Hard SR questions?
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From: rotchm@gmail.com on 6 Aug 2006 10:32 <SNIP> All still confusing.... Do you mean: Ship (A) is at a distance of D=300 000 000 meters wrt E(arth) and aproaches E with speed v=0.99c. Ship continuoulsy sends out pulses (frames) at a rate of 60 fps (60Hz) wrt A [or is that wrt E?] towards E. 1- how much time does it take for A to reach E, wrt E? 2-how much time does it take A to reach E wrt A? 3- etc... etc... ?
From: jt64 on 6 Aug 2006 10:43 I just want to know when first and last frame reached planet. Then we know the "received frame rate" 60/t. This rate is doppler shifted, but it is a bit akward to find out by how much since the transmission from ship is dilated. rotchm(a)gmail.com skrev: > > 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. > > > > The transmission is such that 60 frames will be sent from ship (A) > > during the distance of 300 000 km relative earths inertial frame. > > Thats all not clear and confusing. Rephrase it all. But I will attempt > answer... > v=.99c,b=v/c, d=300 000 000m, N=#frames sent out in d. All this is wrt > Earth. > > > > Now my question. > > > > 1. How long time will it take to travel to planet using (ship point of > > view). > > Planet?, you mean earth?: > > (d/v)*sqrt(1-b2) > > > 2. What is the framerate for the transmission within the ship.(ship > > point of view) > > Unknown bc you have not specified a frame rate. You only specified a > number of frames in d meters, these frames have not been supposed to be > done at a constant rate. Did you mean 60 fps? or 1 frame at beginning > and 59 frames at end of trip? etc... > > > > 3. How long will it take until the front of first frame is received at > > planet.(planet point of view) > > Again, when was the first frame sent out? > > > 4. How long will it take until the ship pass planet.(planet point of > > view) > > t=d/v > > > (*5*) How long was the time span between the first and last > > frame(planet point of view) > > 60 frames in 300 000 000 meters, as posed in the problem, or 60 > frames in (d/v) seconds.
From: jt64 on 6 Aug 2006 10:57 rotchm(a)gmail.com skrev: > <SNIP> > > All still confusing.... > > Do you mean: > > Ship (A) is at a distance of D=300 000 000 meters wrt E(arth) and > aproaches E with speed v=0.99c. Ship continuoulsy sends out pulses > (frames) at a rate of 60 fps (60Hz) wrt A [or is that wrt E?] towards > E. No the ship do not send out pulses at 60fps (60Hz) wrt A, you do not know what the dilated framerate at ship is. You just know how the transmission is sent from within frame of E, you do not know what framerate the transmission is received at earth and not the framerate which it is sent from ship. You only know that ship travels 0.99c and that 60 fps is sent during (1 sec from ship) when (watched,measured) from earth. First bit sent at 300 000 km distance from earth and last bit received at 0.00...1 km distance from earth. It is a very profound question it does not tell the rate transmitted from within ship or rate received from within earth. Only that ship travels in 0.99c and that 60 fps is leaving ship for every 300 000 km as seen from earth. So i want to know the framerate the transmission works with in ship frame. And the framerate that is received in earth frame. *AND* the timespan between the "first and last frame" in earth frame.
From: rotchm@gmail.com on 6 Aug 2006 11:40 > No the ship do not send out pulses at 60fps (60Hz) wrt A, you do not > know what the dilated framerate at ship is. ??? > You just know how the transmission is sent from within frame of E, 'How' as in 'sent by a laser', as in 'by clapping hands'...? Your word 'how' here is confusing and missedused. >you > do not know what framerate the transmission is received at earth and > not the framerate which it is sent from ship. Ok, we do not know the (proper) frequency of the source and we (earth, E) do not know at what frequency we (earth) is receiving it...correct? > You only know that ship travels 0.99c and that 60 fps is sent during (1 > sec from ship) when (watched,measured) from earth. ????????? > First bit sent at 300 000 km distance from earth and last bit received > at 0.00...1 km distance from earth. Ok, that part, I comprehend but does not make sence for the question: *ALL* bits are received by earth at a distance of 0 km from earth!!! The first, as the last bit is received by E at E's location, at 0 km. > It is a very profound question it does not tell the rate transmitted > from within ship or rate received from within earth. > > Only that ship travels in 0.99c and that 60 fps is leaving ship for > every 300 000 km as seen from earth. > > So i want to know the framerate the transmission works with in ship > frame. > And the framerate that is received in earth frame. > *AND* the timespan between the "first and last frame" in earth frame. You seem to want to find the relativistic doppler formulae but you are setting up the problem in a much to confusing way...Do try again... L8r
From: "N:dlzc D:aol T:com (dlzc)" N: dlzc1 D:cox on 6 Aug 2006 13:15
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. > 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 |