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For example, I have a mass, m = 0.1kg and I square root it, giving me m = 0.316 (3s.f.) does the unit still stay as kg, or does it change?

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It can be instructive to flip questions around and see if you already know the answer. In this case consider asking "Do the unit change if you square a quantity?". –  dmckee Apr 30 '13 at 19:28
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@dmckee Well, if I times or divide two quantities together, the units multiply or divide accordingly, so if I square it, it's the same as kg*kg (kg^2) in this example? –  Saras Apr 30 '13 at 19:40
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Saras, yes. And because the square root is the inverse of squaring you can conclude that the units change just as the answers below say. With practice that kind of thinking will become more natural. –  dmckee Apr 30 '13 at 19:43
    
@dmckee But what would k^1/2 mean? Does it make sense? I can't think of any logical explanation for it. I was told to plot a graph of distance against square root of mass, should I record my mass with units of kg^1/2? –  Saras Apr 30 '13 at 19:55
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It wouldn't make any sense at all; unless it was multiplied with another $\mathrm{kg}^{1/2}$. That's a hint. If you get to the end of some computation and the units are screwy that way you may have made a mistake. Go back and check both your physics and you math. –  dmckee Apr 30 '13 at 20:00
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5 Answers

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As the other answers (and dmckee's comments) note, yes, if you take the square root of a dimensional quantity then you need to take the square root of the units too:

$$ \sqrt{4\;{\rm kg}} = 2\;{\rm kg}^{\frac12} $$

And no, I can't think of any meaningful physical interpretation for the unit ${\rm kg}^{\frac12}$ either.

However, in the comments you say that you were "told to plot a graph of distance against square root of mass." What that means is simply that you should scale the mass axis non-linearly, presumably in order to more clearly show the relationship between the two quantities. For labeling the mass axis, you basically have two choices:

  • label the axis $\sqrt m$, with equally spaced ticks at, say, $1\;{\rm kg}^{\frac12}, 2\;{\rm kg}^{\frac12}, 3\;{\rm kg}^{\frac12}, 4\;{\rm kg}^{\frac12}, \dotsc$, or

  • label the axis $m$, with equally spaced ticks at $1\;{\rm kg}, 4\;{\rm kg}, 9\;{\rm kg}, 16\;{\rm kg}, \dotsc$.

While, technically, both of these are valid, I would strongly recommend the latter option. Just compare these two plots and see which one you find easier to read:

Plot with units of kg^(1/2), linear axis scaling $\hspace{60px}$ Plot with units of kg, quadratic axis scaling

Alas, not all plotting software necessarily supports such axis labeling, or at least doesn't make it easy, which is why you sometimes see plots with funny units like ${\rm kg}^{\frac12}$.

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The strange units of the graph on the left makes it harder to read the actual values of m, but the graph on the right has a difficult x-axis scale, which makes left-hand graph better. However, I did a table of measurements labeling square root of m as: √m/kg, would it be sensible to write √m/kg^1/2 then? –  Saras Apr 30 '13 at 21:39
    
Yes. If the unit of $m$ is $\rm kg$, then the correct unit of $\sqrt m$ is $\rm kg^{\frac12}$. That's simply a fact, not a matter of choice. –  Ilmari Karonen Apr 30 '13 at 22:11
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Alternatively, the left-hand abscissa could be labeled $\sqrt{m/\mathrm{kg}}$, so that it actually makes sense, but too seldom is this done. –  Chris White May 1 '13 at 0:55
    
Why you sometimes see plots with funny units is that they are useful. For instance, log or log-log graphs: logarithmic frequency, versus decibels. If the given distance actually varies with the square root of mass, then this square root of mass plot will reveal that by showing a straight line. Plus any deviations from that straight line fit will be more clear. –  Kaz May 1 '13 at 4:36
    
What did you use to make those plots? They look very nice. –  Vibert May 1 '13 at 14:13
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It becomes the square root of the unit. Think of energy:

$$E = \frac{1}{2}mv^{2}$$

If I solve for $v$, I have $v = \sqrt{\frac{2E}{m}}$. Since $\rm 1 J = 1 kg \cdot m^{2}/s^{2}$, we see that the units have to obey the square root, or we will end up with our velocity equalling something other than m/s.

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But what does kg^1/2 represent? Does it even make sense? –  Saras Apr 30 '13 at 19:41
    
It doesn't have to (and doesn't) represent anything physical. –  David Z May 1 '13 at 3:00
    
It represents the square root of a mass. That is all. :D –  Michael Brown May 1 '13 at 3:11
    
What does the square of speed of light represent in $E = mc^2$? Or the square of velocity in the above $\frac{1}{2}mv^2$? It means that energy goes up with the square of speed, that is all. You tell me where the square root of mass comes up and I will come up with an interpretation. –  Kaz May 1 '13 at 3:49
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Yes, the dimension of a quantity changes if it is square-rooted. If $m$ is a mass with dimension $[m]=\textrm{kg}$, $\sqrt{m}$ is not a mass, but another quantity with dimension $[\sqrt{m}] = \textrm{kg}^{1/2}$.

More generally, if $[a] = A$ and if $[b]=B$, then $[a^n b^m] = A^nB^m$ etc.

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Take the root of the unit of area (Eg: 4 m$^2$ )
We get the unit of length (Eg: 2 m) which is the unit for different physical quantity
So it definitely changes

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Let's square root 0.1kg:

  • expressed in kg, we get $\sqrt{0.1}\approx 0.316$.
  • expressed in g, we get $\sqrt{100}=10$.

So obviously the unit changes. If it stayed the same, we'd have $0.316\mbox{kg} = 10\mbox{g}$ which is clearly false.

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