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Power Rules

This isn't quite the same as the rule above. Suppose $y = 3x$. Then:

\begin{displaymath}
e^y = e^{3x}
\end{displaymath}

where $e$ is a constant, in particular the ``natural exponential constant'' (although it would work for $a^y = a^{3x}$ just as well).

In physics (or anything else where the symbols can have units and aren't just pure numbers), this is not generally true unless the arguments are dimensionless. In fact this is a specific example of a general rule that one cannot substitute equalities that carry units into any functional form that has a power-series expansion. On the other hand one can substitute in quantities that in some sense ``look the same'' that are dimensionless. This is easy to understand. Supposed I know that $x$ is a length in meters. I can certainly write down its exponential: $e^x$. But does this make sense?

If I expand $e^x$ in its well-known power series (summarized, derived, defined later):

\begin{displaymath}
e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}...
\end{displaymath}

we see that it is in fact nonsense! How can I add meters to meters-squared to meters-cubed? What is a length plus an area plus a volume?

The only way this can make sense is if we write something like:

\begin{displaymath}
e^{x/\lambda} = 1 + x/\lambda + \frac{(x/\lambda)^2}{2!} +
\frac{(x/\lambda)^3}{3!}...
\end{displaymath}

where $\lambda$ is some other variable or constant with units of length. Then each term in the sum is dimensionless, and there is nothing wrong with setting $\lambda = 1$ meter if that corresponds to the physical length scale of the problem.


next up previous contents
Next: Consistency of Units Up: Power Rules Previous: Taking both sides of   Contents
Robert G. Brown 2009-07-27