The Science of Interstellar

The intersection — maybe the union! — of science and sci-fi geekdom is overcome with excitement about the upcoming movie Interstellar, which opens November 7. It’s a collaboration between director Christopher Nolan and physicist Kip Thorne, both heroes within their respective communities. I haven’t seen it yet myself, nor do I know any secret scoop, but there’s good reason to believe that this film will have some of the most realistic physics of any recent blockbuster we’ve seen. If it’s a success, perhaps other filmmakers will take the hint?

Kip, who is my colleague at Caltech (and a former guest-blogger), got into the science-fiction game quite a while back. He helped Carl Sagan with some science advice for his book Contact, later turned into a movie starring Jodie Foster. In particular, Sagan wanted to have some way for his characters to traverse great distances at speeds faster than light, by taking a shortcut through spacetime. Kip recognized that a wormhole was what was called for, but also realized that any form of faster-than-light travel had the possibility of leading to travel backwards in time. Thus was the entire field of wormhole time travel born.

As good as the movie version of Contact was, it still strayed from Sagan’s original vision, as his own complaints show. (“Ellie disgracefully waffles in the face of lightweight theological objections to rationalism…”) Making a big-budget Hollywood film is necessarily a highly collaborative endeavor, and generally turns into a long series of forced compromises. Kip has long been friends with Lynda Obst, an executive producer on Contact, and for years they batted around ideas for a movie that would really get the science right.

Long story short, Lynda and Kip teamed with screenwriter Jonathan Nolan (brother of Christopher), who wrote a draft of a screenplay, and Christopher eventually agreed to direct. I know that Kip has been very closely involved with the script as the film has developed, and he’s done his darnedest to make sure the science is right, or at least plausible. (We don’t actually whether wormholes are allowed by the laws of physics, but we don’t know that they’re not allowed.) But it’s a long journey, and making the best movie possible is the primary goal. Meanwhile, Adam Rogers at Wired has an in-depth look at the science behind the movie, including the (unsurprising, in retrospect) discovery that the super-accurate visualization software available to the Hollywood special-effects team enable the physicists to see things they hadn’t anticipated. Kip predicts that at least a couple of technical papers will come out of their work.

And that’s not all! Kip has a book coming out on the science behind the movie, which I’m sure will be fantastic. And there is also a documentary on “The Science of Interstellar” that will be shown on TV, in which I play a tiny part. Here is the broadcast schedule for that, as I understand it:

SCIENCE
Wednesday, October 29, at 10pm PDT/9c

AHC (American Heroes Channel)
Sunday, November, 2 at 4pm PST/3c (with a repeat on Monday, November 3 at 4am PST/3c)

DISCOVERY
Thursday, November 6, at 11pm PST/10c

Of course, all the accurate science in the world doesn’t help if you’re not telling an interesting story. But with such talented people working together, I think some optimism is justified. Let’s show the world that science and cinema are partners, not antagonists.

Interstellar Movie - Official Trailer 3

47 Comments

47 thoughts on “The Science of Interstellar”

  1. Ha again I see anti-singularity propaganda.
    Even Martin Bojowald one of the guru of LQG and LQC admitted what everyone knew: the more you try to guet ride of classical singularities the more you get even worse exotic singularities like naked singularities and the same apply to string theory.
    Here is his recent paper http://arxiv.org/abs/1409.3157
    Honestly I can’t see how there can’t be a singularity inside the black hole with our new observational evidence.

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  4. I hated the movie Contact, so having a pretense of science in a movie does not necessarily makes it a good movie.
    I heard that Interstellar has nice CGI of a black hole, but once again that doesn’t by itself make a great story or a great movie.
    South Park doesn’t have nice CGI or anything like that, but some episodes are absolute masterpieces.
    I haven’t seen Interstellar, so at this point I don’t have an opinion on it.

  5. Phil Plait’s review that Sean linked to in another post came very close to putting me off seeing this … until I realised that at least one of Plait’s objections was incorrect. Plait says:

    … the minimum stable orbit around a black hole must be at least three times the size of the black hole itself.

    But that’s only true for a non-rotating hole. For a maximally rotating hole, there are stable orbits right down to the event horizon, and apparently the hole is meant to be rotating.

    Plait’s other objections sound pretty serious, but reading Kip Thorne saying he’s happy with everything except for the size of some ice clouds makes me think the movie might be worth a gamble.

  6. Not only that, for supermassive black holes of sufficient mass, tidal forces aren’t a problem at the event horizon. Trust me though: Plait’s objections are far more reliable than Thorne’s: the movie is not only rotten in science, it sucks as art. Its nothing but yet another Hollywood example of sensationalist tripe.

  7. Hoo-ray for Phil, and boo to science-y sounding hype… irrespective of how much that actual science people get stroked by guys with bags of money.

    Regarding storytelling and drama: Aren’t the kind of theoretical physicists we are urged to pay attention to supposed to have a some sort of notable aesthetic sense? However short of penetrating insight physicists as a group may come as chin-stroking philosophers, pitted up against certain actual philosophers — and I know I’m skating on thin ice there: just passing on a newsbit from a usually reliable source — I must also tip my hat to Phil’s moonlighting as drama critic. These days, offering up a stitch in time given to succinct summary to spare us from 900 (or more) unnecessary words of further analysis just seems like welcome relief. Pretty pictures go for the proverbial dime a dozen any more, so there are less costly ways in terms of both time and money to experience a couple of “Oh, wows!” to visuals. This makes more sense, I know, when one has passed more than six decades on this planet renting atoms.

    By contrast, my November 10 issue of Time magazine assures me on the cover that “Director Christopher Nolan enlists science to explore the soul…” The cover photo employs monochrome photography… so you know this is deep stuff.

  8. P.S.- Before anyone has the chance to miss the point of the preceding (and I know some will), let me append this:

    Have all the fun you want with the Kip stuff and a Sean cameo. Nonetheless, this: movie optional. Trying to make some wiser, better considered choices during a brief spell of consciousness seems like a good idea, given that eternity is a long time to be dead. I don’t think this one will be making too many intelligently conceived “Movies You Must See Before You Die” lists. I like movies; I’m still working my way through the latter sort.

  9. I haven’t had a chance to see the movie yet, but I thought it might be fun to quantify some of the phenomena involved.

    One way to compare the passage of time between a distant observer and an observer close to a black hole is to compute a blue shift for an incoming light wave. But the fact that the observer is orbiting the hole complicates things a bit. In the case of a non-rotating Schwarzschild black hole, you can always figure out the blue shift for a stationary observer at the same distance from the hole (hovering at a fixed point with rocket power) and then multiply that by the purely kinematic time dilation factor between the stationary observer and the orbiting observer.

    But for a rotating black hole, once you’re inside the ergosphere there are no stationary observers! And if you calculate the blue shift directly for the orbiting observer, it will depend on the direction in which the light is coming.

    However, if you decide to look at light that falls straight into the black hole, I’m fairly sure that gives what amounts to a purely “transverse” Doppler shift, due solely to the difference in clock rates.

    For an observer in a circular, equatorial, co-rotating orbit (going around the black hole in the same direction as it’s spinning), at a radial coordinate of ρ M (where M is the mass of the black hole, which becomes a distance when we use geometrical units, and ρ is dimensionless), for an extremal black hole (a black hole with the maximum possible rate of spin), and for incoming light that falls straight in, I get a blue shift factor of:

    B = (1+ ρ^{3/2}) / sqrt(ρ D)

    where

    D = ρ^2 – 3 ρ +2 sqrt(ρ)

    To get the difference in time that Phil Plait cites, of one hour per seven years, we need B=7×365×24=61,320, which corresponds to:

    ρ ≈ 1.00003766170211195

    That’s awfully close to the event horizon, which for an extremal black hole is at ρ=1. Still, in theory for an extremal hole there are stable orbits for any ρ greater than 1.

    What are the tidal accelerations that a body would experience in this orbit? If we define:

    N = 3(ρ-1)^2

    then the three tidal accelerations (per distance from the centre of the orbiting body) are given by:

    A1 = (N/D – 1) / (M^2 ρ^3)
    A2 = -1/ (M^2 ρ^3)
    A3 = (2-N/D)/ (M^2 ρ^3)

    These three add up to zero, as any purely gravitational tidal accelerations in a (near) vacuum must! The first is the radial stretching, the second is the squeezing along a direction aligned with the orbit, and the third is the squeezing into the plane of the orbit, all measured in a frame that moves with the rotating body.

    If we take M to be 100 million solar masses, in geometrical units where the gravitational constant and the speed of light are set to 1 and everything is measured in metres, that comes to about 150 × 10^9 metres, or 150 million kilometres. That’s quite neat, because it’s more or less one astronomical unit!

    If we plug in this value for M and the value for ρ we get in order to match the time dilation required, and if we convert the tidal accelerations into “gees”, or units of Earth’s gravity, per thousand kilometres from the centre of the orbiting body, then I get:

    A1 = 1.22 gees per thousand km
    A2 = -0.41 gees per thousand km
    A3 = -0.81 gees per thousand km

    These are pretty terrifying, when you consider that the Earth’s radius is about 6,000 km.

    The accelerations I’ve given here are purely gravitational. If the planet was spinning, that would add a positive centrifugal component to the first two; for a tidally locked planet, that contribution would be exactly -A2, raising A2 to zero and A1 to 1.63 gees per thousand km.

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  11. I wrote:

    … if you decide to look at light that falls straight into the black hole, I’m fairly sure that gives what amounts to a purely “transverse” Doppler shift, due solely to the difference in clock rates.

    That’s not correct! Sorry.

    Instead of trying to figure out how to get this via a blue shift inside the ergosphere, I should have just stuck to the easier calculation outside the ergosphere, and I would have found that you can get the necessary time dilation there.

    I think the correct time dilation factor (at least outside the ergosphere) for an extremal black hole, and an observer in a co-rotating equatorial orbit, is actually:

    B = ρ (sqrt(ρ) + 1/(ρ-2)) / sqrt(ρ D)

    where ρ and D are defined in my previous comment. The factor of B=61,320 can be achieved with:

    ρ ≈ 2.000025339

    This gives substantially smaller tidal accelerations:

    A1 = 0.134 gees per thousand km
    A2 = -0.051 gees per thousand km
    A3 = -0.083 gees per thousand km

    Still scary, but it makes the whole scenario much more plausible. Sorry about the screw-up in the previous comment!

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  13. I’ve just done a third calculation, which appears to confirm my first one and contradict the second.

    If you directly compute the rate of change of the t coordinate with respect to proper time τ for the observer in orbit, you get:

    dt/dτ = (1+ ρ^{3/2}) / sqrt(ρ D)

    This agrees with the blue shift factor for light seen by the orbiting observer that entered the hole with zero angular momentum, which is what I used for my first calculation.

    In my second calculation, I computed a blue shift for a stationary observer (that is, someone with constant r and ϕ coordinates), and then assumed I could get the factor by which an orbiting observer’s clock rate differed from that of the stationary observer from their relative velocity. That works for a non-rotating black hole; in fact for a non-rotating hole, all three methods give exactly the same answer. But for a rotating hole, curves of constant t are not orthogonal to the world lines of stationary observers, which I think undermines that approach in this case.

    It’s hard to see how dt/dτ can fail to be the right quantity to use … but what that seems to imply about the location of the orbit (r/M ≈ 1.00003766) and the consequently large tidal forces is surprising. Anyway, maybe some expert in the Kerr geometry can settle this definitively.

  14. Having calculated the time dilation factor three different ways, with the first and third methods in agreement, it might be worth summarising why the first and third methods are correct, and the second was mistaken.

    For the first method, imagine a series of concentric circular wavefronts coming in from infinity and converging on the black hole. Assume these wavefronts lie in the equatorial plane, perpendicular to the axis of spin. An observer on a planet that is moving around the hole in a circular, equatorial orbit will measure a certain blue shift for these wavefronts, and because of the rotational symmetry of both the wavefronts themselves and the geometry of the black hole, that blue shift will be the same regardless of where the planet is in its orbit.

    To see why this blue shift tells us how the elapsed time differs for people near or far from the hole, suppose Alice and Bob start out together, far from the hole. Bob keeps his distance, but Alice travels down, spends some time on the orbiting planet, and then flies back out to meet up with Bob again.

    Every incoming wavefront that passes over Bob, in our series of concentric circular wavefronts, will also pass over Alice at some point. So if they both kept count of these events, when they compared notes once they met up again they would agree on the number of incoming wavefronts they’d seen.

    However, when Alice was on the orbiting planet, by her clock these wavefronts would have been striking her at a greater rate, with the difference from the rate Bob measured given by her blue shift factor. So in order for the number of wavefronts both of them count to be the same, Alice must have measured less time passing overall, by that same factor.

    OK, what about the third method, which yields the same answer? For that method, we use a coordinate system that assigns four numbers to every event in space-time around the hole. One common coordinate system for rotating black holes is known as Boyer-Lindquist coordinates, in which the coordinates are usually called t, r, θ and ϕ. Far from the black hole, the last three are just ordinary polar coordinates, and t measures clock time in a frame in which the black hole is at rest. Closer to the hole, though, t can no longer be interpreted that way.

    However, just as Alice and Bob agree about the number of wavefronts that pass over them, they will agree about the number of “surfaces of constant t” that they cross. Since t measures clock time for Bob, Alice can still compare her amount of elapsed time with Bob’s by keeping track of the rate at which she measures the number t changing. So this ought to give the same answer as before, when the two meet up and compare notes.

    And in fact it does. In some ways this method is simpler than the first, because in order to compute the details of the planet’s orbit around the hole, we need to use some coordinate system, and if we choose Boyer-Lindquist coordinates we get the rate at which t changes for an observer on the planet as a part of the description of the planet’s “4-velocity.”

    That said, I think the first method is better in that it makes a statement about physical events that is completely independent of any choice of coordinate system.

    Finally, why was the second method a big mistake? My approach there was to try to split the problem in two: first, compute the blue shift for a stationary observer, and then compare that observer’s clock rate with an orbiting observer whizzing by, using the usual formula for time dilation due to relative motion that applies in special relativity. What could possibly go wrong … ?

    The mistake there was to try to compare an observer on the orbiting planet with a single stationary observer. To split the problem correctly, you would need to consider a whole family of stationary observers, arranged all the way around the orbit, and whose clocks had all been set to zero at the moment they received the same incoming circular wavefront. The catch is, this synchronising wavefront would have arrived “at an angle” to each stationary observer’s world line, rather than orthogonally to them, and any comparison of the orbiting observer’s clock with this family of clocks would need to take account of that, as well as the relative velocity between orbiting and stationary observers.

    So, the wash-up is that a time dilation factor of 61,320 can only be achieved at r ≈ 1.00003766 M, very close to the rotating hole’s event horizon, and even for a black hole of 100 million solar masses, that means tidal forces of the order of one gee per thousand kilometres from the planet’s centre.

  15. The time-slowing effect is due to the fact that there is a limit is speed that matter can achieve. So idea of traveling in time backwards by going faster than the fastest possible speed violates the precondition.

  16. Latverian Diplomat

    I just saw the movie, and I can’t really recommend it, though the performances are excellent, I found much of the storytelling to be unsatisfying.

    But that aside, I noticed a science issue that I haven’t seen remarked on elsewhere.

    Spoilers:

    A major plot point is the catastrophic expedition to Miller’s planet, the time dilated one. They completely know about the time dilation effect, and in fact they can calculate it, but they are not surprised to have years worth of data from Miller about her planet. But once they land, they suddenly realize that Miller would have been down there for only an hour or so (IIRC) so why did they think all that data was real? In fact, since they wisely considered that they should take years of data from all the candidates, why would they even consider a world where it would take millenia (their time) to get years of climate data (planet time)?

    Also, wouldn’t the signals from Miller be red shifted and stretched out over time so that they would have gotten only a few completed messages from her, if they even had the equipment to capture such an extremely red-shifted signal?

    Lastly, the datagrams from the Lazarus probes were clearly non-trivial, since they were comparing the merits of candidates based on the quality of the data received, and so those datagrams would certainly have had a sequence number and/or timestamp. That’s communication protocol 101 stuff. The “echoing” they used to explain all the Miller messages would have been trivial to detect.

    So a major attempt in the film to drive the plot with relativistic physics seems to me to fall apart under scrutiny, I’m sorry to say.

  17. to ride on L.D.’s coat tails; I thought it didn’t make much sense to leave the station outside of time dilated orbit. It saves fuel? How much fuel does it burn living on a space station for 30 years?! That just doesn’t make sense. There didn’t seem to be any logical reason to do that.

  18. Latverian Diplomat

    @Brett:

    To be fair to the film, the original plan would have only taken two years, and IIRC the guy they left behind was going to use that time to make observations of Gargantua that might help with “Plan A”.

    That said, the movie was never very clear about resources available on the Endurance and which activities and choices consumed more or less of them. A few remarks about hydroponics or long lasting power sources like Radio- isotope Thermoelectric Generators would have made some of the handling of resource issues a little more plausible.

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