Physics 374
Fall 2005

HW Set 7
Problem 1

It's given that the standard units for G are ((N (m^2))/(kg^2))(so that (GMm/(r^2))will be in Newtons).  Thus: [G] = ([M][L]/(([T])^2))(((([L])^2)/(([M])^2))) = ((([L])^3)/([M]))[T]^2.
Also, Planck's constant has units of J-sec, so [ℏ] = (\(TraditionalForm\`\([M]\)[L]\^2\/\([T]\)\^2\))([T]) = \(TraditionalForm\`\([M]\)[L]\^2\/\([T]\)\)
and then [c] = (([L])/([T])).

To get a mass, we'll need some conbination of these three such that:
[(G^α)(ℏ^β) (c^γ) ] = (((((([L])^3)/([M]))[T]^2))^α)\(TraditionalForm\`\((\([M]\)[L]\^2\/\([T]\))\)\^β\)((((([L])/([T]))))^γ) = \(TraditionalForm\`\(\(\([L]\)\^\(3  α + 2  β + γ\)\)[M]\^\(\(-α\) + β\)\)[T]\^\(\(-2\) α - β - γ\)\)
					         = [M]
This gives three equations for our three unknown exponents:
			3α+2β+γ = 0           -α+β = 1         -2α-β-γ = 0
			
Solving these simulateous equations gives α = ((-1)/2) , β = γ = (1/2)
So: (Subscript[m, natural]) = \(TraditionalForm\`\(\(ℏ\^\(1\/2\)\) c\^\(1\/2\)\)\/G\^\(1\/2\)\) = (Sqrt[ℏ(c/G)]).

For finding a natural length, the process is the same:

[(G^α)(ℏ^β) (c^γ) ] = \(TraditionalForm\`\((\(\([\)\(L\)\(]\)\)\^3\/\(\([\)\(M\)\(]\)\)[T]\^2)\)\^α\)\(TraditionalForm\`\((\(\([\)\(M\)\(]\)\)[L]\^2\/\(\([\)\(T\)\(]\)\))\)\^β\)\(TraditionalForm\`\((\(\([\)\(L\)\(]\)\)\/\(\([\)\(T\)\(]\)\))\)\^γ\) = \(TraditionalForm\`\(\(\(\([\)\(L\)\(]\)\)\^\(3  α + 2  β + γ\)\)[M]\^\(\(-α\) + β\)\)[T]\^\(\(-2\) α - β - γ\)\)
					         = [L]
And thus       3α+2β+γ = 1           -α+β = 0         -2α-β-γ = 0
Solving this system of equations gives α = β = (1/2), γ = ((-3)/2)
So (Subscript[l, natural])=  \(TraditionalForm\`\(\(G\^\(1\/2\)\) ℏ\^\(1\/2\)\)\/c\^\(3\/2\)\) = (Sqrt[G(ℏ/(c^3))]).

And likewise for the natural time:
[(G^α)(ℏ^β) (c^γ) ] = \(TraditionalForm\`\((\(\([\)\(L\)\(]\)\)\^3\/\(\([\)\(M\)\(]\)\)[T]\^2)\)\^α\)\(TraditionalForm\`\((\(\([\)\(M\)\(]\)\)[L]\^2\/\(\([\)\(T\)\(]\)\))\)\^β\)\(TraditionalForm\`\((\(\([\)\(L\)\(]\)\)\/\(\([\)\(T\)\(]\)\))\)\^γ\) = \(TraditionalForm\`\(\(\(\([\)\(L\)\(]\)\)\^\(3  α + 2  β + γ\)\)[M]\^\(\(-α\) + β\)\)[T]\^\(\(-2\) α - β - γ\)\)
					         = [T]

And thus     3α+2β+γ = 0           -α+β = 0         -2α-β-γ = 1
Solving this system gives α = β = (1/2), γ = ((-5)/2)
So (Subscript[t, natural]) = \(TraditionalForm\`\(\(G\^\(1\/2\)\) ℏ\^\(1\/2\)\)\/c\^\(5\/2\)\) = (Sqrt[G(ℏ/(c^5))])

Notice how tiny these natural scales are:  (Subscript[m, natural]) = 2 x (10^(-8))kg
					       (Subscript[l, natural])=  2 x (10^(-35))m
					       (Subscript[t, natural]) = 5 x (10^(-44))sec
While the mass scale isn't that small (for example, compare to the electron's rest mass of about 9 x (10^(-31)) kg), it is interesting to note that physics is only just beginning to talk about what might be going on at length scales on the order of (10^(-35))m.

Solution by T . Bing