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# On The Science of Interstellar

I greatly debated with myself on whether to write this posting. I have seen Interstellar twice now including the special 70 mm IMAX screening, and am seeing it a third time later today. Simply put, the movie is fascinating. It combines, (yes) accurate science and real depictions of general relativistic effects with a great story as is to be expected from Christopher Nolan.

It was pointed out to me recently that some people have taken to the internet to write extensive articles criticizing the science in the movie, which is very strange. First, I didn’t think too much of it, as Kip Thorne was not only an executive producer, but also a consultant on the film, and has also seen the film. Surely, if there was something wrong from a GR-point-of-view, he would point it out. After all, he did manage to get two original scientific papers out of working on this movie.

The two reviews criticizing the science that I have seen, so far, stem from:

and

2. Roberto Trotta’s article:

They seem to be keen on really nitpicking certain things, which is certainly in their prerogative to do so, but I will just discuss in this article a major flaw in both of their reviews, in which they claim part of the science of Interstellar is wrong. They seem to both have an issue with the time dilation effect as described in the movie of the water planet close to the Black Hole, where it is claimed in the movie that 1 hour in the planet’s reference frame corresponds to 7 years in an observer’s reference frame far from the black hole. The two reviewers then go on to say that this is impossible as:

1. One would have to essentially be a “pinch” from the event horizon of the black hole.

2. The planet would not be in a stable orbit, and would spiral and crash into the black hole’s singularity point.

These are their two grand assumptions, but simply put, these assumptions are very, very wrong! They are basing their assumptions on the Schwarzschild solution of General Relativity:

This metric tensor describes the local geometry of the spacetime outside the region of a static, non-rotating, and spherically symmetric black hole/astrophysical body.

Notice how I emphasized non-rotating. If one uses this geometry as Plait and Trotta have, one will deduce all sorts of wrong conclusions. In truth, as has been said by both Thorne and Nolan during the special features videos posted on YouTube and I believe by the characters in the film, the black hole in the movie is spinning very, very fast and therefore, its angular momentum cannot be neglected. One therefore needs to at minimum use the Kerr metric:

where

J denotes the angular momentum and is absolutely key to understanding that the effect depicted in the movie is indeed very plausible.

Now we deal with the claimed time dilation effect of 1 hour = 7 years as described earlier. It can be shown that the time dilation equation derived from the Kerr metric takes the form:

Substituting for d\tau = 1 hour, and dt = 7 years, one obtains the following relation:

This equation fully describes a black hole of mass M, rotating with angular momentum J, as observed by an observer at radial coordinate r, and angular coordinate theta. The fraction on the right-hand-side of the equation fully depicts the 1 hour = 7 years dilation effect. For the Kerr metric, unlike the Schwarzschild metric, there are several stable orbits that can occur. Plait’s article took issue with the fact that for a stable orbit, the orbital radius should be 3 times the Schwarzschild radius, but as I said, it is due to him assuming the incorrect geometry. In actuality, the Kerr metric allows for three possible stable orbits, which were derived in the paper by Bardeen, Press, Teukolsky: Astrophysical Journal, Vol. 178, pp. 347-370 (1972):

1. For zero angular momentum: innermost stable orbit = 3 x Schwarzschild radius

2. For angular momentum a = M, corresponding to corotational behaviour: innermost stable orbit = 0.5 * Schwarzschild radius

3. For angular momentum a = M, corresponding to retrograde motion: innermost stable orbit = 9/2 * Schwarzschild radius

It turns out that the only way to satisfy the equation above is by considering case #2 here: One obtains two solutions:

or

Therefore, as this shows, it is completely possible to have a rotating black hole with an observer outside of it that experiences such time dilation effects while still exhibiting a stable orbit, that is, it never crashes into the black hole!

Just for fun, let’s plug in some numbers. Let us consider a very massive black hole that has a mass of 2000 Solar Masses, applying the above formulas we see that:

I chose to fix the angular coordinate for demonstration purposes only.

Therefore, this calculation shows that the time dilation effect in the movie is perfectly reasonable and accurate. I will write more about the other aspects of people’s reviews later which on a first reading seem to also be based on incorrect assumptions, but at the present moment, I don’t have the time!

Update: So, an article was just released detailing the science of Interstellar: http://www.space.com/27692-science-of-interstellar-infographic.html

In it, it is said that the mass of the black hole is 100 million solar masses. With this now, I can properly work out the example above, I made up numbers before, because I did not have this information before today! So, here it is re-worked:

For a very massive black hole, in the movie it is stated that M = 100 million times the mass of the sun. With this information, substituting into the equation above, we get:

This will no doubt please anyone who noticed the orbit in the first example seemed too small!

Update: A Comment on Tidal Forces

Also, by popular request, some have claimed that the planet close to the black hole should be completely destroyed by tidal forces, since it is so close to the black hole. This is not so. For this discussion, I will revert back to the Schwarzschild metric, since the mathematics is simpler, but the discussion can of course be extended to the Kerr metric. Consider the planet in question (the water planet) at a radial position r. The tidal forces felt by the planetary body are measured by the orthonormal components of the Riemann curvature tensor. If we consider a static orthonormal frame as is done in Misner, Thorne and Wheeler, we have:

At this radial position, we obtain for the Riemann curvature tensor components:

Now, we can transform over to the planet’s frame by applying a Lorentz boost in the radial direction with velocity:

One sees that all components of the the curvature tensor are completely unaffected by this boost! One therefore sees that none of the components of the curvature tensor in the planet’s reference frame become infinite at the gravitational radius. Moreover, as the planet/observer approaches the horizon, as can be seen from the Riemann components, the tidal forces are finite, and do not tear anything apart, at least when the mass M is very large (as is the case in the film). However, let us see the curvature invariants, for a Schwarzschild metric we have:

This is invariant, and so is a singularity in every reference frame. Indeed, as r -> 0, the tidal forces become infinite. So, only past the horizon, very close to the singularity, do we have to worry about tidal forces from the black hole breaking anything up!

Now, the astro community are largely mistaken on this whole tidal force ripping up the planet. All the papers they use are citing the whole idea of using the Roche limit. This can’t be done for several reasons. As I outlined for another astronomer (who will remain unnamed for this posting), the problem is as follows:

I am a stickler for mathematical form, and I refuse to acknowledge the validity of the Roche limit in General Relativity. Here are my reasons:

1. Even if I was to conclude that a spherical body orbiting a Kerr black hole will break up because of the tidal forces as described by the Roche limit, this conclusion is highly questionable without a 2-body GR approach because: you are assuming from the onset that the Kerr black hole remains spherical, and the mass in question has no effect on the Kerr black hole, so you are implicitly using a far-field approximation from the onset.

2. Newtonian gravity is linear, GR is not. Since there’s no 2-body problem analytic solution in GR, there is simply NO GR equivalent of the Roche limit.

3. The Roche limit is simply a result of Lagrange points in 2-body orbital Newtonian mechanics, and I prefer to leave it there. Adding GR corrections is not good enough.

4. In the Roche limit and the governing Newtonian regime, pressure does not generate any gravitational field, but, as you well know, in GR, pressure does contribute to the En.Mom tenor, and as a result the gravitational attraction. In fact, if collapse happen sufficiently far, the pressure growth goes exponentially and it is far more important than the rest-mass density.

5. The real way to do this problem aside from considering a 2-body problem in GR, and getting an analytic solution, is to consider an internal Schwarzschild geometry in an external Kerr geometry background. But, because of the cross-term in the Kerr metric that one cannot transform away because of any coordinate transformation, the matching conditions are impossible to derive. If I on the other hand assume an external Schwarzschild geometry (which is not relevant for this problem, but…) then one obtains the well-known TOV equation. The TOV equation is essentially how one obtains the collapse conditions properly.

6. The Roche limit is a Newtonian result, and because of the linearity of Newtonian gravity, and the lack of pressure contributing to gravity, prevent any such effect in Newtonian physics.

7. The Roche limit arguments are always weak-field effects, which will not give you an accurate answer especially in this regard.

It does raise an interesting question though. Why do you insist on using the Roche limit if the pressure influencing spacetime curvature (which would be significant because of the magnitude of the tidal forces) cannot be accounted for in this approach? It is in fact worse than this. If I have a significant pressure as implied by the Roche limit, then the En-Mom tensor is no longer non-zero, and one does not even have a Schwarzschild/Kerr or any other vacuum solution. This now goes into the domain of cosmology, which makes this problem, much, more difficult.

Finally, there are also issues having to do with causality, the fact that the governing structural equations in the Roche limit approach are elliptic PDEs (the Poisson equation) and the heat equation, which is a parabolic PDE. Both are acausal, in the case of elliptic PDEs, all solutions are spacelike, and no physical body would move along spacelike hypersurfaces.

Therefore, for all these mathematical points, I refuse to acknowledge the validity of the Roche limit in this situation, and prefer a non-Newtonian GR approach, and it is the only correct way to do this problem. But, like I said, we’re approaching this from different points-of-view, Phil and the astro community seem to be satisfied with approximate solutions! -:) (Thanks to GFE and CCD for pointing some of these points out in an interesting discussion on the mathematical formulation of Einstein’s equations!)

Update: On the whole issue of those giant waves in the movie

A lot of folks have been asking whether the situation of those giant waves that are observed on the water planet near the black hole are feasible in the movie. My honest answer is that I have to solve some equations to find out, but ironically those equations are not astrophysical or general relativistic in nature, they are purely dependent on standard Navier-Stokes theory. If you noticed from the film, the wavelength of the water waves was much, much greater than the depth of the water itself, this situation is ripe for the shallow-water equations, which are obtained by applying the Navier-Stokes equations to such a problem. For those who are interested, any reasonably advanced-level fluid mechanics textbook discusses these equations at length. In any event, these equations are three coupled, nonlinear partial differential equations, and have no analytic solution in general, they look like:

where u is the x-direction velocity, v is the y-direction velocity, h is the height deviation of the pressure surface from the mean height H (i.e., how high the wave will be), H is the height of the pressure surface, g is the local acceleration due to gravity, f is the Coriolis coefficient that is determined from the rotation of the planet, and b denotes viscous drag forces. For the situation in the movie, we are told that the acceleration due to gravity on the planet is 130% that of Earth’s, which means that g = 9.81 x 1.30 = 12.753 m/s^2, f will be reasonably influenced by internal forces in planet’s structure, combined with interestingly enough the Lens-Thirring effect/frame dragging from the rotating black hole which will also cause the planet to precess. Solving these equations must be done numerically, and has been well-studied in the scientific literature, indeed, many simulations have been done. Here are some examples:

Update: A Comment on Hawking Radiation

Some have also claimed that the radiation emitted, namely, Hawking radiation from the massive black hole should be enough to kill the nearby observers. This is also a misconception of what Hawking radiation is. Hawking radiation is a quantum effect, and is given by the equation:

That is, this gives you the temperature of the electromagnetic radiation emitted from a black hole. Let us do some calculations for the black hole in question. For a Kerr black hole, we have that the surface gravity is (see the discussion in Hervik and Gron):

The other constants in the temperature formula above are the well-known Planck’s constant, Boltzmann’s constant, and speed of light. Putting these two equations together and substituting the numbers that we derived in the previous section, we see that the temperature of EM radiation emitted from the giant black hole in the movie is approximately:

which is extremely, extremely negligible! Therefore, no one will die from the EM radiation emitted from the massive spinning black hole!

UPDATE: BY POPULAR REQUEST

COMMENTS: ON THE LAST ACT OF THE MOVIE

So, I have received some requests to discuss the scientific accuracy of the last part of the film, where the main character, Cooper travels through the black hole and reaches a five-dimensional universe.

The key point of understanding why this is possible is to recall once again, that we are using the Kerr metric that depicts spinning black holes, that is everything. In a non-rotating black hole, once someone passes the event horizon, he will have no choice but to continue towards the singularity meeting his eventual death. This is not so for a Kerr black hole. Let us see why:

Much of my discussion is based on the great G.R. books from Hawking and Ellis, Misner, Thorne, Wheeler, Gron and Hervik, and Wald.

Let me write the Kerr metric in a slightly more different form that will be practical for this discussion:

where we have defined per the conventions in Gron and Hervik,

Recall that this metric describes the spacetime outside a rotating black hole with mass M and angular momentum J = Ma. Notice that in this form the singularities of the metric are easily observable. Namely, where \Delta = 0 and \Sigma = 0. The \Delta = 0 equation describes the horizon, and folks familiar with general relativity know that this is coordinate singularity. By a suitable coordinate transformation, one can ‘transform away’ this singularity. However, the \Sigma = 0 denotes in fact a real physical singularity, given by the set of points that satisfy

As can be confirmed the solution to this equation is a two-dimensional ring:

Therefore, while the singularity for a regular Schwarzschild metric is a point singularity from which nothing can escape, the singularity for a Kerr rotating black hole is a ring, which in fact, is avoidable!

To see this, let us dive a bit further into the structure of the Kerr metric. Following Hawking and Ellis’ remarkable text,

From this figure, we note the following remarkable property of the Kerr metric. One passes through the ring singularity in the rotating Black Hole by going from the (x,z) plane on the left to the (x’,z’) plane on the right of the diagram. It can be shown that, because of this complicated topology, closed timelike curves exist in the neighbourhood of the ring singularity. (For those that are interested, a complete discussion involving Killing vectors are detailed in Wald’s GR text). The significance of the existence of closed timelike curves is an observer traversing along these curves can violate causality, and thus go backwards in time by an arbitrary amount. Note that there are some issues regarding stability that I have not detailed here as they are much more technical than what is covered in this posting.

Now, connecting all of this to the movie. The structure above allows one (as has been reported in the literature) to use the Kerr black hole as a wormhole itself. It is therefore plausible that Cooper’s character avoids the singularity of the rotating black hole and transports to another region of the universe. In the movie it is depicted that he ends up in a 5-D universe, 4 spatial dimensions and 1 time. Again, this is perfectly theoretically possible. Purists might argue against it, but like I said, it is theoretically possible. For example, the wormhole could transport you to a region of spacetime where the geometry locally is 5-D Minkowskian and has the metric:

However, as human beings can only perceive of 3 spatial dimensions and 1 time dimension, these four-dimensional spatial sections have to be embedded in a 3-dimensional setting for us to visualize them. These four-dimensional spatial sections are completely Euclidean, and one can think of a tesseract with the following domain:

In fact, in relativity theory, time flows “upwards”. One can foliate the above metric tensor into a 1+4 split, and obtain the following dynamical picture of how an observer “moves through” such a five-dimensional spacetime. Each spatial slice is taken to be 4-D, but since we can’t perceive of 4 spatial dimensions, this 4-D surface is embedded into three-dimensional space to produce the tesseract as in the film:

These are both depicted in the movie. So, once again, anyone saying these are pure fiction/fantasy out of Nolan’s mind are mistaken. There are technical arguments involved giving mathematical conditions showing where these conditions would fail to work, but that is becoming too technical for a science fiction movie. The point is that in large, the theory of General Relativity supports these ideas, and it is all based on the idea of using a rotating black hole in the movie, it is the true centre of the plot!

Update: The cool part of all this is that I obtained this image from the Interstellar website showing Dr. Brand’s blackboard:

Note that the metric tensor on the bottom left-hand-corner is exactly the Kerr metric I described earlier. It seems that indeed the “quantum data” that is to be obtained from the black hole singularity actually is obtained from when TARS falls into the black hole and goes through the ring singularity. What’s interesting is that Thorne’s depiction here (which he drew according to the special features of the movie) actually show you where the quantum data would be with respect  to the singularity in the above Penrose diagram.

## By Dr. Ikjyot Singh Kohli

Sikh, Theoretical and Mathematical Physicist, main research in the structure and dynamics of Einstein's field equations.

## 124 replies on “On The Science of Interstellar”

Nice. I didn’t see the movie yet, but I read people arguing about the habitability of the planet.

“I didn’t see the movie yet…”
Why are you purposely reading spoilers then? Are you a communist?

Thedoodsays:

Well done, I love it when the beauty of math shuts up a dumb critic.

Dumb Criticsays:

Too bad he’s wrong. Hint: Compare the radius of the orbit he calculated with the radius of Earth.

You misunderstood. That is not the radius of the orbit. That is the radius of a stable orbit, one of many possible stable orbits that an occur for a Kerr black hole!

By the way, you’d also notice, the radius of the orbit I calculated was based on the black hole having a mass of 2000 solar masses. But, I have no idea how massive the black hole is in the movie. As you can see from the original equation, a more massive M would allow a greater r, but it was just for demonstration purposes.

Dumb Criticsays:

Ok, it’s the radius of the minimum stable orbit. What’s the radius of the orbit with the time dilation observed in the movie (roughly 60000:1)?

Hi. So, this has all been updated in the post. As I explained in the update, the mass of the black hole was just released in an article yesterday to be 100,000,000 solar masses, much,much,much greater than the 2000 solar masses in my contrived example. With this number, the orbital radius is 10^11 metres, much more realistic! Please see the update.

Dumb Criticsays:

10^11 meters is inside the ergosphere, surely a stable planet is not possible there?

I don’t see why not. First, the equation does indeed describe the innermost stable orbit. Now, don’t forget about the effects of inertial frames. In Boyer-Lindquist coordinates, an observer at infinity observes the infalling object with an angular velocity. But, this infalling object carries an inertial frame which is dragged around the Kerr source. But, one can show that the momentum in the direction of the angular variable is zero, and the infalling object is a zero angular momentum object. Therefore, the object would only experience an intertial dragging effect from the Kerr source.

Dumb Criticsays:

Again, you’re considering the stable orbit for a particle. A planet has radius, which means a very different effect on one side of the planet than the other side, which is the tidal force. Do not assume that it is negligible.

But….. In G.R. objects that have a mass much much less than the Kerr source are test particles. They move along geodesics around the black hole. The geodesic equation/equation of motion is:
x^a”(t) + \Gamma^{a}_{bc} x^b'(t)x^c'(t) = 0,
where the \Gamma^{a}_{bc} are the connection coefficients, which with this metric in a coordinate basis are Christoffel symbols. You will note that this fundamental equation of motion contains nothing about radius, mass, etc…

Dumb Criticsays:

It doesn’t matter that GR treats them like point particles for certain purposes, you can’t extrapolate that to imply that tidal forces don’t exist!

Yes, I put this on firmer footing as well. Please see the update to the post. Thanks.

myth bustersays:

Tidal forces are proportional to the mass of the attracting object times the inverse cube of the distance. This is why the moon causes greater tidal force on the Earth than the sun, despite being much smaller.

The Schwartzschild radius of a black hole is proportional to its mass, which means that the tidal forces experienced at any multiple of the Schwartzschild radius is inversely proportional to the square of the black hole’s mass. Therefore, the tidal forces experienced by a planet orbiting a 10^8 solar mass black hole are indeed negligible.

Thedoodsays:

Hey, you are a dumb critic!
And you owe me royalties for the use of that nick!

Sat Sriakaal sir ji, Thanks for the article. I love the movie

SideshowBobsays:

Could you translate your solution into an actual orbital radius and velocity for the planet, and rotational speed for the surface of the black hole?

Well, I already computed the angular momentum for the black hole, which is J in the above calculations. Because the planet in this case is treated as a test particle in the General Relativistic sense that orbits the black hole, the Einstein equations, and more importantly the geodesic equations do not allow one to compute the orbital radius of the planet. I suppose one could in theory compute the velocity of the planet by integrating the geodesic equation once with respect to time, but I have to think about this a bit more.

Basmothsays:

Just a few questions for a question layman:

1) Time dilation is a result of relative velocity or gravity. To experience such extreme dilation, wouldn’t the gravitational forces essentially tear the planet apart?

2) While the rest were on the planet, Romilly was on a ship that was merely orbiting it. How would such extreme time dilation even apply to him?

1) No, this is a misconception. Gravity is not a force, it is manifestation of spacetime curvature. There are tidal forces, but once can see from the Riemann curvature tensor components, that outside the black hole, these are not significant enough to tear the planet apart.

2. If you notice the time dilation equation above has 4 important variables: the proper time, the time of an observer as measured outside the gravitational field, and the radial and angular coordinates that denote the position of the observer outside of the black hole. Clearly, the equation has solutions for any such coordinates.

Truth Seekersays:

We still don’t really know WHAT gravity is, just know its effects. Still don’t see where the time dilation comes from, or ho fast the planet must be going. Explain a little more clearly. Is it really possible to have a massive black hole spinning so rapidly? What effect would that have on Hawking radiation, if any?

But anyway, what could survive anywhere near any kind of black whole? The radiation would be tremendous.

Hawking radiation would be very, very negligible. The temperature of this radiation is almost zero. I updated my post in this regard, please have a look.

Basmothsays:

But what are the solutions for Romilly not experiencing the same amount of dilation? Surely the difference in distance between him and Cooper would be far too small for Romilly to experience such a vast difference?

myth bustersays:

Hawking radiation wasn’t the issue. It’s the ionizing radiation emitted by the accretion disk, which is super-heated by the gravity-induced adiabatic compression as it falls into the black hole that renders nearby planets uninhabitable.

Yes. I was trying to dispel the claim that some were making that H-radiation would be significant.

Jesse M.says:

Kip Thorne apparently disagrees with you that “outside the black hole”, the tidal forces “are not significant enough to tear the planet apart”–in his book “The Science of Interstellar”, he actually calculates the minimum mass Gargantua needs to have in order for a planet very near its horizon (with an assumed density of 10,000 kg/m^3) not to be ripped apart by tidal forces, which was apparently part of why he chose to have Gargantua’s mass be so large (though he says he actually rounded down from the answer he got, since his calculation suggested it needed to be at least 3.4 * 10^38 kg, whereas 100 million suns is 1.99 * 10^38 kg). This was in the “Some Technical Notes” section at the back of the book, in the notes on chapter 6.

His basic method is to calculate the tidal acceleration between the two sides of the planet, and say that the planet can avoid being ripped apart as long as its own inward gravitational acceleration at the surface is larger than this. Quoting from the book:

“On the faces of the planet farthest from Gargantua and nearest to it, Gargantua’s tidal gravity exerts a stretching acceleration (difference of Gargantua’s gravity between the planet’s surface and its center a distance r away) given by g_tidal = (2GM/R^3)r. Here R is the radius of the planet’s orbit around Gargantua, which is very nearly the same as the radius of Gargantua’s horizon. The planet will be torn apart if this stretching acceleration on its surface exceeds the planet’s own inward gravitational acceleration, so g_tidal must be less than g: g_tidal < g."

Hi. Thanks for your comment.
I don’t know why you say Kip disagrees with me about the fact that “outside the black hole”, “tidal forces are not significant enough to tear the planet apart”, and then precisely cite a case for which my statement was true! -:)

Perhaps, I should have been more precise in my statement as follows:
“Given a black hole of mass M and an object orbiting this black hole with mass m, one can always find an optimal M and m such that the planet is not torn apart, as you correctly say: ” the planet can avoid being ripped apart as long as its own inward gravitational acceleration at the surface is larger than this.”

The essence of my comment was to refute some other people’s comments who were under the impression that because it is a black hole, anything within its vicinity will be torn to shreds, which is clearly not the case.

Jesse M.says:

Sorry, I misunderstood, I thought you were arguing that since the Roche limit doesn’t apply and we don’t know how to do the exact 2-body problem in GR, we’d have no basis for confidently predicting a planet orbiting outside the event horizon of a black hole would be torn apart in any specific example.

No problem. Quite the contrary, that is why I showed the calculations of the Riemann curvature components, which is actually the proper way to predict how and when an orbiting body would be torn apart. Tidal “forces” in that sense, are proportional to the Riemann curvature components.

But, there is still the issue, which Thorne doesn’t address, is that the pressure related to the tidal force, must actually be negligible, which seems contradictory. The whole point of having a black hole is that it is a vacuum solution of the Einstein equations. However, if the pressure becomes too high, by definition, the energy-momentum tensor is non-zero, and you no longer have a space time that contains a black hole. Perhaps, in this case it is a Swiss-cheese or other type of spatially inhomogeneous universe, but that is why even in the book, such calculations involving tidal forces, are at best an approximation.

Jesse M.says:

Romilly wasn’t orbiting the planet, Kip Thorne says in ch. 6 of “The Science of Interstellar” that the Endurance was orbiting Gargantua itself at a distance of “five Gargantua radii”, so about 5 times further out than Miller’s planet which was very close to the event horizon. This means that the distance between the orbit of Miller’s planet and the orbit of Endurance was about 3.9 times the distance between the Earth and the Sun. At this distance, I calculate that for every hour experienced by Romilly, about 1.45 hours are going by on Earth, so it’s a pretty minor difference compared to the seven years on Earth for every hour on Miller’s planet.

SideshowBobsays:

Aslo, have you considered the radius of the planet itself, it seems like one surface would be below the event horizon if it were anywhere near the size of earth.

Actually. I did consider it, but this is kind of pointless. The reason is the way General Relativity works. The most massive object in the gravitational system is the one that curves spacetime the most, in this case, it is the black hole. Everything else moves along geodesics around this black hole, including the planet. That is, everything else is treated like a test particle, because of the immense difference in exhibition of spacetime curvature. This is the same for our solar system as well. For example, to understand the precession of Mercury, one essentially treats the Sun as the Schwarzschild mass, and everything else including our Earth moves along geodesics around the sun.

SideshowBobsays:

Yes, that “immense difference in exhibition of spacetime curvature” is exactly my point. It’s the reason there is a Roche limit, even for much smaller primaries than a black hole. You can treat a planet as a test particle to calculate something like precession, but not to calculate whether the planet can exist close to a black hole. The roche limit for an earthlike planet around a 2000 solar mass black hole is well outside the range where it even matters whether it’s rotating or not.

Truth Seekersays:

How can you have massive curvature of spacetime in a region of space, without massive gravitational field in that same region?

What is a gravitational field in the concept of GR?

g2-ae6105c3b0a24b724f973ab36810c62asays:

Hello! Thank you for doing this math. You’re correct, of course, I did only consider a static black hole. I’m waiting to see what Thorne wrote about this in his book; for one thing he may give the mass of the black hole he was considering. Assuming 2000 solar masses, as you did, the orbital radius is only 3000 km, which makes for a tiny planet. Also, there are still tides to consider…

g2-ae6105c3b0a24b724f973ab36810c62asays:

Oops, WordPress didn’t include my name. This is Phil Plait. 🙂

Hi Phil. Thanks for the reply. Indeed, I used the 2000 solar masses just as in example. The correct mass was released to the public yesterday at 100 000 000 solar masses, and I have since updated the calculation, in which the orbit is much larger. You will note though, the mathematical formulation was flexible, I gave r as a function of M!

SideshowBobsays:

You’re updated radius is still only half the Roche limit for an Earthlike planet around a 100000000 solar mass black hole.

Yes, but the problem is the Roche limit you are citing to my recollection is derived from Newtonian gravity, which doesn’t obviously apply in this situation. For spacetime curvature, where Newtonian-G breaks down, one needs to consider the Roche limit in the context of Riemann curvature, so, I don’t believe your results are accurate in this case.

Dumb Criticsays:

Shouldn’t theta be pi/2, not pi?

No. If you see from the original solutions, \theta cannot be \pi/2 to obtain a unique solution to the set of equations.

myth bustersays:

And the meaning behind this is that the orbital plane cannot contain the black hole’s axis of rotation, because the black hole isn’t rotating at all relative to that plane, and so the Schwatzschild metric would apply.

Neovatarsays:

I was completely awestruck by the concepts in the movie, after getting back home from the cinema all I wanted to do was to see how scientifically accurate the depictions were. From one review to another by scientists, getting a little disappointing at one point or another, but your explanation is by the the best, and cleared stuff up. More than one theoretical model exits and hence the argument.

One other thing, if you could talk about the gravity equation Murph finally completed to address the mass evacuation of humankind, my guess is that is outside known physics. Also encoding the data into the hand of the wristwatch I’m assuming is totally fictional, I mean how much amount of data could be encoded into a repeating tick.

Anyway amazing movie, awesome storytelling melting science emotions, and scientists writing about the science behind the movie even makes it greater. Thanks. This is the supplement to the flick.

Hi. I agree with much of what you say.
On a third watching, I actually tried to take a screenshot of the board where Murph writes down the “gravity equation”, but, there was too much security in the IMAX theatre!

However, from what I did see, she writes a Einstein-Hilbert action integral for spacetime of 10 dimensions. If you recall, one of the reasons why Prof. Brand’s theory was incomplete was that it was unable to reconcile general relativity and quantum mechanics. When Murph writes down the 10-dimensional Einstein-Hilbert action, it seems to suggest that the reconciliation of the two principles lies in superstring theory. That is the only place I am aware of where such an integral would show up. So, the movie seems to take the side that the fundamental theory of nature is string theory, which is certainly a possibility!

Juan D.says:

Bravo! great post, thanks for taking the time to write this up Dr. Ikjyot !

Willsays:

Good job on correcting their mistakes! I had noticed some logical flaws in their articles, especially the one about considering the black hole to be a non-rotating one, contrary to what is said in the movie and to what Thorne and Nolan have stated. Nice to see the maths!

Kit Gamessays:

That’s interesting and all. But I’m still wondering who would be stupid enough to even bother checking out a planet that’s practically being sucked into a black hole for seeding and/or relocating the human race? Trading one disaster for another is damn dumb.

xgenesays:

LOL

No!! It’s not being sucked into a black hole. That is not how black holes work. Things only get “sucked in” guaranteed once an object passes the event horizon of a non-rotating Black hole.

Can you explain your assertion that the Riemann curvature not diverging means the planet is stable? I’ve calculated the Roche limit including GR corrections (from http://arxiv.org/pdf/gr-qc/0501084.pdf) and the planet does not appear remotely stable based on those calculations. How did you calculate the comparison between the tidal force and the internal forces holding the planet together (and keeping it roughly spherical etc)?

Hi.
So, a couple of points here. The Roche limit you calculated including the GR corrections from the paper you cited are problematic from the onset. If you recall from my work, one can only have unique solutions to the time dilation equations if and only if \pi/2 < \theta \leq \pi, or for the other case 0 \leq \theta < \pi/2, that is, the geodesic motion is NOT in the equatorial plane of the Kerr black hole. The paper you cited assumes from the onset that \theta = \pi/2, so those specific corrections would not apply. There is also the additional problem in the paper that they assume a Newtonian potential which satisfies the Poisson equation. This is problematic, since such a PDE as you well-know is elliptic, and admits acausal solutions. In the vicinity of large space-time curvature, such a solution in terms of its characteristics on a pseudo-Riemannian manifold cannot be infinitely propagating.

Therefore, one must consider the more general case. In the Bardeen, Press, Teukolsky paper, one considers the Roche limit in the following sense: For a spherical body of mass m and radius r, the criterion for not experiencing Roche breakup at radius R is approximately M/R^3 \neq m/r^3. There is an infinite number of solutions to this inequality. For example, for our case we have that:
61834.9 \neq m/r^3. How many possible m and r values can you find that will satisfy this inequality and thus showing that the planet is stable. In addition if I consider the naive case, where m << M (as is reasonably depicted in the movie), and that r <= 10m, then, the Roche limit is well within the horizon, an outside observer would never observe the breakup per the Roche limit. Also, there is the larger part of timelike geodesics. The timelike geodesic of the planet point inwards towards the singularity after the horizon, it cannot just breakup, that is the point of the Black hole, as long as the Riemann components don't diverge, that region of spacetime is regular and continuous, and nothing will prevent the planet/spherical body from falling inwards towards the singularity, again, assuming that m << M.

The final point is I don’t know why you are insisting on Roche limits, they were not designed for black hole calculations. Their entire derivation stems from Newtonian gravity. To apply it to relativitistic situations, you have to use Taylor expansions to tack on the relativistic terms. The only real way to do this problem is to solve the 2-body problem in GR, which can’t be done, only approximately, but it can’t be the Newtonian Roche limit.

myth bustersays:

That’s incorrect. Everything, even the nuclei of atoms, is torn apart by tidal forces as one approaches the singularity. The singularity’s radius is zero, but its circumference is infinite. Particles are theoretically stretched and compressed into one-dimensional lines as they approach the singularity, but the truth is that we have no way of predicting what happens beyond the Planck scale.

Reblogged this on Quantumbox and commented:
Science Of Interstellar

Hi,

Thanks for your response. I’m confused by your reading of the Bardeen, Press and Teukolsky paper. I believe you’re looking at equation (3.17), which merely states the point at which the tidal force of the BH overwhelms the self-gravity of the orbiting object. It’s a dimensional argument, and it’s not a statement that “unless the relationship is exactly this the disruption doesn’t happen.” It should really be \lesssim not \sim, if you want to use it as a break-up criterion. And as far as I can tell, the planet orbiting at the ISCO would break up based on that criterion (the LHS is greater than the RHS). The simple approximation they then give discussing whether or not a neutron star will break up inside or outside the horizon is not really applicable because we’ve already calculated that (according to the calculation they suggest, which is just the Newtonian case), it will break up by the time it’s at the ISCO.

We’re not talking about part of the planet escaping the BH. We’re just talking about whether the planet will be able to orbit totally intact, or whether it’ll be stretched to the point of breaking by the tidal forces. We see that happen to things near black holes all the time. Roche limit calculations were *certainly* meant to apply to black holes — that’s one of their most important applications in astrophysics! (See e.g. http://scitation.aip.org/content/aip/magazine/physicstoday/article/67/5/10.1063/PT.3.2382)

Unless there’s something really amazing about Kerr black holes that prevents tidal forces from acting on objects in orbit, I’m still not sure how a planet could survive as a nice pretty spherical habitable world so close to the event horizon.

PS — If the planet is not orbiting in the equatorial plane, it’s crashing through the accretion disk twice an orbit, which seems like a bad thing for a planet to do.

[Numbers: For the planet, I’ve started with the 130% Earth gravity and assumed the density is the same as Earth (rocky planet with oceans), which gives me a mass of 1.7e25 kg and a radius of 8.3e3 km. For the BH, I’m using a mass of 1e8 solar masses with an ISCO at GM/c^2, appropriate for Kerr geometry, which is 1.5e8 km.]

Hi Katie.
I don’t have enough time to go back-and-forth, but I think we are approaching this problem from two different perspectives. You are approaching it from an astrophysics perspective, and I am approaching it from a mathematical physics perspective, which is fine, it is our respective backgrounds. I am a stickler for mathematical form, and I refuse to acknowledge the validity of the Roche limit in General Relativity. Here are my reasons:
1. Even if I was to conclude that a spherical body orbiting a Kerr black hole will break up because of the tidal forces as described by the Roche limit, this conclusion is highly questionable without a 2-body GR approach because: you are assuming from the onset that the Kerr black hole remains spherical, and the mass in question has no effect on the Kerr black hole, so you are implicitly using a far-field approximation from the onset.
2. Newtonian gravity is linear, GR is not. Since there’s no 2-body problem analytic solution in GR, there is simply NO GR equivalent of the Roche limit.
3. The Roche limit is simply a result of Lagrange points in 2-body orbital Newtonian mechanics, and I prefer to leave it there. Adding GR corrections is not good enough.
4. In the Roche limit and the governing Newtonian regime, pressure does not generate any gravitational field, but, as you well know, in GR, pressure does contribute to the En.Mom tenor, and as a result the gravitational attraction. In fact, if collapse happen sufficiently far, the pressure growth goes exponentially and it is far more important than the rest-mass density.
5. The real way to do this problem aside from considering a 2-body problem in GR, and getting an analytic solution, is to consider an internal Schwarzschild geometry in an external Kerr geometry background. But, because of the cross-term in the Kerr metric that one cannot transform away because of any coordinate transformation, the matching conditions are impossible to derive. If I on the other hand assume an external Schwarzschild geometry (which is not relevant for this problem, but…) then one obtains the well-known TOV equation. The TOV equation is essentially how one obtains the collapse conditions properly.
6. The Roche limit is a Newtonian result, and because of the linearity of Newtonian gravity, and the lack of pressure contributing to gravity, prevent any such effect in Newtonian physics.
7. The Roche limit arguments are always weak-field effects, which will not give you an accurate answer especially in this regard.

It does raise an interesting question though. Why do you insist on using the Roche limit if the pressure influencing spacetime curvature (which would be significant because of the magnitude of the tidal forces) cannot be accounted for in this approach? It is in fact worse than this. If I have a significant pressure as implied by the Roche limit, then the En-Mom tensor is no longer non-zero, and one does not even have a Schwarzschild/Kerr or any other vacuum solution. This now goes into the domain of cosmology, which makes this problem, much, more difficult.

Finally, there are also issues having to do with causality, the fact that the governing structural equations in the Roche limit approach are elliptic PDEs (the Poisson equation) and the heat equation, which is a parabolic PDE. Both are acausal, in the case of elliptic PDEs, all solutions are spacelike, and no physical body would move along spacelike hypersurfaces.

Therefore, for all these mathematical points, I refuse to acknowledge the validity of the Roche limit in this situation, and prefer a non-Newtonian GR approach, and it is the only correct way to do this problem. But, like I said, we’re approaching this from different points-of-view, you seem to be satisfied with approximate solutions! -:)

(Thanks to Charles Dyer and George Ellis for discussing some of the aforementioned points in an interesting discussion on spacetime geometry!)

Jesse M.says:

Even if you don’t think the Roche limit is the right way to approach the question, surely you’d agree that some large but finite amount of tidal stretching would be sufficient to break a planet apart? If so, and if the correct GR calculation for the radius at which this would happen is too complicated for you to work out here, I don’t understand why you make the definite assertion in the post that “So, only past the horizon, very close to the singularity, do we have to worry about tidal forces from the black hole breaking anything up!”

Hi. It’s not that it’s too complicated, it’s that GR does not really allow for such a calculation, no one knows how to do it properly, because no one has solved analytically the 2-body problem in GR.

Aside from the reasons I stated above, things breaking up is about length scales. If the black hole is much much larger than the planet, than the probability of break up is really quite low, except for when the components of Riemann diverge which occur past the horizon, not necessarily for Kerr by the way, but certainly for Schwarzschild.

Jesse M.says:

If your argument is that no one has solved the problem exactly and we can’t trust approximations that have been made (like whatever approximations were used in the paper at http://arxiv.org/abs/gr-qc/0501084 which Phil Plait linked to in his blog entry), I don’t understand how you can assert with any confidence that “the probability of break up is really quite low” at any point outside the horizon. Unless you have an exact way to calculate the “probability of break up” then your own statement must be based on approximations of some kind, no? Also, even if no analytic solution has been found, I wonder if anyone has attempted a computer simulation to model tidal breakeup in general relativity, using one of the types of numerical models discussed at http://en.wikipedia.org/wiki/Numerical_relativity

JRKimsays:

Dr, May I ask two questions out of pure curiosity?

1. It is said in the movie that gravity can effect time and space (even the past?).
In the last scenes, Cooper sends messages to the PAST, using binomial coding of ‘gravity’.(cf- the message “STAY” was sent to Murph before the point of time Cooper leaves for NASA – which is backwards in time) I’ve heard that gravity can effect time to the extent of slowing it down, but can it effect the PAST, also?

2. This is a curiosity I’ve had since high school.
According to G.R, velocity gives you time dilation. If I were to spin myself in a spinning chair, would that give me any time dilation? The movement of a spaceship has velocity, and as far as I know circular motion also has velocity, although the direction changes constantly.

I just want to know if it does have a positive time dilating effect, even if it is merely picoseconds, or something more minute.

Hello.
1. Einstein’s theory of general relativity allows for the possibility via special solutions to Einstein’s field equations for observers to travel along what are known as closed timelike curves. This can be also achieved by going through the singularity of a Kerr black hole as explained above. Wormholes can theoretically also transport you to different instances in time, not only just space.

2. It’s not so much the velocity in G.R., it is the amount of spacetime curvature that really gives you time dilation, at least in G.R. For a spinning chair, it is complicated, since this is now special relativity, since you are dealing with the velocity of your chair, but because it is spinning, every point is moving at a different speed. With regards to the spaceship, yes, time dilation is possible, but, you would have to be moving very close to speed of light.

GeorgeHMsays:

hello Dr. Kohli,

>1. Einstein’s theory of general relativity allows for the possibility via >special solutions to Einstein’s field equations for observers to travel >along what are known as closed timelike curves.

Could you elaborate on this a bit more? I was thinking about how the co-ordinates for the NASA station sent in binary from Coop (in 5D) are read by Murph and causes Coop to go to space in the first place.. One led to another and the loop never ends .. but i do not think there was any violation of causality .. What do you think?

One more question. (copying JRKim’s comment)

>It is said in the movie that gravity can effect time and space (even the >past?).
>In the last scenes, Cooper sends messages to the PAST

Does the theory say that the gravity at the Kerr black hole is so strong that it warps space time to such an extent that even the past (for an earth observer) can be affected ..

myth bustersays:

Inside a black hole, time is a space-like dimension, and it is the radius of the test particle that has time-like properties. That’s the Schwartzschild description; the Kerr-Newman description is much more complicated because the angular position of the particle in the direction of rotation is also time-like inside the ergosphere.

Jesse M.says:

It depends not only on the type of black hole (uncharged, non-rotating Schwarzschild black hole vs. charged, rotating Kerr-Newman black hole, for example), but also on what type of coordinate system you use. In Schwarzschild coordinates it’s true that the “time” coordinate is space-like inside the horizon of a Schwarzschild black hole, but this is really just a peculiarity of the coordinate system, not any sort of objective physical fact; if you instead use Kruskal-Szekeres coordinates to describe a Schwarzschild black hole, for example, the time coordinate remains time-like inside the horizon, and the space coordinates remain space-like. And it’s also true that even in flat Minkowski spacetime, one could potentially define a “weird” coordinate system where a given coordinate switches from being time-like to space-like past some boundary.

Yes, the grand lesson is that the horizons in both cases are coordinate singularities that can be easily transformed away. K-S coordinates for example arise from maximally extending the metric in a smooth way.

myth bustersays:

Under extreme enough conditions, a strongly curved spacetime can be used as a time machine into the past. The caveat is that you can’t actually CHANGE the past with such a time machine. The quantum properties of whatever is sent into the past will be restricted such that the act of sending it into the past make sending it into the past inevitable, no matter how improbable this action would be without that constraint.

You can probably answer the following questions as yes or no and you may win for yourself singularity as the prize !

1. Blackholes / Wormholes have been predicted to exist.
2. Black have definitely been observed in reality ?
3. Wormholes have definitely been observed in reality ?
4. Unicorns have definitely been observed in reality ?
5. We have perfect simulations for blackholes and wormholes ?
6. Blackholes can be created by the will of super beings ?
7. Scientists can predict the weather ?
8. Scientists know what happens when we fall into a blackhole because of computer simulations and spreadsheets ?
9. Can scientists discover brightholes by pure mathematics ?
10. Emotions are higher dimensions ?
11. Scientists have discovered gravitons and megatons ?
13. CGI = Science ?
14. Archimedes can bend space ?
15. The mass of a physics textbook is 20 times a Telephone book ?
12. Savvy ?

You win 0111010101010 !

Robert Müllersays:

Dear Dr. Kohli,

Please answer me two questions: Given your calculations, what is the minimum distance from the water planet that has to be reached in order to reduce the time dilation to let’s say 1 hour = 1 year? And how long would it take to reach that distance with a conventional rocket engine?

Yours Robert Müller

Hi!

Thanks for doing the calculation for the spinning black hole – I had assumed a static black hole and a Schwarzschild metric in my estimate.

Even if you can make it work (just!) in GR, Miller’s planet remains an astrophysical impossibility. For example:

– This is an extended object we are talking about, not a point particle. Tidal effects will be severe. Also, you’d have appreciable differences in the time dilation fraction from one side of the planet to the other — perhaps even from the top of the atmosphere to the surface! This close to the black hole, you can’t neglect those differences in position.

[incidentally: if this is a supermassive black hole we are talking about, wouldn’t you expect it to fill a much bigger portion of the sky as seen from the planet? As I remember the movie – but you have seen it more recently and multiple times, so you can correct me – it looked relatively small, which would be implausible given its mass and the closeness of the planet?]

– Deviations from the 1/r potential mean that there is no closed orbit, even if it is stable: this *guarantees* that the planet will hit the accretion disk, which will incinerate the atmosphere. Ooops.

There are further aspects we could quibble about, but stepping back and looking at the bigger picture here: It seems to me that nobody would ever consider looking for a “habitable” planet around a black hole! This is simply not where the habitable zone is (we could talk at length about where the planet gets its heat from, etc). It’s simply bad science to suggest that Miller’s planet could ever be considered humanity’s best bet for survival. So this crucial plot aspect hinges on poor science, even if you can make the time dilation factor kind of work.

As a science communicator, I’m delighted to see so much interest for the science hinted at by Interstellar. But to claim that the movie is “scientifically accurate” is a stretch of the imagination! Much better to say that it is “inspired” by real physics, and leave it at that.

Robert Müllersays:

Interestingly, the planets orbiting the black hole were not supposed to be habitable in the original script by Christopher Nolan’s brother Jonathan and Kip Thorne.

myth bustersays:

And with good reason; aside from the reasons mentioned in the movie, there’s the fact that the accretion disk of the black hole makes a lousy sun. Sure, it shines plenty of light down on the planet, but it also shines an enormous amount of ionizing radiation down on it. The only reason Mann survived as long as he did was because those thick, frozen clouds make a good radiation shield, at the expense of making the closest thing the planet has to a surface too cold for anything to grow on.

Without sounding snide, your comment: “Even if you can make it work (just!) in GR, ” demonstrates the whole darn problem with all of these astro-based comments. If you have a black hole, and a massive one, you have massive spacetime curvature, period. If that is the case, you have G.R., period. Newtonian mechanics will not work.

Now, you can talk about astrophysically due to tidal forces, how the planet should break up, and Newtonian mechanics will give you at best an approximate, far-field solution. You are contradicting your own argument. If you have a Newtonian collapse under significant tidal forces, that you have significant pressure, and here, my friend lies the problem. You started off with a black hole assuming Einstein’s vacuum field solutions: R_{ab} – 1/2g_{ab}R + \Lambda g_{ab} = 0, with the energy-momentum tensor, T_{ab} = 0. The instant you talk about significant pressure, T_{ab} at minimum must assume the form of a perfect fluid T_{ab} = (\mu + p)u_{a} u_{b} + p g_{ab}. This changes the entire metric solution, and you no longer have a spacetime that simply encompasses black holes. So, you can’t have it both ways. You can’t have a black hole spacetime, and also have a significant pressure collapse. Your Newtonian methodology will only work in a very rough, approximate sense.

myth bustersays:

No, it means that there is no SIMPLE closed orbit. The planet’s orbit is closed, but it processes, only returning to its original position after many cycles. In the case of Mercury, the procession rate due to general relativity is 41 seconds of arc per Earth century, which means that Mercury returns to its original position approximately once every 3.2 million years, during which it will orbit the sun about 13 million times. Or rather, this is what would happen if Mercury were the only planet in the solar system; its orbit is perturbed by the gravity of the other planets, making it process much faster than from general relativity alone.

Hello Dr. Kholi,

A couple of days ago you posted your blog on my sister’s (Sophia Nasr) facebook wall, she is currently studying astrophyics at York University. Shortly after, I noticed that Phil Plait commented in this blog that you corrected his error (still available above), which I mentioned to my sister who then commented on her wall.

Today, you’re no longer on Facebook and my sister sent me Phil Plait’s follow-up article on slate.com admitting to his error and originally cited you (and possibly your blog). I checked his article today, that reference has been pulled.

I am very curious as to why you’ve systematically removed yourself from her wall and facebook and possibly Plait’s follow-up article?

Thank you sir!

Dr. Kohli, apologies, I misspelled your name.

Hi Dylan.

Thanks for your message.

Sophia wrote a very beautiful piece on my blog post, but that attracted a lot of attention. I got something like 500 facebook requests, which made me a bit uncomfortable, since people were starting to focus on me, rather than the article, and since I’m not into self-promotion, I temporarily deactivated my FB account until things settle down.

Now, with respect to Plait’s article, I don’t know why the link to my post was removed, or if it was ever there to begin with. As you can see, my post is very much active and still online.

Sincerely,
Ikjyot Singh Kohli

Tylersays:

I was hoping you could comment on the idea behind “Plan A.” What exactly was the “problem of gravity” and why was it a limiting factor for evacuating earth. The centrifuge tech was already available. Could they not lift the installation into space via rocket propulsion or simply build it in space like they did the ISS? Also, and this is less scientific and more something I just didn’t understand, why did the earth refugees not end up on the planet Amelia is on? They just seemed to be happy chilling out in their installation by Saturn.

Dr Kohli,

Thank you for your reply. I tip my hat to your humbleness!

Both my sister and I are very grateful for your work. Sophia hold’s you in high regard.

I also want to say, thank you very much for your work on the film’s physics because this is one of the rare occassion where Hollywood (with the help of Thorne and people like you) created something “almost true” when it comes to science.

I walked away with a better understanding of what you all do, and am that much more proud of my sister!

I hope, at the very least, the movie will inspire our future generation to follow in the footsteps of all you very bright physicists and continue the amazing work that you all do.

All the best!

Dylan

Reblogged this on Entre Nous and commented:
Interstellar 😮

Great read! Quick question – how would the time dialation affect the large waves on the water planet? Wouldn’t the top of the wave be experiencing time faster than the bottom/surface?

myth bustersays:

It’s not the planet’s gravity that’s causing the time dilation. It’s the planet’s proximity to a black hole, which is not appreciably changed by the height of the wave, that is responsible for the time dilation. An Earth-like planet with an escape velocity of 11 km/s has a time dilation factor less than 1+10^-10, which is negligible.

John SMithsays:

Nice article.
However, it doesn’t cover the the following points:

1. Speed of accretion disk rotation. Some claim it is such that the doppler effect would be prominent.
2. X-Ray radiation from the accretion disk should affect characters.
3. Tidal forces would hardly cause the waves of such magnitude.
4. [not as obvious] The craft must have have some new wonder-fuel to have the delta-v it exhibited

X-ray radiation may have been beamed along the accretion plain or rotation axis, but, why wasn’t the same wonder-fuel used in the first liftoff from earth? or to launch all of humanity to Saturn?

Krissays:

That is nice analys. Thanks!

Just wondered. You took into account time dilatation due to observer being deep inside gravity potencial of black hole. What about taking time dilatation due to orbital velocity as well? I can imagine it might be some big fraction of “c”.

Trailblazersays:

Dr. Kohli,
Thanks for your wonderful explanation which arose my curiosity. I shared the link to your blog wherever possible to let people know the right physics. Delving deeper, some questions which are bugging me are:

1. Can radio signals pass inside a black hole, specifically in a rotating, supermassive black hole of this kind??

2. Since Miller’s planet is located close to the black hole has a gravity which is 130% that of earth, is the escape velocity from it too high, say 1000 km/s as some people have pointed out?

3. Any theories on how Cooper exited the black hole?? Since the very premise of the word “black hole” is that nothing can escape from it, not even light, even if it were a rotating one I would think it would satisfy at least this basic property of a black hole even if it is quite different from normal black holes.

Thanks again for your delightful explanations!!

Wow, a comprehensive rebuttal!
I absolutely loved the movie. But I have a doubt regarding the bootstrap paradox, how does it work out?
And why/how does the 5-D world collapse after he transmits the message? Also, is there any wormhole through which he is transported to Saturn?

Mariosays:

Thank you very much. This is an amazing post! I wonder one thing about the water planet, though. By watching the movie (which I love regardless of possible inaccuracies, because it’s a cool story) I thought that the BH around which it was rotating would have been a stellar-mass BH. I thought so, purely from a plausible formation scenario point of view. The fact that it is a SMBH (so I understand from one of your updates), makes the GR numbers much more plausible, but it might make the astrophysics of this planet-BH configuration a bit more unlikely. Maybe there once was a star, with its own planetary system, which happened to pass by the SMBH, loosing its planets, falling into the SMBH, and leaving the planets in stable orbits. Also, if it is a SMBH, the real problem of life on the water planet would be the AGN activity 😉 Thanks again, I loved remembering a bit of the GR I’ve lost in the past 10 years!

Mariosays:

Oh, and another little point on time dilation on the water planet: I guess that the guy waiting on the ship while in orbit around the planet should have experienced a similar effect as the people on the surface, right?

Thank you for this article. And thanks to Interstellar, I learned a lot more about black holes, e.g. how enormous their mass can indeed be.

What still astonishes me, however, is the alleged speed of rotation.

The infographic at space.com and several other sources claim that “Gargantua rotates at an astounding 99.8 percent of the speed of light”, which seems rather impressive. Googling, I found an older (?), unrelated article citing none other than Kip Thorne stating that “most black holes would rotate with a speed that is 99.8% their mass”: http://www.gothosenterprises.com/black_holes/rotating_black_hole.html

That coincidence made me thinking, is it possible someone just mixed up “99.8% their mass” with “99.8% of the speed of light”? How would that change the fate of Miller’s planet, if at all? Is such a fast rotation conceivable? Apparently, RX J1131 rotates at roughly half the speed of light, but I didn’t find anything faster.

HRsays:

Conservation of angular (rotational) momentum – even a slowly rotating star has enormous momentum. Just like a spinning ice-skater, who’s spin gets faster as they pull their arms in, the concentration of the mass of a star to a black hole increases the rate of rotation to very large fractions of the speed of light.

myth bustersays:

Except such a star could not collapse into a black hole that way. During a collapse into a black hole, most of the star’s angular momentum is carried off by the supernova debris, as the rotational velocity combined with the neutron recoil ejects most of the star’s mass.

If Gargantua is a galactic-scale black hole with a “soft” enough horizon to allow for survivable tidal effects, wouldn’t that also imply that the spatial reach of its GR effects would be huge? “Parking” the ship “outside” the well and descending down to the gamma=60,000 level would have to mean a pretty long journey, right? Not to mention the ranger would have arrived at the planet with about 60,000 times its rest mass in kinetic energy, and it would have had to produce that same delta-v again to escape back out of the well.

And to think they ostensibly had too little fuel on board to visit both of the other planets! :o)

There’s plenty of facepalm-worthy science bloopers in that movie. Phil Plait and the others are right to roll their eyes at it, even if not every last aspect of it is implausible.

Without being snide, the facepalm-worthy science blooper is in what you just said, “Not to mention the ranger would have arrived at the planet with about 60,000 times its rest mass in kinetic energy,” are you seriously claiming the validity of E = mc^2 \gamma, where \gamma is the Lorentz boost factor in this case? Do you realize that that is for SR, not GR? That formula is only valid when you flat Minkowski space-time, and 10-D isometry group, there is no such availability for a Kerr metric!

Robert Müllersays:

If Gargantua is a galactic-scale black hole with a “soft” enough horizon to allow for survivable tidal effects, wouldn’t that also imply that the spatial reach of its GR effects would be huge? “Parking” the ship “outside” the well and descending down to the gamma=60,000 level would have to mean a pretty long journey, right?

I would like to know this as well. It has been asked several times, but I did not see any answer.

Hello,
I kind of did reply back to it Christian in earlier post. There are two confusions here:
1. \gamma = 60,000 is the special relativistic Lorentz factor, which does not have any meaning in this situation as we are not in local Minkowski spacetime.
2. There is no such thing as a gravity well in G.R. The closes thing is Flamm’s paraboloid which is an embedding diagram of the Schwarzschild geometry, which also would not apply here as there is no coordinate transformation that can transform away the differential cross term in the Kerr metric.

Therefore, the very question has to be reconsidered as in this context, has no meaning.

ChrisJsays:

Thank you for the interesting and thorough analysis. One question I had about the movie was — how did they get off the water planet? If it had 130% of Earth’s gravity wouldn’t that also require 130% of the thrust required to reach escape velocity? They needed a huge rocket to get off Earth in the first place — how did they get off the water planet with only a little dinky shuttlecraft?

Dear Dr. Kohli,

no, I was not aware of that, thank you for the clarification.

However, my argument was not based on the exact value of the energy expenditure. Surely the gravitational potential difference between an “outside” reference point and a place down the gravity well with a time dilation factor of 60,000 with respect to that reference point would still have to be of order c^2 (give or take a few orders of magnitude) with proper GR treatment?

Cheers

Ok. By how is it a spacecraft that required a large rocket with multiple stages to leave Earth’s gravity well can just fly away a planet with 130% the gravity? There is also the black hole’s gravity well and that little spaceship. We know they weren’t using any super powerful new technology because discovering that technology was half the point of the film.

Ahhhh!!! No such thing as a gravity well!!!!!

Sorry for the improper wording — let me try to wrap my head around this without using the word “gravity well”. I’m basing the following assumptions on what I just read on the Kerr metric on Wikipedia, so please stop me where I’m going wrong.

(1) Even in the treatment of Kerr vacuum, the terms “escape velocity” and “event horizon” appear to be used, so they must still make some kind of sense.

(2) Escape velocity is a monotonous function of the radial coordinate at least down to the event horizon, where it reaches c.

(3) Time dilation is also a monotonous function of the radial coordinate down to the event horizon, where it reaches infinity.

(4) The functional forms of escape velocity and time dilation as a function of radial coordinate are related.

(5) Time dilations one would consider “relativistic” in the SR sense co-occur with escape velocities one would consider “relativistic” in the SR sense.

(6) A location with a time dilation factor of 60,000 with respect to the “flat” exterior space has an escape velocity very close to c.

(7) Descending to, coming to a halt at, and returning from such a location, with a starting point in “flat” exterior space, requires energy expenditures measured in ship rest masses rather than in tonnes of hydrazine.

Has Interstellar fallen to the grandfather paradox?

Here’s my theory- * In the whole movie only Cooper refers ‘they’ as humans. None of the other characters mentioned humans as the source of wormhole. So we should stop getting paranoid over who were ‘they’. ‘They’ can be super intelligent aliens…

[…] the equations he used to come to that conclusion were for a non-rotating black hole. It was shown here that when one uses the correct equations for a spinning black hole and the correct mass of the […]

How scientifically correct is the movie “Interstellar”?

@https://ikjyotsinghkohli24.wordpress.com/2014/11/07/on-the-science-of-interstellar/

[…] находится очень близко к Гаргантюа, согласно ОТО. В этой статье приводится  радиус орбиты, равный 9/2 гравитационного […]

[…] entrada está basada en Ikjyot Singh Kohli, “On the science of Interstellar,” Dr. I. S. Kohli’s Blog, 07 Nov 2014. Para las partes menos técnicos he leído a Rowan Hooper, “Spoiler-free guide to […]

[…] those who saw this movie, check this out. The science behind the movie is solid. love it. On The Science of Interstellar | Dr. Ikjyot Singh Kohli's Blog Do we have any fluid dynamic animals in the house that can make sense of these equations?? […]

[…] Bloggers such as Science magazine’s Daniel Clery, are talking up Interstellar’s scientific accuracy. Most commentators reckon it realistically portrays some aspects of general relativity, and plausibly speculates about things scientists are mulling over. One commentator pooh-poohed the movie’s black hole, but retracted their comment when reminded the black hole in the movie is a Kerr black hole. Mostly when we think about black holes we imagine a Schwarzschild black hole, which does not rotate. A Kerr black hole spins, and that makes all the difference. Better yet, the singularity in a Kerr black hole is not a tiny point: It’s a two-dimensional ring. […]

[…] Jorko universiteto profesorius Dr. Ikjyor Singh Kohli teigia, kad su mokslu kaip tik viskas yra labai gerai. Šitame blogo įraše yra labai daug labai baisios matematikos, nes jis nagrinėja mokslinius […]

Amazing article with difficult math..

nileshsays:

Nice article…
I’m quite confused about, whether I ask question here or not, but I found this blog more informative than others…soo I am asking here…
( My questions sounds like silly… )

1. I was confused about , why did they choose the planet which is situated near black hole ? soon or letter than planet will get eaten up by black-hole… isn’t it ?

2. How he comes into his daughters room’s time span ,when he fall into black hole ? means , why exactly in his room ? why not somewhere else like in other universe, other planet, somewhere on earth ?

No such thing as a silly question.
1. No, black holes don’t eat up anything in general. This will only happen past the black hole’s event horizon (at least for a stationary black hole). For a rotating black hole, it is still not clear what happens for an object moving past the horizon.

2. Why “not” his daughter’s room? It is a science fiction *movie*.

Helmutsays:

Wow, what a treat to read your comments ikjyotsinghkohli24, after stumbling through various sites displaying the usually frustrating pompous arrogance of mostly people who are more eager to force their limited “facts” onto others in order to make it through the day and find the strength to live with their insecurity of knowing not very much about this world.

I have been studying film for a long time and this film presents a new horizon of story telling, it incorporates elements which in the 21st century for the first are becoming “Reality” scientifically, existentially and philosophically.

I have watched the movie twice so far and to me one of the center elements of the plot was gravity. So far not too much has been said about it. I am not a physicist but I have listened to Lisa Randall and her theories and attempts to understand Gravity and she points out that Gravity is a force which is so weak because it may possibly acting upon us while affecting a variety of dimensions. Which would mean that Gravity is not limited to our dimension but can in fact be seen as a traveller which can pass in and out of our dimension. It appears the film was based on exactly that theory when he/ the ghost communicates with the daughter from the 5 D Universe or Multiverse.
The reason why he encounters the daughters room after his journey through the black hole is Love. The writers several times tried to establish Love as a force of existential meaning other than simply fluffy romance. The attempt is made to see Love as a force which plays part in the ordering of universal events and therefore allows the main character to find his daughters room in all of the potential data.
These are all themes which are essential to our lives as humanity is trying to survive itself. Its all part of the rage, to not go calmly into the darkness but to live on and explore.

Joshuasays:

Hey! This has been a great read through the article and comments, but I’m still a bit confused on several aspects. I’m fairly new (1st year) to studying astronomy and physics, so sorry if some of these questions are ridiculous. I’m very curious!

1. How is it possible to enter and escape a black hole? Wouldn’t Cooper be pulled apart or vaporized while passing through the event horizon into the singularity? I don’t really understand the characteristics if a soft singularity with regards to a normal one, and if it affects the horizon’s strength. Additionally, when the black hole collapses around him, shouldn’t the pressure/energy from the collapse destroy him?

2. Why does the black hole collapse, and how did it store the past anyway? I know the 5th dimension is essentially a giant web of infinite possibilities available from a specific moment but that’s about it.

3. I don’t really understand the wormhole. I’ll look into it further myself, but how is it possible to strategically place and stabilize them? Wouldn’t space have to be curved/folded in extremely difficult ways with the help if negative matter? I think I get the concept, but just can’t visualize a curved plane in an omnidirectional void.

Thanks for your help, greatly appreciated! 🙂

P.S. Did you notice anything in the film that’s theoretically impossible, either logically or from a purely mathematical/physical standpoint?

Hi. No such thing as ridiculous questions, except when I ask them.

1. You are thinking of a Schwarzschild black hole, i.e., one that is static and not moving. In this case, if you travel past the horizon, you will be as they say “spaghettified”. For a spinning black hole, i.e., a Kerr black hole, things are a bit different. The Kerr metric actually has two horizons, and one can safely go past one, and also safely travel through the singularity, which for a Kerr metric, is a ring, not a point. There have even been some papers that have shown that travelling through the Kerr singularity, one can essentially traverse a closed timelike curve, which means you can use the Kerr metric as another type of “wormhole”/time-travelling device! I don’t think the black hole collapses around him, where in the movie did you see this?

2. Again, I didn’t see the black hole collapse, he did traverse the horizon to go through the singularity. The whole 5th dimension thing from a physics-standpoint is perfectly reasonable, the fact that he ended up at the moment of in space-time is obviously science fiction. But, regardless, you can think of it like this. In “regular” space-time, we have 3 spatial dimensions and 1 time dimension. Our “events” in our life move up along each one of our timelike curves, i.e., “time flows up”, and we don’t have any control over this. In 5-D, this time vector loses its prestige, and becomes a traversable dimension just like any spatial dimension. So, we can “move through time” in 5-D, just like we can move through space in 4-D.

3. Yes! A wormhole is predicted by vacuum solutions to Einstein’s field equations, but as you say, they are considered physically impossible because they require matter that has negative energy density to hold them together. In cosmology, at least, we are currently not aware of any ordinary matter that has negative energy density, it is highly speculative, but still possible. In principle, it is not difficult to obtain a wormhole from Einstein’s equations, an example of a “non-traversable” wormhole is the regular-old Schwarzschild metric, which is a stationary vacuum spherically symmetric solution of Einstein’s equations.

I didn’t notice anything in the film that is theoretically impossible, from a mathematical physics perspective, which I believe is the same point of reference that Kip Thorne was writing the movie’s science from.

Jesse M.says:

On #1, what you say about the Kerr metric is true, but Kip Thorne actually offered a different explanation for how Cooper could get the “quantum data” from the singularity without being killed in The Science of Interstellar. The difference has to do with the fact that a Kerr metric is an idealized solution that describes an eternal black hole which has remained in the same state infinitely far back in the past, and will continue to do so infinitely far into the future; but in a more realistic rotating black hole that forms at some finite time from the collapse of rotating matter, the structure of the singularities is expected to be different. As you said, the eternal black hole has a ring singularity at the center, which is “timelike”, meaning that it resembles an object in a discrete region of space that people inside the black hole could navigate around and avoid colliding with. But for a more realistic rotating black hole, Kip Thorne says that although physicists don’t have a definitive answer, various attempts at modeling this type of spacetime suggest that instead you would get a “BKL singularity” at the center, which is spacelike just like the singularity in a Schwarzschild black hole, meaning it’s more like a moment in time that lies at all points in space inside the black hole, and is in the inevitable future of everyone inside. The BKL also has a pattern of oscillating tidal forces where the oscillations get arbitrarily large as you get closer to it in time, meaning that you’d be “spaghettified” as you approached it. Thorne discusses all this in ch. 26 of The Science of Interstellar, and he writes that ‘we are not absolutely certain that the singularity inside a black hole’s core is a BKL one … more sophisticated simulations are needed to confirm that the BKL patterns of humongous stretch and squeeze do actually occur in the core of a black hole. I’m almost sure the result of those simulations will be “yes, they do occur.” But I’m not completely certain.’

However, in a realistic rotating black hole that is dynamical rather than eternal, there are also expected to be two other singularities aside from the one at the center, at least one of which is associated with the second “inner” event horizon that you mentioned. For a dynamical black hole there can be gravitational and electromagnetic waves falling into the black hole, and it turns out that these would get infinitely blueshifted at the inner horizon (i.e. the wavelength would get squeezed and approach zero at the horizon, which means the energy density of the wave goes to infinity there), leading to a type of singularity. Thorne refers to this in Ch. 26 as the “infalling singularity”, and describes how in spite of the fact that tidal forces go to infinity here, they reach large magnitudes for only a brief period of time, so that the actual shift in the position of particles making up a falling object (like Cooper) may be small enough that passing through this singularity could be survivable–hence the notion that this might be a “gentle” singularity. Or to quote Thorne, ‘the tidal forces grow so swiftly (Poisson and Israel deduced) that, if they hit you, they will have deformed you by only a finite amount at the moment you reach the singularity … Because your body has been stretched and squeezed by only a finite net amount, when you reach the singularity, it is conceviable you might survive. (Conceivable but unlikely, I think.) In this sense, the infalling singularity is far more “gentle” than the BKL singularity.’ As he points out though, we can’t know what would really happen here without a theory of quantum gravity, since general relativity predicts the energy density would reach the Planck scale where quantum gravity effects are expected to become important.

Finally, there is another singularity which Thorne calls the ‘outflying’ singularity, which was discovered as a theoretical possibility only recently. It was published in a 2012 paper which can be read at http://arxiv.org/abs/1109.5139 — I can’t understand much, but Fig. 4 on p. 17 depicts both the “infalling” and “outflying” singularities on a Penrose diagram, with the infalling singularity labeled the “mass inflation singularity”, and the outflying singularity labeled the “fictitious shock”. I don’t have a very good conceptual grasp on how this singularity forms, so I’ll just quote Thorne’s comments:

‘In the 1990s and 200s, we physicists thought this was the whole story: A BKL singularity, created when the black hole is born. And an infalling singularity that grows afterward. That’s all.

‘Then in late 2012, while Christopher Nolan was negotiating to rewrite and direct Interstellar, a third singularity was discovered by Donald Marolf (University of California at Santa Barbara) and Amos Ori (The Technion, in Haifa, Israel). It was discovered, of course, via an in-depth study of Einstein’s relativistic laws and not via astronomical observations.

‘In retrospect, this singularity should have been obvious. It is an outflying singularity that grows as the black hole ages, just like the infalling singularity grows. It is produced by stuff (gas, dust, light, gravitational waves, etc.) that fell into the black hole before you fell in … A tiny fraction of that stuff is scattered back upward toward you, scattered by the hole’s warpage of space and time, much like sunlight scattered off a curved, smooth ocean wave, which brings us an image of the wave. The upscattered stuff gets compressed, by the black hole’s extreme slowing of time, into a thin layer rather like a sonic book (a “shock front”). The stuff’s gravity produces tidal forces that grown infinitely strong and thence become an outflying singularity. But as for the infalling singularity, so also for this outflying one, the tidal forces are gentle; They grow so quickly, so suddenly, that, if you encounter one, your net distortion is finite, not infinite, at the moment you hit the singularity.’

Thorne goes on to say in Ch. 28 that the idea was that Cooper and TARS passed through the outflying singularity, not the infalling one. The reason for this decision had to do with a plot constraint: even after the tesseract rescued him from the black hole interior and took him on a journey through the “bulk” dimension, Christopher Nolan wanted to impose a rule that only gravitational signals could actually travel back in time, whereas Cooper inside the tesseract could only view the past but not interact with it except via gravitational signals. So when the tesseract returned him to normal spacetime it couldn’t be at a point that was in Cooper’s past at the moment before he was rescued by the tesseract (i.e. it had to be outside of his past light cone at that moment). And after falling through the infalling singularity, someone’s past light cone would encompass the entire future history of the universe, but this wouldn’t be true for someone who just fell through the outflying singularity, so it would be consistent with this self-imposed rule for the tesseract to drop Cooper off near Saturn at a point only slightly in his future.

Jesse M.says:

Minor correction: I said in my last comment that in the paper I linked to, ‘Fig. 4 on p. 17 depicts both the “infalling” and “outflying” singularities on a Penrose diagram, with the infalling singularity labeled the “mass inflation singularity”, and the outflying singularity labeled the “fictitious shock”.’ Reading it again, I see they say it’s called “fictitious” because it is “not actually accessible to any late-infall observer” (which makes sense in terms of the rule on Penrose diagrams that slower-than-light worldlines can’t be inclined by more than 45 degrees from the vertical, so no such curve starting from the exterior region could actually cross the ‘fictitious shock’ line, which is at exactly 45 degrees). In fact, the outflying singularity would actually be the red solid line labeled “shockwave” on that diagram, while the infalling singularity would be the red dotted line.

Also, about #3 and wormholes, as Dr. Singh said these would ordinarily require negative energy densities, though these are actually thought to be possible in a phenomenon known as the “Casimir effect” (but it’s unclear if the Casimir effect would provide negative energy with all the properties required for a traversable wormhole, see the discussion at http://en.wikipedia.org/wiki/Wormhole#Raychaudhuri.27s_theorem_and_exotic_matter ). But in The Science of Interstellar, Thorne says that one of the science-fictional ideas of the movie is that a theory of quantum gravity could allow for manipulation of the gravitational constant G, and that this is at least conceivable in a theory with an extra spatial dimension–specifically, in chapter 25, in the section “Bulk Fields Control the Strength of Gravity”, he writes that ‘If there is no bulk—if the only thing that exists is our four-dimensional universe—then Einstein’s relativistic laws say that G is absolutely constant. The same everywhere in space. Never changing in time. But if the bulk does exist, then the relativistic laws allow this G to change. It might, the Professor speculates, be controlled by bulk fields.” Then later in the same chapter, in the section “Holding the Wormhole Open”, he writes ‘If there is no bulk, then the only way to hold the wormhole open is to thread it with exotic matter that repels gravitationally (Chapter 14). … But there is an alternative, the Professor realizes in my extrapolation of the movie’s story. Bulk fields may do the job. They may hold the wormhole open.’

Deansays:

First of all thanks for the deatailed explanations! Even though some of the math were beyond me..
But I need to ask a question:
How did the future humans put the wormhole near Saturn, if the only way for them to live and evolve is for Cooper to go through the wormhole and then into Gargantua? Is this some kind of paradox? Wouldn’t they go extinct other wise?

Jesse M.says:

Hi Dean, I think the movie was assuming a model of time travel in which all trips through time are constrained to be part of a single self-consistent history (see the discussion of the ‘Novikov self-consistency principle’ at http://en.wikipedia.org/wiki/Novikov_self-consistency_principle ), rather than a model where time travel can “change” the past in some way (or create a new alternate history which exists in parallel with the ‘original’ history). This is the model of time travel that Kip Thorne assumed in his real theoretical work on time travel–I posted a writeup about the relation between Thorne’s theoretical work and the story of Interstellar at http://scifi.stackexchange.com/a/78787/22250 if you’re interested.

Deansays:

So they Novikov principal kicks in when Cooper is saying S.T.A.Y. but because of Novikov his action can’t make him stay so then he decides to give the coordinates to himself do the story won’t change. Right? But how does he give the coordinates in the first place?

Jesse M.says:

Yes, I think he realizes at that point that if he can’t actually change history, he still has a chance to communicate the “quantum data” that TARS gathered during their trip through the black hole, before the tesseract rescued them. After TARS tells him “you even worked out that you can exert a force across spacetime”, Cooper seems to have an aha moment where he says “Gravity. To send a message…gravity crosses the dimensions–including time–and you have the quantum data, now–” Then he first transmits the NASA coordinates to his younger self, and after that transmits the quantum data to Murph via the watch. As for “how does he give the coordinates in the first place”, there’s a line where he actually asks TARS–he says, “TARS, feed me the coordinates of NASA in binary”. So it seems like Nolan and Thorne have taken care to ensure all the information he sends has some kind of conventional origin outside of time loops–he didn’t just remember the NASA coordinates from when he got them in the past, and likewise “They” didn’t just supply Cooper with the quantum data directly, TARS actually had to gather it experimentally.

I think there is some confusion on your part here. a = 0.998 represents the “speed” at which the black hole is rotating, which is taken to be by Kip to be 99.8% the speed of light, i.e., it is a = 0.998*c. The “a” in the calculations for the Kerr metric, is very different. It is a length scale. This can be seen for example by considering a spherical shell that is rotating. It has an angular momentum of J = (2/3) M R^2 $\omega$. The angular momentum for a Kerr black hole is J = M a. One can roughly see that a has the interpretation of being R^2 \omega, which is certainly not a speed!