With expressions such as

$$4\ \mathrm{\frac{m}{s}} \times 2\ \mathrm{kg} = 8\ \mathrm{\frac{m}{s}} \times 1\ \mathrm{kg}$$

We can justify that a $2\ \mathrm{kg}$ mass moving at $4\ \mathrm{m/s}$ has the same momentum as a $1\ \mathrm{kg}$ mass moving at $8\ \mathrm{m/s}$. This might make sense at an intuitive level, but is there a fundamental argument that says units can be manipulated algebraically such that in this case all we've employed is the commutative property?

Or take for instance when we have to do unit conversion and all we do is just cancel stuff according to some proportionalities. We're manipulating units algebraically.

Is this way of doing things blindingly obvious? Or did we have to go out there and find out that it works?

Also, if I have a differential equation such as


I usually solve it as if it were an empty math problem and had no units. That is, I only care about the numbers involved. But how do I prove to myself that this way of doing things is right and won't produce contradictions in terms of its units?

  • $\begingroup$ I dont understand your set of questions. Units are conventions. It is to quantify stuffs. Yes, when you convert units, you do it algebraically, because 1 kg = 1000 g. I dont see the confusion. It is just because it does not make sense to measure the weight of sun in grams or weight of an atom in kg we have different units. Coming back to your second question. Yes, that is the idea behind dimensional homogeneity. If your solution is correct and in "Correct units", you will have the "correct numbers" $\endgroup$ Commented Nov 19, 2014 at 3:20
  • 1
    $\begingroup$ Re the final paragraph, Leibniz notation is designed to make it manifest that when you differentiate or integrate, the dimensional validity of the equation remains valid at every step. For example, in $\int v dt$, the units work out to be distance if you take $dt$ to have units of time. $\endgroup$
    – user4552
    Commented Nov 19, 2014 at 6:29
  • $\begingroup$ @BenCrowell Is there any way to prove that there will always be dimensional validity without me having to go and see (like in the case of Leibniz notation you mention) ? Thanks. $\endgroup$
    – DLV
    Commented Nov 20, 2014 at 22:38

1 Answer 1


I suppose that the best way to argue for this is to consider the units as indicative of the vector space from which the quantities in question originate. The algebra that we have defined in physics is one such that these quantities behave under the natural rules of commutative and associate multiplication, and so when we multiply quantities $m$ and $v$ (to stay consistent with your example above), we have a quantity with units $[m] \times [v]$, which is indicative of that quantity belonging to a different vector space than either of the two original spaces (specified by units $[m]$ and $[v]$). This is my interpretation, however.

It’s somewhat (crudely) analogous to the case of matrix multiplication; that is, you cannot add matrices of unlike dimensions. But, the product of two matrices (not necessarily commutative) can belong to a different vector space than the matrices originated from, in analogy to the above scenario.

Too long, didn’t read: units specify the vector/Hilbert (to be precise) space that to what a quantity belongs and it so happens that we use the unit algebra to specify exactly to what space the quantity in question belongs. We happen to treat them as nifty tools to help keep track of what space we’re landing in when we perform these sorts of operations.

On your second point, I don’t know exactly what you’re asking, but I know that whenever I’m working through a differential equation or any sort of calculation, I frequently check my units to make sure I’m not adding quantities of unlike units (i.e., of different vector spaces, to stay consistent), that I have arguments of transcendental/trigonometric functions that are unitless, etc. If you’re ever in a predicament in which you’re violating some of these rules of sorts, that’s a major red flag that something has gone wrong in your computation.


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