Question about relativity
in [Relativity]
|
Prev: Big Bang
Next: Painus Is Buying AA A New Laptop !
From: BURT on 14 Jul 2010 15:02 On Jul 14, 11:43 am, someone2 <glenn.spig...(a)btinternet.com> wrote: > On 14 July, 19:10, dlzc <dl...(a)cox.net> wrote: > > > > > > > Dear someone2: > > > On Jul 14, 9:40 am, someone2 <glenn.spig...(a)btinternet.com> wrote: > > > > On 14 July, 15:38, dlzc <dl...(a)cox.net> wrote: > > > > On Jul 14, 5:31 am, someone2 <glenn.spig...(a)btinternet.com> wrote: > > > > > > I was wondering if anyone could help me > > > > > understand issues of simultaneity under > > > > > relativity. > > > > > The issues are that positions / distances and > > > > times / durations need to be specified "in > > > > whose frame" they are to be measured. > > > > > > I was thinking of a circular track in space, > > > > > which a vechile moving at 100mph takes 2 > > > > > years to make a circuit, > > > > > Presumably in the track' frame. And the > > > > vehicles isn't really moving fast enough to > > > > be significantly different. > > > > > > and there being a mirror in the middle of > > > > > the track, and an observer 1.5 light years > > > > > from the track, travelling towards it at > > > > > near light speed, > > > > > From above, or the plane of the track? If he > > > > is travelling at 86.6% the speed of light, the > > > > gamma is 2, which is sort of handy... > > > > From above (along a line perpendicular to the > > > plane of the track). We can consider the > > > observer to be travelling at 99.99999% the > > > speed of light. > > > Gamma is 2236 for this speed. > > > > > > while firing a beam of light at the mirror. > > > > > > Would I be correct in thinking that after > > > > > 1 year > > > > > Who's year? > > > > Perhaps we should establish what we mean by a > > > year here. Would you be ok with a year being > > > the event of the earth making one rotation of > > > the sun? > > > Sure. Might be simpler if the racetrack were Earth's orbit, and the > > car the Earth, too. > > > > > > the observer would be 0.5 light years > > > > > from the track, and be receiving back > > > > > a reflection of the fired light beam > > > > > from the mirror, and the light from > > > > > when the vehicle was quarter of the > > > > > way around the track (assumes the > > > > > vehicle to have started its journey > > > > > from the start of the track, > > > > > Circles don't have "start"s. So "from > > > > the position the car was in when the > > > > observer crossed the 1 light year mark". > > > > Tracks can have start lines on though, as > > > in racing tracks, which is what I meant > > > when I referred to the start of the track. > > > > > > as the observer started start its journey > > > > > towards the track & mirror). > > > > > To avoid accelerations, and discussion of > > > > accelerations, it is better to have him at > > > > speed at that point. > > > > Ok > > > > > > If I am not correct, could you please state > > > > > what the correct observations would be > > > > > expected to be. > > > > > The car will not *be seen to* have started > > > > his trip around the track, until some time > > > > after the ship crosses the 1 light year mark. > > > > Some time later than this, light will *be > > > > seen to* have reflected off the mirror. Both > > > > the observer and the car's occupant will > > > > agree on the car's position en passant. Both > > > > will agree the other is "seen to be" moving > > > > more slowly. > > > > 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. > 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. 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? > > > > > > Such that the observation (for the observer > > > travelling towards the track) after 1.125 years > > > travelling would be of the reflection, and the > > > vehicle being seen to have gone 0.325 of the > > > way around the track (a full lap taking 2 years). > > > The Earth will have travelled just over one rotation around the Sun as > > the moving observer crosses the plane of the racetrack. There is a > > full rotation of the Earth in transit to the ship's starting point, so > > the moving observer will see the car start about the halfway point, > > and the Earth make "two" full rotations on his way inbound. Which > > trip will take about 3.9 hours (moving onserver's time) at the gamma > > you've established. > > > > Would you be so kind as to mention how far you > > > think the car would have appeared to have > > > travelled around the track to the observer in > > > the approaching track when first 'seeing' the > > > light reflected off the mirror? > > > I'd have to look up your speeds, and frankly, the answer will confuse > > you more than you are already confused. > > > Let me provide you a helpful link:http://www.physics.adelaide.edu.au/~dkoks/Faq/ > > > And recommend "Spacetime Physics" by Taylor and Wheeler.- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text - If you begin to move objects around you appear to move in the opposite dirction and they shrink in the distance. Mitch Raemsch
From: dlzc on 14 Jul 2010 15:56 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.pdf 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. David A. Smith
From: someone2 on 14 Jul 2010 20:13 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).
From: Mathal on 14 Jul 2010 22:42 On Jul 14, 8:49 am, someone2 <glenn.spig...(a)btinternet.com> wrote: > On 14 July, 14:34, Mathal <mathmusi...(a)gmail.com> wrote: > > > > > > > On Jul 14, 5:31 am, someone2 <glenn.spig...(a)btinternet.com> wrote: > > > > I was wondering if anyone could help me understand issues of > > > simultaneity under relativity. > > > > I was thinking of a circular track in space, which a vechile moving at > > > 100mph takes 2 years to make a circuit, and there being a mirror in > > > the middle of the track, and an observer 1.5 light years from the > > > track, travelling towards it at near light speed, while firing a beam > > > of light at the mirror. > > > > Would I be correct in thinking that after 1 year the observer would be > > > 0.5 light years from the track, and be receiving back a reflection of > > > the fired light beam from the mirror, and the light from when the > > > vehicle was quarter of the way around the track (assumes the vehicle > > > to have started its journey from the start of the track, as the > > > observer started start its journey towards the track & mirror). > > > > If I am not correct, could you please state what the correct > > > observations would be expected to be. > > > After one year from the frame of the mirror the observer would be > > lagging just behind the photons travelling towards the mirror. Neither > > the observer or the light would have arrived yet. > > From the observer's travelling just below light speed's point of view > > a few microseconds have transpired. He can't see the light but it's on > > it's way just a few light micro-seconds ahead of him. > > I thought that light would be moving away from the observer at the > speed of light, regardless of the speed that the observer was moving > at. So it wouldn't be like cars on a motorway where if one car was > going at x, and the other at slightly less that x, that the former > would only be pulling away by the amount the latter was travelling > below x. Light is moving away from the observer at the speed of light. In the frame of the moving observer-travelling at near light speed time is moving slower than in a the other frame that is pertinent to your question- the frame of the mirror. The way you phrased the problem the moving observer is travelling at near light speed with respect to the mirror which can be taken to be the rest frame in this problem. Conversely the mirror can be travelling towards the stationary observer at near light speed and all results would be reversed. Time is moving slower in a faster frame. If you think about it from the mirror frame the photon is released and travels at the speed of light to the mirror. The observer, travelling at near light speed will arrive shortly after the photon. The observer, travelling at near the speed of light is in a frame that operates much slower so that only a minute amount of time has elapsed in this frame when a year has elapsed in the mirror frame. In the observers frame light is travelling away from him at the speed of light. One other thing- the observer could not survive for a moment or two travelling at this velocity. Most particles would collide with him at near light speed and obliterate all traces of his/her existence. The light sent to the mirror would be an absurdly high frequency blowing holes through the mirror with every photon that hit the mirror. There would be virtually no reflection of light. Light travels at the speed of light in both frames. The near light observer and the mirror. The only way this is possible is for the observer to be in a slower time frame. Mathal
From: xxein on 14 Jul 2010 23:08
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). |