I heard an interesting question at one point: "how come, when you throw a ball up on Earth, the parabola is so strongly curved? Spacetime is nearly flat, so how can a straight line become such a steep parabola?"I'll answer this question as I understand it, but I only took four lectures of General Relativity before I gave it up in favour of computability and logic, so if there is a more intuitive and/or less wrong answer out there, please correct me.Intuitive answer: the curve is indeed very gentle, and (e.g.) light will be deflected only very slightly by the curvature; but the ball is moving for a couple of seconds, and that's an eternity. On human scales, the time dimension is much "bigger" than the space dimensions (we're quite big in the time dimension and quite small in the spatial dimensions); the ball moves only a small distance through space but a very large distance through time, amounting to a big distance in spacetime, and so the slight curvature has a bigger effect than you might expect.

 It is much easier to think in two dimensions than three. If you picture the ball following a parabolic trajectory in space then you are really thinking in three dimensions here. It might be easier to think of throwing the ball straight up, after which it eventually falls straight back down. NOW plot this in two dimensions with one of the dimensions being time. As the original comment says, the scaling of the time dimension in seconds is not a good comparison to meters. Physicist like units where the speed of light is 1 (one). In these units, in our two dimensional space-time graph, the ball travels a very far distance while the change in our space dimension is still small. It is very close to straight line.
 Off topic, but in reality, the trajectory of a ball thrown on Earth is not a parabola, but an ellipse [1]:> under the laws of gravity, a parabola is an impossible shape for an object that's gravitationally bound to the Earth. The math simply doesn't work out. If we could design a precise enough experiment, we'd measure that projectiles on Earth make tiny deviations from the predicted parabolic path we all derived in class: microscopic on the scale of a human, but still significant. Instead, objects thrown on Earth trace out an elliptical orbit similar to the Moon.
 It's a parabola in a uniform gravity field, an ellipse in a circular gravity field coming from a point mass.So if you want to be really pedantic, it's never an ellipse because the Earth is not a point mass. It would be equivalent to a point mass if the Earth were a perfect sphere of uniform density, but it isn't. In reality it's a potato like mass blob that's approximated by what geodesists call the "geoid". So in order of approximations the path of a ball thrown on earth is a parabola -> ellipse -> numerical integration of 6-dof initial conditions and spherical harmonics approximation of the earth gravity field.
 > So if you want to be really pedantic, it's never an ellipse because the Earth is not a point mass.This turns out to be not pedantic but very important if you're guiding an ICBM. And when landing on the Moon, Apollo had to deal with irregularities in the Moon's gravity due to mass concentrations, called mascons.(If you're interested in missile guidance, take a look at the book Inventing Accuracy. Among other things, it discusses some of the efforts to map the Earth's gravity field to increase missile accuracy for Trident and Minuteman missiles. I knew a physicist who worked on this.)
 (Ah, yes, pedantry. It has its own gravity. You can't tell me it's not a force. I can't resist its pull.)It wouldn't be an ellipse even if Earth were a point mass. The gravity of the Moon and Sun, the gravitational lumpiness in the sky, has the same effect as the gravitational lumpiness underground. The combined result may be no closer to an ellipse than it is to a parabola.
 But on low earth orbit, the L2 term of earth's oblateness dominates by an order of magnitude compared to the moon and the sun, and the rest of the planets are negligible.Source is a table in the first chapters of Fundamentals of Astrodynamics.
 mcv on Oct 19, 2020 [–] > "Ah, yes, pedantry. It has its own gravity. You can't tell me it's not a force. I can't resist its pull."It's not a force. From the right frame of mind[0], it's just a straight line.[0] Substituting for frame of reference.
 jhoechtl on Oct 19, 2020 [–] To extend the pendantry: Don't we have to account that the onserved object isn't a point mass either?
 Do we have to use this knowledge to toss nukes at each other? Surely there's something else we could be doing instead?
 > Do we have to use this knowledge to toss nukes at each other?ICBMs were never used to toss nukes at people.Pretty sure their existence saved the world from WW3.> Surely there's something else we could be doing instead?The application most useful in everyway life is GPS.
 ICBMs... the black knight and the white knight of the cold war
 avmich on Oct 18, 2020 [–] Spherical mass can be replaced with a point mass. Earth however is not spherical (biggest deviation is polar flattening). And even then, Einstein's model, unlike Newton's, says it's not an ellipse even for a point mass.So, moral of the story - we have to be not too pedantic.
 > Spherical mass can be replaced with a point massI'm not a physicist, but that doesn't sound right. One example is that if you are inside of the sphere, at least some of the mass will be pulling you away from the point at its center. I think what you mean is that a point mass closely approximates a spherical mass, but the degree to which that is true becomes less and less the closer you get to the center. I don't think you have successfully contradicted what the person you responded to said.
 In Newtonian physics, a sphere and point mass are exactly interchangeable as long as you are not inside the sphere. If you are outside the sphere, the equivalence is exact, regardless of distance.Proving this is a classic problem in undergraduate physics.
 Thanks! Shame I never took undergraduate physics but now that you've rung my bell I think we may have discussed this in high school. What an unintuitive result that being even a meter under ground breaks what is, up to that point, a fine model.
 The way the result works is that if you are a meter underground, it is equivalent to standing on top of a sphere with the top meter of mass removed.Or equivalently, if you are inside a shell (sphere outside, hollow sphere inside) then the gravitational effect is zero. This can be explained by analogy with light which also follows the inverse square law--changing your position inside a hollow sphere does not change the fact that you see the sphere all around you. Same reason that there is no electrical field inside a hollow conductive object.
 scotty79 on Oct 19, 2020 [–] Check out gauss law for electromagnetism and gravity. It says that total flux through closed surface is proportional to strength of field sources inside the surface. Flux from outside sources cancels out.You can use this law to see what's the gravity fields as you move under ground and in other very symmetrical cases.
 >What an unintuitive result that being even a meter under ground breaks what is, up to that point, a fine model.It doesn't break it at all. The meter above you can be treated as a hollow shell, which surprisingly has zero net pull, and the solid sphere below can be treated as a point mass just as before.Just remember these two facts, each provable with a simple integral calculation, usually done in high school physics or freshman college physics: a uniform sphere has the same gravitational pull on an object as a point mass at it's center, and the net gravitational pull on an object inside a spherical shell is zero.This all works under perfect spheres, uniform (at the spherical shell level at least) density... There are other cases it works, but this simple case is the basic idea.
 We are inside the atmosphere at least...
 And that brings fluid dynamics into the picture, which produces discrepancies orders of magnitude greater than the parabolic/elliptical/etc. distinction!
 pdonis on Oct 19, 2020 [–] > if you are inside of the sphere, at least some of the mass will be pulling you away from the point at its centerIf the mass is spherically symmetric, this will not be the case; all of the Newtonian forces from the masses further away from the center than you are will cancel out. This is called the "shell theorem", and it turns out to hold even in General Relativity.
 I don’t think this is correct. I think the shell theorem says that the gravitational forces cancel if you are on the inside of a hollow sphere and all mass is on the surface. A perfect sphere of uniform density would not meet the shell theorem assumptions.
 Yes, it would.A solid ball (filled sphere) is just a union of many shells (hollow spheres), so the theorem still applies.en.wikipedia.org/wiki/Shell_theorem
 pdonis on Oct 19, 2020 [–] > I think the shell theorem says that the gravitational forces cancel if you are on the inside of a hollow sphere and all mass is on the surface.No, it's stronger than that. It says that any spherically symmetric distribution of matter outside a certain radius exerts no "gravitational force" on anything inside that radius, whether there is matter inside that radius or not.
 What if my distance to one point on the shell was zero? Would I not feel infinite acceleration towards the massive point on the shell that I was infinitely close to?
 > What if my distance to one point on the shell was zero?Then you are not inside the shell, you are on it. The shell theorem only applies if you are inside the shell.> Would I not feel infinite acceleration towards the massive point on the shell that I was infinitely close to?In General Relativity, matter is not viewed as point particles. It is viewed as a continuum, described by the stress-energy tensor. This is one of those cases where the difference shows up.
 tim333 on Oct 19, 2020 [–] I think how gravity works at really short distances like 10^(-50)m is still a mystery as we don't have a consistent quantum theory of gravity that works there.But as you get close to stuff, say two atoms, the electrostatic and other forces are far greater than the gravitational ones.
 d_tr on Oct 19, 2020 [–] Any point on the shell, whether the shell has zero thickness or not, has zero mass, but a finite mass density is assigned to it. Any finite mass is spread out, so you cannot have zero distance to enough of its points to feel infinite force.
 d_tr on Oct 19, 2020 [–] You and pdonis are correct but you are saying different things. Inside a perfect solid sphere, you would of course feel force everywhere except at the center, but this force would only be due to the mass that is nearer to the center than you.
 The intuition we came up with when we had to solve this issue in undergrad physics was interesting.For each point of any given distance, you can find a disk of points at the plane of the same distance whose gravitational pull will be equivalent to a single point at their center.You do this for all distances x, and you will find an equivalent rod going from the attracted objected to the center of the sphere, of a non-uniform but symmetrical density.We find that the density is linearly correlated to the area of the corresponding section, and we then take the closer half of the rod, multiply the density by 1/r^2, and find that it is constant!For the far half of the rod, we cannot do this, so instead we divide it into two halves of equal pull, then divide those to halves and so on, and find that it this reduces to the equivalent of a point.Now that we have two points in-line with the object, we find the point with the same total mass that exerts and equivalent force, and lo and behold it is at the center.I'm sure there is a much simpler way to intuit it, though :)
 mcv on Oct 19, 2020 [–] > "It would be equivalent to a point mass if the Earth were a perfect sphere of uniform density"Would it? Even of uniform density, the mass would not the at the core, only the center of mass would be. Most of the mass is actually closer to the surface. I'm no physicist, but I'd imagine that would have quite some impact[0] on any trajectory that crosses the surface.And for any ball thrown at human speeds, I'd expect Earth's gravity would be much closer to a uniform gravitational field than one from a point mass.[0] Pun not originally intended but I'm quite happy with it now.
 pdonis on Oct 19, 2020 [–] > It's a parabola in a uniform gravity fieldIn relativity, there is no such thing as a "uniform gravity field", if by that you mean a field where the "acceleration due to gravity" is the same everywhere. The closest you can come is the "gravity field" inside a rocket accelerating in a straight line in empty space, where the acceleration felt by the crew is constant. That kind of "gravity field" has an "acceleration due to gravity" that decreases linearly with height.
 Why would acceleration not be constant? (Assuming the ship isn't moving from the center of the earth outward or something like that.
 > Why would acceleration not be constant?Look up the Bell Spaceship Paradox. In relativity, two spatially separated objects (or two ends of a spatially extended single object) that have the same proper acceleration in the same direction do not stay at rest relative to each other; they move apart, as seen in each of their own frames. This is different from the behavior predicted by Newtonian mechanics. In order to have the two objects (or two ends of a single extended object) remain at rest relative to each other, the one in front has to have a smaller proper acceleration than the one in back. (Rindler coordinates are often used to describe this case.)
 This is crazy, I'm finding it very hard to get an intuitive understanding of this. I guess that's why it's called "paradox".
 > I'm finding it very hard to get an intuitive understanding of this.Yes, it is certainly not intuitive. That's why Bell took the time to write a paper about it years ago.This article might help:
 amelius on Oct 18, 2020 [–] > an ellipse in a circular gravity field coming from a point mass.Even if you consider the point mass to move because of the mutual attraction?
 One foci of the ellipse is the earth+ball center of gravity. The earth is also travelling in a much smaller ellipse around that same foci.Assuming a uniform+spherical earth. And also a uniform+spherical ball I supposed.
 *One focus, of the the two foci.
 > So if you want to be really pedanticBased on what I've seen over the years I've been around here, I'm quite certain that's a primary motivator for a non-trivial number of commenters here.
 > it's never an ellipse because the Earth is not a point mass.At least classically, a sphere is indistinguishable (gravitationally) from a point mass while you're outside it. The earth is pretty sphere-ish, locally speaking.
 > The earth is pretty sphere-ish, locally speaking.Kinda. There are mountains, they have their own gravitational attraction, and it can be measured... even in the 18th century! https://en.m.wikipedia.org/wiki/Schiehallion_experimentIt all depends on how pedantic one wants to be :)
 The first time I heard of the concept, it wasn't actually that, it was how in the 60s, they had to correct for mascons in the moon while orbiting.
 There are very few stable orbits close to the lunar surface. Basically a couple of polar orbits with very specific parameters, and that's it.The rest get so perturbed by gravitational anomalies that they fall out of orbit after a few months or years--faster than low Earth orbit where there is atmospheric drag!
 Sharlin on Oct 18, 2020 [–] Yeah, the moon's gravitational field is quite lumpy compared to Earth's, due to its smaller size and the way it is believed to have formed. Plus you can orbit much closer to the surface due to the lack of atmosphere, so the lumps are steeper.
 Tyr42 on Oct 18, 2020 [–] And a ball thrown in the air follows a parabola, locally speaking. At least, you're better off correcting for air resistance before you sub in the ellipse equation.
 CydeWeys on Oct 19, 2020 [–] Only if the sphere has uniform density. The Earth very much does not have uniform density.
 Koshkin on Oct 19, 2020 [–] Locally speaking, the earth is flat.
 But if you're speaking globally, you've already defined yourself out of calling the earth flat.
 Good point; this becomes more obvious if you imagine throwing the ball up and then immediately collapsing all the mass of the Earth into a single point at the centre. What path does the ball follow now? It's probably following a path we would more usually call an "orbit", and it sure looks a lot like an ellipse. Now just put the mass of the Earth back where it was, and notice that the ball hits the ground before it can trace out very much of its orbit.
 Despite both being conic sections, cutting up an ellipse won't yield you parabolas. An ellipse has two focal points to which the sum of the distances is constant, while a parabola has a focal point and a directrix line to which the difference of the distances is constantly 0. Two different things.
 More symmetrically, Every comic section has one focus and one directrix (and a semi-major axis for scale), eccentricity is the ratio of distance. Ellipse has eccentricity less than 1, and a parabola has eccentricity equal to 1.
 pdkl95 on Oct 19, 2020 [–] Parabolas are just a special case of ellipses where one focus is infinitely far away.Vi Hart has a nice explanation: https://www.youtube.com/watch?v=v-pyuaThp-c
 Koshkin on Oct 19, 2020 [–] Isn’t a conic section made by cutting up, well, a cone?
 A good point to make, but if one is going to be that pedantic one shouldn't call it "impossible". A parabola is possible with the right wind, air resistance, gravity from nearby mountains, etc, and in practice those have a larger impact on almost any suborbital projectile than the difference that turns the trajectory from a parabola to an elliptical arc.
 If you want to get technical, what about air resistance?
 Speak for yourself. My balls reach escape velocity and then make corrections to assume e=1.
 Ellipse, parabola, hyperbola and circles are all the same thing (called a comic). The only difference is if it includes or goes through the point at infinity. The relevant branch of mathematics is called projective geometry.
 On human scales, the time dimension is much "bigger" than the space dimensions...This is really interesting, and it made me wonder how to convert between space and time. I mean, one meter up is equivalent in magnitude to one meter forward, is equivalent to one meter to the right. Is _c_ the conversion between space and time? In other words, is 300 million meters equivalent in magnitude to one second of time?
 This is right, yes. The distance (aka interval) formula for the flat 4D spacetime is`````` dist^2 = (c * seconds)^2 - meters^2 `````` So to compensate increase in 1 second you indeed need _c_ ~= 3e8 meters.
 It is bigger only because you travel slowly in the spacial dimensions. You always travel thorugh spacetime with a constant speed (the speed of light). What happens is that you're usually going with 460 m/s (as Earth revolves around the Sun) and this is not really comparable to your `t` speed in the x/y/z/t coordinate system. So when you are still your speed is something like 230/230/0/299.791.998.
 Or a meter is equivalent to the time that light needs to travel 1 meter.
 Correct, the speed of light is the conversion factor
 I've always wanted to know why a ball doesn't follow a beam of light if they are both following straight lines in spacetime.But even more important, if light beams are reversible under relativity (reflected off a mirror they will backtrack the same path) then light can not enter a black hole because its reversed path could allow it a way out. But then there's that whole thing of objects falling in appear to slow and stop as they approach the event horizon, so maybe light doesnt enter after all.My conclusion is that you cant really understand it without serious study of under someone who already gets it.
 The difference is that light travels through space, but not time (similar to how a vertical line does not travel the x axis, only the y axis). The ball travels through both space and time.The faster you go, the less you travel through time. Thus, if the ball were travelling at the speed of light, it would not travel through time either and would follow the same path as light.
 > The difference is that light travels through space, but not timeThis is a common pop science statement, but it's not correct. A correct statement is that the concept of "speed through spacetime", which is what has to be split into "speed through space" and "speed through time" in the pop science statement, does not apply to a light ray.In more technical language, the tangent vector to the light ray's worldline is not a unit vector, it's a null vector, and the concept of "speed through spacetime" only makes sense for a worldline whose tangent vector is a unit vector.
 I wouldn't say it's incorrect at all. From the point of view of a photon, no time elapses between its the origin and destination endpoints.
 > From the point of view of a photon, no time elapses between its the origin and destination endpoints.No, this is not correct. The correct statement is that the concept of "elapsed time" does not apply to a photon; it only applies to timelike worldlines, not null worldlines.To put it another way, if your statement were true, it would mean that the origin and destination events were the same point in spacetime. But they're not; they're distinct points in spacetime. Which means that, since the spacetime interval along the worldline is zero, you can't use the interval to distinguish points on the photon's worldline. And the concept of "elapsed time" requires that you be able to do that. So the concept of "elapsed time" can't be used for a photon.
 It's impossible for an object with mass to go the speed of light. But you can observe what happens as you approach it:Let's say I put you in a spaceship and accelerate you to 50% the speed of light toward the sun. From an inertial viewer's perspective you are travelling toward the sun at half the speed of light and it takes you ~16 minutes to crash into the sun. But from your perspective it only took ~14 minutes to crash into the sun[0].Repeat the experiment except I accelerate you to .99c. From an inertial viewer's perspective you are travelling toward the sun at nearly speed of light and it takes you ~8 minutes to crash into the sun. But from your perspective it only took ~1 minute to crash into the sun.Repeat the experiment except I accelerate you to .999c. From an inertial viewer's perspective you are travelling toward the sun at nearly speed of light and it takes you ~8 minutes to crash into the sun. But from your perspective it only took 20 seconds to crash into the sun.Repeat the experiment except I accelerate you to .9999c. From an inertial viewer's perspective you are travelling toward the sun at nearly speed of light and it takes you ~8 minutes to crash into the sun. But from your perspective it only took 6 seconds to crash into the sun.Repeat the experiment except I accelerate you to .99999c. From an inertial viewer's perspective you are travelling toward the sun at nearly speed of light and it takes you ~8 minutes to crash into the sun. But from your perspective it only took 2 seconds to crash into the sun.See what's happening? As you approach the speed of light, the amount of time that elapses until you reach your destination approaches zero. So from an inertial observer's point of view, time has completely frozen for travelers approaching light speed.[0] Using time dilation formula from this page: https://www.phy.olemiss.edu/HEP/QuarkNet/time.html
 > you can observe what happens as you approach itYes, but you cannot extrapolate from this to say that the time lapse for a photon would be zero. A photon is not the limit of objects with mass going closer and closer to the speed of light, because "closer and closer to the speed of light" is frame-dependent, but a photon's speed being c is not. I can find an inertial frame in which each of your objects is at rest, and in that frame, you are the one who is "close to c" (in the opposite direction). But that doesn't mean your elapsed time approaches zero. By contrast, it is impossible to find any frame in which a photon is at rest. The two types of objects are fundamentally different.In more technical language, the action of Lorentz transformations on photons is fundamentally different from their action on timelike objects. So it is simply not valid to view photons as some sort of limit "as speed approaches c" of timelike objects.
 >> In more technical language, the action of Lorentz transformations on photons is fundamentally different from their action on timelike objects.I don't believe that, and have never heard it before. There are many ways in which light actually behaves just like particles with mass traveling at speed c. It has to or conservation of momentum is violated.
 lopmotr on Oct 19, 2020 [–] But is there any physical way to distinguish these fundamentally different situations? If not, then perhaps the fundamentalness of it is just an artifact of the formulation.I'm thinking of solar neutrinos which, for a while, we weren't sure if they were massless or not. We had to observe them experiencing a duration of time to conclude they were massive. If we didn't find that, maybe it was just an even shorter duration, not the absence of one and we would never be able to tell the difference.
 > is there any physical way to distinguish these fundamentally different situations?Are you asking if there is a way to distinguish a timelike object from a lightlike object? Of course there is. The fact that, for something that has a very, very small invariant mass, it might be practically difficult does not change the fundamental principle.Also note that the reason it was difficult, for example, to tell whether neutrinos have mass or not is that we can't just do the obvious and straightforward thing and find an inertial frame in which they are at rest (by, for example, taking a rocket and accelerating it in the direction of a neutrino to see if we can bring it to rest relative to the rocket). So we have to resort to indirect methods. But, again, that's a practical limitation that doesn't change the fundamental principle.
 It still doesn't sound physically distinct any more than distinguishing any continuous quantity as being zero or nonzero. If we measure something that looks like 0, we can't be sure if it's just below the sensitivity of our instruments.For neutrinos, even if we accelerated an rocket and somehow checked if a neutrino was at rest relative to it, we might find that it's not. That means we won't know if we need more speed or if it's impossible. I suppose it's a bit easier than that because we only have to accelerate the rocket fast enough that the neutrino's speed becomes measurably less than c, rather than 0. But still, what if we can't even get it to go fast enough for that? No way to prove that it's travelling at c, it seems.I'd like to add that even photons have a nonzero upper bound to their possible rest mass. At least they used to. Is there any way, in principle, to show that it's exactly zero, and thus falls into this distinct category?
 > It still doesn't sound physically distinctIf you try what I described with a light ray, it will be moving away from you at c no matter how much you accelerate in its direction.If you try it with a massive object, even a neutrino with a very, very tiny invariant mass, that will not be the case; its speed relative to you will decrease as you accelerate after it, eventually to zero.There is no continuum between those two possibilities; they are distinct and discrete. The only continuum is in the latter case, where the final speed of the object relative to you will depend continuously on how long you accelerate.> even if we accelerated an rocket and somehow checked if a neutrino was at rest relative to it, we might find that it's not. That means we won't know if we need more speed or if it's impossibleYes, you will know, because you will know if the neutrino's speed relative to you has decreased or not. If it has, it's possible to bring it to rest relative to you. If it hasn't, it's not. See above.> I suppose it's a bit easier than that because we only have to accelerate the rocket fast enough that the neutrino's speed becomes measurably less than c, rather than 0.Exactly.> But still, what if we can't even get it to go fast enough for that?That's basically the position we are in now: we have no way of building a rocket or other device that can accelerate after a neutrino long enough to tell whether its speed relative to the rocket is measurably decreasing. So we have to resort to indirect measurements. But as I said before, that doesn't change the principle.> even photons have a nonzero upper bound to their possible rest massYes, because, as I said, practically speaking we can't run the obvious and straightforward experiment I described, to confirm that a photon moves away from you at c no matter how much you accelerate after it. So we have to resort to indirect measurements, like trying to measure its invariant mass by other means. But that doesn't change the principle.
 > Yes, you will know, because you will know if the neutrino's speed relative to you has decreased or not. If it has, it's possible to bring it to rest relative to you. If it hasn't, it's not. See above.I don't think this quite works because of relativistic addition of velocities. Naively, it seems to. For example, if the object was travelling at 0.999999c relative to you (appears to be 1.0c according to your limited instruments), then you accelerate to 0.50c in its direction, you'd see its speed reduce to 0.50c (same 2s.f. instrument), which would clearly prove it's not massless. But velocities don't add like that relativistically and I think you'd still see it as travelling at 1.0c because it only decreased a tiny amount, below what you instrument can detect. If you use a more precise instrument or a faster rocket, you might measure it as 1.00000c but then you still won't know if it's exactly c or a smidgen less.Maybe I've got my relativistic velocity addition wrong? But it still looks like the same measurement problem as trying to prove a classical object has a speed of exactly 0, which can't be done no matter how accurate our instruments are.
 > Maybe I've got my relativistic velocity addition wrong?For the specific case you give, that depends on what accuracy you are assuming. The relativistic velocity addition would be:v_new = (0.999999 - 0.5) / (1 - 0.999999 * 0.5) = 0.999997.So if your accuracy is, say, 1 part in 100,000, you wouldn't be able to see the difference. But with an accuracy of 1 part in 500,000, you would, even though you wouldn't have been able to see the difference before the acceleration.Also, suppose you accelerated for a second increment equal to the first; you would getv_new = (0.999997 - 0.5) / (1 - 0.999999 * 0.5) = 0.999991.And one more increment:v_new = (0.999991 - 0.5) / (1 - 0.999999 * 0.5) = 0.999973.As you can see, the differences in velocity grow fairly quickly for each equal increment of acceleration; the growth is not at all linear. And, as I said in my other post just now, for any given measurement accuracy, it would be simple to calculate how much acceleration you would need to be able to distinguish moving at exactly c from moving at 0.999999c (or any other speed less than c that you choose) to that accuracy.
 pdonis on Oct 25, 2020 [–] > velocities don't add like that relativistically and I think you'd still see it as travelling at 1.0cNot indefinitely. Sure, if you accelerated for a short enough time in the direction of the neutrino, you might still be within your measurement error and so not have learned anything. But that just means you need to accelerate for a longer time. For any given measurement accuracy, you will be able to calculate how long you need to accelerate, by your clock, to definitely distinguish the two cases. Relativistic velocity addition does not change that fact. All it changes is the details of that calculation; yes, for a given measurement accuracy, you need to accelerate for a longer time, by your clock, to definitely distinguish the cases than you would if velocity addition were linear. But that doesn't mean relativistic velocity addition makes it impossible to distinguish the cases at all, ever. It doesn't.
 With neutrinos we might find that it's not but it'd be impossible to catch a photon as it would always have the same speed of c in our reference frame.> I'd like to add that even photons have a nonzero upper bound to their possible rest mass. At least they used to. Not sure what you're talking about, their momentum? No object with mass can reach the speed of light and we know they're travelling at that exact speed.
 How do we know they're travelling at exactly c? That's my concern. Last I heard, a couple of decades ago, physicists would occasionally measure a new maximum possible rest-mass for photons. It would be very tiny, of course, but they couldn't say it's exactly zero.
 > How do we know they're travelling at exactly c? That's my concern.We don't, strictly speaking. The measurements you refer to aren't even measuring the speed of photons. They're measuring their rest mass.> physicists would occasionally measure a new maximum possible rest-mass for photons. It would be very tiny, of course, but they couldn't say it's exactly zero.Based on just those measurements, no. The most they can say is that the photon rest mass is zero to within some error bar, and the size of the error bar keeps getting smaller. (The current error bar, IIRC, is 10^-52 grams, or about 24 orders of magnitude smaller than the electron mass.)However, we have a ton of indirect evidence that photons are massless; the most extensive body of such evidence is all the evidence for the gauge invariance of electromagnetism. If photons had a nonzero rest mass, that would break electromagnetic gauge invariance. So photons having a nonzero rest mass would be a huge issue for our current theories, in the way that neutrinos having a nonzero rest mass would not; there is no important symmetry coresponding to electromagnetic gauge invariance that is broken by neutrinos having a nonzero rest mass.
 (Shrug) You can go argue with Neil deGrasse Tyson, it's over my pay grade.
 > You can go argue with Neil deGrasse TysonShow me an actual textbook or peer-reviewed paper Tyson has written where he makes this claim. Pop science videos don't count. (Tyson is by no means the only one; Brian Greene is notorious for the same thing.)You won't be able to because there aren't any. No scientist who talks about a photon "experiencing zero time" in informal contexts will try it in a textbook or paper. That's because they know that if they did, other scientists would call them out on it, so they confine such claims to contexts where there are no other experts so there's nobody to call bullshit.Another point is that if this concept were actually scientifically useful, somebody would be using it in a textbook or peer-reviewed paper. The fact that nobody is is a huge clue that the concept is not scientifically useful. It's only useful for selling pop science books or getting views of pop science videos, where, again, there are no other experts around.
 Dude, this is basic theory at the high school level. Here's how it works: spacetime is 4-dimensional, and a vector in spacetime is always c units long. You can restrict your travel solely to X, Y, Z, or time if you like. If you do that, the other three components are going to be zero.Photons put it all into the X, Y, and Z components, leaving nothing for the t component. They experience a change of position in space, but not in time. What's so hard to grasp about this?Another point is that if this concept were actually scientifically useful, somebody would be using it in a textbook or peer-reviewed paper.Seems that a fellow named Maxwell got a lot of mileage out of the concept, even if he didn't know what was really going on.
 > Dude, this is basic theory at the high school level.No, Tyson's claim is not "basic theory at the high school level". It is a particular interpretation of a theory (Special Relativity) that does not work, for the reasons I gave.> Photons put it all into the X, Y, and Z components, leaving nothing for the t component.Wrong. The spacetime vector that describes a photon's trajectory does have a t component.> Seems that a fellow named Maxwell got a lot of mileage out of the conceptDude, if you think the concept Tyson described is the same as the concept that Maxwell got a lot of mileage out of, then you are the one who needs to learn more about "how it works".
 Wrong. The spacetime vector that describes a photon's trajectory does have a t component.Sounds interesting. Where can I read more about this t component?
 If you're actually serious, try any textbook on SR that uses the four-vector formalism.However, this comment has significantly raised my estimate of the probability that you are just trolling.
 phizy on Oct 20, 2020 [–] There is no "point of view of a photon." Photons do not have frames of reference.So you are completely incorrect insofar as what you are saying is physically nonsense.
 Again, take it up with Tyson and others with doctoral-level credentials who've made a career out of explaining these subjects to the unwashed laity. I'm not one of those people. Nobody posting on Hacker News is, as far as can be discerned.If you and others in the thread feel that these popular authorities are spreading misinformation or using inappropriate analogies, doesn't it behoove you to raise an objection with them directly? Or perhaps with the appropriate faculty committees at their institutions?
 > If you and others in the thread feel that these popular authorities are spreading misinformation or using inappropriate analogies, doesn't it behoove you to raise an objection with them directly?If they want to make money by getting "the unwashed laity" to buy their books, why should I stop them? I simply don't buy them myself. If other people want to get told comforting nonsense, that's their problem. Caveat lector.> Or perhaps with the appropriate faculty committees at their institutions?Which would be pointless and absurd, since, as I have already said, the claims in question are not being made in textbooks and peer-reviewed papers.I suspect you are trolling.
 phizy on Oct 27, 2020 [–] So you take no responsibility for spreading misinformation if the prior source is not corrected? Sounds like you're acting in bad faith then.You can win an internet argument anywhere, but if that's all you want to achieve here, then maybe stop trying to debate physics.
 edem on Oct 19, 2020 [–] Light also travels through time it just never makes progress. It still has a 4 vector (x, y, z, t) is is just that its `t` component is zero.
 shahar2k on Oct 19, 2020 [–] does that mean that velocity is constant "speed of light" and only direction in spacetime changes?
 They both follow straight lines but unless you have a really really really strong arm, the baseball is “moving” across far more time than the ray of light. So if it “experiences” much more gravitational effect from the larger swath of space time it covers.Using quotes since terms are more metaphorical than exact.
 > I've always wanted to know why a ball doesn't follow a beam of light if they are both following straight lines in spacetime.Because they're following different straight lines in spacetime. Roughly speaking, if you pick a point in space and a particular direction in space from that point, there is a continuous infinity of possible straight lines in spacetime that point in that direction in space. One endpoint of that continuous infinity is the worldline of a light ray. The ball's path is somewhere in the middle of that continuous infinity.
 It all depends on your frame of reference. From the ligth beam's view nothing happens, it just falls into the black hole. In fact the light beam never experiences time as it travels with the speed of causality.From the reference of the outside observer the light emitted at the event horizon will forever try to leave it, but as spacetime itself casdades into the hole at the speed of light (at the horizon) this light will get redshifted until you can't see it anymore. This doesn't mean that the light you shoot into the hole never reaches the singularity. It just means that the light emitted at the event horizon will struggle forever to get out of the insane warp. Think about this: if you're walking on a conveyor belt with a constant speed `n` in the opposite direction and the belt itself is moving with a constant speed `n` you'll never make progress. This is what happens at the event horizon.Light beams are not reversible. If you use a mirror they won't travel back in time, they will just change course. They will never go back in time.
 > I've always wanted to know why a ball doesn't follow a beam of light if they are both following straight lines in spacetime.Because light is much faster. Throw a ball at the speed of light and you will see exactly the same path.Lets say you shoot rifles with different calibers. Each time the bullet takes one second to hit the ground but the faster bullets travel longer distances. Light is so fast it escapes from the planet but if the planet were big enough even light would hit the ground like a bullet.
 Light is moving so much faster. In the same two seconds it crosses a much larger distance in space.The line is only straight in spacetime, not in space.
 If I’m understanding, a theoretical ball thrown at 1c would follow the same path as a photon?
 Yes, although you can't actually get anything with mass to travel at 1c (subject to our current understanding of physics).
 As I understand it, the only thing that can travel at c are:* photons* spherical, massless cows in a vacuum
 Not sure about cows. I bet a thing traveling at c might not exactly have 3 spatial dimensions.
 Agreed ! The "spherical cow", has to first become a circular cow /s
 Anything massless travels at c, not just photons.
 So this is the core problem with all of the general relativity materials that model it as a rubber sheet causing curvature in spacetime. They always model it with focus on _spatial_ curvature: which is totally able to model an orbit or a hyperbolic trajectory as a geodesic, but it totally cannot model "throwing a ball up" since the geodesic for throwing a ball up is just a straight line.The important thing is that gravitation is a distortion in space-_time_, which is way trickier to model as a rubber sheet because you end up with one dimension of space and one of time. If you distort _those_ (also, they don't distort quite like a ball-in-a-rubber-sheet), you can get the results of a ball being thrown up. It's also possible to visualize this for 2 spatial dimensions with a distorted 3d space, but tricky.
 Great answer!It’s also why I despise the popular portrayal of space time curvature. It looks at space in isolation rather than space time as a whole, and provides no intuition as to why objects traveling at different speeds follow different trajectories.FWIW I think that in general it is better to just teach people that gravity is an acceleration in classical spacetime (as opposed to a force or curvature). It is simply too hard to create intuition for laymen around minkowskian spacetime, and even harder for curved minkowskian spacetime.
 I think treating gravity as an acceleration is too abstract for many high-school students. For all the on-the-Earth problems, you'd have to analyze them in an accelerating reference frame. That means Newton's laws and the equations of motion don't apply. You'd have to use special linear acceleration equations. An inertial reference frame is simpler and more general. Gravity fits that OK if you consider it to be a force distributed in proportion to mass.
 Think about this. You're always travelling through spacetime with a constant speed (the speed of causality == the speed of light). This is your 4 vector because it has 4 components: x, y, z, t. Therefore the faster you go in one dimension (say, x), the slower you go in the other dimensions. The dimension that's affected the most by the warping of spacetime is time (t) in most cases, because you're moving much slower in the other 3. As you go faster your path becomes less warped. This is why the faster your ball is, the less curved its path look like. This also explains why objects with zero x/y/z speed fall straight towards the object that causes the warping: only t remains from your 4 vector so you're essentially moving through spacetime with the speed of causality (light) --> you fall straight down.
 So if light were traveling through "slow glass" where it's speed through the medium was significantly slowed down, we would see it go in a parabola like the ball?
 For example, one of the first observational confirmations of relativity was being able to observe starlight near the sun during an eclipse being slightly out of place because it had been bent by the sun’s gravity.
 > we would see it go in a parabola like the ball?That is a valid path for the light to follow, hypothetically.But if that's what the light was doing, you wouldn't see it. You can only see things when light enters your eyes.
 This is in tune with what happens when you change the timescale in a game engine with proper physics.
 I'm not aware of any game engines that simulate general relativity / 4D spacetime.I think what you may be noticing is that, as you reduce the tick frequency of a newtonian physics simulation, parabolas become less accurate as integration error accumulates.
 Along those lines: during my university years, this is what I saw in my homemade engine before I had a good understanding of numerical stability.
 pishpash on Oct 19, 2020 [–] That's too bad, such a simulator would be great for intuition.
 While we are (unfortunately) lacking general relativity simulators, a few special* relativity simulators exist:"A Slower Speed of Light" http://gamelab.mit.edu/games/a-slower-speed-of-light/"Velocity Raptor" https://testtubegames.com/velocityraptor.html
 I think this is more of an integration method error than an analogous case.
 Are we flat in the time dimension? Or what is our time size?
 It’s something of a philosophical question. We tend to think of distant galaxy’s as if we are viewing them via some FTL means with a single consistent now. NASA for example tracks distant Mars probes like that rather than marking timing based on when the signal was received.Alternatively, you can think of everything that could impact you as something of a now light cone. The second view has the universe existing as a 3D surface in 4D space time which means objects have a temporal width for each observer. That can be a really useful mathematical model.PS: Edited the above several times for clarity.
 I’m reminded of the film Donnie Darko, but I won’t say more as to not spoil anything for anyone.
 > The creatures can see where each star has been and where it is going, so that the heavens are filled with rarefied, luminous spaghetti. And Tralfamadorians don’t see human beings as two-legged creatures, either. They see them as great millepedes—“with babies’ legs at one end and old people’s legs at the other,” says Billy Pilgrim.—Kurt Vonnegut, Slaughterhouse-Five
 Interesting way to phrase the question, the answer is... depends on relative motion. The notion of simultaneity which is the technical term for what you describe as "being flat" in the time-direction is not an absolute thing in relativity. With respect to your own reference system you are essentially "flat" in the time direction [1], but with respect to someone that moves at a certain speed with respect to you you have certain size in the time direction.The best way to picture it is via Minkowski diagrams, you can find some neat visualizations in the wikipedia article: https://en.wikipedia.org/wiki/Relativity_of_simultaneitySimply picture an extended object in the x direction, Lorentz transformations are going to rotate it slightly in the time direction causing it to become extended in the time direction.[1] If you take it to the extreme you may consider that different parts of your body move with respect to each other and thus they are not in the same reference frame. In which case, not even with respect to yourself are totally "flat" in the time direction.
 If you're walking at 1 meter per second with a 1 meter stride. The space-time interval between your footsteps is: sqrt( (299,792,458m)^2 - 1m^2))299,792,458m is for the distance traveled by light in one second. The -1m is for the distance between your feet.
 "All tragedies are finished by a death, all comedies by a marriage." —Lord Byron
 >Or what is our time size?That's just another way of describing the total time you exist.
 Your size in X, Y, Z isn't equal to the total distance you've traveled, so why would your size in t be the total time you've traveled?
 If you consider your body has extent in time, then when you move between two points in space, your body is a long worm connecting those two points. So your size in some way is related to the distance you travel in your life. But even without time, a person's size isn't really well defined anyway. Is a person wider if he stretches his arms out sideways?
 chadlavi on Oct 19, 2020 [–] your size in X, Y, Z, is how much space you occupy at once, i.e., at a fixed point on the time axis. But how would we even make sense of the notion of "size" on the time axis? How much time we take up for a fixed value in one or more of the X, Y, and Z axes?
 We don’t know, because we don’t know if the past or future are objectively real, or if we ride the wave so to speak.
 Eh, that view doesn't make sense since every particle has it's own light cone, hence it's own time cone.PBS spacetime did an episode about this recently.
 Does it exist earlier or later in the time cone ‘already’?
 I never studied this stuff, so the genius of your intuitive explanation is appreciated.My intuitive response to your intuitive explanation: This ball is moving through spacetime relative to the earth, which is in turn also moving through spacetime relative to the sun - and so light is being deflected off of this ball at each point in its position in spacetime relative to the sun for much longer than light is being deflected off of this ball at each point in its position in spacetime relative to the earth - have I got that right?Light is a hitman Perpetual driveby shooter Never missing As we dance through the night
 > have I got that right?Probably not, because "the sun" appears in your response. If you're watching a ball move on Earth, the sun is an irrelevant variable.
 > Intuitive answerYour answer is basically the one given in an early chapter of Misner, Thorne, and Wheeler, which is one of the classic General Relativity textbooks.
 Hm, I think visualizing this would be a nice topic for a 3Blue1Brown's video or something..
 Are we moving through time with constant speed? Or we're constantly accelerating through time?
 Under special relativity, everyone and everything moves at a constant speed `c` through spacetime. If you feel like you're not moving, it's because all your speed is being put towards travelling faster through time. Conversely, if you manage to move very fast through space, the world around you will appear to speed up, because you've had to trade off some of your forward travel through time so as to travel in space; the rest of the world is moving forward in time faster than you are.So you can change your acceleration through the time dimension of spacetime, by dint of changing your acceleration in the spatial ones.
 I've always intuitively understood this to be the reason why it would take infinite force to achieve light speed for a massive object. When we apply a physical force, it is applied in the spatial axes, so it is always perpendicular to the time axis. Acceleration is just rotating some magnitude of your fixed velocity vector out of the time axis and into the spatial axes. When your spatial velocity is apparently zero, then the component of force that is perpendicular to your velocity is large, so you achieve a large deflection. But as you rotate velocity out of time and into space, it becomes more perpendicular to time, so any force applied perpendicularly to time is now more parallel to your velocity, having a smaller component perpendicular to one's velocity. You can't rotate a vector with a parallel force.This is also why you can't travel backwards in time through just acceleration. There is no way to impart a force perpendicular to your velocity vector when it is already perpendicular to time, giving you no way to rotate the vector to have a component that points backward in time.So I've always wondered, whether general relativity allows for forces parallel to time, and we just don't know of any mechanism to actually do so, or if it does not cover such cases because we have no mechanism, or if it disallows it entirely.
 This is a useful way to think about it, but you have to keep in mind that (even flat) spacetime is not Euclidean but Minkowskian: in the distance metric, time has an opposite sign to the spatial dimensions. So when you "rotate" a four-vector, it actually follows the surface of a hyperboloid rather than a sphere, which means that rotation has a discontinuity at Θ = π/4 (in normalized units) and the vector escapes to infinity! Only massless objects can travel "at the speed of π/4", everything else can only approach but never reach that speed.
 ithkuil on Oct 18, 2020 [–] And at the same time, in your new frame of reference, you're still moving at exactly c through the time and 0 through space. But your time axis is no longer parallel with the time axis of the rest of the world.
 Axes being parallel by whose point of view?
 Everyone's.
 Relativity.
 When you're asking if two objects have the same velocity, relativity does not change the answer.
 I think it does if they do not occupy the exact same space. It's only objective that they have the same velocity from their point of view (but then parallel lines concept hardly makes any sense).
 are you talking special relativity or general?
 Special.I mean the whole way looking at it seems wrong to me. There is no "rest of the world" in relativity. Assigning some objective vector to everybody doesn't work. These only make sense from some specific point of view.By the question you asked now I'm assuming, you meant "but hey, without GR..", but even without GR, ignoring that the universe is expanding, assuming flat space time etc. If the universe consisted just of 3 bodies, 1 being you and 2 being rest of the world, then the way of thinking you described still doesn't make sense in context of relativity and may lead to some confusion (apart from it being, to me, incoherent in context of special relativity).But maybe I'm missing something from your picture, I'm happy to read and learn.
 I don't understand.You can certainly imagine a scenario where two objects measure their mutual distance as being constant in time. You can also imagine other scenarios, but I'm asking to imagine the scenario where two objects do measure their distance to be D and then measure it again and it's still D, and measure it again and again and it's still D. That's just the definition of standing still with respect of each other, and when you plot a space time diagram, their space curve is parallel (because their spacial distance doesn't change).Special relativity doesn't make that scenario impossible. It doesn't force things to move. It just describes what happens when things do move (through spacetime).
 Special relativity definitely won't be a problem. Everyone see timings and speeds a little differently, but everyone can also do the calculation from anyone else's point of view. Objects that are in the same reference frame are objectively in the same reference frame in special relativity; everyone agrees. The reference frame is determined only by velocity. And the distortion is determined only by the observer's velocity, so both objects will have the same distortion.GR I'm less familiar with.
 crehn on Oct 19, 2020 [–] Is it similar to how changing FOV in a game makes the movement appear slower/faster, but in both cases the space traversed is identical?
 We're moving through spacetime with a constant speed, so the faster you move through space, the slower you move through time.
 I just had another thought.The word “dimension” in this context is overloaded. We think of space being three dimensions but really it’s only one - velocity relative to a specific reference frame. Thinking of it this way, the word “spacetime” makes sense; it’s a two-dimensional system: “spatial velocity” (S) on one axis and “temporal velocity” (T) on another. Both velocities are always measured against a reference frame, and their sum is c (c=S+T).This would mean that time travel is impossible not because of a “speed limit”, but because c is a dimensionless physical constant.
 By c do you mean the speed of light? How is it dimensionless?
 Whether something is dimensionless is pure convention. By making c dimensionless and, for even more convenience, setting its value to 1, you can measure distance in seconds!
 CydeWeys on Oct 19, 2020 [–] You can simplify some of your wording here. Direction-less (your "spatial") velocity is simply speed.
 I still don’t understand the graphs at the link, but this intuitively makes sense to me. Thank you - I now have a new, apparently accurate mental model of relativity.
 You’re moving tough time at the speed of light.