leaking pen wrote: > So... > > I think i followed all the math on that, very simple math, thank you! > > and, my original didnt start with lights being turned on with the ship > passing the station, but that DOES simplify things. thanks! > > So... What your saying is that if you take into account time dilation, > the light DOES really move the same distance in a set amount of time, > once converted to local time, relative to both. so really, the light > ISN'T traveling at c faster than the ship, it just APPEARS that way to > O' due to time dillation?
You can view it that way, but it's a little hazardous, because time dilation isn't really just a simple number. Thinking of it as a simple ratio leads to a lot of confusion. Time dilation, expressed as a number, is dt/dtau for a particular observer, "A", relative to a particular reference frame, "F". The "dt" value is found by A, by looking at clocks which are stationary in frame F, as A passes them by. The "dtau" value is found by "A" by looking at A's own clock. Note well: "A" uses ONE clock in his/her own frame. "A" uses AT LEAST TWO CLOCKS in frame "F", located at *different* points in frame "F". You can't measure time dilation between two inertial frames without using at least two clocks in one of the frames, because once the observer has passed a clock, it's gone, and they can't see it any more (except at a distance and using a telescope adds unnecessary hair without changing the result). Thus, time dilation actually measures the rate at which time passes along a *particular* *path*. Something that measures a rate of change along a path is a directional derivative, or a "1-form". It's not a simple number. > That makes more sense. But then, that just reinforces to me something > that I feel, and that I've been told is not true. It just seems to me > there should be then a central point, with a central time flow, and > all other things are variants of that, based on their velocity > relative to this fixed point. (center of the universe, if you will) There may be but there doesn't have to be. As far as I know nobody knows for sure if there is. > I mean, if you were to leave a sattelite in space, not orbiting, but > left behind in our orbit, moving just enough so that we come back to > it in the same spot, relative to earth, next year, more time will have > gone by, becuase its not moving as fast, not orbiting round the sun, > yes? Where does it end? what is the most non moving spot? No, the difference is not because the Earth is moving faster. First, let's agree to ignore the Sun's gravity because paying attention to it would throw us into GR. Let's assume the Earth is just tied to a string or something to keep it in orbit. Now, with that assumption, here's the difference: The satellite we dropped is in an inertial frame -- it's not accelerating. The Earth's frame, on the other hand, is not inertial -- it's accelerating the whole time, due to the pull on that string. To deal with acceleration, we don't need GR but we do need some differential geometry and I'm not going to try to write that out in flat ASCII here (and besides I'm too rusty). In simple terms, the distance along any path you might follow through (4-dimensional) space time is called the "interval", and for a particular observer (like the Earth) the "interval" is equal to the elapsed proper time of that observer. So, how far you go, measured as "interval", corresponds exactly to how many seconds pass on your wristwatch. The square of the "interval" between any two fixed points in an inertial frame is, by definition, (ignoring the Y and Z directions) delta_S^2 = delta_T^2 - delta_X^2 It's not hard to use the Lorentz transforms to show that, for an inertial observer in motion with regard to an inertial frame, that definition of "interval" gives us the square of the observer's elapsed proper time between any two events in the frame. (Not hard but I'm not going to do it right here.) It's also not hard to show that the "interval" between any two events is the same, no matter what inertial frame you use to evaluate it. The infinitesimal "interval" traveled by an astronaut "A", from the point of view of an observer "O", is dS^2 = dt^2 - dx^2 and since it's infinitesimal we can use that formula for an astronaut who is *accelerating*. At the infinitesimal scale, where A's velocity hardly varies, we can find the infinitesimal change in A's proper time -- which is to say, how much A's clock will advance by -- from that formula. Then, to find the *total* time A's clock will change on any path, we just integrate it along the path. This is fundamental; it's the definition of the "metric" in special relativity. It's sometimes referred to as the "Minkowksi metric", in reference to Minkowski's development of the 4-dimensional view of relativity, and it's sometimes referred to as the "Lorentz metric", in reference to the fact that it applies to reference frames which are related by the Lorentz transforms. I'm not leading up to taking some horrible integral here. I'm leading up to something else: A "geodesic" is a path followed by an *inertial* observer -- it involves no acceleration. It's simple, straight-line motion. And it can also be shown that, if we look at all possible small variations from geodesic, the path along the geodesic MAXIMIZES the elapsed interval. In other words, if you wander off the geodesic while traveling from event 1 to event 2, your total elapsed proper time (= elapsed interval) will be *smaller* than it is for someone who follows a geodesic between those two events. (This gets into variational calculus and I'm not going to prove it here.) The point is, the Earth in the example is *not* following a geodesic. It's accelerating the whole time. Its non-geodesic motion, then, accounts for the fact that less time elapses for the Earth than for the "stationary" satellite. This is a general principle; if two observers follow different paths between two events -- in other words, they meet *twice* -- at least one of them must have followed a non-geodesic path. So, if one of them followed a geodesic, the *other* one will have seen less time elapse. And so is resolved the "twin paradox" in its usual form, without any reference to a universal rest frame. ****************************** Now this is all fine if the universe is open. Suppose it's not; suppose it's actually a big torus (or something similar). Suppose that, if you went far enough straight up, you'd eventually find yourself coming back up through the floor, without ever having changed direction. In that case, you would have followed a geodesic, *AND* the person who stayed behind would have followed a geodesic, and yet your clocks still wouldn't match. By comparing them, in that case, you could indeed figure out how fast you were going relative to the "universal rest frame" in your universe. It doesn't lead to a contradiction but it most assuredly leads to the breaking of the general "principle of relativity" -- in such a universe there is a distinguished frame of reference and it may not be accurate to assume all physical laws are absolutely identical regardless of velocity.
