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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 ?

sure
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  • possible duplicate of Why do clocks measure arc-length? – ACuriousMind Dec 16 '14 at 13:36
  • this is not really a duplicate. I'm also wondering the relations between clock hypothesis and strong equivalence principle. – sure Dec 16 '14 at 13:45
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    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." – ACuriousMind Dec 16 '14 at 13:55
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    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. – sure Dec 16 '14 at 13:58
  • Neither the epicycle model and aether theories were axioms, they were the results of observations. Both were later disproven by experiments (something you can't actually do with an axiom). – Kyle Kanos Dec 16 '14 at 14:03
  • Both times, Occam's razor is the reason - both theories make superfluous assumptions (hundreds and hundreds of epicycles, or an undetectable aether). As long as you do not produce a theory that reproduces every GR result without the clock postulate, you should stick to using it, since GR is well-tested. As soon as you get a simpler theory without the clock postulate, you should stop believing it. (@KyleKanos: Lorentz aether theory is experimentally indistinguishable from SR) – ACuriousMind Dec 16 '14 at 14:05
  • You shouldn't believe anything in physics, but you should test everything experimentally. The far better question would be: What are the most precise tests for this hypothesis to date? – CuriousOne Dec 16 '14 at 14:11
  • Special relativity and even whole physics since Newton make the assumption that there exists an absolute inertial structure (or even space-time). A theory is not something you can prove or disprove experimentally, please don't be so ignorant about the deep philosophical problems involved in science. Invoking Occam's razor is just one argument that you might want to take into consideration, not necessarily something you should follow blindly. So if you guys believe that to try to understand why some axioms are deeper than the others is irrelevant because its all meaninless, just don't answer. – sure Dec 16 '14 at 14:35
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    @sure: Axioms are something the mathematicians use. Physicists use data. You are talking to the wrong people. Try the math exchange. – CuriousOne Dec 16 '14 at 15:22
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    "please don't be so ignorant" Heh, made me chuckle :) – Danu Dec 16 '14 at 15:29
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    This question appears to be off-topic because the OP evidently is interested in a debate, not an answer to a specific conceptional question. – Alfred Centauri Dec 16 '14 at 17:10

1 Answers1

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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.

John Rennie
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  • Also note that all the measurements of a Rindler observer can be correctly predicted using a purely inertial frame. – Hypnosifl Dec 16 '14 at 15:29
  • 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) – sure Dec 16 '14 at 16:29
  • @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. – John Rennie Dec 16 '14 at 16:33
  • 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 – sure Dec 16 '14 at 16:36
  • 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. – sure Dec 16 '14 at 16:43
  • @sure: you say In this sense, even if to accelerate does not act on space time, it acts on the physics observed and this I agree with. What you observe will indeed vary with your four-acceleration, just as it does for the Schwarzschild horizon. But I don't see why this bears upon the validity or otherwise of the clock postulate. – John Rennie Dec 16 '14 at 16:48
  • Because the physics observed in any linear uniformly accelerated frame is the same, you have here a whole different class of frame as inertial ones. I suspect that the uniformly rotating frames are also a different class of frames that is non equivalent to the two above. I would then believe that the physics perceived has to depends explicitely on the four-acceleration and maybe even its derivatives – sure Dec 16 '14 at 16:57
  • For example, suppose we're in an inertial frame and we observe an object with acceleration $a(t)$. Then, $a . \frac{da}{dt}$ is Galilean invariant in Newtonian mechanics. In particular, if its equal to $0$, then we can easily see that this acceleration is "circular" (or rotating). Acceleration would be linear if $a \wedge \frac{da}{dt} = 0$. Can we translate these conditions in SR ? (Clearly, if we go inside the tangent inertial frame, then the condition transposes to $a^\mu . \frac{da_\mu}{d \tau} = 0$ for the first one (which is indeed lorentz invariant). – sure Dec 16 '14 at 17:07
  • @sure: possibly, though this sounds like a different question. My point is that assuming the clock postulate is true you can handle acceleration in SR with only minor swearing and by doing so you would indeed predict different observables in different accelerating frames. So you can't turn the argument round and say that having different observables in different accelerating frames invalidates the clock postulate. – John Rennie Dec 16 '14 at 17:13
  • Yes I agree that clock postulate shows you that to accelerate is still non trivial. This can also be interpreted as "even if you make such strong assumption, acceleration changes whole physics" and in this sense, this questions the assumption you just made, don't you think so ? – sure Dec 16 '14 at 17:18
  • @sure: no I don't think so. If you could come up with an experimental (even only theoretical) result that differed from the predictions SR makes with the clock postulate you would have my attention. However I know of no such result. – John Rennie Dec 16 '14 at 17:48
  • http://arxiv.org/pdf/quant-ph/0407115v1.pdf see I.21, I'm not sure to understand this choice of metric yet but it clearly violate clock postulate (and clearly, I do agree with all the demonstration above) – sure Dec 16 '14 at 18:06
  • A paper that wasn't published anywhere and isn't cited anywhere? – John Rennie Dec 16 '14 at 18:11
  • feel free to look at E.R. Caianiello's original paper, [34] in the above article. – sure Dec 16 '14 at 18:15