0
$\begingroup$

I'm used to encounter units that come from product or division of units, such that $m/s$ or $kg\cdot m\cdot s^{-2}$. However, I have never seen a unit defined as a sum of units, for instance, $m+J$. I don't know if the reason if just that it doesn't make physical sense or if there are examples of something like that.

Perhaps something like a sum of units could be considered as a vector, each component for each unit. Does this exist in some physical context?

$\endgroup$
  • 1
    $\begingroup$ For this to work you need a relation between the units. For example, one liter of water weighs one kilogram based on density. You could say, the "vector" amount of water of $(1 liter, 1kilogram)$, but there is no convenience or benefit in doing so. $\endgroup$ – safesphere Jun 8 '18 at 15:25
  • 1
    $\begingroup$ Possible duplicate of What justifies dimensional analysis? $\endgroup$ – Kyle Kanos Jun 8 '18 at 16:05
2
$\begingroup$

It’s because it doesn’t make any sense to add different units. Each unit measures a specific dimension. You give example of a m+J. But what does that mean?

It’s like trying to add 4 oranges and 5 fire trucks. What’s the sum of that? Well, it’s 4 oranges + 5 fire trucks, because those are two totally different things. The same is true for m+J: meter measures distance, the Joule measures energy, so it makes no sense to add them.

It does however make sense to multiply units because of the geometry of areas and volume. A joule is a $N \cdot m$, because it is a force being applied (integrated) through the distance. The energy is the area of a curve determined by the force $F(r)$ as integrated over a range of $r$.

In this way it is possible to have “vectors” like you mention, depending on the units you use, and if they have qualities of linear independence. Newtons, meters, and joules could not be a basis for a unit vector space. However, meters, kilograms, seconds could. But physics is not formulated this way.

$\endgroup$
  • $\begingroup$ The linear independence is not required for the vector representation. The opposite in fact. Independent coordinates would have no relation thus making the vector representation meaningless. A linear dependence however, such as between mass and volume, would make the tangent of the vector angle meaningful as density. $\endgroup$ – safesphere Jun 8 '18 at 17:30
-3
$\begingroup$

One non physical example are complex numbers in the complex plane. But I'm not sure this is the same case.

$\endgroup$
  • $\begingroup$ I study maths, so I know complex numbers, but I have never seen them used in physics (though I know there are uses in electronics, but I have no idea how units are related to the complex numbers). $\endgroup$ – Javi Jun 8 '18 at 15:02
  • 5
    $\begingroup$ It's not a good example, because the complex numbers have a relation between the real and imaginary axes, such as $i^2=-1$ Or that $i$ is an operator of rotation by $90$ degrees in the complex plane. There is no relation in what the OP is referring to. $\endgroup$ – safesphere Jun 8 '18 at 15:17

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.