What is the meaning of a fundamental constant? Dimensioned or dimenionless

[Gerenuk]

How would you define your rod? It's surely something like "it consists of so many atoms". But then the length is derivable again.
I mean you have to be able to communicate it to someone without actually handing over the rod. Otherwise the size of you cupboard would be called a universal constants, which doesn't make sense?

 

[nonequilibrium]

I don't define the rod, it's just a certain rod I chose out and keep close.

Oh I see the problem: when the day after I compare L_c with the rod and it has changed, it might be because my rod has shrunk due to temperature difference? Can't we ignore things like that? (claim same environmental conditions)

 

[Gerenuk]

This notion is not what anyone would consider "fundamental". Fundamental means it's the same everywhere in the universe. Electrons have the same mass everywhere in the universe.

Personally, you might wish to redefine "fundamental" as to encompass the size of everyday object's length. Then absolutely everything will be a "fundamental" constant in your view.
But that's philosophical non-sense and does not yield any useful insights.

 

[Gerenuk]

The only useful way to define "fundamental" is by saying it's non-local and basically the same everywhere without me having to bring my sample rod.

This works for electron mass. I'm fairly sure aliens electrons weigh the same as on earth. So there are many of them. Not just a unique item.

 

[nonequilibrium]

Well I wasn't claiming the rod to be fundamental, just a unit. I was basically talking about the fact that a certain (actual) fundamental constant expressed in a certan unit divided by that unit would be a dimensionless constant, and bringing that into the discussion "observational change <=> change in dimensionless constant"

But okay, I think I get your point: using units that are not universal is silly(?) So you'd like me to for example define a meter as x times the width of a hydrogen atom? That sounds universal. For that unit, a change in beta would imply a change in L_c divided by the Bohr radius, or equivalently a change in $$\frac{L_c m_e c \alpha}{h}$$, right? (I've used the idea of Bohr radius out of DaleSpam's posts he linked to)

Would it have made any difference if I had used the current definition of meter? I.e. in function of c?

 

[Gerenuk]

But okay, I think I get your point: using units that are not universal is silly(?) So you'd like me to for example define a meter as x times the width of a hydrogen atom? That sounds universal. For that unit, a change in beta would imply a change in L_c divided by the Bohr radius, or equivalently a change in $$\frac{L_c m_e c \alpha}{h}$$, right? (I've used the idea of Bohr radius out of DaleSpam's posts he linked to)

It's not silly. It's very practical.

But it has no deep physical meaning because it makes absolutely everything object in the world belong to what you call "universal".

If instead you restrict "universal" to measurements that can be "communicated over telephone", then you single out only very few numbers. If it's only so few, then maybe they have some special meaning.

You may use Bohr radius, but then you are basically referring to the constants in there.

 

[Dale]

1) "So, basically we have just c doubling and the permittivity and permeability halving and no other changes." Well I suppose there could be more changes, right? Would it in principle suffice to check each of the dimensionless fundamental constants for a change?

Yes. If there were some measurable change in physics and we checked all of the dimensionless fundamental constants and found them all unchanged then that would indicate at least one unknown fundamental dimensionless constant.

2) So let's for a moment define meter as the length of a certain rod and a second in a similar way (earth's orbit or something). Now let $$L_c$$ be the distance light travels in one second, expressed in meter. Now define $$\beta = \frac{L_c}{[L_c]}$$, then beta is a dimensionless number! It's equal to the amount of rods light passes in a certain fraction of Earth's orbit. What if say $$\beta' = 2 \beta$$? Surely this is an observational difference. So does that imply there must be a change in one of the 26(I think) fundamental dimensionless constants? I suppose it does, right? Because if it doesn't, then there wouldn't be an observational difference, although there certainly is. How do you know what dimensionless constants changed?

You would have to measure each and see which had changed. The length of a rod is governed by the fine structure constant, and the radius of Earth's orbit is governed by the gravitational coupling constants, so those would be the first two to check. After that you would probably check the various mass ratios.

3) If I want to focus on changing k, does it mean that I have to find a dimensionless constant with k in it, so I know what has to change for it to be observable? The weird thing is, k has Kelvin as units (I mean, it has that unit in its unit), so there probably won't be any dimensionless constant for it! Unless N_A (Avog. number) = R/k can be seen as a dimensionless constant, but Baez didn't count it as one, and it might just be a definition? And if one argues "well then maybe k can't be involved in any observational difference", but surely that is wrong if we see PV = kNT => k = PV/NT. If I don't change the definitions involving the right hand side, then a change in k most definitely will be observational?

Yes, but I am not sure which dimensionless fundamental constants (if any) can be expressed in terms of k. I think that the dimensionless fundamental constant for k may in fact be Avogadro's number, but I am far from certain. It certainly was not included in Baez's list and I don't know if that is an oversight or deliberate.

 

[nonequilibrium]

"Yes, but I am not sure which dimensionless fundamental constants (if any) can be expressed in terms of k. I think that the dimensionless fundamental constant for k may in fact be Avogadro's number, but I am far from certain. It certainly was not included in Baez's list and I don't know if that is an oversight or deliberate."
Oh shoot, N_A has a unit: 1/mol. Well $$\frac{R}{kN_A} = 1$$, but that doesn't look like a real fundamental constant due to it being exact...

Hm, I suppose the same trick as before is applicable: define $$\gamma := \frac{k}{[k]}$$, then gamma is a dimensionless universal constant, am I right? (as long as k is universal, which -I think- means "as long as water starts freezing at the same pressure and temperature everywhere"(?)). So if I change gamma, something in the dimensionless constants changes... Is it me or is what I'm doing fishy?

EDIT: in fact: if k is determined by choosing 273K to be the freezing temperature of water, then in essence shouldn't k be already in the fundamental laws that determine chemistry? If so then it would be contradictory to say "change k and keep all other constants the same"... But then again, I wonder if -somehow- you can't say the same for e.g. $$c$$, which would make the previous doubtful.

 

[Gerenuk]

Hm, I suppose the same trick as before is applicable: define $$\gamma := \frac{k}{[k]}$$, then gamma is a dimensionless universal constant, am I right? (as long as k is universal, which -I think- means "as long as water starts freezing at the same pressure and temperature everywhere"(?)). So if I change gamma, something in the dimensionless constants changes... Is it me or is what I'm doing fishy?

That's not the right wording. Of course the boiling point of water is universal. But that is not the question.

k will be the same as long as you define your temperature scale (K) the same way. So k is purely a human invention. You could set k=14 J/X, then appropriately use X as your new unit of temperature, and the physics would be all the same.

It's the same as if the following scenario: You go to your garage and build a funky electronic device. When you spill water on it, it shows 15. You arbitrarily call the unit "Vodka". Now would 15 Vodka be qualified for a universal constant now?
I think it doesn't make much sense. But that's exactly what happened with k! Someone build a so-called thermometer for which he made up some *own units*. k only depends on this choice of units.
That's why k never appears in any dimensionless constants.

EDIT:
Hmm, or maybe another view is more appropriate! k is actually calculable! If only you could solve the quantum mechanics of boiling water, then you could calculate k. Right?

 

[Dale]

Hmm, or maybe another view is more appropriate! k is actually calculable! If only you could solve the quantum mechanics of boiling water, then you could calculate k. Right?

I suspect that is correct and is probably why no dimensionless constants related to thermodynamics made Baez's list. In principle they are not fundamental, but simply due to the application of the fundamental rules to complicated systems. Similarly with the dimensionless constants in fluid mechanics.

 

[Gerenuk]

I'm surprised myself, but it's most probably true that k can in theory be calculated from the fine structure constant and the mass ratio of proton to electron alone.

 

[D H]

I'm surprised myself, but it's most probably true that k can in theory be calculated from the fine structure constant and the mass ratio of proton to electron alone.

I disagree with Boltzmann's constant being calculable. I do agree that it is not fundamental. It is the constant of proportionality that converts temperature to thermal energy, $E_{\text{thermal}} = 3/2 k T$ (for an ideal monatomic gas).

Some of the base units in the SI system are not really "base units" in the eyes of a physicist. Consider luminous intensity (SI unit=candela) and luminous flux (SI unit=lumen). Baez' list of fundamental constants does not contain anything that pertains to these units. These units relate a concept in physics (power) to human physiology (how the eye works). Temperature is, in a sense, another of those non-basic units. The basis for temperature is energy. If we instead measured temperature in units of joules/molecule or joules/mole there would be no need for Boltzmann's constant or the ideal gas constant.

Another way to look at it: Force used to be a base unit in and of itself. Remnants of this concept are still around: The English pound-force, for example. Newton's second law is F=ma in SI units but it is F=kma in English units. The k in F=kma is just a constant of proportionality based on an artificial rather than fundamental definition of units. (Note well: The k in F=kma is not Boltzmann's constant).

While the SI system did do away with force as a base unit (the Newton is instead a derived unit), it did not do away with all non-essential units. Temperature and luminosity remain in SI as units akin to the pound-force.

 

[Gerenuk]

I disagree with Boltzmann's constant being calculable. I do agree that it is not fundamental. It is the constant of proportionality that converts temperature to thermal energy, $E_{\text{thermal}} = 3/2 k T$ (for an ideal monatomic gas).

I'm sure you can. I'll try to explain how.
I take melting and boiling water and put a gas thermometer in contact with it. I put a mark 273 and 373 on the gas thermometer. Next I take any experiment that yields the k. I simulate this experiment in the computer and calculate what will be the results and I'm done.

Refering to your example I would determine the numerical value of the temperature with my gas thermometer setup. I add the unit Kelvin to this number. Next I measure the velocity of the molecules in the gas thermometer. Therefore I can directly determine k.
OK?

EDIT: and I could do these experiments in a quantum mechanical simulator which only knows about the fine structure constant and atomic masses.

 

[nonequilibrium]

I also agree that k has is derivable, of course! The value of k is basically determined by using the point where water freezes as a reference point. Surely such a point must be calculable out of the fundamentals. It's what I tried to say earlier with "in fact: if k is determined by choosing 273K to be the freezing temperature of water, then in essence shouldn't k be already in the fundamental laws that determine chemistry?"

But about k's (un)fundamental nature: but take for example charge: its unit can be put into a combination of L,T,M; so charge Coulomb is also a superfluous definition, isn't it? Yet we don't mind calling e fundamental. Okay, I don't know if the value of e is derivable even using only L,T,M; but the argument given was that T was a superfluous definition because we had energy, my point is that charge seems to be of the same category?

I also read an interesting paper where they said "mass" is a superfluous definition, making G the same status as k... (the author is called N V Studentsov, and it was published). A short quote out of it, of which I'm not entirely sure what it means:

The two coefficients of proportionality in Newton's laws may be equated to one, and then mass and force become derived physical quantities. Unit mass is the mass producing unit acceleration at unit distance. However [bla bla superfluous definition]

I admit it sounds like crackpottery on first sight, but the rest of the paper is really "sane" and professional (it's about the more general matter of defining constants for SI and other systems), so I believe what he says about mass must make sense.

 

[Gerenuk]

But about k's (un)fundamental nature: but take for example charge: its unit can be put into a combination of L,T,M; so charge Coulomb is also a superfluous definition, isn't it? Yet we don't mind calling e fundamental. Okay, I don't know if the value of e is derivable even using only L,T,M; but the argument given was that T was a superfluous definition because we had energy, my point is that charge seems to be of the same category?

Yes, actually. You couldn't make aliens understand you value of e, because it depends on the choice of units. So as I said in the end only the fine structure constant is really fundamental. Morever, to my knowledge there is no theory yet, that can predict mass ratios.

Well, and going deeper into particle physics you probably get the other constants from Baez's lists.

Hmm, I wonder how much the value of one single particle is a fundamental. I could take two particles, let them attract and measure their velocity...
Hmm, just a thought...

 

[nonequilibrium]

Gerenuk, that doesn't really seem to be the point? With that reasoning, you're basically saying k is equally fundamental as c, e, ...? We were talking about if k is less fundamental than the other fundamental constants. First you said k is not important because it depends on the existence of a superfluous unit. Then I argued that e or G could be of the same type, and then you conclude that all dimensioned constants are equal (and according to you, equal in their "unfundamentalness"). It seems like we're hopping around without knowing what the point is, or maybe that's me, I'm just a bit confused about what is actually being claimed. So are you still saying k is of a lesser nature than e or c? And is e of a lesser nature than c? Or are they all of the same nature?

 

[Gerenuk]

My first guess is that indeed all the constants you mention are on an equal footing. I mean the second is defined by an atomic transition, which is not better than defining temperature by the melting point of water.
So in the end only the dimensionless constants are really fundamental.
Does that make sense?

 

[nonequilibrium]

Well it depends on what you call fundamental, so to avoid arbitrary semantics, I'll define different ways of fundamentality:

prerequisite: constant must be independent of space and time (this assumes the units themselves can also be considered independent of space and time)

A) a constant is fundamental if its value doesn't arise out the theory [in principle, for as far as we know] but has to be put in (this means that what is considered fundamental can change in time, but I see no objection to that)

B) a constant is fundamental if it does not contain a superfluous unit

C) a constant is fundamental if and only if a change in its value corresponds with an observational change in our universe

It is clear that only dimensionless constants have c-fundamentality, and I assume this is what Gerenuk was talking about these lasts posts.

What constants have a- and/or b-fundamentality? If k follows out of chemistry which follows out of QM (which seems very plausible), then k does NOT have type 1-f. I would think c, e, G, h would have type 1-f. I suppose either epsilon or µ too, the other one is then determined by c = 1/sqrt(epsilon µ).

As for type 2-f: NOT e, G, k?

 

[Dale]

Temperature is, in a sense, another of those non-basic units. The basis for temperature is energy. If we instead measured temperature in units of joules/molecule or joules/mole there would be no need for Boltzmann's constant or the ideal gas constant.

This is not quite correct. It only works for an ideal monoatomic gas. The correct definition of temperature is dS/dU=1/T, it just happens to work out that for an ideal gas it becomes proportional to the energy per molecule. For other substances it can be more complicated.

However, your overall point is correct, if mass, length, time, and entropy are considered the basic units then temperature is a non-basic unit. Of course, I always feel like my brain is becoming disordered when I think too much about entropy.

 

[Rap]

Boltzmann's constant is just a conversion factor to convert temperature to an energy. You could have temperature defined in units of energy - every where you see kT, just replace it with T, every where you see T alone, leave it alone. The question of what does that energy represent is rather complicated. You could say that it is twice the energy per molecule per degree of freedom of that molecule, but this breaks down when quantum considerations are introduced which can partially "freeze out" some degrees of freedom.

Dimensionless constants are more fundamental than dimensioned constants and it is true that if the dimensionless constants are fixed, then the dimensioned constants could be varying all over the place, and we would have absolutely no way to detect it - which means that the whole idea of "what if I changed the speed of light" has no meaning if the dimensionless constants are kept the same.

There are many fundamental dimensionless constants - one for every force - the electromagnetic coupling constant (aka the fine structure constant), the gravitational coupling constant, the weak coupling constant, the strong coupling constant. Also the ratio of the mass of any elementary particle to the mass of an electron are fundamental dimensionless constants. Of course by "fundamental" I mean that we presently have no way of deriving them, not that they are by definition underivable. The dimensionless "numbers" (e.g. Reynolds number) of fluid dynamics are not constants, they are different for different situations.