At first it looked like you were up-to-speed on the established body of work, and were simply finding a discrepancy in one aspect. But you've tipped your hand, and admitted that your elementary school education is preventing you from exploring further than a Newtonian classical world. Instead of learning, you simply deny everything you don't understand, down to and including relative velocity. We can help people who are willing to learn, but we cannot do anything about the willfully ignorant. This has stopped being a discussion.
For this now I need to calculate the amount of time dilation the satellite clock time will dilate compared to the earth sea level clock in this theory and see if it matches current values of compensation/dilation. Lets calculate at only c then, as you suggest is correct. Photon a travels 1m + 0.000767188m = 1.000767188m Photon b travels 1m - 0.000767188m = 0.999232812m Photons a speed is c 299792458m Photons b speed is c 299792458m Photon a travels the distance 1.000767188m / 299792458 = 3.338200015692189e-9ns Photon b travels the distance 0.999232812m / 299792458 = 3.333081888270852e-9ns Which theory shows a greater drift the c or the c + v? And you going to say the unrealistic scenario of the following is correct, because its the only thing you can say in your defense. That the path length of the photon to the local observers frame doesn't change, but it changes for the remote observer. Photon a travels 1m Photon b travels 1m Photons a speed is c 299792458m Photons b speed is c 299792458m Photon a travels the distance 1 / 299792458 = 3.33564095198152e-9ns Photon b travels the distance 1 / 299792458 = 3.33564095198152e-9ns
That seems to be a common theme among our cranks, although in some, more common than others. Please Register or Log in to view the hidden image!
So in GR the local observer is going to see this. Photon a travels 1m Photon b travels 1m Photons a speed is c 299792458m Photons b speed is c 299792458m Photon a travels the distance 1 / 299792458 = 3.33564095198152e-9ns Photon b travels the distance 1 / 299792458 = 3.33564095198152e-9ns And the GR remote observer is going to see this. Photon a travels 1m + 0.000767188m = 1.000767188m Photon b travels 1m - 0.000767188m = 0.999232812m Photons a speed is c 299792458m Photons b speed is c 299792458m Photon a travels the distance 1.000767188m / 299792458 = 3.338200015692189e-9ns Photon b travels the distance 0.999232812m / 299792458 = 3.333081888270852e-9ns In GR the time changed because the path length increase for the remote observer only and it says their clock runs faster/slower. I dont think so, but let me explain why the faster clock records a different time. We calibrate the clocks at sea level and the one leaves on the eg.satellite and starts traveling at a greater velocity than the one left on Earth. The greater velocity of the satellite clock causes the path length of the satellite clock photons to change and the detector records the frequency changing. The recorded time between the two clocks desynchronises. The problem I find in GR is it assumes the remote and local observers see different things. To the c + v theory they see exactly the same thing only the detector on the faster moving device takes longer to record the frequency because of the increased path length due to greater velocity. I cant find the data I need to calculate the real satellite dilation. In other words both observers see this for earths (local) clock at c + v Photon a travels 1m + 0.000767188m = 1.000767188m Photon b travels 1m - 0.000767188m = 0.999232812m Photons a speed is c + 230000m = 300022458m Photons b speed is c - 230000m = 299562458m Photon a travels the distance 1.000767188m / 300022458m = 3.335640920587351e-9ns Photon b travels the distance 0.999232812m / 299562458m = 3.335640983423898e-9ns And both observers see this for the faster (remote) clock at c + v. This is an example and not real values... Increasing values buy 20% for the example. Photon a travels 1m + 0.0009206256m = 1.0009206256m Photon b travels 1m - 0.0009206256m = 0.9990793744m Photons a speed is c + 276000m = 300068458m Photons b speed is c - 276000m = 299516458m Photon a travels the distance 1.0009206256m / 300068458m = 3.335640914314293e-9ns Photon b travels the distance 0.9990793744m / 299516458m = 3.335640989718168e-9ns The difference in the counted time between the Earth at 230000m/s and faster remote clock at 276000m/s would be 1 sec on earth is 0.999847 sec on the faster moving clock. Is that close to relativity predicted values?
Where have you stated the speed of the satellite? Without knowing that, one cannot determine the dilation factor. The factor you list correlates with a relative speed of 3250 miles per second, which is about 420 times too fast for an orbiting satellite. Note that you are only considering SR, and not taking GR into account. GR is actually a laarger contributing factor to time dilation than SR.
This is correct. No that is incorrect. The remote viewer would say that both photon a and b traveled 1meter. Right in m/sec Since we now know they traveled the same distance this question is moot.
Just to be clear both photons would move 1 meter in 3.33564095198152e-9ns by the remote viewers clock because both photons have a speed of c.
That is for the local clock to complete the distance of 1 cycle for the remote observer. The speed is set to c = 299792458.
if photons only moved at c then the local and remote observer would see this \ instead of this | and this /, and the photon would miss the detector if the clock had great enough velocity and we would detect a drift. The only time you are ever going to see this \ is if the local observer is constantly accelerating when the photon leaves the detector. Which proves that when a photon is emitted it is no longer associated to the local frame of reference, because when the photon is in on route and we accelerate the photon doesn't inherit the new velocity.
Not completely sure I get your way of explaining this. But I think using you method this is how to explain it. The local observer sees this | and the remote viewer sees this /. By remote viewer I mean someone who is in a different inertial frame.
Source What what? You catch on quick, wow, Im so proud of you... if photons only moved at c then the local and remote observer would see this \ instead of this | and this /, and the photon would miss the detector if the clock had great enough velocity and we would detect a drift. if photons only moved at c then the local and remote observer would see this \ instead of this | and this /, and the photon would miss the detector if the clock had great enough velocity and we would detect a drift. if photons only moved at c then the local and remote observer would see this \ instead of this | and this /, and the photon would miss the detector if the clock had great enough velocity and we would detect a drift. if photons only moved at c then the local and remote observer would see this \ instead of this | and this /, and the photon would miss the detector if the clock had great enough velocity and we would detect a drift. did you really ask that? Please Register or Log in to view the hidden image! (c-v)+v ?
This is an unclear statement: That is for the local clock to complete the distance of 1 cycle for the remote observer. The speed is set to c = 299792458. No need to be a dick. Each one of those scenarios is wrong. The local observer sees this |and the remote observer sees this /. That of course assumes the local frame is moving to the right relative to the remote frame.
You are stuck in a 19th century classical Newtonian universe, where space and time are absolutes. All of your assumptions are more than a century out-of-date. False. Do some reading. False. Do some reading. False. Do some reading. False. Do some reading.
