Why speed of light is constant?

[russ_watters]

For further info on one-way tests, see
http://www.kevin.harkess.btinternet.co.uk/oneway/oneway.html

From your link:

Nobody has carried out this simple one-way light speed test using a laser and two clocks, as many consider it unnecessary.

Would a radio signal and two clocks work equally well?

Wisp, such an experiment is relatively easy to conduct. I suspect it hasn't been done simply because by the time a physics student reaches grad school (where they could do it for their thesis), they consider it unnecessary. That means something to me.

However, it doesn't stop an ether theorist from doing it on their own.

 

[geistkiesel]

Absolutely - every theory has very specific meanings for the terms. You can't just plug in random numbers pulled out of the air. Put quite simply, SR requires that the "frequency" you use fit the definition of "frequency." Only #4, which can't be measured because it doesn't exist. Using them where? In an existing theory? You can't arbitrarily change the definition of the variables in an equation to suit your wishes. You can't mix and match values from different reference frames.
Huh? Once again, you can't change reality simply by assuming its different from what is actually observed.

You keep ignoring this, but I'll say it again for the benefit of others: If you send an accurate clock into orbit, leaving its synchronized twin on the ground, and then bring them back together, they will no longer be synchronized. This has been verified experimentally. It is real, whether you want it to be or not.

Similarly, the speed of light has been measured thousands(millions?) of times and every time, the value measured is the same to within the limits of the accuracy of the measurement.

Al the thousands of experiments you referred to have some common elemnents, some explicit some implicit, some overt some "hidden". The story isn' even near the end of the book.

The direction suggested by grounded is going to open a lot of eyes and minds. I am much too humble a man to predict the extent of the progress, but just trying something new will always lead to progress.

I sure don't want dogmatic SR Theorists on my space drive design team.
What does your clock synchronization information have to do with the questions in the post?

 

[russ_watters]

Al the thousands of experiments you referred to have some common elemnents, some explicit some implicit, some overt some "hidden".[emphasis added]

"Elements?" You mean flaws: Provide specific examples or retract that. You do realize that you are claiming every experiment ever done that involves Relativity is flawed and every device that depends on it only works by random chance, right? Be careful around your computer: it may spontaneously combust at any time.

 

[swansont]

Does SR in any manner restrict the mode of frequency determination?

Absolutely - every theory has very specific meanings for the terms. You can't just plug in random numbers pulled out of the air. Put quite simply, SR requires that the "frequency" you use fit the definition of "frequency."

But, within that restriction, the theory does not tell you how to do the measurement

 

[swansont]

If you send an accurate clock into orbit, leaving its synchronized twin on the ground, and then bring them back together, they will no longer be synchronized. This has been verified experimentally. It is real, whether you want it to be or not.

Indeed. When the first GPS satellites were to be launched there were some who questioned whether relativity was real and would need to be accounted for. So they synchronized the clocks on the ground, without adjusting the output frequency of a synthesizer on the satellite, and launched it. Lo and behold, the frequency was off by the predicted amount (within measurement error). And so they turned the output frequency synthesizer on. More http://relativity.livingreviews.org/Articles/lrr-2003-1/node5.html .

 

[swansont]

What does your clock synchronization information have to do with the questions in the post?

Perhaps it was in response to

If the GPS system had been designed using the Grounded system relativistic effects would not exist and corrections would not be necessary.

 

[geistkiesel]

Does SR in any manner restrict the mode of frequency determination?



But, within that restriction, the theory does not tell you how to do the measurement

And this would include determining frequency as counting the full (wave length segments of the on rushing photon train passing through the eye)/sec.

 

[wespe]

And this would include determining frequency as counting the full (wave length segments of the on rushing photon train passing through the eye)/sec.

I guess you could see the waveform using an oscilloscope (if the frequency is in its range). When approaching a light source, the frequency would increase and wavelength would decrease, so that their product would remain constant. It should be possible to see this on the oscilloscope. Say for 100Mhz radio wave, wavelength is 3 meters, so it would have to be a big oscilloscope :)
Am I missing something?

 

[russ_watters]

But, within that restriction, the theory does not tell you how to do the measurement

And why should it? If it had to, every theory would have to contain a summary of every other theory ever presented. Like I said before: scientists assume in their theories that the people reading them understand the fundamentals well enough to be able to make use of the theory. Further, even if you don't agree that distance = speed times time (for example), you must stipulate to it for the purpose of the theory. Making up a new definition of speed would require its own theory. We've had other discussions about applying theories incorrectly or outside of their domain of applicability...

Quibbling over the definitions of things like frequency and speed might make for interesting philosophy, but in a scientific argument over the validity of a theory, its pretty unimpressive (not saying that's you).

Indeed. When the first GPS satellites were to be launched there were some who questioned whether relativity was real and would need to be accounted for. So they synchronized the clocks on the ground, without adjusting the output frequency of a synthesizer on the satellite, and launched it.

I heard it was a political decision, ie. some politician didn't believe in Relativity so he ordered that done. Any truth to that?

 

[geistkiesel]

And why should it? If it had to, every theory would have to contain a summary of every other theory ever presented. Like I said before: scientists assume in their theories that the people reading them understand the fundamentals well enough to be able to make use of the theory. Further, even if you don't agree that distance = speed times time (for example), you must stipulate to it for the purpose of the theory. Making up a new definition of speed would require its own theory. We've had other discussions about applying theories incorrectly or outside of their domain of applicability...

russ_waters, I thought you might appreciate a crack at this as you 'figured' out where I was going re Doc Al. This is he same train/station experiment:

All moving frame values are non-primed with the exception of M’, the consistent location of the observer O in the moving frame.

At no time is there an inference that M’ was at the midpoint of the A and B photons emitted in the stationary frame.

To demonstrate the following:

Einstein’s moving train calculation indicating when the oncoming B photon is detected at t1 the A photon was located at a position consistent with –t1. Said in other words, as t1 is determined from t0 which locates M’ at t0, the A and B were equidistant to M’(t0) when t = t1.

Proof:
A moving observer located at M’ on a moving frame passes through the midpoint M of photon sources located at A and B in the stationary frame just as A and B emit photons. M’ is moving along a line connecting A and B, toward B.

At this instant the moving source t = t0. Later the moving observer detects the photon from B at t1, and later the photon from A at t2. The observer has measured her velocity wrt the stationary frame as v. Determine the position of the A photon at tx in terms of t0, t1, t2, and v when the B photon was detected at t1.

The photon from A must reach the position of M’ when t = t2. Therefore, the distance traveled by the A photon during Δt = t2 – t1, is Δtc. This is equal to the distance cΔt = vΔt + vt1 + vtx . Now we rearrange somewhat to arrive at, vtx = vΔt – cΔt + –vt1. Now as vΔt - cΔt is just -vtx - vt1

vtx = -vtx - vt1 – vt1

2tx = -2t1

tx = -t1

Therefore, in the moving frame the photon from A and the photon from B were equidistant from M’(t0) at t1.

 

[geistkiesel]

einstein's little train that couldn't

Perhaps it was in response to

swansont: I though you would like to try this one on for size.
---------------------------------------------------------
All moving frame values are non-primed with the exception of M’, the consistent location of the observer O in the moving frame.

At no time is there an inference that M’ was at the midpoint of the A and B photons emitted in the stationary frame.

To demonstrate the following:

Einstein’s moving train calculation indicating when the oncoming B photon is detected at t1 the A photon was located at a position consistent with –t1. Said in other words, as t1 is determined from t0 which locates M’ at t0, the A and B were equidistant to M’(t0) when t = t1.

Proof:
A moving observer located at M’ on a moving frame passes through the midpoint M of photon sources located at A and B in the stationary frame just as A and B emit photons. M’ is moving along a line connecting A and B, toward B.

At this instant the moving source t = t0. Later the moving observer detects the photon from B at t1, and later the photon from A at t2. The observer has measured her velocity wrt the stationary frame as v. Determine the position of the A photon at tx in terms of t0, t1, t2, and v when the B photon was detected at t1.

The photon from A must reach the position of M’ when t = t2. Therefore, the distance traveled by the A photon during Δt = t2 – t1, is Δtc. This is equal to the distance cΔt = vΔt + vt1 + vtx . Now we rearrange somewhat to arrive at, vtx = vΔt – cΔt + –vt1. Now as vΔt - cΔt is just -vtx - vt1

vtx = -vtx - vt1 – vt1

2tx = -2t1

tx = -t1

Therefore, in the moving frame the photon from A and the photon from B were equidistant from M’(t0) at t1.