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: Hey! It's me. I'm sitting at my desk typing this question.
Am I at rest?

answer

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

I know that sounds crazy. It starts making sense the more you think about it. Although it still doesn't feel very intuitive to me.

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.

When is a person at rest on Earth?

answer

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.

In the simulation above, classical gravitational potential energy is represented as warping a 2-D surface into a 3rd dimension. This isn't exactly the same math as general relativity, but it's a good analogy.

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

Planet	mass (kg)	radius (km)
Sun	2.00 × 10³⁰	695 700
Mercury	3.301 × 10²³	2440
Venus	4.867 × 10²⁴	6052
Earth	5.972 × 10²⁴	6371
Moon	7.346 × 10²²	1737
Mars	6.417 × 10²³	3390
Jupiter	1.899 × 10²⁷	70 000
Saturn	5.685 × 10²⁶	58 232
Uranus	8.68 × 10²⁵	25 362
Neptune	1.024 × 10²⁶	24 622

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 a year passes on Earth, how much time will pass far from Earth's gravity?

Local Massive Objects Data Table

Planet	mass (kg)	radius (km)
Sun	2.00 × 10³⁰	695 700
Mercury	3.301 × 10²³	2440
Venus	4.867 × 10²⁴	6052
Earth	5.972 × 10²⁴	6371
Moon	7.346 × 10²²	1737
Mars	6.417 × 10²³	3390
Jupiter	1.899 × 10²⁷	70 000
Saturn	5.685 × 10²⁶	58 232
Uranus	8.68 × 10²⁵	25 362
Neptune	1.024 × 10²⁶	24 622

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 run 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

$$M = (4.3 \times 10^{6}) \ M _{\bigodot} $$ $$M = (4.3 \times 10^{6}) (2 \times 10^{30} \, \mathrm{kg}) $$ $$M = 8.6 \times 10^{36} \, \mathrm{kg} $$
$$r = \frac{2GM}{c^2} $$ $$r = \frac{2 (6.67 \times 10^{-11}) (8.6 \times 10^{36})}{(3 \times 10^8) ^2} $$ $$r = 1.27 \times 10^{10} \, \mathrm{m} $$
$$\text{radius of the sun = } 6.95 \times 10^{9} \, \mathrm{m} $$

Practice printout.pdf

In case you wanted more practice I used AI to make some more problems. The rest of the site I made by hand, but generating endless problems seemed safe. I did find mistakes in the AI generated problems, and there are probably some I didn't find. Let me know if something could be fixed. I also added a practice problem on each page with no solution. That's intentional. Have fun!

For math problems, assume the objects are non-rotating and roughly spherical unless the problem says otherwise.

Question: In the rocket thought experiment, why does Einstein connect gravity with acceleration?

answer

A person sealed inside the box cannot tell whether the floor pushes upward because the box is sitting on Earth or because a rocket is accelerating through space.

That equivalence suggests that gravity can be understood as something deeper than a normal force pulling objects through space. General relativity describes it as motion through curved spacetime.

Example: A neutron star has mass 2.8 solar masses and radius 12 km. Use 1 solar mass = 2.0 × 10³⁰ kg. If 30 s pass far from the neutron star, how much time passes on the surface?

solution

Convert the mass and radius first.

$$M = 2.8(2.0 \times 10^{30}\,\mathrm{kg})$$ $$M = 5.6 \times 10^{30}\,\mathrm{kg}$$ $$r = 12\,\mathrm{km}\left(\frac{1000\,\mathrm{m}}{1\,\mathrm{km}}\right)$$ $$r = 1.2 \times 10^4\,\mathrm{m}$$

Use gravitational time dilation.

$$\Delta t_0 = \Delta t_f \sqrt{1 - \left(\frac{2GM}{rc^2}\right)}$$ $$\Delta t_0 = (30\,\mathrm{s})\sqrt{1 - \left(\frac{2(6.674 \times 10^{-11})(5.6 \times 10^{30})}{(1.2 \times 10^4)(3.00 \times 10^8)^2}\right)}$$ $$\Delta t_0 = 16.6\,\mathrm{s}$$

Only about 16.6 s pass on the neutron star surface.

Example: A clock on Earth's surface runs slightly slow because of Earth's gravity. Use Earth mass 5.97 × 10²⁴ kg and Earth radius 6.37 × 10⁶ m. If 1.00 year passes far from Earth's gravity, how many fewer seconds pass on Earth's surface?

solution

Use 1.00 year = 31,536,000 s.

$$\Delta t_f = 31536000\,\mathrm{s}$$ $$\Delta t_0 = \Delta t_f \sqrt{1 - \left(\frac{2GM}{rc^2}\right)}$$ $$\Delta t_0 = (31536000\,\mathrm{s})\sqrt{1 - \left(\frac{2(6.674 \times 10^{-11})(5.97 \times 10^{24})}{(6.37 \times 10^6)(3.00 \times 10^8)^2}\right)}$$ $$\Delta t_0 = 31535999.978\,\mathrm{s}$$

Find the difference.

$$\Delta t_f - \Delta t_0 = 0.022\,\mathrm{s}$$

Earth's surface clock loses about 0.022 s compared with a far-away clock over one year.

Question: Why does gravity bends the path of light if photons have no rest mass?

answer

In general relativity, gravity is not treated as a normal force pulling only on mass.

Mass and energy curve spacetime. Light travels through that curved spacetime, so its path can bend even though photons have no rest mass.

Example: Earth has mass 5.97 × 10²⁴ kg. If Earth were compressed until it became a black hole, what would its Schwarzschild radius be?

solution

Use the Schwarzschild radius equation.

$$r = \frac{2GM}{c^2}$$ $$r = \frac{2(6.674 \times 10^{-11})(5.97 \times 10^{24}\,\mathrm{kg})}{(3.00 \times 10^8\,\mathrm{m/s})^2}$$ $$r = 8.85 \times 10^{-3}\,\mathrm{m}$$

Earth's Schwarzschild radius would be about 0.00885 m, or 8.85 mm.

Example: A collapsed star has mass 6.0 solar masses. Use 1 solar mass = 2.0 × 10³⁰ kg. What is its Schwarzschild radius?

solution

Convert solar masses to kilograms.

$$M = 6.0(2.0 \times 10^{30}\,\mathrm{kg})$$ $$M = 1.2 \times 10^{31}\,\mathrm{kg}$$

Find the Schwarzschild radius.

$$r = \frac{2GM}{c^2}$$ $$r = \frac{2(6.674 \times 10^{-11})(1.2 \times 10^{31}\,\mathrm{kg})}{(3.00 \times 10^8\,\mathrm{m/s})^2}$$ $$r = 1.78 \times 10^4\,\mathrm{m}$$

The event horizon radius is about 17.8 km.

Example: A 6.0 solar mass collapsed star has physical radius 15 km. From the previous example, its Schwarzschild radius is 17.8 km. Is the surface inside or outside the event horizon?

solution

Compare the physical radius with the Schwarzschild radius.

$$R = 15\,\mathrm{km}$$ $$r_s = 17.8\,\mathrm{km}$$ $$R < r_s$$

The surface is inside the event horizon, so this object would be a black hole in this simplified model.

Question: The event horizon is sometimes drawn as a dark surface. Is it a solid surface that an astronaut would hit?

answer

No. The event horizon is not a solid surface.

It is a boundary in spacetime. Outside it, light can still escape. Inside it, our simple model says no signal can escape to far-away observers, although spin makes real black holes more complicated.

Example: Astronomers estimate a black hole event horizon radius of 45 km. Assuming it is non-rotating, what mass does the Schwarzschild radius equation give?

solution

Rearrange the Schwarzschild radius equation.

$$r = \frac{2GM}{c^2}$$ $$M = \frac{rc^2}{2G}$$ $$M = \frac{(45000\,\mathrm{m})(3.00 \times 10^8\,\mathrm{m/s})^2}{2(6.674 \times 10^{-11})}$$ $$M = 3.03 \times 10^{31}\,\mathrm{kg}$$

The mass is about 3.03 × 10³¹ kg, or about 15 solar masses.

Example: A clock hovers 60 km from the center of a non-rotating 10 solar mass black hole. Use 1 solar mass = 2.0 × 10³⁰ kg. If 2.0 s pass far from the black hole, how much time passes on the hovering clock?

solution

Convert mass and radius.

$$M = 10(2.0 \times 10^{30}\,\mathrm{kg})$$ $$M = 2.0 \times 10^{31}\,\mathrm{kg}$$ $$r = 60\,\mathrm{km}\left(\frac{1000\,\mathrm{m}}{1\,\mathrm{km}}\right)$$ $$r = 6.0 \times 10^4\,\mathrm{m}$$

Use gravitational time dilation.

$$\Delta t_0 = \Delta t_f \sqrt{1 - \left(\frac{2GM}{rc^2}\right)}$$ $$\Delta t_0 = (2.0\,\mathrm{s})\sqrt{1 - \left(\frac{2(6.674 \times 10^{-11})(2.0 \times 10^{31})}{(6.0 \times 10^4)(3.00 \times 10^8)^2}\right)}$$ $$\Delta t_0 = 1.42\,\mathrm{s}$$

The hovering clock records about 1.42 s.

Question: Why does general relativity predict gravitational waves?

answer

General relativity treats gravity as curvature of spacetime.

If massive objects move in a way that changes that curvature, the change can travel outward as a wave in spacetime. This is why orbiting or merging black holes can produce gravitational waves.

Example: A non-rotating black hole has mass 10 solar masses. How far from its center would a hovering clock need to be for its time to run at half the far-away rate?

solution

A clock running at half the far-away rate means this factor is 0.50.

$$\sqrt{1 - \left(\frac{2GM}{rc^2}\right)} = 0.50$$ $$1 - \left(\frac{2GM}{rc^2}\right) = 0.25$$ $$\frac{2GM}{rc^2} = 0.75$$ $$r = \frac{2GM}{0.75c^2}$$

Use 10 solar masses.

$$M = 2.0 \times 10^{31}\,\mathrm{kg}$$ $$r = \frac{2(6.674 \times 10^{-11})(2.0 \times 10^{31}\,\mathrm{kg})}{0.75(3.00 \times 10^8\,\mathrm{m/s})^2}$$ $$r = 3.96 \times 10^4\,\mathrm{m}$$

The clock would need to hover about 39.6 km from the center.

Example: The Milky Way's central black hole has mass about 4.0 million solar masses. Use 1 solar mass = 2.0 × 10³⁰ kg. What is its Schwarzschild radius in meters and in million kilometers?

solution

Convert solar masses to kilograms.

$$M = (4.0 \times 10^6)(2.0 \times 10^{30}\,\mathrm{kg})$$ $$M = 8.0 \times 10^{36}\,\mathrm{kg}$$

Find the Schwarzschild radius.

$$r = \frac{2GM}{c^2}$$ $$r = \frac{2(6.674 \times 10^{-11})(8.0 \times 10^{36}\,\mathrm{kg})}{(3.00 \times 10^8\,\mathrm{m/s})^2}$$ $$r = 1.19 \times 10^{10}\,\mathrm{m}$$

Convert meters to kilometers.

$$r = 1.19 \times 10^{10}\,\mathrm{m}\left(\frac{1\,\mathrm{km}}{1000\,\mathrm{m}}\right)$$ $$r = 1.19 \times 10^7\,\mathrm{km}$$

The event horizon radius is about 1.19 × 10¹⁰ m, or 11.9 million km.

Question: Why does a full black hole model probably needs both general relativity and quantum mechanics?

answer

General relativity describes gravity and spacetime very well at large scales and strong gravitational fields.

Quantum mechanics describes very small systems. A black hole is both extremely massive and extremely compact, so it pushes into a region where both theories matter. A complete unified theory does not exist yet.

Reading (12 minutes): Read How to Think About Relativity from Quanta Magazine. Then answer these questions.

What was the luminiferous ether supposed to explain, and why did Einstein argue that physics no longer needed it?

answer

Physicists expected a wave to travel through a medium, so they proposed the ether as the substance carrying light waves. Experiments found no evidence for it. Einstein showed that light could be understood without the ether if every inertial observer measures the same speed of light and space and time are allowed to differ between observers.

The article compares the straight-line distance to a restaurant with the distance traveled along streets. How does this analogy help distinguish coordinate time from proper time?

answer

Coordinate time labels when two events occur within a chosen reference system. Proper time is the time recorded by a clock that actually travels along a particular path between those events. Just as different routes can have different lengths, different paths through spacetime can accumulate different amounts of proper time.

How does the article distinguish spacetime in special relativity from spacetime in general relativity?

answer

Special relativity describes a fixed, flat spacetime without gravity. General relativity allows spacetime to be dynamic and curved by mass and energy. What we experience as gravity is the motion of objects through that curved spacetime.