# General Relativity

General relativity, published by Einstein in 1915, extended the scope of special relativity to acceleration, specifically gravitational acceleration.

The theory of general relativity proposed a unified description of gravity as a geometric property of space and time, or spacetime. Mass and energy curve spacetime. Gravitational acceleration is a consequence of this curvature.

### Spacetime is curved by energy. Curved spacetime produces gravity.

The math behind 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 singularity: high gravity leads to black holes
• Gravitational Time Dilation: gravity slows time
• 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 explains that gravity slows down time.

Time dilation from gravity is less complex than special relativity because both observers agree that time is slower in more gravity.

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

$$\Delta t_0$$ = slow-ticking time for an observer inside the gravitational field at distance r from the center of the mass [s]
$$\Delta t_f$$ = fast-ticking time for an observer outside the influence of the gravitational field [s]
$$G$$ = 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 x 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
planet mass (kg) radius (km) density (g/cm 3)
Sun 2.00×10 30 695,700 1.408
Mercury 3.301×10 23 2,440 5.427
Venus 4.867×10 24 6,052 5.243
Earth 5.972×10 24 6,371 5.515
Moon 7.346×10 22 1,737 3.344
Mars 6.417×10 23 3,390 3.933
Jupiter 1.899×10 27 70,000 1.326
Saturn 5.685×10 26 58,232 0.687
Uranus 8.68×10 25 25,362 1.270
Neptune 1.024×10 26 24,622 1.638
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
planet mass (kg) radius (km) density (g/cm 3)
Sun 2.00×10 30 695,700 1.408
Mercury 3.301×10 23 2,440 5.427
Venus 4.867×10 24 6,052 5.243
Earth 5.972×10 24 6,371 5.515
Moon 7.346×10 22 1,737 3.344
Mars 6.417×10 23 3,390 3.933
Jupiter 1.899×10 27 70,000 1.326
Saturn 5.685×10 26 58,232 0.687
Uranus 8.68×10 25 25,362 1.270
Neptune 1.024×10 26 24,622 1.638
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 is too low for my calculator to display the result. To see the effect I subtracted one from the result in the calculator. $$\Delta t_f - \Delta t = 6.951 \times 10^{-10} \, \mathrm{year}$$ The number above represents the difference between a year in Earth and time far from Earth's gravity. $$\Delta t_f - \Delta t = 21.923 \, \mathrm{ms}$$

## Black Holes

Heavy masses can push gravitational time dilation to the point of breaking because 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.

derivation of event horizon $$\Delta t_0 = \Delta t_f \sqrt{1 - \left( \frac{2GM}{rc^2} \right) }$$ $$\sqrt{1 - \left( \frac{2GM}{rc^2} \right) } = 0$$ $$1 - \left( \frac{2GM}{rc^2} \right) = 0$$ $$1 = \frac{2GM}{rc^2}$$ $$r = \frac{2GM}{c^2}$$

# $$r = \frac{2GM}{c^2}$$

$$G$$ = 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 x 10⁸ [m/s]

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. At the event horizon time stops. 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 to energy from 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.

An accurate model of black holes probably needs a unified theory of physics that combines general relativity with quantum mechanics. This grand unified theory of physics doesn't exist yet, but many physicists are actively working on building one.

Example: Black holes are rare and sometimes hard to see for obvious reasons. A possible candidate is XTE J1118+480. It has a mass of 6 solar masses. How large is it's event horizon?
$$M _{\bigodot} = M_{sun} = 2 \times 10^{30} \, \mathrm{ kg}$$
solution $$M = 6 \ M _{\bigodot}$$ $$M = 6 (2 \times 10^{30} \, \mathrm{kg})$$ $$M = 12 \times 10^{30} \, \mathrm{kg}$$
$$r = \frac{2GM}{c^2}$$ $$r = \frac{2 (6.67 \times 10^{-11}) (12 \times 10^{30})}{(3 \times 10^8) ^2}$$ $$r = 1778.6 \, \mathrm{m}$$
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 $$M = (4.3 \times 10^{6}) \ M _{\bigodot}$$ $$M = (4.3 \times 10^{6}) (2 \times 10^{30} \, \mathrm{kg})$$ $$M = 8.6 \times 10^{36} \, \mathrm{kg}$$
$$r = \frac{2GM}{c^2}$$ $$r = \frac{2 (6.67 \times 10^{-11}) (8.6 \times 10^{36})}{(3 \times 10^8) ^2}$$ $$r = 1.27 \times 10^{10} \, \mathrm{m}$$
$$\text{radius of the sun = } 6.95 \times 10^{9} \, \mathrm{m}$$