# Is a uniform gravitational field a gravitational field?

1. Feb 13, 2007

### 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?

2. Feb 13, 2007

### yogi

Good Question

3. Feb 13, 2007

### 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.

4. Feb 13, 2007

### 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.

Last edited: Feb 13, 2007
5. Feb 14, 2007

### pmb_phy

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 [itex]\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).
[/quote]
What effects are these? You're most likely asking about the differences between gravitational effects and tidal effects.

Best wishes

Pete

6. Feb 14, 2007

### pmb_phy

Hi MeJennifer - Please check your PM messages too.

Pete

7. Feb 14, 2007

### masudr

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: Feb 14, 2007
8. Feb 14, 2007

### pmb_phy

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

9. Feb 14, 2007

### masudr

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.

10. Feb 14, 2007

### pervect

Staff Emeritus
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.

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: Feb 14, 2007
11. Feb 14, 2007

### 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.

12. Feb 14, 2007

### pervect

Staff Emeritus
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.

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.

13. Feb 14, 2007

### nrqed

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.

14. Feb 14, 2007

### masudr

Not over an infinitesimally small region no, but there certainly is in any finite region, as discussed above.

15. Feb 14, 2007

### nrqed

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

16. Feb 15, 2007

### Boustrophedon

Pervect wrote:
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.

17. Feb 15, 2007

### MeJennifer

Could we even detect such a field on a pseudo-Riemanian manifold?

18. Feb 15, 2007

### 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.

19. Feb 15, 2007

### MeJennifer

Yes, you are correct, that would be detectable.

20. Feb 15, 2007

### pervect

Staff Emeritus
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: May 2, 2017
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