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It is often assumed in special relativity that the rate of a clock in a non inertial frame does not depend on the proper acceleration of the observer. The point is, Rindler's observer shows us that the "action" of an accelerated observer on space-time is non trivial (there exists a black hole behind a uniformly accelerated observer). This means that there exists a way to discriminate accelerated observers from inertial ones. Moreover, Unruh's radiation explicitly depends on the proper acceleration $a$ of such observer (as a simple application of strong equivalence principle and hawking's radiation). This also shows that accelerating has a non trivial action on the whole physics observed. Proper acceleration being Lorentz invariant, it is also an absolute, so it is totally plausible that it changes whole physics and particularly clock.

Why should we believe in the clock hypothesis ? What would that implies with respect to strong equivalence principle and general relativity ?

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closed as off-topic by Alfred Centauri, Danu, Martin, JamalS, Rob Jeffries Dec 22 '14 at 11:27

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    $\begingroup$ It is not an exact duplicate - but note that the answer to that question essentially tells you that the clock postulate is independent from the rest of the axioms, hence it needs to be added as an axiom a priori. Asking why we should believe axioms has almost never a better answer than "Because they work, duh." $\endgroup$ – ACuriousMind Dec 16 '14 at 13:55
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    $\begingroup$ Then why don't you believe in epicycle model or aether theory ? It works too, so please, don't be so liberal about physics. Physics is about trying to make sense of things, that is to understand the world, not just describing it. $\endgroup$ – sure Dec 16 '14 at 13:58
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    $\begingroup$ @sure: Axioms are something the mathematicians use. Physicists use data. You are talking to the wrong people. Try the math exchange. $\endgroup$ – CuriousOne Dec 16 '14 at 15:22
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    $\begingroup$ "please don't be so ignorant" Heh, made me chuckle :) $\endgroup$ – Danu Dec 16 '14 at 15:29
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    $\begingroup$ This question appears to be off-topic because the OP evidently is interested in a debate, not an answer to a specific conceptional question. $\endgroup$ – Alfred Centauri Dec 16 '14 at 17:10
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Your starting point is incorrect. You say:

The point is, Rindler's observer shows us that the "action" of an accelerated observer on space-time is non trivial (there exists a black hole behind a uniformly accelerated observer).

You're correct that there is a singularity, but it is only a coordinate singularity. The Riemann tensor is everywhere zero in Rindler spacetime i.e. spacetime is flat. In fact the Rindler metric is easily shown to be the same as the Minkowski metric. So acceleration has no action on spacetime.

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  • $\begingroup$ Also note that all the measurements of a Rindler observer can be correctly predicted using a purely inertial frame. $\endgroup$ – Hypnosifl Dec 16 '14 at 15:29
  • $\begingroup$ How do you define the presence of a black hole then ? There's an horizon at least right ? Moreover, this horizon also radiates and there's a temperature coming from it (cf unruh's radiation) $\endgroup$ – sure Dec 16 '14 at 16:29
  • $\begingroup$ @sure: What we call the event horizon is a coordinate singularity, and is a result of the coordinate choice we've made i.e. the Schwarzschild coordinates. Jump into a black hole and you will encounter no horizon - in the coordinates of a freely falling observer there is no coordinate singularity. By contrast the singularity at the centre is a real singularity due to the curvature becoming infinite, and all observers in all coordinate systems will agree that it's there. $\endgroup$ – John Rennie Dec 16 '14 at 16:33
  • $\begingroup$ Yes I agree that the presence of an horizon is not frame invariant, but it is clear that the Rindler's observer will never see anyone crossing it, and so on, as for schwarzchild blackhole. The presence of Unruh's temperature/radiation is also frame dependant, but it exists for some classe of frame (the uniformly linearly accelerated at least) and yet its not just a choice of coordinate that implies that. In this sense, even if to accelerate does not act on space time, it acts on the physics observed $\endgroup$ – sure Dec 16 '14 at 16:36
  • $\begingroup$ All of this makes me think that to accelerate is not something you can forget at all in how you treat the physics. The presence of a horizon, while being relative to a class of frame, is not at all just "syntaxic singularity". I think that it would be a giant mistake to conclude that. $\endgroup$ – sure Dec 16 '14 at 16:43

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