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My question is about the article Swimming in Spacetime.

My gut reaction on first reading it was "this violates conservation of momentum, doesn't it?". I now realize, however, that this doesn't allow something to change its momentum; it only allows something to move (change position) without ever having a nonzero momentum. Since this is relativity, there is no simple relationship between momentum and velocity like p = mv, so this is all well and good. An object can move, with a constant momentum of zero, by changing its shape in a nontrivial cycle.

However, now I'm thinking about a different conservation law and I can't see how "swimming through spacetime" is possible without violating it. The conserved quantity I'm thinking of is the Noether charge associated with Lorentz boosts, which is basically x - (p/E)t, that is, the position of the center of mass projected back to time t=0. If p = 0, then the conserved quantity is simply x, the position of the center of mass. This obviously contradicts the whole swimming idea.

What's going on here? Is swimming through spacetime only possible if the spacetime is curved in some way that breaks the symmetry under Lorentz boosts? Or is there some error in my reasoning?

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Free version of the article: puhep1.princeton.edu/~mcdonald/examples/mechanics/… –  KennyTM Nov 16 '10 at 7:02
    
@KennyTM it requires a password, doesn't seem to be a free version though. I found another here: ftp.ics.uci.edu/pub/wayne0/papers/belgrade/… but i'm not 100% its legal-free, if someone has reasonable reason to think this isn't legal, i will remove the link –  lurscher Apr 6 '11 at 14:59
    
Here are some non-paywalled papers: groups.csail.mit.edu/mac/users/wisdom (may be the same one), arxiv.org/abs/gr-qc/0510054 –  Ben Crowell Aug 11 '11 at 15:45
    
A non-paywalled blog (with video animation of a similar effect realized on a sphere): science20.com/hammock_physicist/swimming_through_empty_space –  Johannes Jan 22 '12 at 1:49
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3 Answers 3

up vote 8 down vote accepted

What's going on here? Is swimming through spacetime only possible if the spacetime is curved in some way that breaks the symmetry under Lorentz boosts? Or is there some error in my reasoning?

That is precisely the case. No error in your reasoning. In the case of a curved spacetime the "center of mass" of an extended body is no longer well-defined w.r.t external - i.e. located in an asymptotically flat region - observers.

In order to "swim" through spacetime one exploits the inhomogeneities of the gravitational field. The presence of these inhomogeneities breaks local Lorentz symmetry which is necessary for the mechanism to work.

In particular the scale of the swimmer and the inhomogeneities should be comparable. This is one reason why, at present, the construction of an actual swimmer is far beyond our technological means.


Edit: For those interested on extended body effects in GR there is are classic papers by Dixon. More recently Abraham Harte has done some amazing work along these lines Extended-body effects in cosmological spacetimes.

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Nice. That's pretty much what I was thinking, but I was really unsure of myself because the Science article didn't say anything like that. –  Keenan Pepper Dec 8 '10 at 2:17
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Hard to tell exactly what the scenario is from that article. From what is shown, my guess is that the swimmer is doing work in deforming the object, which then moves the object. Then, the object having moved, the swimmer deforms the object BACK.

While this cycle would create zero work classically, in the case of relativity, you are now at a point where the gravitational potential has a different value, and therefore, the work you do to restore the object has been "redshifted" to a different value. In essence, the 'swimming' scheme converts gravitational potential energy into kinetic energy.

But that might not be quite what they're doing in this article.

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Well, I hope that my primitive understanding of GR makes for a good non-expert explanation... In GR, the Lorentz group symmetries are generally only valid locally, that is for a given space-time point. If you want to translate a vector to another space-time point, you need to do a parallel transport, which usually introduces correction terms depending on the curvature

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That's true, but what if the spacetime is nice and symmetric under Lorentz boosts? Does that mean it's impossible to "swim"? The paper already explains how it's impossible to "swim" in flat Minkowski space, but there are other nice symmetric spaces, like de Sitter space. Is it possible to "swim" through de Sitter space? –  Keenan Pepper Nov 16 '10 at 16:33
    
@Keenan, great question! can you please edit your question and add this particular subquestion? i find it very important –  lurscher Apr 6 '11 at 18:53
    
It's impossible to swim in any maximally-symmetric space, so Minkowski and de Sitter are out. There is, however, quite a bit of freedom in almost every other spacetime. –  Stingray Apr 6 '11 at 21:48
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