SI World and Planck World - And an IDEA

I was thinking about adding this to the back of the last post, Coupling Constants, but it was already pretty long.  To understand the following though, it is probably useful to read that post first.

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To make my point, I am going to compare the same activity conducted in two different worlds.  The first world I am going to call SI world, in which SI units are used.  The second world is Planck world, in which – unsurprisingly – only Planck units are used.  In Planck world, everyday people are used to talking in terms of “p-mass”, “p-length”, “p-time”, “p-charge” and all of the derived units, such as “p-force” (instead of newtons), “p-energy” (instead of joules), “p-current” (instead of amperes) and so on.  Because they live in Planck world, they know that all the fundamental physical constants are unity (see the table at the end of Coupling Constants) and they do everything in terms of Planck.

The activity is considering the force on a medium sized pineapple at sea level on an Earth-like planet.

There’s the simple way:

And then there’s the complex way, noting that we wouldn’t use the complex way other than as a method to extract a value which we can use to compare the strength of forces:

The complex way provides us with a couple of things, the value of n₁n₂/r², which could be reused if were talking about a conglomeration of charged massive particles (like protons) and, via a little side track, the coupling constants (which for Planck world is, of course, unity).  Note however that the magnitude of the gravitational coupling constants relates only to the square of the mass divided by Planck mass, so the side track isn’t really necessary in either world.

The same sort of process applies if we consider charged objects, imagining the same number of items (protons) for each object and the same distance, and noting that the resultant force will be one of repulsion:

Note that α_Ep is the fine structure constant, α.  Again, it is just the square of the value (the elementary charge) divided by the related Planck value (Planck charge).

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There’s no real reason to use the complex method to calculate the force.  And as mentioned above, there’s no need to use the side track in either case (gravitational or electromagnetic).  If you want to know the strength of coupling associated with any reference mass or charge, just divide the reference value (mass or charge) by the related Planck value (mass or charge) and square the result.  Simple.

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Just out of curiosity, I plugged in different values, namely two objects with a Planck unit each (mass in the first case and charge in the second).  These are the results:

Unsurprisingly, from one perspective at least, the values end up the same.  The question then arises: what is the Planck charge?  One way to think of it as the charge on the smallest possible charged (but non-rotating) black hole.

Such a black hole has a radius of r_s=2GM/c²=l_Pl.  So, M=c²l_Pl/G/2=m_Pl/2. It also becomes “extremal” when 2r_Q=r_s, where r²_Q=Q²G/(4πε₀c⁴), so:

2r_Q=2Q.√(G/(4πε₀c⁴))=l_Pl

Q=l_Pl.c².√(4πε₀/G)/2

Noting that l_Pl=√(ħG/c³) and

Q=√(ħG/c³).c².√(4πε₀/G)/2=√(ħc.4πε₀)/2=q_Pl/2

This means that the ratio of charge to mass in an extremal Reissner–Nordström black hole is unity, and half a Planck charge is the maximal charge on a minimal black hole (of half a Planck mass and radius of one Planck length).  This does at least gives some sort of meaning to the values I used, although it should be noted that it would not be possible to have such masses or charges at a separation of only one Planck length.  Halve them and you might be in business.

Generally, gravitation is labelled as a weak force, while electromagnetism is considered to be strong(er).  This is because, when considering atomic scales, we are talking about quite small masses and multiples of elementary charge.  Consider a hydrogen ion, which is a proton (the other isotopes of hydrogen which have neutrons have the special names deuterium [²H] and tritium [³H]).  This has a mass of 1.673×10^-27 kg, or 7.69×10^-20 units of Planck mass, and one elementary charge of 1.602×10^-19 C, or 0.0854 units of Planck charge.  So, yes, gravitational force between two protons is miniscule in relation to their mutual electromagnetic force of repulsion, but only because the charge is enormous by comparison.

What we could say, perhaps, is that subatomic (and elementary) particles can carry significantly more charge than mass.  Why is that?  I don’t know if anyone has even asked the question before.  There is fact that particles have relativistic mass, which varies between frames with different speeds, but not relativistic charge.

So, the question that comes to me is, what speed (in a given frame) would a proton need to reach to have the same magnitude of Planck mass as it has in Planck charge.

The relativistic mass equation is (where m₀ is the rest mass in a given frame):

m=m₀/√(1-v²/c²)

Rearranging for v:

v =√(1- m₀²/m²).c

We want to get the magnitude of the mass of the proton (m₀=m_p) in units Planck mass to equal the charge (e) in units of Planck charge (||m||=||e||):

v =√(1- (7.69×10^-20)²/(0.0854)²).c=√(1- 8.09×10^-37).c≈(1-4.05×10^-35).c

In other words, a proton achieves the relativistic mass equivalent in magnitude to its charge only when extremely close to the speed of light.  There is plenty of wriggle room in there, for sure, but the bottom line is that in order that the relativistic mass doesn’t get bigger than the charge (within realistic speeds), the rest mass of a proton must have a magnitude that is significantly lower than the magnitude of the charge. Other subatomic and elementary charged particles are available, but they all have significantly less mass than the proton (and charge within one order of magnitude).

The results in the tables above might make one wonder if the Planck force is a maximum.  It could be, at least in a sense.  It is the force required to accelerate one unit of Planck mass to the speed of light in a period of one Planck time, which would require one unit of Planck energy.  It would basically be instantaneous, especially if Planck time is fundamental, since it would be across a distance of one unit of Planck length.  Remember that one unit of Planck length per unit of Planck time is the speed of light.

The Falcon Heavy rocket has a maximum thrust of 22.8×10⁶ N. NASA’s Space Launch System (Block 1) has a maximum thrust of 39×10⁶ N.  The Starship rocket (still in development) has a projected maximum thrust of 89×10⁶ N.  These are a long way short of 1.21×10⁴⁴ N, but of course we talking about a completely different scale.

For comparison, we can consider the strong force.  Note that in the second paragraph of the related Wikipedia page, the forces are expressed in terms of a proton at an approximate distance of 10^-15 m.  We know the mass and charge of a proton, and can assume that the interaction is between two protons (being bound into a nucleus).  Here are the gravitational and electromagnetic forces involved (just the simple calculations, since it can be seen above that the force and coupling constants don’t change):

The strong force is approximately 100 times stronger than electromagnetic force so, in this case (between two protons), in the order of 20 kN.  See also here (per Bdushaw – I don’t know how accurate this is, or whether it is merely supposed to be representative):

Curiously, this value is within one order of magnitude of the force due of two Planck black holes (m=m_Pl/2) or indeed two Planck masses (m=m_Pl) separated by the same distance – namely 10^-15 m.

The forces are 8 kN in the first instance and 31 kN in the second or, in both cases, approximately 10⁴ N.  Given that this is based on a distance of approximately 10^-15 m and a force that is approximately 100 times stronger than the electromagnetic force, the results being approximately the same is remarkable.

Coincidence?  Probably.  I can’t think of a reason why the maximum strong force at a femtometre should be equivalent to the gravitational force between two individual Planck masses that are separated by a femtometre (and the force distance relationship would certainly not follow the curve shown above).

Well … not a good reason.  Here’s a completely bonkers reason.  Say we had an Intelligent Design and Engineering Agent (IDEA) who has committed to making the background of the universe pretty much like ours, a Planck universe in which the fundamental physical constants resolve to unity and the minimum divisions of space and time are Planck values.  The IDEA has a vague plan for the proton, but hasn’t set a size, mass or charge on it yet, just knowing that protons and neutrons will be bound together in a nucleus by some sort of strong force.  In the worst case, two protons will have to be bound together at a distance of a femtometre, but not for very long (note that ²He is extremely unstable, with a half-life of about 10^-9s).  The IDEA wants to leave open a wide range of values for the mass and charge of the proton (but knows that the proton will be very small), so what value of the strong force at about a femtometre will be required?

In summary:

•    The proton can be any size (smallest being a Planck volume)
•    The proton can be any mass (although the maximum mass in a Planck volume is half a unit of Planck mass)
•    The proton can have any charge (although the maximum charge in a Planck volume is half a unit of Planck charge)
•    Separation between two bound protons in worst case is ~1fm
•    Worst case is maximum charge, minimum mass (effectively zero mass)
•    The IDEA selects, for some reason, the Planck volume as the reference size of the yet to be determined proton

So the IDEA would need a strong force that just about counteracts the repulsive force of two half units of Planck charges:

Therefore, a bit under 10kN.  But the IDEA is not just a designer, it is also an engineer which means that it is going over-engineer the design, so we can easily double that – bump the strong force up to about 20kN.

(Note that, in some explanations this doesn't actually seem to be required.  In those explanations, the apparent factor of 2 appears to be due merely to vague specifications.  For example, the strong force is sometime defined as 137 times stronger [or approximately so] than the electromagnetic force for an elementary charge.  If that were so, the would mean that the strong force is precisely strong enough to account for two charges acting on each other that are q_Pl/e greater than e, or (q_Pl/e)²=1/α stronger.  I have my doubts about this, in part because the strong force as commonly referred to is actually the residual strong force, being the force that is left over from the actual strong force that is holding the constituent quarks together.  It would surely be nice to be able to say that the strong force is has a strength of unity in some sense, but things are not quite that simple.)

This over-engineering would mean that if protons were particles with minimal mass and the maximal charge on a Planck black hole (half a unit of Planck charge), then the nucleus would still hold together with the strong force.  Once this value had been set, the IDEA would be limited to a low proton mass value, because the ²He nucleus would become stable if protons were significantly heavier.

(Stability of ²He nuclei would affect the availability of free protons to be fused into isotopes of hydrogen on the path to other isotopes of helium and maybe prevent stars from forming, meaning that shining stars, planets, pond scum and, ultimately, IDEA worshipping humans would then have to be created using magic and/or miracles rather than physics).

Of course, I don’t think that this is the explanation.  But it is interesting that the forces involved are in approximately the same order of magnitude.