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:
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]
\( M = 4.02 \times 10^{30} \, \mathrm{ kg} \quad \quad r = 13 \, \mathrm{km}\)
solution
$$ \Delta t_0 = \Delta t_f \sqrt{1 - \left( \frac{2GM}{rc^2} \right) } $$ $$ \Delta t_0 = 1 \sqrt{1 - \left( \frac{2(6.674 \times 10^{-11})(4.02\times 10^{30})}{(13000)(3 \times 10^8)^2} \right) } $$ $$ \Delta t_0 = 1 \sqrt{1 - 0.4586 } $$ $$ \Delta t_0 = 0.7358 \, \mathrm{s} $$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
$$ \Delta t_0 = \Delta t_f \sqrt{1 - \left( \frac{2GM}{rc^2} \right) } $$ $$ 10 = \Delta t_f \sqrt{1 - \left( \frac{2(6.67408 \times 10^{-11})(2\times 10^{30})}{(6.957 \times 10^{8})(3 \times 10^8)^2} \right) } $$ $$ 10 = \Delta t_f \sqrt{1 - 4.2637 \times 10^{-6}} $$ $$ 10 = \Delta t_f (0.999997868) $$ $$ \Delta t = 10.00002132 \, \mathrm{ s} $$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.
This radius is called the Schwarzschild radius, or the event horizon.
derivation of event horizon
$$ \Delta t_0 = \Delta t_f \sqrt{1 - \left( \frac{2GM}{rc^2} \right) } $$ $$\sqrt{1 - \left( \frac{2GM}{rc^2} \right) } = 0 $$ $$1 - \left( \frac{2GM}{rc^2} \right) = 0 $$ $$1 = \frac{2GM}{rc^2} $$ $$r = \frac{2GM}{c^2} $$$$r = \frac{2GM}{c^2} $$
\(G\) = 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.
\( M _{\bigodot} = M_{sun} = 2 \times 10^{30} \, \mathrm{ kg}\)
solution
$$M = 6 \ M _{\bigodot} $$ $$M = 6 (2 \times 10^{30} \, \mathrm{kg}) $$ $$M = 12 \times 10^{30} \, \mathrm{kg} $$$$r = \frac{2GM}{c^2} $$ $$r = \frac{2 (6.67 \times 10^{-11}) (12 \times 10^{30})}{(3 \times 10^8) ^2} $$ $$r = 1778.6 \, \mathrm{m} $$
solution
$$M = (4.3 \times 10^{6}) \ M _{\bigodot} $$ $$M = (4.3 \times 10^{6}) (2 \times 10^{30} \, \mathrm{kg}) $$ $$M = 8.6 \times 10^{36} \, \mathrm{kg} $$$$r = \frac{2GM}{c^2} $$ $$r = \frac{2 (6.67 \times 10^{-11}) (8.6 \times 10^{36})}{(3 \times 10^8) ^2} $$ $$r = 1.27 \times 10^{10} \, \mathrm{m} $$
$$\text{radius of the sun = } 6.95 \times 10^{9} \, \mathrm{m} $$