scrumbledober 5 years ago

Scott Manley did a great quick video about this

https://www.youtube.com/watch?v=Smp7TqccTpY

  • gibolt 5 years ago

    After watching the video, I feel like there should be software that gets regular updates of planetary surfaces and highlights drastic changes.

    Probably exists somewhere, but would be neat to see historical changes

    • dotancohen 5 years ago

      If you can persuade some congressional committee to fund NASA to develop, launch, and maintain the probes then I'm sure you'll find no shortage of people willing to pay the taxes for it all.

      I happen to agree with you, and would be happy to see taxes allocated towards further exploration of the solar system. But resources are limited.

      • geggam 5 years ago

        If only we got to choose what to pay taxes for...

        • ajuc 5 years ago

          In Poland you can choose one non-profit organization to pass 1% of your taxes for last year. If you choose nothing it goes to the state budget.

          It's a nice idea, even if some organizations abuse it (they give out free tax-filling software that has their organization hardcoded).

          • pawelk 5 years ago

            I wouldn't call this abuse - it's a marketing tactic, and it's rarely hard coded, more likely selected by default - at least that was my experience.

            Just to add some numbers for perspective. In 2018 there was 763M PLN (~$200M USD) to split among 9000 approved organizations. The largest one got 166M with another 100 orgs getting 1M+.

          • baud147258 5 years ago

            It doesn't work exactly this way in France, but any donation to a registered non-profit gets you a 66% rebate of that amount to your income taxe (within a 20% limit of your taxable income).

            • ajuc 5 years ago

              Yeah, it's the same in the result, but the barrier of entry is significantly lower (easier to fill 1 field in tax form once a year than to make a payment and then get a refund).

              • baud147258 5 years ago

                > get a refund

                We don't get a refund (perhaps I misused a word in my last comment): when filling my tax, there's a box for amount donated, which is then used when calculating the amount of money I'll have to pay as taxes.

                But yeah, the barrier is higher, but I think the choice of organizations is wider in our case, like the organization to which I gave money the last two years wouldn't be on the list for example.

    • azernik 5 years ago

      We already do this for the starfield (that's how, for example, most Near Earth Object searches work); it's just harder to get high-resolution total-coverage imagery for planetary surfaces. As far as I understand, we only are getting repeat imagery from the LRO because it did one fast low-resolution pass and is now doing a lower-altitude, high-resolution pass.

    • InclinedPlane 5 years ago

      There are automated systems that monitor the moon for impacts regularly, however they don't watch during a full moon because it's too bright (except during a lunar eclipse).

      • A2017U1 5 years ago

        > they don't watch during a full moon because it's too bright

        Have little knowledge of astronomy but from a photography perspective this is quite amusing, usually it's the opposite problem for many photos.

    • simcop2387 5 years ago

      The biggest problem with this is getting the images. Projects like the LRO do this already I believe, but most changes are otherwise uninteresting.

    • ridgeguy 5 years ago

      I'll speculate that NRO and similar agencies have such software. Doubt it's OSS, tho'.

  • ghilston 5 years ago

    I can't read his name without hearing it in his voice:

    > Scott Manley here

    • tyfon 5 years ago

      My wife always comes to sit in the couch when I turn on something from him just to hear the dialect.

      His way of saying Mun is the best ever and I'm not sure if it's his dialect or if it comes from playing Kerbal Space Program too much :)

      • beezischillin 5 years ago

        I used to watch the ever living daylights out of his channel, but I kinda forgot about Kerbal Space Program for a while. I'm gonna have to pick it up again now get to binging his videos again.

        Him saying "Mün" is just one of the best things ever.

        • tehduder9 5 years ago

          he does a lot of space news and rocket videos now; they're very informative

      • quakeguy 5 years ago

        What kind of dialect is it if i may ask? Hearing him for the first time(a year ago or so), i thought he is from England. Really curious as his speech is quite unusual and also pleasant in a way.

        • 19870213 5 years ago

          He is from Scotland, and if I recall correctly, currently a greencardholder living in San Fransisco.

          • mnw21cam 5 years ago

            Aye. (For Pete's sake, don't call him English!)

        • zrobotics 5 years ago

          Can't find the video, but I believe he's specifically from Edinburgh, as there are many variants of Scottish accents.

        • arethuza 5 years ago

          As a Scot he definitely sounds Scottish to me but with an overlay of an American accent - can't tell where in Scotland but I would guess East coast. A few of his words sounded fairly broad to me.

        • gowld 5 years ago

          Sounds Scottish.

    • JshWright 5 years ago

      You missed the "Hullo"

sizzzzlerz 5 years ago

I know its just a coincidence that the meteor hit during a lunar eclipse while millions of people might have been watching (and probably not seeing it). But, it gives me chills to think that lump of rock has been orbiting through the solar system for possibly billions of years with its unknown fate already sealed. All it took was for the Sun and all the planets and moons to go through their cycles an uncountable number of times to bring it to its ultimate conclusion.

Now, unknown asteroid, your watch is done.

  • jessriedel 5 years ago

    > for possibly billions of years with its unknown fate already sealed

    Probably not for quite that long. The path of a meteor is only going to be deterministic on a timescale given roughly by the largest Lyapunov exponent associated with its orbit. On longer timescales, even the very tiny irreducible quantum uncertainty in its initial state would be enough to cause it to miss the Moon.

    Not sure what the Lyapunov time is for these meteors, but typical time scales for planets are 2M-200M years. For small things like meteors, there are likely other sources of noise (e.g., solar wind pressure) with quantum uncertainty that lead to non-determinism even faster.

    • nonbel 5 years ago

      >"The path of a meteor is only going to be deterministic on a timescale given roughly by the largest Lyapunov exponent associated with its orbit. On longer timescales, even the very tiny irreducible quantum uncertainty in its initial state would be enough to cause it to miss the Moon."

      My understanding is that chaotic systems are still deterministic, the state at some later time is just very sensitive to the initial conditions (which it may be impossible to know with much precision). "Irreducible quantum uncertainty" is something else that has nothing to do with Lyapunov time.

      I also see Lyapunov died in 1918,[1] while the uncertainty principle was introduced in 1927.[2]

      [1] https://en.wikipedia.org/wiki/Aleksandr_Lyapunov [2] https://en.wikipedia.org/wiki/Uncertainty_principle

      • Someone 5 years ago

        ”the state at some later time is just very sensitive to the initial conditions”

        I am sure it has zero impact in practice, but theoretically, there’s a random component that isn’t part of the initial conditions.

        Our current understanding of radioactive decay is that it is 100% random, and that the law of conservation of momentum also holds at quantum scale. If so, radio-active decay that emits a particle that escapes from an orbit around the meteor (relatively easy, given the lack of atmosphere and low mass of the meteor) has a (minute) random effect on the velocity vector of the remaining larger part.

        Because individual decay events are uncorrelated, adding up all the effects of radio-active decay over time will make the already extremely tiny effect a lot smaller (by the square root of the number of decays, if I’m not mistaken, but there may the timing of decays gives earlier decays a larger effect on the eventual course, so it may be slightly less dramatic).

        That’s why I doubt it has any practical impact.

        • tomp 5 years ago

          > Our current understanding of radioactive decay is that it is 100% random

          Isn't it only locally random? I.e. there could still be some global "initial conditions" that make it deterministic (yet unknowable for us)... As in, given a sequence of numbers, there's no way to tell if it was generated randomly or just pseudorandomly (e.g. by RSA/SHA).

          • db48x 5 years ago

            That would be called a "hidden variable", and it's already proved that no theory involving hidden variables can be correct.

      • jessriedel 5 years ago

        > My understanding is that chaotic systems are still deterministic,

        Classical chaotic systems are indeed deterministic in principle, and only effectively unpredictable due to finite measurement precision. But the world isn't classical, it's quantum, and quantum systems are fundamentally indeterministic.

        (You can make a philosophical move where you say the wavefunction is deterministic even if the results of our measurements are indeterministic, and the wavefunction is the "real" thing, but that's orthogonal to the point I'm making. All I'm saying is that the meteor's path, on long enough timescales, is random in the same sense as radioactive decay.)

        > "Irreducible quantum uncertainty" is something else that has nothing to do with Lyapunov time.

        I think I'm going to have to pull rank on you and just say you're wrong on this one :) There's a whole field of quantum chaos that explores the relationship between these things in depth. Here's one of my favorite papers:

        https://arxiv.org/abs/gr-qc/9402006

        > I also see Lyapunov died in 1918,[1] while the uncertainty principle was introduced in 1927.[2]

        I'm discussing the implications of chaotic dynamics for quantum systems, and scientists generally do not live to see the implication of all their ideas. Maxwell realized that light is just vibrations in the electromagnetic field, and he died in 1879, yet we still start by writing down Maxwell's equations when learning how light operates under the laws of quantum electrodynamics.

      • nine_k 5 years ago

        Solar wind can be strong close to Sun during flares. If you can predict flares on the scale of millions of years, then likely you could reduce uncertainty to just quantum noise which is way weaker.

      • adrianN 5 years ago

        Quantum effects have an influence on macroscopic objects, including asteroids. We can't really see them because of decoherence. Here's a paper that looks at the tumbling of Hyperion from a quantum viewpoint:

        https://arxiv.org/abs/quant-ph/0503170

    • pdonis 5 years ago

      > The path of a meteor is only going to be deterministic on a timescale given roughly by the largest Lyapunov exponent associated with its orbit.

      No, it's only going to be predictable on that timescale. The uncertainty involved has nothing to do with quantum uncertainty; it's purely classical uncertainty because we can only make observations with finite accuracy.

      • tlb 5 years ago

        The Lyapunov time scale defines the time for any variation of position or momentum to be amplified by a factor of e. The exponential growth applies equally to a measurement error and a quantum event affecting the object's momentum. Quantum events such as alpha decay (shooting out a helium nucleus with momentum of about 1e-21 Ns) start off small but after 50 Lyapunov times can get huge.

    • jefurii 5 years ago

      This is why I read Hacker News.

    • nickysielicki 5 years ago

      Somehow, that is even more bone-chilling.

    • bellerose 5 years ago

      Quantum uncertainty doesn’t change the outcome from being deterministic.

    • gerdesj 5 years ago

      >> for possibly billions of years with its unknown fate already sealed

      >Probably not for quite that long.

      No, you are right "Probably not for quite that long" - way longer.

      Parent was looking deep in wonder at something and frankly the bits that make up the meteor are as old as they are old. Those atoms have been hanging around the place for quite a while but I'm not too sure that anyone can really decide when they should be considered as "existing" as a meteor.

      • azernik 5 years ago

        GP was talking about determinism (chaos theory), not the Ship of Theseus problem. i.e. the same meteor has existed for a long time, but orbital dynamics in the real solar system aren't really deterministic on such long time scales.

  • anonytrary 5 years ago

    If you think about it, the Earth and Moon and every other lump of rock in the universe are not fundamentally different than this one. Eventually their watches will end, too.

    • balabaster 5 years ago

      Yeah, it's weird like the light from some of the stars takes so long to get to us that when it left home, the dinosaurs were still roaming the earth... and then after some 65 million years, it hits us allowing us to see that star which was once there in its place in the sky, and one assumes still is, but we can't be sure, and that's that. Done. Over. Gone.

      Though, I supposed given that energy can neither be created nor destroyed it somehow changes form and is absorbed in some fashion.

      Crazy for something to spend so long travelling in a straight(ish) line and the BAM! Finished.

      • sethammons 5 years ago

        The one that gets me: in the future, the expanded universe will be so spread out that there will be no detectable anything outside of your immediate solar neighborhood. No stars. No nothing. There will be no reason to assume the rest of the universe even exists. We live at an amazing time in our universe's history where we can see stars and learn from them.

        • dredmorbius 5 years ago

          The expansion constant is weaker than gravity, such that gravitationally-bound structures, including everything from asteroids up through galactic clusters, will survive.

          Beyond the local galactic cluster though -- quite possibly nothing.

          • ValleyOfTheMtns 5 years ago

            > Beyond the local galactic cluster though -- quite possibly nothing.

            Do you mean the Local Group?

            https://en.wikipedia.org/wiki/Local_Group

            Or the Local Supercluster?

            https://en.wikipedia.org/wiki/Virgo_Supercluster

            • dwaltrip 5 years ago

              The Local Group is gravitationally bound, but that isn't the case with the local supercluster. I believe that most (all?) superclusters are not gravitationally bound.

              In fact, the entire local group will eventually be a single galaxy I believe, mostly composed of the contents of the Milky Way and Andromeda galaxies.

      • warent 5 years ago

        Light that traveled across trillions of trillions of kilometers, passing endless galaxies, nebulae, planets, (possibly extraterrestrial lifeforms) and other breathtaking interstellar objects, all so that it can end its journey on the bum of a random tourist at on a beach. What a world!

        • JCharante 5 years ago

          Fortunately for the light, its trip was pretty instantaneous.

          • balabaster 5 years ago

            It's a hard concept to wrap your head around. Eons of time have passed for us, planets have formed, entire species have evolved, flourished and become extinct, ice ages have happened, wars, humanity has grown up and has reached a point where we're on the cusp of escaping the boundaries of this rock we call home... but for the light, it blinked into existence and out again, in a moment, time and distance completely irrelevant to its existence... it's as if instead of traveling across our space and time, it is merely passing through from some other dimension.

          • mark-r 5 years ago

            Relatively speaking.

      • harshulpandav 5 years ago

        Seeing dinosaurs could be possible if a 4 light year wide telescope is sent via a worm hole to one of the galaxies in the Virgo cluster (65m light years away) and live imagery of Earth (not via worm hole) is sent back via the worm hole to Earth.

        • balabaster 5 years ago

          I think this would give us all kinds of insight we're totally wrong about at the moment.

      • gowld 5 years ago

        Light travels in perfectly straight lines (geodesics in spacetime). The trip takes 0 time and 0 distance for the photon, from emission til absorption.

        • omarchowdhury 5 years ago

          Isn't the emission from a star, and the absorption within an eye or a telescope? There is clearly distance and time between the two here, how are you reducing it to 0?

          • sk5t 5 years ago

            Time passes for us, but not for the photon traveling at lightspeed.

            • bookofjoe 5 years ago

              It's always now for a photon.

            • omarchowdhury 5 years ago

              Would you explain why this is so? The speed of light is finite, not infinite.

              • zaarn 5 years ago

                A photon is moving at the speed of light. Things moving at the speed of light do not experience distance or time (and have no mass), for them their creation and destruction are the same event.

                This effect is also valid for objects with actual mass, the faster you go the slower your time moves relative to everyone else and the shorter distances become. You can't hit the speed of light but you can get fairly close.

                • omarchowdhury 5 years ago

                  > A photon is moving at the speed of light. Things moving at the speed of light do not experience distance or time (and have no mass), for them their creation and destruction are the same event.

                  How does this follow? There is still a traversal of spacetime at a limited speed between the creation and destruction. If the speed of light was infinite then yes I would agree with the proposition that no time or space experienced.

                  You are offering a description without the explanation.

                  • zaarn 5 years ago

                    >There is still a traversal of spacetime at a limited speed between the creation and destruction.

                    From your point of view, that is correct.

                    From the hypothetical point of view of a photon, an object moving at the speed of light, time does not pass and the distance gets squished into zero.

                    You can observe a partial effect by looking at cosmic rays. We can detect several decay products of cosmic rays on the ground despite these decay products being far too short lived and too slow to have actually reached the ground. However, due to their high speed, their internal clock runs "slower" compared to ours and the distance between the top of the atmosphere where they are created in a collision or decay and the ground is much shorter than they can travel at their speed.

                    A photon experiences the same effect times infinity; distance and time no longer have a reasonable meaning other than being 0.

                    • omarchowdhury 5 years ago

                      I'm sorry, I still don't understand and if you can provide me with further material to research I would appreciate it. From my point of view, for there to be 0 time and 0 distance is equivalent to saying the speed of light is infinite, ie. the point of emission is equivalent to its annihilation, but that is traversed at the limited quantity of c.

                      • zaarn 5 years ago

                        It's a bit of a hard subject to digest, I admit that.

                        I guess one thing would be to accept that distance and experienced time aren't objective.

                        If you move in a train, the time you spend inside the train moving and the distance you cover from your perspective will differ from the time that is shown on the clock at the station as well as the distance between the stations. By a very tiny amount.

                        This process is exponential, so if you double the speed you travel at (as a number relative to four) the effect on your time and distance covered will quadruple.

                        When you travel at C, then this effect becomes infinite. Time passes infinitely slow to the point that the time passed becomes zero. Same for distance.

                        It also doesn't mean the speed of light is infinite, rather, from the perspective of the photon, ignoring the time part, the universe is exactly 0 in size. In turn that means travelling to some point in the universe, from the perspective of that photon, is instant. From the perspective of the outside observer, the photon is not traveling 0 distance, it's moving at the speed of light to a far away destination. From the photon's perspective, it's speed is not infinite, it's not moving at all, it's created and annihilated at the same point in space and time. From ours it's moving at the speed of light for a long time.

                        Because these are different frames of reference, it's not as easy to compare speeds experienced in one to the other, especially if one is moving at C relative to the other.

                        (Simplified) The entire affair is necessary because only things not having mass can move at C (and in turn, must move a C, a massless particle like a photon must always move at C). The entire rest of the universe has mass. From the photons frame of reference, everything else is moving at C. But that isn't allowed because that stuff has mass. So space itself is compressed into a 0 sized point to stop things from moving at C (from the perspective of the photon). Now, the photon however wouldn't be moving because in 0 sized space, you can't be moving at the speed of light. So in turn, time is experienced instantly and the photon is annihilated instantly so it can actually move at C (because it didn't move at all, it was destroyed instantly). In short; the photon doesn't experience time or distance because if it did it would be slower than C, it must always move exactly at C, therefore the universe arranges a situation in which the photon doesn't have to move at all, from it's own perspective.

                        • omarchowdhury 5 years ago

                          Thank you. What would you recommend for me to study to learn more about this behavior of photons, specifically?

                          • db48x 5 years ago

                            Minute Physics has some good videos about relativity.

                        • raattgift 5 years ago

                          We can parameterize any curve in 4-d spacetime with a smooth and monotonic function that maps points on the curve to real numbers. In fact, you're doing that in the text above without realizing it: the function in question is proper time.

                          It's fine -- commonplace, even, especially in the uncurved spacetime of Special Relativity -- to use proper time to parameterize a timelike geodesic. However, those are far from the only curves in a metric-equipped spacetime.

                          In particular, for a null geodesic -- a freely-falling path taken by a photon -- proper time won't do as a parametrization because proper time is zero everywhere along it. However, there's nothing particularly unusual about a photon's worldline: it's just a curve where at every point along it the tangent vector is a light-like vector.

                          On the other hand, along such curves we can use an affine parameter. Wald does this in his _General Relativity_ textbook this way: a geodesic is a curve whose tangent vector satisfies T^{a}\nabla_{a}T^{b} = 0 \RightArrow T^{a}\nabla_{a}T^{b} = \alpha T^{b}, where \alpha is an arbirary function on the curve. We can always reparameterize the second version into the first -- any parameter satisfying the first version is an affine parameter. We can equivalently say g_{\mu\nu}\frac{dx^\mu}{ds}\frac{dx^\nu}{ds} = 0 for all s, where g is the metric tensor and s is an affine parameter along the geodesic x(s). There's some greater depth starting at https://en.wikipedia.org/wiki/Geodesic#Affine_geodesics

                          Affine parameterization lets one calculate the behaviour of photons moving through curved spacetime. If we define momentum with respect to the affine parameter derivative like this: k^{\mu} = \dot{x}^{\mu} then we can take the affine parameter value at one point on the geodesic and use that value's derivative at a different point on the geodesic, giving us a momentum difference between the points that is physically the gravitational redshift or blueshift. Additionally, if \alpha = f(x) then k^{\mu}(x) encodes the photon's momentum components, and from them we can extract the relation E = pc.

                          It doesn't matter that the photon has no proper time because the proper time isn't really physically meaningful. We can still slice up the photon's spacetime into time-indexed 3d spacelike hypervolumes and (for a Eulerian observer at rest in this time-indexing) see that the photon is more red in the slice at time t and more blue in the slice at time t'. We can moreover, thanks to the affine parameterization, calculate how much redder it is at time t and at time t' for arbitrary observers, just like we can calculate the length contraction of a massive object or the time dilation of a clock using their respective proper times.

                          > the universe arranges a situation in which the photon doesn't have to move at all, from it's own perspective

                          No, you've chosen to use concepts from Special Relativity in a limit in which you get yourself into trouble. The photon has a non-zero-length worldline, and you can slice it arbitrarily; you're not restricted to the one unique slicing in which you are least able to talk about its evolution.

                          • zaarn 5 years ago

                            Sadly, you've done what I wanted to avoid; making this subject extremely complicated.

                            When you simplify, you need to drop a lot of accuracy and I'm not totally familiar with the field so even more gets lost.

                            This is above my understanding, not to mention that of someone I'm trying to explain this in very basic and simple terms.

                            • raattgift 5 years ago

                              > above my understanding

                              Let's see if we can expand your understanding a little, if you like. I'll also try to make some sense in case someone else sees this in the future.

                              The tl;dr (cf final paragraph below) is that even though it is hard to make sense of a photon-with-wristwatch (does it ever tick?), General Relativity is a theory in which we can calculate the behaviours of an object travelling at the speed of light interacting with other things travelling at the same speed or slower. Moreover, in General Relativity light can redshift and blueshift due to local gravitational influences and by travelling cosmological distances through the metric expansion of space. We may need to know how much redder it is at some point in spacetime compared to some other point in spacetime, and to some extent that question depends on the photon and its history. We can make sense of the history part as follows:

                              First think four-dimensionally. Rather thinking of the propagation of than an object with some spatial extent from point A in space to point B in space at a later time, let's start with a pointlike object, literally of dimension zero.

                              Let's promote this 0-d object in 3-d space to a 1-d object in 4-d spacetime. That promoted object is a worldline. Points A and B are now simply two different points on the worldline.

                              Now we need a distance function along the worldline. As we're interested in the distance between A and B we want to choose any arbitrary function that gives a monotonic value along the worldline, for example starting with 0 at A and ending at some larger number at B, with every point in between having a value between 0 and the value at B.

                              If our worldline is everywhere timelike then we can simply extract the proper time at each point on the worldline from the line element of the spacetime's metric.

                              However, for a photon, the worldline is everywhere lightlike, and the proper time is undefined at every point, so we can't use it. Attempting to reason about this, by for example setting the proper time to the same value (e.g. zero) everywhere on the worldline, leads to incorrect conclusions in your explanation a couple postings back.

                              Instead, we can choose an arbitrary function. The requirement, again, is that the value at A is less than the value of B and every point inbetween along the worldline. We can without difficulty do better than using a function that returns an undefined or identical value at A, B and points inbetween.

                              Ideally we can choose a function that gives a number at every point on the worldline, even extending beyond A or B, and which we can relate to the the curvature of the spacetime in which we find our worldline. If the interesting part of our worldline is always on a single null geodesic, then there is a good choice of function: the affine parameter. It satisfies the geodesic equation, and gives the right tangent vector for anywhere along the geodesic.

                              Decomposing back into our 0-d particle, this means that at any point in anyone's time, one can use the affine paramater on the 0-d particle's trajectory to figure out how its vectors parallel evolve between two points on that trajectory. Indeed, an observer from her or his perspective can predict where the particle will go, and where it has come from.

                              An object with spatial extent goes from a worldline to a worldtube, but the principle is the same: each point on the worldtube can be distinguished using an appropriate function. For a photon, we still use the affine parameter, because photons travel on null geodesics. They just freely-fall through curved spacetime until they have some direct interaction; that's all being on a null geodesic means.

                              In a spacetime with timelike worldtubes and lightlike worldtubes we have a network of intersections in spacetime, and we are interested in the behaviours at those intersections. Anyone can apply an arbitrary system of coordinates -- or distances along each intersecting worldtube -- and calculate physical quantities at the intersection points in spacetime. Because these intersections are in spacetime there is no "I got to that point in space first and just missed it"; time really doesn't matter -- the point is that at the same point in spacetime, two worldtubes interact.

                              General Relativity (in our 4-d universe) almost always lets us build a small region of locally flat spacetime around such an intersection, such that we can then use the Special Relativity background for calculations. The Standard Model, Quantum Electrodynamics, and other relativistic theories describing light all work in this flat-spacetime Special Relativity bubble, even if the wider spacetime is curved.

                              Alternatively, we know how to foliate spacetime along well-chosen timelike axes, which also slices all worldtubes into objects of spatial extent moving from one spatial slice of spacetime to the next. This is a 3+1 formalism on General Relativity, but it's important that it's still General Relativity with worldtubes on curved spacetime (which has a metric for which we can find lightlike and timelike geodesics). But even in such approaches we don't have the behaviour photons making no sense against the chosen timelike axis. Indeed, we can look at the lapse function which generates a proper time increment even for photons: \delta\tau = \alpha(t,x)\delta t.

                              In physical cosmology, we can slice up our universe along a time axis called the scale factor, and then we find a lapse function (and shift vector) for everything in the spacetime. A free-streaming photon's worldtube has properties (e.g. wave-vector) that are well defined at each scale factor.

                              Whether a cosmic microwave background photon can ask itself whether it feels tired (redder) as we take the scale factor closer to our present day, or otherwise check a "scale factor wristwatch" or look out the window to see where the horizon is, is really a problem for metaphysics. We can calculate it in General Relativity, and work out that it is redder today than it was at the surface of last scattering. Moreover, we can calculate counterfactuals: a CMB or quasar photon that freely streams to us along a path that never takes it near a galaxy (other than ours) versus a CMB or quasar photon that takes a trajectory that brings it near a massive galaxy or cluster will have different redness on arrival here. Compare the redness to a clock-on-spaceship: along the first trajectory nowhere near massive objects, we get one reading of the clock on arrival, but along the trajectory which passes near the massive galaxy or cluster we will have a different (earlier) reading. The second clock ticked slower because it went near a massive object. The second photon is redder because it went near a massive object.

        • peanutz454 5 years ago

          0 time due to time dilation, but 0 distance?

          • jstanley 5 years ago

            I think the reason it experiences 0 time is actually because of length contraction.

            The photon experiences the entire universe contracted to 0 length, so of course it travels along its path in 0 time.

            • Retra 5 years ago

              "For the photon" and "the photon's experience" are nonsense concepts. There is no such frame of reference, so it makes no sense to postulate what things would look like from it.

              • warent 5 years ago

                Unless you have a better theory of spacetime and gravity than Einstein, it's safe to say frames of reference are a fundamental aspect of the nature of the universe.

                https://en.m.wikipedia.org/wiki/Observer_(special_relativity...

                • Retra 5 years ago

                  I didn't say frames of reference don't exist, I said frames of reference don't exist for photons. I'm not sure Einstein or his theories indicate any disagreement on that point.

                  • warent 5 years ago

                    You're right, I misread and jumped the gun. That being said, can you cite photons having no frame of reference? Not denying it to be true, but would like to know more about it.

                    • Retra 5 years ago

                      Frames of reference are identified by the objects which have zero velocity relative to them. A photon cannot have zero velocity, so there is no frame of reference for a photon.

                      You can tell how much nonsense it is by reading the conclusions of the posters above: that 'time doesn't pass' and 'length is contracted to zero', which are just funny ways of saying spacetime doesn't exist and that's not a position any physicist will take seriously.

                      • pixl97 5 years ago

                        So if you are on planet A and want to get to planet B that is 1 lightyear away. You have a magical space ship that can accelerate to c in one minute planet A relative time.

                        How long does it take to to planet A on the travelers watch. No copping out and saying they've been destroyed by inertia.

                        I am going to assume it takes one year and and one minute-ish compared to synced clocks in the AB pair.

                        • Armisael16 5 years ago

                          Your assumption is not supported by another current theories I’m aware of. Our physics doesn’t give any answer to that question - they throw a divide-by-zero error.

                          The fundamental fact that led Einstein to special relativity is that light in a vacuum appears to be traveling at c regardless of the observer’s frame of reference.

                          If the observer is a photon, however, it must be able to observe a photon traveling at 0 m/s relatice to it (is, the photon itself). This removes the cornerstone of special relativity and the whole thing comes apart. Using it to make predictions at this point is pointless.

                          There may be a way to describe the perspective of a photon - we haven’t discovered all of physics yet - but none of our current theories do so.

                        • thaumasiotes 5 years ago

                          > You have a magical space ship that can accelerate to c in one minute planet A relative time.

                          > How long does it take to to planet A on the travelers watch.

                          You're the one who discovered magic; the answer to this question depends completely on how magic works.

                          But without magic, it is impossible for an object with mass to accelerate to c. Why do you believe the question has an answer?

                        • mnw21cam 5 years ago

                          We can approach this problem while avoiding all those annoying "you can't"s by considering what would happen if your magical space ship accelerated to almost c in one minute.

                          The answer is that the distance from A to B would be length-contracted, and would take (from the traveller's point of view) very little time to traverse. In the limit, once the traveller has reached the speed of light, the entire universe in front of them contracts to zero length, and they (from their point of view) can cross the entire universe in no time. Like a photon, they would experience no time between achieving the speed of light, and hitting something.

                          From the point of view of planet A and B, it still takes a whole year for the traveller to make the journey.

                          • Armisael16 5 years ago

                            No. Not like a photon. Why do would taking the limit give you an accurate answer?

                            This is like saying that if you put two protons in exactly the same place they would repel each other with infinite Coulombic force.

                      • samatman 5 years ago

                        Real question: aren’t photons in coherent light in the same frame of reference?

                        If not, why not?

                        • thaumasiotes 5 years ago

                          Based on Retra's earlier comments, the answer to "if not, why not?" is pretty simple -- two photons are not in the same frame of reference because neither photon is in any frame of reference, and therefore there is no frame of reference which both photons are in.

                          If you were willing to assume that one photon did belong to a frame of reference, you would see that that photon was moving at 0 (impossible, but necessary for it to be part of the frame of reference) and all other photons were moving at c (since the velocity of a photon is c in any inertial frame of reference). Thus, all other photons would not be part of the same frame of reference as the reference photon; no two photons can belong to the same frame of reference.

          • 12298765 5 years ago

            Yeah, due to length contraction

        • balabaster 5 years ago

          Light's path is affected by gravity of the bodies it passes in space, so technically it doesn't travel in perfectly straight lines at all.

          • saagarjha 5 years ago

            The parent commenter mentions this when discussing "geodesics in spacetime".

      • quakeguy 5 years ago

        Some light that reaches us was sent out when earth wasn't even formed... And some light will never reach us, the universe is expanding at a higher rate, future astronomers will find the sky empty. That's crazy as well.

      • swebs 5 years ago

        Energy is created all the time in the form of Dark Energy.

  • noonespecial 5 years ago

    Something the size of a football hit the moon so hard we saw it happen 238,900 miles away...

    That's a spectacular end if you ask me.

  • SiempreViernes 5 years ago

    Well, it is not that much coincidence that it was seen during an event where millions paid extra close attention and the moon itself was less bright.

  • samstave 5 years ago

    Why dO we not have a continuous high def stream of the moon period

    One should be able to go to moon.com and literally see the state of the moon at that exact moment i. Real time, i. High def

    • justinator 5 years ago

      I think we do it's just called, "the sky"

      • samstave 5 years ago

        If I run really fast, I can keep the moon in my vision continuously...

        Why are your legs so weak?

        • SmellyGeekBoy 5 years ago

          What are you planning exactly, just checking up on the moon during the day? Get back to work! ;)

        • justinator 5 years ago

          Why are your legs so weak?

          I get that this is a joke, but it's funny you're asking someone like me about being out of shape/weak. I'm laughing, just in a different way than you may have expected.

    • dotancohen 5 years ago

      Who is "we"? If you mean your local space agency, then ask whoever is funding them why they don't fund a continuous moon stream.

  • lholden 5 years ago

    Meteor impacts happen way more frequently than you might realize... and they hit Earth all the time too. :)

    Heck, last year we had a notable meteor come down where I live in Michigan.

    Obviously though... having significantly more people watching the moon at the time lends itself to this kind of thing being a lot more likely to be noticed.

  • IAmGraydon 5 years ago

    Well, technically every atom in your body has been waiting around for billions of years to become an animated meatbag. Everything has an impressive history if you think about it.

  • cwkoss 5 years ago

    ASTEROID is evolving...

    ...

    ...

    BOOM!

    ASTEROID is now CRATER. triumphant music

  • miyhhbftdx 5 years ago

    Dr. Brian Greene steadfastly believes that all future events are already scripted.

    There is no free will. ^_^

wizzard 5 years ago

> The so-called “super wolf blood moon” was eagerly watched by millions of people around the world, mostly via live streaming video.

How do they know most people watched via live stream as opposed to just looking out the window or a telescope?

  • SmellyGeekBoy 5 years ago

    On the subject of the "super wolf blood moon" I wonder how many more random words they can stuff in there to keep the interest of tabloid newspaper readers and Facebook aunts.

  • gowld 5 years ago

    It was visible from all of North and South America, and Western Europe, so most people couldn't see it.

    • wizzard 5 years ago

      Even if millions of people from Asia and Australia were interested enough to watch a live stream of it, I don’t believe they can categorically state that the majority of the people on the entire planet watched via live stream.

      I’m mostly arguing because I find the idea of everyone watching the eclipse on a video screen kind of sad.

      • Already__Taken 5 years ago

        Why, I got up in the UK, found it cloudy and went back to bed. That I could have streamed it live is amazing if you think about it.

    • jstanley 5 years ago

      * if you were lucky enough not to have complete cloud cover for the duration.

      It got so cloudy where I am shortly after the Earth's shadow began drifting across the moon that I couldn't even see which direction the moon was in. So I watched a live stream instead.

      • jefurii 5 years ago

        I checked ClearDarkSky[1] and was lucky enough to find a place to see it withing driving distance. There were three other carfuls of people from Southern California to see it. The clouds rolled in just after the point of total eclipse.

        [1] http://cleardarksky.com [edit: https->http]

        • jstanley 5 years ago

          Looks like the URL needs to be plain-http, not https.

          And unfortunately it thinks North America is the entire world. :)

  • quizme2000 5 years ago

    What is interesting is that the flash may have not been visible to the naked eye, but most DSLR type camera that were used to stream the eclipse would pick it up.

    Also NASA was (and currently) shutdown during the eclipse.

  • rishav_sharan 5 years ago

    Okay i will bite. Why "super wolf"?

    • russdill 5 years ago

      The first full moon in January is nicknamed the wolf moon. A moon that is full at perigee is nicknamed a super moon.

Syzygies 5 years ago

Wouldn't a meteorite impact be even more visible in the dark part of a new moon? Why did researchers go out of their way to observe one during a lunar eclipse?

(In elementary school during NASA's Mercury program, a joke made the rounds that a certain Eastern Bloc country's space program would land on the Sun by "flying at night".)

  • dylan604 5 years ago

    Yes, the new moon is when they are most active, however, one of the guys in the program specifically has been trying to see an impact during an eclipse for quite some time now. The shadow of the Earth causes the new moon phase, and it is the same shadow that causes the eclipse. Why wouldn't they take the free bonus 2ish hours of dark moon time to search?

    • davidmurdoch 5 years ago

      Meteorites aren't more active during a new moon; they're just easier to see.

      The shadow of the Earth does not cause the new moon phase. If it did, it'd be a lunar eclipse.

    • logfromblammo 5 years ago

      The shadow of the Moon itself causes the new-phase.

      The only eclipse possible during the new-phase of the Moon is a solar eclipse. Lunar eclipses only happen during the full-phase.

  • macintux 5 years ago

    The project that captured this one also actively monitors around new moons and previous eclipses.

  • chabes 5 years ago

    It sounds like it was simply observed by multiple separate casual observations. I doubt anyone would “go out of their way to observe one during a lunar eclipse”

swebs 5 years ago

Did the impact produce light, or was that kicked-up dust reflecting sunlight?

jelliclesfarm 5 years ago

“Agent”!!! Seveneves by Neal Stephenson!

bitwize 5 years ago

So it was a SuperBloodWolfMeteorImpactMoon. Awesome.

diggernet 5 years ago

Looking forward to high resolution before and after photos from LRO.

bsherrill 5 years ago

I’m curious are we sure it was an asteroid that hit the surface?

  • LandR 5 years ago

    As opposed to ?

demarq 5 years ago

still amazed something the size of a football could cause that

cpeterso 5 years ago

This story reminded me of Neal Stephenson's 2015 novel Seveneves, where the Moon, struck by some unknown object, breaks apart. Most life on Earth is destroyed except for a few thousand people who can escape. Humanity has one year to prepare civilization for 5000 years in orbit before they can safely return to Earth again. Very interesting thought experiment on the technological and genetic outcomes!

https://en.wikipedia.org/wiki/Seveneves

  • quizme2000 5 years ago

    Spoiler Alert! Yes in the first pages there is a moon event, the rest of your comment just gives away the plot.

    • cmpaul 5 years ago

      Agreed, could probably have avoided giving away the plot here. The book is definitely worth the read!

      • cpeterso 5 years ago

        I see what you mean, but I don't think I gave away too much of the plot. The back of the book itself provides all those same details, other than the word "Moon" which is revealed in the first sentence of the book.

        https://images-na.ssl-images-amazon.com/images/I/515blU3eIzL...

        • detritus 5 years ago

          I have this book on my shelf to read and, as with most books, I try not to read synposes or cover details in advance, preferring instead to have it be as much of an unknown journey as possible, so ... thanks for giving away what sounds like a fundamental part of the book and arguing against perhaps editing your original post.

  • Tepix 5 years ago

    Did you really have to put a spoiler in the first sentence? Sheesh!

    • _whiteCaps_ 5 years ago

      > The moon blew up without warning and for no apparent reason. It was waxing, only one day short of full. The time was 05:03:12 UTC. Later it would be designated A+0.0.0, or simply Zero.

      These are the first few sentences of the book. The rest of the comment are spoilers though.

      • cpeterso 5 years ago

        "Spoilers" that are revealed on the back of the book itself:

        https://images-na.ssl-images-amazon.com/images/I/515blU3eIzL...

        All those same details are there, other than the word "Moon" which is revealed in the first sentence of the book. FWIW, I would edit or remove my original comment to remove spoilers but HN no longer allows me to edit that comment. :(

  • doh 5 years ago

    I have a crazy deja vu. I picked up this boom yesterday night and read the first chapter before going to sleep. And then this came along. I though for a second that I had to dream the plot.