Does really m^3/m yields m^2?

Question

Apart the evidence that such assumption works in physics formulae, how can someone make a statement like that? Maybe the formuale works just because are worked around that assumption?

Why phyicist do not object something so apparentily simple that could actually be wrong?

To measure a meter you use a reference sample (say a stick).

To measure a cube meter you use a reference sample (a mini cube).

What you actually see is that ther's a relation between number of sticks and number of cubes, but you have no evidence other than that that things would not work otherwise. It seems that actual measure system is just the simplest enumerable set that explain our observations.

The most interesting point comes from a comment on this question

Just saw this one. Interesting. But beware: Who says time and space are the deepest axioms? I can construct a program in which space, at least, is an emergent property of rules that restrict how a globally accessible set of points interact with each other. That sounds pretty abstract, but if you look carefully at how quantum mechanics works, there are some uncomfortable similarities in it to just that sort of situation. Also, defining time when you are always in "now" requires you to model of the past, not touch it directly. That implies intelligence just to define t meaningfully.

Answer 1

It is correct if not equivalent in most contexts. For example, if you have a cuboid tank for water, $m^3$ would be the unit used to denote its volume, $m$ the depth of water. Then, when you try to denote the ratio of volume of water to depth with the unit $m^3/m$, you suddenly realize what you are trying to mean is equivalent to its horizontal cross-sectional area, which has the unit $m^2$.

In another example in meteorology, precipitation is measured in $mm$, the rate $mm/a$. By converting the $a$ (annum or year) into SI, you will find that the rate of precipitation has the unit of velocity. This at first glance seems rather odd, and it may seem illogical to shorten $mm^3/(mm^2*a)$ to $mm/a$. However, further thought will reveal that indeed the rate of precipitation is equivalent to a speed quantity, that is the average speed at which the water level rises as a result of precipitation, not accounting for evaporation. This is true no matter the surface area of the water body, as long as the water body remains still and waveless. Therefore, yet again the shortened unit is in fact meaningfully equivalent to another quantity, therefore such usage will not pose any problem.

Now to show a bad example, where such shortening of units may be frowned upon. Lets say you have a spring, and for each meter you extend the spring you will be rewarded $1 m^3$ of soda. In this case, the reward ratio of the game would have the unit $m^3/m$, and turning it into $m^2$ would be nonsense if not esoteric. In this case, it might be useful to add subscripts to differentiate between units: $m^3_{drink}/m_{spring} $ though I have only seen this notation once before, in a McGraw Hill chemistry textbook where each $mol$ confusingly referred to the amount of a different chemical.

All I have talked about above only matters if you care about the meaning of units. If not, then such shortening is always correct, since it would then serve only the purpose of ensuring dimension and magnitude consistency. However, I personally prefer to emphasize the meaning and significance of units, since it helps me make sense of what I am calculating.

Answer 2

Ever since Einstein formulated relativity we describe spacetime as a manifold equipped with a metric. Then we can use all the machinery of differential geometry to calculate properties of our spacetime. In particular, volume is defined using the volume element, $dV$, and we get volumes by integrating $dV$.

For a flat space with Cartesian coordinates the volume element is:

$$ dV = dx\,dy\,dz $$

and I hope it's obvious that this leads to the conclusion that volume of a cuboid is the area of one of it's sides times the length of the edge at right angles to that face. Hence we find $l^3/l = l^2$.

All this is by definition, and it's an adequate answer if spacetime really is a manifold equipped with a metric i.e. if a measured volume really is the same as the volume defined above. Presumably your question is asking if this is the case. If so there is no answer. The only way we have of testing a mathematical model of reality is to compare the predictions of the model with experiment, and so far experiment has failed to disprove the model of spacetime as a manifold plus metric.

For completeness we should note that in the real universe, i.e. curved spacetime, the volume element is given by:

$$ dV = \sqrt{g} dx\,dy\,dz $$

where the metric tensor $g$ is a function of $x$, $y$ and $z$. That means in curved spacetime the area of a cube is not its volume divided by the length of an edge.

Answer 3

$${m^3\over m} = {m^3\over m^1} = m^{3-1}=m^2$$

Units work the same way as algebraic quantities, at least with integer powers. A volume divided by a length will give something with the same dimensions as an area.