## Fun with dimensional analysis 6 – Two basic, important rules

Here are two very simple rules when it comes to evaluating an expression in terms of its dimensions. So simple that they might seem blindingly obvious, but if you haven’t thought about them before then you might not have consciously employed them:

1. You can’t add apples to oranges. Meaning, any summation you perform must have each individual term be dimensionally equivalent. To see why, consider the counter-question “What is $\displaystyle 5 \ m + 5 \ kg$?” It’s nonsense. This rule is also useful when evaluating integrals, since an integral can be thought of as a sum of a bunch of differential elements, so the dimensions of the integral must be the same as the integrand (and remember, a differential dx is thought of as “a little bit of x” and therefore has the same dimensions as x).
2. Exponents are dimensionless. I recall years ago in a physics lab getting some value in an exponent to have dimensions, say in units of meters, and asking the TA what to do about it. He replied that the resultant equation should have units of $\displaystyle e^{ \mbox{[meters]} }$. That is incorrect, and I later found out I had made a mistake. If I had thought about it or known previously that exponents must be dimensionless, I would have had no such difficulty (and I assume he had no prior experience either). To see why, consider $\displaystyle 2^6$ which is just 2 times itself 6 times, resulting in just some number. Now consider $\displaystyle e^6$. Again, e is just some number and so multiplying it by itself 6 times should also just give some number. Now what if we had some actually formula in the exponent, like $\displaystyle e^{ -E / k_B T}$. Since it is still just e times itself some number of times, we must have the exponent $\displaystyle -E / k_B T$ be dimensionless, as we surely do in this case. Note that I am strictly talking about when we have some real number as the base, given how exponents are defined. Obviously if you had something like $\displaystyle d^2$ where d is some distance, the resultant dimensions would be $\displaystyle \lbrack L \rbrack^2$.