Does the magnitude of a physical quantity have units? For example, if a velocity vector is $36\ \mathrm{m\,s^{-1}}\ \hat{u}$, is its magnitude $36\ \mathrm{m\,s^{-1}}$ or just $36$? Also why?


The magnitude has units. In your example, it's physically how fast you're going, which is measured with units. It doesn't make sense to say you're going "36", and so it doesn't make sense to say the magnitude of your velocity vector is 36.

Saying that the magnitude is 36 is a bad idea, because if you measured in cm/s instead, the magnitude would be 3600, and the magnitude would change depending on what units you had. Instead, we attach units to the magnitude so it can be expressed as 36 m/s or 3600 cm/s, but these are the same quantity, so the magnitude doesn't change with different units. It's a property of the vector, not an accident of the units chosen.

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    $\begingroup$ Incidentally, we can also arrive at the same result from the opposite side: After decomposing into magnitude and direction vector, the latter should have magnitude 1, but 1 is dimensionless. It wouldn't make sense to have units in your direction vector, so they have to live in the magnitude instead. $\endgroup$ – Kevin Aug 14 '18 at 23:29
  • $\begingroup$ @Kevin I'm actually not so sure about that. I think you could develop a perfectly good mathematical system where units are kept in the direction vector (so it would really be a "direction-unit vector"). The problem is just that it would be less useful than a system where the units are kept separate, because you'd need different unit vectors for different types of quantities. $\endgroup$ – David Z Aug 15 '18 at 5:37
  • $\begingroup$ @Kevin - I agree. Geometry => the science of measuring the world. To measure the world we use various "yardsticks". A measured distance is a scalar multiple of some agreed upon yarstick and the scalar multiple doesn't carry the units, the yardstick does. It is not just a random stick, it is of a specific measure. $\endgroup$ – ja72 Aug 15 '18 at 13:37
  • $\begingroup$ Can "speed" be a unit? $\endgroup$ – BruceWayne Aug 15 '18 at 17:05

It makes sense to assign the units to the magnitude and not the direction vector, but it would work either way.

Consider the position vector denoted by $$ \boldsymbol{r} = \pmatrix{ 3\, {\rm m}\\ 2\, {\rm m}\\ 6\, {\rm m}} = \pmatrix{3\\2\\6} {\rm m} $$

The magnitude of the vector is $ \| \boldsymbol{r} \| = 7 {\rm m} $, but to decompose it into magnitude and direction we have a choice:

$$ \boldsymbol{r} = (7 {\rm m}) \,\pmatrix{ \frac{3}{7} \\ \frac{2}{7} \\ \frac{6}{7} } = (7) \,\pmatrix{ \frac{3}{7}\,{\rm m} \\ \frac{2}{7}\,{\rm m} \\ \frac{6}{7}\,{\rm m} } $$

  • The first being the distance $7 \,{\rm m}$ in the direction $\left(\frac{3}{7} , \frac{2}{7} , \frac{6}{7} \right)$. This is the span interpretation when one spans "x" distance along a particular line.
  • The second being 7 times the distances $\left(\frac{3}{7}\,{\rm m} , \frac{2}{7}\,{\rm m} , \frac{6}{7}\,{\rm m} \right)$. This is the ruler interpretation, where one moves "x" number of ticks in more direction and each tick has units.

In fact, there are cases where both the magnitude and unit vector may contain units. For example, consider a planar force acting along a line. Now combine the force components with the equipollent torque at the origin.

$$ \boldsymbol{f} = \left[ \matrix{ \vec{F} \\ \tau } \right] = \left[ \matrix{6.9282\,{\rm N} \\ 4.0 \,{\rm N} \\ 20.0 \,{\rm N\,m}} \right] = (8.0\,{\rm N}) \left[ \matrix{ 0.866 \\ 0.5 \\ 2.5\,{\rm m} } \right] = F \,\hat{\ell} $$

The magnitude of the force is $F = 8\,{\rm N}$, acting along a line with equation $a y - b x + c =0$ where $\hat{\ell} = (a,b,c) = (0.866, 0.5, 2.5)$ and the direction vector $(a,b)$ has unit magnitude $\sqrt{0.866^2+0.5^2}=1$ and thus $c=2.5$ is a distance quantitiy (for $ay-bx+c=0$ to be dimensionally accurate).

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    $\begingroup$ Why did you pick out "7 times the distances $\left(\frac{3}{7}\,{\rm m} , \frac{2}{7}\,{\rm m} , \frac{6}{7}\,{\rm m} \right)$"? It's no more special than "14 times the distances $\left(\frac{3}{14}\,{\rm m} , \frac{2}{14}\,{\rm m} , \frac{6}{14}\,{\rm m} \right)$" $\endgroup$ – JiK Aug 14 '18 at 20:31
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    $\begingroup$ Stating @JiK's point more strongly, I would disagree with this answer. The magnitude is $7\rm{m}$, and that doesn't become negotiable when we "decompose it into magnitude and direction". As such, a direction vector should always have unitless magnitude $1$. (A caveat: I am not a physicist, so it's possible that this is something where physics conventions differ from what would be common in math, so I am refraining from voting.) $\endgroup$ – Mark S. Aug 15 '18 at 0:37
  • $\begingroup$ @JiK I'm guessing the reason for choosing $\left(\frac{3}{7}\,{\rm m} , \frac{2}{7}\,{\rm m} , \frac{6}{7}\,{\rm m} \right)$ is that it's a "unit vector" - not in the usual sense, but in the sense that its magnitude is equal to the chosen unit of length, namely $1\ \mathrm{m}$. On the other hand $\left(\frac{3}{14}\,{\rm m} , \frac{2}{14}\,{\rm m} , \frac{6}{14}\,{\rm m} \right)$ doesn't have the same property. I also disagree with this answer along similar lines to what Mark said, but I do think I see the logic behind it. $\endgroup$ – David Z Aug 15 '18 at 5:34
  • $\begingroup$ @JiK - The function of a ruler is to describe the scalar multiple of a unit dimension. Such as "three and a half times one inch" for $3.5\,{\rm in}$. Similarly, the notation $7\;\left(\frac{3}{7}\,{\rm m} , \frac{2}{7}\,{\rm m} , \frac{6}{7}\,{\rm m} \right)$. $\endgroup$ – ja72 Aug 15 '18 at 13:29
  • $\begingroup$ @DavidZ - when I first thought of this issue, I was like you, in that it was preposterous the assign units to basis vectors. But after some work in projective geometry, I realized that mathematically the two interpretations are identical (dual to each other). $\endgroup$ – ja72 Aug 15 '18 at 13:32

You would quote the speed as $36 \, \rm m\,s^{-1}$ because it is $36$ times $1 \, \rm m\,s^{-1}$.
If you wrote just $36$ what would that mean?
$36$ times what ????

Now what about $36 \, \rm m\,s^{-1}\, \hat u$?
All you have now is extra information about the direction of the velocity and there is no extra information about the magnitude (speed) which is still $36 \, \rm m\,s^{-1}$.


Some people here seem to be getting hung up on vectors, though they really don't have anything to do with the question. There just happens to be a vector used as part of the use case example, but that is extraneous information which is irrelevant to the question and its answer.

Yes, a quantity has units.

I went to the store earlier and I bought 4.

Four what?

Oh, sorry... I bought 4 apples.

Now an example that is closer to yours:

Do you know why I stopped you?

No sir, I'm not sure. The sign reads "Speed limit: 45" and my speed was 41.

That is not true. You were doing 65.

[much arguing and time later...]

So you see sir, I was doing 45 miles per hour. If your city dislikes the units that the rest of the country uses, you need to state on your signs that the speed limit is 45 kilometers per hour.

That last example also leads us into a good (but false) counter point. Quantities are often given without units. In reality, even a sign which reads only "Speed Limit 65" is providing units even though it does not look like it.

Whenever we are talking, listening, reading, or writing there is practically always a context. Speed limits in the United States are known and understood to be provided in miles per hour even if not specified on the sign. Even if not written the "Speed Limit 65" sign still has units of miles per hour, so the units are provided.

Now let's jump over to a super technical science lab setting. A pair of workers are taking measurements, and one of them asks "How high is it this time?" The other responds "121." The first one writes down "121". Both have omitted the units, but the units are still there, merely assumed. Later, when their boss reads the note left on a nearby work bench, he says to someone nearby "That's great that we got it up to 121 kilowatts today."

For your specific use case, $36\ \mathrm{m\,s^{-1}}\ \hat{u}$, insisting that the magnitude of the velocity is merely 36 and is unit-less would be false. In fact, that does not even make any sense. In fact, this does not even need to be a math or physics question - it is basic English language. A vector is magnitude and direction, and 'm/s' is part of the magnitude.

If you come to me with a formal report in a formal setting that suggests the speed of a velocity is 36, then that report is utterly useless to everyone. I could ask you "Did you measure it in banana-lengths per millennium?" In fact, the report (again, assuming it is completely formal) is in fact complete gibberish and not something that anybody can use for anything. In fact, the report would probably be trashed and you would be asked to write a new one... or at least fix the current one.


A velocity vector is some 'arrow' in a mathematical space (Euclidean) which we denote with $\mathcal{R}^3$. The velocity vector is a physical object so therefore we use three dimensions (assuming non-relativistic). This means that we can write:

$v=\sum^3_{i=1}v^ie_i=v^1e_1+v^2 e_2+v^3e_3 \in \mathcal{R}^3$ (the symbol $\in$ means that $v$ is an element of $\mathcal{R}^3$) where $v^i$ is the component in one of the directions so for example the x-direction, which we denote by $v^1=v_x$ and $e_i$ the basis vector in that direction (a standard basis consists of orthonormal vectors, this means that the vectors are geometrically orthogonal and that they have unit length). So what you are doing is write the velocity vector in terms of these basis vectors, and of course in three dimensions you have three basis vectors (imagine up,right and straightforward).

However you are talking about numbers, this means that you talk about the length of the vector. The length of a vector is given by the following equation:

$||v||=\sqrt{(v^1)^2+(v^2)^2+(v^3)^2}=\sqrt{v_x^2+v_y^2+v_z^2} \in \mathcal{R}$. (1)

$\mathcal{R}$ denotes that it is a (real) number. So once you know for example that the velocity of an object is 36 (=$||v||$). This could really mean anything. The units are arbitrary in mathematics. But of course not in physics. 36 here just means that the numbers corresponding to $v_x,v_y,v_z$ without mentioning units (however we do assume the same units) add up to this number according to equation (1). Once you know for instance that $v_x=10m/s,v_y=1m/s,v_z=0$. For $v_z$ I ignore the units since it is zero for every choice of (velocity) units. In this case the total velocity (see equation (1)) will be $||v||\approx 10.05m/s$ since $\sqrt{(10m/s)^2+(1m/s)^2}=\sqrt{10^2+1^2+0^2}m/s\approx 10.05 m/s$.You see of course that we could just say $||v||=\sqrt{10^2+1^2+0^2}\approx 10.05$ and look for the units afterwards, this is the math way (and advanced physics) of dealing with units.

So take away from this that a vector is not a number/scalar but an object which has a length which is a scalar.


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