|
|
Thread Tools | Display Modes |
#51
|
|||
|
|||
Why does SRT and LET have identical Math???
jem wrote: kenseto wrote: "Sam Wormley" wrote in message news:y4aoh.341693$1i1.296817@attbi_s72... kenseto wrote: There is no experimental support that B will predict that A's clock is running slow. Seto is wrong again! Clocks A and B are separated by some non zero distance. There is a non zero constant velocity between clocks A and B, such that |v| 0 and dv/dt = 0 . The observer in the frame of the clock (A) measures the clock (B) running slow (time dilation as modeled by SR). Empirically so! The observer in the frame of the clock (B) measures the clock (A) running slow (time dilation as modeled by SR). Empirically so! Hey idiot assertion is not emperical. GPS clock is B and ground clock is A: From A's point of view: The SR effect on B as predicted by A is 7 us/day running slow. From B's point of view: The SR effect on A as predicted by B is 7 us/day running fast. SR's /reciprocal/ time dilation effect applies only in the case of *Inertial* clocks. How many times have you been told this? 50? 100? How many more times will you have to be told? Since no frame on earth is inertial then SR must be wrong on earth. How many time you have been told that?? |
#52
|
|||
|
|||
Why does SRT and LET have identical Math???
kenseto wrote:
jem wrote: kenseto wrote: "Sam Wormley" wrote in message news:y4aoh.341693$1i1.296817@attbi_s72... kenseto wrote: There is no experimental support that B will predict that A's clock is running slow. Seto is wrong again! Clocks A and B are separated by some non zero distance. There is a non zero constant velocity between clocks A and B, such that |v| 0 and dv/dt = 0 . The observer in the frame of the clock (A) measures the clock (B) running slow (time dilation as modeled by SR). Empirically so! The observer in the frame of the clock (B) measures the clock (A) running slow (time dilation as modeled by SR). Empirically so! Hey idiot assertion is not emperical. GPS clock is B and ground clock is A: From A's point of view: The SR effect on B as predicted by A is 7 us/day running slow. From B's point of view: The SR effect on A as predicted by B is 7 us/day running fast. SR's /reciprocal/ time dilation effect applies only in the case of *Inertial* clocks. How many times have you been told this? 50? 100? How many more times will you have to be told? Since no frame on earth is inertial then SR must be wrong on earth. How many time you have been told that?? The theory itself (i.e. what it says) has nothing /whatsoever/ to do with how well the theory duplicates the measurements it's intended to predict. In SR, when a clock orbits an Inertial clock (an approximation to the GPS situation), time dilation is not reciprocal. So there's the first additional time you needed to be told. |
#53
|
|||
|
|||
Why does SRT and LET have identical Math???
"jem" wrote in message ... kenseto wrote: jem wrote: kenseto wrote: "Sam Wormley" wrote in message news:y4aoh.341693$1i1.296817@attbi_s72... kenseto wrote: There is no experimental support that B will predict that A's clock is running slow. Seto is wrong again! Clocks A and B are separated by some non zero distance. There is a non zero constant velocity between clocks A and B, such that |v| 0 and dv/dt = 0 . The observer in the frame of the clock (A) measures the clock (B) running slow (time dilation as modeled by SR). Empirically so! The observer in the frame of the clock (B) measures the clock (A) running slow (time dilation as modeled by SR). Empirically so! Hey idiot assertion is not emperical. GPS clock is B and ground clock is A: From A's point of view: The SR effect on B as predicted by A is 7 us/day running slow. From B's point of view: The SR effect on A as predicted by B is 7 us/day running fast. SR's /reciprocal/ time dilation effect applies only in the case of *Inertial* clocks. How many times have you been told this? 50? 100? How many more times will you have to be told? Since no frame on earth is inertial then SR must be wrong on earth. How many time you have been told that?? The theory itself (i.e. what it says) has nothing /whatsoever/ to do with how well the theory duplicates the measurements it's intended to predict. In SR, when a clock orbits an Inertial clock (an approximation to the GPS situation), time dilation is not reciprocal. But you said that the GPS situation is due to that the orbiting clock is not inertial. Since you now deny that so what is the cause that the SR time dilation effect on the GPS is not reciprocal??? Ken Seto |
#54
|
|||
|
|||
Why does SRT and LET have identical Math???
kenseto wrote:
"jem" wrote in message ... kenseto wrote: jem wrote: kenseto wrote: "Sam Wormley" wrote in message news:y4aoh.341693$1i1.296817@attbi_s72... kenseto wrote: There is no experimental support that B will predict that A's clock is running slow. Seto is wrong again! Clocks A and B are separated by some non zero distance. There is a non zero constant velocity between clocks A and B, such that |v| 0 and dv/dt = 0 . The observer in the frame of the clock (A) measures the clock (B) running slow (time dilation as modeled by SR). Empirically so! The observer in the frame of the clock (B) measures the clock (A) running slow (time dilation as modeled by SR). Empirically so! Hey idiot assertion is not emperical. GPS clock is B and ground clock is A: From A's point of view: The SR effect on B as predicted by A is 7 us/day running slow. From B's point of view: The SR effect on A as predicted by B is 7 us/day running fast. SR's /reciprocal/ time dilation effect applies only in the case of *Inertial* clocks. How many times have you been told this? 50? 100? How many more times will you have to be told? Since no frame on earth is inertial then SR must be wrong on earth. How many time you have been told that?? The theory itself (i.e. what it says) has nothing /whatsoever/ to do with how well the theory duplicates the measurements it's intended to predict. In SR, when a clock orbits an Inertial clock (an approximation to the GPS situation), time dilation is not reciprocal. But you said that the GPS situation is due to that the orbiting clock is not inertial. Since you now deny that so what is the cause that the SR time dilation effect on the GPS is not reciprocal??? In my example, the /orbiting/ clock is not Inertial - the /orbited/ clock is. |
#55
|
|||
|
|||
Why does SRT and LET have identical Math???
"jem" wrote in message ... kenseto wrote: "jem" wrote in message ... kenseto wrote: jem wrote: kenseto wrote: "Sam Wormley" wrote in message news:y4aoh.341693$1i1.296817@attbi_s72... kenseto wrote: There is no experimental support that B will predict that A's clock is running slow. Seto is wrong again! Clocks A and B are separated by some non zero distance. There is a non zero constant velocity between clocks A and B, such that |v| 0 and dv/dt = 0 . The observer in the frame of the clock (A) measures the clock (B) running slow (time dilation as modeled by SR). Empirically so! The observer in the frame of the clock (B) measures the clock (A) running slow (time dilation as modeled by SR). Empirically so! Hey idiot assertion is not emperical. GPS clock is B and ground clock is A: From A's point of view: The SR effect on B as predicted by A is 7 us/day running slow. From B's point of view: The SR effect on A as predicted by B is 7 us/day running fast. SR's /reciprocal/ time dilation effect applies only in the case of *Inertial* clocks. How many times have you been told this? 50? 100? How many more times will you have to be told? Since no frame on earth is inertial then SR must be wrong on earth. How many time you have been told that?? The theory itself (i.e. what it says) has nothing /whatsoever/ to do with how well the theory duplicates the measurements it's intended to predict. In SR, when a clock orbits an Inertial clock (an approximation to the GPS situation), time dilation is not reciprocal. But you said that the GPS situation is due to that the orbiting clock is not inertial. Since you now deny that so what is the cause that the SR time dilation effect on the GPS is not reciprocal??? In my example, the /orbiting/ clock is not Inertial - the /orbited/ clock is. ROTFLOL.....so you can assign *inertial* and *not inertial* to a clock to suit your theory? |
#56
|
|||
|
|||
Why does SRT and LET have identical Math???
In sci.physics.relativity, kenseto
wrote on Tue, 9 Jan 2007 06:56:34 -0500 : "The Ghost In The Machine" wrote in message ... In sci.physics.relativity, kenseto wrote on Mon, 8 Jan 2007 13:04:41 -0500 : "The Ghost In The Machine" wrote in message ... In sci.physics.relativity, kenseto wrote on Mon, 8 Jan 2007 11:19:30 -0500 : "Sam Wormley" wrote in message news:gatoh.342690$1i1.309253@attbi_s72... kenseto wrote: Hey idiot.....he selected the frame that sees all the clocks moving wrt the observer are running slow. That means that the observer is in a preferred frame. No, Seto, no "preferred" frames required. Special relativity predicts that clocks in relative motion wrt the observer will measure t' = t * gamma. This is routinely observed. ROTFLOL.....wormy doesn't even understand SR. In SR the observer will measure t'=t/gamma. You're both wrong. :-) SR predicts t' = sqrt(1-v/c)/sqrt(1+v/c), since both sides are using lightspeed Dopper measurements; there are no other methods of measurement. Sigh....you are incredibly stupid. Doppler measurement is not the rate of an observed clock. It is as far as the observer's concerned. How else is he supposed to know what the rate is, jump frames and whip out a slide rule? :-P Hey idiot you don't measure the rate of a moving clock. You use the SR time dilation to predict its rate. That's not a measurement; that's a prediction. -- #191, Does anyone else remember the 1802? -- Posted via a free Usenet account from http://www.teranews.com |
#57
|
|||
|
|||
Why does SRT and LET have identical Math???
"The Ghost In The Machine" wrote in message ... In sci.physics.relativity, kenseto wrote on Tue, 9 Jan 2007 06:56:34 -0500 : "The Ghost In The Machine" wrote in message ... In sci.physics.relativity, kenseto wrote on Mon, 8 Jan 2007 13:04:41 -0500 : "The Ghost In The Machine" wrote in message ... In sci.physics.relativity, kenseto wrote on Mon, 8 Jan 2007 11:19:30 -0500 : "Sam Wormley" wrote in message news:gatoh.342690$1i1.309253@attbi_s72... kenseto wrote: Hey idiot.....he selected the frame that sees all the clocks moving wrt the observer are running slow. That means that the observer is in a preferred frame. No, Seto, no "preferred" frames required. Special relativity predicts that clocks in relative motion wrt the observer will measure t' = t * gamma. This is routinely observed. ROTFLOL.....wormy doesn't even understand SR. In SR the observer will measure t'=t/gamma. You're both wrong. :-) SR predicts t' = sqrt(1-v/c)/sqrt(1+v/c), since both sides are using lightspeed Dopper measurements; there are no other methods of measurement. Sigh....you are incredibly stupid. Doppler measurement is not the rate of an observed clock. It is as far as the observer's concerned. How else is he supposed to know what the rate is, jump frames and whip out a slide rule? :-P Hey idiot you don't measure the rate of a moving clock. You use the SR time dilation to predict its rate. That's not a measurement; that's a prediction. Right.....there is no way to measure the actual rate of a moving clock. You can confirm your prediction after the clocks are reunited. Measuring the doppler shift is not the actual rate of the clock. Why? Because the rate of a moving clock doesn't change when it is in a state of uniform motion. |
#58
|
|||
|
|||
Why does SRT and LET have identical Math???
In sci.physics.relativity, kenseto
wrote on Thu, 11 Jan 2007 08:55:51 -0500 : "The Ghost In The Machine" wrote in message ... In sci.physics.relativity, kenseto wrote on Tue, 9 Jan 2007 06:56:34 -0500 : "The Ghost In The Machine" wrote in message ... In sci.physics.relativity, kenseto wrote on Mon, 8 Jan 2007 13:04:41 -0500 : "The Ghost In The Machine" wrote in message ... In sci.physics.relativity, kenseto wrote on Mon, 8 Jan 2007 11:19:30 -0500 : "Sam Wormley" wrote in message news:gatoh.342690$1i1.309253@attbi_s72... kenseto wrote: Hey idiot.....he selected the frame that sees all the clocks moving wrt the observer are running slow. That means that the observer is in a preferred frame. No, Seto, no "preferred" frames required. Special relativity predicts that clocks in relative motion wrt the observer will measure t' = t * gamma. This is routinely observed. ROTFLOL.....wormy doesn't even understand SR. In SR the observer will measure t'=t/gamma. You're both wrong. :-) SR predicts t' = sqrt(1-v/c)/sqrt(1+v/c), since both sides are using lightspeed Dopper measurements; there are no other methods of measurement. Sigh....you are incredibly stupid. Doppler measurement is not the rate of an observed clock. It is as far as the observer's concerned. How else is he supposed to know what the rate is, jump frames and whip out a slide rule? :-P Hey idiot you don't measure the rate of a moving clock. You use the SR time dilation to predict its rate. That's not a measurement; that's a prediction. Right.....there is no way to measure the actual rate of a moving clock. You can confirm your prediction after the clocks are reunited. Measuring the doppler shift is not the actual rate of the clock. Why? Because the rate of a moving clock doesn't change when it is in a state of uniform motion. As opposed to what, a state of non-motion? Be very precise here, and include the acceleration: [1] the clock as measured by someone by its elbow, as the clock accelerates from 0 m/s (relative to the "stationary" clock) to the final velocity). [2] the "stationary" clock measurements of the moving clocks' light signals, a combination of Doppler effect and SR gamma correction. [3] the calculated value of the moving clocks' time ticks, correcting for the Doppler effect. My computations suggest that [1] indicates "no change", [2] will indicate a clock speedup or slowdown because of the Doppler (though for most problems of this type it's a slowdown), and [3] will always indicate a slowdown in SR, if done properly. -- #191, fortune: not found -- Posted via a free Usenet account from http://www.teranews.com |
#59
|
|||
|
|||
Why does SRT and LET have identical Math???
"The Ghost In The Machine" wrote in message news In sci.physics.relativity, kenseto wrote on Thu, 11 Jan 2007 08:55:51 -0500 : "The Ghost In The Machine" wrote in message ... In sci.physics.relativity, kenseto wrote on Tue, 9 Jan 2007 06:56:34 -0500 : "The Ghost In The Machine" wrote in message ... In sci.physics.relativity, kenseto wrote on Mon, 8 Jan 2007 13:04:41 -0500 : "The Ghost In The Machine" wrote in message ... In sci.physics.relativity, kenseto wrote on Mon, 8 Jan 2007 11:19:30 -0500 : "Sam Wormley" wrote in message news:gatoh.342690$1i1.309253@attbi_s72... kenseto wrote: Hey idiot.....he selected the frame that sees all the clocks moving wrt the observer are running slow. That means that the observer is in a preferred frame. No, Seto, no "preferred" frames required. Special relativity predicts that clocks in relative motion wrt the observer will measure t' = t * gamma. This is routinely observed. ROTFLOL.....wormy doesn't even understand SR. In SR the observer will measure t'=t/gamma. You're both wrong. :-) SR predicts t' = sqrt(1-v/c)/sqrt(1+v/c), since both sides are using lightspeed Dopper measurements; there are no other methods of measurement. Sigh....you are incredibly stupid. Doppler measurement is not the rate of an observed clock. It is as far as the observer's concerned. How else is he supposed to know what the rate is, jump frames and whip out a slide rule? :-P Hey idiot you don't measure the rate of a moving clock. You use the SR time dilation to predict its rate. That's not a measurement; that's a prediction. Right.....there is no way to measure the actual rate of a moving clock. You can confirm your prediction after the clocks are reunited. Measuring the doppler shift is not the actual rate of the clock. Why? Because the rate of a moving clock doesn't change when it is in a state of uniform motion. As opposed to what, a state of non-motion? Be very precise here, and include the acceleration: [1] the clock as measured by someone by its elbow, as the clock accelerates from 0 m/s (relative to the "stationary" clock) to the final velocity). The rate of the accelerated clcok would run slower than the stay at home clock. [2] the "stationary" clock measurements of the moving clocks' light signals, a combination of Doppler effect and SR gamma correction. Doppler effect is not the rate of the observed clock. [3] the calculated value of the moving clocks' time ticks, correcting for the Doppler effect. The rate of an observed clcok does not change if it is in uniform motion (no acceleration). Ken Seto My computations suggest that [1] indicates "no change", [2] will indicate a clock speedup or slowdown because of the Doppler (though for most problems of this type it's a slowdown), and [3] will always indicate a slowdown in SR, if done properly. -- #191, fortune: not found -- Posted via a free Usenet account from http://www.teranews.com |
Thread Tools | |
Display Modes | |
|
|
Similar Threads | ||||
Thread | Thread Starter | Forum | Replies | Last Post |
Math for Astrophysics | Steve Willner | Research | 0 | November 7th 06 11:17 PM |
could anyone send me figures showing the mechanism of a identical docking system,thank u | [email protected] | Space Shuttle | 1 | August 4th 05 05:19 AM |
mystic math 2 | Ian Beardsley | Amateur Astronomy | 14 | July 9th 04 07:42 AM |
mystic math | Ian Beardsley | Amateur Astronomy | 34 | July 4th 04 02:43 AM |
For the math wizards here | Don | Misc | 0 | March 27th 04 05:11 AM |