2

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
  • 220,844
stancallewier
  • 31

1 Answers1

6

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
  • 16,916