md65536

It's not filled with long waves that occupy space. Light is quantized into packets that have a point location. Wherever you measure "where" light energy is, it will be measured as individual photons each with a single point location. A sea of photons in space would be entirely empty space with photons in it, taking up no volume. Another way it's not like a sea or a volume filled with matter is that the photons don't interact. They won't spread out to fill a space, they'll just go where they're going. I would not agree with this (as an amateur), though I think many people would. The wave-particle duality doesn't mean that light "is" a wave and it "is" a particle, only that it consistently has properties of either, and it only certainly has those properties when measured. So, when you're not measuring the particle properties of a photon, there's really no point in saying that it "is" a particle even where it's not being measured. You could easily just say that yes, if they were measured there would be photons all over between Sun and Earth at all times. But a sure statement about photons in between measurements can cause problems. For example, the sun is about 8 minutes away, so you could say that there are a bunch of photons traveling at different points along an 8-minute journey between the Sun and you. But if you accelerate to near c you can shorten that journey to say 1 minute. Then there is only 1 minute worth of photons, but they're still traveling at the same speed. What happened to the other 7 minutes worth? You might imagine relative time being "adjusted", so those en-route photons are brought back into the Sun and now haven't left yet. BUT I think this is pointless. This is an example of a confusing interpretation based on the assertion that the photons must exist as particles along their entire (8-minute) journey. On the other hand, if you don't measure any photons in between Sun and Earth, you only have to say that they "are" where they're measured, and the existence of photons beyond how they are measured becomes irrelevant philosophy. So I would not agree or say anything certain about the existence of photons anywhere between where they're emitted and detected (absorbed or measured). Or consider a different example. Suppose two people agree that there are photons constantly traveling along straight lines from Sun to Earth. Now put a double-slit between Sun and Earth. Suppose one person says "Each photon must go through either one slit or the other" while the other disagrees and says "Each photon must go through both slits simultaneously." It depends on interpretation, and neither interpretation is meaningfully measured. Similarly, agreeing on the behavior of photons without the double-slit depends on interpretation, even if it's easier to agree due to no obvious problems with intuition.

In addition to the equations not making sense, they are consistent with everything else that's known about photons, and there is no way in which such a reference frame makes physical sense. Think of what it means to have an observational frame of reference, and I suspect that anything you think of will not apply to a photon. All lengths contract to zero in the direction of travel, so they can't measure length. All trips would take zero time, so there is no way to measure travel time. A photon's state doesn't change from emission to absorption, ie. it doesn't age, ie. it measures no passage of time. A photon doesn't absorb other photons, so it can't observe anything while traveling. To accelerate to c requires infinite energy, unless you have no mass, but if you have no mass then you have no rest energy, so you can't be at rest and you can't have a rest frame. All these things fit together, and there's nothing that describes a photon being able to observe. The math models physical reality, and the math doesn't work for a photon's frame of reference, while there is also no such thing known or theoretical that makes sense to model.

I think I'm starting to get it. And you say this works if the separation is in any direction, it doesn't matter if it's in the direction of V? Say, as per OP, a separation only in the y-axis, that's the same as a separation only in the x-axis? Or in other words, your math with one spatial dimension works the same as with 2 or 3, and it doesn't matter how the separation vector and the velocity vector are oriented relative to the other?

Is the x-axis separation of the photons equal to 0 for all observers, including those whose velocity relative to the S frame is not in the same direction as V? Yes, if your equations are complete then I'm wrong. But don't your equations also hold that [math]\gamma(V)t_0=0[/math] in the case that x_0 = 0 and t_0 = 0? Which implies that if the photon emissions are simultaneous in frame S they're simultaneous in all frames?

The latter, because I'm having trouble following your math. I suspect that you're answering a different question but I might be wrong. If there is a y-axis separation, and x_0 = 0, and letting t_0 = 0 (the photons are emitted simultaneously in S frame), then the x separation is always 0 in your formula. Is that true for all observers, even ones that aren't limited to motion in the x direction? I think this is relevant, because it is fairly common for people to claim that there is an absolute frame of reference by ignoring any frames where their premises don't hold. I believe that the premise implied by OP that the x-separation is 0 in all frames is wrong, but your math seems to support it.

But what if the separation in space is not in the direction of V, such as in OP's setup?

In post #1 the photons are described as moving parallel to each other. The only separation is along the y-axis, not in the direction of travel. Good for you for using math, I admit once again that I'm deficient. However the math is useless if it describes something entirely different from the prose. If you wish to continue discussing your variation instead of OP's, why not start a new thread?

Do the photons have a comoving frame? Delta1212 understands what I was getting at. I was speaking only of OP's premise that the photons are "always" side-by-side. I agree it's not an essential part of the answer. If OP's thought experiment requires the photons to be separated and side-by-side in all frames, then it's a problem, but the experiment can be restated without that requirement.

Are the photons emitted simultaneously in all frames? If no, are they side-by-side in all frames? If no, is OP's premise in the thought experiment flawed? Isn't that relevant? Do photons have a comoving frame? Edit: I guess you could put the photons together and thus have them emitted simultaneously in every frame, and still come to the same conclusions as OP did if you try to consider an invalid frame in which the photons' relative speed is 0. So RoS isn't a necessary part of the answer and it's possible to have the photons stay together in every frame. I guess this would only come into play if you used the distance between the photons in a calculation that derives a contradiction if absolute simultaneity is assumed, which I don't think has been done here, so you're right that RoS need not be considered. That's a good point. No matter how distorted the bar is by relativistic effects, it remains "together" in any frame. OP could use that if "the photons remain side-by-side" was replaced with "the photons remain close to each other." However the bar is only an analogy, since the bar has a rest frame but photons don't. (The bar ends don't have an invariant "single speed" but the photons do.)

No, I'm not. You're right that they have the same speed, but if they're not emitted simultaneously then they won't be "several photons moving parallel to each other, side-by-side, always staying abreast to each other," as per OP. Unless their separation is negligible, they can't be emitted simultaneously in all frames. This is a correct application of RoS. For example, if there is separation and you allow observers arbitrarily close to the speed of light, you can shorten the travel time of light arbitrarily (due to length contraction), and contrive an observer for whom one photon reaches its destination before the other photon is emitted. In this case the photons don't travel together in any way.

Because they are only together in some reference frames. If they are "side-by-side" implying separated by some distance and emitted simultaneously, you must realize that in some other frames they are not emitted simultaneously and thus not "together" in all frames. If they are not separated by any distance then the argument is vacuous. Otherwise, you have to falsely assume absolute simultaneity in order to conclude that they're "together" for all observers.

Here's how I imagine the internets becoming intelligent on the order of hundreds of years. Already, Google is using AI to solve problems using solutions that humans don't even understand. http://www.cringely.com/2014/04/15/big-data-new-artificial-intelligence/ I wouldn't call that intelligent yet. I'm sure it would be possible to analyze the computer's behavior and break everything down into an (unintelligent) algorithms. But computers currently do things that humans don't yet understand. Say we apply this to networks, and have some part of the internet set up to configure its own layout and connections efficiently. It would use AI to create better designs than humans could directly program it to do. Now say you have several different instances of this running, and some perform better than others, and humans replicate the better systems elsewhere. Now we have an element of "survival of the fittest". Suppose after some decades the systems are complicated enough that the selection and deployment is automated and also done by some AI. For example say that you have some self-configured subnetworks that become unstable or fail from time to time, and some other that is more resilient, and thus performs better. Whether by human or machine choice, the stable system might be reproduced. If by random coincidence a system favors self-preservation, it might end up being better and so preferentially selected. This can happen even if the humans never programmed self-preservation as a goal, if it's only a side effect of the stated goal of "efficiency". Now you have machines selecting for self-preservation and it's a short mental leap to Skynet. Fast-forward a hundred years. With more AI-designed systems in use in all sorts of applications, on multiple layers (eg. a vehicle's partially evolved AI is connecting with evolved wide area networks, and the interaction between them is also being figured out by computers), the collection of systems is becoming more complex than humans want to understand, and we're letting the computers do more on their own while we're concerned only with the results, not the design. (Of course we'd try to build-in safety to prevent the computers taking over and killing us, so let's just suppose that we end up doing that adequately.) As well, the networks could become immensely more dense, say with smart materials and stuff. Imagine for example every thread in a shirt being able to network information, and processing and sensors distributed throughout a material, processing stuff that I couldn't even conceive of now. Then we have a many orders of magnitude bigger network, evolving itself to process information more efficiently, evolving goals that humans never gave it and maybe couldn't even relate to or understand. If at some point the whole distributed system had a number of connections comparable to the human brain, and/or iterated on itself a large number of times comparable to the number of mutations and changes involved in evolution and growth of the human brain, it's possible that some form of consciousness might evolve. For sure evolved intelligence (without consciousness) would be useful. It might even be possible that some other form of thought that may or may not be considered "consciousness" evolves, but is just as advanced or more, perhaps something we will never imagine until it happens. And if a computer is able to evolve and replicate itself (eg. the Singularity), it could get always get more sophisticated. If after 100 years and a network covering Earth with more processing power than all the human brains combined is not conscious, then what about a thousand years later or several trillion times more processing? Finally, if the internet remains a part of the connection to whatever might develop from technological innovation, it could be part of the creation of a consciousness.

We're getting off topic here, but "proof by example" is meaningless. Your examples are showing your lack of experience. There are better examples that are a lot closer to real intelligence, in games as well as other areas.

Your entire arguments are basically "I assume everything is this way and I can't imagine anything else." Computer game AI isn't all yes/no. Biology can sometimes accept programming. I've seen attempts to mimic human intelligence in games by programming some sort of human-like decision process, and it usually makes the characters look dum because they inevitably encounter some situation they haven't been programmed to "think" their way out of, and then it's clear that they have no brain. Some of the best AI results I've seen (in my limited experience) involve programming very simple actions, but which produce complex behaviors. Ants are an excellent analogy. There are several different simple behaviors that ants have, and each ant may be acting out some behavior in a thoughtless "programmed" way, but the behavior of a colony together can seem intelligent. This is how I prefer game AI: simple instructions with complex results, as if the behavior of a character or group has an emergent intelligence. The human brain is made of cells, and evolved from lower life forms that at some point in the past must not have been "intelligent". It's conceivable that intelligence can emerge from the interaction of many individually unintelligent parts. In this way, a network might be able to emerge intelligence if it is able to self-evolve or produce complex behaviors from the interaction of its parts, which might not necessarily be programmed in. Computer intelligence that you know of is a clever illusion, but that doesn't mean other things are too (like neural networks and machine learning systems), and it also doesn't mean that "real intelligence" is not also a clever illusion. What's possible isn't limited to what's already happened (in the past one might have argued "flight requires organic biology", "it's impossible for a machine to beat a human at chess", etc), and it certainly shouldn't limit imagination. Do you consider something like "calculate direction to move a game object" a yes/no decision?

Thanks imatfaal, I think this should clear up a lot of issues. However I think something needs to be decided about the interaction of the test masses. Whether they're allowed to intersect or stick together gravitationally should make a difference. For example, here is a situation that approaches swansont's conditions while keeping xyzt's condition that the masses can't intersect. Place Earth at the origin (0,0). Place the heavier test mass at (x,r), and the lighter test mass at (x,-r). According to swansont's solution, the motion of the test masses in the y direction is irrelevant, and their x position should remain the same. They should remain at a fixed offset relative to each other, gravitationally stuck together. Can we not conclude then that the Earth must end up with a positive y-value, and thus make contact with the heavier mass first? For example if one test mass has a mass equal to Earth's while the other's is near zero, Earth will end up around (x/2,r/2). What assumptions (other than allowing the test masses to intersect) could change this result? Or how could swansont's solution work without some different assumption? If they're allowed to intersect they can be placed on the x-axis with no interaction (shell theorem), and certainly impact Earth simultaneously.

No, you're wrong. It's true that swansont and xyzt are speaking of different cases that rule out the possibility of the other's. If you choose to use your reasoning alone, then you must also factor in (among other possible complications) that the lighter test mass has a different distance to travel to the center of gravitation of the COMBINED mass, than the initial distance to Earth. For example, with theta=180 degrees, the "combined mass" shrinks in size over time as its two components attract, increasing the total distance the lighter mass needs to fall. There are cases where they hit at the same time, and cases where the heavier mass hits first, and it should be possible to contrive cases where the lighter mass hits first.

If the masses are the same size and at the same location they will hit at the same time.

With 3 bodies in general the bodies don't accelerate toward the center of mass. The nearer of two bodies will have a greater pull per unit mass. I don't follow your line of reasoning here. I never said that the "mass in a shell" masses would hit the Earth at the same time. It was only mentioned to correct your statement: So I'm not sure what you're correcting, but sure, I'll accept your correction anyway. Thanks. Do you accept my correction too?

Yes, exactly. As swansont said, as the Rabinowitz paper points out, the answer depends on how the question is stated and how the experiment is set up. I hope we can all agree on that. Yes, in this case the larger object hits first, not the more massive object. The point is that they have the same acceleration with respect to Earth, and that it is physically reasonable to allow the test masses to be close together without interacting gravitationally.

You can't go over a bridge while in your car. You have to get out and fall side-by-side (see post #51). IF they are initially separated, "they would separate (radially)" does not refer to separation from each other, but an increase in the difference of their respective distances to Earth. The initial "radial separation" is 0. The increasing "radial separation" comes from the fact that the Earth accelerates slightly more toward the heavier of the falling masses (and then it might also give the heavier mass a faster acceleration because it's closer). Note that if the falling objects are initially separated and equidistant from Earth, the lighter one is initially farther from the center of mass of the system. I think that xyzt's right about what would happen in that set up, but wrong about it being the only possibility (I don't see why you can't fall inside a car while being separate from the car). I think swansont's right when the separation of the falling objects is negligible. To be fair, the original question didn't say negligible separation or side-by-side, but with impossibly great separation, "far away from each other so that the marble sized object with greater mass did not have a gravitational effect on the lesser mass marble sized object". However that doesn't apply to every case discussed in this thread. Edit: Actually, I think that the only way to satisfy the conditions of the original question would be if the objects could occupy the same space, and were exactly co-located.

No net attractive force. It's called the shell theorem. http://en.wikipedia.org/wiki/Shell_theorem It's true that you can create a different setup that gives a different answer than other setups. You can also set it up to avoid those extra complications, and have two masses centered at the same place without a net attractive force between them. As mentioned, the paper explains this. If you have more questions, have a go at trying to read the paper.

I realize you don't need to read the paper to know that everyone else is wrong, but try anyway: http://arxiv.org/ftp/physics/papers/0702/0702155.pdf Your assertion that they must interact gravitationally is incorrect. Edit: Fixed link? (Yes, Mordred it's the Rabinowitz paper, I'd copied the link from your post too, haha)

It's not just a simplification but a specification. It depends on what you're asking about. If you only want to know about the gravitational acceleration then it's reasonable. If you want to know how a system will behave in general then you need to use the more complicated setup. However, more complicated won't necessarily answer the question you're asking. If you have a general 3-body system, then you will have lateral acceleration, which means you have to include inertial pseudo forces, right? So xyzt's complicated setup is still not complicated enough. For example, if we have 3 point masses, Tiny, Large, and Huge, with Tiny and Large very near so that tiny accelerates mainly toward Large (instead of to Huge), then... doesn't the acceleration direction change over time due to the changing location of the other body, so Tiny would end up orbiting Large? And then couldn't you adjust things, eg. place Huge closer or farther away so that the time of minimal distance between the bodies (ie. "impact") can happen with Tiny at a different point of its orbit? In other words, in general the question isn't answerable and depends on the specifics. So the complicated answer is not useful unless you're considering a specific case, and you set up the specific case by choice to avoid such complications. I might be wrong.

Sounds right, I see no fault. It's easier if you consider two arbitrary masses. Yes, from the outside perspective the mouse has a higher speed on impact but only relative to the outside observer. The closing speed of the two masses is greater for the elephant. By analogy, imagine accelerating a car into a parked truck, vs. letting the truck accelerate toward the car from the same spot, starting at the same time. The speed of the car relative to the ground will be greater if the truck is parked, but the relative impact speed will be greater if the truck also accelerates. Also... we're talking "point mouse" and "point elephant" here, right? Earth's gravitational gradient over the height of an elephant would surely be more significant than the gravitational pull of the elephant. Also also... just to avoid confusion this is the case where the elephant and the mouse are dropped separately, not simultaneously as swansont and xyzt are discussing...

Skimming the thread I didn't see this last question answered. Yes, the elephant falls faster because the Earth also falls toward the elephant. But don't forget the caveat "If you measure the rate of fall assuming the the Earth is stationary", in which case the answer is No. They fall at the same rate. This makes more sense in context of the other thread where the falling mass is of the same magnitude as Earth's, where it more significant. Consider two large masses, and you're keeping one of them stationary, so that it is resisting the gravitational attraction of the other. What is the force that would be required to keep it stationary? (Ans: it's equal and opposite of the fictitious force of the other mass's gravitational pull on the stationary object.) Suppose that you kept the one mass stationary using rockets. What is the acceleration provided by those rockets? (Ans: Equal and opposite of the gravitational acceleration toward the other mass.) If you keep one mass stationary, and sum up the accelerations, you find that the other mass indeed accelerates at a rate independent of its own mass (though the force involved in keeping the first mass stationary depends on it). In your example, the increased closing acceleration due to the elephant's mass would be negated by the acceleration provided by the rockets (or the nail on which Earth is pinned, or the acceleration of your frame of reference,* or whatever means you use to keep Earth stationary). * Err... nah... I guess you'd have to physically counteract the pull of gravity to negate the increased closing acceleration. But if you can keep Earth stationary by calling the pull on it "negligible", then the increased closing acceleration is negligible.