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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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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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  • $\begingroup$ 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. $\endgroup$ Dec 8, 2010 at 2:17
  • $\begingroup$ In particular, you cannot define a center of mass because in a curved manifold, points cannot be interchanged with vectors and then averaged, like you can in a flat one. $\endgroup$ Sep 3, 2021 at 0:21
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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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    $\begingroup$ 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? $\endgroup$ Nov 16, 2010 at 16:33
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    $\begingroup$ @Keenan, great question! can you please edit your question and add this particular subquestion? i find it very important $\endgroup$
    – lurscher
    Apr 6, 2011 at 18:53
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    $\begingroup$ 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. $\endgroup$
    – Stingray
    Apr 6, 2011 at 21:48
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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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Is swimming through spacetime only possible if the spacetime is curved in some way that breaks the symmetry under Lorentz boosts?

It's impossible in Minkowski spacetime, and anything else breaks the global Lorentz symmetry. It doesn't even have to be curved. For example, in the cylindrical spacetime obtained from Minkowski spacetime by identifying $(t,x,y,z)$ and $(t,x{+}1,y,z)$, the curvature is zero everywhere, but you can change your $x$ coordinate by tossing a ball in the $+x$ direction and catching it when it returns from the $-x$ direction. The argument from Noether's theorem doesn't rule this out because this spacetime isn't invariant under boosts (except in the $yz$ plane).

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.

You can change your momentum by "swimming". Even in Newtonian gravity, if you're at rest in a nonuniform gravitational field, and remain at rest because it happens to integrate to zero over your mass, you can generally change the value of the integral by redistributing your mass, and thereby accelerate. This doesn't violate conservation of momentum because there's a backreaction on the sources of the field. In general relativity, instead of saying that the field is nonuniform, you say that spacetime is curved, but it's a still a gravitational field, and you can "swim" for essentially the same reason as in Newtonian gravity. It's difficult to say what conservation of momentum should mean in GR, but you presumably can define a pseudomomentum that is conserved if you don't neglect the backreaction.

This paper of course neglects the backreaction. In fact, nothing about the paper makes sense to me from a physical perspective. It starts with a discussion of swimming in fluids at low Reynolds number, where friction is so high that inertial motion is effectively impossible, and uses that to motivate a discussion of motion in vacuum, where not only is there zero resistance to motion, but the distinction between motion and rest doesn't even make sense. It's inevitable that you can accelerate and not just change position by gravitational swimming because it's impossible to even distinguish a state of zero gravitational acceleration in a generally covariant way.

The paper by Harte that's mentioned in the top-voted answer also ignores the backreaction, and is effectively invalidated by that. Its conclusion says:

Even in the presence of linear and angular momentum conservation laws, it was shown that bodies can control the magnitudes of their center-of-mass acceleration and spin using purely internal processes.

The bodies he studied are (as he notes) surrounded by, and permeated by, a fluid with exactly the properties of the ideal Hubble fluid. The effects he found are due to gravitational interaction with that fluid. When the body changes its linear or angular momentum, it's imparting an equal and opposite momentum to the fluid. It could propel itself far more effectively in this situation by exploiting a stronger interaction than gravity, using a Bussard ramjet for instance. Even literally swimming in the Hubble fluid would move you by an amount that would certainly be many orders of magnitude larger than the effect that Harte found (though still far too small to be useful).

The basic issue with both papers is that they ignore the Einstein field equations, and just do differential geometry on a fixed spacetime background. The Newtonian equivalent of that is ignoring $F=GMm/r^2$ and just using $F=ma$ and a fixed background force field. Of course you'll find reactionless propulsion in that situation: you assumed it.

This answer is related.

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