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? 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) 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? On Mon, Jun 8, 2009 at 7:39 AM, Stephen A. Lawrence<sa...@pobox.com> wrote: > OK here goes. Response below is to Michael's original message and to > Leaking's response. > > The reasponse to Leaking is lengthy; the response to Michael comes 'way > down at the end, after it. > > leaking pen wrote: >> Since the magnetic field is em radiation of a sort, > > A magnetic field is a magnetic field, ça c'est tout. EM radiation is a > wave in the field. As such they're different. Sound waves are not air, > even though they travel in air. > >> think of it like >> the classic spaceship with a flashlight scenario (which is the ONLY >> thing i have EVER found in physics that i still cannot wrap my mind >> against. I understand what it is saying, my brain just refuses to >> accept it as accurate) >> >> if your on a spaceship going .9 c, and you turn on your headlamps, the >> light will go forward at, to your appearence, c away from you, as if >> you were standing still. Now, someone on the spacestation you're >> passing would see you moving at .9 c, and the light moving at c, not >> at c away from you PLUS your velocity, but simply c away from you, but >> c from their perspective. >> >> now, this means you each see the light reaching different distances at >> the same time, which is where my mind rebels. > > No, on two counts. > > First, you've left out Fitzgerald contraction; the traveler on the > spaceship sees the space station as being squished along the line of > travel. The observers on the space station, OTOH, see the traveler's > spaceship as being squished along the line of travel. (Symmetric, of > course.) So, distance measures in the two frames of reference are > wildly confused to start with, and trying to ask when something reaches > some *distance* is going to result in confusion. Ask, rather, when it > reaches a particular *point*. When we talk about a particular point in > space and time, we call it an "event". So, instead of asking about > distance, let's drop a space beacon into the picture, and say the light > hits the beacon, and let's ask about when and where that happens, rather > than asking about how far the light has gone. > > Second, you've assumed "at the same time" means something, but when > you're discussing two different frames of reference moving at > relativistic speeds, it does *not*. The problem is not just time > dilation, it's clock skew, and failure to ... er ... "grok" clock skew > is the single biggest problem people run into in this area. > > The example as you wrote it is, of course, very fuzzy; it will take a > lot more words to make it precise. To make it into something you can > test (in a gedanken sense) we need to sharpen up the details. We've > already started to do that by adding a beacon; we'll continue with the > necessary sharpening now. > > You seem to have said the headlights are turned on at the moment when > the ship passes the station. OK, let's take that as the origin, in both > reference frames: The lights go on at time 0, at which time the ship is > at location 0, and the station is at location 0, in both frames. > > You didn't specify a direction, but let's say that, as seen from the > space station, the ship is moving along the X axis in the "+" direction, > and the headlights, of course, are also shining along the X axis. So, > we can reduce the problem to 1 spacial dimension and 1 time dimension. > > We need to name our coordinates: > > x = spacial location in the space station frame > t = time in the space station frame > x' = spacial location in the spaceship frame > t' = spacial location in the spaceship frame > > Note that the space station is located at x=0 in its own frame of > reference, and the space ship is located at x'=0 in the ship's own frame > of reference, and those coordinates don't change (you're always > stationary relative to yourself!). > > And of course if we set v=0.9, then the spaceship is moving at velocity > +v=0.9, as seen from the station, and the station is moving at velocity > -v=-0.9, as seen from the ship. > > In the ship's frame, the leading wave front of the light moves along the > X axis at C. At some moment it strikes the beacon we dropped into the > picture. Let's assume there is an observer named "O'" in the > spaceship's frame -- which means, O' is an observer who is traveling in > tandem with the spaceship, who is *stationary* relative to the > spaceship, and who has a clock which is synchronized to the spaceship's > clock, as can be confirmed by use of telescopes by O' and by the folks > on the ship. Assume O' is at (fixed) distance X1' from the ship. Let's > also assume that O' happens to be next to the beacon (passing by) when > the light arrives. We use the reading on the clock of O' to determine > what time the beam hits the beacon in the ship's frame. Call that time > T1'. At that moment, the beacon is observed by O' to be distance X1' > from the ship. > > Similarly, there is an observer named "O" in the station's frame; "O" is > stationary relative to the space station, is at fixed distance "X1" from > the station, and has a clock which is synched to the space station > clock. And "O" also just happens to be passing the space beacon at the > moment when the light arrives! At that moment, "O" checks the time, and > sees that it's "T1", and checks his location, and sees that it's > distance "X1" from the space station. > > Let's start in the station's frame (the unprimed frame). If the beam > moves at C, and it started at time 0 at location 0 (as we assumed) then > we must have > > X1 = C * T1 > > Now I can't do this without making mistakes unless we introduce one > additional simplification: Let's set C=1. (Measure time in seconds, > and distance in light-seconds, then naturally C=1.) Then we have > > X1 = T1 > > Now we need the Lorentz transforms. To change coords from the station's > frame to the ship's frame we have > > g = 1/sqrt(1-v^2) > > x' = (x - v*t)*g > t' = (t - v*x)*g > > And to change back we have > > x = (x' + v*t')*g > t = (t' + v*x')*g > > Now right here we can see something interesting, if we stop and look for > it: At time 0 in the ship's frame, as we move to the right in the > ship's frame -- look at increasing values of x' -- we see that time, in > the station's frame, is moving AHEAD. At time t'=0 in the ship's frame, > time in the station frame (as observed from the ship's frame) is given > by t = v*g*x'. That means that, as the ship charges ahead in space, > it's ALSO charging ahead in *time*. Think about that a bit -- the ship > is not just moving in space, it's actually *shortcutting* through space > *and* time. And so it arrives, not just at a distant point in space, > but at a future point in time, relative to its own clock. And that's > how come the pilot ends up younger than expected, if we look at the > clock in the station's frame when he arrives. This argument can be > carried farther but not just now. > > But back to the main event -- we have the beam hitting the beacon at > time T1, and at location X1=T1, as seen by observer O. Let's transform > those coords to the ship's frame (keeping in mind that X1=T1): > > X1' = (T1 - v*T1)*g = T1*g*(1 - v) > > T1' = (T1 - v*T1)*g = T1*g*(1 - v) > > And we see that T1' = X1', so the beam moved at C in the ship's frame, > as well. > > No contradiction. > > But here's the main point: The clock of "O" and the clock of "O'" do > *NOT* read the same value at the moment when they pass each other and > pass the beacon and the light arrives at the beacon: > > T1' = T1 * g * (1 - v) = T1 * sqrt((1-v)/(1+v)) < T1 > > And the distance it traveled is also different in the two frames: > > X1' = X1 * g * (1 - v) = X1 * sqrt((1-v)/(1+v)) < X1 > > The distance the light traveled, and the time it took, were both > *smaller* in the ship's frame. We can say this, as a sort of English > semi-explanation: The distance traveled was smaller, because the ship > was "chasing" the light, so it didn't go as far, as seen from the ship. > But the time elapsed was smaller, too, because of so-called "time > dilation", and the two effects exactly canceled, leaving the light > traveling at C in both frames. > > I doubt this will help, 'cause the effects are fundamentally > counterintuitive, but it might ;-) > > Now, comment on Michael's post follows, below. > > >> >> (If i have this incorrect, someone PLEASE correct me, as it hurts my head...) >> >> On Sun, Jun 7, 2009 at 8:47 PM, Michael Crosiar<crosia...@yahoo.com> wrote: >>> Hello vortexians, >>> >>> Before I begin, I want to thank all of you. I have been lurking here for >>> years. I have seen the trolls come and go. They amuse for a while, then they >>> get old. But those of you who are of a true vortexian spirit always find new >>> and exciting food for the mind to try out. I don't have the math or science >>> background that you have, and yes, I am jealous. But obviously I do have the >>> interest or I would have gone away a long time ago. I don't post much, guess >>> I'm afraid I'll get shot down - and I know I wouldn't have had the time to >>> follow and respond to my own threads - and that would suck for all of us. >>> But circumstances change and I suddenly find I have much more time than I >>> would like. I've grown a little older and am not so scared to raise my hand >>> in class. So agian, thank you for sharing and thank you for putting up with >>> my incessant lurking :) >>> >>> And if I go astray, please let me know, I have gained a deep respect for all >>> of you. I will not be offended. >>> >>> I have a simple thought experiment I would like your comments on. >>> >>> We create a torroidal magnetic field and rotate it > > Whoops your gedanken just jumped the tracks. You *can't* rotate a field. > > You can rotate an object. You can rotate a frame of reference. You can > rotate your head trying to follow an obscure argument. But you can't > rotate a field, nor move it, nor do anything else which requires > "pushing" on it because you can't get a grip on it. > > A so-called "traveling" field, like EM radiation, is a field which > disappears at one point and appears at another point. If you look at > the equations, and think about what they're telling you, that turns out > to be the only thing they're saying -- there is nothing in there about > the fields actually "moving", nor about the "velocity" of the field > itself. The things that move are the maxima, minima, nodes, and > accompanying values. The field itself just kind of "is". > > But in your gedanken, you're rotating *something*, even if it's not the > field. So, let's back up and ask exactly what you *are* rotating. > Rotation is physical, so you must rotate a physical thing. So, what > physical thing are you thinking of rotating? > > Spell that part out and we can take it from there, and the whole thing > will make much more sense, I'm sure. > > > >>> at relativistic >>> velosities, such that the inside of the torroid would be rotating at near >>> the speed of light. The outside of the field would extand outwards and would >>> have an agular velocity that would be greater, proportional to the increase >>> in circumference. First, is that correct? Clearly nothing can go faster than >>> the speed of light, but as we increase the speed of the rotation, the energy >>> must go somewhere, yes? Would this cause the mass of the field to change? In >>> other words, would it bend space-time inside the field? And could the >>> curvature be negative or positive depending on the direction of rotation >>> relative to the N/S pole? Would time run at a different rate inside the >>> field versus outside the field? If we were to place a radioactive isotope >>> inside the field, could we cause it to decay faster or slower? >>> >>> I'll be anxiously awaiting your insights, >>> >>> C. Michael Crosiar >>> >>> >> > >
