In 1907 Einstein published an important thought experiment. Imagine a person in a box with no way to look outside. That person couldn't distinguish between the acceleration of gravity or the acceleration from a rocket. If both situations look and feel the same, maybe they are the same. Maybe acceleration and gravity are equivalent.

Classical Physics- On Earth, the rocket is at rest and the ball accelerates because of gravity.
- In space, the ball is at rest and the rocket accelerates because of thrust.

Equivalent Acceleration and Gravity Treating gravity and acceleration as equivalent means that both situations are the same. The ball is at rest and the rocket is accelerating. The 9.8 m/s² we feel is caused by the Earth accelerating upward like a rocket!

**Question:**I'm sitting at my desk typing this question right now. Am I at rest? When is a person at rest on Earth?

## answer

I am not at rest. I am accelerating at up at 9.8 m/s² because my chair is pushing me upwards.

The relativistic point of view would be that the space around the Earth is compressing, but the atoms of the Earth resists this compression. This means that the surface of the Earth is constantly accelerating upwards.

On Earth the only way to not accelerate up would be to fall. Falling is when you are at rest.

## General Relativity

The equivalence of acceleration and gravity drove Einstein towards a new theory that would extend the scope of special relativity to include gravity. In 1915 Einstein completed his theory of general relativity. General relativity describes gravity as the curvature of space and time, or spacetime.

### Spacetime is curved by energy. Gravitation occurs when a body tries to move in a straight line through curved spacetime.

Mass is a type of potential energy, so it also curves spacetime.

As bodies travel through curved space they follow the shortest distance between two points, a straight line. The pull of gravity is what a straight line through curved space looks like. Another example of a straight line through a curved space is the great circle routes around a sphere.

A full mathematical explanation of general relativity is beyond the scope of this page, but we will explore some of the predictions:

**gravitational lensing:**gravity bends the path of light

**gravity waves:**moving sources of gravity make waves in spacetime

**gravitational time dilation:**gravity slows time

**black holes:**very dense bodies with gravity so strong that time stops

The theory of general relativity suggests a significant change to fundamental aspects of physics, but there is overwhelming evidence in its favor.

## Gravitational Time Dilation

General relativity predicts that an accelerating reference frame experiences time dilation. This effect is similar, but different from how relative velocity dilates time in special relativity.

You can't feel **time dilation** inside an accelerating frame of reference,
but observers outside the frame see everything inside as slower.
People age slower and move slower. They talk slower, with deeper voices.
Light originating inside a time dilated frame of reference has a lower frequency and a different color when observed from outside.

### $$ \Delta t_0 = \Delta t_f \sqrt{1 - \left( \frac{2GM}{rc^2} \right) } $$

\(\Delta t_0\) = fewer seconds pass for an observer at distance r from the center of the mass\(\Delta t_f\) = more seconds pass for an observer very far from the mass

\(G\) = 6.67408 × 10

^{-11}= universal gravitation constant [N m²/kg²]

\(M\) = mass of gravity well [kg, kilograms]

\(r\) = distance to the center of the mass [m, meters]

\(c\) = speed of light, 3 × 10⁸ [m/s]

**Example:**If one second passes outside the influence of the star's gravity, how much time passes on the surface of the star? Let's use the mass and radius of a super dense neutron star, like PSR J0348+0432.

\( M = 4.02 \times 10^{30} \, \mathrm{ kg} \quad \quad r = 13 \, \mathrm{km}\)

## solution

$$ \Delta t_0 = \Delta t_f \sqrt{1 - \left( \frac{2GM}{rc^2} \right) } $$ $$ \Delta t_0 = 1 \sqrt{1 - \left( \frac{2(6.674 \times 10^{-11})(4.02\times 10^{30})}{(13000)(3 \times 10^8)^2} \right) } $$ $$ \Delta t_0 = 1 \sqrt{1 - 0.4586 } $$ $$ \Delta t_0 = 0.7358 \, \mathrm{s} $$**Example:**If 10 seconds pass on the surface of the Sun, how much time passes far from the Sun's gravity?

## Local Massive Objects Data Table

## solution

$$ \Delta t_0 = \Delta t_f \sqrt{1 - \left( \frac{2GM}{rc^2} \right) } $$ $$ 10 = \Delta t_f \sqrt{1 - \left( \frac{2(6.67408 \times 10^{-11})(2\times 10^{30})}{(6.957 \times 10^{8})(3 \times 10^8)^2} \right) } $$ $$ 10 = \Delta t_f \sqrt{1 - 4.2637 \times 10^{-6}} $$ $$ 10 = \Delta t_f (0.999997868) $$ $$ \Delta t = 10.00002132 \, \mathrm{ s} $$**Example:**How much does the passage of time slow due to Earth's gravitational field. If year passes on Earth, how much time will pass far from Earth's gravity?

## Local Massive Objects Data Table

## solution

$$ \Delta t_0 = \Delta t_f \sqrt{1 - \left( \frac{2GM}{rc^2} \right) } $$ $$ 1 = \Delta t_f \sqrt{1 - \left( \frac{2(6.67408 \times 10^{-11})(5.972 \times 10^{24})}{(6.371 \times 10^6)(3 \times 10^8)^2} \right) } $$ $$ 1 = \Delta t_f \sqrt{1 - \left( 1.39 \times 10 ^{-9} \right) } $$ $$ 1 = \Delta t_f (0.999999999 \dots ) $$The time dilation effect was too low for my calculator to display the result. I was about to give up, but then I tried subtracting one from the result in the calculator. I got a result. The number represents the difference between a year on Earth and a year with no time dilation.

$$\Delta t_f - \Delta t = 6.951 \times 10^{-10} \, \mathrm{year}$$ $$\Delta t_f - \Delta t = 0.021923 \, \mathrm{s}$$## Black Holes

Extremely dense masses can push the gravitational time dilation equation to the point of breaking when you have to take the square root of a negative number. We can find where the equation breaks down by solving for the radius where the time dilation approaches zero.

This radius is called the Schwarzschild radius, or the **event horizon**.

## derivation of event horizon

$$ \Delta t_0 = \Delta t_f \sqrt{1 - \left( \frac{2GM}{rc^2} \right) } $$ $$\text{undefined for square roots at or below zero} $$ $$\sqrt{1 - \left( \frac{2GM}{rc^2} \right) } = 0 $$ $$1 - \left( \frac{2GM}{rc^2} \right) = 0 $$ $$1 = \frac{2GM}{rc^2} $$ $$r = \frac{2GM}{c^2} $$# $$r = \frac{2GM}{c^2} $$

\(G\) = 6.67408 × 10

^{-11}= universal gravitation constant [N m²/kg²]

\(M\) = mass of black hole [kg, kilograms]

\(r\) = radius of event horizon [m, meters]

\(c\) = speed of light, 3 × 10⁸ [m/s]

The behavior of spacetime at distances below r is undefined.

Objects that achieve the high density needed to reach this point are called **black holes**. We can only speculate how black holes might behave, but the time dilation equation suggests very extreme outcomes.
As a body approaches the event horizon, the passage of time approaches zero. From an outsider's perspective, objects fall into black holes and never get out, frozen in time.

Black holes are a possible outcome at the end of the life of a very massive star. Stars convert mass into energy through nuclear fusion. This energy balances the force of gravity and prevents stars from becoming black holes. When stars runs out of nuclear fuel, gravity will dominate and a black hole may form.

At the center of most galaxies there is a super massive black hole. Even our galaxy, the Milky Way, has one with a mass of 4 million Suns.

An accurate black hole model probably needs a unified theory of physics that combines general relativity with quantum field theory. A grand unified theory of physics doesn't exist yet, but many physicists are actively looking for one.

**Example:**Black holes are rare and sometimes hard to see because they trap light. A possible candidate is XTE J1118+480. It has a mass of 6 solar masses. How large is its event horizon?

\( M _{\bigodot} = M_{sun} = 2 \times 10^{30} \, \mathrm{ kg}\)

## solution

$$M = 6 \ M _{\bigodot} $$ $$M = 6 (2 \times 10^{30} \, \mathrm{kg}) $$ $$M = 12 \times 10^{30} \, \mathrm{kg} $$$$r = \frac{2GM}{c^2} $$ $$r = \frac{2 (6.67 \times 10^{-11}) (12 \times 10^{30})}{(3 \times 10^8) ^2} $$ $$r = 17\,786.6 \, \mathrm{m} $$

**Example:**At the center of most massive galaxies exists a supermassive black hole. Our galaxy, The Milky Way, has one with the mass of 4.3 million solar masses. Find the radius of its event horizon.

## solution

$$M = (4.3 \times 10^{6}) \ M _{\bigodot} $$ $$M = (4.3 \times 10^{6}) (2 \times 10^{30} \, \mathrm{kg}) $$ $$M = 8.6 \times 10^{36} \, \mathrm{kg} $$$$r = \frac{2GM}{c^2} $$ $$r = \frac{2 (6.67 \times 10^{-11}) (8.6 \times 10^{36})}{(3 \times 10^8) ^2} $$ $$r = 1.27 \times 10^{10} \, \mathrm{m} $$

$$\text{radius of the sun = } 6.95 \times 10^{9} \, \mathrm{m} $$