IF planck's constant, Gravitational constant & velocity are taken as base units, find the dimensions of mass,length and time in their term. Mam this was the same question asked in our unit test. the same words as I have typed here were there in the question paper. So you are requested to help me out

Asked by  | 21st May, 2012, 10:41: PM

Expert Answer:

Fundamental physical constants
ConstantSymbolDimensionValue in SI units with uncertainties[5]
Speed of light in vacuum c L T??1 2.99792458×108 m s?1
(exact by definition of meter)
Gravitational constant G L3 M?1 T??2 6.67384(80)×10?11 m3kg?1s?2[6]
Reduced Planck constant ? = h/2?
where h is Planck constant
L2 M T??1 1.054571726(47)×10?34 J s[7]
Where, L is the length, M is meter and T is the time
the gravitational attractive force of two bodies of 1 Planck mass each, set apart by 1 Planck length is 1 Planck force. Likewise, the distance traveled by light during 1 Planck time is 1 Planck length.

Planck units are physical units of measurement defined exclusively in terms of five universal physical constants listed below, in such a manner that these five physical constants take on the numerical value of 1 when expressed in terms of these units. Planck units elegantly simplify particular algebraic expressions appearing in physical law. Originally proposed in 1899 by German physicist Max Planck, these units are also known as natural units because the origin of their definition comes only from properties of nature and not from any human construct. Planck units are only one system of natural units among other systems, but are considered unique in that these units are not based on properties of any prototype object, orparticle (that would be arbitrarily chosen) but are based only on properties of free space. The universal constants that Planck units, by definition, normalize to 1 are the:

  • Gravitational constant, G;
  • Reduced Planck constant, ?;
  • Speed of light in a vacuum, c;
  • Coulomb constant,  (sometimes ke or k);
  • Boltzmann constant, kB (sometimes k).

Answered by  | 22nd May, 2012, 11:55: AM

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