}{1,000,000}\frac{{\mu}\text{g}}{\text{cm}^3}=\frac{2}{1,000,000}\frac{{\mu}\text{g}}{\text{cm}^3}\frac{\frac{1}{1,000,000}\text{g}}{\mu\text{g}}=\frac{2}{10^{12}}\frac{\text{g}}{\text{cm}^3}=2\times10^{-12}\frac{\text{g}}{\text{cm}^3}. $$ Again we are allowed to multiply by (1/1,000,000 g)/(1 µg) in the first step since that's just equal to 1.

share|improve this answer
2  
This is a great answer! +1 –  M Turgeon May 29 '12 at 15:31
add comment

You can treat units like ordinary factors.

For example

$2 \mu g/m^{−3}=2 \cdot \mu g/m^{−3}=2\cdot(0.000001g)/(100cm)^{-3}=2\cdot 0.000001/100^{-3}\cdot g/(cm)^{-3}$

Just do plain algebra and replace $m\to 100\cdot cm$ with no further thinking.

EDIT: there is no need to visualize this when you merely want to change units and there are much harder physical constants with more complicating units.

share|improve this answer
1  
I agree you don't have to visualise - but some people learn better when they can visualise things, until they become confident with the notation. I remember when I was ten having a problem understanding the mathematics of acceleration because I couldn't make immediate or intuitive sense of $s^{-2}$. When I did "get" it, it was easy. –  Mark Bennet May 29 '12 at 14:00
    
When trying to understand these simpler units, try taking them apart. For example $1ms^{-2}$, the unit of acceleration, means a change of velocity per time, i.e. $1ms^{-1}/s$. Velocity in turn is distance per time. –  Wormbo May 29 '12 at 16:23
    
I suggest one should visualize equations and not units (unless they are a/b fraction with no power). If you skip making mental essays about what fractions of unit powers mean, you have much more time to think about stuff that is really matters. Maths is made for providing tool so that you get results without having to visualize complicating stuff like high dimensional space. –  Gerenuk May 29 '12 at 17:29
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.