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As is well established, Special Relativity ensures that no observer can ever tell from the experiments he has been doing in his car whether the car is moving or not as long as the car constitutes an inertial frame of reference. This I find in complete accord with the Special Principle of Relativity.

Now, the most fundamental issue Einstein himself had with the Special Relativity was that it still picks out some mysterious inertial frames which we find no reason to exist if we completely embrace the general principle of relativity. There should not be any standard of absolute acceleration. Thus, I thought that General Relativity must be of such a nature that it abolishes the absolute acceleration standards just as Special Relativity abolished the notion of absolute velocity.

Now, imagine this scenario. There is a lift and there is the vacuum in the lift. Thus, $R_{\mu \nu}=0$ in every possible coordinate system. Here, I can choose a coordinate system in which $\Gamma^{\alpha}_{\beta\gamma}=0$ but I can also choose a coordinate system in which $\Gamma^{\alpha}_{\beta\gamma}\neq0$. Now, can't I assert that the frames in which, the Ricci tensor is trivial and yet the connections are not, are accelerated and the ones in wich Ricci tensor is trivial and the connections are also trivial are inertial? This would not be an artifact as there is a definite way of making a distinction. (Whereas in SR, there was not such way of making a distinction between a rest frame and a moving frame.)

Edit: So the problem is that if I can identify a frame as inertial and the others as non-inertial then based on the same I can establish some local standards of non-acceleration or those of acceleration. This is completely against the spirit of the general principle of relativity. According to the general principle of relativity, there should be absolutely no way to tell which object is moving and which is at rest, i.e., I should be able to call $A$ to be at rest and $B$ to be accelerating as well as $B$ to be at rest and $A$ to be accelerating- and do the Physics in any of them the same way. But this essence is spoilt here.

P.S.:

I know that there is a gauge freedom in choosing the components of the metric even if I have been provided with the curvature tensor and this results in multiple possible connections for a single curvature tensor. But this is the mechanism of how this different possibility of connections arises- not a valid way of denying that I can make distinction between coordinate systems which I can, in turn, use to define absolute acceleration.

I also know that the observer which is using the coordinate system with non-trivial connections can attribute these effects to a gravitational field. But this seems like an excuse to me when I imagine a spacetime which is entirely flat and has the stress-energy tensor identically zero everywhere. In such a universe, the introduction of a homogeneous gravitational field to explain the non-triviality of connection in certain coordinate systems would be an utter artifact - in the sense that whom should I ascribe the origin of such a field? And even more strikingly, in a completely empty universe, in one particular set of frames, my symbols will be trivial and in all the rest they won't be. This is a measurable and clear distinction. Put it another way, only in a particular class of coordinates, I will be able to synchronize a non-local array of clocks whereas, in the rest, I won't be able to do so.

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  • $\begingroup$ Theoretically, you can not detect absolute acceleration in GR because any effect such an acceleration might produce can be mimicked by gravity with a fanciful metric tensor. You can't know that the Ricci tensor is trivial because you can't know what comes from inertia and what from gravity. You can still make your distinction, but it will be merely a mathematical fiat. This is the gist of Nogueira's answer, I think. However, such mimicry in general needs matter with some quaint properties, including negative mass density, so barring that some frames are distinguished. $\endgroup$ – Conifold Nov 4 '16 at 20:47
  • $\begingroup$ I can know that the Ricci tensor is trivial. No matter how fancy I make the metric, the Ricci tensor will remain trivial if the space is flat. $\endgroup$ – Feynmans Out for Grumpy Cat Nov 4 '16 at 21:02
  • $\begingroup$ You can not know that the space is flat for the same reason. Here is how Einstein motivated his general relativity principle in 1907:"The gravitational field has only a relative existence... because for an observer falling freely from the roof of a house there exists — at least in his immediate surroundings — no gravitational field... the observer lacks any objective means of perceiving himself as falling in a gravitational field. Rather he has the right to consider his state as one of rest and his environment as field-free relative to gravitation." $\endgroup$ – Conifold Nov 4 '16 at 22:12
  • $\begingroup$ The existence of a gravitational field only means that the Christoffel symbols are non-trivial in the corresponding coordinate system. It doesn't mean that space is curved. Space being curved or flat is a frame invariant fact. One can't make a flat space curved via jumping around different coordinate systems. $\endgroup$ – Feynmans Out for Grumpy Cat Nov 4 '16 at 22:52
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I think I understand what you're asking so I'll answer accordingly. Ignore this answer if I've got the wrong end of the stick.

General relativity tells us that the four acceleration is given by:

$$ A^\alpha = \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} + \Gamma^\alpha_{\,\,\mu\nu}U^\mu U^\nu \tag{1} $$

So there are two contributions, the time dependence of the coordinates and the term in the Christoffel symbols. Since the four-acceleration is a four-vector the norm of the four-acceleration, the proper acceleration, is an invariant so it will be the same in all coordinate systems.

If we consider a freely falling observer in Minkowski spacetime (i.e. your lift) then the norm of the four-acceleration is zero. As you say, we can choose coordinates where $\mathrm d^2x^\alpha/\mathrm d\tau^2=0$ and $\Gamma^\alpha_{\,\,\mu\nu}=0$ and this is what we'd call an inertial frame. Alternatively we could choose accelerating coordinates, like the Rindler coordinates, where neither $\mathrm d^2x^\alpha/\mathrm d\tau^2=0$ nor $\Gamma^\alpha_{\,\,\mu\nu}=0$ but of course the proper acceleration of our freely falling observer would still come out as zero.

I'd guess we agree so far, but where we disagree is that I don't see that there's anything different between GR and SR or indeed classical mechanics. The invariant is the proper acceleration of the observer and that is always unambiguously measurable because the observer just has to weight themselves. The same equation (1) applies to curved spacetime, flat spacetime and indeed to non-relativistic motion where the manifold is Riemannian.

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  • $\begingroup$ I guess it is not clear what I am asking then. I too agree that everyone will agree upon the norm of the four-acceleration. My problem was that in say, Rindler coordinate, I can easily identify my frame to be accelerating. But in SR, one can't identify oneself's frame to be moving. So SR does its job of abolishing the standards of absolute velocity perfectly. But GR leaves the room for identifying some standards of absolute acceleration whereas one of its presumed jobs was to abolish every standard of absolute motion. $\endgroup$ – Feynmans Out for Grumpy Cat Nov 4 '16 at 19:09
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    $\begingroup$ @Dvij: there is already no absolute velocity in classical mechanics. So it is incorrect to say that special relativity abolishes the standards of absolute velocity. In fact SR introduced an absolute velocity which is the speed of light. Given that it is also incorrect to say that GR abolishes the notion of absolute acceleration, rather GR treats all reference systems equivalent and the equations of motion are the same in all reference systems (that is they are absolute). (+1 to John Rennie for the nice explanation) $\endgroup$ – Fabian Nov 4 '16 at 19:31
  • $\begingroup$ @Dvij Rindler coordinates are coordinates on SR or the flat space-time in GR. The acceleration is indeed a measurable quantity to the observer in SR or flat space. Turns out that in GR the space-time can curve and then we lost a background to set up the "intertial frames", the favourite ones. We need now that the fundamental physical laws don't make any distinctions between different coordinate system, otherwise you could use this distinction on the coordinates system to set up a preferred background. $\endgroup$ – Nogueira Nov 4 '16 at 19:49
  • $\begingroup$ Actually, you can "violate" this covariance in a state dependent way, you can fix a particular class of states to the space-time (a particular geometry, say flat for example) and make sense of acceleration as a displacement of the free-fall paths in this geometry, giving the very same thing as in SR (in the case of flat space) $\endgroup$ – Nogueira Nov 4 '16 at 19:49
  • $\begingroup$ @Fabian There was the proposed Absolute Space in Newtonian Mechanics which implied an absolute velocity. Though absolute space remained metaphysical and the laws were indeed invariant under transformation among frames associated with each other via a constant velocity. But Maxwell's laws had this inevitable requirement of an absolute rest frame and they had to be invalid in all the other frames which were even slightly moving wrt it. In that sense, classical mechanics severely introduced absolute velocity and Special Relativity abolished it. Invariant velocity doesn't mean absolute velocity. $\endgroup$ – Feynmans Out for Grumpy Cat Nov 4 '16 at 20:07
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Generally speaking, I think you are right. Indeed general relativity, in spite of its name, preserves preferred local reference frames where laws of physics take their simplest form. They are local inertial reference frames defined around timelike geodesics. Laws take a simpler form there, because the connection coefficients vanish along the geodesic in these coordinates. It is true that you can however write down physical laws into a form formally identical in every coordinate frame using tensor analysis, however we cannot ignore the physical fact that the connection coefficients select preferred local reference frames. This fact is of greatest relevance since it permits us to extend physical laws from special relativity to general relativity, at least for laws including first derivatives at most.

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  • $\begingroup$ Yes. I know that the great success of the entire enterprise of Einstein lies in the fact that it tells us how to compute the outcomes of physical phenomena in any arbitrary coordinate system. We weren't able to do this before. But still, the existence of local inertial frames which can be said to have definite preference over the other local frames is bugging me. Are there any extensions to classical GR which addresses this issue? $\endgroup$ – Feynmans Out for Grumpy Cat Nov 4 '16 at 20:30
  • $\begingroup$ Also classical mechanics can be formulated into a completely covariant form introducing an affine non-metrical connection. Inertial forces are represented by means of connection coefficients. These coefficients vanish in inertial reference frames. To some extent also the classical gravitation can be described by part of this connection...However the complete development of this approach has been reached only within the formulation of GR. $\endgroup$ – Valter Moretti Nov 4 '16 at 20:38
  • $\begingroup$ Sorry, I do not have an answer to the last question in your comment. $\endgroup$ – Valter Moretti Nov 4 '16 at 20:40
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No, the observer that is freely falling can determine that there is curvature by measuring second derivatives of the metric tensor, which would include calculations of the first derivatives of the connection, and obtain a Riemann tensor. If there is no curvature they will all be zero. If there is at least some of them will not be zero. And he can compute various scalars from them, where at least one will not be zero.

The error in the thinking is that it ignores that that 'inertial' frame where the observer is freely falling is only local. The observer will measure deviations from flatness as he goes away from his local area, or equivalently, the connection may be zero at his point, but its derivatives, and with them used to construct the curvature tensor and scalars, which will have some nonzero component. If the observer has a good enough accelerometer, some inches or feet or more from his reference point for his local inertial frame, he will measure an acceleration then. If falling into a black hole, for instance, his body will elongate as the gravitational field increases, even if the observer thought his body was local and was in his inertial frame. He simply was wrong.

Both thenWeak and the Einstein Equivalence Principle state that those inertial frames are only in 'small enough regions of spacetime'. I forget what the Strong Equivalence Principle adds to the Weak, but I think it includes the 'small enough regions' statement or idea. Those inertial frames cannot be extended and stay inertial.

So, no, there is no preferred frame. The local free falling inertial frame is just a good initial approximation where he can ignore the gravitational filed, because near that the accelerations are approximately the same. But it is only an approximation, and when he goes a distance away it fails being inertial.

Einstein's General Relativity captured that Relativity principle perfectly. It's held up to all the conceptual and real physical experiments, like your thought experiment.

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  • $\begingroup$ My thought experiment concerned a case where there indeed is no curvature. $R_{{\mu}{\nu}} = 0$. But in one frame, the connection vanishes whereas, in another, it doesn't. Put it another way, only in a particular class of coordinates, I will be able to synchronize a non-local array of clocks whereas, in the rest, I won't be able to do so. Doesn't this create a distinction among reference frames (which shouldn't be possible according to the principle of relativity)? $\endgroup$ – Feynmans Out for Grumpy Cat May 12 '17 at 19:42
  • $\begingroup$ No, the connection is not invariant and will normally change from one coordinate system (cs) to another. And it can be zero at a point in one cs (called a local inertial frame) and not another. But any time you measure the invariant curvature like R you'll get 0. By the way you condition does not mean the 4 index tensor for R is 0, so other invariants could be nonzero. In that frame your spacetime is Minkowski locally (an infinitesial neighborhood), but using your frame of reference you'll still be able to measure curvatures and notice nonzero curvature invariants, and agree with other frames $\endgroup$ – Bob Bee May 12 '17 at 20:29
  • $\begingroup$ If it vanished everywhere you'd have Minkowski everywhere. Other could use say a rotating frame, do measurements, and still would conclude the spacetime is Minkowski and that they are using a lousy cs, but still valid. Btw, connections are first delivatives but if you know those everywhere you can calculate second and all other derivatives everwhere. They would be zero if the connection is zero everywhere $\endgroup$ – Bob Bee May 12 '17 at 20:31
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    $\begingroup$ @Djiv $R_{\mu\nu} =0$ does not mean zero curvature. It just means no mass energy, i.e., empty. It still can be curved. Gravitational waves are like that, even if very very strong. Also, outside a black hole it's like that also. You'd need the 4 index tensor to be 0 for that. $\endgroup$ – Bob Bee May 12 '17 at 23:24
  • $\begingroup$ Yes, I do get that. But what I mean by zero curvature is actual zero curvature. (I should have written $R_{abcd}=0$.) What you are saying is that in such a case (or in any case actually), every frame will agree upon the vanishing (or non-vanishing) nature of the 4 index curvature tensor. I completely agree. But, in order to violate the principle of general relativity, I need to find just one way to manifest the asymmetry between frames. And that way seems to be the way of variant nature of the connections - zero in some frames and non-zero in others. Don't you think this is problematic? $\endgroup$ – Feynmans Out for Grumpy Cat May 13 '17 at 3:23
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Let's consider a completely empty spacetime with $T_{\mu\nu}=0$ everywhere (in every coordinate set-up). Now, consider two events $A$ and $B$. Now, consider several clocks who all simultaneously tick their zeros at $A$ and then do some motion and again meet at $B$ and stop ticking (assume that the events $A$ and $B$ are of such a nature as to allow such a motion - this is the only qualification on their otherwise generic nature). The numbers on their dials at $B$ are certainly frame-invariant - though these numbers can differ from one another, of course. Now, it's an experimental fact that there exists a limit as to how big this number can be. And only a unique clock out of all the possible clocks that connect $A$ and $B$ displays this number. Now, consider all such possible pairs of events that can be connected by such clocks. And for each pair of events, there exists a unique clock which displays a maximum number. Now, form a set (call $\Lambda$) of all these unique clocks determined by all the possible pairs of events $A$ and $B$. It is an experimental fact that every member of this set is in uniform motion with respect to every other member (i.e. if you attach indefinitely extended rulers of Euclidean kind with any of these clocks (with that clock being at the center) and put extra clocks at every point of the coordinate set-up synchronized with the central clock via a symmetric procedure then the coordinate velocity of all the clocks of the set $\Lambda$ will be constant.) This means that a completely empty space also has a very definite intrinsic structure to it which determines the extremum distance (and corresponding geodesic) between each pair of events and, in addition, has a (global) structure of such a kind that particles on these geodesics see each other going with constant relative velocity. (This second property of its structure seems to be a derivative of the fact that a global synchronization of clocks is possible in empty space. But I am not sure.) Thus, this set $\Lambda$ (determined by the intrinsic frame-invariant metrical structure of the spacetime) creates a local (and global) standard of non-acceleration.

So, two frames (one accelerating with respect to the other and one of them having vanishing Christoffel symbols) differ in a very physical sense that one sets up geodesics for the spacetime whereas the other doesn't. This difference stems from the inherent frame-invariant metrical structure that even the spacetime has. Thus, the fact that in one frame, particles go unaccelerated and in the other, particles experience some acceleration also stems from the fact that one of the frames does have a special status owing to the inherent frame-invariant metrical structure that even the empty spacetime has. So, although the principle of relativity of all the kinds of motion isn't respected in the sense in which the principle of relativity of uniform motion is respected in special relativistic set-ups, we get a reason as to why - the existence of definite geodesics (and their relation to each other) in the empty space.

The fact that timelike geodesics, in full GR with the energy-matter, maximizes the proper time can also be thought of in the perspective described above. The existence of mass-energy-momentum determines specific paths in spacetime called geodesics which maximizes the proper time. And the local standards of acceleration (or non-acceleration) are determined by the particles following these geodesics. There isn't a full symmetry between frames because of this very fact that all the particles following the geodesic (as determined by mass-energy-momentum) are in uniform motion with respect to each other. This makes these particles' frames the local standard of non-acceleration. Similarly, in empty space, one should possibly expect the absence of the standards of acceleration only if there is the absence of geodesics. But that not being the case, there certainly exists the standards of (non) acceleration in empty space as well.

So, what General Relativity does (rather than establishing true relativity of all kinds of motion) is establishing a duality between accelerated motion and gravity through the principle of general covariance. Which is a statement more about the effects of gravity than about the relativity of all kinds of motion. It says that (since the geodesics define free-fall) the effect of gravity is uniquely determined by the laws of coordinate transformations from the local inertial frame which is the frame attached to a particle (locally) in free-fall. The key physical insight to be absorbed (above the most fundamental fact that we can always go to a local inertial frame) is that gravity exhibits all its effects only and only through determining geodesics. Because, once we know the geodesic, we know the local inertial frame and once we know local inertial frame, we can transform to whatever generic frame we want and we will have taken care of describing effects of gravity in that generic frame!

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protected by Qmechanic Nov 4 '16 at 20:26

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