# How to handle dimensional analysis under exponents

Say there is a river of saline solution and I want to measure the flux of salt. That is, I want to measure the miligrams (mg) of salt per second ($\sec$) passing through a square centimeter area (cm$^2$) perpendicular to the flow of the saline solution. (This is a boring straight river with no twists or bends in it.) The quantity I am interested in measuring (call it $f_{salt}$) therefore has units of $\frac{\text{mg}}{\sec \text{cm} ^2}$. I already know the amount of salt per liter of saline solution is $x\frac{\text{mg}}{\text{L}}$, so my strategy for measuring the salt flux is to just measure the flux of the saline solution (call this $y_{sol}\frac{\text{L}}{\sec \text{cm} ^2}$) and then to multiply this by $x_{salt}\frac{\text{mg}}{\text{L}}$. That is,

$$f_{salt} = y_{sol}\frac{\text{L}}{\sec \text{cm} ^2}x_{salt}\frac{\text{mg}}{\text{L}}= y_{sol}x_{salt} \frac{\text{mg}}{\sec \text{cm} ^2}$$

Say I've theoretically derived or want to test the following hypothetical power law relation between salt flux and a certain "salt budget" parameter $w$ which is also measured in units of $\text{mg}$ of salt per ${\sec \text{cm} ^2}$

$$f_{salt} = K\left(w\frac{\text{mg}}{\sec \text{cm} ^2}\right)^{\alpha} \tag{1}$$

I am particularly interested in the case where $w = Af_{salt}$, where $A$ is another dimensionless constant. And I am particularly interested in solving for $y_{sol}$ when this is the case.

At first I naively ignore dimensions; I just plug in $Ay_{sol}x_{salt}$ for $w$ and solve the equation $(1)$ for $y_{sol}$, giving:

$$y_{sol}=\frac{(KA^{\alpha})^{\frac{1}{1 - \alpha}}}{x_{salt}} \tag{2}$$

I know that $y_{sol}$ is measured in units of $\frac{\text{L}}{\sec \text{cm} ^2}$, so I blithely assume that the RHS of this equation $(2)$ is automatically expressed in units of $\frac{\text{L}}{\sec \text{cm} ^2}$ too.

But as I begin to play around with this expression $(2)$ experimentally it dawns on me that something weird is going on. I go back to the drawing board, again plugging in $Ay_{sol}x_{salt}$ for $w$ and solving equation $(1)$ for $y_{sol}$, but this time keeping units explicit:

$$f_{salt} = K\left(w\frac{\text{mg}}{\sec \text{cm} ^2}\right)^{\alpha}$$

$$y_{sol}x_{salt} = K\left(Ay_{sol}x_{salt}\frac{\text{mg}}{\sec \text{cm} ^2}\right)^{\alpha}$$

$$y_{sol}x_{salt} = K\left(Ay_{sol}\frac{\text{L}}{\sec \text{cm} ^2}x_{salt}\frac{\text{mg}}{\text{L}} \right)^{\alpha}$$

$$\frac{y_{sol}}{y_{sol}^{\alpha}} = K\frac{\left(Ax_{salt}\frac{\text{mg}}{L}\right)^{\alpha}}{x_{salt}\left(\frac{\text{mg}}{L}\right)^{\alpha}} \left(\frac{L}{\sec \text{cm} ^2}\right)^{\alpha} \tag{3}$$

$$y_{sol}^{1 - \alpha} = KA^{\alpha}x_{salt}^{\alpha - 1} \left(\frac{L}{\sec \text{cm} ^2}\right)^{\alpha} \tag{4}$$

$$y_{sol} = \frac{(KA^{\alpha})^{\frac{1}{1 - \alpha}}}{x_{salt}} \left(\frac{L}{\sec \text{cm} ^2}\right)^{\frac{\alpha}{1 - \alpha}} \tag{5}$$

So, if these steps are valid, then by using equation $(2)$ I was using an expression for $y_{sol}$ expressed in "weird" units of $\left(\frac{L}{\sec \text{cm} ^2}\right)^{\frac{\alpha}{1 - \alpha}}$. In order to get an expression for $y_{sol}$ in "normal" units, I have to do one more step of taking the $\frac{1 - \alpha}{\alpha}$ root of both sides, giving:

$$y_{sol}^{\frac{1}{\alpha} - 1} = \frac{K^{\frac{1}{\alpha}}A}{x_{salt}^{\frac{1}{\alpha} - 1}} \frac{L}{\sec \text{cm} ^2} \tag{6}$$

which I take to be the equation I should really use if I am interested in $y_{sol}$ expressed in "normal" units.

My question is, did I get the dimensional analysis right? Is $(6)$ the equation I should use? More specifically, I am worried about the step in equation $(3)$ where I make units of $\left(\frac{\text{mg}}{L}\right)^{\alpha}$ appear in the denominator when I divide by $x_{salt}$, so that they will cancel with the units in the numerator. I did this because it sort of makes logical sense to me, plus it was the only way I could get units of $\frac{L}{\sec \text{cm} ^2}$ alone on the RHS, which is required to avoid violating dimensional homogeneity when I get $y_{sol}$ alone on the LHS.

PS - This is a dimensional analysis question. Please do not fixate on the actual physics involved in the equations.

The problematic equation is $f_{salt}=Kw^\alpha$. It is problematic because $w$ has dimensions. The simplest way to solve this problem is to split your constant $K$ into $k/(w_0)^\alpha$ for some arbitrary chosen reference value $w_0$. Then your equation becomes
$$f_{salt}=k\left(\frac{w}{w_0}\right)^\alpha$$
This has two advantages: first, in contrast to the original equation (where the dimensions of $K$ depended on $\alpha$), you now know exactly what dimensions your proportionality constant is supposed to have; namely, $k$ has the same dimensions as $f_{salt}$. Second, now the quantity being exponentiated is dimensionless, which eliminates any effect that $\alpha$ would have on the dimensional analysis.
• Thanks. This then results in $y_{sol}^{1-\alpha} =k \frac{x_{salt}^{\alpha - 1}A^{\alpha}}{y_{sol0}x_{salt0}} \frac{L}{\sec \text{cm}^2}$, so there is still exponentiation happening on the LHS in order to get an expression in "normal" units. Hope that's ok with everyone. – ben Sep 7 '18 at 20:25