Understanding Planck Length Derivation

The Planck length ($L_P$) represents the scale at which the effects of quantum gravity become significant. It can be derived using dimensional analysis from the fundamental constants: the gravitational constant ($G$), the reduced Planck constant ($\hbar$), and the speed of light ($c$).

Dimensional Analysis Setup

We need to find the combination of $G$, $\hbar$, and $c$ that results in dimensions of length [L]. Let's list the dimensions of each constant:

• $G$: $[M^{-1} L^3 T^{-2}]$
• $\hbar$: $[M L^2 T^{-1}]$
• $c$: $[L T^{-1}]$

Assume the Planck length is given by $L_P = G^a \hbar^b c^d$. We equate the dimensions:

$[L] = [M^{-1} L^3 T^{-2}]^a [M L^2 T^{-1}]^b [L T^{-1}]^d$

$[L^1 M^0 T^0] = [M^{-a+b} L^{3a+2b+d} T^{-2a-b-d}]$

Solving for Exponents

Equating the exponents for each dimension (Mass M, Length L, Time T):

1. Mass (M): $-a + b = 0 \implies a = b$
2. Time (T): $-2a - b - d = 0$. Substituting $a=b$, we get $-2a - a - d = 0 \implies -3a - d = 0 \implies d = -3a$
3. Length (L): $3a + 2b + d = 1$. Substituting $a=b$ and $d=-3a$, we get $3a + 2a + (-3a) = 1 \implies 5a - 3a = 1 \implies 2a = 1 \implies a = \frac{1}{2}$

From these results:

• $a = \frac{1}{2}$
• $b = a = \frac{1}{2}$
• $d = -3a = -\frac{3}{2}$

Final Planck Length Formula

Substituting the exponents back into the formula $L_P = G^a \hbar^b c^d$:

$L_P = G^{1/2} \hbar^{1/2} c^{-3/2}$

This can be rewritten as:

$L_P = \sqrt{\frac{G \hbar}{c^3}} = \left(\frac{G\hbar}{c^3}\right)^{1/2}$

This matches Option 3.