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In general relativity, the presence of gravity warps space-time yet clearly me accelerating will not warp space-time. It is hints at in one of the comments of this answer that acceleration bends world lines but does not actually cause the warping of space time. Is this right? If it is then why could we not have acceleration warping and gravity bending world lines. If it is not why not?

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  • $\begingroup$ I think this hinges on your definition of "warp". You can certainly argue that you accelerating away from me will see each other's space-time differently, so isn't that "warping" ? $\endgroup$ – StephenG Jan 21 '17 at 20:23
  • $\begingroup$ @StephenG I am relativity new to GR so my interruption of wrap may not be 'correct'. But I would say 'warping' means that the surface cannot be mapped using a Cartesian set of coordinates. $\endgroup$ – Quantum spaghettification Jan 21 '17 at 20:30
  • $\begingroup$ If I decipher the phrasing correctly the reason is that the curvature of spacetime is a geometric invariant, and so is the same in all frames, while the curvature of a worldline is not, and so can vary across frames. In particular, in accelerating frame the worldline curvature will change, "warping" it, but spacetime curvature will remain the same. $\endgroup$ – Conifold Jan 21 '17 at 21:55
  • $\begingroup$ @Conifold if I throw a ball straight ahead in an accelerated frame such as Earth's gravitational field, the floor (which is straight) and the path of the ball (which is also straight, per conservation of momentum) meet; the only way two parallel straight lines can cross is if the coordinate system itself is curved. This is where we get the notion of curved spacetime. $\endgroup$ – Asher Sep 19 '17 at 22:23
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Although "warping" is probably not an academically well-recognized term, I feel safe to assert that warping means curvature of the spacetime. Curvature of the spacetime is the property of the spacetime being of such a nature that two infinitesimally separated geodesics (paths for which the vector generated by the parallel transport (along the path) of the tangent vector to the path at one point is also the tangent vector to the path at the resulting point) that are parallel at a point do not remain parallel in the infinitesimal neighborhood of that point. As you can easily see, the definition of the curvature is manifestly coordinate invariant. Thus, a frame being accelerated wrt the previous frame does not make the spacetime curved if it were not curved in the original frame. So, the reason acceleration doesn't warp spacetime is because the acceleration of a frame wrt the other has nothing to do with the coordinate invariant properties of the spacetime, rather it is just the relation between two frames.

Now, in some sense, according to the Equivalence principle, gravity and acceleration are the same things. In this context, gravity also doesn't always curve the spacetime. For example, if a box, in the deep empty space, is being pulled with some acceleration with respect to the local inertial frame then there would be gravity in the frame attached to the box but the spacetime is still flat.

But, there is some distinction between this kind of gravity and a rather "genuine" gravity, for example, the kind of gravity produced by the Earth or some other matter-energy distribution. This kind of gravity, called "true gravity" by Weinberg, is defined as the sort of gravity that makes the path followed by two neighboring free particles in the spacetime (who are initially going parallel to each other in their spacetime trajectories) converge. By the Equivalence principle, the space-time trajectories of the particles are precisely the geodesic curves in the spacetime. Thus, it follows that true gravity represents nothing but the curvature of spacetime. Notice that such true gravity can also be present in the perfect vacuum without any source whatsoever, e.g., in the gravitational waves. So, coming back to your question, I would say the curvature of the spacetime is what is represented by "true gravity" rather than saying that gravity causes spacetime curvature. One might be tempted to say, based on the Einstein field equations, that the energy-momentum tensor $T_{\mu\nu}$ causes the curvature of spacetime but as I said, the curvature can be present even without $T_{\mu\nu}$. So, you can not call it the exclusive cause of the spacetime curvature.

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Let's clear up some preliminaries. In general, any matter content described by a stress-energy $T_{\mu\nu}$ will give rise to space-time curvature, described by a metric satisfying,

$$R_{\mu\nu}-\frac12 g_{\mu\nu}R = 8\pi G\, T_{\mu\nu}.$$

Now, for a Minkowski space time, with $T_{\mu\nu} = 0$ everywhere, we have the Minkowski space-time solution,

$$ds^2 = dt^2 - dx^2-dy^2-dz^2$$

Now, if you are in a uniformly accelerating frame of reference, then the metric can be expressed in terms of Rindler coordinates, and omitting certain caveats, takes the form,

$$ds^2 = g^2x^2 dt^2 - dx^2-dy^2-dz^2.$$

The accelerated observers will no longer perceive the space-time as being totally flat, though of course for any neighbourhood on the manifold, it holds by definition that it is homeomorphic to an open subset of $\mathbb R^{1,3}$. Furthermore, just as a black hole has a horizon, there is a so-called Rindler horizon at $x=0$ but this is a coordinate singularity. Moreover, there is an analogue of Hawking radiation known as the Unruh effect. It can be shown that a body with acceleration $a$ measures a temperature $T \propto a$.

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    $\begingroup$ "Now, in the case of a vacuum $T_{μν}=0$ everywhere, we have the Minkowski space-time solution" - This is not true for two reasons: The first (and the minor one) is that you are forgetting (maybe for the sake of simplicity) the cosmological constant. The second (and the important one) is that GR in more than 2 + 1 dimensional always admits gravitational waves which make the spacetime non-Minkowskian even if $T_{\mu\nu}=0$. $\endgroup$ – Dvij Mankad Sep 19 '17 at 12:59
  • $\begingroup$ This is at utterly the wrong mathematical level for the OP. $\endgroup$ – Ben Crowell Sep 19 '17 at 15:43
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    $\begingroup$ @BenCrowell This has been discussed extensively in the meta stack exchange... The answers are not just for the OP, they are for the audience that may read this question, which obviously has a varying background. $\endgroup$ – JamalS Sep 19 '17 at 18:21
  • $\begingroup$ @Dvij It's a pedantic correction; changing the ordering fixes it, but since I am pedantic myself I've made the correction :) $\endgroup$ – JamalS Sep 19 '17 at 18:23
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If I accelerate and declare my system as a reference frame, then the world lines of everything else will start to bend.

Thought experiment: I have a space ship that is stationary (whatever that means in detail), but not accelerating. I toss a ball across the space ship very slowly. Now I start accelerating with the space ship in the orthogonal direction. The ball will start to move on a curve with respect to the spaceship. So acceleration does bend world lines.

This bending comes from the non-linear transformation that transforms into your accelerating frame. And since it is accelerating, it has an explicit time dependence that will bend world lines.

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