## Thursday, May 18, 2017

### Why does so much time pass in Interstellar?

While on a quest to save the human race, astronauts in the film Interstellar travel to a foreign planet orbiting closely around a supermassive black hole.  Due to the strong gravity, time on the planet is distorted, being artificially compressed.  Seven entire years here on Earth are squeezed down to just one hour of time on the planet.

This is one of the many bizarre effects of Einstein's general theory of relativity, referred to as gravitational time dilation.  I had some students from my physics class recently ask me to explain this phenomenon.  So I prepared what I think is a fairly straightforward explanation of the phenomenon, assuming only a knowledge of 1st semester physics and some simple calculus.

A lot of this material I am borrowing from Kip Thorne's book Black Holes and Time Warps, so if you want to get closer to the source, I recommend giving it a read.

Some background.

Back in the time of Newton, light was believed to be made up of "corpuscles", or little particles.  There were good reasons for thinking this; namely, the huge success of geometric optics.  Your eye doctor uses geometric optics to fit your prescription.  In geometric optics, light is treated as rays following straight lines.  The things that move in nice straight lines like this are particles.  Thus, it was believed that light was made up of particles.

Of course, this was later discredited by showing that light was made up of waves, and then re-credited again some hundred years later by showing that light was made up of discrete, particle-like excitations of fields called photons, while also still being made up of waves at the same time.

Light is massless.  However, one of the idiosyncracies of gravity is that all objects move the same way in a gravitational field, regardless of their mass.  This was famously demonstrated by Galileo in his apocryphal Tower of Pisa experiment.  Galileo climbed to the top of the Leaning Tower and dropped two metal balls, one being much more massive than the other.  The two hit the ground at exactly the same time, demonstrating that heavier objects don't fall any faster than lighter objects.

 *Author is not professional artist
This same experiment is repeated millions of times a year in physics classrooms around the world, and you can do it right now if you doubt me.  Take a small steel marble and a lead weight like a barbell.  Hold them at the same height, and drop them.  They will strike at the same time.

The reason for this was later explained by Newton in terms of the ideas of inertia and forces.

Inertia is the tendency of an object to resist changes in its motion.  We use the word in common language analogously; if an idea has inertia behind it, it means that it's become so popular that you have to overcome a lot of pushback to change the public's mind.  Objects have inertia.  Objects with a lot of inertia require a lot of force to make them start moving, or to get them to stop moving.  The inertia of an object is measured by its mass.

The mass of an object measures its resistance to changes in its motion.

Forces are the things that change an object's motion.  Forces come from one object acting on a second object.  The act could be through contact (like a push or a pull), or it could be from a distance (as with magnets or gravity).

The force acting on an object is what overcomes the object's inertia.

The two ideas - inertia and force - can be related with the simple equation, known as Newton's Second Law, $$F = ma$$ where $m$ is the object's mass, $a$ is its acceleration, and $F$ is the total force acting on it.

One force that Newton proposed was gravity.  Of course everyone knew things fell before Newton, but Newton gave the first mathematical formulation for why.  He posited that the strength of the gravitational force between two objects is given by $$F = G \frac{m_1m_2}{r^2}$$ where $m_1, m_2$ are the masses of the two objects, $r$ is the distance between them, and $G$ is a physical constant, equal to $G = 6.67\times10^{-11} m^3/kg/s^2$.  This force is attractive.  It acts to bring the objects together.

For an object of mass $m$, near a planet or star with mass $M$, the force of gravity from the planet, combined with Newton's Second Law, leads to $$ma = -G\frac{Mm}{r^2}$$ This force is pulling $m$ toward the center of the planet, which is represented by the negative sign.  Since $m$ appears on both sides, we can cancel it. So $$a = -G\frac{M}{r^2}.$$ The acceleration of an object in a gravitational field depends on the mass of the planet/sun, and the distance from planet/sun.  It does not depend on the object's mass.  The mass cancels.  Therefore, all objects accelerate the same in a gravitational field, no matter their mass.

Now again, light is massless. Or it has a mass of 0. That would normally suggest that the force of gravity would be $$F = G\frac{0\cdot M}{r^2} = 0,$$ however, since the zero-mass "cancels", we can get the intriguing possibility of light also falling in a gravitational field.

This brings us to dark stars.

Dark Stars

A dark star was a prediction of Newton's gravity and the corpuscular theory of light.  Since light was thought a particle, and since the zero mass could cancel in the gravity equation, then you should perhaps be able to analyze light's motion in a gravitational field the same way you would for any other particle.

If you throw an object up into the air (any object - try it now), it will begin rising, but it will gradually slow down.  The slowing down is because, of course, gravity, which causes an acceleration in the opposite direction, downward toward the center of the Earth.  However, the larger the initial speed the object has going up, the higher it will go.  Throwing a bullet in the air doesn't make it go as high as shooting a bullet into the air.
 *Author is not professional artist
An object with larger initial speed takes longer to slow down to a halt, and in that time it can keep moving upward higher.

Think of this in terms of trying to stop your car.  If you're moving slowly through a neighborhood, you can stop on a dime in case a kid goes chasing a ball.  But if you're speeding on the freeway, even if you floor the brakes you won't come to a complete stop until you've traveled a much longer distance.

When the object is finally slowed to a halt, that is the highest position in its flight.  The object continues being accelerated downward, causing it to then begin falling and return to earh.

Notice that for gravity, the acceleration gets smaller for increased distance.  The equation was $$a = -G \frac{M}{r^2}.$$ As $r$ gets bigger, then $r^2$ gets really big, and we are dividing by bigger and bigger numbers. Dividing by a big number produces a small number. Therefore, $a$ gets smaller and smaller. This means that as we get farther away, the acceleration slowly drops to nothing.

So as we increase the launch speed of the object, it will rise in the air higher and higher before slowing to a stop and falling back down.  Yet, as it rises higher and higher, the acceleration will get smaller and smaller.

Similarly with objects launched from the Earth, eventually we reach a speed where the object rises high enough that gravity will be insufficient to pull the object back.  The object will just keep going, forever.  This speed is called the escape velocity.  Knowing the escape velocity of Earth is important to rocket scientists at NASA when planning launches.

This is where we need calculus.

Consider an object that starts at the surface of a planet of mass $M$ and radius $R$.  Its acceleration is given by the formula $$a = \frac{GM}{r^2}.$$ Acceleration is defined as the time rate of change of velocity, and velocity is the time rate of change of position.  That is, $$a = \frac{dv}{dt},$$ $$v = \frac{dr}{dt},$$ and hence $$a = \frac{d^2r}{dt^2}.$$ Therefore
$$\frac{d^2r}{dt^2} = -G\frac{M}{r^3}.$$ Using the chain rule of calculus, we can rewrite $$\frac{d^2r}{dt^2} = \frac{dv}{dt} = \frac{dr}{dt}\frac{dv}{dr} = v\frac{dv}{dr},$$ which we put into the equation $$v\frac{dv}{dr} = -G\frac{M}{r^2} \Longrightarrow vdv = -GM \frac{dr}{r^2}$$ Now we integrate this expression.  Pay attention to the limits of integration.  The velocity will begin at some initial value $v_e$, then gradually decrease to zero.  The distance from the planet will begin at $R$, the radius of the planet, and go to infinity.  So we throw the object fast enough that by the time it stops, it is infinitely far away, where $a\rightarrow 0.$  That is, $$\int_{v_e}^0 v dv = - GM \int_{R}^\infty \frac{dr}{r^2},$$ which evalutes to
$$\frac{1}{2}v_e^2 = \frac{GM}{R},$$ or $$v_e = \sqrt{\frac{2GM}{R}}.$$ This value, $v_e$, is called the escape velocity.  It tells you the minimum speed an object must have in order to escape the gravtiational pull of another object.

Suppose the objects under consideration were particles of light.  Stars are constantly spitting out light.  Stars and planets obey the same gravitational rules (the only difference is the magnitude of the mass).  So for a star of radius $R$ and mass $M$, the escape velocity is given by the equation above.  Any object trying to escape the star moving slower than this will just fall right back; any object moving faster than the escape velocity will be able to continue moving off to infinity.

Light moves with speed $c = 3.0\times10^8$ m/s.  For any star you've ever seen, the escape velocity of the star is less than this, meaning the light from the star is able to escape and reach you.  That's how you see it.

However, consider what happens to the escape velocity as we pack more mass into a smaller star.  Since the equation is $$v_e = \sqrt{\frac{2GM}{R}},$$ then as $M$ increases (a more massive star) or as $R$ decreases (a more compact star), the escape velocity also increases.

Consider a star of constant mass $M$, but we gradually squeeze it down so that $R$ decreases.  Eventually, $R$ will become small enough that $v_e \geq c$ -- that is, the escape velocity is larger than the speed of light.  This occurs when $$R \leq \frac{2GM}{c^2}.$$ A star with such a radius in the Newtonian picture is called a dark star.  The prediction was, that for such a star, light leaving it would eventually fall back into the star due to its gravity.  The light would be able to leave the star, but it wouldn't be able to travel arbitrarily far.  Eventually, the light corpuscles would reach a maximum height, just like when you throw a ball in the air, and then fall back towards the star.  Since the light wouldn't be able to reach arbirarily far away, such a star would not shine.  It would be dark.  It would be there, but we wouldn't be able to see it.  Any light that could show us the star would be sucked in by gravity and never make it to us.

The radius $2GM/c^2$ is important enough that I am going to give it a special name.  I am going to call it the Schwarzschild radius, and denote it by $$R_\ast = \frac{2GM}{c^2}.$$ Remember this, because it will come up again.

The dark star prediction might sound like a black hole, but it isn't quite the same thing.  Dark stars are a prediction of Newtonian mechanics, but Newtonian mechanics only describes things at speeds much less than the speed of light.  Dark stars also operate under the idea that light is accelerated, when we now know that the speed of light is a constant and can't change.

If we are much further away than $R_\ast$, then Newton describes everything just fine.  For stars and planets, the surface of the star or planet is always well outside of $R_\ast$, so the strange behavior at $r\leq R_\ast$ never occurs.  However, if we want to describe stars or planets where $R<R_\ast$, the speed of light is becoming relevant. To talk about situations where light speed is relevant, we need to turn to Einstein.

By adding in Einstein's insights to gravity and light, we can refine the dark star picture to make it align with our modern understanding, and from this derive black holes.  Once we have black holes, we'll be able to see why time slows down near them.

To go further, then, I need to introduce the concept of relativity.

Brief Intro To Relativity

The word "Relativity" refers to the different perceptions of time experienced by different observers.  However, the relativity of Einstein's theory is not a subjective kind of relativity.

You know how when you're waiting on something, like a package to arrive or a pot to boil or a phone to ring, that time seems to go so slow?  And how when you're playing a video game or reading a book or out with friends, time seems to fly by?

Consider Alice and Bob sitting in the same room.  Alice is engrossed in playing video games, whereas Bob is waiting on an important phone call.  At the end of an hour, Alice feels like only five minutes have passed, while to Bob it felt like an eternity.

That is just a subjective experience, and has nothing to do with Einstein's theory.

Every clock in the room with Alice and Bob will say an hour has passed.  The watch Alice is wearing reports the same time as the watch Bob is wearing.  Their subjective experience differs, but their objective measures of time are exactly the same.

Einstein's theory is about objective measures of time passing.  When Bob is near a black hole, and Alice is back on Earth, it isn't that Alice just feels like 7 years pass for every hour that Bob feels goes by.  It is that Bob's watch, and Alice's watch, when brought together and compared, will likewise show the difference.

And the effect isn't limited to wrist watches.  It could be counted by aging; if Bob and Alice were twins separated at the moment of birth, with Bob sent to the black hole, then Bob would still be a newborn and Alice would be going into the 3rd grade.  The effect could be counted by nuclear decay.  The effect could be counted through microbe growth.  The effect could be counted through any process at all that takes place in time.

This is profoundly different from just "time flies when you're having fun."  The effect I'm talking about is an objective one, and that is what I'm going to try to explain.

What it comes down to is the constancy of the speed of light.  Light moves with the same speed, $c$, no matter how fast you are moving relative to the source of the light.  When you are in front of a stationary car, the light from the headlights reaches you at $3.0\times10^8$ meters per second.  When the car is driving toward you at 10,000 m/s (an impossibly lage speed for a car), the light from the headlights still reaches you at $3.0\times10^8$ m/s.  When the car drives backwards, the light reaches you at $3.0\times10^8$ m/s.  The car's speed has no effect on the speed of the light reaching you.  The velocities don't add.

Light moves with the speed $c=3.0\times10^8$ m/s for all observers, no matter who does the observing.

This is really bizarre, and has some really bizarre impiclations.  One of those implications is that time is relative.  Objectively relative.

Consider a different situation.  Consider a high-speed train moving with a velocity of $v$ to the right.  Alice is riding onboard a cart in this train, holding in her hand a perfect clock with keeps perfectly regular time.  Also onboard is a laser.  The laser shoots tiny bursts of laser light directly upward toward the ceiling of the train cart.  Directly above the laser's position is a mirror; the laser pulse will bounce off of the mirror and return to the laser.  Alice measures the length of time that it takes the pulse of light to make the total trip, using her perfect clock.

Let the height of the train car be $h$.  Then the light moves a total distance of $2h$.  It does so at constant speed $c$, meaning that $$c t_A = 2h,$$
where $t_A$ is the time Alice measures.  In words, the time the light is traveling, multiplied by its speed, equals the distance it travels.  Solving for $t_A$, we find $$t_A = \frac{2h}{c}.$$
Now consider Bob, who is on the ground next to the train.  Bob conducts his own measurement, using his own perfect clock.  From Bob's point of view, the light doesn't just move straight up and down, because the train cart is moving.  If the light takes total time $t_B$, then the light also moves a distance $vt_B$ up to the right at the same time as it moves a distance $h$ up and $h$ down, following along the hypotenuse of a right triangle each segment.  So in Bob's frame of reference, the light moves a total distance of $$d_B = 2 \sqrt{ (2h)^2 + (vt_B)^2},$$ and since the light moves at constant speed $c$, then $$c t_B = 2 \sqrt{ (2h)^2 + (vt_B)^2}.$$
Solving this for $t_B$ is a bit more difficult, since $t_B$ is on both sides of the equation, but with a little algebra, we find, $$t_B = \frac{2h}{\sqrt{c^2-v^2}} = t_A \frac{c}{\sqrt{c^2-v^2}}.$$

Bob and Alice are both talking about the same occurences.  For both Bob and Alice, light begins at one point, and later reaches another point.  And they're both using perfect clocks to measure time. But the amount of time that Bob measures passing between the two events, and the amount of time Alice measures passing betwen the two events, differs by a factor of $c/\sqrt{c^2-v^2}.$

This is truly bizarre.  The time should be the same.  Yet here we see it is not.  The culprit for this oddity is the fact that the same speed of light, $c$, was used for both Alice and Bob, to measure the distance the light traveled.  This effect is called time dilation, because the time that Bob measures is bigger than the time Alice measures.  To Bob, it appears that Alice's clock is running slow.  It leads to a good aphorism in relativity: moving clocks run slow.

Notice that if $v =0$, then $$t_B = t_A \frac{c}{\sqrt{c^2-0^2}} = t_A \frac{c}{\sqrt{c^2}} = t_A,$$ so that if the train is not moving (or, if Bob is on the train with Alice), then they both measure the same amount of time.  It's only because they are not at rest with respect to one another that the time changes. Thus, your perceptions of time and space are relative to your frame of reference.

Notice also, the aphorism is true in reverse.  While Bob thinks Alice is moving with speed $v$ to the right, it is equally true that Alice sees Bob moving with speed $v$ to the left.  Moving clocks run slow, so according to Alice, it is actually Bob whose clock is running slow.  Were Bob to perform the same experiment with light on the ground as Alice on the train, it would be Alice who measures a longer amount of time for Bob's experiment.  This leads to the issue of the twin paradox in relativity.

Let's plug in some real values to this.  Suppose a high speed train is moving at something like 200 mph.  That's really fast.  I only know $c$ in m/s, so to get units to work I need the train's sped in m/s.  A good rule of thumb going from mph to m/s is to just divide it by half; it's not exactly true, but it's close enough to get a feel for the problem.  So 200 mph is about 100 m/s.  Plugging in, $$\frac{c}{\sqrt{c^2-v^2}} = \frac{3.0\times10^8}{\sqrt{(3.0\times10^8)^2 - (100)^2}} = 1.00000000000006.$$ Thus, even on a train going faster than anything you've probably ever seen in normal life, the effect of time dilation is minuscule.  A clock on this train, after an hour, will be off by about 1/5th of a nanosecond ($2\times10^{-10}$ s).  This is why you've never noticed it, and why this prediction seems impossibly strange.  It only becomes relevant when the relative speed, $v$, becomes close to $c$.

This is a demonstration of time dilation: in Einstein's theory of relativity, the time measured by two observers can be very different, objectively, even when they use perfect clocks.

The time can be different, and the distances can be different, depending on who is doing the measuring.  That's not good.  We'd like to find a measurement that doesn't change, but that is constant no matter who measures it.  To do this, we need to amend our understanding of space.  We modify three-dimensional space by tacking on a fourth dimension (time), and henceforth discussing motion in a 4D spacetime.

Geometry of Spacetime

We need a way to talk about measuring distances and times that is invariant -- that is, that doesn't change with reference frames.  When we talk about distances in normal life, we use the Pythagorean theorem:$$\Delta s^2 = \Delta x^2 + \Delta y^2.$$

If we were working in three dimensions instead of two, then $$\Delta s^2 = \Delta x^2 + y^2 + \Delta z^2.$$
In four or higher dimensions, we tack on a $\Delta u^2$ then $\Delta v^2$ then a $\Delta w^2$ until we run out of letters in the alphabet.  For Relativity, what we do is similar to this, but not exactly like this.  The difference is that we need to somehow take into account the fact that nothing moves faster than light.

We know that, in a given amount of time $\Delta t$, light will move a set distance $\Delta x = c\Delta t$.  Another (and more circuitous) way to say this is $$0 = c^2\Delta t^2 - \Delta x^2.$$
Nothing moves faster than light, so for anything, its speed is $v < c$, and hence in a time $\Delta t$ it moves a distance $$\Delta x = v\Delta t < c \Delta t.$$  That is to say for a massive object, $$c^2\Delta t^2 - \Delta x^2 > 0,$$ where, again, $\Delta x$ is the distance measured and $\Delta t$ is the duration of time measured. This is going to be our distance measurement in space-time, $\Delta s^2$, which we give the special name spacetime interval. Considering movement in all three spatial directions, we the interval is $$\Delta s^2 = c^2\Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2.$$

If we consider the above situation with Alice, and think about her own movement, then in her own frame of reference she doesn't move at all, and an amount of time $\Delta t_{A}$ just passes uneventfully. That is to say, for Alice, the spacetime interval she travels is $$\Delta s^2 = c^2\Delta t_A^2 - \Delta x_A^2 = \Delta t_A^2.$$ In this case, we give Alice's measurement of time the special name proper time, and call it $\Delta \tau = \Delta t_A$. The proper time is the duration of time measured by an observer in her own reference frame; in this case, it's the duration of time that Alice measures on her clock.

To show that the spacetime interval is truly invariant, let's consider the situation from Bob's perspective.

In an amount of time $\Delta t_B$, Bob observes Alice moving a total distance $\Delta x_B = v\Delta t_B.$ Thus, according to Bob, Alice's spacetime interval is $$\Delta s^2 = c^2\Delta t_B^2 - v^2\Delta t_B^2 = (\sqrt{c^2 - v^2}\Delta t_B)^2,$$ but we showed earlier that Bob and Alice's clocks disagree by a factor of $c/\sqrt{c^2-v^2}$, and therefore $$\Delta s^2 = (\sqrt{c^2 - v^2}\Delta t_B)^2 = (c \Delta t_A)^2 = c^2\Delta \tau^2.$$ Therefore this spacetime interval is invariant -- it doesn't change with the observer. We can specify the invariant spacetime interval traveled by any object moving by having an observer moving alongside the object just record the time that passes on a stopwatch. This time (multiplied by $c$) gives the invariant spacetime interval.

It is worth showing that for the case of the laser beam, we would arrive at the same answer. In Alice's frame, the beam moves a vertical distance $\Delta y = 2h$ over a time $\Delta t_A$. Thus, $$\Delta s_A^2 = c^2\Delta t_A^2 - (2h)^2 = 0.$$ In Bob's frame, the beam moves a vertical distance $2h$, but also moves a horizontal distance $v\Delta t_B$, and thus $$\Delta s_B^2 = c^2\Delta t_B^2 - v^2\Delta t_B^2 - (2h)^2 = c^2\Delta t_A^2 - (2h)^2 = 0,$$ so even for a more complicated object that is moving with respect to both Alice and Bob in their own respective frames, both observes will make difference measurements of the amount of time $\Delta t$, and different measurements of the distance traveled, but both will measure the exact same interval $\Delta s$.

While this is confusing, reconsider the situation of normal spatial lengths of something normal, like a broom. Let the broom have a length $L$. If you lay the broom flat on the floor, then $\Delta x = L, \Delta y = 0$, and $$\Delta s^2 = \Delta x^2 + \Delta y^2 = L^2 + 0 = L^2.$$ If you take the exact same broom and stand it straight up, then $\Delta x = 0, \Delta y = L$, and $$\Delta s^2 = \Delta x^2 + \Delta y^2 = 0 + L^2 = L^2.$$ If you set the broom at an angle of 45 degrees, then $\Delta x = \frac{L}{\sqrt{2}}, \Delta y = \frac{L}{\sqrt{2}}$, and $$\Delta s^2 = \Delta x^2 + \Delta y^2 = \frac{L^2}{2} + \frac{L^2}{2} = L^2.$$ You can try for any angle you want - you will always get a length of $L$ (obviously). This is true even though we get different values of $\Delta x$ and $\Delta y$, depending on how the broom is oriented.

The issue of time dilation (and length contraction) is the same thing in special relativity. You have an interval of a fixed length, $\Delta s$, but depending on your speed, you will view that interview at different "angles", and hence make different measurements of $\Delta t$ and $\Delta x.$  For Alice, in traveling along the tracks, the entire interval is along the time direction, meaning she measures it on a clock.  This is like the broom standing up.  For Bob watching Alice move on the train tracks, Alice is moving in both space and in time, similar to our broom at 45 degrees.

This demonstrates for us time dilation in Special Relativity. What happens in Interstellar is still time dilation, but of a completely different sort. There is it not due to relative velocities, but due to the curvature of spacetime. And to talk about that, we will need to get into General Relativity, and examine the curvature of spacetime around a black hole.

Now, on to black holes.

As we just demonstrated, when we make the jump to spacetime (which is necessary when we consider speeds near the speed of light), we switch from the usual Pythagorean way of measuring distances, and instead focus on the spacetime interval $$\Delta s^2 = c^2\Delta\tau^2 = c^2\Delta t^2 - \Delta x^2 -\Delta y^2 - \Delta z^2.$$ Suppose we are close to a star, at a fixed distance $r$ away. As we saw, we experience a force $$F = ma = m\frac{d^2 r}{dt^2} = -\frac{GMm}{r^2}$$, and thus $$-\frac{GM}{r^2} = \frac{d^2r}{dt^2} = \frac{dr}{dt}\frac{d}{dr}\left(\frac{dr}{dt}\right) = v \frac{dv}{dr}.$$ Ignore your screaming Calculus teacher, and just multiply both sides by $dr$, then integrate, to find again $$\frac{1}{2} v^2 = \int vdv = \int -\frac{GM}{r^2} = \frac{GM}{r},$$ or $$\left(\frac{dr}{dt}\right)^2 = \frac{2GM}{r}$$, or even more succinctly, $$dr^2 = \frac{2GM}{r}dt^2.$$
If we stick this into the equation for the interval, then $$c^2\Delta\tau^2 =c^2\Delta t^2 - \Delta r^2 = c^2\Delta t^2 - \frac{2GM}{r^2}\Delta t^2$$ and dividing by $c^2$, this reduces to $$\Delta\tau = \sqrt{1 - \frac{2GM}{c^2r}}\Delta t = \sqrt{1 - \frac{R_\ast}{r}}\Delta t.$$ Our old friend $R_\ast$ from the dark star has returned! But now, it is truly describing a black hole. The term on the left above describes the proper time -- that is, the time that passes on a perfect clock in the reference frame of an observer hanging out at constant distance $r$ from a star. The term on the right is a complicated function that you will notice depends on exactly how close we are to the star. Thus, our objective measurement of time, on perfect clocks, will be changed the closer we get to heavily gravitating objects. (The way we got this equation, by the way, is supremely suspicious, but you should look up the Schwarzschild metric to double check that we got the right answer)

If $\Delta \tau$ is what our clock measures, then what is $\Delta t$? It is called coordinate time. The coordinate time will be the time measured by someone who is unaffected by the curvature of spacetime; in this case, the time measured by someone infinitely far away. We can see this by setting $r \rightarrow \infty$. Then we just get $\Delta \tau = \Delta t$, so that the coordinate time corresponds to the proper time of an observer infinitely far away. As has been mentioned, for a normal star, the radius of the actual star will always be $R \gg R_\ast$. However, if we compress a star further and further, so that its radius becomes smaller and smaller, then we get $R < R_\ast$, and that is when we have a black hole.

Suppose you are in a spaceship very, very far away from the black hole-- so far, we'll pretend $r \approx \infty$, so that gravitational time dilation has no affect. You then get into a landing vessel and travel to a radius to a fixed distance $r = 5 R_\ast$ away from the black hole, and wait for $\Delta \tau = 100$ seconds. Ignore any effects of you traveling back and forth. Back on the spaceship, the time that passed during this 100 s interval was $$\Delta t = \frac{\Delta \tau}{\sqrt{1 - \frac{1}{5}}} = \frac{100}{0.8944} = 111.8$$ seconds. So they count off slightly more time than you do; or, in their terms, your clock is running slow. This is gravitational time dilation.

The closer you get to the black hole, the stronger this effect will be. Suppose instead of $r = 5R_\ast$, you now go down to $r = 3R_\ast$ and again wait for 100 s. Then the time that passes onboard the ship is $$\Delta t = \frac{\Delta \tau}{\sqrt{1 - \frac{1}{3}}} = \frac{100}{0.816} = 122.47$$ seconds. At $r = 2R_\ast$, this becomes, $$\Delta t = \frac{\Delta \tau}{\sqrt{1 - \frac{1}{2}} }= \frac{100}{0.707} = 141.44$$ which is nearly half-again as long. At $r = 1.1 R_\ast$, the time that passes on the ship during your 100 s has jumped to $$\Delta t = \frac{\Delta \tau}{\sqrt{1 - \frac{1}{1.1}}} = 331.6.$$
In the case of Interstellar, you can find out how close they were to the black hole. One hour was seven years, and $$\Delta t = 7 \text{ years} \times \frac{365\text{ days}}{\text{year}}\times\frac{ 24\text{ hours} }{\text{day} } = 61320 \text{ hours}$$, so, with some algebra, $$r = R_\ast \frac{\Delta t^2}{Delta t^2 - \Delta \tau^2} = R_\ast \frac{61320}{61320-1} = 1.000016 R_\ast,$$ which is awfully close to the black hole.

The question that naturally arises is -- what happens if you go all the way down to $R_\ast$?

You'll notice that at $r= R_\ast$, a zero appears in the denominator. Taking limits as we get closer and closer, $$\Delta t = \lim_{r\rightarrow R_{\ast}} \frac{\Delta \tau}{\sqrt{1 - R_\ast/r}} \rightarrow \infty.$$ If you spend just 100 s at $r = R_\ast$, then an eternity of time will have passed in the universe behind you. This special area, a distance $r = R_\ast$ away from the black hole, is what is called its event horizon. You can get arbitrarily close to it and still come back (albeit with severe time dilation). However, once you cross the $r = R_\ast$ mark, there is no going back. All of the time of the universe behind you will have passed away by the time you turn around; there is nothing left to go back to.

Of course, for an event horizon it isn't just that you wouldn't want to return. You actually physically can't. The interval between points just inside and points just outside becomes infinite; you would have to travel at infinite speeds to escape in finite time. Since the fastest you can move is light speed, and since light speed is finite, then not even light is able to escape. The gravitational time dilation becomes so extreme that nothing can escape from the inside of a black hole.

And that's how gravitational time dilation works, in a nutshell.