Is a uniform gravitational field a gravitational field?

[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?)

 

[pmb_phy]

Come on guys, surely this is a case of semantics.

Its a case of terminology. Boustrophedon and myself have fundamentally different viewsa on an issue or two. I had meant that to get through in my previous post but I guess I did a lousy job at that. I suggest to Boustrophedon that we agree to disagree since it is impossible to say the other is wrong if their source of disagreement is what they accept as terminology.

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).

One has to understand what they're talking about really well if they are to use the non-vanishing of affine connection to determine the presence of a gravitational field. E.g. one can choose spatial polar coordinates in Minkowski spacetime in an inertial frame and all the affine connection components will not vanish. One has to understand that in the chosen frame of reference in locally Cartesian coordinates the non-vanishing affine connections mean a non-vanishing g-field.

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?)

I'm trying my best to avoid those discussions which are about terminology. There is always someone who says something as an absolute, i.e. if you believe differently then you don't truly understand GR. If I correct that statement once then two more of the same kind of statements or simple rejections with nothing more than "No it isn't". Its very frustrating to know when to end it, especiallay when most people refuse to agree to disagree.

Note: Boustrophedon - I am not talking about you per se in the above. This has happened elsewhere hundreds of times to me. I believe that you and I can agree to disagree. By that I mean that I understand your view and you understand my view and we each understand that the other understands their views but we agree that this is the case and leave it at that.

Kind regards

Pete
Pete

 

[pmb_phy]

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.

I fail to understand how the term uniform/homogeneous could be confusing to you. The equivalence principle tells you what a uniform/homogeneous g-field is. I've only seen one relativist get this wrong out of many journal articles I've read on this subject.

Pete

 

[Boustrophedon]

I acknowledge with pmb_phy that we have differing opinions on terminology of gravitation on the one hand and how readily things simply 'defined' may be said to 'exist' on the other. I quite accept that these are issues that cannot be nailed down and would be fruitless to pursue.

I would, however, like to distinguish 'uniform' from 'homogeneous'. I cannot think of any usage, scientific or otherwise, of the term 'homogeneous' where it means anything other than constant, equal or ' the same' in all parts or in all directions. A homogeneous g-field would then be one where the 'g-force' felt would be identical at whichever spatial position it's measured. Forward, back, up down, sideways, anywhere.

'Uniform' is a more slippery term and does not have a consistent meaning in general usage. In physics the term 'uniform g-field' has come to mean something quite different from 'homogeneous'. By far the most common usage is to mean a g-field where the felt 'g-force' varies in inverse linear proportion to distance, either in the direction of the g-force or against it, whilst remaining constant in any perpendicular plane. Test particle trajectories in a uniform g-field are usually represented as various hyperbolae belonging to the same vertex.

 

[vanesch]

'Uniform' is a more slippery term and does not have a consistent meaning in general usage. In physics the term 'uniform g-field' has come to mean something quite different from 'homogeneous'. By far the most common usage is to mean a g-field where the felt 'g-force' varies in inverse linear proportion to distance, either in the direction of the g-force or against it, whilst remaining constant in any perpendicular plane.

Yes, but to even be able to SAY what it means "varies in inverse linear proportion to distance" on a general manifold, you need to know the metric, (or at least have a connection). Moreover, the concept of "perpendicular plane" is also given by the metric.

 

[vanesch]

I fail to understand how the term uniform/homogeneous could be confusing to you. The equivalence principle tells you what a uniform/homogeneous g-field is. I've only seen one relativist get this wrong out of many journal articles I've read on this subject.

Initially, I also, like you, understood "uniform gravitational field" as the equivalent of an accelerated observer in a flat space. This simply comes about because in Newtonian gravity, there's not much doubt about what is the gravitational field: it is the vector field of gravitational acceleration in a Newtonian "inertial frame" (no matter how that is defined!), over the Euclidean space. So a uniform gravitational field in Newtonian physics is simply the physical situation where we have g = constant vector over all of space and time, understanding that we are working in an "inertial frame" (one in which Newton's laws are valid).

However, to the remark "uniform" could mean other things than this, one needs indeed to be more specific. What is "gravitational field" ? Is it the connection ? Is it the metric tensor ? Is it the Riemann tensor ? All of them are related of course, but depending on which one exactly one is going to require to be "uniform" (is this not also a coordinate-dependent notion ?), there can be different choices, leading to different physical situations.

 

[Boustrophedon]

We're not talking about a general manifold - this is 'flat' spacetime. I am reporting what I perceive as the commonly held usage of 'uniform g-field' in the literature, not that I necessarily agree with it. The 'inverse linear proportion' comes from a requirement that distance measures remain constant for a free falling observer. See also 'uniformly accelerating reference frames' for the equivalent relationships.

 

[JesseM]

A while ago pervect posted some explanation and links to info on what is meant by "uniform gravitational field" in GR, in post#9 of this thread. From what I gather, a uniform gravitational field is actually not uniform in terms of the G-forces experienced at different heights, but it is equivalent to what would be measured in flat spacetime if you were using a coordinate system whose position coordinates were determined by rulers undergoing Born rigid acceleration. Here was pervect's post:

If two observers are undergoing Born rigid acceleration each observer (the front and back observer) will measure a different acceleration with his or her local accelerometer.

See for instance http://www.mathpages.com/home/kmath422/kmath422.htm

Quote:
Trailing sections of the rod must undergo a greater acceleration in order to maintain Born rigidity with the leading end, and the required acceleration is inversely proportional to the distance from the pivot event
http://arxiv.org/abs/physics/9810017 also discusses this. This is the simplest peer reviewed English language reference I've been able to find on the topic. (The mathpages article is also pretty good IMO and may be easier to follow though of course it is not peer reviewed).

What the literature calls a "uniform gravitational field" isn't actually uniform as measured by local accelerometers.

See for instance http://arxiv.org/PS_cache/physics/pdf/0604/0604025.pdf for an example of this usage.

While perhaps the naming choice is unfortunate, it appears to be what the literature uses :-(.\
The metric is defined not by setting the "felt acceleration" to a constant, but by setting the Ricci tensor to zero. I.e one is looking for a vacuum solution to Einstein's equations, this is the sort of result one gets in an accelerating spaceship, where as we've already seen the acceleration as measured with a local accelerometer depends on position (is not uniform) as long as the spaceship is "rigid".

Note that one can apply the notion of "radar rigidity" as well as "Born rigidity" to define a "rigid spaceship" - while radar distance isn't equal to the distance as defined by the Lorentz interval, in an accelerating frame an object with a constant radar distance will also have a constant distance as measured by the Lorentz interval. Thus both distance measures will agree as far as rigidity goes, even though they are not exactly equivalent.

 

[pmb_phy]

A while ago pervect posted some explanation and links to info on what is meant by "uniform gravitational field" in GR, in post#9 of this thread. From what I gather, a uniform gravitational field is actually not uniform in terms of the G-forces experienced at different heights, but it is equivalent to what would be measured in flat spacetime if you were using a coordinate system whose position coordinates were determined by rulers undergoing Born rigid acceleration. Here was pervect's post:

Actually the literature uses the term "uniform" to mean, in part, "no tidal forces (zero Riemann tensor)". That's about as simple of a definition that you'll find

Pete

 

[pervect]

Actually the literature uses the term "uniform" to mean, in part, "no tidal forces (zero Riemann tensor)". That's about as simple of a definition that you'll find

Pete

I agree with Pete about the standard usage in the literature, a uniform gravitational field is usually taken to mean one with a zero Riemann tensor from what I've read.

My disagareement with Boustrophedon should be reasonably well known by now. If I haven't responded, its not because I've changed my mind on the issue or think his arguments have any merit, it's basically because I'm short on time, have already given numerous references to the literature, and don't see the point of repeating myself endlessly like a broken record.

Something that I probably should have mentioned before but never got around to is that equating components of the Riemann tensor with tidal forces really only works for observers following a geodesic, as the derivation of this equivalence comes from the geodesic deviation equation.

This means that the zero Riemann tensor isn't quite the same as "no tidal forces" for an accelerating observer, if one interprets the tidal force as the difference in proper accelerations between two ends of a rigid bar (which is how I would define and measure a tidal force).

Next up is probably a long argument on how to define a tidal force :-).

 

[pmb_phy]

Next up is probably a long argument on how to define a tidal force :-).

I know you're just kidding but for the benefit of those who don't know ...


In Newtonian mechanics

http://www.geocities.com/physics_world/mech/tidal_force_tensor.htm


In General Relativity; Multiply Eq.(22), the 4-tidal-acceleration through by proper mass amd that gives you the tidal force on one of the particles in reference to the other particle.

http://www.geocities.com/physics_world/gr/geodesic_deviation.htm


Pete

 

[MeJennifer]

This means that the zero Riemann tensor isn't quite the same as "no tidal forces" for an accelerating observer, if one interprets the tidal force as the difference in proper accelerations between two ends of a rigid bar (which is how I would define and measure a tidal force).

Well for clarity's sake, a tidal force is not a force.
But more importantly, widening the definition of 'tidal forces', by including effects in flat space-time, creates more confusion than that it clarifies things.

Even in flat space-time there is a difference in proper acceleration between two ends of a rigid bar that is aligned in the direction of motion. Such effects, however, have nothing to do with forces or curvature, they are the consequence of the non-positive definite signature of the metric, which result in hyperbolic as opposed to euclidean space relationships.

On the other hand the curvature of the Riemann tensor is observer independent it is either curved or it is not and it is, arguably, not related to the non-positive definite signature of the metric.

 

[pmb_phy]

In the first place nobody in the GR community defines tidal force as the difference in proper accelerations between two ends of a rigid bar. If the spacetime is Minkowskian then there will be no stress in the rod and that is indicative of the presence of a 4-vector.

As Kip S. Thorne states in Black Holes & Time Warps on page 110

...tidal gravity is a manifestation of spacetime curvature.

Thus it is impossible to have tidal forces without spacetime curvature. And I pesonally trust Kip Thorne in anything he writes about GR!

Well for clarity's sake, a tidal force is not a force.

On the contrary, tidal force is a 4-force and thus a force under any definition. See Eq. (22) which shows the relative 4-acceleration of one particle with respect to another particle. Multiply this acceleration 4-vector by the proper mass of the first particle and you will have the 4-force on that particle.

http://www.geocities.com/physics_world/gr/geodesic_deviation.htm

But more importantly, widening the definition of 'tidal forces', by including effects in flat space-time, creates more confusion than that it clarifies things.

I agree with ya.

On the other hand the curvature of the Riemann tensor is observer independent it is either curved or it is not and it is, arguably, not related to the non-positive definite signature of the metric.

Quite correct.

Boy MJ. You really picked a topic which drew a lot of attention. Quite an intelligent and very important question at that too. Is it possible to e-mail an article to you? I want to show you what can happen when someone doesn't understand what you now understand. I'm referring to an article from the American Journal of Physics that I scanned into my computer. If you don't/can't receive attachments then, if there is some interest in this article, I will post it on my web page anbd let it remain there for several days, long enough for people with interest to download it.

Pete

 

[robphy]

On the contrary, tidal force is a 4-force and thus a force under any definition. See Eq. (22) which shows the relative 4-acceleration of one particle with respect to another particle. Multiply this acceleration 4-vector by the proper mass of the first particle and you will have the 4-force on that particle.

In this construction of "geodesic deviation", you deal with two geodesics, each with zero worldline-curvature, i.e. zero 4-acceleration. It seems to me that the relative-acceleration between these two geodesics [which is defined by two geodesics] is different from the 4-acceleration [the worldline-curvature] of a single worldline. In my understanding, Newton's Law says that the net 4-force on a particle is proportional to the particle's worldline-curvature [particle's 4-acceleration].

So, it seems to me that "tidal force" is not a 4-force in the sense I just described.

Can you provide a "standard reference" for the claim that "tidal force is a 4-force"?

 

[pmb_phy]

In this construction of "geodesic deviation", you deal with two geodesics, each with zero worldline-curvature, i.e. zero 4-acceleration. It seems to me that the relative-acceleration between these two geodesics [which is defined by two geodesics] is different from the 4-acceleration [the worldline-curvature] of a single worldline. In my understanding, Newton's Law says that the net 4-force on a particle is proportional to the particle's worldline-curvature [particle's 4-acceleration].

So, it seems to me that "tidal force" is not a 4-force in the sense I just described.

But you're referring to a single geodesic whereas geodesic deviation deals with two geodesics. Each geodesic has zero 4-acceleration but the relative 4-acceleration of two particles on two geodesics which are deviating has a non-zero value.

Can you provide a "standard reference" for the claim that "tidal force is a 4-force"?

Nope. I also can't find the operating manual for a wheel either.

I'll ask around and see what kind of answers I get. But it seems that the term "tidal force" has a real meaning and the one I defined seems to be the only one that can fit this meaning. Do you have an alternate defininng relation in equatioon form?

Pete

 

[robphy]

But you're referring to a single geodesic whereas geodesic deviation deals with two geodesics. Each geodesic has zero 4-acceleration but the relative 4-acceleration of two particles on two geodesics which are deviating has a non-zero value.

Yes, that's essentially what I wrote.

Nope. I also can't find the operating manual for a wheel either.

:zzz:

I'll ask around and see what kind of answers I get. But it seems that the term "tidal force" has a real meaning and the one I defined seems to be the only one that can fit this meaning. Do you have an alternate defininng relation in equatioon form?

Pete

Let me clarify my concern about "tidal forces" (with quotes suggesting it is idea that might not be taken literally [as a force]).

I don't have a problem with the concept of "tidal forces" and their GR interpretation via geodesic deviation (using the Riemann curvature tensor).

I do have a problem with your claim that the "tidal force is a 4-force" [until I see a standard reference that clearly says that it is].
Let me amplify my concern [which you paraphrased] with the following example. Since each worldline carries its own accelerometer, each geodesic carries an accelerometer that reads zero. So, neither geodesic particle experiences a 4-force [or else its accelerometer would be displaced from its zero position]. This interpretation seems to be at odds with your description

Multiply this acceleration 4-vector by the proper mass of the first particle and you will have the 4-force on that particle.

 

[Boustrophedon]

The discussion seems to have retreated into meaningless abstraction. I'm sure pmb_phy will recognise the following quote: "In so far as mathematics refers to reality, it is not certain, and in so far as it is certain, it does not refer to reality".
What people are forgetting is that a 'uniform gravitational field' is supposed, first and foremost, to be 'equivalent' to a uniformly accelerating frame. Considerations of geodesic deviation and which curvature tensor to use are at best secondary, if even relevant. A uniform gravitational field is defined in the context of SR, that is to say in 'flat' spacetime.

If, and it's a big if, one accepts that a uniformly accelerating frame must have a greater 'felt' g-force at the 'back end' such that the 'force' seems to increase in the direction it's pointing, then it would seem that an equivalent 'uniform gravitational field' should have the same characteristic.
However, there are two ways of establishing a uniform gravitational field, one is by indistinguishability from uniform acceleration to a 'closed room' observer as just hinted at, but the other is indistinguishability from inertial motion for a free-falling 'closed room'. For the latter case we find again a variation in g-force along the room - but in the opposite direction - this time decreasing in the direction the g-force points.

Defining a uniform gravitational field as that field such that a freely falling 'closed room' seems inertial to a co-falling observer, then the field must be 'weaker' lower down. That is to say the g-force increases 'higher' in the field, in reverse to the 'uniformly accelerating frame' itself.
Thus the description of a 'uniform gravitational field' as usually presented would seem to be incorrect. If the 'g-force' decreases 'higher' in the field then a falling frame ( falling elevator ) will not be inertial, and if the 'g-force' increases 'higher' in the field it will be quite different ( opposite ) from the experience of 'uniform acceleration'.

 

[pmb_phy]

The discussion seems to have retreated into meaningless abstraction. I'm sure pmb_phy will recognise the following quote: "In so far as mathematics refers to reality, it is not certain, and in so far as it is certain, it does not refer to reality".

Sure. Its one of my favorite Einstein quotes.

Pete

 

[pmb_phy]

I don't have a problem with the concept of "tidal forces" and their GR interpretation via geodesic deviation (using the Riemann curvature tensor).

Yes. I understood that when you first wrote it. But consider a extended body in curved spacetime. Due to tidal forces the body will be under stress. This stress is due to some force. Do you have a better way to describe this force. I'm not stuck on that 4-force thing regarding geodesics. I was tossing the idea around. I guess I should have been more clear on that.

Pete

 

[pmb_phy]

Well for clarity's sake, a tidal force is not a force.

Its more accurate to say that the force of gravity is an inertial force rather than a 4-force.

Pete

 

[JesseM]

Actually the literature uses the term "uniform" to mean, in part, "no tidal forces (zero Riemann tensor)". That's about as simple of a definition that you'll find

So would this mean the G-forces experienced in any accelerating coordinate system in flat spacetime could be treated as the result of a uniform gravitational field, even if the coordinate system was accelerating in a non-uniform way? The G-forces measured by an accelerometer in a uniform gravitational field can be changing with time as well as space, in other words?