# I Planck Mass

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1. Aug 31, 2016

### Figaro

What is the difference between these reduced Planck masses?

$M_p = \sqrt \frac{ħc}{8πG}~$ and $~M_p = \sqrt \frac{1}{8πG}$

I see people use either, but what is the difference? It can't be that ħ and c are set to 1 because G is also a Planck unit and it would be nonsense to set a planck mass if this would be the case.

2. Aug 31, 2016

### vanhees71

There is no difference. It's only that most of the time, one sets indeed conveniently $\hbar=c=1$ (the socalled "natural units") since these are anyway only conversion factors for a very arbitrary and unnatural set of units used in experimental physics and everyday life (of course, it's important to have these units accurately defined and reproducible, I'm talking only about theoretical physics, where one can use other units with more ease).

With these natural units you measure lengths and time intervals in the same units (in HEP usually in fm, were $1 \; \mathrm{fm}=10^{-15} \mathrm{m}$). In relativistic physics it's nearly absurd to measure space and time in different units, and indeed even in the SI the corresponding conversion factor, the speed of light in vacuum, is fixed exactly to some value. Since time can be measured very accurately the second as the unit of time invervals is defined via a frequency of a hyperfine transition in cesium, and the meter from that by giving $c$ this value (which was chosen such as to keep the meter as accurately the same as it was defined by previous definitions). Since further the rest energy of a particle is related to its mass by $E_0=m c^2$, also masses and energies are measured in the same units (in HEP usually in MeV or GeV). Since further the dimensions energy and momentum are only different by a quantitity with dimension velocity they are also measured in the same units.

Now you set in addition $\hbar=1$. It's of the dimension of an action, i.e., of the dimension momentum times length or energy times time. Since these quantities are now dimensionless instead of the energy-momentum-mass units and the time-lengths units you need only one of them. In practice one usually keeps however MeV or GeV for the former set of quantities and fm for the latter. The only thing to convert between them in natural units is the value $\hbar c= 0.197 \;\mathrm{GeV} \, \mathrm{fm}$. Then you can always easily get the quantities in SI units, if needed, just from dimensional analysis. These units are very convenient in HEP physics, and that's why they are used.

On the other hand, it's not the end of the story, because you still have one arbitrary unit left (like GeV for mass, energy, momentum). But taking into account gravity (i.e., the General Theory of Relativity) you have one more conversion factor which is just due to choice of arbitrary units, and that's the gravitational constant $G$ or (equivalently) the Planck mass! The Planck mass thus is a natural unit for mass, energy, and momentum. Now you can measure all these quantities in terms of Planck masses, and thus setting $M_p=1$ eliminates this last arbitrariness. Now all quantities are measured in terms of dimensionless numbers and that's a true natural system of units. That's why Planck introduced these units and that's why they're named after him. The only reason why they are almost never used is that the numbers for everyday issues and even in HEP physics (note that the LHC provides the largest men-made energies for protons and heavy ions and this is just at the TeV scale), because the Planck mass is simply huge. In HEP units it's $1.220910 \cdot 10^{19}\, GeV$. In SI units it's tiny, namely $2.176470 \cdot 10^{-8} \; \mathrm{kg}$. So in everyday life you'd have huge numbers to by a pound of meat, and in HEP physis you'd have tiny numbers for masses of elementary particles and available beam energies.

The good thing is that the choice of units is in principle completely irrelevant since physics is independent of the choice of units. No matter in which units you measure things, the natural laws stay always the same. Which units you choose is just a question of convenience for the physics you are interested in.

3. Aug 31, 2016

### Figaro

Thank you for the further clarification, but is it really natural to set $~ħ,c =1~$ only and not G, I mean, why bother set some to 1 while not G, why not just write it all out or not since they are all planck units. Sorry, because it's just hard to digest.