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I'm having trouble understanding the rules for dimensional and dimensionless constants. In dimensional analysis, you can only add or subtract quantities with the same dimension. For example, if $f=12t+2t^2$, then $12=\frac{[f]}{[t]}$ and $2=\frac{[f]}{[t]^2}$ are dimensional constants.

However, another rule I've come across is that the argument of things like $f(x)=e^{x}$ must be dimensionless, because $e^x=1+x+\frac{x^2}{2!}+...$, and you can't add $[x]$ to $[x]^2$ and so on. But using the previous logic with dimensional constants, can't we let $x$ have dimensions $[x]$, and just demand that $1=[x]$, $[x]=[x]$, $[\frac{1}{2}]=\frac{1}{[x]}$, etc? That would seem to indicate $f(x)=e^x$ does have dimensions?

Similarly, if you want to use dimensional analysis to check whether your solution is correct, can't you just arbitrarily assign units to a constant to make your equation consistent? For example, $A=2\pi r^3+2\pi r^2h$ where $[A]=ft^2$ and $[r]=[h]=ft$. The right hand side has dimensions $ft^3$, but the left has dimensions $ft^2$, so this should indicate the equation is not correct. But couldn't you just say, well, I'll make $[2\pi]=\frac{1}{ft}$ so the dimensions on both sides are consistent?

Qmechanic
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    Your statement $12=[f]/[t]$ makes no sense. $12$ is a number, it cannot have dimensions. Consequently, your equation $f=12t+2t^2$ also does not make sense. Similarly, $1/2$ does not have dimensions and $2\pi$ does not have dimensions. They are numbers. – Prahar Jan 20 '24 at 08:40
  • @Prahar If you consider g=9.81, this is a fixed constant with units m/s^2. – user124910 Jan 20 '24 at 09:20
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    No, $g=9.81~\text{m/s}^2$ is a constant with units $\text{m/s}^2$. On the other hand, $g=9.81$ is also a constant with NO units. – Prahar Jan 20 '24 at 09:22

5 Answers5

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Pure numbers like 1, 2, π, etc. always are “lifted” into the algebra as dimensionless.

So to be formal, the dimensional algebra concerns items which are a pair of a real number $\mathbb R$ and seven rational numbers $\mathbb Q^7$. (At least in SI units. There's no formal reason it had to be seven.)

A constant like the speed of light $c$ is thus the pair $(299~792~458, (0,1,-1,0,0,0,0)).$ Meanwhile the fundamental charge $e$ is the pair $(1.602~176~634\times10^{-19},(0,0,1,0,1,0,0))$.

In terms of algebra, we define that we can add $(r_1,u_1)+(r_2,u_2) = (r_1+r_2, u_1),$ if and only if $u_1=u_2$. This means that $c + e$ is a type error, those numbers did not have the same units and were not eligible to be added. Meanwhile we are much more free to multiply, $(r_1,u_1)\cdot(r_2,u_2) = (r_1\cdot r_2, u_1+u_2),$ and so $c\cdot e$ is fully allowed and is approximately $(4.803~204~71\times10^{-11},(0,1,0,0,1,0,0))$.

Using these definitions, some other things “lift” into the algebra. This just means that those things can be easily defined in a way that works intuitively as if they were their former selves. So the real numbers “lift” as $r \mapsto (r, (0,0,0,0,0,0,0)),$ an easy embedding , and then this is intuitive because with these rules $3c$ is straightforward multiplication while $3+c$ is another type error, and one straightforwardly has things like $c+3c=4c$ without caring much about it.

One useful lift is division, this inherits the problems of division by zero.

Another useful lift is all rational powers, $(r, u)^q = (r^q,q~u).$ However, this cannot be extended to a dimensionful exponent in any easy ways, nor to a real exponent (at least, not without switching from $\mathbb Q^7$ to $\mathbb R^7$). So $\sqrt{c}$ is well-defined but $c^\pi$ and $c^c$ are type errors.

Some other useful constants are:$$\begin{align}\text{kg} &= (1,(1,0,0,0,0,0,0))\\ \text{m} &= (1,(0,1,0,0,0,0,0))\\ \text{s} &= (1,(0,0,1,0,0,0,0))\\ \text{K} &= (1,(0,0,0,1,0,0,0))\\ \text{C} &= (1,(0,0,1,0,1,0,0))\\ \text{mol} &= (1,(0,0,0,0,0,1,0))\\ \text{cd} &= (1,(0,0,0,0,0,0,1)).\\ \end{align}$$ These just make things much easier to read, like $$ \sqrt{c}\approx17314.5 ~~\text{m}^{1/2}/\text{s}^{1/2}.$$

And finally, the thing that you can't lift—the arbitrary function. We have a standard procedure which we use to lift functions to other algebraic domains. So for example $e^M$ where $M$ is a square matrix, we define this using the standard power series of $e^x$, because we have addit and multiplication among square matrices. But because you get a sum of terms (pure number) times $x^n$, and the pure number is dimensionless, the only way to avoid the type error is if $x$ is also dimensionless.

This means that the only arbitrary functions in the dimensional algebra have the form $y=U~f(x_1, x_2, x_3, \dots)$ where $U$ is some multiplication that has the right units ($y=(r_1,q)$ and $U=(r_2, q)$ for the same $q$), and $x_1, \dots$ are all unitless multiplications ($q=(0,\dots,0)$). If you like this can be thought of as the “fundamental theorem of dimensional analysis,” you write out all of the dimensional constants and initial conditions that could possibly pertain to the problem, you write out all of the dimensionless constants you can make with them $x_1, \dots$, and then when you run into a problem that requires a force as its answer, say, you use any combination of dimensional constants to get a force $(1,1,-2,0,0,0,0)$ and multiply by an arbitrary function of the $x_i,$ this is the most general possible expression. This procedure thereby non-dimensionalizes the problem, you have $y/U$ as a pure number and the $x_i$s as pure numbers.

CR Drost
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    Ah, that's a nice way to think of it. Yes, it seems similar to what Prahar is saying, I was mistaking numbers (projections in $\mathbb{R}\oplus\mathbb{Q}^n$ onto the first coordinate, or $\mathbb{R}\oplus (0,...,0)$) with constants (elements of $\mathbb{R}\oplus\mathbb{Q}^n$). – user124910 Jan 20 '24 at 09:56
  • Yep! Added a quick last paragraph because actually when you make this confusion very deliberately, the dimensional algebra steps entirely out of your way and gives you your physics-problem with just pure real numbers, which is always nice. – CR Drost Jan 20 '24 at 18:48
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The confusion is arising due to the sloppiness of notation that physicists love to employ (myself included).

Suppose we start with an equation of the form (I am taking @JohnRennie's example) $$ s = u t + \frac{1}{2} a t^2 $$ and we have $u=12$ m/s and $a=4$ m/s$^2$. The CORRECT way to proceed is then to write $$ \tag{1} s = (12~\text{m/s})\, t + (2~\text{m/s$^2$})\, t^2 $$ Of course, repeatedly writing out all the units explicitly is quite tedious. We therefore become a bit SLOPPY and instead write $$ s = 12 t + 2 t^2 \tag{2} $$ However, we should always remember that (1) is the actual correct way to write the equation, and (2) is just sloppy notation.

That being said, it is important to note that it is OK to use sloppy notation (we are not mathematicians) as long as there is no confusion. For instance, in equation (2), we know that $[s]=\text{m}$ and we know that $[t]=\text{s}$, so we can immediately deduce that the 12 actually means 12 m/s and the 2 means 2 m/s$^2$.

It is NOT OK to use sloppy notation when there can be confusion. For instance, if we set $s=10$ m in (2) and blindly drop all the units, we get $$ 10=12t+2t^2 \tag{3} $$ Now, there is confusion regarding the units of this equation. Given (3) by itself, there is no way for us to figure out what units we should assign to 10, 12, and 2.


To summarize: The CORRECT way of writing physics equations is (1). Despite its sloppiness, (2) is OK because given (2), we can deduce (1). (3) is not OK.

Prahar
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I guess you are thinking of equations like:

$$ s = ut + \tfrac12at^2 $$

So when you write:

$$ f=12t+2t^2 $$

you are saying $u = 12~\textrm{m/s}$ and $a=4~\textrm{m/s}^2$. The point is that the equation is defined as having dimensionful constants in it and the constants only look dimensionless because you are replacing them by their numeric values.

But in the expansion of $e^x$ the constants are all defined as dimensionless and you cannot just casually redefine them to have dimensions.

John Rennie
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  • How do you determine when an equation is defined as having dimensional constants? To me, it seems like the condition "f has dimensions [f], and t has dimensions [t]" forces the the constants to have units. I would say, f and t have 'defined' dimensions, and the dimensions of the constants is a consequence. Shouldn't it work that way with the expansion of $e^x$? Define dimensions for $e^x$ and $x$, and the dimensions of the constants is a result. Could you elaborate a bit more on what you mean by the constants in $e^x$ are defined as dimensionless? – user124910 Jan 20 '24 at 09:19
  • To be clear (for the OP), I am guessing what you meant is that it is simply not correct to write $f = 12 t + 2 t^2$. What one should write is $f = (12\text{m/s})t + (2\text{m/s$^2$})t^2$. This is the only correct version of that equation. – Prahar Jan 20 '24 at 09:20
  • @user124910 - constants are not generically numbers. Constants can have units. Numbers cannot have units. $12~\text{m/s}$ is a constant. It has a unit. $12$ is also a constant. It has no unit. – Prahar Jan 20 '24 at 09:20
  • That seems more reasonable to me. I came across this question https://math.stackexchange.com/questions/3261593/second-derivative-and-dimensional-analysis – user124910 Jan 20 '24 at 09:27
  • In it, they are asked to do essentially the same thing. Find the dimension of two different 16's, where the consensus seems to be that [16]=[f]/[t] for the first 16, but you're saying it should be understood as [16 [f]/[t]]=[f]/[t]? – user124910 Jan 20 '24 at 09:32
  • yes, that's what I'm saying. – Prahar Jan 20 '24 at 09:44
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Dimensional analysis is applied to formulae before values are substituted for terms. To borrow an example from another answer, the expression

$f = 12t + 2t^2$

cannot be dimensionally consistent unless one or both of the coefficients $12$ and $2$ actually has dimensions because it is a value substituted for another term. However, when we see formula such as

$A=2\pi r^3+2\pi r^2h$

then we can assume that two occurrences of $2\pi$ are dimensionless constants in the formula, so if $R$ and $h$ have dimension $L$ and $A$ has dimensions $L^3$ then the formula is dimensionally consistent.

gandalf61
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Rather than use dimensions I will choose a system of units.

Here is one way of looking at an equation.

$s = ut + \tfrac12at^2$ is an equation which involves numbers but all the numbers must refer to the same thing, ie when counting you cannot add oranges to apples to must stick either to counting all oranges or all apples.

Thus, $s$ in you equation is a number which tells you how many times a length is greater than a chosen unit of length, say $1\,\rm metre$, ie $[s]\times 1\,{\rm m}$.

Looking at your first equation,
$([s]\times 1\,{\rm m}) = ([u]\times 1\,{\rm m\,s^{-1}})\times ([t] \times 1\,{\rm s})+ \left(\frac 12 \right )\times ([a]\times 1\,{\rm m\,s^{-2}}) \times ([t^2]\times 1\,{\rm s^2})$
$\Rightarrow ([s]\times 1\,{\rm m}) = ([v\,t][\times 1\,{\rm m})+([\frac12\,a\,t^2]\times 1\,{\rm m})$

The $\frac 12$ has no units, it is a pure number which in this case multiplies a length measured in metres, ie halves the number of metres evaluated from $a\,t^2$.

With $e^x$ the index $x$ has no units/dimensions.
For example, $x= t/\tau$ where $t$ is a time measured in seconds and $\tau$ is a time constant measured in seconds.

In you example $A=2\pi r^3+2\pi r^2h$, a definition of $\pi$ is from a ratio of two lengths and so it is a dimensionless / unit less constant and cannot get away from that with $A,\,r^3$ and $r^2\,h$ having the same dimensions/units.

Note that another way of writing equations is instead of $([s]\times 1\,{\rm m})$, using $(\dfrac{s}{1\,{\rm m}})$ or $(\dfrac{s}{\rm m})$ or $s/\rm m$.

$\dfrac {s}{\rm m} = \dfrac {v}{\rm m\,s^{-1}} \cdot\dfrac {t}{\rm s} +\dfrac {1/2}{1}\cdot\dfrac {a}{\rm m\,s^{-2}}\cdot\dfrac {t^2}{\rm s^{2}}$ and many people do this to check the consistency of their calculation.

More on Rules and Style Conventions for Expressing Values of Quantities.

Related - Does a squared unit require squaring the value?

Farcher
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