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Bell's paradox has in the past been the topic of quite heated discussions. It is posed in the context of a silk thread connecting two identical rockets whose engines are ignited at the same instant. The paradox has revolved around whether or not the silk thread breaks as viewed by different observers. The answer is that the thread does break but that is does so for apparently different reasons according to different observers.

I am curious what happens if the two rocket engines are replaced by a uniform gravitational field and the rockets are in freefall. I assume that all the observers agree that the thread does not break? Both rockets are in the same inertial frame and cannot even recognise their common acceleration. A stationary observer will surely see the Lorentz contraction of the thread but I'm less sure about how the observer would measure the distance between the two rockets. Also, both rockets are falling into a gravitational well. Does gravitational time dilation play any role in the arguments?

[clarification: at a time instant, t=0, both rockets and the 'stationary' observer have zero relative velocity. The stationary observer is immersed in the same gravitational field as the rockets but remains stationary by virtue of a force that exactly counteracts the gravitational acceleration]

Roger Wood
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    Gravitational time dilation depends on the potential, so if the gravitational field is truly uniform then both rockets will experience the same dilation (and their clocks will remain in sync). Presumably the spatial analogue is why the thread will not break. But one might argue that a truly uniform gravitational field is unphysical. – Eric Smith Apr 28 '21 at 00:57
  • What do you mean by uniform gravitational field? In particular, are you thinking of Newton's model of gravity, or the curved-spacetime model of gravity? – Chiral Anomaly Apr 28 '21 at 01:35
  • @Chiral Anomaly uniform as, for example, far from a very massive black hole where both M and r^2 approach infinity but M/r^2 is finite and the length of the silk thread is finite. – Roger Wood Apr 28 '21 at 03:14
  • @EricSmith The gravitational field is uniform and the rockets and thread are oriented along the same direction as the field. If the rockets were stationary, the leading rocket would be at a lower potential and its clock would run more slowly. However, both rockets are in freefall cannot sense the gravitational field or potential. – Roger Wood Apr 28 '21 at 03:19
  • Okay, that's a good clarification. And by "far" (from a very massive black hole), do you mean that tidal effects are completely negligible? In other words, the length of the string is negligible compared to the characteristic scale over which spacetime curvature is noticeable? (Beware that this condition can be satisfied even on the event horizon of a large black hole, so "far" doesn't need to be far at all.) – Chiral Anomaly Apr 28 '21 at 05:30
  • @ChiralAnomaly Yes, we can consider the field constant in magnitude and direction over a volume that is large compared with the two rockets and the space that they might travel into during the experiment. – Roger Wood Apr 28 '21 at 06:07
  • If tidal effects are negligible, then so is the curvature of spacetime, so the spacetime is flat. That's where my comments were coming from: in the curved-spacetime model of gravity, a uniform gravitational field is physically equivalent to no gravitational field at all. This is the so-called equivalence principle. Since the tidal effects are not exactly zero in your scenario, but only very small, I suppose you could get cumulative long-term material effects like this answer describes, but I assume that's not what you're asking about. – Chiral Anomaly Apr 29 '21 at 03:18
  • @ChiralAnomaly (wrong link, maybe?) – Roger Wood Apr 29 '21 at 05:16
  • The link is correct, because it explains why some real materials could break after being subjected to a very small tension for a very long time. I assume that's not what you're asking about, though. – Chiral Anomaly Apr 29 '21 at 13:37
  • @ChiralAnomaly got it, thanks. I should clarify the question. The scenario I've posed seems clear to me. The thread doesn't break - neither for the rocketeers nor the stationary observer. Now I wonder why it breaks in Bell's paradox? The lack of simultaneity seems to be part of the argument. – Roger Wood Apr 29 '21 at 16:32
  • In the usual paradox, the rockets are accelerating in an absolute sense, meaning that objects sitting inside the rockets have nonzero weight. Their accelerations are equal and unchanging, meaning that a 1-kilogram mass always has the same weight in both rockets. This definition of acceleration doesn't rely on the rocket's motion relative to anything else. (By the way, if the rockets were both in free-fall, then their accelerations would both be zero in this absolute sense. That's the equivalence principle again.) (part 1/3) – Chiral Anomaly Apr 30 '21 at 02:03
  • (part 2/3) The fact that their accelerations are equal tempts us to think their velocities must also be equal at any given instant. But "instant" is observer-dependent. Suppose they both turn on their rocket engines at the same initial time from their own point of view. Then, at every instant thereafter according to an observer inside one of the rockets, the difference in the rockets' velocities is nonzero, and observers inside both rockets agree about the sign of the difference. This strongly suggests that the string must break, and with a little more effort we can turn it into a proof. – Chiral Anomaly Apr 30 '21 at 02:03
  • (part 3/3) Here's another approach: Say 3 people all start at rest. Two of them turn on equal rocket engines in equal directions. The 3rd person remains at rest. According to the 3rd person, at any given time, the rockets have equal speeds, so the string's length seems to remain the same. But the string's velocity is increasing, so if it really did have constant length (in the rest-frame of one of its ends), then it should keep getting shorter according to the 3rd observer (length contraction). It doesn't, so it must be getting stretched in the rest fram of either of its ends. $\to$ It breaks. – Chiral Anomaly Apr 30 '21 at 02:03
  • @Chiral Anomaly I think I understand the resolution of Bell's paradox. Under conditions of acceleration, the leading rocket sees the trailing rocket as suffering time dilation and thus accelerating more slowly, So the rockets drift apart. Conversely, the trailing rocket sees the opposite time dilation and sees the leading rocket drift forward. – Roger Wood Apr 30 '21 at 17:31
  • @ChiralAnomaly Now I'm suddenly wondering what is meant by a constant gravitational field? Is it one that produces a certain acceleration for the local observer or one that produces a certain acceleration for an independent observer at a fixed location? An acceleration measured deeper in the field would seem smaller to a distant observer due to gravitational time dilation. – Roger Wood Apr 30 '21 at 17:34
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    In Newton's model, a constant gravitational field is a force that accelerates everything equally ("accelerates" in the relative-motion sense). But that situation is locally indistinguishable from no acceleration at all, because things are weightless either way. Thus Newton's model is artificial: the formalism distinguishes between two situations that are physically indistinguishable. GR is more natural: the formalism of GR makes no distinction at all between uniform gravitational field and no gravitational field. In GR, spacetime can be flat (no tidal effects) or curved (tidal effects). – Chiral Anomaly May 01 '21 at 14:44
  • @ChiralAnomaly Yes, if you added a uniform gravitational field to the entire universe, nobody would notice. If you added two different values of uniform field to two different regions, they could tell the difference between those values, but still not the absolute values. – Roger Wood May 01 '21 at 19:41

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A falling rod length-contracts without any stress. When the coordinate speed of the rod is 0.86 * coordinate speed of a light pulse next to the rod, then the length of the rod is half of its rest-length.

Accelerating observer observing a rod, observes the rod length-contracting without any stress. When the coordinate speed of the rod is 0.86 * coordinate speed of a light pulse next to the rod, then the length of the rod is half of its rest-length.

Accelerating observer observing a fast moving rod, observes the rod length-contracting fast without any stress. The rear of the rod moves x percent faster than the front. The absolute speed difference is large when the speed of the rod is large.

Also an observer observing a rod moving fast downwards, observes the rod length-contracting fast without any stress. The rear of the rod moves x percent faster than the front. The absolute speed difference between the rear and the front is large when the speed of the rod is large.

stuffu
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  • thanks for the answer. I'm afraid I don't understand the answer or its correspondence with the question, but the statement about zero stress in the rod seems very relevant. Could you perhaps cast the answer in terms of three observers, one in each rocket and one that remains stationary at the position where the rockets start. – Roger Wood Apr 28 '21 at 06:57
  • @RogerWood Just replace "rod" with "fleet". Fleet of two rockets falling. And fleet of two rockets floating in space, observed by an accelerating observer. I try to make my post clearer later. – stuffu Apr 28 '21 at 07:07
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Let's say we "drop" two electrons in an uniform electric field. The distance between electrons does not change as the electrons "fall". For many people that have studied relativity this is just very weird, and impossible to understand. This is the Bell's spaceship paradox.

Let's say we drop two masses in an uniform gravity field. The distance between the masses changes as the masses fall. Now for those aforementioned people this should be just normal, and easy to understand. Right? Or maybe not?

Let's say we drop two masses inside an accelerating rocket. The distance between the masses changes as the masses fall. This is jut normal Lorentz-contraction. Right?

Well anyway, in the first paragraph there was a weird absence of Lorentz-contraction. While in the second paragraph there was a Lorentz-contraction, which was equivalent to the Lorentz-contraction in the third paragraph, as the equivalence principle requires.

stuffu
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  • I'm not really following, but are you arguing that the thread will break? – Roger Wood Apr 29 '21 at 16:35
  • 1: Joe and a thread are accelerating in a rocket. Then the thread stops accelerating, what happens to the thread according to Joe? 2: Bob is standing on a planet feeling an acceleration, one hair falls off of Bob's head, what happens to the hair according to Bob? Equivalence principle says that Joe and Bob observe the same things. .... Next we might attach some masses to the threads. – stuffu Apr 30 '21 at 10:17
  • @RogerWood The thread does not break, and gravitational time dilation has a lot to do with it. – stuffu Apr 30 '21 at 10:53