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Let's say I have a tube, of large radius (about 5 - 7 meters in diameter), with traversable wormholes at the ends. The wormholes are arranged as such that if something falls inside one hole from inside of the tube, it will come out at the other end still inside the tube. Now, let's say I empty all air from the tube (to make a "vacuum tube," if you will), set it upright and somehow manage to get a rock (or whatever other object) in there. My question now is, does the situation described above rule out the existence of traversable wormholes?

Or, if not, since the rock is falling through the wormhole over and over again, will it always be accelerating at the same rate? Or will its velocity only always be approaching light speed? In either case, would the rock's mass increase to the point that it overpowers Earth's gravity, or even collapses into a singularity? Or is there something that would prevent that from happening?

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    $\begingroup$ I actually meant you should edit the other one, and then I'd reopen it. I'm only seeing this now, otherwise I would have gotten to it earlier. Anyway, I'll delete the original since this replaces it. $\endgroup$
    – David Z
    Commented Feb 26, 2012 at 21:51
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    $\begingroup$ Sure, no problem. For future reference, you should never repost a question, just edit the original and ask for it to be reopened. You can flag it for moderator attention to speed that process along. (By the way, I will mention that although I'm not going to close this, your second paragraph is pretty close to the boundaries of what is acceptable on this site.) $\endgroup$
    – David Z
    Commented Feb 26, 2012 at 21:55
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    $\begingroup$ No problem. We're trying to be (1) consistent about the rules, (2) aggressive about enforcing them and (3) nice. Sometimes it hard to get all the bits right at one time. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Feb 26, 2012 at 21:55
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    $\begingroup$ The problem is that it isn't (or wasn't, I suppose) specific enough. The presence of a wormhole in a gravitational field may mean that it's impossible to define a gravitational potential function, in which case everything you've learned about gravity goes out the window; or if not, the potential will be very strangely behaved, in which case, again, everything you've learned about gravity goes out the window. Either way, it's necessary to use some advanced general relativity to answer this question. We can only do that because we know that this wormhole of yours is governed by GR, (cont.) $\endgroup$
    – David Z
    Commented Feb 26, 2012 at 22:18
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    $\begingroup$ and in turn we only know that because you have used the buzzword "traversable wormhole," which specifies that you are talking about a reasonably well defined (if not well studied) structure of spacetime. In the earlier version when you were talking about "portals," that could have been anything. Certainly portals in the style of the video game are not realistically possible, for example. That's why your question fell under the fictional physics clause of the FAQ. $\endgroup$
    – David Z
    Commented Feb 26, 2012 at 22:20

2 Answers 2

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I found an entry in a Wormhole FAQ that seems to address your thought experiment:

"Is a wormhole whose mouths are arranged vertically in a gravitational field a source of unlimited energy?

No. The argument in favor of such a wormhole being an energy source is this: An object falls from the upper mouth, gains kinetic energy as it falls, enters the lower mouth, reemerges from the upper mouth with this newly acquired kinetic energy, and repeats the cycle to gain even more kinetic energy ad infinitum. The problem with this is that general relativity does not permit discontinuities in the metric – the descriptor of the geometry of spacetime. This means that the gravitational potential of an object at the lower mouth must continuously rise within the wormhole to match the potential it had at the upper mouth. In other words, this traversal of the wormhole is “uphill” and therefore requires work. This work precisely cancels the gain in kinetic energy."

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    $\begingroup$ But how does the wormhole 'know' how to match the gravitational potentials? This is certainly true for non-wormhole GR, but just baldly stating that it works for wormholes as well without giving some kind of rationale seems unjustified. $\endgroup$
    – Peter Shor
    Commented Mar 3, 2012 at 23:45
  • $\begingroup$ +1: for the link, but I agree with Peter Shor that there is no argument in the linked page. $\endgroup$
    – Ron Maimon
    Commented Mar 4, 2012 at 1:57
  • $\begingroup$ Yes, while this is a nice answer (from a presumably trustworthy source), I would like to see some reasoning as well. Thanks a bundle for the link though. $\endgroup$
    – ThisIsNotAnId
    Commented Mar 4, 2012 at 6:13
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    $\begingroup$ @Peter Shor: The "gravitational potential" is a Newtonian concept that does not actually exist in GR, which is a purely metric theory. However, one can consider the quantity -ln(-g00)/2, which becomes the Newtonian potential, where gravity is weak. Notice that this quantity will only be discontinuous if the 00 component of the metric is discontinuous. Also, it's not the integral of the Newtonian potential about a loop that is zero. It's the line integral of its gradient. This requirement is tantamount to the continuity of the potential and therefore the metric. $\endgroup$
    – Belizean
    Commented Mar 31, 2012 at 1:52
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    $\begingroup$ @Jerry Schirmer: Even if you were to permit a discontinuous metric, you would find that as the object emerges from the upper mouth (in the gravitational field of some planet or star), its gain in potential energy would have been stolen from the wormhole’s mass. After sufficiently many trips through the wormhole, the wormhole would become unstable to the traversals of the object, whose mass would now be comparable to, if not exceeding, that of the shrinking wormhole. $\endgroup$
    – Belizean
    Commented Mar 27, 2017 at 11:16
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I'm no GR whiz, so this answer may be wrong. Comments appreciated.

There are three things I see here. The first issue is that wormholes do not (may not) accelerate particles. Yes, while falling in a particle will accelerate; but while popping out, the reverse will happen, giving rise to a zero average acceleration. The exit of a wormhole is just like the entrance; so it will gravitationally attract you inwards.

The second issue is that if the rock does manage to accelerate, it will obtain the energy from the gravitational field. In other words, it's own gravitational effects will counter the wormhole, reducing the 'energy' of the wormhole's field (GR gravitational PE is not well defined, though). What may happen is this:(speculation) A wormhole's throat has a negative energy density. If the energy density becomes positive, the wormhole necks off like a piece of taffy, giving rise to two black holes or just expanding outwards (can't remember which). Popping a rock inside increases the energy density. If the rock keeps accelerating, it will get enough energy at some point or the other to destroy the wormhole.

The third issue is trivial to the paradox but worth mentioning. In relativity, we cannot have a 'constant acceleration' if acceleration is $\frac{d\vec{v}}{dt}$. Since force is $\frac{d\gamma m_0\vec{v}}{dt}$, even with a proportional-to-mass force (like gravity), the acceleration will not be constant and will slow nearly to a standstill as the velocity approaches lightspeed. $\gamma v$ will increase at a constant rate, ($\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$ becomes very large, near lightspeed)

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  • $\begingroup$ I don't know if it's just me, but your Latex isn't processing right. Other than that, quite an interesting answer. I haven't formally taken any course on modern Physics yet, but I followed along with some of it. $\endgroup$
    – ThisIsNotAnId
    Commented Feb 27, 2012 at 2:37
  • $\begingroup$ I can see the Latex now. $\endgroup$
    – ThisIsNotAnId
    Commented Feb 27, 2012 at 2:41
  • $\begingroup$ @ThisIsNotAnIn neither have I taken any formal courses. My modern physics knowledge is picked up from here and there. So its sometimes patchy. Oh, and the latex thing happened because I accidentally submitted the answer before finishing it. $\endgroup$
    – Manishearth
    Commented Feb 27, 2012 at 2:44
  • $\begingroup$ @ThisIsNotAnId whoops typo in the reply $\endgroup$
    – Manishearth
    Commented Feb 27, 2012 at 13:39
  • $\begingroup$ The last issue is not really relevant--- constant acceleration in relativity is defined as constant momentum gain per unit time, or (nontrivially) equivalently, constant acceleration in the rest frame of the accelerating particle. The accelerations caused by the neck of the wormhole naively scale out, because the wormhole gravity has the same mass at top and bottom, so falling in from far away should be compensated by falling out at the other end. But this might not be true, so you might have found the resolution, I have to think about it. $\endgroup$
    – Ron Maimon
    Commented Mar 4, 2012 at 2:36

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