According to Einstein, if you could ride a beam of light and look at your surroundings it would seem as if everything around you has stopped. Even time. If this is the case how come there is a time delay from the suns rays reaching earth? 8 minutes I believe. From the photons point of view time on earth has stopped from the time the photon left the sun and therefore from OUR perspective it would seem as if the light had reached earth instantly. --RedSingularity
Actually, Einstein realized that wondering what it would be like to ride a beam of light is meaningless because it's impossible.
Because you're not riding the beam of light. Observers in different reference frames see different things. From the photon's point of view, time has stopped and therefore it gets to earth instantly. From our point of view, obviously, it takes 8 minutes.
FAQ: What does the world look like in a frame of reference moving at the speed of light? This question has a long and honorable history. As a young student, Einstein tried to imagine what an electromagnetic wave would look like from the point of view of a motorcyclist riding alongside it. But we now know, thanks to Einstein himself, that it really doesn't make sense to talk about such observers. The most straightforward argument is based on the positivist idea that concepts only mean something if you can define how to measure them operationally. If we accept this philosophical stance (which is by no means compatible with every concept we ever discuss in physics), then we need to be able to physically realize this frame in terms of an observer and measuring devices. But we can't. It would take an infinite amount of energy to accelerate Einstein and his motorcycle to the speed of light. Since arguments from positivism can often kill off perfectly interesting and reasonable concepts, we might ask whether there are other reasons not to allow such frames. There are. One of the most basic geometrical ideas is intersection. In relativity, we expect that even if different observers disagree about many things, they agree about intersections of world-lines. Either the particles collided or they didn't. The arrow either hit the bull's-eye or it didn't. So although general relativity is far more permissive than Newtonian mechanics about changes of coordinates, there is a restriction that they should be smooth, one-to-one functions. If there was something like a Lorentz transformation for v=c, it wouldn't be one-to-one, so it wouldn't be mathematically compatible with the structure of relativity. (An easy way to see that it can't be one-to-one is that the length contraction would reduce a finite distance to a point.) What if a system of interacting, massless particles was conscious, and could make observations? The argument given in the preceding paragraph proves that this isn't possible, but let's be more explicit. There are two possibilities. The velocity V of the system's center of mass either moves at c, or it doesn't. If V=c, then all the particles are moving along parallel lines, and therefore they aren't interacting, can't perform computations, and can't be conscious. (This is also consistent with the fact that the proper time s of a particle moving at c is constant, ds=0.) If V is less than c, then the observer's frame of reference isn't moving at c. Either way, we don't get an observer moving at c.
Lets say that i had a "beam rifle". This rifle is equipped with a high magnification optical scope. Now i aim this rifle at someone standing 20 light seconds away and pull the trigger. Will the person be able to walk out of the way of that beam of light before he/she is hit with it? According to russ_watters from the photons point of view it reaches the person instantly and therefore the person will not be able to move out of the way from the time the trigger is pulled. --RedSingularity
So then only a human/observer has the right to a point of view. In this case the shooter and the target. Does that mean that the target would have time to move out of the beams path from the time the trigger is pulled? (20 seconds to be exact) --RedSingularity
russ_watters was being a bit misleading, a photon does not literally have its own inertial frame to define its own "point of view", see bcrowell's post. You might also take a look at this previous thread on the subject, where in post #21 I pointed out that the closest you can come is to think about what a fast-moving observer sees in the limit as his speed relative to the rest of the world approaches the speed of light. If we imagine your rifle fires a bullet at a speed very close to the speed of light, then it's true that in the bullet's own frame, the distance between you and the person you're firing at is shrunk down to almost zero because of the length contraction effect. It's also true that because of the time dilation effect, in the bullet's frame both your clock and the clock of the guy you're firing at are running very slow, almost stopped (these effects are reciprocal, so in your frame a clock traveling along with the bullet would be almost stopped too, and the length of the bullet would be contracted to nearly zero in the bullet's direction of motion). However, one other effect to keep in mind is the relativity of simultaneity (see this article as well), which says that if two events happened "simultaneously" in one frame (both events have the same time-coordinate in that frame), they did not happen simultaneously in another frame. So, suppose in your frame your clock and the clock of the guy you're firing at are synchronized, and you fire when your clock reads t=0, so in your frame the event of your firing is simultaneous with the event of the other guy's clock also reading t=0 seconds. If the bullet is traveling very close to the speed of light, then in the frame where you and the other guy are at rest it will take just over 20 seconds for the bullet to reach his position, so even if he waits until t=19 seconds on his clock to step aside, he can avoid being hit. But in the frame of the bullet, the event of the bullet being fired from your gun is not simultaneous with the event of the other guy's clock reading t=0 seconds, instead it is simultaneous with the other guy's clock reading just a tiny bit less than 20 seconds! So in the bullet's frame, the other guy has already stepped aside at the moment the bullet is being fired, thus even though it takes close to zero time to reach the position that guy was formerly standing, the guy isn't standing there any more so the bullet won't hit him.
Interesting. Would there be the same effect on a "laser" beam coming from the rifle instead of a "bullet" which has mass? The laser beam would have no mass and therefore would not suffer the same consequences of luminal speeds as an object with mass.
Nope, that isn't what #4 says. See the paragraph beginning with "Since arguments from positivism...," which doesn't say anything about observers. The point of that initial phrase is to explicitly say that the previous paragraph's discussion of observers may not be a reliable guide, so we should carry out reasoning that is independent of the idea of an observer. No, if he has no other source of information than the arrival of the beam on his chest. Yes, if has some other such source of information that the beam will soon arrive in that location.
Which effect for the bullet are you asking about? Again, light doesn't have its own inertial rest frame (it would violate the basic postulates of relativity if it did, since one of the basic postulates says that light moves at c in every inertial frame), so you can't talk about time and distance and simultaneity "from the photon's point of view" like you can for the bullet. But again, you are free to consider what happens from the bullet's point of view in the limit as the bullet's speed approaches c.
Sure, if the distance is 20 light-seconds then in the frame where the shooter and target are at rest it will take 20 seconds for the laser to reach the target, so if the shooter fires at t=0 on his clock, and the target has a clock which is synchronized with the shooter's clock according to the definition of simultaneity in their own rest frame, then the target can step away at any time before t=20 on his clock (say, at t=19 as in my earlier post) and avoid being hit.
Forgive me if I am not catching on right away but the photons in that beam of light are moving at exactly "c" which would mean that everything around those photons has stopped moving including the target.
"The Lorentz transformations tell us that time stands still for an object moving at the speed of light. From the point of view of the photon, of course, it is everything else that is rushing past at the speed of light. And under such extreme conditions, the Lorentz-Fitzgerald contraction reduces the distances between all objects to zero. You can either say that time does not exists for an electromagnetic wave, so that it is everywhere along its path (everywhere in the Universe) at once; or you can say the distance does not exist for an electromagnetic wave, so that it "touches" everything in the Universe at once."
When you say "everything around those photons has stopped moving", are you imagining things in a "frame" where the photon is at rest? If not, do you understand that all claims about how fast different clocks are ticking depend on choosing a coordinate system (i.e. you are discussing how much time elapses on the clock in a short increment of coordinate time), there is no coordinate-independent physical notion of the rate a clock is ticking? Anyway, if we return to the idea of considering the limit as a bullet approaches c, it's true that in this limit, the clock of the shooter and target both approach a rate of 0 ticks per second of coordinate time in the bullet's frame. But it's also true that in this limit, the time on the target's clock when the gun is fired approaches a reading of 20 seconds, so if the target stepped aside when his clock read 19 seconds, he has already stepped aside when the bullet is fired. Thus it if we understand that we are really talking about limits rather than an actual rest frame for the photon, it is approximately correct to say that "from the photon's point of view", both the shooter and the target are frozen, but while the shooter's clock is frozen at 0 seconds, the target's clock is frozen at a reading of 20 seconds at a position that's out of the line of fire of the gun.
JesseM's suggestion to consider a bullet going at just less than c is a good one. It answers what you are essentially asking (a question about simultaneity) without getting into non-physical nonsense like a photon's frame. If you absolutely insist on talking about a photon's perspective then you need to provide the coordinate transformations from an inertial frame to the photon's frame.
When you're going to quote something like this, you might want to give the source. It seems to be Schrodinger's Kittens, by John Gribbin. I was able to look at the relevant pages of the book using Amazon's Look Inside feature. The relevant things to note from the context are: -On p. 77, he explains (in an indented block of text) the standard point of view -- that there is no such frame of reference. -Your quote, from p. 79, is taken from a context on pp. 79-80 that makes it clear that he admits he is presenting a nonstandard view. He feels that other physicists haven't analyzed this deeply enough, and he believes that his own point of view on this explains all kinds of otherwise inexplicable mysteries about quantum mechanics.
Who is the author of this quote? From a google books search it may be John Gribbin in the book 'Schroedinger's Kittens', a nontechnical book for a popular audience. Like I said, in some approximate sense you can talk about the "point of view of a photon" in terms of what would be seen in the limit as a slower-than-light inertial observer approached a speed of c relative to his surroundings, but this isn't really rigorous and I don't think you'll find any textbooks or scientific papers talking about the "point of view" of a photon. edit: beaten to it by bcrowell!