Is a uniform gravitational field a gravitational field?

[MeJennifer]

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?

 

[yogi]

Good Question

 

[pervect]

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]

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.

 

[pmb_phy]

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 ...&quot;<br /> <br /> 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<br /> <br /> <b>Principle of Equivalence,</b> F. Rohrlich, Ann. Phys. 22, 169-191, (1963), page 173 (aailable by request in e-mail).<br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Or in other words how do we explain the seemingly gravitational effects in a space-time region with zero intrinsic curvature? </div> </div> </blockquote>[/quote]<br /> What effects are these? You&#039;re most likely asking about the differences between gravitational effects and tidal effects.<br /> <br /> Best wishes<br /> <br /> Pete

 

[pmb_phy]

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

 

[masudr]

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

 

[pmb_phy]

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

 

[masudr]

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]

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.

 

[Boustrophedon]

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]

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]

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.

 

[masudr]

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]

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

 

[Boustrophedon]

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.

 

[MeJennifer]

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?

 

[Boustrophedon]

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.

 

[MeJennifer]

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]

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

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.

 

[Haelfix]

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 that's 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]

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?

 

[MeJennifer]

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]

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.

 

[MeJennifer]

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]

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?"

 

[pmb_phy]

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]

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]

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]

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

<br /> 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<br />

It has a zero Riemann.

\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.

[add]
I should probably add, though, that even in flat space-time with the usual Minkowski coordinates, it's possible to have a set of geodesics which depart linearly away from each other, due to differences in initial velocity.So perhaps this may have been a long exercise in semantics.

 

[pmb_phy]

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 wasa responding to that comment about an aggregate of particles in free fall but I made the mistake of assuming that the poster was referring to a uniform g-field. I was actually responding to this comment:

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.

As far as "real" gravitational field. What is "real" is what you define it to be. When Einstein created GR he defined the gravitational field such that a uniform gravitational field was equivalent to a uniformly accelerating frame of reference. That means that, according to Einstein, the quantity which determines the presense of a gravitational field is the non-vanishing of the affine connection and not the non-vanishing of the Riemann tensor.

I recall an article written in the Am. J. Phys. by someone who associated gravity with spacetime curvature. He started off with a wrong definition. So when his derivation showed that the uniform g-field had spacetime curvature he didn't question his result, he strutted with pride that he proved that a uniform g-filed has tidal forces.

Pete

 

[Boustrophedon]

I don't think you'd get many takers for the notion that "what's real is what you define it to be". I think Einstein actually used the term "homogeneous" which is less ambiguous. A homogeneous gravitational field would indeed have no tidal effects and be equivalent to an accelerating system but since no distribution of matter can produce such a field it is not "realistic" or not "real".
Whilst mathematics is indispensable for describing the physical world, it does not follow that anything describable in mathematics necessarily has a counterpart in the "real" world.

 

[pmb_phy]

I don't think you'd get many takers for the notion that "what's real is what you define it to be".

Do you know when o = sqrt(-1) came into existence? When someone said "That number when multiplied by itself equals -1" - And then it existed. I didn't make that to be a formal definition which works in all concievable cases. It applies in defining terminology in physics more or less.

The rest seems to be about semantics/definitions and I loathe semantics.

Pete

 

[Boustrophedon]

Hmmm, and at the time it was called "imaginary". When Gauss reduced it to just algebraic properties of number pairs it became "complex". So what ? Existence in mathematics has quite a different implication from its meaning in physics. As J.L.Synge once wrote:
"A cube is after all a cube, and not merely a set of number triads subject to three linear inequalities".

 

[JesseM]

I don't think you'd get many takers for the notion that "what's real is what you define it to be". I think Einstein actually used the term "homogeneous" which is less ambiguous. A homogeneous gravitational field would indeed have no tidal effects and be equivalent to an accelerating system but since no distribution of matter can produce such a field it is not "realistic" or not "real".

But a uniform gravitational field is just what is seen in flat spacetime in a certain type of accelerating coordinate system--since it's flat spacetime, you just need a complete absence of matter.

 

[pmb_phy]

Hmmm, and at the time it was called "imaginary". When Gauss reduced it to just algebraic properties of number pairs it became "complex". So what ? Existence in mathematics has quite a different implication from its meaning in physics. As J.L.Synge once wrote:
"A cube is after all a cube, and not merely a set of number triads subject to three linear inequalities".

You're kidding right?? I was giving an example of what it can mean for something to exist. So what if I used an example from math. The example of a gravitational field is a perfect example of something that exisists or not depending on how "gravitational field" is defined.

 

[Boustrophedon]

Not really. The EP merely states an 'equivalence' between acceleration and gravitation - it does not say that an accelerating frame "is" a gravitational field any more than it says a gravitational field "is" an accelerating system.
The term 'gravitational' means a field created by, and only by, the presence of matter. I seem to remember the phrase "matter tells space how to curve and curved space tells matter how to move" being a recurring motif in Misner, Thorne and Wheeler.

 

[pmb_phy]

Not really. The EP merely states an 'equivalence' between acceleration and gravitation - it does not say that an accelerating frame "is" a gravitational field any more than it says a gravitational field "is" an accelerating system.

To be exact, the equivalence principle states

A uniform gravitational field is equivalent to a uniformly accelerating frame of reference

Let's take a look at Einstein's wording on this.

From The Foundation of the General Theory of Relativity A. Einstein, Annalen der Physik, 49, 1916;

Let K be a Galilean system of reference, i.e. a system of relatively to which (at least int he four-dimensional region under consideration) a mass, sufficiently distant from other masses, is moving in uniform motion in a straight line. Let K' be a second system of reference moving relatively to K in a uniformly accelerated translation. Then, relatively to K', a mass sufficiently distant from other masses would have an accelerated motion such that its acceleration and direction of acceleration are independant of the material composition and physical state of the mass.
Does this permit an observer at rest relatively to K' to infer that he is on a
"really" accelerating system of reference? The answer is in the negative; for the above-mentioned relation of freely movable masses to K' may be interpreted equally well in the following way. The system of reference K' is unacelerated, but the space-time territory in question is under the sway of a gravitational field, which generates the accelerated motion of the bodies relatively to K'.
[...]
It will be seen from these reflexions that in pursuing the general theory of relativity we shall be led to a theory of gravitation, since we are able to "produce" a gravitational field merely but changing coordinates.

In Einstein's book Relativity: The Special and the General Theory Einstein writes on page 172 (skipping over the redundant part given in the above paper, Einstein uses the systems S₁ an S₂ instead of K and K', respectively

Relative to S₂, therefore, there exists a state which, at least to the first approximation, cannot be distinguished from a gravitational field. The foloowing concept is thus compatible with the observed facts: S₂ is also equivalent to an "inertial system"; but with respect to S₂ a (homogeneous) gravitational system is present (about the origin of which one does not worry in this connection).

In Einstein's book The Meaning of Relativity he writes on page 57

Let now K be an inertial system. Masses whch are sufficiently far from each other and from other bodies are then, with respect to K, free from acceleration. We shall also refer to these masses to a system of co-ordinates K', uniformly accelerated with respect to K. Relatively to K' all the masses are have equal and parallel direcetions of acceleration; with respect to K' they behave just as if a gravitational field were present and K' were unaccelerated. Overlooking for the present the "cause" of such a gravitational field, which will occupy is later, there is nothing to prevent our conceiving this gravitational field as real, that is, the conception that K' is "at rest" and a gravitational field is present and we may consider as equivalent to the conception that only K is an "allowable" system of coordinates and no gravitational field is present. The assumption of the complete equivalence of of the system of coordinates, K and K', we call the "principle of equivalence.

The term 'gravitational' means a field created by, and only by, the presence of matter.

If your name is Sir Issac Newton then I'd agree with you. But if your name was Albert Einstein then I'd disagree with you. In any case nobody is saying that there is no source. Dennis Sciama formulated a scheme that had a source to this "produced" gravitational field. Its somewhere in Peacock's "Cosmological Physics" but at the moment I can't locate it. A more concrete example is that of a sphere with uniform mmass density with a cavity cut out from within the sphere. Inside this cavity there will be a uniform gravitational field. If you're falling from rest, i.e. relative to the sphere itself, then you are in flat spacetime (i.e. withing the cavity spacetime is flat) and can consider yourself at rest in an inertial frame, that of the falling frame. However if you now change from your falling frame to one which is uniformly accelerating relative you the inertial (falling) frame in such a way that you are at rest with respeect to the cavity then there is a uniform gravitational field present and there is a definite source to this field.

Pete

 

[Boustrophedon]

Your "exactly" stated equivalence principle is incorrect. There are various different ways of expressing the weak and strong EP and yours is not a good one since it implies that only the uniform cases are equivalent.
The point is that "uniform" acceleration is just that, whereas gravitational fields never are - which is why it's always possible to distinguish them by detecting the tidal effects.

I can't see much mileage in century-old quotes from Albert E. Is it the case that no progress has been made, no clearer or better understanding achieved subsequently that reference must be made to original writings, as in religious scripture ? His was the first word, not the last - he often changed his mind and did not see his theory in the precision that has been worked on it since.

I don't think I agree with your spherical cavity being perfectly uniform (and non-zero ). In any case I don't see any inconsistency with the last quote about "...created by, and only by...". An 'ordinary' gravitational field would be equivalent to a certain non-uniformly accelerating system but it does not mean that acceleration per se 'creates' gravitation. The indistinguishability is highly local - investigation of more distant environment easily determines which is which.

 

[pmb_phy]

Your "exactly" stated equivalence principle is incorrect. There are various different ways of expressing the weak and strong EP and yours is not a good one since it implies that only the uniform cases are equivalent.
The point is that "uniform" acceleration is just that, whereas gravitational fields never are - which is why it's always possible to distinguish them by detecting the tidal effects.

I can't see much mileage in century-old quotes from Albert E. Is it the case that no progress has been made, no clearer or better understanding achieved subsequently that reference must be made to original writings, as in religious scripture ? His was the first word, not the last - he often changed his mind and did not see his theory in the precision that has been worked on it since.

I don't think I agree with your spherical cavity being perfectly uniform (and non-zero ). In any case I don't see any inconsistency with the last quote about "...created by, and only by...". An 'ordinary' gravitational field would be equivalent to a certain non-uniformly accelerating system but it does not mean that acceleration per se 'creates' gravitation. The indistinguishability is highly local - investigation of more distant environment easily determines which is which.

From hereon in I doubt that anything new will be said on either side so I will humbly bow out of this discussion.

In any case the GR course at Harvard uses the very text I just quoted from, i.e. The Meaning of Relativity.

Pete

 

[Boustrophedon]

Lots of stuff gets recommended as 'backround reading'...

 

[pmb_phy]

Lots of stuff gets recommended as 'backround reading'...

That text is the main text used for the course. It is not background material. That was the way it was when I spoke to the guy who taught GR at Harvard anyway (2003?). I don't know what has been happening since.

Why do you have such a distaste for Einstein's original work? Do you believe somehow people have changed GR from its day of creation? I think it was MTW who claimed it has remained unchanged since then. Text like that often quote Einstein.

I see I let myself get looped into responding to a thread I chose not to participate anymore. The reason I chose not to respond is that each response I give you will be met with rejection when none is warrented. That's very boring and those conversations never go anywhere. There's always two people who believe that they can "prove" the other wrong or that they can change the other person's mind. I've never seen this happen in my lifetime so I don't see it happening here.

Bye bye

Pete

 

[tehno]

Einstein didn't consider gravitational effects as the consequence of any force field ,but simply as effects of changed space-time metric upon bodies,didn't he?If that view is supported ,uniform or nonuniform gravitational "fields" are both nonexistent...

 

[Boustrophedon]

I don't recall having expressed any distaste for Einstein's original work, yet I would be genuinely surprised if Einstein's book was used as the main text in a Harvard course. There are a good few 'classic' texts by Pauli, Eddington, Weyl etc. as well as Einstein but it is not showing 'distaste' to point out that they are seriously out of date for all but subsidiary reading in a modern GR course.

It is simply not valid to state the EP as "a uniform gravitational field is equivalent to a uniformly accelerating frame of reference". Whilst we may be able to agree on what a uniformly accelerating frame is, we are left with the obvious question - "What is a uniform gravitational field ?" that started this thread. If you have to define it as "the field that is created by, or experienced in a uniformly accelerating frame" then the given statement of the EP becomes a meaningless tautology.

The EP can be expressed in a weak form to do with equivalence of inertial and gravitational mass and in this form it is exact. It can also be expressed in a strong form in which it is inexact, namely that the laws of physics are unchanged in a freely falling reference frame. The inexactness of course being due to the tidal effects of inherently non-uniform gravitational fields.

 

[pmb_phy]

I don't recall having expressed any distaste for Einstein's original work, yet I would be genuinely surprised if Einstein's book was used as the main text in a Harvard course. There are a good few 'classic' texts by Pauli, Eddington, Weyl etc. as well as Einstein but it is not showing 'distaste' to point out that they are seriously out of date for all but subsidiary reading in a modern GR course.

It is simply not valid to state the EP as "a uniform gravitational field is equivalent to a uniformly accelerating frame of reference". Whilst we may be able to agree on what a uniformly accelerating frame is, we are left with the obvious question - "What is a uniform gravitational field ?" that started this thread. If you have to define it as "the field that is created by, or experienced in a uniformly accelerating frame" then the given statement of the EP becomes a meaningless tautology.

The EP can be expressed in a weak form to do with equivalence of inertial and gravitational mass and in this form it is exact. It can also be expressed in a strong form in which it is inexact, namely that the laws of physics are unchanged in a freely falling reference frame. The inexactness of course being due to the tidal effects of inherently non-uniform gravitational fields.

I don't comprehend why you're making comments to me whenI told you that I wouldn't be posting again in this forum. I certainly hope that its not just a way to get the last word in. Perhaps you felt you needed to post a rebuttle?

In any case I've been reviewing my personal posting habits and the decesions I've made because of them. It led to a lot of people being blocked so as to minimize reading insults, jibes etc. towards me. The moderators do a lousy job in this area when it comes to me.

If you so desire I will continue on with this thread. But I can almost promise you that it will reduce to "Yes it does!" and "No it doesn't!" responses.

To begin with I'll answer one of your questions, i.e. "What is a uniform gravitational field ?" This is no mystery and never has been. A uniform g-field is a g-field in which the Riemann tensor vanishes in the spacetime domain in which the field is held to be uniform. You'll find this in articles such as

[1] Principle of Equivalence, F. Rohrlich, Ann. Phys. 22, 169-191, (1963), page 173

[2] Radiation from a Uniformly Accelerated Charge, David G. Boulware, Ann. Phys., 124, (1980), page 174

[3] Relativistic solutions to the falling body in a uniform gravitational field, Carl G. Adler, Robert W. Brehme, Am. J. Phys. 59 (3), March 1991

Then there is your supposition 'It is simply not valid to state the EP as "a uniform gravitational field is equivalent to a uniformly accelerating frame of reference'".

This kind of statement will required additional questions which the poster could head off. In this case it is on the poster to prove his claim that the equivalence principle as I stated above is wrong. This statement is a postulate and cannot be determined to be wrong by reasoning alone. Experimental results can only prove a postulate to be correct. On could actually define the uniform g-field as the spacetime described by the metric which is obtained when transformed to a uniformly accelerating frame of reference.

Warning: This seems to be one of those arguements which don't have an ending. It will most likely end with you and I going back and forth saying the same thing.

I don't mean any of the above comments that I wrote to be anything but professional. In that tone it may have come across like I'm being a pain in the rump. But I assure you that I'm a kind and generaous person who has an infinite amount of patience when I'm not being slammed and I'm not constantly repeating myself over and over and over again. At that point I do not respond in kind. I simply stop responding. Okay? I mean I didn't want us to get off on the wrong foot.

Best wishes

Pete

 

[vanesch]

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.

But in a uniform gravitational field, they wouldn't move apart.
Take a uniform gravitational field in Newtonian physics (the only place where such a notion makes sense in fact).

We start with particles at x1_0,x2_0,x3_0 and at rest in the "inertial frame" (Newtonian inertial frame) and have constant g in the x-direction.

At t=0, x1(t=0) = x1_0 ; x2(t=0) = x2_0 ; x3(t=0) = x3_0.

At another t, we have:

x1(t) = g/2 t^2 + x1_0
x2(t) = g/2 t^2 + x2_0
x3(t) = g/2 t^2 + x3_0


so x1-x2 remains x1_0 - x2_0
etc...

In other words, in a uniform gravity field in Newtonian mechanics, there is no change in distance between the particles.

Now, in GR, a uniform gravity field is just a funny choice of coordinates, which can be transformed away in a Minkowski metric. There also, the distance between the test particles (in the Minkowski coordinates) remains constant if they were "at rest" at x^0 = 0.

Don't confuse particles at different positions in a uniform gravity field, with particles DROPPED AT DIFFERENT TIMES in a uniform gravity field.
THOSE particles separate in distance of course. But that is because at the moment where both of them are "free" (and hence follow geodesics), they are not both at rest (only the second one, who is just dropped, is).

 

[Boustrophedon]

A uniform g-field is a g-field in which the Riemann tensor vanishes in the spacetime domain in which the field is held to be uniform.

I don't think this avoids both pitfalls of tautology or triviality. If Riemann is zero, how is the g-field supposed to exist ? It must either be zero or ...have arisen from uniform acceleration !

While Einstein may not have considered a non-vanishing Riemann tensor to be the criterion for 'gravitation', it has become, I think, the prevalent modern view. The Adler & Brehme paper is just one example that commits what I think is called an 'abuse of terminology' in ascribing the term 'gravitational field' to a region of flat spacetime devoid of matter.

The concept of 'uniform gravitational field' amounts to "that gravitational field that would be equivalent to a uniformly accelerated frame". The metric is just the metric of uniform acceleration as it is represented. However, I feel that such a policy creates both a redundancy and a confusion. We have a redundancy of two names for exactly the same thing, and we have a confusion over how gravitation can be supposed to exist in flat spacetime in the absence of matter.

If authors that like to refer to "in a uniform gravitational field" would simply amend their terminology to "in a uniformly accelerating reference frame", nothing would be lost while gaining greater consistency in the definition of gravitation and spacetime curvature.

Vanesch writes:

But in a uniform gravitational field, they wouldn't move apart.

I covered this in an earlier post. "Uniform" is ambiguous. Things can increase or decrease uniformly. Authors sometimes use uniform to mean homogeneous or isotropic (as you are) and sometimes (more often) they use 'uniform gravitational field' to mean merely 'linear' in the sense that the field may increase or decrease longitudinally but be constant laterally. I would regard your isotropic field as equivalent to uniform acceleration but because of peculiar 'Lorentz contraction' beliefs, many authors think a uniformly accelerated frame must have a g-force that decreases toward the 'front'. Consequently test particles do move apart in their kind of equivalent 'uniform fields'. Needless to say I think both tyes of 'uniform gravitational field' are unnecessary fictions.

 

[pmb_phy]

I don't think this avoids both pitfalls of tautology or triviality. If Riemann is zero, how is the g-field supposed to exist ?

If you had asked Einstein (or read his work on this subject) then you'd get a clear answer - The non-vanishing of the gravitational field does not require the non-vanishing of the Riemann tensor. It onlyu requires the non-vanishing of the affine connection. See page 467 of MTW (No \Gamma's mean no "gravitational field"...)

While Einstein may not have considered a non-vanishing Riemann tensor to be the criterion for 'gravitation', it has become, I think, the prevalent modern view.

Einstein was aware of this view and he rejected it.

Note: If this conversation has turned to the meaning and quantities which are required to nont vanish when there is a gravitational field present then I have zero problem with Einstein's view. This so-called "modern view" is based on the some peoples desire that the existence of something is an absolute. I have no such qualms and Einstein didn't either. Einstein was a massiovely shapr dude and I haven't seen an aruement which has proved anynotion he held fir, to have ever been proven wrong (changing from Affine connection to Riemann tensor for existence of grvitational field is not a proof that Einstein's view was wrong). If that is all we are going to talk about then why rehash the same issues? Do you actually think you could change my mind? Do you think you've presented an argument I haven't heard many times before? Do you think I haven't researched this to death? Well I don't think that of you and I believe that your mind cannot be changed since you are set in your views just as well as I am set in mine.

So what do you wish to talk about now without having to rehash or repeat something we went over already?

Pete

ps - Please note that I'm not trying to be a wise acre. I'm rally a kind person at heart even if I don't come across it over the internet. I just don't want you to get the idea that I'm paying short shrift to what you're saying to me. I believe that you believe what you've said and there is a large force in physics backing you up. However I don't hold an opinion because I've taken a poll and have gone with the most popular opinion. I hold an opinion for the same reason as you - We believe it is the most logical. Do you disagree with this?

 

[vanesch]

I covered this in an earlier post. "Uniform" is ambiguous.

Sure, but I thought it meant what it means in Newtonian physics, like, for small creatures like us, in small laboratories, approximately what is observed "at the surface of the earth" on scales way way below the Earth radius.

Once you start thinking about what a uniform gravitational field might be, independent of an observer frame, you are right: it can probably mean different things. For instance, a constant Ricci scalar...

Things can increase or decrease uniformly.

Mmmm, that's tricky if you want to be observer-independent !

And if you allow for observer dependence, then the Rindler observer which has constant acceleration, does also the trick...

 

[masudr]

See page 467 of MTW (No \Gamma's mean no "gravitational field"...)

Come on guys, surely this is a case of semantics. You can call a gravitational field whatever you want really; as long as you are precise about what you mean. In any case, it's probably better to stick to mathematical language and say straight out: the connection vanishes (or doesn't) or the Riemann tensor vanishes (or doesn't).

The components of the Levi-Civita connection can be made zero by a suitable coordinate transformation. The components of the Riemann tensor cannot. I know we all know this. In this sense, the mathematical language is precise.

However, I'd just add, that my opinion on it is that we shouldn't attribute physical reality to things that may just be an artifact of the particular coordinate system we have chosen (see for example, the problem that Eddington-Finkelstein coordinates resolve in the Schwarzschild metric). In this sense, using the connection to define the existence of a gravitational field is counter-intuitive to the development of general relativity; which aims to move away from coordinate dependent description (general covariance, anyone?)