I understand this is a very elementary question, but I haven't been able to come up with any elementary reason why it should work. Also, why should quantities in an exponential be dimensionless?
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$\begingroup$ Possibly not elementary, but I found this post by Terrance Tao very insightful $\endgroup$– By SymmetryCommented Jan 24, 2023 at 17:20
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$\begingroup$ Possible duplicates: physics.stackexchange.com/q/364771/2451 , physics.stackexchange.com/q/13060/2451 , physics.stackexchange.com/q/109995/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Jan 24, 2023 at 17:38
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$\begingroup$ Hi Neeladri Reddy, Please only ask 1 question per post. $\endgroup$– Qmechanic ♦Commented Jan 24, 2023 at 17:43
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$\begingroup$ Just to say, quantities not just in an exponential but in any function at all should be dimensionless, with the possible exception of the simple form $f(x) = x^p$ for some power $p$, because then the units are easy to attach. $\endgroup$– Andrew SteaneCommented Jan 24, 2023 at 19:16
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2 Answers
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Let's look at what dimensional analysis tells us:
- If two quantities have different units, they're not the same. For example, kinetic energy can't be $mv$ or $mv^3$. (It could be $mv^2$; in Newtonian physics it's actually $\frac12mv^2$, but DA can't tell you that.) This makes some algebra mistakes easy to spot.
- Some problems of the form "which products of powers of these variables could be that variable?" have a unique solution; some don't, but we still get some constraints.
- Some such problems have no solution, which means there must be some other important variable you're missing.
- If $x+y$ makes sense then $x,\,y$ have the same dimension. If $ax^m+bx^n$ makes sense with dimensionless $a,\,b$ for $m\ne n$ then $x$ must be dimensionless. Therefore, series such as $e^x=1+x+x^2/2+\cdots$ require $x$ to be dimensionless.
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When you say "dimensional analysis works" I assume you mean to ask, why does it make sense that both sides of an equation involving physical units should always have the same dimension. Well, if for example you know that $distance=speed\cdot time$ or concisely $x=v\cdot t$ then it follows that the units of speed and time should be so defined as to provide you with units of distance when multiplied: $$L=\frac{L}{T}\cdot T$$ Note that while dimensional analysis can guide you about the general form of the equation, it doesn't enable you to always guess correctly at how an equation would look like precisely. For example also $x=2\cdot v\cdot t$ is dimensionally correct (since 2 is dimensionless) but this equation is clearly wrong.
Finally, it is not true that quantities in the exponent are always dimensionless. An example is the generic wave equation: $$\psi{(x)}=e^{i(kx-wt)}$$
While the quantity in the exponential on the whole is dimensionless, as it must be because radians are dimensionless (Thanks John Alexiou for correction) quantities with dimension such as x and t can appear there
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2$\begingroup$ The exponent in the wave function is dimensionless as
radianis unitless. $\endgroup$ Commented Jan 24, 2023 at 17:04 -
1$\begingroup$ Quite right! As a whole it must be dimensionless. I wanted to clarify however that quantites with dimension can appear there (and indeed be undone by appropriate multipliers) $\endgroup$– AmitCommented Jan 24, 2023 at 17:06
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$\begingroup$ "...units of speed and time should be so defined as to provide you with units of distance when multiplied..." This is such a nice way to answer the question. +1 $\endgroup$ Commented Jan 24, 2023 at 17:18