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If you have a number with a dimensional quantity in its exponent, what is the dimensionality of this number then? For example when you have $e^{(4J)}$ or $2^{(4N)}$, with $J$ and $N$ respectively Joules and Newton.

I know that you can't take the logarithm of a dimensional quantity, but I don't really understand the case with exponents.

Qmechanic
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stancallewier
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1 Answers1

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For $\mathcal e^C \in \mathbb R$ to be valid, $C$ itself must belong to the real set. However, it does not, because it's a measurement unit (type) . Hence initial proposition is invalid.

In general, all mathematical mappings in some sets $\mathbb R, \mathbb C$ which expects input to be a number and returns element in the same set, - does not work with dimensions, because it's a type. So, $x^C, e^C, \ln(C), \sin(C) $ and similar does not lead to a number, but rather to a new type, which is ill-defined in physics, because has no analog in measurements.

However, other type mappings like $\sqrt[n]{C} $ may be valid, as for example $\sqrt[3]{m^3} = m$ (length) which has analog in measurements and real world.

So, the final thought is that for dimensional mapping $C_1 \to C_2$ to be valid, - both units $C_1, C_2$ must have physical analog in the real world, otherwise we will end up with dimensions which simply doesn't make any sense.

BONUS

Why do math operations like $+-\times /$ work on non-numbers, dimensions ?

Examples:

Addition

We can make addition like $4J+2J$, because it reassembles equation of the form $aC+bC$ and by distributive property of binary operations we can re-write it to be $(a+b)C$, which leaves us adding plain real numbers and scaling the result by a measurement unit $C$.

Division

$\frac {4J}{2J}$ makes sense only because it makes form $\frac {aC}{bC}$, which is equivalent to $\frac ab$, so just a division of plain real numbers and result is dimensionless, because dimensions cancel-out.

Subtraction

Validity is same as for addition because $aC-bC = aC+(-b)C$

Multiplication

And pity, here dimensional magic ends, because $aC \times bC = (a \times b) C^2$ does not guarantee by itself that new dimension $C^2$ is physically valid, i.e. we get an element which does not apply to an old set and thus must be validated separately. So, $4J \cdot 2J = 8J^2$ does not make sense, but $4m \cdot 2m = 8m^2$ - does, because it's an area.

Agnius Vasiliauskas
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  • So if you should attribute a dimension to $e^{4J}$, it would be $e^J$? But since that isn't a dimension of a measurable quantity, that is kind of meaningless? – stancallewier Dec 29 '23 at 12:57
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    Yes, it's equivalent to $\left (e^4\right) ^J$. $e^4$ part is calculable, but dimension $x^J$ has no physical correspondence to a measurements. – Agnius Vasiliauskas Dec 29 '23 at 14:33
  • Joules-squared makes sense and is a perfectly valid unit. It even occurs in a famous formula, but that’s irrelevant; no combination of units is “physically invalid”. – Ghoster Dec 29 '23 at 22:08
  • @Ghoster, So you say that unit $e^J$ makes sense to you? Ok, can you share more info what this unit represents? And formulas doesn't validate types, experiments and measurements do. I can make artificially thousands of meaningless types, like $\sin(kg)$ and can even create formula for you using that type, like $m=\sin^{-1} ( \sin(m)) $, but it will be meaningless mumbo-jumbo, no? – Agnius Vasiliauskas Dec 30 '23 at 10:40
  • No, I do not say that. I wasn’t talking about $e^J$, which is dimensionally inconsistent. I was talking about $J^2$, $J^{19/3}s^{7/8}$, etc., which have no inconsistency, make sense, and are valid. Your statement that $J^2$ does not make sense but $m^2$ does is simply wrong. – Ghoster Dec 30 '23 at 18:03
  • You said that "no combination of units is physically invalid." And $e^J$ is basically a "combined unit", so you did say that actually. Oh my, and what $J^{19/3} $ unit measures or let alone joule squared? Dimensional consistency is not everything, derived unit must have grounds also in measurements. I haven't heard anybody measuring $J^2$, but it's plenty of resources with energy measurement. – Agnius Vasiliauskas Dec 30 '23 at 18:53
  • And I don't see $J^2$ in SI derived units- (https://en.wikipedia.org/wiki/SI_derived_unit), so I stand by myself, - that $J^2$ is not very common thing in Physics. Btw, same for $(m/s)^2$, it's used in many places like kinetic energy, statistical physics and others, but by itself, only velocity makes sense. To make sense of $v^2$ you have to combine it with some addition unit, for getting derived SI unit from velocity. Like I said, there's infinite possibilities for mangling units, but only several have been used in Physics. Why? Think about it. – Agnius Vasiliauskas Dec 30 '23 at 19:16
  • Dimensional consistency is not everything, derived unit must have grounds also in measurements. That is completely false, but I’m not going to argue with you further. I’ve downvoted and I’m done. – Ghoster Dec 31 '23 at 20:44
  • No need to be angry. If you can explain what $J^2$ means - do so, and I'll replace it in my post with other meaningless unit which must be plenty, like $kg^3$ or $s^e$. I understand that in theory you can do whatever you like with units, like assign 1 to all units or use arbitrary unit system. However, Physics describes fundamental laws of nature, which doesn't have billions of independent parameters and so real unit system reflects that, - only couple base SI units and several derived units and it's enough for physics laws mapping. – Agnius Vasiliauskas Jan 01 '24 at 09:17