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What are some illustrative examples of dimensional analysis at work? Especially, where can it be used to significantly reduce the difficultly of a computation?

4 Answers4

4

Period of a Harmonic Oscillator

This derivation is, in my opinion, fairly spectacular.

Required is a formula for the period $T$ of oscillation in terms of mass $m$, spring constant $k$, and gravitational acceleration $g$. The units of $T$ are $[\text{time}]$, and the three independent variables have units, respectively, $$m:[\text{mass}],\quad k:[\text{mass}/\text{time}^2],\quad g:[\text{dist}/\text{time}^2].$$

We cannot include $g$ because the $[\text{dist}]$ unit is impossible to cancel. Therefore we take $\sqrt{m/k}$ with units $$\sqrt{\frac{[\text{mass}]}{[\text{mass}/\text{time}^2]}}=[\text{time}].$$

Hence it follows that $$T=\kappa\sqrt{\frac{m}{k}}$$ for some constant $\kappa$, and we have the formula.

3

Calculating the yield of a nuclear weapon

Source: http://en.wikipedia.org/wiki/Nuclear_weapon_yield#Calculating_yields_and_controversy

Nuclear weapons tests often had photographs of the test showing the time after denotation and a scale describing the radius of the shock wave.

Noting that the shockwave follows Newtonian physics and the laws of fluid mechanics, the radius of the explosion depends on:

  • The energy of the explosion: $\mathcal{M}\mathcal{L}^2\mathcal{T}^{-2}$;
  • The time after the detonation: $\mathcal{T}$;
  • The density of air: $\mathcal{M}\mathcal{L}^{-3}$;
  • The radius of the explosion: $\mathcal{L}$.

Solving for a dimensionless constant, we have

$$\pi = \left(\mathcal{M}\mathcal{L}^2\mathcal{T}^{-2}\right)^a\mathcal{T}^b\left(\mathcal{M}\mathcal{L}^{-3}\right)^c\mathcal{L}^d.$$

This gives us a system of equations:

$$\begin{align*} a+c &= 0, \\ 2a-3c+d &= 0, \\ -2a +b &= 0. \end{align*}$$

In matrix representation, we want to compute the kernel of

$$M = \begin{pmatrix} 1 & 0 & 1 & 0 \\ 2 & 0 & -3 & 1 \\ -2 & 1 & 0 & 0\end{pmatrix}.$$

Doing so, we find $a = -1, b = -2, c = 1, d = 5$.

Therefore,

$$E = k\left(\frac{r^5\rho}{t^2}\right).$$

Emily
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  • +1 Excellent! I like that you actually solved for the constants algebraically, very neat. –  May 20 '14 at 15:05
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There's an interesting take on this at http://1ucasvb.tumblr.com/post/79629813053/ - sections 'Dimensions vs. Units' and 'Dimensions vs. Concepts' - in response to the question 'What are your thoughts on the pi v. tau debate?' What the author is saying is that if we use tau instead of pi (which does simplify calculations) we should also affix a unit (the radian) to tau because this would simplify them even further.

LucasVB
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Vector Projection of $u$ onto $v$

This is not so much an answer as an example, for which I would like additional input or explanation.

Consider the length of the vector projection $$\text{proj}_\mathbf{v}\mathbf{u}$$ of $u$ onto $v$. Say we (magically) know that the formula should involve $\mathbf{u}\cdot\mathbf{v}$. Would it be correct to say that because $\mathbf{u}\cdot\mathbf{v}=||u||\,||v||\cos\theta$, the dot product has these units? $$[\text{length}]^2[\text{unitless number}]=[\text{length}]^2$$ The final answer should be a length, so we have to divide out one dimension of length to obtain the true answer. By checking a few simple cases (such as projection of a zero vector, where we should not divide by its magnitude), we conclude that of the two obvious lengths which present themselves, it is appropriate to divide by that of $\mathbf{v}$. $$||\text{proj}_\mathbf{v}\mathbf{u}||=\frac{\mathbf{u}\cdot\mathbf{v}}{||\mathbf{v}||}$$

Is this valid reasoning? Putting aside the issue that many of these steps are hand-wavy, is it correct to consider a dot product to have units of length squared, or is it (as I think most people would intuitively guess, including myself if I had not thought up this example) a unitless scalar?

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    If vectors are spatial, then each component represents length, so the dot product is the sum of square lengths. So the dot product is length squared. – Emily May 22 '14 at 01:00
  • @Arkamis Can this same idea be used to derive the formula for the magnitude of a vector? Probably not I think, but at least it checks (because it is the square root of a dot product). Can dimensional analysis help here at all? –  May 22 '14 at 01:20
  • Sure. The magnitude is the square root of the sum of squares: $||x||=\sqrt {x_1^2+x_2^2+x_3^2} \propto (\mathcal {L}^2)^{1/2} = \mathcal {L} $. – Emily May 22 '14 at 01:55