To find the proper combination of physical constants to find the Planck length, we can use a procedure called dimensional analysis. In this procedure we use the dimensions (or units) of physical quantities to find how they must be combined in order to give us another physical quantity of interest to us. This is a technique used by scientists when they do not have a theory to describe some phenomenon.
As an aside, the technique can (and should) be used by students of science to check the answers to computations they make, because if the units of their answer come out wrong, the answer itself must be wrong. Dimensional analysis is a quick and easy self-check, although it does not guarantee an answer is correct just because the units are.
Generally in dimensional analysis square brackets surrounding a quantity indicate that we should consider only its dimensions (or units), not its numerical value. The numerical value will be used at the end, but first we consider just the units. So, in determining the Planck length, we have a relation that says:
[l] = [ha cb Gc] = [h]a [c]b [G]c
We will use the SI units for each dimension to simplify the analysis. Of course, this procedure will work with other units as well; it is only the dimension, whether length, time, mass, etc., that matters. We pick SI for convenience and specificity.
The units for the three quantities we have are:
[h] = J s
[c] = m/s
[G] = N m2 / kg2
Substituting these units into the above expression, we get:
m = (J s)a (m/s)b (N m2/ kg2)c
We have to break down joules and newtons into their equivalent combinations of meters, kilograms, and seconds to proceed. These are
J = kg m2/s2
N = kg m/s2
And now we substitute these in.
m = ( (kg m2/s2) (s) )a (m/s)b ( (kg m/s2) (m2/ kg2) )c
We have to combine like terms. This gives us
m = ( kg m2/s )a (m/s)b (m3/kgs2 )c
and finally
m = ( kga–c) (m2a+b+3c) (s–a–b–2c)
We have derived three expressions, and these can be used to relate the exponents to each other. By comparing the units on either side of the equation, we find that
a – c = 0 since there are no kilograms to the left of the equal sign.
–a–b–2c = 0 since there are no seconds to the left of the equal sign.
2a+b+3c = 1 since there is only one power of meters to the left of the equal sign.
This gives us three equations in three unknowns, and we can solve them in the normal way of substitution. We find the values a = c = 1/2, b = –3/2.
This leads us to the Planck length:
\begin{equation} l_{planck}= h^{1/2}G^{1/2}c^{-3/2} = \left( \frac{h G}{c^3}\right)^{1/2} \end{equation}
If instead of a length scale we had wanted to find a mass scale, we could have used the same procedure, except that the left-hand side of our equation would have had one power of kilograms instead of one power of meters. Everything else would have been the same. In that case we would have found the Planck mass:
\begin{equation} m_{planck}= h^{1/2}c^{1/2}G^{-3/2} = \left( \frac{h c}{G^3}\right)^{1/2} \end{equation}
If you are interested, see if you can figure out how this expression is determined. The procedure is identical to the previous one, but for the minor difference we have already pointed out. The Planck mass will be important in our consideration of the earliest moments of the Universe, when its age was of order the Planck time. If you like you can also try to work out the Planck time this way and compare it to the value mentioned in the main text—they should be the same.
The German physicist Max Planck first undertook dimensional analysis of this sort in 1899. He thought of these units, lplanck, tplanck, mplanck, etc., as natural units, in that they are independent of any particular theory or particle. The basic units of length, mass, and time can be combined in various ways to derive expressions for energy, force, pressure, etc., so we can also compute a Planck energy, a Planck force, and so on.
In addition to the three mechanical constants we have used here, Planck also used the Coulomb constant (like the gravitational constant G, but for the electrical force) and the Botlzmann constant, which is important in thermodynamics and statistical physics. These are used when we wish to venture beyond mechanics and include electromagnetic and thermal phenomena.
Planck units are just one set of many other natural units that are possible, but in some sense they are the most fundamental because, as mentioned already, they do not depend on the properties of any particular particle or theory. They depend only on physical constants related to space, time, and the interactions between particles.