Question

The dimensions of the area \( A \) of a black hole can be written in terms of the universal constant \( G \), its mass \( M \), and the speed of light \( c \) as \( A = G^\alpha M^\beta c^\gamma \). Here

• A. \( \alpha = -2, \beta = -2, \gamma = 4 \)
• B. \( \alpha = 2, \beta = 2, \gamma = -4 \)
• C. \( \alpha = 3, \beta = 3, \gamma = -2 \)
• D. \( \alpha = -3, \beta = -3, \gamma = 2 \)

Hint:

In dimensional analysis, always express each physical quantity in terms of fundamental dimensions and solve for unknowns by equating powers of \( M \), \( L \), and \( T \).

Answer 1

The Correct Option is B

Step 1: The area of a black hole's event horizon is the surface area of a sphere of the Schwarzschild radius, \( r_s = \dfrac{2GM}{c^{2}} \).
Step 2: The area is \( A = 4\pi r_s^{2} = 4\pi \left( \dfrac{2GM}{c^{2}} \right)^{2} = 16\pi \, G^{2}M^{2}c^{-4} \).
Step 3: Comparing this with \( A = G^{\alpha}M^{\beta}c^{\gamma} \), the exponents are read off directly as \( \alpha = 2 \), \( \beta = 2 \), and \( \gamma = -4 \).
\[ \boxed{\alpha = 2,\ \beta = 2,\ \gamma = -4} \]

Answer 2

Matching the dimensions of \( [M] \), \( [L] \), and \( [T] \) on both sides of \( A = G^{\alpha}M^{\beta}c^{\gamma} \) gives three equations: \( -\alpha+\beta=0 \), \( 3\alpha+\gamma=2 \), and \( -2\alpha-\gamma=0 \). We can solve these in a different order to see which set of values holds. From the mass equation, \( \beta = \alpha \). From the length equation, \( \gamma = 2-3\alpha \). Substituting this into the time equation gives \( -2\alpha-(2-3\alpha)=0 \), which simplifies to \( \alpha - 2 = 0 \), so \( \alpha = 2 \). This gives \( \beta = 2 \) and \( \gamma = 2-3(2) = -4 \).

1. Option (1) \( \alpha=-2,\beta=-2,\gamma=4 \): Checking against \( \beta=\alpha \): \( -2=-2 \) holds, but checking \( \gamma=2-3\alpha \): \( 2-3(-2)=8 \neq 4 \). This option does not satisfy the derived relations.
2. Option (2) \( \alpha=2,\beta=2,\gamma=-4 \): Here \( \beta=\alpha \) gives \( 2=2 \), and \( \gamma=2-3\alpha \) gives \( 2-6=-4 \), which matches exactly. This is the consistent set of exponents.
3. Option (3) \( \alpha=3,\beta=3,\gamma=-2 \): Here \( \beta=\alpha \) holds (\(3=3\)), but \( \gamma=2-3(3)=-7 \neq -2 \), so this option fails.
4. Option (4) \( \alpha=-3,\beta=-3,\gamma=2 \): Here \( \beta=\alpha \) holds (\(-3=-3\)), but \( \gamma=2-3(-3)=11 \neq 2 \), so this option fails.

Only option (2) is consistent with both derived relations simultaneously.

Therefore, the correct answer is \( \alpha = 2, \beta = 2, \gamma = -4 \).