Calculating the Evaporation of an Earth-Mass Black Hole via Hawking Radiation

Black holes, fascinating and mysterious objects in the universe, exhibit a unique and incredible property – they can evaporate. This process, known as Hawking radiation, results from the quantum effects around the black hole’s event horizon. However, not all black holes evaporate within a reasonable timeframe. In this article, we will delve into the calculation of the evaporation time for an Earth-mass black hole, highlighting the significance of Hawking radiation and addressing common misconceptions.

Understanding Black Hole Evaporation

Black holes come in many sizes, from microscopic to the Sagittarius A* black hole at the center of the Milky Way. The size at which a black hole will evaporate is vastly smaller than most. A blue whale, with a mass of around 300,000 kilograms, can be crushed into a black hole that would evaporate in just 1 second. However, the Earth's mass, which is approximately 5.972 x 1024 kilograms, would form a black hole with an event horizon of just 1/3 of an inch or about 9 millimeters. This black hole, while incredibly dense, will still undergo evaporation over an immense timescale.

The Role of Hawking Radiation

Hawking radiation is a theoretical process first proposed by Stephen Hawking, suggesting that black holes emit particles and energy. The rate of energy loss due to Hawking radiation is proportional to the mass of the black hole, with the smaller the black hole, the faster it evaporates. The equation for estimating the evaporation time of a black hole is given by:

t ≈ frac{5120 pi G^2 M^3}{hbar c^4}

Calculating Evaporation Time for an Earth-Mass Black Hole

Let's apply this to an Earth-mass black hole. Plugging in the values, we get:

Step-by-Step Calculation

First, calculate M3:

M^3 ≈ (5.972 times 10^{24})^3 ≈ 2.11 times 10^{74} frac{mathbf{kg}^3}

Substitute the values into the formula:

t ≈ frac{5120 pi (6.674 times 10^{-11})^2 (2.11 times 10^{74})}{(1.055 times 10^{-34})(3.00 times 10^8)^4}

This yields:

t ≈ 8.64 times 10^{67} text{ seconds}

Converting this to years:

t ≈ frac{8.64 times 10^{67}}{60 times 60 times 24 times 365.25} ≈ 2.74 times 10^{60} text{ years}

Interpreting the Results

The results show that an Earth-mass black hole would take approximately 2.74 x 1060 years to evaporate via Hawking radiation. This is far beyond the current age of the universe, which is around 13.8 billion years. This calculation underscores the incredible duration of black hole lifespans under the effects of Hawking radiation.

Addressing Misconceptions

A common misconception is that particles or information actually leaves a black hole through Hawking radiation. In reality, it is a quantum effect where particles attached to the black hole get separated as they approach the event horizon. One particle falls into the black hole, while the other is ejected as radiation. This process does not involve the particles leaving the black hole, but rather a quantum entanglement effect occurring at the event horizon.

While the evaporation of an Earth-mass black hole is an extremely slow process, understanding the physics behind it offers profound insights into the fundamental nature of black holes and the universe. The study of Hawking radiation continues to be a crucial area of research in theoretical physics and cosmology.