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I've been trying to understand how a ratio of two of the same units gives a dimensionless quantity. It makes no sense to me. For example, density in terms of mass and volume:

$$d = \frac{m}{V}$$

That I can interpret from the definition of division that on every unit of volume comes $n$ units of mass. But here's something that bugs me. If I wanted to discertain what is the mass in $5$ units of volume (whatever they are, laws of physics should be independent from the underlying units if they are correctly defined), that gives a ratio of $5V/1V$ and gives me a dimensionless number $5$.

So, that can be stated as: On every unit of volume there comes $5$... What? $5$ what? How can it be dimensionless? I understand that units have to be defined in terms of pointing a finger to something (such as a meter, they cannot be described as a singular number, that would make no sense). So, by assuming they are the same, we can state that is equal to 1, or that the value is dimensionless (unit of 1). But how to interpret that intuitively?

My primary question is: "So, that can be stated as: On every unit of volume there comes $5$... What? $5$ what?" If we drop the units, it seems to me that this can be replaced by anything that is dimensionless ($Pa/Pa$, $kg/kg$, $m^2/m^2$).

This seems to fall into dimensional analysis, but I can't create the tag.

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Density is not a dimensionless constant, as mass has different units than volume. –  Alex Becker May 18 '12 at 23:05
@AlexBecker That's not the point. In order to discertain how much mass goes into 5 units of volume, you have to multiply against the density and that introduces the volume over volume division which results in a dimensionless number. And that's what bugs me. –  ProblemFinder May 18 '12 at 23:07
Hence, this gives the question: "On every unit of volume, there comes $n$... What? $n$ what?" Because $n$ is dimensionless, it wouldn't be a problem if the math hadn't "killed off" the $V/V$ dimension –  ProblemFinder May 18 '12 at 23:09
No I agree with Alex. You are using improper terminology. Density is not dimensionaless. If you took the ratio of the area of a square to the are of the inscribed circle then you would have a dimensionless constant and where would be yuor problem? i don't think there is. Because if you doubled the sides of the square you would quadruple the area of the square and also the ares of the inscribed circle. The ratio is the same even though the scale of the 2-dimensional objects changed. Density is mass per unit volume. Suppose we have a rubber ball for radius 2 with uniform density. –  Michael Chernick May 18 '12 at 23:20
@MichaelChernick Who said that density is dimensionless? I used it as an example of something that makes sense (dimension preserving). And then I pointed to what doesn't make sense to me, that is the "cancel-out dimensions" in a ratio. Mr. Hardy is spot on what interested me. Thanks, all! –  ProblemFinder May 18 '12 at 23:26

1 Answer 1

up vote 2 down vote accepted

A dimensionless quantity has a numerical value that does not depend on which units of measurement are used.

If you say that in certain circumstances one volume will always be 5 times another, then the number 5 does not depend on what units of measurment you use. If change your units of measurement of volume from cubic feet to cubic yards, the number of units of volume will be only $(1/3)^3$ times what it was before. But the $5$ above in no way depends on what units of measurement are used.

If you're wonder how to express the price of widgets when you can get three of them for $\$6$, you've got $\$6/3 = \$2$ and you say they're $\$2$ apiece, and that comes with a unit of measurement. But if you ask how many widgets you can get for $\$6$ and it costs $\$2$ per gallon, then you've got $\$6/\$2$ and that's $3$ with _no "${\$}$". That makes sense since the $2$ doesn't depend on whether you're measuring the price in dollars or in cents or whatever. Hence dimensionless---no "$\$$".

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Thank you very much, Mr. Hardy. I thought there was more to it, I simply had to ask. –  ProblemFinder May 18 '12 at 23:28
My comment was similar to Michael Hardy's but he said it much better. –  Michael Chernick May 18 '12 at 23:37

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