On physical units . note

Perspective $\blacktriangleright$ On physical units . note $\blacktriangleright$ Its about time . note
$\blacktriangleright$ On electronics . note


Abstract notes on unit-typing

Here are some ideas capturing units as a type system. We'll use it to restrict operations between structures that exhibit the same algebraic rules. E.g. a positions $x_1, x_2, \cdot$ and temperatures $T_1, T_2, \cdot$ may be reals and we may want for $T_1+T_2$ to make sense, but we want to forbid considering e.g. $T_1+x_1$.

Call $U$ and $E$ the unit-universe resp. expression-universe.



(the above two just say $[-]$ maps from $E$ to $U$)


For a mulitplication $\cdot$ of expression, we define an enodmorphism of units via

$[a][b]:=[a\cdot b]$

Remark on two bijective maps

The map $a\mapsto a^s$ (whenever there's a structure such that “$a^s$” makes sense, and for $s\ne 0$) as well as actions $a\mapsto e\cdot a$ (multiplication by $e:[e]$, and when $e$ isn't the zero of the multiplication) are of particular interest to us.

Any bijective polymorphic function $p_{-}$ is a natural transformation for the following functor it induces:


$F(f) : FA \to FB$

$F(f) := p_{B} \circ f \circ p^{-1}_{FA}$

Remark on maps

We generalize the unit algebra of multiplication:

If $f$ is a function, then we may set $[f]:=[f(x)/x]$ so that $[f(x)]=[f][x]$. With this $f(x):=x^3$ has unit $[f]=[x^2]$. We're actually used to this convention for linear operators, as in $Ax=y$.

With this,

$\frac{{\mathrm d}}{{\mathrm d}x}a\,:\,[x^{-1}a]$

$\frac{{\mathrm d}}{{\mathrm d}x}\,:\,[x^{-1}]$

$\int_{x_0}^{x_1}a\,{{\mathrm d}x} \,:\, [x\cdot a]$

$\int_{x_0}^{x_1}{{\mathrm d}x} \,:\, [x]$

For a theory with a rich algebraic structure, if $e$ is the “main unit”, then $[e^{-1}]$ is also very relevant.

For any dynamical theory, that will be

$$ \require{AMScd} \begin{CD} {[t]} \\ @VV{s\,=\,(-1)}V \\ {[\omega]} \end{CD} $$

The maps from additional $[a]$ to $[e]$ resp $[e^{-1}]$ are obtained by multiplications as in

$$ \require{AMScd} \begin{CD} {[x]} @>{e\,=\,c^{(-1)}}>> {[t]} \\ @V{s\,=\,(-1)}VV @VV{s\,=\,(-1)}V \\ {[k]} @>>{e\,=\,c}> {[\omega]} \end{CD} $$

Relevant pyhsical theories

Neither the general framework of Quantum mechanics nor Newtonian mechanics provide a finite constant of time, i.e. a singled out unit of time. The defining equations of may be written

  • $h_A\,\left(\tau_t\dfrac{{\mathrm d}}{{\mathrm d}t}\right)\rho = {\mathrm i}^*[A,\rho]$.
  • $h_A\,\left(\tau_t\dfrac{{\mathrm d}}{{\mathrm d}t}\right)\Psi = {\mathrm i}^*A\,\Psi$.
  • $h_F\,\left(\tau_t\dfrac{{\mathrm d}}{{\mathrm d}t}\right)^2x=l_x\,F(x,\frac{{\mathrm d}}{{\mathrm d}t}x,t)$

$ h_A\,\tau_t^2\,\Delta_{t_1}^{t_2} x' = l_x\,\int A(x(t), x'(t), t)\,{\mathrm d}t $

where $\tau_t$ is a time constant characteristic for the system and $h$'s other parameters, possibly to be fixed in experiment. The usage of all the constants isn't standard. The indices like in $h_A$ mostly denotes that they share unit with that index expression. $f=1/\tau_t$ is a characteristic rate or frequency (with $f=\tau_\pi\omega$ where $\tau_\pi\equiv{2\pi}$ and with a characteristic circle frequency $\omega$) such as the first gap $f_1-f_0$. It makes sense to call it “frequency” as soon as we deal with an expression or system that soon enough returns to its previous state, like ${\mathrm e}^{-2\pi\cdot{\mathrm i}ft}$ does.

$A$ is a linear Hermitean operator and $F$ some function and all constants may be absorbed into them. We may call $A$ the arousal of a state (it's often $\propto$ the occuptation and we may naturally choose it to have units of action). In quantum mechanics in general, but in particular if we pull out the frequency of the Hamiltonian like that, we want to think of the state occupation numbers as the main extensive quantity. The one that can be traded between subsystem.

The vector $x$ is restricted to ${\mathbb R}^n$.

Independent units may be associated with $\Psi$, although the fact that $[\Psi]$ cancels in the former equation suggests to not assign it a unit (I'd be interested to track it though, e.g. let $[\Psi]^2$ be a unit of “observable” - after all, each measureable value $\langle \Psi|A|\Psi\rangle$ has units $[\Psi]^2[A]$).

todo: rewrite the following

I'm not gonna write down the standard presentation of the equation here, but assume those in the following. We must have

  • $[H]=\dfrac{[h_G]}{[t]}$
  • $[F]=\dfrac{[m_F][x]}{[t]^2}$

Those two frameworks are a priori independent. In the following, we will discuss the implications for different $H$ and $F$.

Remark on the proportionality constants/coefficients:

I'm personally for setting up both of the equation with neither $h_G$ nor $m_F$.

Of course, in the case of the Newton, this means an object doesn't excert oce force $F$, but instead we deal with “forces” that are different for each thing they act on.

Compare: In the standard presentation, the $H$ is also chosen new for every $\Psi$ and $h_G$ is taken to be one single constant.

The unit $\dfrac{[H]}{[h_G]}$ must be $\dfrac{1}{[t]}$, i.e. one over time (e.g. Hertz).

For the classical system, there is a natural time conserved quantity of with units $[F][x]=\dfrac{[m_F][x]^2}{[t]^2}$, called the energy of the system. Since multiplies of the eigenvalues of $H$ are what's conserved for static systems, $h_G$ is chosen so that $[H]$ has units of energy as well. So conventionally, one gives a new name to $[m_F]$ (e.g. “grams”) and relates $[h_G]$ to $[x]$, $[t]$ and $[m_F]$.

If $h_G$ is instead chosen to be unitless, eigenvalues of $H$ are frequencies. Then, if the concept of inert mass is introduced via a relationship akin to “$m_F\cdot\frac{{\mathrm d}^2}{{\mathrm d}t^2}\langle x\rangle = \nabla \langle H\rangle$”, one gets $[m_F]=\dfrac{[t]}{[x]^2}$.

For the Hamiltonan $H \propto A\cdot\Delta$ (a theory ansatz which results in the classical quadratic dispersion relation), one has $[H]=\dfrac{[A]}{[x]^2}$, so that the Schrödinger equation requires $[A]=[h_G]\dfrac{[x]^2}{[t]}$. If $[h_G]$ is chosen to be an action $\dfrac{[m][x]^2}{[t]}$, any mass $\mu$ provides a natural candidate $\dfrac{h_G^2}{\mu}$ for $A$.

The unit of time is further discussed in Its about time . note.

We discuss the two fundamental classical forces.

The constants $G$ and $\varepsilon_0$ and $k_B$ merely make it possible to choose convenient units for gravitational mass, electrical charge and temperature, respectively.

The speed of light $c$ is a constant popping up in Maxwell's equations that is fixed independently of the three notions above.


Newtonian force is defined as

  • $F_G=-G\cdot\dfrac{m_G\, M_G}{x^2}$,

with $[m_G]=[M_G]$. Thus

  • $[m_G]=\sqrt{\dfrac{[m_F][x]^3}{[G][t]^2}}$.

The equivalence principle suggests setting $[m_G]=[m_F]$. This then implies $[m_F]=\dfrac{[x]^3}{[G][t]^2}$. If we chose to give a name to $[m_F]$ (e.g. “grams”), this in turn fixes $[G]$


In the electrostatic theory, another particle property akin to the inert mass is introduced, namely the electrical charge $q_x$.

  • $F_E = q_x\cdot\,E$

so that

  • $[E]=\dfrac{[m_F]}{[q_x]}\dfrac{[x]}{[t]^2}$


The field $E$ is governed by Maxwell's equations

  • $\nabla E = k_E\cdot\rho$

with $[\rho]=\dfrac{[q_x]}{[x]^3}$, by definition.

  • $\nabla B=0$
  • $\dfrac{\partial}{\partial t}E = c^2\,\nabla B + \dots$
  • $\dfrac{\partial}{\partial t}B = - \nabla E$

The first implies $ [q_x]=\sqrt{\dfrac{[m_F][x]^3}{[k_E][t]^2}}$. In the SI system, the new name $A$ for “Ampere” is introduced and $[q_x]$ is chosen to be $A\,[t]$.

The second implies nothing.

The last two imply $[c]=\dfrac{[x]}{[t]}$. Fixing scales for $x$ and $t$ fixes the numerical value for $c$ experimentally. This can in turn be used to chose new scales that simplify the equations.

Remark on Rescaling of the field:

Working with the quantity $E':=\dfrac{E}{c}$ such that $F_{E'} = q_x\cdot c\cdot E'$ and $[E']=\dfrac{[m_F]}{[q_x]}\dfrac{1}{[t]}$ simplifies units and also $[E']=[B]$. Moreover using $\nabla':=c\,\nabla$ with $[\nabla']=\dfrac{1}{[t]}$ makes it so that all $c$'s disapper from the four equations.

P-planck time is the smallest amount of time

[citation needed]

There's no strong reason to believe that any of those particular numbers represent a least observable quantity.

Planck units are just a meme without physical significance.

You can look at them as follows:

The Schrödinger equation tells us how fast a quantum mechanical state oscillates. For an Eigenstate with energy E, it reads

$ \dfrac {d} {d t} \psi = (-i) \dfrac{ E }{ \hbar } \psi $

The rate of change of the state is determined by its eigen-frequency $ \omega = \dfrac{ E }{ \hbar } $.

Consider now the Einstein equation, which tells us how spacetime is affected by matter

$ G_{ \mu \nu } = 8 \pi \dfrac { G} { c^4} T_{ \mu \nu } $

Here c is the speed of light, G is the gravitational constant, $ T_{ \mu \nu } $ is the matter energy density tensor and $ G_{ \mu \nu } $ is the Einstein tensor. The effect of basic matter on spacetime curvature is not large, because $ c^4 $ is big. In fact, $ \dfrac { c^4} { G} = F_P $ is the Planck force of 10^40 Newton, and so things only get crazy when the pressure $ T_{ \mu \nu } $ get accordingly high.

We can rephrase this: If g is the metric tensor, the primary quantity determining spacetime, the time component of $ G_{ \mu \nu } $ expands as

$ G_{00} = \dfrac { 1} {c^2 } \dfrac { d^2} {dt^2 } g $

Now let $ L $ by any characteristic length scale, $ T $ the time that light needs to travel L, and consider the energy estimate $ E = L^3 T_{00} $.

With the Planck length $ l_P = \sqrt{ \dfrac { G \hbar }{ c^3 } } $, we may write 00 of the Einsten equation as

$ \left( T \dfrac { d} {dt} \right) \dfrac { d} {dt} g = 8 \pi \left( \dfrac { l_P } { L } \right)^2 \dfrac { E} {\hbar } $

So even if T is small, there is a quadratic comparison of the (energy along the) length to the Planck length.

For notes on units in electronics, see On electronics . note.

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