# light propagates at c + v?

Discussion in 'Alternative Theories' started by BdS, Nov 21, 2015.

1. ### DaveC426913Valued Senior Member

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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.

3. ### BdSRegistered Senior Member

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422
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

Last edited: Jan 23, 2016

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21,803
That seems to be a common theme among our cranks, although in some, more common than others.

7. ### BdSRegistered Senior Member

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422
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?

8. ### DaveC426913Valued Senior Member

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11,108
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.

Last edited: Feb 14, 2016
9. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

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10,738
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.

10. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

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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.

11. ### BdSRegistered Senior Member

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422
That is for the local clock to complete the distance of 1 cycle for the remote observer. The speed is set to c = 299792458.

Last edited: Feb 17, 2016
12. ### BdSRegistered Senior Member

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422
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.

Last edited: Feb 17, 2016

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10,738
What?

14. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

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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.

15. ### DaveC426913Valued Senior Member

Messages:
11,108
FOR THE LOVE OF ALL THAT IS HOLY - FALSE.

!&@#&* 271 posts of this $!&#$!*@

16. ### BdSRegistered Senior Member

Messages:
422
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.

(c-v)+v ?

Last edited: Feb 20, 2016
17. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

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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.

18. ### DaveC426913Valued Senior Member

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11,108
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.