How do I determine what physical constants can be combined to derive a set of natural units?

Question

I am looking into natural units (units of measurement based only on universal physical constants). Different systems of natural units use different physical constants as their defining constants. If I want to create a system of natural units, how do I determine what constants can be combined to give meaningful results?

For example, let's say I want to create a system of natural units with base units of the following dimensions:

• Mass $M$
• Length $L$
• Time $T$
• Electric Current $I$
• Temperature $ϴ$

Let's also say I want to use the following physical constants as defining constants:

• Speed of light in a vacuum $c$ (dimension $L⋅T^{−1}$)
• Reduced Planck constant $ħ$ (dimension $M⋅L^2⋅T^{−1}$)
• Elementary charge $e$ (dimension $T⋅I$)
• Boltzmann constant $k$ (dimension $M⋅L^2⋅T^{−2}⋅ϴ^{−1}$)

To derive the base units, I know I need to combine powers of the defining constants, like $c^α × ħ^β × e^γ × k^δ$. Dimensionally, that expands to:

$$(M^0 L^1 T^{−1} I^0 ϴ^0)^α × (M^1 L^2 T^{−1} I^0 ϴ^0)^β × (M^0 L^0 T^1 I^1 ϴ^0)^γ × (M^1 L^2 T^{−2} I^0 ϴ^{−1})^δ$$

Multiplying through and collecting terms gives the expression:

$$M^{0 α + 1 β + 0 γ + 1 δ} × L^{1 α + 2 β + 0 γ + 2 δ} × T^{−1 α − 1 β + 1 γ − 2 δ} × I^{0 α + 0 β + 1 γ + 0 δ} × ϴ^{0 α + 0 β + 0 γ − 1 δ}$$

To find the combination of base constants needed to derive any unit with some combination of the five base dimensions, I then need to solve the system of equations represented by the exponent expressions, set equal to the desired exponent value. For example, isolating the unit for mass gives the system of equations:

$$ 1 β + 1 δ = 1, \\ 1 α + 2 β + 2 δ = 0, \\ −1 α − 1 β + 1 γ − 2 δ = 0, \\ 1 γ = 0, \\ −1 δ = 0 $$

However, this system of equations is not solvable. I know that the specified defining constants plus the gravitational constant $G$ (dimension $M^{−1}⋅L^3⋅T^{−2}$) lead to a solvable system of equations, as that gives the Planck units plus an electrical dimension. And I know that the Planck units still work if you replace the elementary charge constant $e$ (dimension $T⋅I$) with the vacuum electric permittivity constant $ε_0$ (dimension $M^{−1}⋅L^{−3}⋅T^4⋅I^2$), even though their dimensions are completely different.

What determines which constants combine to give solvable equations? Is there a way to determine what constants will work with each other, besides just guess-and-checking?

Answer 1

You've got 5 base dimensions here, so you can associate any physical dimension with a $5$-tuple of real numbers. For example, we could write $$M^\alpha L^\beta T^\gamma I^\delta \Theta^\epsilon\rightarrow \pmatrix{\alpha\\\beta\\\gamma\\\delta\\\epsilon}\in \mathbb R^5$$

When two quantities are multiplied or divided, the $5$-tuples representing their dimensions are added or subtracted. Similarly when a quantity is raised to a power, the corresponding $5$-tuple is multiplied by that power. For instance,

$$L/T \rightarrow\pmatrix{0\\1\\-1\\0\\0}, \qquad ML/T^2 \rightarrow \pmatrix{1\\1\\-2\\0\\0}$$ In this sense, physical dimensions can be understood as elements of the vector space $\mathbb R^5$. In order to come up with a system of "natural units," you just need five quantities whose dimensions are linearly independent of one another.

Your example fails because you only provide four - their dimensions obviously don't span $\mathbb R^5$, so there will be some dimensions which cannot be expressed as a combination of them. You can make it work by choosing one additional constant whose dimension is linearly independent of the four you have already provided.

---

If we use the four physical constants I give in my question, plus the Coulomb constant K (dimension M¹ L³ T⁻⁴ I⁻² ϴ⁰) to get a full set of five, the resulting system of equations is unsolvable, but I don't see how any of the five constants are linearly dependent with another.

Let's make the following identifications

$$[c]=\pmatrix{0\\1\\-1\\0\\0}\equiv \vec v_1, \quad [\hbar]=\pmatrix{1\\2\\-1\\0\\0}\equiv \vec v_2, \quad [e]=\pmatrix{0\\0\\1\\1\\0}\equiv \vec v_3$$ $$[k_B]=\pmatrix{1\\2\\2\\0\\-1}\equiv \vec v_4, \quad [k_C]=\pmatrix{1\\3\\-4\\-2\\0}\equiv \vec v_5$$

The series of equations obtained by writing $\vec v_5 = A\vec v_1+B\vec v_2 + C\vec v_3 + D\vec v_4$ is $$\matrix{B+D=1\\A+2B+2D=3\\-A-B+C+2D = -4\\C = -2 \\-D = 0}$$

This set of equations can be solved easily to yield $(A,B,C,D)=(1,1,-2,0)$, implying that $\vec v_5 = \vec v_1+\vec v_2-2\vec v_3$. In other words, the dimensions of $k_C$ are equal to the dimensions of $c\hbar/e^2$ - namely $$\left[\frac{c\hbar}{e^2}\right] = \frac{L}{T}\frac{ML^2}{T} \frac{1}{I^2T^2} = \frac{ML^3}{T^4I^2} = \left[k_C\right]$$