In dimensionless analysis, coefficients of quantities which have the same unit for numerator and denominator are said to be dimensionless. I feel the word dimensionless is actually wrong and should be replaced by "of dimension number". For example, the Mach number is of dimension one.

Many people write, for this case:

Mach-Number | Dimension: "-" | Unit: "1"

As mentioned before, I would say 'Dimension: "1"' in this place. But what about the unit? $\text m/\text s$ divided by $\text m/\text s$ is equal to one. But is the number one a unit by definition? Or should one say that the Mach number has no unit and therefore 'Unit: "-"'?

  • 1
    $\begingroup$ Related answer: physics.stackexchange.com/a/60007/17609 $\endgroup$ – Keep these mind Sep 17 '13 at 14:03
  • 1
    $\begingroup$ You can think of an expression such as 9.8 m/s2 as being an element of a certain group. The elements of this group can be thought of as ordered pairs $(n,u)$, where $n$ is a real number and $u$ is an element of a group representing the units. In the group that $u$ belongs to, there has to be an identity element. $\endgroup$ – user4552 Sep 17 '13 at 15:48

This is analogous to the definition of an empty product in mathematics. For a finite non-empty set $S=\{s_1,\ldots,s_n\}$, the product over $S$ can be defined as $$\prod_{s\in S}s=s_1\times \cdots\times s_n.$$ For such a product you'd want disjoint unions to map into products: if $R\cap S=\emptyset$, then you want $\prod_{x\in R\cup S}x=\left(\prod_{s\in S}s\right) \times \left(\prod_{r\in R}r\right)$, but for this to make sense you want to be able to handle the empty set, and the only way to make the rules consistent is to set $$\prod_{s\in\emptyset}s=1.$$ This essentially says: if there's nothing to multiply, the result is one. (Similarly, empty sums are defined to be zero, for the same reason.) In the case in hand, you could simply say if there are no units to multiply, then you get one. As Luboš points out, this is the harmless only consistent choice, as multiplying by one does not change the quantity.

Moreover, this empty-product intuition can be carried out to a full formalization of physical dimensions and units as a vector space. The whole works is in this answer of mine, but the essential idea is that positive physical quantities form a vector space over the rationals, where "addition" is multiplication of two quantities and "scalar multiplication" is raising the quantity to a rational power. This vector-space formalism is precisely the reason why dimensional analysis often boils down to a set of linear equations. Moreover, in this vector space the 'zero' is the physical quantity and unit $1$ - neither vector space makes sense unless $1$ is both a quantity and a unit.

Ultimately, of course, it boils down to convention, so people can just say "I'm going to do this in this other way" and they won't be "wrong" as such. However, in general, the consistent way to assign things is to say that dimensionless quantities have dimension $1$ (modulo whatever square bracket convention you're using) and unit $1$.

To back this up a bit, for those that care about organizational guidance, the BIPM publishes the International Vocabulary of Metrology, which states (§1.8, note 1) that

The term "dimensionless quantity" is commonly used and is kept here for historical reasons. It stems from the fact that all exponents are zero in the symbolic representation of the dimension for such quantities. The term "quantity of dimension one" reflects the convention in which the symbolic representation of the dimension for such quantities is the symbol 1 (see ISO 31-0:1992, 2.2.6).

This is essentially the same in the ISO document, which has been superseded by ISO 80000-3:2009 (paywalled, but free preview available), which has an essentially identical entry in §3.8.

Finally, and as a response to some of the comments by Luboš Motl, this applies to the term "physical dimension" as understood by the majority of physical scientists.

There is also an alternative convention, used in high-energy contexts where you work in natural units with $\hbar=c=1$, in which you're left with a single nontrivial dimension, usually taken to be mass (=energy). In that context, it is usual to say a quantity or operator has "dimension $N$" to mean that it has mass dimension $N$ i.e. it has physical dimension $m^N$, but since there's only ever mass as the base quantity it often gets dropped. However, this is very much a corner case with respect to the rest of physical science, and high-energy theorists are remiss if they forget that their "dimension $N$" only works in natural units, which are useless outside of their small domain.

  • $\begingroup$ Dear @Larifari, dimension 0 operators are standard things. In this terminology "dimension N", N is the exponent, usually above the mass. See e.g. these dozens of papers explicitly containing "dimension 0 operator": scholar.google.com/… $\endgroup$ – Luboš Motl Sep 18 '13 at 14:39
  • $\begingroup$ @LubošMotl Yes, but that is in natural units with $\hbar=c=1$, and you only have a single dimension, mass ($M$). In that case you can, as a shorthand, drop the M^ when you have dimension $M^N$ operators, and call them dimension $N$, but "dimension 0" operators really have dimension $M^0=1$. This doesn't extend to non-natural units, unless you're willing to have "dimension (1,1,-2)" quantities (i.e. force). $\endgroup$ – Emilio Pisanty Sep 18 '13 at 14:57
  • 1
    $\begingroup$ Well, @Emilio, "your" convention for giving the units their name also changes when you switch to different units, doesn't it? For example, in Planck units, everything becomes "dimension 1" in your language, i.e. dimensionless in the normal physicist's terminology. As long as one understands dimensional analysis and logarithms, this is a purely terminological debate and "dimension 1" for dimensionless is very unusual among physicists while "dimension N operators" for the exponent is an important part of the modern physics terminology. $\endgroup$ – Luboš Motl Sep 20 '13 at 4:48
  • $\begingroup$ @Luboš Heads up on an edit of this answer. Indeed the "name" for units would change in different units, in the same sense that $1\:\rm C = 1\: A\:s$ - it's just different coordinate representations of the same quantity, and not at all a problem. Saying "dimension $1$" for dimensionless is uncommon because there's often no call for it, but if it's in a context where consistency matters (like a table of quantities with their dimensions) then it's the only consistent choice. "Dimension $N$ operators", on the other hand, do matter in HEP theory but they don't matter anywhere else. $\endgroup$ – Emilio Pisanty Aug 29 '16 at 12:08

The number $1$ may be linguistically described as "unity". This very number is the original source of various words in the terminology, like the "unit matrix" (a matrix behaving like the number $1$).

It is a convention to write down that dimensionless quantities like the Mach number have units $1$ because the multiplication by $1$ changes nothing about the result – this is the counterpart of the multiplication by another unit like ${\rm m/s}$.

It just looks more coherent to write the unit $1$ into the tables. But verbally, one may also say that quantities with "this unit" have no units whatsoever. They are dimensionless. As long as one understands the logic, there's no problem in following these somewhat inconsistent conventions in which we sometimes say the units to be $1$ and sometimes we say that the units aren't there.

In the tables, the "unit" column means "the ratio of the full quantity and its numerical value". With this definition, the result may be calculated as $1$ without any problems. It's similar to the task to compute budget deficits as the difference of revenue and expenses. If the latter two are equal, the difference is just $0$. One may write $0$ although he could also write it as $-$ and say that the difference "doesn't exist". The numbers $0$ and $1$ play the role of the "neutral objects" for addition and multiplication, respectively.

  • $\begingroup$ Thank you, actually, the word dimensionless is false by definition of the big standards. So I think, "1" should be the choice for both cases. Mach-Number, Dimension: "1"; Unit: "1". I will wait a little time, if someone has a prove. Thank you though, for an interesting answer. $\endgroup$ – LaRiFaRi Sep 17 '13 at 13:20
  • 5
    $\begingroup$ Hi LaRiFaRi, the word dimensionless means that the units of the quantity are $1$. There is nothing wrong about the word and the word is important. It's the same like saying that the budget is without a deficit if the deficit is zero. A zero deficit may be said to exist: the number zero exists - but equivalently, we may say that it doesn't exist because "zero" means that it is not there. For multiplication, 1 plays exactly the same role. $\endgroup$ – Luboš Motl Sep 17 '13 at 13:26

DIN EN ISO 80000-1:2012-10 on p. 26

Chapter 6.5.5: The Unit One

Translated from German by me:

The coherent SI-unit for every value of dimension number is the unit one, symbol 1. This unit is, in general, not written, if such a value is given by its number. E.g. number of turns in a coil $N = 25 \cdot 1 = 25$

Translated and summarized by me:

Special namings [I think that would be rad, degree, ...] for the unit one may be combined with SI-prefixes. The symbols % and pro-mil are part of the coherent unit one. The unit symbol 1 itself should not be combined with prefixes but then be written as powers of 10.

The standard recommends, not to use the word dimensionless or dimension 1 but "dimension number". The other definitions are "outdated". As an example, the amount of substance 5 mmol/mol is given, where the value is 5, the unit 1 and the dimension the number 0.001.

  • $\begingroup$ 1) Please quote carefully. Your answer (v1) seems to contain more quote than actually marked as such. Use > before each paragraph. If you want to place a comment of your own in a quote, use [ ], not ( ). 2) Is the original text freely available on the Internet? Then please provide a link. 3) What is the status of this text? Is it a proposal/a draft/ratified/accepted/etc.? $\endgroup$ – Keep these mind Sep 17 '13 at 18:16
  • $\begingroup$ Thanks for your tips. Corrected that. The source is not free. The original belongs to me and I am not allowed to upload that. It is a draft version of an already existing standard. The mentioned chapter is not marked as new though. (changes are discussed in the beginning of this paper) $\endgroup$ – LaRiFaRi Sep 17 '13 at 19:27

If $6$ and $7$ each have units $1$, and if this unit behaves like other units, then $6\times 7=42$ would have units $1^2$. My height is measured in metres, but half my height is in $\mathrm{m}1^{-1}$. If you multiply this by two you get back my height in metres, but if you add it to itself you get a quantity that's numerically equal to my height, but has units $\mathrm{m}1^{-1}$.

I think it would be rather difficult to make this consistent, so we have to conclude that if $1$ is a unit, it behaves rather differently from other units. And if that's so, why should we call it a unit at all?

To clarify: I do actually think it's fine to say the units of a quantity are $1$. This answer is only referring to the part of the question that says "is the number $1$ a unit by definition", which I took to mean "is it a base unit, like $\mathrm{m}$ or $\mathrm{K}$?"

  • $\begingroup$ hehe, interesting post. But the last sentence "1=1/1 has no units at all" is not correct as it would have the unit 1/1 which is 1. It is always allowed to unite units together. With the typical mathematical rules. As 1^2 is 1 and 1^-1 is 1, we are still fine with the definition, that one is a unit. Mathematically seen. I understand your concerns, that's why I asked in the first place. $\endgroup$ – LaRiFaRi Sep 17 '13 at 15:11
  • 1
    $\begingroup$ Ok, I'll concede that if $\frac{1\:\mathrm{m}}{1\:\mathrm{m}}$ has units $1$ then $\frac{1\:\mathrm{1}}{1\:\mathrm{1}}$ should also have units $1$ -- I've updated my answer. $\endgroup$ – Nathaniel Sep 17 '13 at 15:38
  • $\begingroup$ The use of units in physical quantities can be formalized by making units a group, equivalent to a vector space over the complex numbers (see e.g. Wikipedia), for which the identity (or zero vector) is essential and equal to 1. If you are going to allow algebra with units (such as $3\,\text s\times 2\,\text m/\text s=6\,\text m$), then surely you need to allow such basic algebra as $1^2=1$ (and that follows from $1 x=x$ for any $x$, which is why you introduce it to begin with). $\endgroup$ – Emilio Pisanty Sep 17 '13 at 21:04
  • $\begingroup$ @EmilioPisanty I think everyone's agreed it's a linguistic thing, and my post was intended as a humorous way to approach that. The point is that if we treat $1$ as a base unit then it has to behave like other base units, which means not obeying such algebraic identities - the whole point was it's not going to be consistent. I would say that if dimensions are equivalent to a vector space then the base units are the basis vectors. It would be a mistake to refer to the zero vector as a basis vector, and I think it's a mistake to refer to 1 as a unit. $\endgroup$ – Nathaniel Sep 18 '13 at 1:45
  • 1
    $\begingroup$ @EmilioPisanty I can see that I had confounded the concept of "units" with "base unit" - I've added a clarification to my answer. $\endgroup$ – Nathaniel Sep 18 '13 at 1:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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