Question about relativity
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From: xxein on 15 Jul 2010 19:24 On Jul 15, 12:36 am, Jonathan Doolin <good4us...(a)gmail.com> wrote: > > It does. But *in the ship's frame*, the rest distance of 1 light year > > can be measured (by the ships' instruments) by many different methods > > to be about 3.9 light hours. > > That sounds about right. > > with 99.99999% of the speed of light, gamma = 2236, so 1 year comes > out to be 3.912 hours. > > It might be helpful to consider the apparent superluminal motion, > though. Let us consider for a moment exactly 100 million meters. > > As the earth approaches us at 99.99999% of the speed of light, it > releases some light, which comes at us at exactly 100% of the speed of > light. At the moment that that light reaches us, we will see the > earth as it was then, exactly 100 million meters away. > > However, during the time since the earth released that light, it has > traversed 99.99999% of that distance itself, and in fact, is now > exactly 1 meter away. In effect, this means that by all appearances, > the earth is traveling toward us at about 100 million times the speed > of light. Quite troubling. > > I think if you reason it out, properly, you'll realize that events > around and about the earth will be similarly running in fast-forward. > This superluminal effect should be considered along with the time- > dilation effect, though it overwhelms the time dilation effect in this > situation. (The time-dilation factor being around 2236, while this > effect being on the order of 100 million.) > > I think, in answer to your question, then, someone2, is that the > observer will see the car racing very quickly around the mirror during > the 3.9 hours the track is approaching. xxein: You don't know gamma, do you?
From: someone2 on 15 Jul 2010 19:37 On 15 July, 15:29, dlzc <dl...(a)cox.net> wrote: > Dear someone2: > > On Jul 15, 3:02 am, someone2 <glenn.spig...(a)btinternet.com> wrote: > > > > > > > On 15 July, 05:08,dlzc<dl...(a)cox.net> wrote: > > > On Jul 14, 5:13 pm, someone2 <glenn.spig...(a)btinternet.com> wrote: > > > > On 14 July, 20:56,dlzc<dl...(a)cox.net> wrote: > > > > > On Jul 14, 11:43 am, someone2 <glenn.spig...(a)btinternet.com> wrote:> On 14 July, 19:10,dlzc<dl...(a)cox.net> wrote: > > > > > > > On Jul 14, 9:40 am, someone2 <glenn.spig...(a)btinternet.com> wrote: > > > > > > wrote: > > > > > ... > > > > > > > > > I was thinking that since although the light > > > > > > > > would be travelling away from the observer at > > > > > > > > the speed of light, and thus be 0.75 light years > > > > > > > > in front of the observer after 0.75 of a year > > > > > > > > ... just not the observer's year. > > > > > > > > > (and thus have reached the mirror by the time > > > > > > > > the observer is about 0.75 light years from the > > > > > > > > mirror), > > > > > > > > No. > > > > > > > > > it would only have travelled 0.25 light years > > > > > > > > back towards the observer, after the 1 year. > > > > > > > > At the speed you have indicated, the moving > > > > > > > observer is just some short distance from the > > > > > > > mirror (a few million miles, say) when the > > > > > > > light hits. > > > > > > > Ah this might seem to be a misunderstanding of > > > > > > mine then. I thought that unlike a car on a > > > > > > motorway where if car A was travelling at speed > > > > > > x-y, and car B was travelling at speed x, car B > > > > > > would only seem to pull away from car A at y mph, > > > > > > with light I thought, the idea was that even if > > > > > > the space ship is going at near the speed of > > > > > > light, the fired beam would still appear to travel > > > > > > away from it at the speed of light. > > > > > > It does. But *in the ship's frame*, the > > > > > rest distance of 1 light year can be > > > > > measured (by the ships' instruments) by many > > > > > different methods to be about 3.9 light hours. > > > > > > > Likewise if the observer was travelling towards > > > > > > an incoming light beam at the near the speed of > > > > > > light, the light would only appear to travel > > > > > > towards the observer at the speed of light. > > > > > > Correct. > > > > > > > Are you saying that this is not the case, and > > > > > > that if a beam of light is fired from an object > > > > > > moving at velocity x in the same direction as > > > > > > the light beam, that the light beam will only > > > > > > appear to move away from that object at c-x? > > > > > > No. What I am saying is that "time dilation" and > > > > > "length contraction" are two faces of the same > > > > > coin. Please look at the link I provided, > > > > > at minimum. Or here, if you have the stomach for > > > > > it. > > > > <snip link now broken by Google.Groups> > > > > > > Everyone sets themselves up for "frame jumps", > > > > > since "common sense" says that measurements by > > > > > different observers "must be" the same for > > > > > all frames. It is sometimes close enough where > > > > > our "common sense" is trained, but it is a > > > > > serious mistake at speeds close to c. > > > > > > It is even detectable at the speed of aircraft > > > > > (special relativity), and/or altitudes of a few > > > > > hundred feet (general relativity), if you can > > > > > integrate the effects over time. > > > > > The question is about the frame of reference of > > > > the observer that is approaching the tracke. I'm > > > > not sure where this frame is ever left, so I'm > > > > not clear on where you are suggesting frame > > > > jumping is taking place. I thought you were > > > > saying that the observer approaching the track > > > > would find themselves only a small distance away > > > > from the track as the beam of light they fired > > > > reached the mirror, since you said: > > > > > "At the speed you have indicated, the moving > > > > observer is just some short distance from the > > > > mirror (a few million miles, say) when the > > > > light hits." > > > > ... the "few million miles" in the frame of the > > > track and car, of course. > > > > > Though if they were, then I'm not clear how > > > > the light appeared to them to be leaving them > > > > at the spead of light, since it is only a little > > > > bit ahead of them. > > > > ... because their clock is running slow, and > > > their lengths contracted, they can measure c > > > to be the same friendly constant... > > > > > I am assuming that they will be using the vehicle > > > > going around the track as their clock, > > > > Why would they do that? Even at constant speed, > > > the rate of vehicle motion will be extraordinarily > > > fast when approaching the track, and > > > extraordinarily slow when departing from the other > > > side. Much easier just to carry your own clock with > > > you. Like "heartbeats" if nothing else. > > > > > taking into account how long it takes the light > > > > to reach them at their given position. Or is it > > > > that if they did this, that the light would not > > > > seem to be leaving them at the speed of light, > > > > but only if they used a clock on board their > > > > ship, such that measured by that onboard clock > > > > the journey appears to have taken less time, > > > > Bingo. > > > Ah ok. So if they used the track as the clock, > > and took into account the time they would have > > expected the light to have reached them from > > the track, > > "the time they would have expected" > ... is that rest time, or knowing the the tools of relativity time? > I meant it in the sense that if an outpost sent out a horse rider every hour along a path, and the riders travelled at a set speed, that if one was to be coming down the path towards the outpost a certain speed, one could work out the expected interval between meeting a rider, and still use the meeting of a rider as a clock, though wouldn't consider there to be an hour interval between each when you'd expect to meet each rider. In the same sense that if a picture of a horse was beamed in light each hour, that heading towards the light source beaming the pictures, you could still calculate the interval you'd expect between each picture, but you wouldn't consider the interval to be an hour, since you were moving towards the source. Likewise with the track, and the interval between each lap. > > it wouldn't seem that light was travelling at c. > > Only with a frame jump implied with using measurements from diffferent > frames. > > > It would simply be that the laws of physics > > would propergate slower at high velocity (and > > perhaps under high gravity, I don't know), > > Know that gravity need have nothing to do with this. It creates > orders of magnitude more math, for a correction of fractions of a > percent. > Ok, we can ignore gravity. > > such that the tick of a clock on board the > > ship might synchronise with a lap of the > > vehicle on the track while the ship is > > stationary, but when approacing the track at > > a higher speed it might synchronise with 5 > > laps of the vehicle on the track for example. > > Yes, given the correct closing speed and/or track length / speed. > > > Since the propergation rate of the laws is > > referred to as t (time), > > Laws don't propagate. Laws aren't really laws. Theories describe > sets of observations, including propagation. > I wasn't suggesting that laws themselves propergated, just that they contained a propergation rate referred to as t (time). Though perhaps I shouldn't have referred to t as the propergation rate, and instead of referred to it as an event synchronisation indicator. > > they talk about time going slower or faster > > in particular frames of reference. Have I > > understood this correctly? > > Not quite. I think you have the concept, but the language is loose. > Everything in a frame appears to be self-consistent, not "time going > slower or faster". The differences in length and "time rate" occur > when viewing *another* frame with enough difference in velocity, that > relativity's effects are larger than the error-bars of measurement. > > I'm not sure that I have got the concept. I was thinking at the high velocity that the observed rate of propergation on the ship will have slowed relative to the track. Such that radioactive decay on the ship would take place slower than on the track. To hopefully indicate what I mean, if everything in an alternate universe things propergated twice as fast, then observations within it should still fit with same 'laws of physics'. Since the propergation is measured in terms of synchronisity. The clock which measured a second would turn twice as fast, while the ball dropped twice as fast, so it would still drop at Xm/s in both universes. The variable t would still indicate the synchornisation of events. It seems to me that what you are referring to as 'frames' are areas where the propergation rate is deemed to be uniform. I might not be communicating this very well, do you feel that you are able to understand what I am saying? > > > > > > and therefore the slight distance the light > > > > beam was ahead of them, given the time shown > > > > on the onboard clock, > > > > ... and the contracted distance ... > > > > > would indicate light travelling at c. > > > > Correct. > > > > > Purely from the frame of reference of the > > > > observer that fires the beam of light towards > > > > the track, what will they be expected to see > > > > at the time the light is reflected back to > > > > them (how far will the vechile have appeared > > > > to have gone around the track)? > > > > You said the vehicle traveled 100mph, presumably > > > in its own frame, and it travels really close to > > > exactly one year, so: > > > 100 * 24 * 365.25 (just averaging the year) = > > > 876,600 miles, in just under 2 hours. The > > > illusion of traveling FTL. Outbound (the next > > > light year), it slows to an apparent 0.1 mph so > > > the next 3.9 hours takes it a light year away, > > > and the car travels less than a mile. > > > "The illusion of traveling FTL." > > > For it to be an illusion, wouldn't that mean the > > acceptance that they hadn't really travelled that > > distance in two hours, and that it was simply that > > the laws of physics propergated slower while they > > were at speed? > > No "propagation of laws". > > SHIP... > 3.9 hours "really" passes onboard the ship. Only they travel 3.9 > light hours to pass the mirror. And they measure c as c. > > TRACK... > 1 year passes on the track. They (might) see the distant ship leave > the "1 light year point" about 3.9 hours before the ship blows by > them. The light is 1 year in transit, and the ship just that little > bit longer. And c is measured as c in the track's frame. > > > Else why would they not be considering themselves > > to ahave observed FTL travel. > > Because it involves a frame jump. Measurements made in different > frames when combined "prove" all sorts of thoughtless things. Just > keep track of who measures what. > > By the way, you think your ship is "really fast"? Check this out...http://www.fourmilab.ch/documents/OhMyGodParticle/ > I took a quick look at the link, and it seemed that the particle only appeared to be going near to light speed according to an observer in another frame. An observer on the particle would have considered it to be travelling 14705200000c if I have understood it correctly. The people in the ship measure the distance they are about to travel (to the track) while at rest, then measure with an on board clock how long they took to travel it. Where was the frame jump?
From: BURT on 15 Jul 2010 19:38 On Jul 15, 4:24 pm, xxein <xxxx...(a)gmail.com> wrote: > On Jul 15, 12:36 am, Jonathan Doolin <good4us...(a)gmail.com> wrote: > > > > > > > > It does. But *in the ship's frame*, the rest distance of 1 light year > > > can be measured (by the ships' instruments) by many different methods > > > to be about 3.9 light hours. > > > That sounds about right. > > > with 99.99999% of the speed of light, gamma = 2236, so 1 year comes > > out to be 3.912 hours. > > > It might be helpful to consider the apparent superluminal motion, > > though. Let us consider for a moment exactly 100 million meters. > > > As the earth approaches us at 99.99999% of the speed of light, it > > releases some light, which comes at us at exactly 100% of the speed of > > light. At the moment that that light reaches us, we will see the > > earth as it was then, exactly 100 million meters away. > > > However, during the time since the earth released that light, it has > > traversed 99.99999% of that distance itself, and in fact, is now > > exactly 1 meter away. In effect, this means that by all appearances, > > the earth is traveling toward us at about 100 million times the speed > > of light. Quite troubling. > > > I think if you reason it out, properly, you'll realize that events > > around and about the earth will be similarly running in fast-forward. > > This superluminal effect should be considered along with the time- > > dilation effect, though it overwhelms the time dilation effect in this > > situation. (The time-dilation factor being around 2236, while this > > effect being on the order of 100 million.) > > > I think, in answer to your question, then, someone2, is that the > > observer will see the car racing very quickly around the mirror during > > the 3.9 hours the track is approaching. > > xxein: You don't know gamma, do you?- Hide quoted text - > > - Show quoted text You can say the road is moving instead of the car in Relativity. Mitch Raemsch
From: dlzc on 16 Jul 2010 01:28 Dear someone2: trimming down... On Jul 15, 4:37 pm, someone2 <glenn.spig...(a)btinternet.com> wrote: > On 15 July, 15:29,dlzc<dl...(a)cox.net> wrote: > > On Jul 15, 3:02 am, someone2 <glenn.spig...(a)btinternet.com> wrote: > > > Ah ok. So if they used the track as the clock, > > > and took into account the time they would have > > > expected the light to have reached them from > > > the track, > > > "the time they would have expected" > > ... is that rest time, or knowing the the tools > > of relativity time? > > I meant it in the sense that if an outpost sent > out a horse rider every hour along a path, and > the riders travelled at a set speed, that if one > was to be coming down the path towards the outpost > a certain speed, one could work out the expected > interval between meeting a rider, and still use > the meeting of a rider as a clock, though > wouldn't consider there to be an hour interval > between each when you'd expect to meet each rider. Your expectation is correct, in that: 1) you are moving towards the outpost, and so you will encounter riders in less time than they depart the outpost, and 2) your time is dilated, so you'd not get riders every hour anyway. > In the same sense that if a picture of a horse > was beamed in light each hour, that heading > towards the light source beaming the pictures, > you could still calculate the interval you'd > expect between each picture, but you wouldn't > consider the interval to be an hour, since > you were moving towards the source. *And* time is dilated for you. > Likewise with the track, and the interval > between each lap. > > > > it wouldn't seem that light was travelling at c. > > > Only with a frame jump implied with using > > measurements from diffferent frames. > > > > It would simply be that the laws of physics > > > would propergate slower at high velocity (and > > > perhaps under high gravity, I don't know), > > > Know that gravity need have nothing to do with > > this. It creates orders of magnitude more math, > > for a correction of fractions of a percent. > > Ok, we can ignore gravity. > > > > such that the tick of a clock on board the > > > ship might synchronise with a lap of the > > > vehicle on the track while the ship is > > > stationary, but when approacing the track at > > > a higher speed it might synchronise with 5 > > > laps of the vehicle on the track for example. > > > Yes, given the correct closing speed and/or > > track length / speed. > > > > Since the propergation rate of the laws is > > > referred to as t (time), > > > Laws don't propagate. Laws aren't really laws. > > Theories describe sets of observations, > > including propagation. > > I wasn't suggesting that laws themselves > propergated, just that they contained a > propergation rate referred to as t (time). > Though perhaps I shouldn't have referred to t > as the propergation rate, and instead > of referred to it as an event synchronisation > indicator. Time is a fundamental property of a system, and one way or another, we don't get time "contained in" anything. Time spans all theories... > > > they talk about time going slower or faster > > > in particular frames of reference. Have I > > > understood this correctly? > > > Not quite. I think you have the concept, but > > the language is loose. Everything in a frame > > appears to be self-consistent, not "time going > > slower or faster". The differences in length > > and "time rate" occur when viewing *another* > > frame with enough difference in velocity, that > > relativity's effects are larger than the > > error-bars of measurement. > > I'm not sure that I have got the concept. > > I was thinking at the high velocity that the > observed rate of propergation on the ship will > have slowed relative to the track. Such > that radioactive decay on the ship would take > place slower than on the track. Radioactive decay, chemical clocks, spring clocks, heartbeats, even hard boiling eggs, all will be slowed by the same amount. > To hopefully indicate what I mean, if > everything in an alternate universe things > propergated twice as fast, then observations > within it should still fit with same 'laws of > physics'. Why should that be the case? Adding additional Universes to this only muddies up matters. > Since the propergation is measured in terms > of synchronisity. No, it is not. > The clock which measured a second would turn > twice as fast, while the ball dropped twice > as fast, so it would still drop at Xm/s in both > universes. The variable t would still indicate > the synchornisation of events. Doesn't follow. Why don't you stick to *this* Universe, destroy whatever pre-copnceived notions you have about "synchornisation" and "propergation", and get "Spacetime Physics"? > It seems to me that what you are referring to > as 'frames' are areas where the propergation > rate is deemed to be uniform. They are defined as collections of objects / points that have a constant, uniform, non-accelerated motion. > I might not be communicating this very well, > do you feel that you are able to understand > what I am saying? See, I thought you were asking questions, not selling your preconceived, imagined meanings of words. If I have to learn your language before I can talk to you, this will take too long. Get "Spacetime Physics", and see what the language of relativity is. .... > > > > You said the vehicle traveled 100mph, presumably > > > > in its own frame, and it travels really close to > > > > exactly one year, so: > > > > 100 * 24 * 365.25 (just averaging the year) = > > > > 876,600 miles, in just under 2 hours. Note to self. The speed of light is 186,000 miles per second, not per hour. No observation of apparent FTL is expected. .... > > By the way, you think your ship is "really fast"? Check > > this out... http://www.fourmilab.ch/documents/OhMyGodParticle/ > > I took a quick look at the link, and it seemed that > the particle only appeared to be going near to light > speed according to an observer in another frame. An > observer on the particle would have considered it to > be travelling 14705200000c if I have understood it > correctly. No, this requires a frame jump. Duration in the particle's frame, and distance in the rest frame. > The people in the ship measure the distance they > are about to travel (to the track) while at rest, one frame (similar to the track and car's frame) > then measure with an on board clock how > long they took to travel it. Second frame, one that is separated from the other by *massive* acceleration. > Where was the frame jump? Distance in the rest frame, and duration in the moving frame. If they measure distance while moving, they get the same speed of 99.99999%c, just as you specified, for the trip. They can use parallax, subtended size of the track, intensity of source, and all will yield the distance as just a tad over 3.9 light hours away (as they cross the "rest frame 1 light year away" mark. David A. Smith
From: Bruce Richmond on 16 Jul 2010 03:40
On Jul 15, 7:37 pm, someone2 <glenn.spig...(a)btinternet.com> wrote: > On 15 July, 15:29, dlzc <dl...(a)cox.net> wrote: > > > > > > > Dear someone2: > > > On Jul 15, 3:02 am, someone2 <glenn.spig...(a)btinternet.com> wrote: > > > > On 15 July, 05:08,dlzc<dl...(a)cox.net> wrote: > > > > On Jul 14, 5:13 pm, someone2 <glenn.spig...(a)btinternet.com> wrote: > > > > > On 14 July, 20:56,dlzc<dl...(a)cox.net> wrote: > > > > > > On Jul 14, 11:43 am, someone2 <glenn.spig...(a)btinternet.com> wrote:> On 14 July, 19:10,dlzc<dl...(a)cox.net> wrote: > > > > > > > > On Jul 14, 9:40 am, someone2 <glenn.spig...(a)btinternet.com> wrote: > > > > > > > wrote: > > > > > > ... > > > > > > > > > > I was thinking that since although the light > > > > > > > > > would be travelling away from the observer at > > > > > > > > > the speed of light, and thus be 0.75 light years > > > > > > > > > in front of the observer after 0.75 of a year > > > > > > > > > ... just not the observer's year. > > > > > > > > > > (and thus have reached the mirror by the time > > > > > > > > > the observer is about 0.75 light years from the > > > > > > > > > mirror), > > > > > > > > > No. > > > > > > > > > > it would only have travelled 0.25 light years > > > > > > > > > back towards the observer, after the 1 year. > > > > > > > > > At the speed you have indicated, the moving > > > > > > > > observer is just some short distance from the > > > > > > > > mirror (a few million miles, say) when the > > > > > > > > light hits. > > > > > > > > Ah this might seem to be a misunderstanding of > > > > > > > mine then. I thought that unlike a car on a > > > > > > > motorway where if car A was travelling at speed > > > > > > > x-y, and car B was travelling at speed x, car B > > > > > > > would only seem to pull away from car A at y mph, > > > > > > > with light I thought, the idea was that even if > > > > > > > the space ship is going at near the speed of > > > > > > > light, the fired beam would still appear to travel > > > > > > > away from it at the speed of light. > > > > > > > It does. But *in the ship's frame*, the > > > > > > rest distance of 1 light year can be > > > > > > measured (by the ships' instruments) by many > > > > > > different methods to be about 3.9 light hours. > > > > > > > > Likewise if the observer was travelling towards > > > > > > > an incoming light beam at the near the speed of > > > > > > > light, the light would only appear to travel > > > > > > > towards the observer at the speed of light. > > > > > > > Correct. > > > > > > > > Are you saying that this is not the case, and > > > > > > > that if a beam of light is fired from an object > > > > > > > moving at velocity x in the same direction as > > > > > > > the light beam, that the light beam will only > > > > > > > appear to move away from that object at c-x? > > > > > > > No. What I am saying is that "time dilation" and > > > > > > "length contraction" are two faces of the same > > > > > > coin. Please look at the link I provided, > > > > > > at minimum. Or here, if you have the stomach for > > > > > > it. > > > > > <snip link now broken by Google.Groups> > > > > > > > Everyone sets themselves up for "frame jumps", > > > > > > since "common sense" says that measurements by > > > > > > different observers "must be" the same for > > > > > > all frames. It is sometimes close enough where > > > > > > our "common sense" is trained, but it is a > > > > > > serious mistake at speeds close to c. > > > > > > > It is even detectable at the speed of aircraft > > > > > > (special relativity), and/or altitudes of a few > > > > > > hundred feet (general relativity), if you can > > > > > > integrate the effects over time. > > > > > > The question is about the frame of reference of > > > > > the observer that is approaching the tracke. I'm > > > > > not sure where this frame is ever left, so I'm > > > > > not clear on where you are suggesting frame > > > > > jumping is taking place. I thought you were > > > > > saying that the observer approaching the track > > > > > would find themselves only a small distance away > > > > > from the track as the beam of light they fired > > > > > reached the mirror, since you said: > > > > > > "At the speed you have indicated, the moving > > > > > observer is just some short distance from the > > > > > mirror (a few million miles, say) when the > > > > > light hits." > > > > > ... the "few million miles" in the frame of the > > > > track and car, of course. > > > > > > Though if they were, then I'm not clear how > > > > > the light appeared to them to be leaving them > > > > > at the spead of light, since it is only a little > > > > > bit ahead of them. > > > > > ... because their clock is running slow, and > > > > their lengths contracted, they can measure c > > > > to be the same friendly constant... > > > > > > I am assuming that they will be using the vehicle > > > > > going around the track as their clock, > > > > > Why would they do that? Even at constant speed, > > > > the rate of vehicle motion will be extraordinarily > > > > fast when approaching the track, and > > > > extraordinarily slow when departing from the other > > > > side. Much easier just to carry your own clock with > > > > you. Like "heartbeats" if nothing else. > > > > > > taking into account how long it takes the light > > > > > to reach them at their given position. Or is it > > > > > that if they did this, that the light would not > > > > > seem to be leaving them at the speed of light, > > > > > but only if they used a clock on board their > > > > > ship, such that measured by that onboard clock > > > > > the journey appears to have taken less time, > > > > > Bingo. > > > > Ah ok. So if they used the track as the clock, > > > and took into account the time they would have > > > expected the light to have reached them from > > > the track, > > > "the time they would have expected" > > ... is that rest time, or knowing the the tools of relativity time? > > I meant it in the sense that if an outpost sent out a horse rider > every hour along a path, and the riders travelled at a set speed, that > if one was to be coming down the path towards the outpost a certain > speed, one could work out the expected interval between meeting a > rider, and still use the meeting of a rider as a clock, though > wouldn't consider there to be an hour interval between each when you'd > expect to meet each rider. > > In the same sense that if a picture of a horse was beamed in light > each hour, that heading towards the light source beaming the pictures, > you could still calculate the interval you'd expect between each > picture, but you wouldn't consider the interval to be an hour, since > you were moving towards the source. > > Likewise with the track, and the interval between each lap. > > > > it wouldn't seem that light was travelling at c. > > > Only with a frame jump implied with using measurements from diffferent > > frames. > > > > It would simply be that the laws of physics > > > would propergate slower at high velocity (and > > > perhaps under high gravity, I don't know), > > > Know that gravity need have nothing to do with this. It creates > > orders of magnitude more math, for a correction of fractions of a > > percent. > > Ok, we can ignore gravity. > > > > such that the tick of a clock on board the > > > ship might synchronise with a lap of the > > > vehicle on the track while the ship is > > > stationary, but when approacing the track at > > > a higher speed it might synchronise with 5 > > > laps of the vehicle on the track for example. > > > Yes, given the correct closing speed and/or track length / speed. > > > > Since the propergation rate of the laws is > > > referred to as t (time), > > > Laws don't propagate. Laws aren't really laws. Theories describe > > sets of observations, including propagation. > > I wasn't suggesting that laws themselves propergated, just that they > contained a propergation rate referred to as t (time). Though perhaps > I shouldn't have referred to t as the propergation rate, and instead > of referred to it as an event synchronisation indicator. > > > > they talk about time going slower or faster > > > in particular frames of reference. Have I > > > understood this correctly? > > > Not quite. I think you have the concept, but the language is loose. > > Everything in a frame appears to be self-consistent, not "time going > > slower or faster". The differences in length and "time rate" occur > > when viewing *another* frame with enough difference in velocity, that > > relativity's effects are larger than the error-bars of measurement. > > I'm not sure that I have got the concept. The slowing of a moving clock is caused by how you measure its rate and the effects of relative simultanity. Same deal with the length of a rod. All the clocks are identical and tic at the same rate, as measured in their rest frame. Move the clock to a different rest frame and it will tick at the same rate as the clocks already in that frame. Same deal with a meter rod. It is the same length as all other meter rods in the frame, and when moved to another frame will be the same length as the meter rods in that frame. To set up your coordinate system put a clock at each end of a rod. To sync the clocks flash beams of light back and forth between the clocks. Since light travels the same speed in either direction, when the clocks are adjusted so the light takes the same time to travel the length of the rod in either direction the clocks are in sync. Now switch to a second frame that is moving relative to the first. You set up a series of rods and clocks and sych them. Now watch as the first rod passes by and a light flashes along its length. The light is moving at c as measured by your stationary clocks. Because the first rod is moving relative to you, you calculate that the light is traveling at c+v relative to it. Note that it is a calculation, not a direct measurement. We call this a closing speed. Flash the light in the opposit direction and you get c-v. If we measure how long it takes the light to travel the lenght of the rod in opposit directions we get different times. But we know that an observer moving with the rod measures the light to take the same time in either direction. The reason he is able to do that is because, from our perspective, his clocks are out of sync. Remember, they were set based on light traveling at c relative to the moving rod. So both frames measure the speed of light to be c in their own frame while calculating it to be c+/-v in the other frame. All the clocks in the moving frame are out of sync relative to ours. If they measure the tick rate of one of our clocks, each time the clock ticks it will be next to a different clock, to which it will be compared to determine its tic rate. So the clocks being out of sync causes a difference in the measured tic rate of the passing clock. Finally they measure the length of our meter rod by noting where the ends of the rod are *at the same time*. As far as we are concerned their clocks are out of sync, so they mark where the ends of the rod are at different times, giving them a false measurement of the rod's length. Two events seperated by distance are said to be simultaneous when they happen at the same time. Since the clocks in one frame are out of sync with the clocks in the other, if the events happen at the same time in one frame they must happen at different times in the other frame. That is Relative Simultanity. > I was thinking at the high velocity that the observed rate of > propergation on the ship will have slowed relative to the track. Such > that radioactive decay on the ship would take place slower than on the > track. > > To hopefully indicate what I mean, if everything in an alternate > universe things propergated twice as fast, then observations within it > should still fit with same 'laws of physics'. Since the propergation > is measured in terms of synchronisity. The clock which measured a > second would turn twice as fast, while the ball dropped twice as fast, > so it would still drop at Xm/s in both universes. The variable t would > still indicate the synchornisation of events. > > It seems to me that what you are referring to as 'frames' are areas > where the propergation rate is deemed to be uniform. > > I might not be communicating this very well, do you feel that you are > able to understand what I am saying? > > > > > > > > > > and therefore the slight distance the light > > > > > beam was ahead of them, given the time shown > > > > > on the onboard clock, > > > > > ... and the contracted distance ... > > > > > > would indicate light travelling at c. > > > > > Correct. > > > > > > Purely from the frame of reference of the > > > > > observer that fires the beam of light towards > > > > > the track, what will they be expected to see > > > > > at the time the light is reflected back to > > > > > them (how far will the vechile have appeared > > > > > to have gone around the track)? > > > > > You said the vehicle traveled 100mph, presumably > > > > in its own frame, and it travels really close to > > > > exactly one year, so: > > > > 100 * 24 * 365.25 (just averaging the year) = > > > > 876,600 miles, in just under- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -... > > read more » |