# I Is there a unit system with $\hbar=G=c=\mu_0=1$?

#### JuanC97

Hello, I've been told that 'natural units' ensure $\hbar=G=c=\mu_0=1$ but...
when I look for it in Wikipedia I find that there are (mainly) two kinds of natural/geometrized units: The Planck ones and the Stoney ones,
(https://en.wikipedia.org/wiki/Geometrized_unit_system)

Also, some variations of the CGS system and Heaviside-Lorentz unit system share some of the desired magnitudes normalized to one
(https://en.wikipedia.org/wiki/Centimetre–gram–second_system_of_units#Various_extensions_of_the_CGS_system_to_electromagnetism)
(https://en.wikipedia.org/wiki/Lorentz–Heaviside_units)

but none of these systems seem to be compatible with $\hbar=G=c=\mu_0=1$ so...
The question is: Does exist a unit system with those specific magnitudes set to one?

Related Special and General Relativity News on Phys.org

#### PeterDonis

Mentor
Does exist a unit system with those specific magnitudes set to one?
No. It's not possible to set $\hbar$, $G$, and $c$ all to one. (Setting $c = 1$ requires setting $\mu_0 = 1$, or at least I've never seen units that don't do that, so I'm leaving $\mu_0$ out for the rest of this post.) Basically, you have two choices:

(1) Set $G = c = 1$. This amounts to measuring mass/energy and length/time in the same units. However, if you do this, you can't set $\hbar = 1$, because the ordinary units of $\hbar$ are energy times time, so the units of $\hbar$ in a system where $G = c = 1$ would be energy/mass/length/time squared. (I've never actually seen these units used for $\hbar$, because I've only seen units with $G = c = 1$ used in the context of GR, which is not a quantum theory so $\hbar$ never appears.)

(2) Set $\hbar = c = 1$. This amounts to measuring mass/energy and length/time in inverse units (i.e., mass/energy is inverse length/time, and vice versa). However, if you do this, you can't set $G = 1$, because the ordinary units of $G$ are energy times distance divided by mass, so the units of $G$ in a system where $\hbar = c = 1$ will be energy/mass squared, or inverse length/time squared.

#### sweet springs

ℏ=G=c=μ0=1
If so consequently $\epsilon_0=1$ also.

$$\hbar=G=c=k_B=1$$ is usually used. All the physical quantities are expressed by one of dimensions T,L and M. I do not think we can reduce more.

#### PeterDonis

Mentor
$\hbar=G=c=k_B=1$ is usually used.
This is impossible; as I've already pointed out, you can't set $\hbar$, $G$, and $c$ all to 1. Where are you getting this from?

#### sweet springs

For an example,
----------------
The five universal constants that Planck units, by definition, normalize to 1 are:
the speed of light in a vacuum, c,
the gravitational constant, G,
the reduced Planck constant, ħ,
the Coulomb constant, ke = 1/4πε0
the Boltzmann constant, kB
-----------------
https://en.wikipedia.org/wiki/Planck_units

I find $\mu_0=4\pi$ here. This may partly satisfy OP and would satisfy him fully by slight modification of unit charge definition or ke=1/4$\pi$.

Last edited:

#### PeterDonis

Mentor
The five universal constants that Planck units, by definition, normalize to 1
The Wikipedia page is misleading here. The numeric values are "1", but the units of the constants are not all dimensionless. It's not possible to make all of those constants dimensionless, for the reason I gave before. I am understanding the OP to be asking about the possibilities for setting constants to the dimensionless value 1.

#### sweet springs

Thanks. I understand 1 here means normalization not dimensionless.
I have thought that making any physical constant having dimension dimensionless value is impossible.

#### PeterDonis

Mentor
I have thought that making any physical constant having dimension dimensionless value is impossible.
Of course it's possible. The most common example is setting $c = 1$ (dimensionless) in relativity. That amounts to measuring length and time in the same units. The reason this is possible is that space and time are not separate things; relativity tells us that they are linked, so we can measure them in the same units.

#### sweet springs

In usual unit 30 cm gives us an idea that it is a length. c=1(dimensionless) means that 30 cm has double meanings, 30 cm of length or 30 cm of time ,i.e. 1 nano second in usual unit. Or complex like $\sqrt{t^2-x^2}$.
In c=1(dimensionless) we have to add dimension L, T to a value to distinguish what we mean by the value .

Last edited:

#### PeterDonis

Mentor
c=1(dimensionless) means that 30 cm has double meanings, 30 cm of length or 30 cm of time ,i.e. 1 nano second in usual unit.
Yes.

In usual unit 30 cm gives us an idea that it is a length.
That's an artifact of human history and has nothing to do with physics. We could also use "time" nomenclature. For example, in astronomy contexts distance is often measured in years--we just call them light-years. We could just as easily call the 30 cm of time in the case you describe 30 light-cm of time (in fact, Taylor & Wheeler's textbook Spacetime Physics takes exactly this approach, except they use meters instead of centimeters). None of this affects the physics at all; it's just a matter of nomenclature.

We have to add dimension L, T to a value to specify what we mean by the value .
No, we don't. Spacelike and timelike intervals are perfectly distinguishable without putting different "dimension" labels on them. You just look at the sign of the squared interval.

#### sweet springs

Spacelike and timelike intervals are perfectly distinguishable without putting different "dimension" labels on them. You just look at the sign of the squared interval.
Interesting. So in $\sqrt{x^2-t^2}$ convention of interval, real 30 cm means 30 cm length and imaginary 30i cm means 1 nano second of time.

#### sweet springs

The question is: Does exist a unit system with those specific magnitudes set to one?
------
The Planck charge is defined as:[1][2]
${\displaystyle q_{\text{P}}={\sqrt {4\pi \epsilon _{0}\hbar c}}={\frac {e}{\sqrt {\alpha }}}\approx 1.875\;5459\times 10^{-18}}$coulombs,
where
${\displaystyle c\ }\$ is the speed of light in the vacuum
${\displaystyle \hbar }$ is the reduced Planck constant
${\displaystyle \epsilon _{0}\ } \$ is the permittivity of free space
${\displaystyle e\ }$ is the elementary charge
${\displaystyle \alpha \ } \$ is the fine structure constant.
------
https://en.wikipedia.org/wiki/Planck_charge

I hope OP would modify Planck unit system by making $\sqrt{4\pi}q_P$ be a new unit charge to get the unit he wants, in case he mentions the physical constants be not dimensionless but just normalized.

#### stevendaryl

Staff Emeritus
The Wikipedia page is misleading here. The numeric values are "1", but the units of the constants are not all dimensionless. It's not possible to make all of those constants dimensionless, for the reason I gave before. I am understanding the OP to be asking about the possibilities for setting constants to the dimensionless value 1.
I see the distinction you're making, but I'm wondering whether there is an objective notion of what a quantity's units are. If someone always measured distances in light-seconds, it might not occur to them that time and distance have different units. Is there some objective sense in which they do? We can certainly choose to measure time and distance in different units. But we can also choose to measure depths in the ocean with different units (fathoms) than we use for measuring horizontal distances (yards, say). There would be a universal constant, $K$, that converts horizontal distance to vertical distance. Using meters for both amounts to setting the constant to 1.

#### PeterDonis

Mentor
If someone always measured distances in light-seconds, it might not occur to them that time and distance have different units. Is there some objective sense in which they do?
I don't think so. But I think there is an objective sense in which you can't make all quantities unitless; for example, in post #2, the argument I made for why you can't set $\hbar$, $G$, and $c$ all to the dimensionless number $1$ is also an argument for why you can't make mass/energy/length/time unitless; those quantities have to have units. But you can choose whether they all have the same units, or whether one pair (mass/energy) has inverse units from the other pair (length/time).

#### Mister T

Gold Member
Is the distinction between unit and dimension relevant here?

For example, the radian is a unit, but it is not a dimension. If you have, say, an angle that measures 1.3 radians, that is a pure (as in dimensionless) number.

If the speed of light is 1 light year per year, is that a dimensionless number? I've noticed that people can be passionate about this business of measuring distance and time in the same unit. They don't like it because distance and time have different dimensions.

I can't quite sort out their objection. We used Taylor and Wheeler early on in my education, so I find the notion of using the same unit to measure both distance and time perfectly acceptable, but as an aside I do remember that it frustrated me when I was first exposed to it,

#### PeterDonis

Mentor
If the speed of light is 1 light year per year, is that a dimensionless number?
I would say yes, but I understand that many people find that interpretation difficult. I would say, however, that you can treat this speed (or any speed, in the "natural" units of relativity) as a dimensionless number, and still accurately model all of the physics. Someone who insists that a light-year and a year are "different units" is not wrong, exactly, but they are adding something to their model that is entirely superfluous, and would be scraped right off by Occam's Razor if they were willing to wield it.

#### sweet springs

I have heard some mathematicians saying that they do not understand why physics people care of dimensions of value. It seems nonsense to them. Maybe we have to follow them.

#### stevendaryl

Staff Emeritus
I have heard some mathematicians saying that they do not understand why physics people care of dimensions of value. It seems nonsense to them. Maybe we have to follow them.
Partly, it's a sanity check on calculations. If you're calculating an energy, for instance, and your answer has the wrong units, then you know that you made a mistake somewhere.

I remember as a kid thinking about Einstein's formula: $E = mc^2$. I had a moment of vertigo when it occurred to me that depending on whether you measure $c$ in meters per second or miles per second, you get wildly different answers for how much energy is associated with a particular amount of mass. I don't know why that bothered more than the simpler paradox that the length of an object was different depending on whether you were using inches or meters.

#### stevendaryl

Staff Emeritus
There's also a nice benefit to keeping $\hbar$ and $c$ in the expressions for calculations, because then we know that a way to get a classical or nonrelativistic limit is to look at expressions where those constants drop out (or terms involving them become negligibly small compared with other terms).

#### Mister T

Gold Member
There's also a nice benefit to keeping $\hbar$ and $c$ in the expressions for calculations, because then we know that a way to get a classical or nonrelativistic limit is to look at expressions where those constants drop out (or terms involving them become negligibly small compared with other terms).
An atomic physicist demonstrated that a lot of atomic behavior can be modelled without relativity or quantum physics. He would say there's not a c or an h anywhere in these equations.

#### JuanC97

No. It's not possible to set $\hbar$, $G$, and $c$ all to one [...] Basically, you have two choices:

(1) Set $G = c = 1$. This amounts to measuring mass/energy and length/time in the same units. [...]
(2) Set $\hbar = c = 1$. This amounts to measuring mass/energy and length/time in inverse units [...]
This was really useful Peter, just as you pointed out is not possible to set $G=c=\hbar=1$.

I hope OP would modify Planck unit system by making √4πqP4πqP\sqrt{4\pi}q_P be a new unit charge to get the unit he wants, in case he mentions the physical constants be not dimensionless but just normalized.
I thought about this too. Redefining or the Planck charge or the electric charge $e$ you could have $\epsilon_0=1$, then $\mu_0=1$ is satisfied if the unit system is natural ( $\hbar=c=1$ ).

-------------------------------
Finally, I double-checked and I found I dind't need to have $G=1$ so the unit system that was refered as 'natural' in my context (particle physics/cosmology) was simply a system with $\hbar=c=\mu_0=1$
http://www.phys.ufl.edu/~korytov/phz4390/note_01_NaturalUnits_SMsummary.pdf

Even more, I think cosmologists and quantum field theorists can maintain $\hbar=c=\mu_0=1$ and set $K_B=1$ in order to simplify equations related to statistics and temperature. (naively, I guess that's because $\hbar=1$ and $c=1$ affect kinetic quantities related to angular momentum and velocity, $\mu_0=1$ just affects electric quantities, therefore, one should be allowed to modify $K_B$ - affecting kinetic quantities related to temperature - in order to normalize the 3 kinetic quantities and the electric one that were mentioned at the very beginning of the link I just posted).

Last edited:

#### PeterDonis

Mentor
Even more, I think cosmologists and quantum field theorists can maintain $\hbar=c=\mu_0=1$ and set $K_B=1$ in order to simplify equations related to statistics and temperature.
This is commonly done in quantum field theory and quantum statistics, yes. In cosmology, I think it's more common to set $G = 1$ instead of $\hbar = 1$ (but the rest would be the same).

#### sha1000

I understand that one can play with the units. But what about the dimensions? Does time really exist for example? I'm confused.

#### SiennaTheGr8

I would say yes, but I understand that many people find that interpretation difficult. I would say, however, that you can treat this speed (or any speed, in the "natural" units of relativity) as a dimensionless number, and still accurately model all of the physics. Someone who insists that a light-year and a year are "different units" is not wrong, exactly, but they are adding something to their model that is entirely superfluous, and would be scraped right off by Occam's Razor if they were willing to wield it.
For me it's as simple as this:

Our universe has a speed limit. Therefore every speed can be expressed as is a dimensionless fraction of it. I view $\beta$ as primary, not as shorthand for $v/c$.

#### PeterDonis

Mentor
Does time really exist
This question is too vague to answer as it stands. But if you operationalize it, it has a clear answer: time is what clocks measure.

"Is there a unit system with $\hbar=G=c=\mu_0=1$?"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving