Hard SR questions?
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From: The Ghost In The Machine on 6 Aug 2006 23:00 On Sun, 06 Aug 2006 06:18:01 -0700, jt64 wrote: > Ship (A) travel 0.99c approaching earth. > At a distance of 300 000 km as seen from planet (B) inertial frame the > ship start a framed TV transmission. Considering that the planet probably won't know the ship is even there prior to the transmission there are some issues with the above statement. However, one can of course ask the question as to when the ship should start broadcasting in order to get 60 frames in before it passes by or impacts the planet. The answer is obvious: (0,-1)_S. (In other words, 1 second before flyby or impact -- assuming the ship knows when it will flyby or impact, ship time.) Because we're using SR, and therefore the Lorentz, this transforms into (-7.0179239, -7.088812)_E Since Earth is incapable of seeing things 7 light-seconds from its position, we need to adjust this a little, by adding abs(x_E) to t_E (basically, the light needs to move a little); therefore, we get the event (0,-0.0708881)_E as the start of the transmission. Those 60 frames will compress into a packet of 0.0708881 second, for a frequency of 846.4 frames per second, as observed by Earth. Note that this is not limited to frames; if the ship is trying to broadcast at a frequency of 100 MHz, the Earth will receive it at the frequency 1.4107 GHz. As the ship flies by the Earth the situation shifts, and everything slows wa-y-y-y-y down, as observed from Earth. This can be computed by looking at the event (0,1)_S which is after the ship has broadcast 20 frames. By the Lorentz this transforms into (7.0179239, 7.088812)_E and now the second cluster of 60 frames will occupy 14.1067 seconds of Earth time for an effective frequency of 4.2533 frames/second and a broadcast frequency of only 7.0888 MHz. > > The transmission is such that 60 frames will be sent from ship (A) > during the distance of 300 000 km relative earths inertial frame. Not quite that simple, sport. > > Now my question. > > 1. How long time will it take to travel to planet using (ship point of > view). This question is extremely confused, as you've not yet established precisely where/when the ship is. Using my assumptions, of course, the first wavefront will take 7.0179239 seconds from ship to Earth as one might measure from Earth were it at all possible, but the time skew screws things up so badly there's no way to know what happened *prior* to the 0.0708881 seconds before impact, as determined from Earth. The ship might observe it as taking 0.99 seconds, but it can't determine that easily either. Everyone is locked into his respective origin. > 2. What is the framerate for the transmission within the ship.(ship > point of view) 60 Hz, of course. Unless you want the spacecraft to actually try to slow down its transmissions...? [rest snipped] Please clarify your assumptions. -- #191, ewill3(a)earthlink.net It's still legal to go .sigless.
From: jt64 on 7 Aug 2006 04:56 The Ghost In The Machine skrev: > On Sun, 06 Aug 2006 06:18:01 -0700, jt64 wrote: > > > Ship (A) travel 0.99c approaching earth. > > At a distance of 300 000 km as seen from planet (B) inertial frame the > > ship start a framed TV transmission. > > Considering that the planet probably won't know the ship is even there > prior to the transmission Well it will there is a pod there at rest that register when the ship pass by, using doppler shift. >there are some issues with the above statement. > However, one can of course ask the question as to when the ship should > start broadcasting in order to get 60 frames in before it passes by or > impacts the planet. The ship inertial frame do not send 60 frames/hz The planets inertial view do not receive 60 frames/hz Between 300 000 to 0 km the ship send out 60 frames/hz as measured from the fram of earth. Let us say it is digital 1 bit sent 300 000 km away, last bit sent 0,00...1 km from planet. > The answer is obvious: (0,-1)_S. (In other words, 1 > second before flyby or impact -- assuming the ship knows when it will > flyby or impact, ship time.) > > Because we're using SR, and therefore the Lorentz, this transforms into > > (-7.0179239, -7.088812)_E > > Since Earth is incapable of seeing things 7 light-seconds from its > position, we need to adjust this a little, by adding abs(x_E) to t_E > (basically, the light needs to move a little); therefore, we get the event > > (0,-0.0708881)_E > > as the start of the transmission. Those 60 frames will compress into a > packet of 0.0708881 second, for a frequency of 846.4 frames per second, as > observed by Earth. > > Note that this is not limited to frames; if the ship is trying to > broadcast at a frequency of 100 MHz, the Earth will receive it at the > frequency 1.4107 GHz. > > As the ship flies by the Earth the situation shifts, and everything slows > wa-y-y-y-y down, as observed from Earth. This can be computed by looking > at the event > > (0,1)_S > > which is after the ship has broadcast 20 frames. By the Lorentz this > transforms into > > (7.0179239, 7.088812)_E > > and now the second cluster of 60 frames will occupy 14.1067 seconds of > Earth time for an effective frequency of 4.2533 frames/second and a > broadcast frequency of only 7.0888 MHz. > > > > > The transmission is such that 60 frames will be sent from ship (A) > > during the distance of 300 000 km relative earths inertial frame. > > Not quite that simple, sport. Well it is look above, they will be sent from ship not from within the ships inertial frame. > > > > Now my question. > > > > 1. How long time will it take to travel to planet using (ship point of > > view). > > This question is extremely confused, as you've not yet established > precisely where/when the ship is. Using my assumptions, of course, the > first wavefront will take 7.0179239 seconds from ship to Earth as one > might measure from Earth were it at all possible, but the time skew screws > things up so badly there's no way to know what happened *prior* to the > 0.0708881 seconds before impact, as determined from Earth. The ship pass *exactly* a earth synchronised pod 300 000 km away. The exact moment ship pass by is measured by doppler shift. I would say it will take 1.01 seconds from earths point of view and then take on the calculation from there. To travel there the first frame is received *exactly* after one second and the last frame must be received after 1.0999... just before the ship pass. Then you apply the Lorentz transformation to find out how it look from the ship point of view. > The ship might observe it as taking 0.99 seconds, but it can't determine > that easily either. Everyone is locked into his respective origin. No i do not think it would you were more or less right above. > > 2. What is the framerate for the transmission within the ship.(ship > > point of view) > > 60 Hz, of course. Unless you want the spacecraft to actually try to slow > down its transmissions...? No it is absolutly not the framerate from the inertial ship frame. > [rest snipped] > > Please clarify your assumptions. > > -- > #191, ewill3(a)earthlink.net > It's still legal to go .sigless.
From: The Ghost In The Machine on 11 Aug 2006 02:00 In sci.physics.relativity, jt64(a)tele2.se <jt64(a)tele2.se> wrote on 7 Aug 2006 01:56:49 -0700 <1154941009.075265.131900(a)m73g2000cwd.googlegroups.com>: > > The Ghost In The Machine skrev: > >> On Sun, 06 Aug 2006 06:18:01 -0700, jt64 wrote: >> >> > Ship (A) travel 0.99c approaching earth. >> > At a distance of 300 000 km as seen from planet (B) inertial frame the >> > ship start a framed TV transmission. >> >> Considering that the planet probably won't know the ship is even there >> prior to the transmission > > Well it will there is a pod there at rest that register when the ship > pass by, using doppler shift. That pod is going to have its own issues, even assuming that it can remain motionless with respect to the center of the Earth (as opposed to orbiting it). > >>there are some issues with the above statement. >> However, one can of course ask the question as to when the ship should >> start broadcasting in order to get 60 frames in before it passes by or >> impacts the planet. > > The ship inertial frame do not send 60 frames/hz Depends on who's doing the measurement. The ship's equipment is indeed sending at 60 Hz. (BTW, that's more correctly represented "60 frames/second".) > The planets inertial view do not receive 60 frames/hz Correct, because of the time compression. > Between 300 000 to 0 km the ship send out 60 frames/hz as measured from > the fram of earth. > Let us say it is digital 1 bit sent 300 000 km away, last bit sent > 0,00...1 km from planet. Frames are fine; it wouldn't make any difference. > >> The answer is obvious: (0,-1)_S. (In other words, 1 >> second before flyby or impact -- assuming the ship knows when it will >> flyby or impact, ship time.) >> >> Because we're using SR, and therefore the Lorentz, this transforms into >> >> (-7.0179239, -7.088812)_E >> >> Since Earth is incapable of seeing things 7 light-seconds from its >> position, we need to adjust this a little, by adding abs(x_E) to t_E >> (basically, the light needs to move a little); therefore, we get the event >> >> (0,-0.0708881)_E >> >> as the start of the transmission. Those 60 frames will compress into a >> packet of 0.0708881 second, for a frequency of 846.4 frames per second, as >> observed by Earth. >> >> Note that this is not limited to frames; if the ship is trying to >> broadcast at a frequency of 100 MHz, the Earth will receive it at the >> frequency 1.4107 GHz. >> >> As the ship flies by the Earth the situation shifts, and everything slows >> wa-y-y-y-y down, as observed from Earth. This can be computed by looking >> at the event >> >> (0,1)_S >> >> which is after the ship has broadcast 20 frames. By the Lorentz this >> transforms into >> >> (7.0179239, 7.088812)_E >> >> and now the second cluster of 60 frames will occupy 14.1067 seconds of >> Earth time for an effective frequency of 4.2533 frames/second and a >> broadcast frequency of only 7.0888 MHz. >> >> > >> > The transmission is such that 60 frames will be sent from ship (A) >> > during the distance of 300 000 km relative earths inertial frame. >> >> Not quite that simple, sport. > > Well it is look above, they will be sent from ship not from within the > ships inertial frame. And if the ship is not within its own inertial frame, where is it? :-) > >> > >> > Now my question. >> > >> > 1. How long time will it take to travel to planet using (ship point of >> > view). >> >> This question is extremely confused, as you've not yet established >> precisely where/when the ship is. Using my assumptions, of course, the >> first wavefront will take 7.0179239 seconds from ship to Earth as one >> might measure from Earth were it at all possible, but the time skew screws >> things up so badly there's no way to know what happened *prior* to the >> 0.0708881 seconds before impact, as determined from Earth. > > The ship pass *exactly* a earth synchronised pod 300 000 km away. The > exact moment ship pass by is measured by doppler shift. It will take 1 second for the Earth to know where the ship is from this beacon. > > I would say it will take 1.01 seconds from earths point of view and > then take on the calculation from there. To travel there the first > frame is received *exactly* after one second and the last frame must be > received > after 1.0999... just before the ship pass. > > Then you apply the Lorentz transformation to find out how it look from > the ship point of view. You can do that, if you like. In order for Earth to receive 60 frames at 60 Hz, one calculates the following (0,-1)_E = (-7.0179239, -7.088812)_S In order for a beam to reach this point it must start 7.0179239 seconds earlier from the ship's origin, or -14.106736 seconds. Ergo, the ship sends using a frame rate of about 4.25329 Hz, so that Earth can properly receive it. [rest snipped] -- #191, ewill3(a)earthlink.net Windows Vista. Because it's time to refresh your hardware. Trust us.
From: jt64 on 13 Aug 2006 10:09 The Ghost In The Machine skrev: > In sci.physics.relativity, jt64(a)tele2.se > <jt64(a)tele2.se> > wrote > on 7 Aug 2006 01:56:49 -0700 > <1154941009.075265.131900(a)m73g2000cwd.googlegroups.com>: > > > > The Ghost In The Machine skrev: > > > >> On Sun, 06 Aug 2006 06:18:01 -0700, jt64 wrote: > >> > >> > Ship (A) travel 0.99c approaching earth. > >> > At a distance of 300 000 km as seen from planet (B) inertial frame the > >> > ship start a framed TV transmission. > >> > >> Considering that the planet probably won't know the ship is even there > >> prior to the transmission > > > > Well it will there is a pod there at rest that register when the ship > > pass by, using doppler shift. > > That pod is going to have its own issues, even assuming that it can > remain motionless with respect to the center of the Earth (as opposed to > orbiting it). Well it is really not earth this is a very small planet inhibited by ants. But the have very nice transmiting and receiving technologies for signals. > > > >>there are some issues with the above statement. > >> However, one can of course ask the question as to when the ship should > >> start broadcasting in order to get 60 frames in before it passes by or > >> impacts the planet. > > > > The ship inertial frame do not send 60 frames/hz > > Depends on who's doing the measurement. The ship's > equipment is indeed sending at 60 Hz. (BTW, that's more > correctly represented "60 frames/second".) No it is not ->i told you<-, the observers of planet know that during the 300 000km journey towards earth the ship leave 60 frames.(*there are thousands of tv receivers along the way*) That do not equal a framerate of 60 hz at the ship there is some dilation *and doppler* to consider. > > The planets inertial view do not receive 60 frames/hz > > Correct, because of the time compression. Mainly because relativities<->"doppler effect" is one of the most ugly mathematical artefacts i ever seen. You see if the ship traveled really close to c. The doppler effect slowly would reach towards infinity. But unfortunatly that would also mean that no frames would be sent out due to the time dilation, it is the most ugly artefact of Lorentz transformation ever ;)... > > Between 300 000 to 0 km the ship send out 60 frames/hz as measured from > > the fram of earth. > > Let us say it is digital 1 bit sent 300 000 km away, last bit sent > > 0,00...1 km from planet. > > Frames are fine; it wouldn't make any difference. Oh it would you, just do not understand the example, you see frames as sent out from ship(frame) and frames as measured at planet(frame) during a second. *Really differs* from frames that leave the ship during 300 000 km travel towards planet(frame) and time planet(frame) as registered by the thousands of receivers placed along the way. Hint:There is some doppler and dilation to consider. > > > >> The answer is obvious: (0,-1)_S. (In other words, 1 > >> second before flyby or impact -- assuming the ship knows when it will > >> flyby or impact, ship time.) > >> > >> Because we're using SR, and therefore the Lorentz, this transforms into > >> > >> (-7.0179239, -7.088812)_E > >> > >> Since Earth is incapable of seeing things 7 light-seconds from its > >> position, we need to adjust this a little, by adding abs(x_E) to t_E > >> (basically, the light needs to move a little); therefore, we get the event > >> > >> (0,-0.0708881)_E > >> > >> as the start of the transmission. Those 60 frames will compress into a > >> packet of 0.0708881 second, for a frequency of 846.4 frames per second, as > >> observed by Earth. > >> > >> Note that this is not limited to frames; if the ship is trying to > >> broadcast at a frequency of 100 MHz, the Earth will receive it at the > >> frequency 1.4107 GHz. > >> > >> As the ship flies by the Earth the situation shifts, and everything slows > >> wa-y-y-y-y down, as observed from Earth. This can be computed by looking > >> at the event > >> > >> (0,1)_S > >> > >> which is after the ship has broadcast 20 frames. By the Lorentz this > >> transforms into > >> > >> (7.0179239, 7.088812)_E > >> > >> and now the second cluster of 60 frames will occupy 14.1067 seconds of > >> Earth time for an effective frequency of 4.2533 frames/second and a > >> broadcast frequency of only 7.0888 MHz. > >> > >> > > >> > The transmission is such that 60 frames will be sent from ship (A) > >> > during the distance of 300 000 km relative earths inertial frame. > >> > >> Not quite that simple, sport. > > > > Well it is look above, they will be sent from ship not from within the > > ships inertial frame. > > And if the ship is not within its own inertial frame, where is it? :-) Yes the *ship* is but the *signal* travel through relativities doppler frame, and then there are some time dilation on ship to consider. And i told you it is the most awful, ugly mathematical artefact i ever seen. > > > >> > > >> > Now my question. > >> > > >> > 1. How long time will it take to travel to planet using (ship point of > >> > view). > >> > >> This question is extremely confused, as you've not yet established > >> precisely where/when the ship is. Using my assumptions, of course, the > >> first wavefront will take 7.0179239 seconds from ship to Earth as one > >> might measure from Earth were it at all possible, but the time skew screws > >> things up so badly there's no way to know what happened *prior* to the > >> 0.0708881 seconds before impact, as determined from Earth. > > > > The ship pass *exactly* a earth synchronised pod 300 000 km away. The > > exact moment ship pass by is measured by doppler shift. > > It will take 1 second for the Earth to know where the ship is > from this beacon. No it actually pass earth after 1.01 seconds the clock at pod and earth are synchronised and clock of earth-pod actually show 1.01 this is real values. > > > > I would say it will take 1.01 seconds from earths point of view and > > then take on the calculation from there. To travel there the first > > frame is received *exactly* after one second and the last frame must be > > received > > after 1.0999... just before the ship pass. > > > > Then you apply the Lorentz transformation to find out how it look from > > the ship point of view. > > You can do that, if you like. In order for Earth to receive 60 fr
From: jt64 on 13 Aug 2006 14:04
N:dlzc D:aol T:com (dlzc) skrev: > Dear jt64: > > <jt64(a)tele2.se> wrote in message > news:1154870281.617016.89810(a)m79g2000cwm.googlegroups.com... > > Ship (A) travel 0.99c approaching earth. > > Presumably earth = B? > Presumably the 0.99c is as determined by B? > > gamma = ~7 > let c = 300,000 km/sec > > > At a distance of 300 000 km as seen from > > planet (B) inertial frame the ship start a > > framed TV transmission. > > ... assumed the transmission is periodic, continuous, and the > last frame is completed as A passes B. > > > The transmission is such that 60 frames will > > be sent from ship (A) during the distance of > > 300 000 km relative earths inertial frame. > > > > Now my question. > > ... Question*s* ... > > > 1. How long time will it take to travel to > > planet using (ship point of view). > > "It"? Do you mean the transmission or the ship? It is a > continuous data stream, whose last frame exits A's transmitter > just as A passes B, right? > > The distance travelled during this transmission is (300,000 / 7) > km. The ship also measures B moving at 0.99c. So the duration > is > > 1/7 = (1 + 0.99)*t > t = 0.07 sec. > > > 2. What is the framerate for the transmission > > within the ship.(ship point of view) > > To meet your givens: > 60 / 0.07 = 840 frames per second. No that would be 420 fps, with perfect certainty, because as you remember... >From ship point of view time actually slow down by a factor of 7, their second to travel to earth sending out the 60 frames, actually just take 0,142857.... as measured from within ship. And you can not argue with that, you could argue with the distance to earth but that is a total other question. So from the ship point of view the framerate 60/0,142857 =420 fps for the actual *distance* regardless the length as measured from within ship. So you are actually wrong the framerate will be 420 fps from within ship frame. Because 60 fps leaves ship between point 300 000 km and 0 km and all of the frames will have travelled to earth within one second. So there is no doubt that during the *distance* no more than actually 420 fps leave the ship because that would increase the dilation by a ratio >7/1 It is easy to check just multiply the distance with 7 to see how many frames actually sent within one second from ship. (But then again within SR distances not always what they seem to be) Maybe you think about a stationary ship or a ship who does not suffer from doppler? If the framerate actually was 840 fps then the distance would be 14*300 000=4200000km for a light second, from within the planet frame of view which would not correspond to the contracted distance. 300 000/0,141 is plain wrong...... > > > 3. How long will it take until the front of first frame > > is received at planet.(planet point of view) > > How about the frame previous to the one that started emitting at > 300,000? The math is cleaner. The one that just finished at > 300,000 km arrived 1 second later. > > > 4. How long will it take until the ship pass planet. > >(planet point of view) > > 1 / 0.99 = 1.01 seconds > > > (*5*) How long was the time span between the > > first and last frame(planet point of view) > > From the *start* of the first frame... > > 0.01 seconds. > framerate: 60 / 0.01 = 6000 Hz > checking to see if we get ship's gamma: > 6000 / 7 = 857 Hz ... the difference between this and 840 Hz is > roundoff error. > > No one expects you to *like* relativity. But you will find that > it is self-consistent. > > David A. Smith |