SR/GR use absolute time to synchronize the GPS clocks with the ground clock.
in [Relativity]
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From: kenseto on 29 Mar 2010 13:59 On Mar 28, 7:10 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) wrote: > kenseto <kens...(a)erinet.com> writes: > >On Mar 28, 9:39 am, moro...(a)world.std.spaamtrap.com (Michael Moroney) > >wrote: > >> As I said, X is the conversion factor between absolute time and observed > >> time. If they're the same. X=1.0. > >No there is no conversion factor between absolute time and observed > >time. The A observer predicts that an interval of absolute time in his > >frame such as his clock second represented by a clock reading of (1/ > >gamma_ab second) on the B clock. > > These two sentences contradict each other. First, you say there is no > conversion factor, then in the very next sentence, you say the conversion > factor is (1/gamma_ab). I said that there is no constant X conversion factor as you asserted. A's conversion factor is 1/gamma_ab and B's conversion factor is 1/ gamma_ba. Also you seem to think that the conversion factor is converting clock time to absolute time. That is wrong....it is used to predict the clock reading on an observed clock for a specific interval of absolute time (such as a clock second) on the observer's clock. > > >> >If B is the observer he will say that his clock second represents a > >> >specific amount of absolute time. This amount of absolute time is > >> >predicted to have a clock reading of 1/2 second on the A clock. > >> >The rest of your post is due to your misundertanding of absolute time.. > > >> First, all that makes the "B" frame special, specifically absolute. > >No...that does not make the B frame absolute. It only says that the B > >second will contain a specific amount of absolute time. > > If the clock time of the A frame is half that of the B frame and the > conversion factor is also derived from the same factor 1/gamma_ab, it > has to be special. Well that's what SR says....it says that every frame is equivalent including the absolute rest frame and that's why every SR observer assumes that he is in a state of absolute rest. Also that's why every SR observer asserts that all the clocks moving wrt him are running slow. In IRT an IRT observer does not assume that he is in a state of absolute rest and that's why he says that a clock moving wrt him can run slow or fast compare to his clock. From that you can see that SR is a subset of IRT. > > >> SR says there are no absolute frames. =A0I also stated the problem > >> so that there were no "special" frames, not even accidentally. > >SR doesn't say no absolute frame. SR says that all frames are > >equivalent, including the absolute rest frame. > > Again, these two sentences contradict each other. Actually the second > sentence contradicts itself, the part before the comma (all frames are > equivalent, there's a "different" frame that's the absolute rest frame) There is no contracdiction....LET assumes the existence of an absolute rest frame and it uses that rest frame to derive the math. SR says that all frames are eqiuivalent so the SR observer call the inertial frame as the absolute rest frame and use it to derive its math. That's why SR and LET have the same math. > > > That's why every SR > >observer the absolute rest frame to do calculations. > > This sentence no verb. > > > THat's why every > >SR observer claims the exclusive properties of the absolute rest > >frame....that all the clocks moving wrt him are running slow and all > >th erulers moivng wrt him are contracted. > > You appear to have confused the phrase "absolute rest frame" and > "reference frame". No....only the absolute rest observer can claim that all the clocks moving wrt him are running slow. Both SR and LET claims the properties of the absolute rest frame to derive the math. > > Maybe, instead of "absolute time" you really mean "reference time" of > some sort? (yes, I know) No...absolute time is real and the rate of passage of absolute time is constant in all frames of reference. The purpose of SR/GR and IRT is to calculate the clock value on an observed clock for a specific interval of absolute time in the observer's clock. > > >> Second, if I do the exact same thing swapping A and B, I get the same > >> result, except frame "A" is special, and I do believe you'd say that > >> 1/2 clock second on the B clock represents the same amount of absolute > >> time as 1 second on the A clock. > >Here's your problem....A will measure B to have different velocity > >than B will measure A. Why? Because A's clock second contains a > >different amount of absolute time than B's clock second. That means > >that: 1/gamma_ab =/= 1/gamma_ba > > How could A and B see each other as having anything other than equal and > opposite velocities? Because 1 A sec. contain a different amount of absolute time than 1 B second. That means that the A will measure a different velocity for B and B will measure a different velocity for A. > > >> So.... if 1/2 second on the B clock is the same amount of absolute time > >> as 1 second on the A clock, > >Yes according to the SR observer A. > >> AND if 1/2 second on the A clock is the same > >> amount of absolute time as 1 second on the B clock, > >No B will measure A to have different velocity than .866 c and thus he > >predicts that B's clock second is repensented by (1/gamma_ba second) > >on the A clock. > > Well, I guess that would mean that if B saw A as having a conversion > factor of 1/gamma_ba, then A would have to see B as having a conversion > factor of gamma_ba, so that gamma_ab = 1/gamma_ba. NO, NO....B saw A as having a conversion factor of 1/gamma_ba and A saw B as having a conversion factor of 1/gamma_ab. >That's the only way > for there to be a consistent conversion from absolute time to observed > time in both frames. Sigh...you don't convert clock time to absolute time. The observer A clock second represents a specific amount of absolute time and this amount of absolute time will have a clock reading of 1/gamma_ab second on the B clock. > > Now tell me, what velocity would correspond to a gamma less than 1? Why says that gamma can be less than 1? > > Or, tell me, if B sees A as moving at 0.866 c, what velocity does A see > B have? A will have to measure B's velocity using his clock second. Ken Seto
From: BURT on 29 Mar 2010 15:50 On Mar 29, 10:59 am, kenseto <kens...(a)erinet.com> wrote: > On Mar 28, 7:10 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) > wrote: > > > kenseto <kens...(a)erinet.com> writes: > > >On Mar 28, 9:39 am, moro...(a)world.std.spaamtrap.com (Michael Moroney) > > >wrote: > > >> As I said, X is the conversion factor between absolute time and observed > > >> time. If they're the same. X=1.0. > > >No there is no conversion factor between absolute time and observed > > >time. The A observer predicts that an interval of absolute time in his > > >frame such as his clock second represented by a clock reading of (1/ > > >gamma_ab second) on the B clock. > > > These two sentences contradict each other. First, you say there is no > > conversion factor, then in the very next sentence, you say the conversion > > factor is (1/gamma_ab). > > I said that there is no constant X conversion factor as you asserted. > A's conversion factor is 1/gamma_ab and B's conversion factor is 1/ > gamma_ba. Also you seem to think that the conversion factor is > converting clock time to absolute time. That is wrong....it is used to > predict the clock reading on an observed clock for a specific interval > of absolute time (such as a clock second) on the observer's clock. > > > > > >> >If B is the observer he will say that his clock second represents a > > >> >specific amount of absolute time. This amount of absolute time is > > >> >predicted to have a clock reading of 1/2 second on the A clock. > > >> >The rest of your post is due to your misundertanding of absolute time. > > > >> First, all that makes the "B" frame special, specifically absolute. > > >No...that does not make the B frame absolute. It only says that the B > > >second will contain a specific amount of absolute time. > > > If the clock time of the A frame is half that of the B frame and the > > conversion factor is also derived from the same factor 1/gamma_ab, it > > has to be special. > > Well that's what SR says....it says that every frame is equivalent > including the absolute rest frame and that's why every SR observer > assumes that he is in a state of absolute rest. Also that's why every > SR observer asserts that all the clocks moving wrt him are running > slow. > In IRT an IRT observer does not assume that he is in a state of > absolute rest and that's why he says that a clock moving wrt him can > run slow or fast compare to his clock. From that you can see that SR > is a subset of IRT. > > > > > >> SR says there are no absolute frames. =A0I also stated the problem > > >> so that there were no "special" frames, not even accidentally. > > >SR doesn't say no absolute frame. SR says that all frames are > > >equivalent, including the absolute rest frame. > > > Again, these two sentences contradict each other. Actually the second > > sentence contradicts itself, the part before the comma (all frames are > > equivalent, there's a "different" frame that's the absolute rest frame) > > There is no contracdiction....LET assumes the existence of an absolute > rest frame and it uses that rest frame to derive the math. SR says > that all frames are eqiuivalent so the SR observer call the inertial > frame as the absolute rest frame and use it to derive its math. That's > why SR and LET have the same math. > > > > > > That's why every SR > > >observer the absolute rest frame to do calculations. > > > This sentence no verb. > > > > THat's why every > > >SR observer claims the exclusive properties of the absolute rest > > >frame....that all the clocks moving wrt him are running slow and all > > >th erulers moivng wrt him are contracted. > > > You appear to have confused the phrase "absolute rest frame" and > > "reference frame". > > No....only the absolute rest observer can claim that all the clocks > moving wrt him are running slow. Both SR and LET claims the properties > of the absolute rest frame to derive the math. > > > > > Maybe, instead of "absolute time" you really mean "reference time" of > > some sort? (yes, I know) > > No...absolute time is real and the rate of passage of absolute time is > constant in all frames of reference. The purpose of SR/GR and IRT is > to calculate the clock value on an observed clock for a specific > interval of absolute time in the observer's clock. > > > > > >> Second, if I do the exact same thing swapping A and B, I get the same > > >> result, except frame "A" is special, and I do believe you'd say that > > >> 1/2 clock second on the B clock represents the same amount of absolute > > >> time as 1 second on the A clock. > > >Here's your problem....A will measure B to have different velocity > > >than B will measure A. Why? Because A's clock second contains a > > >different amount of absolute time than B's clock second. That means > > >that: 1/gamma_ab =/= 1/gamma_ba > > > How could A and B see each other as having anything other than equal and > > opposite velocities? > > Because 1 A sec. contain a different amount of absolute time than 1 B > second. That means that the A will measure a different velocity for B > and B will measure a different velocity for A. > > > > > >> So.... if 1/2 second on the B clock is the same amount of absolute time > > >> as 1 second on the A clock, > > >Yes according to the SR observer A. > > >> AND if 1/2 second on the A clock is the same > > >> amount of absolute time as 1 second on the B clock, > > >No B will measure A to have different velocity than .866 c and thus he > > >predicts that B's clock second is repensented by (1/gamma_ba second) > > >on the A clock. > > > Well, I guess that would mean that if B saw A as having a conversion > > factor of 1/gamma_ba, then A would have to see B as having a conversion > > factor of gamma_ba, so that gamma_ab = 1/gamma_ba. > > NO, NO....B saw A as having a conversion factor of 1/gamma_ba and A > saw B as having a conversion factor of 1/gamma_ab. > > >That's the only way > > for there to be a consistent conversion from absolute time to observed > > time in both frames. > > Sigh...you don't convert clock time to absolute time. The observer A > clock second represents a specific amount of absolute time and this > amount of absolute time will have a clock reading of 1/gamma_ab second > on the B clock. > > > > > Now tell me, what velocity would correspond to a gamma less than 1? > > Why says that gamma can be less than 1? > > > > > Or, tell me, if B sees A as moving at 0.866 c, what velocity does A see > > B have? > > A will have to measure B's velocity using his clock second. > > Ken Seto When the train passes the station and sees the stations clock what is it doing? They both will not be able to see the others clock going slower. If that were the case they would be equal in there slowness. The train accelerated and slowed its time rate while the station did not. Mitch Raemsch
From: Michael Moroney on 29 Mar 2010 17:06 kenseto <kenseto(a)erinet.com> writes: >On Mar 28, 7:10 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) >wrote: >> >No there is no conversion factor between absolute time and observed >> >time. The A observer predicts that an interval of absolute time in his >> >frame such as his clock second represented by a clock reading of (1/ >> >gamma_ab second) on the B clock. >> >> These two sentences contradict each other. First, you say there is no >> conversion factor, then in the very next sentence, you say the conversion >> factor is (1/gamma_ab). >I said that there is no constant X conversion factor as you asserted. >A's conversion factor is 1/gamma_ab and B's conversion factor is 1/ >gamma_ba. OK so you think they can be different. I don't see how that could ever possibly be true. If A sees B as moving at velocity V, B seeing A as moving at any velocity other than -V would throw all physics as we know it out the window. Particle accelerators, radar, all kinds of things simply wouldn't work. Anyway, the Michelson-Morley Experiment and all its followups should have detected our velocity around the sun relative to this "absolute frame". > Also you seem to think that the conversion factor is >converting clock time to absolute time. That is wrong....it is used to >predict the clock reading on an observed clock for a specific interval >of absolute time (such as a clock second) on the observer's clock. Well, if I can find the clock reading by multiplying the absolute time by X (or 1/gamma_ab if you prefer, I can find the absolute time by dividing the clock reading by the same number. Algebra 101. >> >> >If B is the observer he will say that his clock second represents a >> >> >specific amount of absolute time. This amount of absolute time is >> >> >predicted to have a clock reading of 1/2 second on the A clock. >> >> >The rest of your post is due to your misundertanding of absolute time. >> >> >> First, all that makes the "B" frame special, specifically absolute. >> >No...that does not make the B frame absolute. It only says that the B >> >second will contain a specific amount of absolute time. >> >> If the clock time of the A frame is half that of the B frame and the >> conversion factor is also derived from the same factor 1/gamma_ab, it >> has to be special. >Well that's what SR says....it says that every frame is equivalent >including the absolute rest frame No it doesn't. In fact, Special RELATIVITY does not allow for any absolute rest frame whatsoever. That's why they used the word RELATIVITY! If you are going to modify or disprove SR, you're going to have to understand SR first. > and that's why every SR observer >assumes that he is in a state of absolute rest. No he doesn't. He picks a reference frame for the observer, usually one which simplifies the math. Besides, how can one pick the absolute frame? Isn't it, like, ABSOLUTE? > Also that's why every >SR observer asserts that all the clocks moving wrt him are running >slow. No, that's not true. SR asserts that an observer in any frame will see the clocks in objects moving relative to that frame as running slow. Back to my example: A sees his own clock as normal, but sees B's clock as running slow. B sees his own clock as normal, but sees A's clock as running slow. >In IRT an IRT observer does not assume that he is in a state of >absolute rest and that's why he says that a clock moving wrt him can >run slow or fast compare to his clock. When has any such thing ever been observed? >> > THat's why every >> >SR observer claims the exclusive properties of the absolute rest >> >frame....that all the clocks moving wrt him are running slow and all >> >th erulers moivng wrt him are contracted. >> >> You appear to have confused the phrase "absolute rest frame" and >> "reference frame". >No....only the absolute rest observer can claim that all the clocks >moving wrt him are running slow. Both SR and LET claims the properties >of the absolute rest frame to derive the math. OK, C is stationary in this absolute frame. C's clock runs at the same rate as absolute time. (correct?) A and B are moving in opposite directions at v, according to C. C sees A's and B's clocks running slow by gamma_ca and gamma_cb, which are equal. (Correct?) What velocity does A see B moving at? What rate does A see B's clock run? What velocity does A see C moving at? What rate does A see C's clock run? What velocity does B see A moving at? What rate does B see A's clock run? What velocity does B see C moving at? What rate does B see C's clock run? >>That's the only way >> for there to be a consistent conversion from absolute time to observed >> time in both frames. >Sigh...you don't convert clock time to absolute time. The observer A >clock second represents a specific amount of absolute time and this >amount of absolute time will have a clock reading of 1/gamma_ab second >on the B clock. Again, if I can convert from absolute time to clock time by multiplying by 1/gamma_ab, I can find absolute time from clock time by dividing by 1/gamma_ab. Simple algebra.
From: kenseto on 29 Mar 2010 17:53 On Mar 29, 5:06 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) wrote: > kenseto <kens...(a)erinet.com> writes: > >On Mar 28, 7:10 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) > >wrote: > >> >No there is no conversion factor between absolute time and observed > >> >time. The A observer predicts that an interval of absolute time in his > >> >frame such as his clock second represented by a clock reading of (1/ > >> >gamma_ab second) on the B clock. > > >> These two sentences contradict each other. First, you say there is no > >> conversion factor, then in the very next sentence, you say the conversion > >> factor is (1/gamma_ab). > >I said that there is no constant X conversion factor as you asserted. > >A's conversion factor is 1/gamma_ab and B's conversion factor is 1/ > >gamma_ba. > > OK so you think they can be different. I don't see how that could ever > possibly be true. If A sees B as moving at velocity V, B seeing A as > moving at any velocity other than -V would throw all physics as we know it > out the window. Particle accelerators, radar, all kinds of things simply > wouldn't work. Anyway, the Michelson-Morley Experiment and all its > followups should have detected our velocity around the sun relative > to this "absolute frame". No you are putting up bogus arguements. Everything will work as before. Each observer will measure V with his clock but the V_a is not the same as V_b because an A second is not equal to a B second. > > > Also you seem to think that the conversion factor is > >converting clock time to absolute time. That is wrong....it is used to > >predict the clock reading on an observed clock for a specific interval > >of absolute time (such as a clock second) on the observer's clock. > > Well, if I can find the clock reading by multiplying the absolute time > by X (or 1/gamma_ab if you prefer, I can find the absolute time by > dividing the clock reading by the same number. Algebra 101. No...you already know that the amount of absolute time involved is represented by 1 of A's clock second. That same amount of absolute time is represented by 1/gamma_ab second on the B clock. > > >> >> >If B is the observer he will say that his clock second represents a > >> >> >specific amount of absolute time. This amount of absolute time is > >> >> >predicted to have a clock reading of 1/2 second on the A clock. > >> >> >The rest of your post is due to your misundertanding of absolute time. > > >> >> First, all that makes the "B" frame special, specifically absolute. > >> >No...that does not make the B frame absolute. It only says that the B > >> >second will contain a specific amount of absolute time. > > >> If the clock time of the A frame is half that of the B frame and the > >> conversion factor is also derived from the same factor 1/gamma_ab, it > >> has to be special. > >Well that's what SR says....it says that every frame is equivalent > >including the absolute rest frame > > No it doesn't. In fact, Special RELATIVITY does not allow for any > absolute rest frame whatsoever. That's why they used the word > RELATIVITY! If you are going to modify or disprove SR, you're going > to have to understand SR first. If SR does not allow for any absolute rest frame then how come the SR observer claims the exclusive properties of the absolute rest frame? > > > and that's why every SR observer > >assumes that he is in a state of absolute rest. > > No he doesn't. He picks a reference frame for the observer, usually > one which simplifies the math. Besides, how can one pick the absolute > frame? Isn't it, like, ABSOLUTE? Yes he does...he selects the absolute frame to simplify the math....it allows the SR observer to claim that all the clocks moving wrt him are running slow. > > > Also that's why every > >SR observer asserts that all the clocks moving wrt him are running > >slow. > > No, that's not true. SR asserts that an observer in any frame will > see the clocks in objects moving relative to that frame as running > slow. Back to my example: A sees his own clock as normal, but sees > B's clock as running slow. B sees his own clock as normal, but sees > A's clock as running slow. That's exactly what I said: Every SR observer assumes that he is in a state of rest and the observed clock is doing the moving. > > >In IRT an IRT observer does not assume that he is in a state of > >absolute rest and that's why he says that a clock moving wrt him can > >run slow or fast compare to his clock. > > When has any such thing ever been observed? From the ground clock point of view: The SR effect on the GPS clock is 7 us/day running slow compared to the ground clock. From the GPS clock point of view: the SR effect on the ground clock is 7 us/day running fast compared to the GPS clock. > > >> > THat's why every > >> >SR observer claims the exclusive properties of the absolute rest > >> >frame....that all the clocks moving wrt him are running slow and all > >> >th erulers moivng wrt him are contracted. > > >> You appear to have confused the phrase "absolute rest frame" and > >> "reference frame". > >No....only the absolute rest observer can claim that all the clocks > >moving wrt him are running slow. Both SR and LET claims the properties > >of the absolute rest frame to derive the math. > > OK, C is stationary in this absolute frame. C's clock runs at the same > rate as absolute time. (correct?) This statement has no meaning. C's clock second will contain a specific amount of absolute time. This amount of absolute time is the least amount of absolute time compared to any clock second that is not in a state of absolute rest. > A and B are moving in opposite > directions at v, according to C. C sees A's and B's clocks running slow > by gamma_ca and gamma_cb, which are equal. (Correct?) No it depends on the value of gamma_ca and gamma_cb. > > What velocity does A see B moving at? A will have to use his clock second to measure the velocity of B. >What rate does A see B's clock > run? What velocity does A see C moving at? What rate does A see C's > clock run? According to SR A will see B's clock running at 1/gamma_ab. according to IRT A will see B's clock running at 1/gamma_ab or Gamma_ab > > What velocity does B see A moving at? What rate does B see A's clock > run? What velocity does B see C moving at? What rate does B see C's > clock run? According to SR B will see A's clock running at 1/gamma_ba. according to IRT B will see A's clock running at 1/gamma_ba or Gamma_ba n Seto> >>That's the only way > >> for there to be a consistent conversion from absolute time to observed > >> time in both frames. > >Sigh...you don't convert clock time to absolute time. The observer A > >clock second represents a specific amount of absolute time and this > >amount of absolute time will have a clock reading of 1/gamma_ab second > >on the B clock. > > Again, if I can convert from absolute time to clock time by multiplying > by 1/gamma_ab, I can find absolute time from clock time by dividing > by 1/gamma_ab. Simple algebra.
From: BURT on 29 Mar 2010 17:56
On Mar 29, 2:06 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) wrote: > kenseto <kens...(a)erinet.com> writes: > >On Mar 28, 7:10 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) > >wrote: > >> >No there is no conversion factor between absolute time and observed > >> >time. The A observer predicts that an interval of absolute time in his > >> >frame such as his clock second represented by a clock reading of (1/ > >> >gamma_ab second) on the B clock. > > >> These two sentences contradict each other. First, you say there is no > >> conversion factor, then in the very next sentence, you say the conversion > >> factor is (1/gamma_ab). > >I said that there is no constant X conversion factor as you asserted. > >A's conversion factor is 1/gamma_ab and B's conversion factor is 1/ > >gamma_ba. > > OK so you think they can be different. I don't see how that could ever > possibly be true. If A sees B as moving at velocity V, B seeing A as > moving at any velocity other than -V would throw all physics as we know it > out the window. Particle accelerators, radar, all kinds of things simply > wouldn't work. Anyway, the Michelson-Morley Experiment and all its > followups should have detected our velocity around the sun relative > to this "absolute frame". > > > Also you seem to think that the conversion factor is > >converting clock time to absolute time. That is wrong....it is used to > >predict the clock reading on an observed clock for a specific interval > >of absolute time (such as a clock second) on the observer's clock. > > Well, if I can find the clock reading by multiplying the absolute time > by X (or 1/gamma_ab if you prefer, I can find the absolute time by > dividing the clock reading by the same number. Algebra 101. > > >> >> >If B is the observer he will say that his clock second represents a > >> >> >specific amount of absolute time. This amount of absolute time is > >> >> >predicted to have a clock reading of 1/2 second on the A clock. > >> >> >The rest of your post is due to your misundertanding of absolute time. > > >> >> First, all that makes the "B" frame special, specifically absolute. > >> >No...that does not make the B frame absolute. It only says that the B > >> >second will contain a specific amount of absolute time. > > >> If the clock time of the A frame is half that of the B frame and the > >> conversion factor is also derived from the same factor 1/gamma_ab, it > >> has to be special. > >Well that's what SR says....it says that every frame is equivalent > >including the absolute rest frame > > No it doesn't. In fact, Special RELATIVITY does not allow for any > absolute rest frame whatsoever. That's why they used the word > RELATIVITY! If you are going to modify or disprove SR, you're going > to have to understand SR first. > > > and that's why every SR observer > >assumes that he is in a state of absolute rest. > > No he doesn't. He picks a reference frame for the observer, usually > one which simplifies the math. Besides, how can one pick the absolute > frame? Isn't it, like, ABSOLUTE? > > > Also that's why every > >SR observer asserts that all the clocks moving wrt him are running > >slow. > > No, that's not true. SR asserts that an observer in any frame will > see the clocks in objects moving relative to that frame as running > slow. Back to my example: A sees his own clock as normal, but sees > B's clock as running slow. B sees his own clock as normal, but sees > A's clock as running slow. > > >In IRT an IRT observer does not assume that he is in a state of > >absolute rest and that's why he says that a clock moving wrt him can > >run slow or fast compare to his clock. > > When has any such thing ever been observed? > > >> > THat's why every > >> >SR observer claims the exclusive properties of the absolute rest > >> >frame....that all the clocks moving wrt him are running slow and all > >> >th erulers moivng wrt him are contracted. > > >> You appear to have confused the phrase "absolute rest frame" and > >> "reference frame". > >No....only the absolute rest observer can claim that all the clocks > >moving wrt him are running slow. Both SR and LET claims the properties > >of the absolute rest frame to derive the math. > > OK, C is stationary in this absolute frame. C's clock runs at the same > rate as absolute time. (correct?) A and B are moving in opposite > directions at v, according to C. C sees A's and B's clocks running slow > by gamma_ca and gamma_cb, which are equal. (Correct?) > > What velocity does A see B moving at? What rate does A see B's clock > run? What velocity does A see C moving at? What rate does A see C's > clock run? > > What velocity does B see A moving at? What rate does B see A's clock > run? What velocity does B see C moving at? What rate does B see C's > clock run? > > >>That's the only way > >> for there to be a consistent conversion from absolute time to observed > >> time in both frames. > >Sigh...you don't convert clock time to absolute time. The observer A > >clock second represents a specific amount of absolute time and this > >amount of absolute time will have a clock reading of 1/gamma_ab second > >on the B clock. > > Again, if I can convert from absolute time to clock time by multiplying > by 1/gamma_ab, I can find absolute time from clock time by dividing > by 1/gamma_ab. Simple algebra. For how long does the stations clock run slow to the passing train? If the station ages more? Mitch Raemsch |