If $1$cm = $.01$m then shouldn't the square root of $1$cm = the square root of $.01$m but the square root of $1$ = $1$ while the square root of $0.01$ = $0.1$ So my dilemma, is the square root of $1$ centimeter = $0.01$ meters or $0.1$ meters?
4 Answers
I think you need to think of the square root of a distance as a new unit. Just as a ${m^2}$ is an area, not a distance. So the following works:
$\sqrt {1m} = \sqrt {100cm} $
$\sqrt m = 10\sqrt {cm} $
$\sqrt {1cm} = \sqrt {cm} $
and
$\sqrt {.01m} = .1\sqrt m = .1\left( {10\sqrt {cm} } \right) = \sqrt {cm} $
So it works out either way.
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The square root of $0.01\,\mathrm{m}$ is $0.1\,\mathrm{m}^{1/2}$.
Such quantities whose units have fractional exponents don't have a ready physical interpretation, but can sometimes be useful as intermediates in calculations anyway.
From this calculation we can also see the necessary relation between the units $\mathrm m^{1/2}$ and $\mathrm{cm}^{1/2}$, namely $1\,\mathrm{cm}^{1/2} = 0.1\,\mathrm{m}^{1/2}$.
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"don't have a ready physical interpretation, but can sometimes be useful as intermediates" - sounds the same as complex numbers (where btw, physical interpretation lagged the algebra). – alancalvitti Jan 31 '15 at 17:09
In Quantum Mechanics the square root of the inverse meter arises as the unit of the value of (1D) wave function. Since the square of the wave function is the probability of finding a particle per unit length, the wave function itself has the unit of inverse unit length. And the physical meaning of the square root of an inverse meter itself is who knows what.
The side of a square of area $0.01 \text{ m}^2$ is $0.1 \text { m}$. The side of a square of area $1 \text{ cm}^2$ is $1 \text { cm}$. There is no square root of a centimeter-there is (aside from fracals) nothing of dimension $\frac 12$, which is what would be required.
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In materials engineering, it is not unheard of to have units of MPa(m)^1/2, or $\textrm{MPa}\sqrt{\textrm{m}}$. – Emily Nov 14 '13 at 19:34
But as $1$ cm $=0.01$ m $\implies1$ cm$^2=(0.01)^2$m$^2$
– lab bhattacharjee Nov 14 '13 at 19:29