So, its not velocity that causes time dillation, thats simply a convenient way of reffering to it.
Its the difference in actual space traveled during the interval compared to going in a geodesic, or straight line? which, honestly, is a sum of the velocities of the trip of the non geodesic object, yes? (damn, i think i reconfused myself) Thank you very very much btw for taking the time on this Stephen! On Mon, Jun 8, 2009 at 9:23 AM, Stephen A. Lawrence<sa...@pobox.com> wrote: > > > 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. > >
