If the velocity of light c, the constant of gravitation G, and Plank's constant h are chosen as fundamental units, find the dimensions of mass, length, and time in terms of c, G, and h.

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In physics, the choice of fundamental units can vary depending on the field of study. In certain advanced fields, such as theoretical physics, it can be useful to use a system of natural units in which the velocity of light in a vacuum (c), Newton’s gravitational constant (G), and Planck’s constant (h) are chosen as the base units. These are known as Planck units. By setting these constants to 1, other units and constants can be expressed in terms of these fundamental quantities.

Property	Description
Context	The article explores expressing other physical quantities in terms of c (velocity of light), G (gravitational constant), and h (Planck’s constant) as fundamental units.
Reasoning	These constants are fundamental and have well-defined values, allowing for a consistent system of units.
Challenge	We need to derive the dimensions (powers of c, G, and h) for other physical quantities.

Contents hide

1 Explanation:

2 Derivation Steps:

3 Final Answer:

Explanation:

Dimensional analysis involves using the dimensions of physical quantities to understand their nature and the relations between them. The dimensions of any physical quantity can be expressed in terms of the fundamental quantities of mass (M), length (L), and time (T). When c, G, and h are taken as fundamental units, the dimensions of mass, length, and time can be derived by expressing the units of these constants in terms of M, L, and T and then solving the resulting system of equations.

The known dimensions of c, G, and h are:

c (velocity of light) has dimensions $[LT^{-1}]$ .
G (gravitational constant) has dimensions $[M^{-1}L^3T^{-2}]$ .
h (Planck’s constant) has dimensions $[ML^2T^{-1}]$ .

By setting each of these to unity, one can solve for the dimensions of M, L, and T in terms of c, G, and h.

Derivation Steps:

Start by creating a ratio that will isolate the dimension of mass (M). This is done by dividing [h][c] by [G], which yields:
$\dfrac{[h][c]}{[G]} = \dfrac{[ML^2T^{-1}][LT^{-1}]}{[M^{-1}L^3T^{-2}]} = [M^2]$
This means the dimensions of mass squared can be expressed as the product of the dimensions of h and c divided by G.
Taking the square root of both sides gives the dimension of mass (M) as:
$[M] = [h^{1/2}c^{1/2}G^{-1/2}]$
Next, to find the dimension of length (L), take the ratio [h]/[c] which yields:
$[L] = \dfrac{[h]}{[c][M]} = \dfrac{[ML^2T^{-1}]}{[LT^{-1}][M]}$
$[L] = [ML]$
Solving for L gives:
$[L] = \dfrac{h}{c[M]} = \dfrac{h}{c(h^{1/2}c^{1/2}G^{-1/2})}$
$[L] = [h^{1/2}c^{-3/2}G^{1/2}]$
Finally, to find the dimension of time (T), use the dimension of c:
$[c] = [LT^{-1}]$
This implies:
$[T] = \dfrac{[L]}{[c]} = \dfrac{[h^{1/2}c^{-3/2}G^{1/2}]}{[c]} = [h^{1/2}c^{-5/2}G^{1/2}]$