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If you have $y = s^2 e^{-t}$ , what would the units be if $s$ is in meters and $t$ is in seconds, does $y$ even have units? Also what about $y=s^2 \ln(t)$, would you get the same answer ?

mvw
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user341469
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  • Units of $s$ must be $m^{-1/2}$ – caverac Nov 13 '16 at 19:12
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    If $t$ is measured in seconds, there must be an implicit constant with units of "per second" in the exponent: The exponential series $\sum_{k} t^{k}/k!$ contains terms of every non-negative degree. Are you sure $e^{-t}$ doesn't refer to $1/e^{t}$, with $e^{t}$ in seconds, or something of that type? – Andrew D. Hwang Nov 13 '16 at 19:20
  • I'm voting to close this because it is not about mathematics. – Matt Samuel Nov 13 '16 at 19:22
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    @MattSamuel Dimensional analysis is very much part of mathematics. – Robert Israel Nov 13 '16 at 19:31

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Transcendental functions such as $\exp$ and $\ln$ operate on and return pure numbers. So $e^{-t}$ with $t$ in seconds doesn't really make sense; it should be $e^{-ct}$ where $c$ is in $\text{sec}^{-1}$. Perhaps its numerical value happens to be $1\; \text{sec}^{-1}$. And then $s^2 e^{-ct}$ would be in $\text{meters}^2$.

Robert Israel
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  • Technically, if you treat various units as gradations in a graded algebra, $e^{-t}$ with $t$ in seconds does make sense, although it is not homogeneous. I'm not a physicist, so I'm not sure if they even entertain dimensionally non-homogeneous expressions, but mathematically there is nothing wrong (or unusual, really) with it at all. – tomasz Nov 13 '16 at 20:55