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Natural Science Forum / Physics / General Physics / April 2007[GR] Which clock is ahead?
Precise clock A is fired from a cannon vertically upward. Another twin clock B remains at rest near the cannon. After clock A lands, which clock is ahead? Clocks are insensitive to acclerations. Air resistance is absent. I think this is the same as with the twins paradox. One twin remained at home, and the other traveled up and then down. Therefore the clock B must be older, that is B is ahead. But the answer key says otherwise. > Precise clock A is fired from a cannon vertically upward. > Another twin clock B remains at rest near the cannon. > After clock A lands, which clock is ahead? > Clocks are insensitive to acclerations. > Air resistance is absent. It depends on how fast the traveler is going and how far he went. > I think this is the same as with the twins paradox. It isn't. > One twin remained at home, and the other traveled up > and then down. Therefore the clock B must be older, > that is B is ahead. But the answer key says otherwise. It depends. > Precise clock A is fired from a cannon vertically upward. > Another twin clock B remains at rest near the cannon. [quoted text clipped - 4 lines] > and then down. Therefore the clock B must be older, > that is B is ahead. But the answer key says otherwise. The one that undergoes motion solely on its inertia and gravity is always ahead. That is: the one which is shot up in a cannon and falls back to the ground is ahead. That's A, not B. The difference is the same as it is where in 0-gravity A remains stationary and B undergoes 1-G acceleration moving at the same initial velocity A had been shot upward in the other version: moving away from A, continuously slowing down by 1-G, then moving back to A. >> Precise clock A is fired from a cannon vertically upward. >> Another twin clock B remains at rest near the cannon. [quoted text clipped - 7 lines] > The one that undergoes motion solely on its inertia and gravity is > always ahead. Not necessarily... gravitational potential is a player so speed and distance must be considered. > > The one that undergoes motion solely on its inertia and gravity is > > always ahead. > > Not necessarily... gravitational potential is a player so > speed and distance must be considered. Think action principle, Sam. The action integral of a particle is its proper time. It's optimized for inertial motions. This applies, through the principle of equivalence, to geodesic motions as well. >>> The one that undergoes motion solely on its inertia and gravity is >>> always ahead. [quoted text clipped - 4 lines] > proper time. It's optimized for inertial motions. This applies, > through the principle of equivalence, to geodesic motions as well. Thanks. >>> The one that undergoes motion solely on its inertia and gravity is >>> always ahead. [quoted text clipped - 4 lines] > proper time. It's optimized for inertial motions. This applies, > through the principle of equivalence, to geodesic motions as well. Thanks Rock! -Sam > Precise clock A is fired from a cannon vertically upward. > Another twin clock B remains at rest near the cannon. [quoted text clipped - 6 lines] > and then down. Therefore the clock B must be older, > that is B is ahead. But the answer key says otherwise. SR and GR carefully balance each other to give Newtonian answers. Only the arithmetic is wrong because computers do not know how to use a minus sign and Dorks can't count to 2. http://users.telenet.be/vdmoortel/dirk/Physics/TwinsEvents.html "We use 3 inertial reference frames" [because Dorks can't get the result they want in two]. "In neither of these frames any form of acceleration is felt" [neither one of all three]. "In order for the travelling twin to make HIS trip, SHE must be in frame S' while going away". "if T = 5 years and v = 0.8c, then the stay at home twin will have aged 10 years". "Your conclusion is dead wrong. Start over, but skip the first part and the conclusion." -- Dork Van de local village idiot. "You made a mistake" -- Dork Van de psychotic fumble mumbler. "The man is a malicious troll" - Dork Van de merde. "What is this? Some kind of quote of some post? An introduction to the sh.t you produce later on? sh.t that you expect someone will bother reading?" - Dork's response to Einstein's paper, quoted by Kellerher. Belgium is where the farts blow. ASSistant professor Paul B. Andersen, tusseladd: "That is, we can reverse the directions of the frames which is the same as interchanging the frames, which - as I have told you a LOT of times, OBVIOUSLY will lead to the transform: t = (tau-xi*v/c^2)/sqrt(1-v^2/c^2) x = (xi - v*tau)/sqrt(1-v^2/c^2) or: tau = (t+xv/c^2)/sqrt(1-v^2/c^2) xi = (x + vt)/sqrt(1-v^2/c^2)" Steve Lawrence (sal) the hypocrite: "In general I'd say there are three main reasons people ignore questions in this forum. 1) They accidentally overlook the question (not really ignoring it) 2) They think it's boring and don't want to discuss it 3) They can't answer it and don't want to admit that." Ref: news:pan.2007.01.14.14.05.37.850117@nospam.org. 4) They know the answer but don't want to answer it because that would be embarrassing to them. How far is it from A to A and how long will it take to get there? http://www.androcles01.pwp.blueyonder.co.uk/Analemmae/Analemmae.htm#JanPB >> Precise clock A is fired from a cannon vertically upward. >> Another twin clock B remains at rest near the cannon. [quoted text clipped - 60 lines] > > http://www.androcles01.pwp.blueyonder.co.uk/Analemmae/Analemmae.htm#JanPB Never mind. In Androcles math a train leaves London, straight up to Birmingham, returns to London, and then Androcles calculates the travel distance: London - London = zero. And the train did it in no time, traveltime (L to B) - traveltime (B to L ) = zero. And Androcles average speed was 0 / 0 w. > Precise clock A is fired from a cannon vertically upward. > Another twin clock B remains at rest near the cannon. [quoted text clipped - 6 lines] > and then down. Therefore the clock B must be older, > that is B is ahead. But the answer key says otherwise. Your problem is not the same as the twin paradox because gravity is involved. But you can make it into a twin paradox problem by applying the Equivalence Principle: The results would be the same if the experiment was performed in zero gravity in an elevator accelerating upward. In the equivalent elevator experiment, clock A is at rest or moving at constant velocity, and clock B is accelerating along with the floor of the elevator. -- Jim E. Black >> Precise clock A is fired from a cannon vertically upward. >> Another twin clock B remains at rest near the cannon. [quoted text clipped - 17 lines] > -- > Jim E. Black Why would anyone want to create a problem? > On Apr 17, 5:51 pm, sasha semenov <c...@yahoo.com> wrote: >> Precise clock A is fired from a cannon vertically upward. >> Another twin clock B remains at rest near the cannon. >> After clock A lands, which clock is ahead? >> Clocks are insensitive to acclerations. >> Air resistance is absent. Clock A is at all times above B, and is therefore always in a milder gravitational environment during which it will be gaining time vs B. Its velocity has nothing to do with it. JP >> I think this is the same as with the twins paradox. >> One twin remained at home, and the other traveled up [quoted text clipped - 11 lines] > -- > Jim E. Black Why would anyone want to create a problem? |
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