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Is power of only a number or an exponential function is dimensionless? If power of any other thing can also be dimensionless then please explain with examples.


marked as duplicate by Kyle Kanos, Brandon Enright, jinawee, DavePhD, BMS Apr 11 '14 at 5:35

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  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – user80551 Apr 10 '14 at 7:41
  • $\begingroup$ if i had understood that, i would have never posted this question. i want an answer which can easily be understood.@user80551 $\endgroup$ – elle Apr 10 '14 at 7:48
  • $\begingroup$ One way to think about it is this: Mathematical identities for functions like the exponential function are about numbers, so in order to sensibly use these functions we must use numbers only as their input. $\endgroup$ – Danu Apr 10 '14 at 8:10
  • $\begingroup$ Related (dupe?) physics.stackexchange.com/q/7668 Partially related physics.stackexchange.com/q/13060 $\endgroup$ – user80551 Apr 10 '14 at 8:55
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    $\begingroup$ Another way of seeing clearly why an exponential's argument should be dimensionless is to Taylor expand: $\exp(x) = 1 +x + x^2 /2 + ...$ Every term has a different dimension if $x$ is dimensionful, and hence cannot be summed. See: damtp.cam.ac.uk/user/tong/relativity/three.pdf for detailed notes on dimensional analysis. $\endgroup$ – JamalS Apr 10 '14 at 9:53

To consider an example, take the case of exponential decay

$$N=N_\circ e^{-\lambda t}$$

We can write this as

\begin{eqnarray*} N & = & \frac{N_{\circ}}{e^{\lambda t}}\\ & = & \frac{N_{\circ}}{\underbrace{e\times e\times e\times e\times\ldots \times e}_{\lambda t\text{ times}}} \end{eqnarray*}

So $\lambda t$ must be a dimensionless term that is telling how many times we should multiply $e$ by itself. Thus, $\lambda t$ must be dimensionless "overall". Individually, $\lambda$ has the dimensions of $[T^{-1}]$ which cancels with $t$ to give a net dimensionless quantity.

$\underbrace{e\times e\times e\times \ldots}_{10 \text{ meters times}}$ makes no sense mathematically.

We could have taken a dimensional quantity instead of $e$ but the exponent $\lambda t$ would still be dimensionless. eg in the kinematical equation $s=ut + \frac 12 at^2$, $t^2$ has the dimensions of $[T^2]$ but the exponent $2$ is dimensionless.

The same applies to transcendental functions i.e. logarithmic, trigonometric, etc.

  • $\begingroup$ it is getting impossible for me to bring the alphabets in superscript $\endgroup$ – elle Apr 10 '14 at 10:26
  • $\begingroup$ @elle It's mathjax, see meta.math.stackexchange.com/questions/5020/… $\endgroup$ – user80551 Apr 10 '14 at 10:28
  • $\begingroup$ In $e^{\lambda t}$ ‘λt’ is dimensionless, so we say that ‘$e^{\lambda t}$ ’ (whole) is dimensionless. Now in $e^{\lambda t}$ ‘λt’ is dimensionless, so would $e^{\lambda t}$ (whole) be dimensionless? $\endgroup$ – elle Apr 10 '14 at 10:35
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    $\begingroup$ @elle The exponent ($\lambda t$) is dimensionless but if the base ($e$) is not dimensionless, then as a whole, $e^{\lambda t}$ won't be dimensionless (unless the exponent is zero of course.) Eg $\text {meter}^2$ has the dimensions $[L^2]$. Here, the exponent $2$ is dimensionless. $\endgroup$ – user80551 Apr 10 '14 at 10:42

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