# Is a uniform gravitational field a gravitational field?

Is a "uniform gravitational field" a gravitational field?

Is a "uniform gravitational field" a gravitational field?

Or in other words how do we explain the seemingly gravitational effects in a space-time region with zero intrinsic curvature?

Good Question

pervect
Staff Emeritus
The term "gravitational field" is rather ambiguous. I think that most people think of a gravitational field as what's measured by an accelerometer "at rest".

What is measured by an accelerometer is actually a path curvature (and in certain circumstances can be described as a Christoffel symbol). It's basically the invariant norm of a 4-acceleration.

This is the sense of "gravitational field" used in the famous "elevator experiment".

Other less commonly used definitions of "gravitational field" have been proposed and used. The Riemann curvature tensor (certain components of which have a physical interpretation as tidal forces) is one of them. This is probably not what most people see as a "gravitational field", but it is a true tensor quantity. Of course, this particular usage is not compatible with the "elevator experiment".

I believe I've also seen the metric of space-time itself as the "gravitational field".

There's a chapter in MTW that talks about all of these as possible interpretations of the term "gravitational field". None of them is singled out for special treatment as "the true gravitational field" however. Rather, the term "gravitational field" is recognized as being rather vague, and when one wants to be precise, one is advised to avoid this term in favor of something better defined.

D H
Staff Emeritus
How can something be "at rest" in a "gravitational field" unless some force other than gravity is acting to keep it "at rest"? The accelerometer measures the acceleration caused by this force other than gravity, not gravity itself.

Gravity is the one thing an accelerometer cannot measure. The navigational software in an aircraft or spacecraft equipped with accelerometers includes a mathematical model of the gravitational field precisely because the accelerometer doesn't measure the acceleration due to gravity.

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Is a "uniform gravitational field" a gravitational field?
That depends on who you were to ask. There are some camps which refer to regions of spacetime in which the spacetime was curved as being a region in which a gravitational field exists. Then there were others, such as Einstein, who said that a gravitational field could be produced by a change in spacetime coordinates. In the former the Riemann tensor didn't vanish in that region in those coordinates (hence it didn't vanish in any system of coordinates. Then there is the affine connection. If given a system of coordinates and the affine connection doesn't vanish then there is a gravitational field in that region (which could be transformed away). As Misner, Thorne and Wheeler say in Gravitation page 467 "No $\Gammas[\itex] no gravitational field ..." Some authors actually demand that the spacetime for a uniform gravitational field have zero Riemann tensor in that region which the curvature vanishes. If you have the chance see Principle of Equivalence, F. Rohrlich, Ann. Phys. 22, 169-191, (1963), page 173 (aailable by request in e-mail). Or in other words how do we explain the seemingly gravitational effects in a space-time region with zero intrinsic curvature? [/quote] What effects are these? You're most likely asking about the differences between gravitational effects and tidal effects. Best wishes Pete Is a "uniform gravitational field" a gravitational field? Or in other words how do we explain the seemingly gravitational effects in a space-time region with zero intrinsic curvature? Hi MeJennifer - Please check your PM messages too. Pete Is a "uniform gravitational field" a gravitational field? Or in other words how do we explain the seemingly gravitational effects in a space-time region with zero intrinsic curvature? I assume that by gravitational field you mean a gravitational field that cannot be transformed away. A real gravitational field will have the so-called tidal forces. This is a physical effect that cannot be transformed away. i.e. since absolute accelerations can be transformed away to zero, absolute accelerations are not important; instead the relative accelerations (which appear as tidal forces) are the important things. In fact, one of the ways to get to Einstein's equation is to consider the geodesic deviation of a cloud of particles, and then compare that to the equivalent tidal force that would be apparent from Newtonian gravitation. Basically: if there is zero curvature then you can always transform to Minkowski coordinates. If there isn't, then you can transform to Minkowski coordinates locally (i.e. you can make all the components of the connection vanish) but you can't get rid of the tidal forces (i.e. you can't make the Riemann tensor zero!) N.B. I may have used a non-standard usage of the term Minkowski coordinates; to be precise, by Minkowski coordinates, I mean a system in which the metric takes the form $$g_{ab} dx^a dx^b = ds^2 = dt^2 - dx^2-dy^2-dz^2$$ Last edited: I assume that by gravitational field you mean a gravitational field that cannot be transformed away. If you choose a choice of the term "gravitational" that requires spacetime to be curved then the term "uniform gravitational field" is a contradiction in terms. Its very definition requires flat spacetime in sime global neighborhood. Pete If you choose a choice of the term "gravitational" that requires spacetime to be curved then the term "uniform gravitational field" is a contradiction in terms. Its very definition requires flat spacetime in sime global neighborhood. Pete Indeed. That is what I was hoping to point out by the brief discussion on relative accelerations. Of course, these wouldn't be present in a uniform gravitational field. Perhaps it would have been better to explicitly state that. pervect Staff Emeritus Science Advisor How can something be "at rest" in a "gravitational field" unless some force other than gravity is acting to keep it "at rest"? The accelerometer measures the acceleration caused by this force other than gravity, not gravity itself. When you have a static spacetime, the space-time itself (or rather the symmetries of the space-time) defines a notion of "at rest". Very roughly speaking, an object "at rest" has constant metric coefficients, while a moving object will see varying metric coefficients. This trick really only works with a static space-time, though. This is discussed a bit in Wald, IIRC. If you want to get really technical, "At rest" means timlelike orbits of the Killing vector field. (And I think that one has to add that these orbits have an orthogonal space-like hypersurface, though Wald doesn't mention this requirement specifically.) I do agree that it's the acceleration by the force required to keep the object at rest that one measures on the accelerometer. Gravity is the one thing an accelerometer cannot measure. The navigational software in an aircraft or spacecraft equipped with accelerometers includes a mathematical model of the gravitational field precisely because the accelerometer doesn't measure the acceleration due to gravity. In a sense, gravity as a force doesn't really exist at all. It's better modeled as a curved space-time, through which objects follow a geodesic path (one that extremizes proper time). However, under some special conditions it is possible to consistently treat gravity as if it were a force. Viewing gravity as curved space-time leads to the idea of the Riemann tensor, a measure of the curvature of space-time, as being the 'gravitational field' as I mentioned earlier. However, this usage is not compatible with the "elevator experiment" AFAIK. As mentioned by some other posters, the space-time in the elevator is perfectly flat, the Riemann tensor is zero everywhere. So IMO popular usage (as illustrated by the so-called gravity in the elevator experiment) really does view gravity as that quantity which is measured by an accelerometer. Last edited: I think Matsudr and pmb_phy are on the right track. Here is an extremely good link on this subject: http://arxiv.org/ftp/physics/papers/0204/0204044.pdf It is not necessarily authoritative but covers the issues involved very well. Misner, Thorne and Wheeler make it plain that "spacetime curvature" is necessary and sufficient for a gravitational field to exist and that all realistic gravitational fields will have "tidal effects". Since in our actual universe a gravitational field is perfectly uniform only when it is zero, one could reasonably argue that uniform gravitational fields do not exist. The reason that the Equivalence Principle only holds exactly at a point ( the term "locally" often used implies "to a sufficiently good approximation"), is because any gravitational field is always distinguishable from an accelerating system by the tidal forces that are always detectable over any finite distance. Thus a "uniform" gravitational field could be regarded as the fictional gravitational analogue of an accelerating system if the equivalence principle were true over finite regions. pervect Staff Emeritus Science Advisor I think Matsudr and pmb_phy are on the right track. Here is an extremely good link on this subject: http://arxiv.org/ftp/physics/papers/0204/0204044.pdf Of course that link is authored by pmb_phy (and is not peer reviewed). This particular paper of his I don't have any serious disagreements with, however, except perhaps for certain matters of emphasis. It is not necessarily authoritative but covers the issues involved very well. Misner, Thorne and Wheeler make it plain that "spacetime curvature" is necessary and sufficient for a gravitational field to exist and that all realistic gravitational fields will have "tidal effects". Huh? The following quote from Pete's paper taken from MTW is fairly representative of their position, and is in fact the one I alluded to earlier. … nowhere has a precise definition of the term “gravitational field” been given --- nor will one be given. Many different mathematical entities are associated with gravitation; the metric, the Riemann curvature tensor, the curvature scalar … Each of these plays an important role in gravitation theory, and none is so much more central than the others that it deserves the name “gravitational field.” nrqed Science Advisor Homework Helper Gold Member Or in other words how do we explain the seemingly gravitational effects in a space-time region with zero intrinsic curvature? Which gravitational effects do you have in mind? In free-fall, there are absolutely no effects at all. So any "seemingly" gravitational effect is just an artifact of the choice of frame. Which gravitational effects do you have in mind? In free-fall, there are absolutely no effects at all. Not over an infinitesimally small region no, but there certainly is in any finite region, as discussed above. nrqed Science Advisor Homework Helper Gold Member Not over an infinitesimally small region no, but there certainly is in any finite region, as discussed above. I meant that in a uniform gravitational field there is no effect at all in free fall, not even tidal forces. Even over a finite region. That's what I meant. I agree that in a realistic gravitational field (eg near the Earth), even in free fall there are some effects over any finite region. Regards Patrick Pervect wrote: Huh? Presumably denoting scepticism. .....But huh? ....I don't see why since the quote from MTW merely lists a few different ways of representing "spacetime curvature" - consistent with what I said. It might also be worth making a distinction between lateral and longitudinal tidal effects. All real gravitational fields have both and they only disappear when the field diminishes to zero eg. infinitely far from an isolated body or at the midpoint between two identical masses etc. The term "uniform gravitational field" is often used to mean a "linear" field where lateral tidal effects are absent but longitudinal variation is present. Such a fictional artifact would not be a "homogeneous" field - a term sometimes used interchangeably with "uniform" in this context. The term "uniform gravitational field" is often used to mean a "linear" field where lateral tidal effects are absent but longitudinal variation is present. Such a fictional artifact would not be a "homogeneous" field - a term sometimes used interchangeably with "uniform" in this context. Could we even detect such a field on a pseudo-Riemanian manifold? Why not ? An aggregate of free falling test particles would move apart in the direction of motion but would not move together across the same direction. Why not ? An aggregate of free falling test particles would move apart in the direction of motion but would not move together across the same direction. Yes, you are correct, that would be detectable. pervect Staff Emeritus Science Advisor The whole point that I'm trying to make is that the term "uniform gravitational field" is so vague that one has to read the paper in question to find out what the author considers to be the "gravitational field" and in what sense it is "uniform". When I say "most people" consider the gravitational field to be the proper acceleration, I did not intend to imply that "most authors of articles on relativity" consider the gravitational field to be proper acceleration! I can see that I wasn't very clear in this remark - "most people" has to be defined in some sample space, and the sample space I was talking about was the sample space of forums like these. Basically, people in forums like these for the most part still use the Newtonian definition of "gravitational field". This is important in communication in forums like these, because if I start talking about the gravitational field as a Riemann tensor, or some variant therof, and my audience is thinking about in terms of basically Newtonian concept of the gravitational field as a vector, the "force per unit mass", communication is unlikely to occur. As far as the sense in which this term is used in the literature, a quick google suggests http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000057000012001121000001&idtype=cvips&gifs=yes [Broken] I'll put it on my list of things to get from the library, I haven't found a publicly available source. Desloge has written a lot of similar papers, it would also be worthwhile to do a better literature search to see if other authors also use the term to mean the same thing. Last edited by a moderator: Haelfix Science Advisor Yea, its often possible to redefine away various notions of curvature, so I like that definition less. I personally subscribe more to anything that outputs a nonzero tidal force as that seems to me to have nonambiguous physical meaning, alternatively something that outputs gravitational radiation (though thats less general). Nevertheless there is still a chicken and egg syndrome. Assume you have a box of pure EM stuff. This will have of course have nonzero stress energy and will cause a gravitational field. But 'gravity' strictly speaking didn't induce anything. So to see first causes one sort of has to go back to linearized gravity and track down the components outputing the global curvature or tidal tensor quantities that we wish to use as our reference for what we define a 'gravitational field' to be. Messy. And sort of irrelevant. JesseM Science Advisor Why not ? An aggregate of free falling test particles would move apart in the direction of motion but would not move together across the same direction. For a "uniform gravitational field", wouldn't that only be true in certain choices of coordinate systems? Since the spacetime is flat, you could always transform into a different coordinate system where the very same test particles would both be moving inertially, in straight lines at constant speeds, right? Since the spacetime is flat, you could always transform into a different coordinate system where the very same test particles would both be moving inertially, in straight lines at constant speeds, right? What do you mean "since the space-time is flat"? How can you have a space-time that is (Riemann) flat where you have diverging geodesics in the direction of motion? JesseM Science Advisor What do you mean "since the space-time is flat"? How can you have a space-time that is (Riemann) flat where you have diverging geodesics in the direction of motion? From previous discussions, I remember it was mentioned that if the spacetime curvature is zero in one coordinate system, it will be zero in all coordinate systems. Also, in posts #64 and #69 here pervect says that a universe totally empty of matter can be modeled either as an expanding universe with negative spatial (not spacetime) curvature, or as an ordinary static Minkowski universe with no spatial curvature. From the description on the MIT course page pervect linked to, it appears that in the first type of coordinate system, test particles seem to diverge due to the expansion of space, in the other they diverge simply because they started out with velocity vectors pointing in different directions, so they move apart on straight lines: As an interesting aside, we might ask why the Milne model has k = 1. Since there is no matter, there shouldn’t be any general relativity effects, and so we would ordinarily expect that the metric should be the normal, flat, Minkowski special relativity metric. Why is this space hyperbolic instead? The answer is an illustration of the subtleties that can arise in changing coordinate systems. In fact, the metric of the Milne universe can be viewed as either a flat, Minkowski metric, or as the negatively curved metric of an open universe, depending on what coordinate system one uses. If one uses coordinates for time and space as they would be measured by a single inertial observer, then one finds a Minkowski metric; in this way of describing the model, it is clear that special relativity is sufficient, and general relativity plays no role. In this coordinate system all the test particles start at the origin at time t = 0, and they move outward from the origin at speeds ranging from zero, up to (but not including) the speed of light. On the other hand, we can describe the same universe in a way that treats all the test particles on an equal footing. In this description we define time not as it would be measured by a single observer, but instead we define the time at each location as the time that would be measured by observers riding with the test particles at that location. This definition is what we have been calling “cosmic time” in our description of cosmology. One can also introduce a comoving spatial coordinate system that expands with the motion of the particles. With a particular definition of these spatial coordinates, one can show that the metric is precisely that of an open Robertson-Walker universe with R(t) = t. So, I'm assuming something similar would be true in the case of a "uniform gravitational field", you could transform into a coordinate system with no gravitational field where the divergence of test particles that was previously explained in terms of the field will now just look like test particles diverging because of initial velocities pointing in different directions. I think this would be implied by the equivalence principle, since the effect of test particles diverging due to the "uniform gravitational field" could also be observed in a very small box over a short period of time (so tidal forces were negligible) at rest on the earth, but the equivalence principle says these observations should differ negligibly from similar ones made in an accelerating box in deep space. This section of the twin paradox FAQ also says that phenomena observed by a uniformly accelerating observer can be recast in a coordinate system where the observer is at rest and there is a uniform gravitational field. Last edited: Whether a given space-time is (Riemann) curved does not depend on the chosen coordinates, coordinates are simply a map to describe a region of space-time they obviously do not make a manifold curved or flat. I would be very careful with metrics with cross terms, they are useful for particular calculations but more confusing than helpful in getting an understanding of the properties of space-time, at least that is what I think. JesseM Science Advisor Whether a given space-time is (Riemann) curved does not depend on the chosen coordinates, coordinates are simply a map to describe a region of space-time they obviously do not make a manifold curved or flat. Yes, that's what I meant when I said that "if the spacetime curvature is zero in one coordinate system, it will be zero in all coordinate systems." I was just saying that, depending on your choice of coordinate system, the geodesic paths of test particles could be diverging due to gravitation or due to different initial directions in the absence of gravity, in response to your question "How can you have a space-time that is (Riemann) flat where you have diverging geodesics in the direction of motion?" Yes, that's what I meant when I said that "if the spacetime curvature is zero in one coordinate system, it will be zero in all coordinate systems." I was just saying that, depending on your choice of coordinate system, the geodesic paths of test particles could be diverging due to gravitation or due to different initial directions in the absence of gravity, in response to your question "How can you have a space-time that is (Riemann) flat where you have diverging geodesics in the direction of motion?" Geodesic deviation is also not frame dependant. If it were then the Riemann tensor would also be frame dependant. A geodesic is a geometric object whose nature does't change. Tidal forces exist if and only if a spherical object placed in the field which is subject to no external sources will become deformed do to the non-vanishing Riemann tensor. The fact that the coordinate acceleeration varies with height does not mean that there is geodesic deviation. In fact if you were to calculate the deviation you'd find it to be zero. Pete JesseM Science Advisor Geodesic deviation is also not frame dependant. If it were then the Riemann tensor would also be frame dependant. A geodesic is a geometric object whose nature does't change. Tidal forces exist if and only if a spherical object placed in the field which is subject to no external sources will become deformed do to the non-vanishing Riemann tensor. The fact that the coordinate acceleeration varies with height does not mean that there is geodesic deviation. In fact if you were to calculate the deviation you'd find it to be zero. Pete I wasn't thinking in terms of any technical term such as "geodesic deviation", when MeJennifer talked about "diverging geodesics in the direction of motion" I thought this was just a reference to Boustrophedon's earlier comment that "An aggregate of free falling test particles would move apart in the direction of motion but would not move together across the same direction." I think this was meant as a reference to the fact that the strength of a "uniform gravitational field" actually varies spatially, so that something higher in the field would have a lower rate of acceleration than something lower in the field, so in this coordinate system the distance between dropped test particles would increase as they fell. My point was just that any phenomenon that you explain in terms of a "uniform gravitational field" can be transformed into an inertial coordinate system where there is no gravity present, and must therefore be explainable in terms of the ordinary inertial motion of test particles in this coordinate system (assuming the distances between the test particles was still changing in the inertial coordinate system, it would just be because they had different initial velocities). pervect Staff Emeritus Science Advisor From previous discussions, I remember it was mentioned that if the spacetime curvature is zero in one coordinate system, it will be zero in all coordinate systems. Also, in posts #64 and #69 here pervect says that a universe totally empty of matter can be modeled either as an expanding universe with negative spatial (not spacetime) curvature, or as an ordinary static Minkowski universe with no spatial curvature. Yep. I also give the metric for this case in https://www.physicsforums.com/showpost.php?p=754243&postcount=78 It's related to the Milne cosmology, so I called it the Milne metric. The moral of the story is that distance measures depend on the notion of simultaneity one adopts. pervect Staff Emeritus Science Advisor Geodesic deviation is also not frame dependant. If it were then the Riemann tensor would also be frame dependant. A geodesic is a geometric object whose nature does't change. Tidal forces exist if and only if a spherical object placed in the field which is subject to no external sources will become deformed do to the non-vanishing Riemann tensor. The fact that the coordinate acceleeration varies with height does not mean that there is geodesic deviation. In fact if you were to calculate the deviation you'd find it to be zero. Pete True, but consider the Milne metric from https://www.physicsforums.com/showpost.php?p=754243&postcount=78 $$ds^2 = -dt^2 + t^2 d \chi^2 + t^2 sinh(\chi)^2 d \theta^2 + t^2 sinh(\chi)^2 sin(\theta)^2 d \phi^2$$ It has a zero Riemann. [itex]\chi=\theta=\phi$ = constant are geodesics - this can be seen from the fact that the above metric is an example of a FRW metric.

Becauses the scale factor is a(t)^2 = t^2, we have the scale factor a(t)=t and thus nearby geodesics do not accelerate away from each other. Thus there is no geodesic deviation, as one would expect from a metric whose Riemann is zero and the geodesic deviation equation.

However, while the geodesics do not accelerate away from each other, they do not maintain a constant distance either. Because the geodesics are given by $\chi$= constant, the distance between them increases proportionally to the scale factor a(t), i.e. the distance between geodesics is proportional to time and not constant.

The distance measure used with the Milne metric is different from the distance measure used with the flat Minkowski metric, even though there is a variable transformation that maps one into the other.

i.e. if you substitute

$t1 = t*cosh(\chi)$, $r1 = -t*sinh(\chi)$

into the Minkowski metric

$-dt1^2 + dr1^2 + r1^2(d \theta^2 + sin(\theta)^2 d \phi^2)$

you get the Milne metric.

This is an example of how distance measures are coordinate dependent - different notions of simultaneity give rise to different notions of distance.