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No real scientist cares about what you consider a "valid" experiment.

Good luck with that.

So, what is the "eternal question"?

Assume the ridiculous, you get ridiculous conclusions. It is a process called GiGo. What is the "eternal question"?

Yes Light is Electro-Magnetic waves, as such, it Always travels at c, as evidenced by the E-M wave equation. No.

You cannot associate a frame of reference to photons. I am quite sure you have been told this before.

Yep

Yes, you are wrong. Two events that are simultaneous and co-located in one frame are simultaneous and co-located in ALL frames. This is why [math](x_0,t_0)=(0,0)[/math] makes the vents co-located in all frames. Proof: [math]t'=\gamma(V)(t-vx/c^2)[/math] Therefore the temporal separation between events, [math]\Delta t'[/math] satisfies: [math]\Delta t'=\gamma(V)(\Delta t-v \Delta x/c^2)[/math] If [math](\Delta t, \Delta x)=(0,0)[/math] then [math]\Delta t'=0[/math] in ALL frames.

Yes, you finally got it. Your question doesn't make sense, try re-posting it. What gives you this idea? If [math]x_0=0[/math] then basic algebra teaches you that [math]\gamma(V)x_0=0[/math].

I already explained, make [math]x_0=0[/math] in my formulas.Do you think that you can do that or I need to do it for you?

Not necessarily, I write the theory of my experiments before I perform the experiments. It can go either way. Bottom line, in both cases you need to be able to write down the math.

Right. And when you don't have the math tools, you can't claim to be doing physics.

Then you can make [math]x_0=0[/math] in my post. If you know how to do basic algebra you will find that the "photons are side by side" in ALL frames. I gave out the GENERAL solution, you can derive all the particular solutions from it. At one point, "Delta1212" was asking about photons that were separated in the x direction, so I posted the general solution. . What if the photons are not emitted simultaneously in frame S? Then , one needs to change the equations as follows: In a frame S, there are two photons being emitted simultaneously, [math]x_0[/math] apart in space and [math]t_0[/math] apart in timr. The equations of motion (not "events") are: [math]x_1=ct[/math] [math]x_2=x_0+c(t+t_0)[/math] Their separation is [math]x_2-x_1=x_0+ct_0[/math] and their relative speed is [math]\frac{dx_2}{dt}-\frac{dx_1}{dt}=0[/math]. In ANY other frame S' moving with speed V wrt S: [math]x'_1=\gamma(V)(x_1-Vt)[/math] [math]x'_2=\gamma(V)(x_2-Vt)[/math] The photon separation in frame S' is [math]x'_2-x'_1=\gamma(V)(x_0+ct_0)[/math] and their relative speed is [math]\frac{dx'_2}{dt}-\frac{dx'_1}{dt}=0[/math].

The problem with you two is that you write prose instead of math. The language of physics is math. Why don't you try to put the prose above into math?

Huh? There is no relative motion between the two particles . In any frame. There are no "events", just two particles with no relative motion. Since the language of physics is math, let me explain this to you mathematically: In a frame S, there are two photons being emitted simultaneously, [math]x_0[/math] apart. The equations of motion (not "events") are: [math]x_1=ct[/math] [math]x_2=x_0+ct[/math] Their separation is [math]x_2-x_1=x_0[/math] and their relative speed is [math]\frac{dx_2}{dt}-\frac{dx_1}{dt}=0[/math]. In ANY other frame S' moving with speed V wrt S: [math]x'_1=\gamma(V)(x_1-Vt)[/math] [math]x'_2=\gamma(V)(x_2-Vt)[/math] The photon separation in frame S' is [math]x'_2-x'_1=\gamma(V)x_0[/math] and their relative speed is [math]\frac{dx'_2}{dt}-\frac{dx'_1}{dt}=0[/math].

Any two particles that have a comoving frame, are comoving in all frames. This is the point that both of you are missing. The fact that the distance between the particles is different (due to length contraction) is utterly irrelevant.

You know the answer to this question: no. Yet, as I explained, RoS is irrelevant in this case. The relative speed of the photons is zero in all frames.

The speed of the two particles is 0 in their comoving frame. Their relative speed is zero. The speed of the two particles is V in any frame moving at speed V wrt their comoving frame. Their relative speed is....0! Nothing to do with RoS.

Any two particles that have a comoving frame, are comoving in all frames. The reason is simple, they have the same speed wrt of all frames. You are misapplying RoS.

Because this is what you have been doing <shrug> You mean that they share the same reference frame, you don't even have your basic terminology right. This doesn't mean anything about the existence of an "absolute" reference frame.

Correct Because it doesn't follow, the photons "move together" in ALL reference frames. Hence GiGo.

Actually, it is. From a wrong premise you can draw any conclusion. This is known under "GiGo" (garbage in, garbage out).

The correct statement is: IN VACUUM ONLY, photons move AT THE SAME SPEED © with respect TO ALL REFERENCE FRAMES. Now , Robin, what would that "absolute frame of reference" be?

Easy: [latex] \displaystyle d \tau_2= d \tau_1 \sqrt{\frac{1-r_s/r_2-\frac{(v_2/c)^2}{1-r_s/r_2}-(\frac{r_2 \omega_2}{c})^2}{1-r_s/r_1-\frac{(v_1/c)^2}{1-r_s/r_1}-(\frac{r_1 \omega_1}{c})^2}} [/latex] Integrate over one sidereal day the RHS and you get [math] \tau_2[/math]. The "gravitational deviation" is not 45 721.203ns. Find another hobby, "proving relativity wrong" has been a failure.

Yes, the radial motion is very small but it is still accounted for by the wonderful people who designed GPS. The adjustment is table driven based on the ephemeris.