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There is no two observers in the example in post 2, there is only one observer. You can stop negging my posts, if you do not understand the basics, just ask.

....because you get the same exact effect if you consider the earthling observer co-moving with the Earth , while the Earth is moving wrt. Andromeda. Sure , Andromeda's speed wrt the Earth is bigger than the pacing speed of the earthling, I already mentioned that but that is irrelevant. This is basic relativity....

You do understand that motion is relative, right? You do realize that earthling fixed on Earth moving wrt Andromeda is the same thing as earthling moving wrt. Earth fixed wrt Andromeda. The amusing thing is that you insist on the latter, when, in fact, Andromeda is moving wrt the Earth.

If you continue to take things straight, without making the effort of understanding, you will never learn. Your choice. In relativity , motion is ... relative.

You do understand that motion is relative, right? You do realize that earthling fixed on Earth moving wrt Andromeda is the same thing as earthling moving wrt. Earth fixed wrt Andromeda. The amusing thing is that you insist on the latter, when, in fact, Andromeda is moving wrt the Earth.

You can have one description or you can also go by the Wolfram description, they are both valid. I cited the Wolfram page and I explained the Wolfram page <shrug>

So is the scenario I pointed out from the Wolfram page in post 2: "Suppose an Earthling is slowly strolling in the direction of Andromeda. Then events on Andromeda that occur, in concept, simultaneously in the Earthling's frame of reference, depend rather sensitively on his or her walking speed. Roughly, an increase in walking speed of one foot per second corresponds to a simultaneous event occurring on Andromeda about an Earth day later! This can be deduced from the Lorentz-transformation equation , where is the time advance on Andromeda, which can be considered simultaneous with an event on Earth occurring at , most conveniently set equal to 0. Note that the observer on Earth can only "infer" what is happening simultaneously on Andromeda. He would not actually "see" what is occurring until 2.5 million years later." Once again, I do hope that you realize that motion is relative: earthling fixed on Earth moving wrt Andromeda is the same thing as earthling moving wrt. Earth fixed wrt Andromeda. The amusing thing is that you insist on the latter, when, in fact, Andromeda is moving wrt the Earth.

This is the problem with your insistence of "plugging in numbers" instead of using symbolic calculations. I defined the variables in my post quite clearly, [math]v[/math] is the speed between the Earth and Andromeda (or any other galaxy). assuming that the galaxy moves away from the Earth, you get the same exact effect with the earthling fixed wrt. the Earth. Both scenarios are a direct consequence of the same Lorentz transform , [math]t'=\gamma(t+vx/c^2)[/math]. Like I said, you realize that motion is relative, right? The fact that Andromeda is moving towards the Earth (with a higher speed than "pacing") is also inconsequential, just replace it with a galaxy that is moving away from the Earth, you will get a much stronger effect than the one with the two earthlings as in the original paradox. Both scenarios have the same exact (simple) explanation. As long as I defined the variables correctly (and I did), the example stands. The OP question was "why does the distance play a role?". I explained that. This is clearly incorrect due to a basic mistake in your arithmetic, you forgot to square the speed of light: [math]t = \frac{0+(1.3)(2.3 \times 10^{22})/299792458^2}{\sqrt{1-1.3^2/(299792458)^2}}[/math] The result is very far off from 3.8 days, it is more like 1ms. Ain't arithmetic a bitch?

You realize that the effect is exactly the same, don't you?

...or the earthling is stationary on the Earth and Andromeda is moving away from the Earth. This is exactly what the Lorentz transform I showed reflects. I am quite sure that you know that motion is relative.

Here is a very good explanation as to why the effect depends on the distance to Andromeda. Basically, it all stems from the Lorentz transform for time: [math]t_a=\gamma(t_e+vx_e/c^2)[/math] Above, [math]t_e[/math] represents "when" the Andromeda effect happened as measured in the Earth frame and [math]x_e[/math] represents "where" the Andromeda effect happened as measured in the Earth frame. [math]t_a[/math] represents "when" the Andromeda effect happened as measured in the Andromeda frame. [math]v[/math] represents the relative speed between the Earth and Andromeda. In general, [math]vx_e<<c^2[/math] but, for cosmological distances, like in the case of Earth-Andromeda, [math]x_e[/math] becomes large enough such that [math]vx_e[/math] is comparable with [math]c^2[/math] such that the "advancement effect" [math]vx_e/c^2[/math] is no longer negligible. The reason that there was (and still is) so much controversy around the effect in the physics and philosophy circles is that the effect isn't measurable.

I already answered that, you need to go back and read the answer. Once again, SR forbids "accelerating observers", this is basic special relativity, only inertial frames (observers) are allowed. As to mistakes, I am quite sure Iggy realized by now the mistake in his two misguided attempts at showing that "equal" acceleration produces different effects on elapsed proper time (he is using the same scalar for acceleration but different profiles or different vectors, depending on the scenario). I have debunked both attempts and I will continue to debunk any further attempt because such attempts simply violate mainstream physics, as explained long ago by Markus, at the beginning of your thread. One day, you might understand it as well. I'd love to see your calculations for the above. Like I told you several times, physics isn't English composition, you need to be able to express your thoughts in a mathematical formalism. So, far,we have yet to see that from you, rest assured that I will take less than 30 minutes from your post to point out your mistakes. You can also stop the cowardly act of negging my posts trying to help you learn physics.

true

In SR objects cannot be reference frames, reference frames are geometric constructs. They have infinite extents in terms of space. An object at rest in a certain frame has all its points at rest in that frame. I hope this answers your questions (and that it corrects some of your misconceptions).

I already did, twice. I showed you that [math]\tau_B=\tau_C[/math] "Time dilation" is the incorrect term, we are discussing "elapsed proper time", please try to learn the correct terminology. Your own calculations contradict your claim: So, according to you, the elapsed time does not depend on acceleration, right? What does [math]"a"[/math] represent in the above formula? This is the third time I am asking you this question. Please answer. Excellent, so according to your own scenario: [math]\tau_A \ne \tau_B \ne \tau_C[/math]. Also according to you, [math]\tau_B,\tau_C[/math] depend clearly on the proper acceleration and on the fact that the acceleration profiles are different. So, congratulations! you just proved yourself wrong.

We've been over this already, you agreed to stop the sleigh of hand and apply the acceleration tangentially , remember? ...because you are using a sleigh of hand that I have already exposed as a cheap cheat. Several times. You are running around in circles, excuse the pun You did it obviously incorrectly , as already pointed out, you "forgot" that the time, as measured by A, needs to be the same for both B and C and that you needed to have the same acceleration profile. Using a different acceleration profile gives (predictably) different answers. I already anticipated that you'll try this sleight of hand in a prior post. Even with all this cheap cheats, the answers still depend on acceleration, so you fail to make any valid point. Now, try to do this honestly, by applying the acceleration profile of B to C's circular motion. What's the answer?

Do you have a reading difficulty ? I have explained to you (repeatedly) that the situation is the one described in your post 297, whereby observer C goes in a circle. I have also explained to you (repeatedly) that the speed profiles are the same for both B and C. Last time I explained that to you was right before yours. Yep, I explained this to you about 4 times as well, observer A measures B and C start together and come back together after elapsed time T on A's clock. Do you have so much difficulty following simple explanations? Besides, the guy "pacing" ( C ) must pace REPEATEDLY, until B returns, so your calculation, under your own scenario, is dead wrong. Finally, with all your mistakes piling up, what are you trying to prove? The results (if you eventually learn how to do the calculations correctly) depend on acceleration.

You have just reduced yourself to trolling my posts. The number is correct, I showed you the formalism, you know better than to tell me that I don't know. Congratulations for applying the formula correctly, now can you do the same for a circular trajectory? The C twin goes in a circle, following the same profile, so how much longer do you plan to keep up this charade? You already admitted that acceleration is important, so where do you plan to go with this game? So, according to you, [math]\tau_B[/math] does not depend on acceleration, right? Or does it? Or you can't make up your mind?

Not really, you seem to have difficulty in understanding the basic fact that [math]\tau_B=\tau_C=\int_0^T{\sqrt{1-v^2}dt}[/math] for identical [math]v[/math] profiles. Of course, you will attempt another sleigh of hand and say that you made the [math]v[/math] profiles to be different, right? Even if you do that, the answer is, contrary to your oft repeated claims, a function of acceleration and its duration. This is basic stuff, how much do you plan to keep up this charade? So, according to you, [math]\tau_B[/math] is not a function of the proper acceleration, [math]a[/math], right?

Then we are done For the second time, I already did it for you. Can you look up?

This is a simple "yes" or "no" question, so what is the answer? I already did it for you. Look up.

I really don't like your continuing intent on using numerical calculations, physicists use symbolic calculations. It is very simple , really: A watches B and C following the same profile, for time [math]T[/math] (accelerate, coast, decelerate with the same proper acceleration on each segment of the trip). B makes a linear trip while C runs in circles. (your scenario from #297) . Therefore [math]\tau_B=\tau_C=\int_0^T{\sqrt{1-v^2(t)}dt}[/math] Please stop evading and answer my question: Acceleration is: A. Important in the twins paradox B. Not important in the twins paradox.

Yep, the distance plays no role. Something else does, do you know how to do this calculation all by yourself? You still haven't answered my basic question: Acceleration is: A. Important in the twins paradox B. Not important in the twins paradox. Please do so. Yet, Minguzzi's answer is a direct function of proper acceleration. Fancy that! The explanation is that, in SR, "accelerated observer" and "acceleration" are not one and the same thing: "accelerated observers" are prohibited in SR, by definition, whereas SR deal with acceleration just fine. You can't perform physics by cherry picking quotes, you need to learn it. So far, you are just cherry picking quotes. You can stop the cowardly act of negging me, I am just trying to teach you physics.

This is false, I already debunked this claim. So did Markus. I did, I already pointed out your mistakes. Here it is again, for the fourth time: I have already proven you wrong on this subject, if the 2g acceleration is tangential , instead of your sleigh of hand attempt at making it normal to the direction of motion, the twins age, contrary to your incorrect claims, the same. The total elapsed times for the twins will be : [math]\Delta \tau1 = 2 T_c / \sqrt{ 1 + (a1 \ T_a/c)^2 } + 4 c / a1 \ \text{arsinh}( a1 \ T_a/c )[/math] [math]\Delta \tau2 = 2 T_c / \sqrt{ 1 + (a2 \ T_a/c)^2 } + 4 c / a2 \ \text{arsinh}( a2 \ T_a/c )[/math] Making [math]a1=a2=2g[/math] one gets: [math]\Delta \tau1=\Delta \tau2[/math] ---------------------------------------------------------------------------------------------------- *Vesselin Petkov, I know this guy, he is a well known crackpot, a not so subtle relativity denier. Congratulations, you found a really good reference!

Doesn't matter, you took the trouble to explain it in mathematical terms (albeit not very well) and I debunked it by pointing out your misconceptions. So, please answer this simple question: Acceleration is: A. Important in the twins paradox B. Not important in the twins paradox. Which one is it, Iggy? A or B? Give a straight answer, stop hiding behind inapplicable quotes. Is the claim in the title of this thread . I have already shown the fallacy in your above statement, in simple words , you are using a cheap trick, very easy to debunk, you try to apply one acceleration normal to the direction of motion while you are applying the other acceleration tangential to the motion, so their effects are obviously different. From the above sleigh of hand you declare that the effect "doesn't depend on acceleration. " I have already proven you wrong on this subject, if the 2g acceleration is tangential , instead of your sleigh of hand attempt at making it normal to the direction of motion, the twins age, contrary to your incorrect claims, the same. The total elapsed times for the twins will be : [math]\Delta \tau1 = 2 T_c / \sqrt{ 1 + (a1 \ T_a/c)^2 } + 4 c / a1 \ \text{arsinh}( a1 \ T_a/c )[/math] [math]\Delta \tau2 = 2 T_c / \sqrt{ 1 + (a2 \ T_a/c)^2 } + 4 c / a2 \ \text{arsinh}( a2 \ T_a/c )[/math] Making [math]a1=a2=2g[/math] one gets: [math]\Delta \tau1=\Delta \tau2[/math] Not when you try to calculate total elapsed time, see here (you can also check the references). Your error of thinking becomes even more obvious when the calculation is done from the perspective of an accelerated twin. See here. You know, earlier you have suggested in a very arrogant manner that I submit to peer reviewed journals my "findings". I did much more than that, I published not one but three papers on related subjects in mainstream physics journals (in addition to writing the two wiki paragraphs that I keep referring you to). So, perhaps you could drop the attitude and start really taking in the information that is being provided on the subject.