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Yes, it does. Google "photon orbits".

It is good to see that I managed to convince you.

Their proper distance wrt to what? Each other? It doesn't appear so since [math]x_i(\tau)=x_{i0}+\frac{c}{a_i^2}(ch \frac{a_i \tau}{c}-1)[/math] so [math]x_2-x_1 \ne x_3-x_2[/math] . can you explain what you mean? In the example I worked for you, the distance from the rocket center to the aft and fore point were equal. The other thing I'd like you to explain is how all your scenarios relate to the title of this thread. The coordinate - independent light speed is not affected by the gravitational field, neither is it affected by accelerated frames, so what is with all these scenarios?

The total elapsed proper time is given by the integral [math]\tau=\int_C{\sqrt{1-(v(t)/c)^2}dt}[/math]. The only thing you need to know is how does v vary as a function of the coordinate time t and the path C followed by the twin. Plug in these two and you will get the answer. Depending on how you make v to vary over time , you can get any result you want. While the result does not depend explicitly on the acceleration (t depends explicitly on v), it surely depends implicitly since v depends on the acceleration, as you found out in your exercise above. Now, if we remember that [math]v(t)=\frac{at}{\sqrt{1+(at/c)^2}}[/math] it follows immediately that [math]\tau=\frac{c}{a} arcsinh \frac{at}{c}[/math] The takeaway is that the total elapsed proper time, depends on the acceleration (profile).

Nope, it is a very simple scenario, two pulses of light is emitted from the center of an uniformly accelerated rocket , one towards the front and the other one towards the rear. Find the elapsed times when the two pulses hit the two rocket ends, as expressed in what the OP calls "the lab" (i.e. the inertial frame of the Earth). This was repeated quite clearly throughout the thread by the OP author and by me. The math I posted describes his scenario, did you read the whole set of exchanges? The discussion has nothing to do with rigidity.

In the opposite direction, the light front "chases" the front of the rocket that recedes from it, so the equation is: [math]\frac{c^2}{a}(\sqrt{1+(at/c)^2}-1) = ct+L/2[/math] Solving for t, you get that the fronts of light hit the front and the aft of the rocket at different times.

No, it doesn't. I showed you how to do the calculations symbolically. Try doing the same thing (please stop using tables and numbers) and you might find your error. This is not equivalent with sending a pulse from the center of one rocket towards its front/rear end. You are dealing with TWO accelerations, this is much more complicated than what you set to solve, not to mention that your equations do not reflect the above. Try solving the simple case (one rocket) first.

What "rear ship"? There is only one ship and you are sending a signal towards the front and one signal towards the rear of the same ship.

This is not quite right, the units don't match and the equation you ended up with is not what you want. The correct equation is: [math]\frac{c^2}{a}(\sqrt{1+(at/c)^2}-1) + ct=L/2[/math] The above has a very nice solution: [math]t=\frac{L}{2c} \frac{1+aL/(4c^2)}{1+aL/(2c^2)}[/math] So, the answer should be very , very close to [math]\frac{L}{2c}[/math], so your t=0.375 doesn't seem correct.

I make a habit of not following any numerical calculation (because it is not tractable). Can you put what you just said in the form of a symbolic calculation, please?

Yes, I showed you the calculations. I answered this question as well: we don't know until we calculate the travel times. You know the (unequal) travel times in the inertial (what you call "lab") frame. Using the hyperbolic motion formulas, you can calculate the travel times in the accelerated frame of the rocket.

There is no such thing as "light goes faster" (or "slower"). In the Shapiro delay , light speed is c, constant, do not perpetrate the incorrect claims Iggy has been making. Light follows geodesics, in the presence of gravitational bodies, such geodesics are not lines in the Euclidian sense. See this illustration. Mathematically: [math]c \Delta t_{12}=X_1+X_2+r_s ln \frac{X_1 X_2}{R^2}[/math] Note the "c" in the LHS? This denotes the (invariant) speed of light. See the extra term [math]r_s ln \frac{X_1 X_2}{R^2}[/math] in the RHS? This denotes the "deviation" of the light path (null geodesic) from an Euclidian (straight) line.

I do not know why you insist in posting your crank stuff, over and over. You need to understand the light speed is an invariant, it definitely isn't gx as you keep maintaining. You are utterly confused in your attempt at pushing fringe ideas. What you derived is nothing but the coordinate speed of light in an accelerated frame. As I explained earlier, the coordinate speed of light can be ANYTHING, it is a MEANINGLESS expression. BTW, your muddled derivation can be done CORRECTLY in just a few lines: -start with the Rindler metric [math]ds^2=(\frac{gx}{c^2})^2 (cdt)^2 -dx^2[/math] -light follows null geodesics, [math]ds=0[/math] so [math]\frac{dx}{dt}=\pm \frac{gx}{c}[/math]. -if you use the alternate form of the metric : [math]ds^2=(1+\frac{gx}{c^2})^2 (cdt)^2 -dx^2[/math] you will get a different value for the coordinate speed of light. [math]\frac{dx}{dt}=\pm c (1+ \frac{gx}{c^2})[/math]. Done. It shows clearly that what you get is the coordinate speed of light, totally useless. Mathematically much more rigorous and more general than what you did, physically, totally useless. Light speed is an invariant, c, you need to learn that.

In other words, you have no falsifiable predictions. This means you have no viable theory. Come back when you have a mathematical formalism, being an antirelativist is not reason enough to consider your prose as science.

Mainstream textbooks sharply disagree with the above. Circular motion is characterized by the velocity vector changing direction. Linear motion is exactly the opposite.

Not only that the units were wrong but also the relationship [math]c=xg[/math] is wrong as well. The speed of light is constant, c. [math]xg[/math] is not constant for the mere reason that x is variable. How about the other gross, error, the one about the ever increasing speed of light in the Shapiro delay? You never addressed your other error on the subject. The crackpottery posted insistently by Iggy is getting a little much. There are two natural frames in which to describe a uniformly accelerating system. They are the accelerating coordinates (i.e. the proper (rest frame) time [math]\tau[/math] and related spatial coordinate [math]X[/math] ) and the coordinates of one inertial frame that matches the accelerating frame at one instant (as represented by [math](x,t)[/math]). The transformation between the coordinate systems is given by: [math]x=-\frac{c^2}{g}+(X+\frac{c^2}{g})cosh(\frac{g \tau}{c})[/math] [math]ct=(X+\frac{c^2}{g})sinh(\frac{g \tau}{c})[/math] where g is the uniform 3-acceleration magnitude. At the pivot point [math]X=- \frac{c^2}{g}[/math] time must be frozen to zero, [math]ct=0[/math] and [math]x=-\frac{c^2}{g}=X[/math], (thus the name pivot point). This doesn't mean in any shape or form that [math]c=xg[/math] as Iggy keeps claiming. Light speed, contrary to Iggy's crank claims, does NOT VARY in relativity, no matter if the motion is accelerated or not, it is a constant. If it did, the whole theory would fall apart.

But this is not what you wrote, what you wrote is the nonsense [math]c=xg[/math]. Not to mention the wrong things you posted about Shapiro delay.

I am quite familiar, thank you. Trouble is, the errors that you posted have nothing to do with natural units, have to do with the straight nonsense of this caliber:

Couldn't possibly be correct since [math]xg[/math] is measured in [math]\frac{m^2}{s^2}[/math]. What you wrote is total nonsense. This is incorrect as well, Shapiro delay is due to the fact that the light path is "lengthened" by the amount [math]r_s ln \frac{X_1 X_2}{R^2}[/math] from the "straight line" [math]X_1+X_2[/math]. See Rindler, "Relativity Special , General and Cosmological" page 237, for example: [math]c \Delta t_{12}=X_1+X_2+r_s ln \frac{X_1 X_2}{R^2}[/math] Contrary to your fringe claim, the speed of light is NOT "slowed down", the trajectory follows a geodesic, so the distance traveled is larger than the Euclidian distance. See here , for a nice illustration of the effect.

The math that I wrote for you in the previous post can be used to calculate exactly what happens in the lab frame. You can use the equations for the times [math]t_{up}, t_{down}[/math] that I derived for you. Yes. You don't know that, don't be so sure, the elevator frame is accelerated, you would need to do a transformation from the Earth frame (inertial) into the elevator frame (accelerated) in order to find out the strike times. Do you know the equations for hyperbolic motion? Be careful, the elevator frame is not an inertial frame (it is accelerated wrt the inertial frame of the Earth). I recommend that you do all calculations in the inertial frame, as I did, it is much easier.

Plot [math]v(t)=\frac{at}{\sqrt{1+(at/c)^2}}[/math] as a function of [math]t[/math] for various values of the parameter [math]a[/math] (acceleration) on the same graph paper.

It doesn't, it goes faster. You need to replace the source going towards the EH with a source going away from the EH. You need to replace the observer at infinity with the one hovering just above the EH. In effect, you need to swap the positions of the observer and the light source and redo the math as I will do for you in the next section. Let's fix the ideas, using the Equivalence Principle we replace the gravitational body with an accelerated rocket, flying in an area devoid of any gravitational field. Put the light source on the "floor" and shine it towards the "ceiling". Light appears to traverse the length of the rocket in time [math]ct_{up}=L+0.5at^2_{up}[/math] because, as the light moves "up", the ceiling "runs away" by the distance [math]0.5at^2_{up}[/math]. So, [math]t_{up}=\frac{c-\sqrt{c^2-2aL}}{a}[/math] Now' let's swap the roles of the source and the observer, i.e. put the source on the "ceiling" and the observer on the "floor". The equation becomes: [math]ct_{down}+0.5at^2_{down}=L[/math] because the floor is accelerating towards the light pulse, so the distance L is covered from both ends, giving: [math]t_{down}=\frac{-c+\sqrt{c^2+2aL}}{a}[/math] An astute observer notices that [math]t_{down}<t_{up}[/math]. It is as if the light moves faster "down" the gravity well.

Check out the Pound-Rebka experiment.

Both analogies are wrong. The correct analogy is done in terms of energy not time-distance. The closer the light source is to the EH , the lower the energy of the emitted photons, so they "climb" the gravity well towards the distant observer "slower" (and slower).

Nope, it is [math]v=\frac{at}{\sqrt{1+(at/c)^2}}[/math]. You rolled multiple errors in one formula. Because the formula E2=m2c4+p2c2 is a trivial consequence of [math]E=\gamma mc^2[/math] and [math]p=\gamma mv[/math]. When v=0 [math]\gamma=1[/math] and [math]E=mc^2[/math].