Hard SR questions?
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From: *"N:dlzc D:aol T:com (dlzc)" N: dlzc1 D:cox on 13 Aug 2006 14:30 Dear jt64: <jt64(a)tele2.se> wrote in message news:1155492275.314190.173840(a)m73g2000cwd.googlegroups.com... > > 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... Bull. .... >> > (*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. Please note that 420 Hz is very far from 840 Hz, which calculation you accepted without comment. David A. Smith
From: jt64 on 13 Aug 2006 14:49 jt64(a)tele2.se skrev: > 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...... Well maybe i should explain a bit more careful the distance contraction will have a ration of 300 000/0,141 that mean that the ship will find the distance to be 2127659 km when travel in 0,99 c And as you see your proposed length is *ACTUALLY THE DOUBLE*, and that you see is just 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. > > > > > > > >Please note that 420 Hz is very far from 840 Hz, which > >calculation you accepted without comment. No i never did it is just plain wrong and i told you so it will be 420 Hz transmission from the inertial frame during a second. (According to relativity) and Lorentz it is easy to see when you check my figures for the actual length contraction above. Nothing else, it is just wishful thinking on your behalf. JT > > No one expects you to *like* relativity. But you will find that > > it is self-consistent. > > > > David A. Smith
From: jt64 on 13 Aug 2006 15:06 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. > > > 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. And then again 840/60=14 and 14*300 000=4200000km and yet 300 000/0,141=2127659km as expected by the contraction. Seem you somehow managed to *DOUBLE THE FREQUENCY* Kenny > > David A. Smith
From: *"N:dlzc D:aol T:com (dlzc)" N: dlzc1 D:cox on 13 Aug 2006 15:17 Dear jt64: <jt64(a)tele2.se> wrote in message news:1155494971.336801.33070(a)p79g2000cwp.googlegroups.com... .... > (According to relativity) and Lorentz it is easy to see > when you check my figures for the actual length > contraction above. Nothing else, it is just wishful > thinking on your behalf. Your figures are mutually inconsistent. Care to try for second prize? David A. Smith
From: jt64 on 13 Aug 2006 16:13
N:dlzc D:aol T:com (dlzc) skrev: > Dear jt64: > > <jt64(a)tele2.se> wrote in message > news:1155494971.336801.33070(a)p79g2000cwp.googlegroups.com... > ... > > (According to relativity) and Lorentz it is easy to see > > when you check my figures for the actual length > > contraction above. Nothing else, it is just wishful > > thinking on your behalf. > > Your figures are mutually inconsistent. > > Care to try for second prize? (*AS STATED IN THE SETUP*) Yes i will try during the "0,14285714285714285714285714285714 sec" *CALCULATED* that pass from within ship travelin "0,99c"*SETUP* and a distance of "300 000 km"*SETUP* (from start point until it reach earth) the ship deliver "60 frames"*SETUP*. IF THE DILATION HAVE A RATIO of roughly 7 to 1 at 0,99c then it follows that the ship will have a framerate of 60*7 if the measured 60 frames is delivered during "0,14285714285714285714285714285714 sec". This is not math it is basic understanding of relations a 8 year old kid would know that if you split an orange in 7 parts it takes seven parts to make an orange not 14........ You see 1 orange!=2 orange Try it, it's not that hard.... > > David A. Smith |