Question about relativity
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From: dlzc on 14 Jul 2010 11:55 Dear someone2: On Jul 14, 8:46 am, someone2 <glenn.spig...(a)btinternet.com> wrote: > On 14 July, 14:45, "Inertial" <relativ...(a)rest.com> wrote: > > >"someone2" wrote in message > > >news:ace972e7-fb08-4d6a-9866-5d96e5fc22eb(a)b35g2000yqi.googlegroups.com.... .... > > > while firing a beam of light at the > > > mirror. > > > > Would I be correct in thinking that > > > after 1 year > > > According to whom? An observer on the > > vehicle? An observer at the mirror > > at 0,0,0? The moving observer you mentioned? > > The moving observer mentioned. So the "light year" initial distance is determined by this observer while in motion, or at some rest state? Are track size, and vehicle speed also determined by this moving observer? Relativity is about mapping measurements in one frame, to measurements made in another frame. So it is really important not to "frame jump" around using measurements from different frames, else you arrive at "paradoxes". If you are wanting to understand, this is *key*. David A. Smith
From: someone2 on 14 Jul 2010 12:40 On 14 July, 15:38, dlzc <dl...(a)cox.net> wrote: > Dear someone2: > > 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. > > 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? > > 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 (and thus have reached the mirror by the time the observer is about 0.75 light years from the mirror), it would only have travelled 0.25 light years back towards the observer, after the 1 year. 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). 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?
From: someone2 on 14 Jul 2010 13:07 On 14 July, 16:55, dlzc <dl...(a)cox.net> wrote: > Dear someone2: > > On Jul 14, 8:46 am, someone2 <glenn.spig...(a)btinternet.com> wrote: > > > On 14 July, 14:45, "Inertial" <relativ...(a)rest.com> wrote: > > > >"someone2" wrote in message > > > >news:ace972e7-fb08-4d6a-9866-5d96e5fc22eb(a)b35g2000yqi.googlegroups.com... > ... > > > > while firing a beam of light at the > > > > mirror. > > > > > Would I be correct in thinking that > > > > after 1 year > > > > According to whom? An observer on the > > > vehicle? An observer at the mirror > > > at 0,0,0? The moving observer you mentioned? > > > The moving observer mentioned. > > So the "light year" initial distance is determined by this observer > while in motion, or at some rest state? Are track size, and vehicle > speed also determined by this moving observer? > > Relativity is about mapping measurements in one frame, to measurements > made in another frame. So it is really important not to "frame jump" > around using measurements from different frames, else you arrive at > "paradoxes". If you are wanting to understand, this is *key*. > Can we consider that the observer was at the track (positioned between the earth and the sun), and that the earth is seen to rotate around the sun two times for every lap the vehicle makes of the track. The observer then leaves the track in a space ship, and flies off at what it considers to be a speed of 1 millionth the speed of light, and after it has flown for two million years (the observer calculates that they would have expected the track vehicle to have made 1 million laps of the track). Then while stationary they fire a beam of light at the mirror in the centre of the track, and after the period of time when they thought the track vehicle would have made 2 million laps, the reflected light returns (are you ok that this could happen?). They then speed up towards the track again, and on covering 1/4 of the distance to the track while at maximum velocity (near the speed of light), fire a beam of light at the mirror in the middle of the track. Does this seem ok?
From: dlzc on 14 Jul 2010 14:10 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. > 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. David A. Smith
From: someone2 on 14 Jul 2010 14:43
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. > |