Question about relativity
in [Relativity]
|
Prev: Big Bang
Next: Painus Is Buying AA A New Laptop !
From: Inertial on 14 Jul 2010 09:45 >"someone2" wrote in message >news:ace972e7-fb08-4d6a-9866-5d96e5fc22eb(a)b35g2000yqi.googlegroups.com... >On 14 July, 13:47, harald <h...(a)swissonline.ch> wrote: >> On Jul 14, 2:31 pm, someone2 <glenn.spig...(a)btinternet.com> wrote: >> >> > I was wondering if anyone could help me understand issues >> > of simultaneity under relativity. >> >> Please allow me first to test: Do you understand that "simultaneity" >> in relativity theory is largely a matter of convention? Its not really .. unless you want clocks showing the wrong times. SR says that, for two bodies A and B at rest in some inertial frame, the light takes the same time to travel from A to B as it does from B to A. That means you can use light to synchrnoise the clocks. That is a very precise meaning. Any other setting would mean the clocks would not be showing the speed of light as isotropically c, and so would be 'wrong'. >> > >I hadn't really thought about it, though am not suprised. > >> > 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, >> >> You mean NOT on the track? Then where? >> > > >If you were to think of a graph where there is a circle (the track) >around the centre point x=0, y=0, and z=0 (where the mirror is), such >that for all points on the circle, z = 0 and x-squared + y-squared = >radius-squared and the circumferance is such that a vechile moving at >100mph would take 2 years to make a circuit. According to whom? An observer on the vehicle? An observer at the mirror at 0,0,0? The moving observer you mentioned? >The observer is approaching along the z axis where x=0 and y = 0. > > >> > travelling towards it at near light speed, According to whom? An observer on the vehicle? An observer at the mirror at 0,0,0? The moving observer you mentioned? >> 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 observer would be >> > 0.5 light years from the track >> > 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). You need to specify more clearly
From: dlzc on 14 Jul 2010 10:38 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... > while firing a beam of light at the mirror. > > Would I be correct in thinking that after > 1 year Who's 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, Circles don't have "start"s. So "from the position the car was in when the observer crossed the 1 light year mark". > 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. > 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. David A. Smith
From: harald on 14 Jul 2010 11:42 On Jul 14, 3:45 pm, "Inertial" <relativ...(a)rest.com> wrote: > >"someone2" wrote in message > >news:ace972e7-fb08-4d6a-9866-5d96e5fc22eb(a)b35g2000yqi.googlegroups.com.... > >On 14 July, 13:47, harald <h...(a)swissonline.ch> wrote: > >> On Jul 14, 2:31 pm, someone2 <glenn.spig...(a)btinternet.com> wrote: > > >> > I was wondering if anyone could help me understand issues > >> > of simultaneity under relativity. > > >> Please allow me first to test: Do you understand that "simultaneity" > >> in relativity theory is largely a matter of convention? > > Its not really .. unless you want clocks showing the wrong times. "Wrong" is "relative" here - see below. ;-) > SR says > that, for two bodies A and B at rest in some inertial frame, the light takes > the same time to travel from A to B as it does from B to A. Here you put the cart in front of the horse. It's much better to say it like this: 'We assume that the clocks can be adjusted in such a way that the propagation velocity of every light ray in vacuum - measured by means of these clocks - becomes everywhere equal to a universal constant c, provided that the coordinate system is not accelerated.' For special applications we may decide to proceed a little different: for example with GPS one defines the speed of light as c *relative to the ECI "frame"*, making it by the same definition (c-v) relative to the "moving" surface of the earth ("Sagnac" correction). > That means you > can use light to synchrnoise the clocks. That is a very precise meaning. > Any other setting would mean the clocks would not be showing the speed of > light as isotropically c, and so would be 'wrong'. It would be "right" according to the same convention relative to a system that is moving at a certain velocity relatively to the one that you have in mind! Therefore: "we establish by *definition* that the ``time'' required by light to travel from A to B equals the ``time'' it requires to travel from B to A." Cheers, Harald
From: someone2 on 14 Jul 2010 11:46 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.... > >On 14 July, 13:47, harald <h...(a)swissonline.ch> wrote: > >> On Jul 14, 2:31 pm, someone2 <glenn.spig...(a)btinternet.com> wrote: > > >> > I was wondering if anyone could help me understand issues > >> > of simultaneity under relativity. > > >> Please allow me first to test: Do you understand that "simultaneity" > >> in relativity theory is largely a matter of convention? > > Its not really .. unless you want clocks showing the wrong times. SR says > that, for two bodies A and B at rest in some inertial frame, the light takes > the same time to travel from A to B as it does from B to A. That means you > can use light to synchrnoise the clocks. That is a very precise meaning. > Any other setting would mean the clocks would not be showing the speed of > light as isotropically c, and so would be 'wrong'. > > > > > > > > >I hadn't really thought about it, though am not suprised. > > >> > 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, > > >> You mean NOT on the track? Then where? > > >If you were to think of a graph where there is a circle (the track) > >around the centre point x=0, y=0, and z=0 (where the mirror is), such > >that for all points on the circle, z = 0 and x-squared + y-squared = > >radius-squared and the circumferance is such that a vechile moving at > >100mph would take 2 years to make a circuit. > > According to whom? An observer on the vehicle? An observer at the mirror > at 0,0,0? The moving observer you mentioned? > > >The observer is approaching along the z axis where x=0 and y = 0. > > >> > travelling towards it at near light speed, > > According to whom? An observer on the vehicle? An observer at the mirror > at 0,0,0? The moving observer you mentioned? > > >> 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. > >> > the observer would be > >> > 0.5 light years from the track > >> > 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). > > You need to specify more clearly >
From: someone2 on 14 Jul 2010 11:49
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. |