Is a uniform gravitational field a gravitational field?

[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?

 

[pmb_phy]

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?

There can still be a g-force present but the field won't be uniform. A rotating frame is a good example of this.

Pete

 

[JesseM]

There can still be a g-force present but the field won't be uniform. A rotating frame is a good example of this.

Pete

But you said "uniform gravitational field" in GR just meant "no tidal forces"--I thought that tidal forces could only occur in curved spacetime by definition (see the comment here for example), while I was just talking about the "gravitational field" experienced by an non-uniformly accelerating observer in flat spacetime. Am I mistaken about the definition of tidal force, or did I misunderstand your comment?

 

[Boustrophedon]

From what I gather, a uniform gravitational field is actually not uniform in terms of the G-forces experienced at different heights

How about whether it increases or decreases with height ? I don't see how you can have it both ways - and the snag is: it needs to increase with height for a free-falling frame to be inertial and yet it needs to decrease with height to resemble 'uniform acceleration'

 

[pmb_phy]

But you said "uniform gravitational field" in GR just meant "no tidal forces"--I thought that tidal forces could only occur in curved spacetime by definition (see the comment here for example), while I was just talking about the "gravitational field" experienced by an non-uniformly accelerating observer in flat spacetime. Am I mistaken about the definition of tidal force, or did I misunderstand your comment?

A uniform g-field requires a zero Riemann tensor. But this is not the only requirement. Just a necessary one.

Pete

 

[JesseM]

A uniform g-field requires a zero Riemann tensor. But this is not the only requirement. Just a necessary one.

OK, so was I wrong in inferring from pervect's post that an additional requirement is that a uniform g-field is equivalent to what would be measured in a coordinate system whose measuring-rods are undergoing Born rigid acceleration?

 

[pmb_phy]

OK, so was I wrong in inferring from pervect's post that an additional requirement is that a uniform g-field is equivalent to what would be measured in a coordinate system whose measuring-rods are undergoing Born rigid acceleration?

Hmmm! I never heard of Born rigid acceleration. That's what I just love about forums like these. Things come up where they've never come up before.

What is Born rigid acceleration?

Pete

 

[coalquay404]

Hmmm! I never heard of Born rigid acceleration. That's what I just love about forums like these. Things come up where they've never come up before.

What is Born rigid acceleration?

Pete

Born rigidity is a well-known concept, particularly in the context of the rigid rotating disk in special relativity. See, for example, Vol. 1 of Held.

I haven't read the thread since it seems to be an exercise in pedantry, but presumably that's what JesseM is talking about.

 

[JesseM]

Hmmm! I never heard of Born rigid acceleration. That's what I just love about forums like these. Things come up where they've never come up before.

What is Born rigid acceleration?

Pete

There are some links in pervect's post which I quoted in my post #57 earlier. This one looks useful, for example. In the case of linear acceleration, my understanding is that Born rigidity means the different parts of the object are accelerating in such a way that the length of the object in the instantaneous inertial reference frame of any part of it will remain constant from one moment to another.

 

[pmb_phy]

There are some links in pervect's post which I quoted in my post #57 earlier. This one looks useful, for example. In the case of linear acceleration, my understanding is that Born rigidity means the different parts of the object are accelerating in such a way that the length of the object in the instantaneous inertial reference frame of any part of it will remain constant from one moment to another.

As I recall, Born rigid acceleration is a requirement for either a uniformly acceleration of a uniform gravitational field. If the metrics you end up with are not identical then something has gone wrong with the derivation. Only experimentation can tell the two apart and to do such an experiment would require the use of a uniform gravitational field.

I'll try to scan in Mould's derivation for a uniformly accerating frame of reference but I'm extremely busy now. Is there anyone who would want to read it? Otherwise I see no sense in posting the section on my website in PDF format.

Kind Regards

Pete

ps - When I get some time later I'll read pervects post in more detail to see exactly what he was saying.

 

[Boustrophedon]

my understanding is that Born rigidity means the different parts of the object are accelerating in such a way that the length of the object in the instantaneous inertial reference frame of any part of it will remain constant from one moment to another.

This is an exact and precisely correct definition of "Born rigid (accelerated) motion". What cannot be relied upon, however, are the various other 'interpretations' of it and 'constructions' put upon it - such as that the acceleration 'g-force' should diminish towards the forward end.

Earlier in this thread it was stated that "uniform acceleration" is the same thing as what has just been defined as "Born rigid acceleration" and this is quite correct.
However, one property that a uniformly accelerating frame (or Born rigid frame) absolutely must have is that it should be perfectly inertial to co-moving observers when freely falling in a "uniform gravitational field". Any version of the equivalence principle makes this mandatory.

Unfortunately, if the g-force of a uniform field diminishes with increasing height (by "indistinguishability" equivalence), the free falling frame by the same token will have to experience diminishing g-forces towards its 'forward' end, which will be 'lower in the field'.
So the g-forces existing and the g-forces required (for uniform acceleration) decrease in opposite directions and cannot possibly cancel across any finite sized falling frame.

Thus the equivalence principle itself forces the conclusion that the only tenable definition of either a 'uniform gravitational field' or of 'uniform (or Born rigid) acceleration' is such that the g-force and the acceleration is precisely constant across and in all parts of a given frame.

 

[pervect]

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.

I told you we'd have a big argument over what a tidal force is :-).

The point I was trying to make is that by at least some definitions of the term "tidal force", the Riemann tensor isn't always the same as the tidal force, specifically for observers not following geodesics.

The geodesic deviation equation only tells us that the Riemann tensor gives us the tidal force for an observer following a geodesic - it says nothing about accelerated observers.

The Riemann tensor is independent of coordinates. (More specifically, the Riemann transforms in a certain known manner under changes of coordinates that imply that it can be interpreted as a coordinate-independent geometric entity.)

This means that in a frame-field, the Riemann tensor as an abstract entity must be independent of the motion of the frame field. Or in coordinate dependent language, it means that for any two observer in relative motion at the same point, that the components of their Riemann are related by the appropriate Lorentz transform.

But consider a static observer near a black hole - we know that the tidal force for such an observer approaches infinity as the observer approaches the event horizon, for it requires infinite acceleration to hold station at the horizon and a finite acceleration to hold station just above the horizon.

Now consider an observer free-falling through the event horizon of a black hole. The tidal force for such an observer is finite and independent of his velocity. (I can provide a reference if needed).

One can achieve similar results by considering an observer on rigid bar. If the bar is not accelerating, there is no tidal force. If the bar accelerates, there is a (usually very small) tidal force. But there is no Riemann tensor.

This demonstrates that the concept of the tidal force is not quite the same as the coordinate-independent concept of the Riemann - for the tidal force for the stationary observer is not the same as the tidal force for the free-falling observer.

Usually this difference between the Riemann and the tidal force doesn't matter - unfortunately, sometimes it does, and this particular issue (the accelerated rod) is one of those cases where it does matter.

 

[Jorrie]

Free-falling through a black hole?

Now consider an observer free-falling through a black hole. The tidal force for such an observer is finite and independent of his velocity. (I can provide a reference if needed).

Great post, as usual. I'm sure that you meant "free-falling through the event horizon"

Jorrie

 

[pervect]

Great post, as usual. I'm sure that you meant "free-falling through the event horizon"

Jorrie

Oops, yep - I fixed it.

 

[pmb_phy]

I told you we'd have a big argument over what a tidal force is :-).

That was understood as soon as I read read your comment predicting this. Howewever I have no intention to include myself in such a discussion. I guess you would say that I've already recognized it as an arguements in semantics and ha ve decided to steer myself clear away from it.

Pete

 

[MeJennifer]

Unfortunately, if the g-force of a uniform field diminishes with increasing height (by "indistinguishability" equivalence), the free falling frame by the same token will have to experience diminishing g-forces towards its 'forward' end, which will be 'lower in the field'.

Looks like you have it upside down Boustrophedon.
The free falling frame will encounter an increasing g-force when it approaches the center of mass.

 

[MeJennifer]

But consider a static observer near a black hole - we know that the tidal force for such an observer approaches infinity as the observer approaches the event horizon, for it requires infinite acceleration to hold station at the horizon and a finite acceleration to hold station just above the horizon.

Now consider an observer free-falling through the event horizon of a black hole. The tidal force for such an observer is finite and independent of his velocity. (I can provide a reference if needed).

One can achieve similar results by considering an observer on rigid bar. If the bar is not accelerating, there is no tidal force. If the bar accelerates, there is a (usually very small) tidal force. But there is no Riemann tensor.

I am not sure what you are trying to demonstrate here.
What has your example with the accelerating bar to do with the curvature in a gravitational field?
Furthermore the suggestion that the tidal forces are infinite at the event horizon is simply a coordinate effect.
A coordinate independent interpretation is that all future paths are curved towards the singularity at the event horizon, but that does not mean that the curvature is infinite there.

This demonstrates that the concept of the tidal force is not quite the same as the coordinate-independent concept of the Riemann - for the tidal force for the stationary observer is not the same as the tidal force for the free-falling observer.

Sorry but mentioning acceleration and space-time curvature together in one posting does not demonstrate anything.

I have the impression that for some reason or another you like people to think that accelerated movement is in some way the same as curved space-time.
Einstein never suggested that!
And it is not what the equivalence principle is saying.

Is it just a personal claim you make that "a tidal force is the difference in proper accelerations between two ends of a rigid bar" or do you claim that this is globally accepted as such? If so, would you care to provide at least one reference to a published book on relativity?

 

[Boustrophedon]

Looks like you have it upside down Boustrophedon.
The free falling frame will encounter an increasing g-force when it approaches the center of mass.

I haven't got it 'upside down'. I meant "...experience diminishing inertial g-force towards its forward end..." (ie. oppositely directed to the 'pull' of the g-field) I am pointing out a contradiction in the "decreasing g-force" definition of a 'uniform' field' or 'uniform acceleration'.
If you care to check what I have written it should now be clearer that I take the incorrect definition of 'uniform acceleration' as having diminishing 'inertial' g-force towards its forward-moving end, and then...

(a) Draw the conclusion that an 'equivalent' uniform 'gravitational' field must have g-force diminish with height since it should be 'indistinguishable' to inhabitants/travellers.

(b) Apply the principle that a free-falling frame (accelerating 'downwards') in a uniform gravitational field should be perfectly inertial.

(c) Show immediate contradiction that: the free-falling frame inertial g-force will have to be smaller at the 'forward' end - which is the end lower in the field since it's falling - while the gravitational pull is greater lower in the field from (a).

Thus the gravitational pull and the inertial g-force (which point in opposite directions) can never cancel over any vertical distance greater than zero and the finite frame can never be inertial ! [Try drawing a diagram if it's still not clear.]

It follows easily that the only 'uniform g-field' that satisfies both the indistinguishability condition and the inertial free-fall condition is a field having constant g-force with height.
It also follows that a 'uniformly accelerating frame' must have a constant g-force (accelerometer reading) along the direction of travel.

 

[Jorrie]

Furthermore the suggestion that the tidal forces are infinite at the event horizon is simply a coordinate effect.
A coordinate independent interpretation is that all future paths are curved towards the singularity at the event horizon, but that does not mean that the curvature is infinite there.

My understanding of what pervect was illustrating is that since the static acceleration at the event horizon of a hole diverges, while the free-fall acceleration remains finite there, the tidal gravity for static and free-fall frames cannot be the same everywhere. The Riemann tensor is however coordinate independent.

This may have implications for the definition of a 'uniform gravitational field', which I think pervect is best at explaining.

 

[MeJennifer]

It follows easily that the only 'uniform g-field' that satisfies both the indistinguishability condition and the inertial free-fall condition is a field having constant g-force with height.

Looks like you are changing tunes now Boustrophedon.
And, you are confused about what you call "the indistinguishability condition".

Einstein never claimed that the equivalence principle works "at a distance", it only applies locally.

 

[MeJennifer]

My understanding of what pervect was illustrating is that since the static acceleration at the event horizon of a hole diverges, while the free-fall acceleration remains finite there, the tidal gravity for static and free-fall frames cannot be the same everywhere.

All those "conclusions" are coordinate dependent observations. Tidal effects are variations of the curvature of space-time in a given region. See for instance Schutz: "A First Course in General Relativity". If I recall correctly (I am currently in Thailand and have no access to books) his definition is quite literal using the Earth and moon as examples.

 

[Boustrophedon]

I don't quite see in what way I've 'changed tunes' - perhaps you might amplify ? What I have abbreviated to the 'indistinguishabilty condition' is that the equivalence of a uniformly accelerated reference frame and a 'uniform gravitational field' is such that a laboratory in either situation should be unable by any experiment to determine which condition is the source of the g-forces that are experienced. This means that if uniform acceleration involves a g-force that is measured higher as one checks over a distance in the direction the g-force is pointing (as proposed) then the 'indistinguishable' uniform gravitational field should be identical, ie. have (downward) g-force diminishing with height.
The reason that the equivalence principle only applies 'at a point' is that 'real' gravitational fields are all non-uniform and thus have 'tidal effects'. A hypothetical perfectly uniform gravitational field would be one where such tidal effects do not occur so that the equivalence principle would hold over significant distances. From the arguments given it appears that the only sense of 'uniformity' that satisfies these criteria is where the g-force (gravitational or inertial) is constant along the direction of its action.

 

[Jorrie]

Really?

All those "conclusions" are coordinate dependent observations. Tidal effects are variations of the curvature of space-time in a given region.

Do you mean that the 'g-force' felt by a static observer just outside of a Schwarzschild event horizon (wasting an enormous amount of fuel to stay static) is coordinate dependent?

Is the fact that an observer attempting to remain static at such an horizon needs an infinite force (-> infinite acceleration required) coordinate dependent?

I somehow don't think so, but I would like to know the real answer.

 

[pmb_phy]

Do you mean that the 'g-force' felt by a static observer just outside of a Schwarzschild event horizon (wasting an enormous amount of fuel to stay static) is coordinate dependent?

That is one way to stay static but not the only way. I'm sitting in a chair in my living room and I'm at rest in the gravitatiomal field.

Pete

 

[pervect]

Do you mean that the 'g-force' felt by a static observer just outside of a Schwarzschild event horizon (wasting an enormous amount of fuel to stay static) is coordinate dependent?

Is the fact that an observer attempting to remain static at such an horizon needs an infinite force (-> infinite acceleration required) coordinate dependent?

I somehow don't think so, but I would like to know the real answer.

Given a particular path through space-time the magnitude of the 4-acceleration along that path is a geometric (i.e. coordinate independent) quantity - i.e. the 4-acceleration is a 4-vector, and like other 4-vectors, its length is an invariant.

This scalar quantity has a physical interpretation as the magnitude of the felt g-force.

Treating the acceleration 4-vector as a 3-vector does involve some coordinate dependence - basically, one is defining the time \partial / \partial tto point along the path, and furthermore one is requiring space to be orthogonal to time.

 

[MeJennifer]

Do you mean that the 'g-force' felt by a static observer just outside of a Schwarzschild event horizon (wasting an enormous amount of fuel to stay static) is coordinate dependent?

See Pervect's comments.

Is the fact that an observer attempting to remain static at such an horizon needs an infinite force (-> infinite acceleration required) coordinate dependent?

See Pervect's comments.

I somehow don't think so, but I would like to know the real answer.

Well we agree, so you are fighting the straw man.

But the notion, as some present here, that the tidal force is infinite at the event horizon for an accelerating observer is not correct IMHO.

Tidal forces appear when a gravitational field is inhomogeneous, or in other terms, in a region where the Riemann curvature tensor is not constant. Tidal forces are curvature differentials in space-time not some property of the state of an observer's acceleration.

If you wish to believe that tidal forces depend on the state of acceleration instead of being a property of space-time, or even, talking about confusion, a combination of that, then that is of course entirely up to you. And then you, consequently, can state that, when approaching the event horizon, the tidal forces are infinite for an accelerating observer while they are not infinite for an observer in free fall.
But I frankly fail to see how one could develop a better understanding of general relativity by such an idiosyncratic view.

My advice in developing a better understanding of general relativity is to learn how the properties and shape of space-time determine how observers see their world, as opposed to understanding the observer's world through fancy metrics with cross terms and coordinate dependent views or in attempt to somewhat equate the hyperbolic properties of flat Minkowski space-time with space-time curvature.

Anyhow, don't be surprised that many will disagree with such a notion of what tidal forces are.

 

[Boustrophedon]

Tidal forces appear when a gravitational field is inhomogeneous, or in other terms, in a region where the Riemann curvature tensor is not constant.

You mean - ...in a region where the Riemann curvature tensor is not zero, surely !?

 

[Jorrie]

Anyhow, don't be surprised that many will disagree with such a notion of what tidal forces are.

Yep, it seems that tidal gravity is as difficult to define as "uniform gravitational field"!

 

[Jorrie]

That is one way to stay static but not the only way. I'm sitting in a chair in my living room and I'm at rest in the gravitatiomal field.

Pete

Your favorite armchair would not work so nicely just outside of an event horizon - well, maybe with a BIG rocket attached.

I believe one can actually enjoy a 'static 1g' just outside of a Schwarzschild hole's horizon if it (the hole) is BIG, really BIG.

 

[George Jones]

Do you mean that the 'g-force' felt by a static observer just outside of a Schwarzschild event horizon (wasting an enormous amount of fuel to stay static) is coordinate dependent?

Is the fact that an observer attempting to remain static at such an horizon needs an infinite force (-> infinite acceleration required) coordinate dependent?

No.

I somehow don't think so, but I would like to know the real answer.

An observer static at r experiences a 4-acceleration (and thus a 3-accleration, since 4-acceleration is orhtogonal to 4-velocity) given by

\left( 1 - \frac{2M}{r} \right)^{-\frac{1}{2}} \frac{M}{r^2},

which blows up at the event horizon of a black hole of any mass.

Tidal stresses at the event horizon decrease as the mass of the black hole increases.

 

[Jorrie]

No.
Tidal stresses at the event horizon decrease as the mass of the black hole increases.

Thanks George, I'm comfortable with that (and with the equation you gave).

What's not clear to me is how the 'tidal forces' experienced by 'static' and 'free-falling' observers differ in the vicinity of the event horizon of a hole, irrespective of its mass. It appears that pervect and MeJennifer have different views.

However, I also understand that there may be different definitions of "static observers" and 'tidal forces' at play here...

Jorrie