Question about relativity
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From: dlzc on 15 Jul 2010 00:08 Dear someone2: 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. > 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. > (I took a look at the link you provided btw to > the FAQ, and it is a useful link, I'm not clear > what I should have looked at for the answer > to the question though, perhaps if you just gave > me the answer and explained how you arrived at it, > it would help). You might search for: "relativistic doppler shift" .... for your next answers. David A. Smith
From: Jonathan Doolin on 15 Jul 2010 00:36 > 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.
From: Androcles on 15 Jul 2010 02:16 "Jonathan Doolin" <good4usoul(a)gmail.com> wrote in message news:4867b12d-5d59-47b6-b599-ee93e5c1c194(a)j13g2000yqj.googlegroups.com... > 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. ============================================ a distance of 1 ly = 3.9 hours ? You are either lying or fuckin' insane, Droolin' Doolin. Show me your calculation on a spreadsheet.
From: someone2 on 15 Jul 2010 05:43 On 15 July, 04:08, xxein <xxxx...(a)gmail.com> wrote: > On Jul 14, 8:13 pm, someone2 <glenn.spig...(a)btinternet.com> wrote: > > > > > > > On 14 July, 20:56, dlzc <dl...(a)cox.net> wrote: > > > > Dear someone2: > > > > 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. > > > >http://www.motionmountain.net/mmdownload.php?f=motionmountain-volume2... > > > > 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." > > > 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. I am assuming that they will be using the vehicle > > going around the track as their clock, 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, and therefore the slight distance the light beam was > > ahead of them, given the time shown on the onboard clock, would > > indicate light travelling at c. > > > 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)? > > > (I took a look at the link you provided btw to the FAQ, and it is a > > useful link, I'm not clear what I should have looked at for the answer > > to the question though, perhaps if you just gave me the answer and > > explained how you arrived at it, it would help).- Hide quoted text - > > > - Show quoted text - > > xxein: To understand this better, don't allow gravity to have an > affect. It will confuse your issue of particular understanding. > Worry about gravity later. I can help you there. > > Lorentz was very capable to handle inertial frames. Einstein tried to > make a shortcut math to make it simpler to compute. But Einsein left > out the physic of it all. Lorentz did not say that the speed of light > was a constant c in all frames. There are moving frames. What does > that mean for observers? It means that if observers (in the simplest > terms) are moving at the same velocity along the same line, that they > will observe each other as if their shared inertial frame was moving > or not. That is all there is to the observational aspect of it. > > Now. There is a physic of how this happens. It is contained in the > Lorentz formulas. Einstein thought there was a math shortcut that > superceded the physic. Silly man (or his wife). But everyone liked > it. Just as the Einstein followers you read here. > > c is a physical constant --- not a math constant just because it is > observed that way. A few posts tried to tell you that you can observe > light going away from you. No such thing. You can only observe light > coming directly toward you. How else would light come to your > observation? How else could you travel at a speed that was > differentiated from the speed of light? These guys aren't real > physicists. They are math beleivers - just like Einstein. Just for > fun, ask them to describe the evolution of a universe from energy > released to what is now. They have no coherent explanation. It might > be an unfair question but ask them how the energy got released and/or > where it came from. > > Nevertheless, the term of 'inertial frame' is ill-defined except for > the purpose of a theory. Just because someone believes a partiular > theory, it does not mean that the definition is physically a correct > one. > > But let's get back to what you want to understand. Observers A and B > and/or what will A observe if moving wrt B (and vv). See? We already > have moving frames. They are in a physical respect to c already. How > can any moving frame claim that c is c to him? Only by what each > observes. It becomes paradoxical to a physic, but not to a physics > made from observation. So? Do we have knowledge of the physic or > only a belief in a math theory to make a physics? The dilemna is that > the math theory tells us what we will physically measure but not > physically why it happens that way. > > Lorentz puts out the physics and Einstein puts out a short math. > Please learn the difference. > > I'm too lazy right now to look up my old notes to show how this works, > but please learn that a math expression, written to give the same math > result but with different form, can give the same value. Which form > is correct to express the physical meaning? One describes this and > the other describes that. That the values are the same does not mean > they describe the physical process as the same in it's proper logical > and/or physical context. Perhaps I'll get unlazy some day and give > you a concrete example. > > In the meantime, study Lorentz and wonder why gravity exists (even > though he didn't know or addres it).- Hide quoted text - > Thanks for the reply. It might be useful if you could say from the frame of reference of the observer that fires the beam of light towards the track, what they would 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)
From: someone2 on 15 Jul 2010 06:02
On 15 July, 05:08, dlzc <dl...(a)cox.net> wrote: > Dear someone2: > > 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, it wouldn't seem that light was travelling at c. It would simply be that the laws of physics would propergate slower at high velocity (and perhaps under high gravity, I don't know), 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. Since the propergation rate of the laws is referred to as t (time), they talk about time going slower or faster in particular frames of reference. Have I understood this correctly? > > 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? Else why would they not be considering themselves to ahave observed FTL travel. > > (I took a look at the link you provided btw to > > the FAQ, and it is a useful link, I'm not clear > > what I should have looked at for the answer > > to the question though, perhaps if you just gave > > me the answer and explained how you arrived at it, > > it would help). > > You might search for: > "relativistic doppler shift" > ... for your next answers. > > David A. Smith- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text - |