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All my life the approach has been as follows:

In math class I learn the rules and almost always deal with purely numerical problems.

In physics class I apply the things learned in math class but this time our quantities have units. Now, once the equation is well put and has dimensional homogeneity the problem is magically reduced to a purely numerical problem like the ones dealt with in math class. You might say that units do not actually vanish and it's actually just for practical reasons that teachers choose not to write them down, but I would like to see an explanation as to why this is OK? (Ignoring the units).

For instance, in my mathematical physics class I learn about vector calculus and vector analysis. First I learn it abstractly in a mathematically rigorous manner mostly dealing with no numbers because we derive the equations and theorems via the manipulation of letters. Now, I'm supposed to apply the exact same theorems and the same reasoning for problems that are dimensionful and this tears me apart for some unknown reason.

I need a profound reason or something that could prove to me that units are shouldn't bother me at all. I kind of see dimensionless and dimensional math as somehow separate, so I need a good argument that tells me this shouldn't be my way of seeing things.

I don't know if my question even makes sense, so thank you if you've read this far.

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  • $\begingroup$ When you multiply and divide numbers in $\mathbb{R}$, does the potential existence of $\mathbb{C}^n$ worry you? $\endgroup$
    – Emily
    Jan 17, 2015 at 1:49
  • $\begingroup$ Well, no. But are you saying dimensional stuff belong to the real numbers or something like that? I don't think I see your point. Thanks. $\endgroup$
    – ben ari
    Jan 17, 2015 at 1:52
  • $\begingroup$ The question at physics.stackexchange.com/questions/98241/… might be helpful. $\endgroup$
    – KCd
    Jan 17, 2015 at 2:22

6 Answers 6

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We can think of units in a formal sense: they can be multiplied and divided, but different units cannot be added. In one way, this is like working in a space of real numbers where each axis represents a power of a dimension, those axes are closed under addition, and that multiplication sends us to a different axis (i.e. you can do meters + meters, but not meters + meters-squared). So in a sense, you could form a basis from the fundamental units and their powers -- actually, this is pretty much what is happening in the Buckingham Pi theorem! Of course, we have a unitless basis element as well.

Nevertheless, it's not really worth worrying about. As I said, we needn't worry about $\mathbb{C}^5$ if we're trying to compute $22 \times 58$. So as far as profound reasons go, that's about as profound as I can make it -- the terms dimension and unit are used for precisely that reason. Operating on them allows us to work in a dimension with a unit basis where certain very-familiar rules hold.

Another analogy: if you walk 5 meters in one direction, you walk 5 meters. If you walk 5 meters in a different direction, you walk 5 meters. It may be eventually important to know whether you're walking east or south, but for the scope of simply wanting to know how far you've walked, we need not even care that east or south exist!

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Consider the set of $1\times1$ matrices with real entries. We know how to add these. $[a]+[b]=[a+b]$. We know how to multiply them. [a][b]=[ab]. We can do division, subtraction, exponentiation, etc. It turns out they behave just like the real numbers! It seems that the only difference is that we put some brackets around them. (Turns out they're isomorphic fields.)

Now what if I asked you to compute $5+[8]$? The question doesn't make sense. You're adding a real number to a matrix. We haven't defined a way to do that.

Units are a bit like the brackets on a $1\times1$ matrix; they're bits of notation that don't change the underlying logic. And just as it makes no sense to add a matrix to a scalar, it makes no sense to try to add $5m+8s$.

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  • $\begingroup$ Hi. I think an objection to the fact that "they behave like numbers" could be this: Consider drawing a position vector in R^3 how would you really know how to draw it if you don't know what [meters] is? The solution might be that [meters] is not any number at all. I don't know if this is stupid, but I thought it could be worth thinking about. $\endgroup$
    – ben ari
    Jan 24, 2015 at 1:17
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I expect it's involved (as well as counterproductive to dispelling your sense of unease) to give a general answer, and impossible to give a non-controversial account. But assuming I've understood your question, here's one way to think of things:

There's a mathematical notion of the real number system (a complete ordered field), whose properties are independent of measurements or physics (in the sense that mathematicians specify axioms and deduce properties; of course, the axioms themselves are motivated by physics/experience).

There are physical notions (length/spatial displacement along a particular direction, area, mass, duration, ...) corresponding to measurable quantities that, in an idealized sense, behave like real numbers, e.g., can be added/concatenated/agglomerated or compared; or under appropriate conditions can be multiplied, as when multiplying two orthogonal lengths to obtain an area.

A choice of units for a particular quantity is a "mapping" or "single-valued association" from a physical notion (e.g., a length) to a mathematical one (a real number). (A single ruler may be both $12$ inches and $30.48$ centimeters in length.)

Now, in order to get meaningful physical interpretations from formulas, the units associated to terms have to be chosen compatibly. After all, you cannot add $x$ inches to $y$ centimeters and get $(x + y)$ of some length unit independent of $x$ and $y$. (We could agree that $5$ inches plus $3$ centimeters is $8$ "bugs", but this "definition" of "bug" would depend on the particular addends, not just on the particular units; $5$ inches plus $4$ centimeters would not be $9$ bugs.)

Similar considerations of unit compatibility hold whenever you compute physical quantities using mathematical theorems. Mathematical operations correspond to physical operations when, and only when, units are chosen compatibly.

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You can view units as being transcendental elements adjoined to the field of real numbers. This is why you can't add two different units in the real-world unless some fundamental relationship is known. For example if we didn't know that energy is somehow related to mass, distance and time, but could measure amount of kinetic energy in some way, it would be utterly mysterious to attempt to add say kinetic energy of an object to $mgh$ where $m$ is its mass, $g$ is the acceleration due to gravity and $h$ is the height above ground (for small $h$), but once we figure out that these two are somewhat the same thing up to some constant factor, then it will make sense to be able to add them together. In fact we have defined $\text{J} = \text{kg m$^2$ s$^{-2}$}$ today.

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If you wish, you can regard all physical statements to be only about numbers and units only to be a means to formulate the statements, which do not relate to the mathematics involved. For example, we could reformulate

If we weld a wire of length $x$ to a wire of length $y$, the resulting wire has length $x+y$. (1A)

such that it does not use any arithmetics involving units as follows:

If we have one wire of length $x∈ℝ^+$ in reference to some unit length and another wire of length $y∈ℝ^+$ in reference to the same unit length, and we weld those wires together, the resulting wire has the length $x+y$ in reference to that unit length. (1B)

 In another example, we could reformulate:

If a wire has a lenth $x$ at temperature $t$, there is a constant $α$ such that its length at any temperature $s$, the wire has a length of $x·(1+α(s-t))$. (2A)

A wire has a length $x∈ℝ^+$ in reference to some unit length at a temperature $t∈ℝ^+$ in reference to some unit temperature. Then there exists an $α∈ℝ$ (which depends on the unit length and unit temperature, but not on $x$, $s$ or $t$) such that at any temperature $s∈ℝ^+$ in reference to the aforementioned unit temperature, the wire has a length $x·(1+α(s-t))$ in reference to the above. (2B)

Of course the above only works for reasonable definitions of “a wire has length $x∈ℝ^+$ in reference to some unit length” and similar, which complies with the arithmetic of real numbers. For example, we need:

If a length $L$ is $x∈ℝ^+$ in reference to some unit length $A$ and $y∈ℝ^+$ in reference to some unit length $B$, then the length $A$ is $x/y$ in reference to the unit length $B$.

Without this, the statement 1B would not be universally true for all unit lengths.

That such definitions exist is an extremely well established empirical fact, but nothing more. (Historically it’s of course rather that the real numbers were abstractions of the arithmetic properties of physical values and one could take the point of view that mathematics is justified by the very well established empirical observation that the axioms of some prominent algebraic structures comply with reality.)

Making use of this observation, we can introduce an arithmetics for physical values (i.e., real numbers equipped with a unit) and, by equipping constants such as $α$ in 2B to appropriate units, we can arrive at the usual formulation of physics which uses these arithmetics as a notational shortcut, if you so wish.

If we moreover only use units that are expressed in terms of some algebraically independent base units (such as the SI base units), we can now omit writing down the units and can deduce the unit of our results from their dimension. E.g., if we strictly work in the SI system, we know that a velocity has the unit $\frac{\text{m}}{\text{s}}$ (as this is the only unit for velocities in this system). This is prominently done in high-energy physics, where usually $\hbar$, $c$ and $\text{MeV}$ are used as base units and $\hbar$ and $c$ are omitted (which is usually written as the cringeworthy $\hbar=c=1$).

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If we have a dimensionful equation like $$s = vt$$ we can make it dimensionless by dividing by an appropriate unit: \begin{eqnarray*} s &=& vt \,\,\,\,\,\,|:\text{m}\\ \frac{s}{\text{m}} &=& \frac{vt}{\text{m}}\\ \frac{s}{\text{m}} &=& \frac{v}{\frac{\text{m}}{\text{s}}} \frac{t}{\text{s}}\\ s^* &=& v^*t^*\\ \end{eqnarray*} The quantities $s^*$, $v^*$ and $t^*$ now are dimensionless numbers. $s^*$, for example, is a certain number of meters, not a physical quantity with the dimension of length like $s$.

If you want to make a numerical calculation with a computer, you are implicitly using the latter form of the equation.

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