DanMP

This is good, because I expected a smaller x Can you calculate the correct value?

True, but in your equation (#27), the angular speed, [math]\omega[/math], seemed to be of a test clock orbiting the Earth, which is not the case. Anyway, I finally got the time to do & now to write the math: Your (corrected) equation is: [latex]\frac{f®}{f(x)}=\sqrt{\frac{1+2\Phi(x)/c^2-((d-x) \omega/c)^2}{1+2\Phi®/c^2-((d-R) \omega/c)^2}}[/latex] where: - d is the distance between the Sun and the Earth - x is the distance between the center of the Earth and the test clock on the Earth-Sun line - [latex]\omega[/latex] is the angular speed of the Earth around the Sun - R is the Earth radius so [latex]f(x)=f®\sqrt{\frac{1+2\Phi®/c^2-((d-R) \omega/c)^2}{1+2\Phi(x)/c^2-((d-x) \omega/c)^2}}[/latex] and we need to find the point where [latex]f'(x)=0[/latex] This leads to: [latex]\frac {2\Phi'(x)}{c^2}+\frac {2(d-x)\omega^2}{c^2}=0[/latex] (1) As in #24: [math]\Phi(x)=-\frac{GM}{d-x}-\frac {Gm}x[/math] and [math]\Phi'(x)=-\frac{GM}{(d-x)^2}+\frac {Gm}{x^2}[/math] So (1) become: [math]-\frac{GM}{(d-x)^2}+\frac {Gm}{x^2}+(d-x)\omega^2=0[/math] or [math]\frac{GM}{(d-x)^2}=\frac {Gm}{x^2}+(d-x)\omega^2[/math] Meaning that the point where a test clock going from the Earth's surface towards the Sun would switch from increasing its tick rate to decrease it again is exactly where the gravitational pull of the Sun on the clock is cancelled by the gravitational pull of the Earth and the centrifugal force (in the above equation the mass of the clock is 1 kg). This happens at x=1482493941.5 m (approx. 1,5 million km), probably on the Hill sphere, as I kind of suggested in #24. Interesting. This can and should be tested using 2 atomic clocks linked with fibre optic cable, the first pulling the other with a few km long cable, spiraling upwards from the Earth.

Then the equation should be: [latex]\frac{f®}{f(x)}=\sqrt{\frac{1+2\Phi(x)/c^2-((d-x) \omega/c)^2}{1+2\Phi®/c^2-((d-R) \omega/c)^2}}[/latex] where: - d is the distance between the Sun and the Earth - x is the distance between the center of the Earth and the test clock on the Earth-Sun line - [latex]\omega[/latex] is the angular speed of the Earth around the Sun

Why? And what is R?

Yes, they do cancel. Anyway, thank you very much for your replies.

Ok, thank you very much. I think I got it. Like here we can express the gravitational potential at the point of interest as: [math]V(x)=-\frac{GM}{R-x}-\frac {Gm}x[/math] where x is the distance from the Earth to the point of interest and R is the Sun-Earth distance. [math]\frac {d V(x)}{dx}=-\frac{GM}{(R-x)^2}+\frac {Gm}{x^2}[/math] The point of interest is where the above is 0, so: [math]\frac{M}{(R-x)^2}=\frac {m}{x^2}[/math] and [math]x=\frac{R\sqrt{\frac m M}}{1+\sqrt{\frac m M}}= 258,775 km[/math] This is the point where Sun/Earth gravitational forces cancel each other. And if we include the orbital rotation the cancellation of forces takes place much higher, on the Hill sphere. So where is the real point in which a clock going towards the Sun would reverse its ticking rate? Do we have some experimental data?

This value (about 252,850 km above Earth's surface) looks much better Tank you! Still, I don't quite understand your calculation. Please be kind and explain it again, beginning with the origin of this: [latex]\frac{f®}{f(x)}=\sqrt{\frac{\Phi(x)}{\Phi®}}[/latex]

I don't think that there is covered the case of a unit mass located between 2 massive objects ... Also there is no mention of centrifugal forces. More of it, you did't use the gravitational time dilation equation + speed related time dilation (due to orbital speeds) ... Without them one can propose the solution as being on the Earth's Hill sphere, between Sun and Earth ...

Please post a link to one of them.

I don't think that your result is corect, because gravitational potential is defined: and on the Earth - Sun line the work that needs to be done by an external agent to move the unit mass is affected by both the Sun and the Earth ... and by the centrifugal force ... Also, it's not clear (at least for me) why/how did you reach the conclusion that the point I'm looking for is exactly there. The equation for gravitational time dilation is: Maybe we should use it ... The point where the clock would stop increasing its tick rate and then start to decrease again. Yes, remaining on the Earth-Sun line. I don't know how to solve the problem. That's why I asked here if there is a proper calculation/solution.

Yes, and we see/measure the slowing of time in a very rapid moving rocket from our frame of reference also. In the rocket's frame time is running normal. So, in a modern view how we explain the propagation of light / photon travel in vacuum?

We know (measured) that a clock at sea level "ticks"slower than a clock on a mountain. As we get higher, the clock runs even faster. Now, if we go much higher, towards the Sun, the clock should begin, at some point, to run slower, influenced by Sun's gravitation. Where is that "point", at what distance from Earth? It was calculated and/or tested experimentally?

Ok, thank you

Right, so it is outside SR bounds. Ok, but then what theory can be applied in order to compare the time of the photon with our time? I remember reading that for a photon there is no time, no past, no future ... but if there is change, it should also be time.

If a rocket is traveling with almost the speed of light, the static observer "can" see that the time in that rocket is dilated, i.e. things are happening in a slower rate. Very very close to c, time dilation gets bigger and, from the static observer, nothing appear to happen in the rocket. It's like everything freezes. Well, photons are travelling with the speed c but they appear to exhibit/produce rapid oscillations/variations: a change in the electric field creates a changing magnetic field that in turn creates another electric field and so on. How is this possible? At that speed it should be no change at all, according to special relativity.

I thought you realized that it's no ether wind ... In a rigid assembly the shift is much smaller. Look at this redo: As I wrote above, general relativity can explain it. Maximum shifts occurred when one mirror was at its top position and the other at its lowest, 2 times per cycle, because they reverse position. Space-time is warped near Earth, so the upper and lower regions are different. Maybe Strange can explain better. I want to add that the refractive index may also play a role since light travel slower in denser air (below) ... They should try in vacuum or water. The time of travel is also affected by the gravitational time dilation (a clock above the orizontal beamsplitter ticks faster then a clock below). In the link I suggested: you can see the math showing that speed changing effects are much bigger than gravitational effects but still smaller then the measured values. I think that this problem (noon/midnight difference) can be solved by having 2 accurate atomic clocks, one at South pole and the other at North pole, during few weeks in mid summer/winter.

So you think that there is no speed related difference or that this difference was miscalculated? You are welcome. About vertical M&M, well, the flexing was addressed (read below the video link I offered): At the beginning I also thought that flexing was the "culprit", but then, the fact that maximum shift occurred when the beamsplitter was horizontal made me think that there is something real going on ... One beam is going up and the other down ... I'm sorry to disappoint you but this can be explained by general relativity. Near a massive object (Earth) the "space" is warped (see gravitational wells) and the "upward" path becomes smaller than "downward" path, although the assembly is rigid. Just think about clocks ticking at different paces at different altitudes. The daytime shift, about 0.5 fringes, may be a problem for Einstein's relativity (if confirmed). I think that it may have something to do with noon/midnight time difference discussed above. Madmac, I think that aether wind is nonsense, but I can and will offer you a better, intuitive, relativity, one that you'll like It will take few weeks/months (I have to rewrite it) but you will be pleased.

Alzetta (in 3.6 Clock bias on and near Earth - pg 35) wrote: So it seems that speed difference does not compensate the gravitational difference. And, if I get it right, in order to have the measured (it was repeated after Hill?) difference you have to be on Equator and the spin axle to be vertical (not tilted) ... It may be a problem ... Timekeeping and the speed of light - new insights from pulsar observations was published by C.M. Hill in 02/1995. And 300 ps is 3 x 10-10s. I wonder if there are more recent measurements. About Cahill: On the same page there is an interesting video with M&M vertical exp. (here are more info), with moving fringes ... Can you explain it?

Ok, I agree, but there is or there is not a midnight/noon difference? It should be?

I found the articles: Clocks and the Equivalence Principle by Ronald R. Hatch (you may download the PDF): + Report on the “Clocks and the Equivalence Principle” by Ronald R. Hatch by ALZETTA Jonathan + Those scandalous clocks by Ronald R. Hatch I didn't understand why it is a "missing effect" if "it is found that earth-based clocks actually do run at different rates at midnight and noon" ...

If particles are "field excitations", what (and how) is creating those excitations? Bogie, your ideea is interesting. I have a somehow similar one, but I will not present it now because I have to post at least 2 other theories/ideas first (the background).

My urologist told me something similar. He said that I should have sex at least 5 times per week, for a healthy prostate, etc.. When I asked why, he said that in old times part of the treatment for prostatitis was the massage, performed similar with the rectal exam ... Of course it was at least awkward ... Through statistics they found that men with more than 5 ejaculations per week enjoyed a lower risk of prostate cancer, so, he said, probably the ejaculation is similar to a prostate massage. BUT, if this is true, than mountain-biking should be also good/healthy for the prostate, because off-road you get the ultimate prostate massage

so the "GW signal" is from the BBH as a whole. Where you, swansont, considered a Shapiro delay than? I remind you for the last time: I was talking about how gravitational information / pull is Shapiro delayed, creating gravitational fluctuations (waves) in LIGO observers: Why 100ms and not 300ms, or even 10s? Let's consider t = the time needed for gravitational information from BH1 and BH2, when side by side, to reach the Earth observer (EO), and T = the orbital period for the last orbit before BH1 and BH2 merged. 1. At aprox. t + T/4 , gravitational information from BH1 is received by EO, while the one from BH2 (behind BH1) is Shapiro delayed with more than T, so it's on its way towards EO. 2. At t +T/2 we receive gravitational information from BH1 and BH2 (again side by side). The one from BH2 in step 1 is still on its way. 3. At aprox. t + 3T/4 , gravitational information from BH2 (now in front of BH1) is received by EO, while the one from BH1 (now behind BH2) is Shapiro delayed with more than T. 4. At t + T the merge is complete and the gravitational information from BH2 in step 1 and BH1 in step 3 are still on their way to EO. Of course, things are not that simple (we still need GR for the full solution), but it is a good way to see how gravitational information (and gravitational pull) fluctuate at EO with 2 times the orbital frequency, due to Shapiro delay.

I sent you to the post on purpose, because I'm tired to write again and again the same things just because you (almost all of you) choose to ignore/forget them. And wouldn't it have been easier to just search "10 times" in the thread? I even mentioned wikipedia and dark mater and you answered: "Where is Wikipedia wrong? And I wouldn't tell them (whoever "they" are) I would correct it." ... And about supporting the claim, I also wrote: (this implies that DM is very important for GR, in any amount) and There you'll see that around BHs is always a lot of DM. So, if you really want answers from me, don't make me re-post again and again what I wrote, just because you are to lazy to search (or think?). Wait for the theory. I can understand and accept hostility, as long as you are correct, but here you are not. "Shapiro (or Shapiro-like) delay of GW signals" implies that it's about the delay of the GW signal from BBH by a third massive object. I never discussed such an idea because is irelevant. If you add incorrectness to hostility and laziness I may decide to end my dialog with you.

I was talking about how gravitational information / pull is Shapiro delayed, creating gravitational fluctuations (waves) in LIGO observers: Why 100ms and not 300ms, or even 10s? Let's consider t = the time needed for gravitational information from BH1 and BH2, when side by side, to reach the Earth observer (EO), and T = the orbital period for the last orbit before BH1 and BH2 merged. 1. At aprox. t + T/4 , gravitational information from BH1 is received by EO, while the one from BH2 (behind BH1) is Shapiro delayed with more than T, so it's on its way towards EO. 2. At t +T/2 we receive gravitational information from BH1 and BH2 (again side by side). The one from BH2 in step 1 is still on its way. 3. At aprox. t + 3T/4 , gravitational information from BH2 (now in front of BH1) is received by EO, while the one from BH1 (now behind BH2) is Shapiro delayed with more than T. 4. At t + T the merge is complete and the gravitational information from BH2 in step 1 and BH1 in step 3 are still on their way to EO. Of course, things are not that simple (we still need GR for the full solution), but it is a good way to see how gravitational information (and gravitational pull) fluctuate at EO with 2 times the orbital frequency, due to Shapiro delay.