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In physics, the Planck mass, denoted by mP, is the unit of mass in the system of natural units known as Planck units. It is approximately 0.02 milligrams (roughly the mass of a flea egg[1]). For comparison, this value is of the order of 1015 (a quadrillion) times larger than the highest energy available to contemporary particle accelerators.[2]

It is defined as:

${\displaystyle m_{\text{P}}={\sqrt {\frac {\hbar c}{G}}},}$

where c is the speed of light in a vacuum, G is the gravitational constant, and ħ is the reduced Planck constant.

Substituting values for the various components in this definition gives the approximate equivalent value of this unit in terms of other units of mass:

${\displaystyle 1\ m_{\mathrm {P} }}$1.220910×1019 GeV/c2 = 2.176470(51)×10−8 kg[3] = 21.76470 μg = 1.3107×1019 u.[4]

Particle physicists and cosmologists often use an alternative normalization with the reduced Planck mass, which is

${\displaystyle M_{\text{P}}={\sqrt {\frac {\hbar c}{8\pi G}}}}$4.341×10−9 kg = 2.435×1018 GeV/c2.

The factor of ${\displaystyle 1/{\sqrt {8\pi }}}$ simplifies a number of equations in general relativity.

## Derivations

### Dimensional analysis

The formula for the Planck mass can be derived by dimensional analysis. In this approach, one starts with the three physical constants ħ, c, and G, and attempts to combine them to get a quantity whose dimension is mass. The formula sought is of the form

${\displaystyle m_{\text{P}}=c^{n_{1}}G^{n_{2}}\hbar ^{n_{3}},}$

where ${\displaystyle n_{1},n_{2},n_{3}}$ are constants to be determined by equating the dimensions of both sides. Using the symbols ${\displaystyle {\mathsf {M}}}$ for mass, ${\displaystyle {\mathsf {L}}}$ for length and ${\displaystyle {\mathsf {T}}}$ for time, and writing [x] to denote the dimension of some physical quantity x, we have the following:

${\displaystyle \,[c]\,={\mathsf {L}}{\mathsf {T}}^{-1}\ }$
${\displaystyle [G]={\mathsf {M}}^{-1}{\mathsf {L}}^{3}{\mathsf {T}}^{-2}}$
${\displaystyle \,[\hbar ]\,={\mathsf {M}}{\mathsf {L}}^{2}{\mathsf {T}}^{-1}\ }$.

Therefore,

${\displaystyle [c^{n_{1}}G^{n_{2}}\hbar ^{n_{3}}]={\mathsf {M}}^{-n_{2}+n_{3}}{\mathsf {L}}^{n_{1}+3n_{2}+2n_{3}}{\mathsf {T}}^{-n_{1}-2n_{2}-n_{3}}.}$

If one wants this to equal ${\displaystyle {\mathsf {M}}}$, the dimension of mass, using ${\displaystyle {\mathsf {M}}={\mathsf {M}}^{1}{\mathsf {L}}^{0}{\mathsf {T}}^{0}}$, the following equations need to hold:

${\displaystyle ~~~~~~~~~-n_{2}+~~n_{3}=1\ }$
${\displaystyle ~~~n_{1}+3n_{2}+2n_{3}=0\ }$
${\displaystyle -n_{1}-2n_{2}-~~n_{3}=0\ }$.

The solution of this system is:

${\displaystyle n_{1}=1/2,n_{2}=-1/2,n_{3}=1/2.\ }$

Thus, the Planck mass is:

${\displaystyle m_{\text{P}}=c^{1/2}G^{-1/2}\hbar ^{1/2}={\sqrt {\frac {c\hbar }{G}}}.}$

To be sure, dimensional analysis can only determine a formula up to a dimensionless multiplicative factor. There is no a priori reason for starting with the reduced Planck constant ħ instead of the original Planck constant h, which differs from it by a factor of 2π.

### Elimination of a coupling constant

Equivalently, the Planck mass is defined such that the gravitational potential energy between two masses mP of separation r is equal to the energy of a photon (or graviton) of angular wavelength r (see the Planck relation), or that their ratio equals one.

${\displaystyle E={\frac {Gm_{\text{P}}^{2}}{r}}={\frac {\hbar c}{r}}.}$

Isolating mP, we get that

${\displaystyle m_{\text{P}}={\sqrt {\frac {\hbar c}{G}}}}$

Note that if, instead of Planck masses, the electron mass were used, the equation would require a gravitational coupling constant, analogous to how the equation of the fine-structure constant relates the elementary charge and the Planck charge. Thus, the Planck mass can be viewed as resulting from absorbing the gravitational coupling constant into the unit of mass (and those of distance/time as well), like the Planck charge does for the fine-structure constant.