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A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of
Fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (~15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number(or any other interesting number you can think of ). If its coincidence with the age of
the universe were significant, what would this imply for the constancy of fundamental
constant. 


straight A space relation space which space gives space us space the space age space of space universe space is space given space by comma space

straight t space equals space open parentheses fraction numerator straight e squared over denominator 4 πε subscript straight o end fraction close parentheses squared space cross times space fraction numerator 1 over denominator straight m subscript straight p straight m subscript straight e squared straight c cubed straight G space end fraction

where comma space

straight t space equals space age space of space universe comma

straight e space equals space 1.6 space cross times 10 to the power of – 19 space end exponent straight C comma space is space the space charge space of space electron comma

straight epsilon subscript straight o space equals space absolute space permittivity space

straight m subscript straight p space equals space 1.67 space cross times space 10 to the power of – 27 end exponent space kg comma space si space the space mass space of space proton

straight m subscript straight e space equals space 9.1 space cross times space 10 to the power of – 31 end exponent space kg comma space mass space of space electron

straight c space equals space speed space of space light space equals space 3 space cross times space 10 to the power of 8 space straight m divided by straight s

straight G space equals space 6.67 space cross times space 10 to the power of 11 space Nm squared space kg to the power of – 2 end exponent space
Putting these values in the above equation, we get

straight t space equals fraction numerator space left parenthesis 1.6 cross times 10 to the power of negative 19 end exponent right parenthesis to the power of 4 space cross times space space left parenthesis 9 space cross times space 10 to the power of 9 right parenthesis squared over denominator left parenthesis 9.1 space cross times space 10 to the power of negative 31 end exponent right parenthesis to the power of 2 space end exponent cross times space 1.67 space cross times space 10 to the power of negative 27 end exponent space cross times space left parenthesis 3 space cross times space 10 to the power of 8 right parenthesis cubed space cross times space 6.67 space cross times space 10 to the power of negative 11 end exponent space end fraction

space space equals space open square brackets fraction numerator left parenthesis 1.6 right parenthesis to the power of 4 cross times 81 over denominator 9.1 space cross times space 1.67 space cross times space 27 space cross times space 6.67 cross times 365 cross times 24 cross times 3600 end fraction close square brackets cross times 10 to the power of 18 space years space

space space equals space 6 space billion space years
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