Merged Relativity+ / Farsight

[ctamblyn]

Continued from another thread, as it didn't really belong there and seems to be an as yet unexplored part of John Duffield / Farsight's unusual perspective on physics, i.e. the topic of this thread:

No, tidal force is where the force of gravity is different at your feet than your head. The magnitude needn't be greater. Remember, force is a vector!

Geddoutofit.

...

No, really. And I must say that your implied claim to be an authority on this topic is rather undermined by the fact that you avoided even acknowledging what ought to be two very easy questions which get to the very heart of this matter. I repeat them here for your convenience:

1. Let's say you have a purely radial gravitational field where the magnitude of the field strength, g, is a function only of the radial coordinate r. Show me how you would compute the tidal accelerations at a distance r from the origin (i.e. find a formula for the accelerations in terms of g(r); you can probably even look it up).

2. Following on from the above, suppose we have a field in which g(r) is a constant. Remember, g(r) is only the magnitude of the field, the direction is always radially inwards. Using your answer to part 1, compute the tidal accelerations at distance r in this special case.

I'd have thought that you'd see this is a great opportunity to support your argument rigorously, rather than hand-waving past the issue and appealing to (misinterpreted) authorities. If you avoid answering them yet again, I suppose I'll have to take it as a sign that either (a) you don't know how to answer them, undermining your claim to understand this stuff or (b) you know the answers, but are unwilling to post them because you know they contradict your claims.

ETA: This is in response to Farsight's erroneous claim in the other thread, that a spherically symmetric gravitational field in which the field has constant magnitude g everywhere but points towards the origin is (a) uniform and (b) free of tidal forces. The discussion started around here.

 

[Farsight]

No. It's unnecessary. Because as I said, in a gravitational field, you fall down. Funny that you edited that out of your reply. Pah, you need to work on your sophistry. Now as I said, I'm off. Talk among yourselves, but do try to talk physics eh?

 

[ctamblyn]

No. It's unnecessary.

Actually, it's quite necessary if you want to make a serious argument for your position. I think it's pretty safe to conclude that no such serious argument actually exists.

 

[ben m]

No. It's unnecessary. Because as I said, in a gravitational field, you fall down. Funny that you edited that out of your reply.

He didn't edit that out, it's right there--- "the acceleration is radially inwards."

 

[ctamblyn]

In some ways, the misunderstanding John Duffield is suffering regarding his field configuration is related to the misunderstandings people sometimes have when they first learn about circular motion in physics. The velocity of a particle in uniform circular motion is constant in magnitude, but not in direction. As a result, the acceleration is non-zero even though the speed constant. Vectors can change without changing length.

John Duffield's artificial example is a bit more complicated, but his error is essentially the same. We have a gravitational field - which is a vector field - of constant magnitude throughout space, but having a variable direction (since it must always point at the origin in John Duffield's set-up). If you look at the forces at two neighbouring points in space (that don't lie on a straight line connecting the origin), you will find that they pull in very slightly different directions, and so a small but extended body will experience different forces on different parts. In fact, you can easily see from the way the field lines converge at the origin that the body not only accelerates towards the origin, but gets squeezed inwards perpendicular to the radial direction. If you were falling feet first, these forces would be trying to make you thinner.

In more realistic scenarios, where the magnitude of the force also increases as the origin is approached, the above squeezing effects are still present, but there is an additional stretching which acts along the radial direction. The combined effect of these two types of force is the "spaghettification" effect John Duffield mentioned in the other thread.

In general, whenever (and only when) you have a non-uniform gravitational field, you will find that extended bodies in free fall not only experience an overall acceleration, but get squeezed and/or stretched in various directions. The corresponding forces are referred to quite generally as tidal forces.

 

[Roboramma]

Two apples fall toward the center of this uniform gravitational field. At first they are separated by some distance. Say, five feet, and are at rest both relative to each other and the center of the gravitational field. Because the acceleration to toward the center, they will move in that direction. Assume further that both apples are at an equal distance from the center.

Thus when they reach that center they will do so simultaneously and bump into each other. They will experience a tidal force toward each other.

Farsight, are you actually trying to deny this??

 

[Farsight]

Robo: No, but that's not what Ray was referring to on page 20 of http://arxiv.org/abs/physics/0204044, and it isn't the tidal force associated with Riemann curvature. See the picture on the right of the wiki Riemann curvature tensor article, then for a clearer picture of "the rubber sheet analogy" look at the Newtonian gravitational potential plot here.

The curvature you can see depicts tidal force which is the second derivative of potential. It's associated with Riemann curvature. (Also see Ricci curvature and Weyl curvature). Riemann curvature is "the defining feature" of a gravitational field because without it your plot can't get off the flat and level.

A uniform gravitational field would look like an inverted cone. It has no Riemann curvature, it doesn't diminish with distance, and isn't a real gravitational field. The tidal force you're talking about is pretty irrelevant. Ctamblyn is trying to bamboozle you I'm afraid.

 

[edd]

Gravitational potentials do not represent the shape of the underlying space. They are not embedding diagrams.

 

[ben m]

A uniform gravitational field would look like an inverted cone. It has no Riemann curvature, it doesn't diminish with distance, and isn't a real gravitational field. The tidal force you're talking about is pretty irrelevant. Ctamblyn is trying to bamboozle you I'm afraid.

"look like" in what sense? In the rubber sheet analogy, the analogy which is used to attempt to explain gravity to 12-year-olds in science museums? It's better than nothing, but it's just an analogy, and contains numerous explicitly-misleading features which lead amateurs to incorrect predictions in anything other than trivial cases.

In the correct rubber-sheet analogy, you can make a sheet shaped like a cone (there is not "inverted" or "right side up" in the more accurate version of the analogy---your use of the word "inverted" is what clued me into the fact that you're thinking of the misleading kiddie version). In this cone, there is no gravitational force except at a singularity in the center. An object left at rest on the "cone" will not accelerate towards the center. There are no geodesics that orbit the center. You are correct that such a cone has no Riemann curvature, but you're incorrect to say that a cone has a "uniform gravitational field". A conical sheet has the same Newtonian gravitational fields as flat sheet does---none at all.

Or perhaps you meant that the gravitational potential energy is a cone, with constant slope and therefore constant gravitational acceleration? Draw the GR version of that and you'll see that it DOES have Riemann curvature. Go ahead, try it.

 

[Vorpal]

Robo: No, but that's not what Ray was referring to on page 20 of http://arxiv.org/abs/physics/0204044, and it isn't the tidal force associated with Riemann curvature. ...

The essence of the Newtonian intuition...

If you look at the forces at two neighbouring points in space (that don't lie on a straight line connecting the origin), you will find that they pull in very slightly different directions, and so a small but extended body will experience different forces on different parts. In fact, you can easily see from the way the field lines converge at the origin that the body not only accelerates towards the origin, but gets squeezed inwards perpendicular to the radial direction. If you were falling feet first, these forces would be trying to make you thinner.

carries over into GTR: if you have two geodesics with four-velocity u and a nearby geodesic connected with n connecting points of equal affine parameter/proper time, then the geodesic deviation^WP between them is related to the Riemann tensor through
[1] D²n/dλ² = R(u,n)u = -E,
where E is called the tidal tensor or the gravitoelectric tensor (cf. Bel decomposition^WP or the Riemann tensor). A vanishing tidal tensor does not necessarily imply that the spacetime is flat, though that's not directly relevant here.

Computing geodesic deviation, in, say, the Schwarzschild spacetime makes it very clear that a radially freely infalling body is both stretched in the radial direction and squeezed in the transverse directions. Thus, spaghettification is not just to radial stretching. Moreover, ctamblyn is also correct that a radially directed constant-magnitude acceleration would give a pure transverse squeeze and would not be a "uniform gravitational field", regardless of whether one considers Ray to be is correct or incorrect about uniform gravitational fields in GTR.

 

[W.D.Clinger]

Two apples fall toward the center of this uniform gravitational field. At first they are separated by some distance. Say, five feet, and are at rest both relative to each other and the center of the gravitational field. Because the acceleration to toward the center, they will move in that direction. Assume further that both apples are at an equal distance from the center.

Thus when they reach that center they will do so simultaneously and bump into each other. They will experience a tidal force toward each other.

Farsight, are you actually trying to deny this??

Robo: No, but that's not what Ray was referring to on page 20 of http://arxiv.org/abs/physics/0204044, and it isn't the tidal force associated with Riemann curvature.

Au contraire. Roboramma's example is indeed a manifestation of the tidal force associated with Riemann curvature.

It isn't surprising that Farsight got this wrong. The components of the Riemann tensor are calculated from the Christoffel symbols, which are determined by the components of what Einstein called the fundamental (metric) tensor, and Farsight has never understood what those g_µν were about.

The curvature you can see depicts tidal force which is the second derivative of potential. It's associated with Riemann curvature. (Also see Ricci curvature and Weyl curvature). Riemann curvature is "the defining feature" of a gravitational field because without it your plot can't get off the flat and level.

Farsight's right about that.

Farsight has often complained that no one ever says "Farsight's right about that." I'm happy to take advantage of this rare opportunity to say he's right.

In the sentence I highlighted, Farsight is advocating what Peter M Brown, in the preprint cited by Farsight, refers to as the view taken by "MGR (Modern General Relativity)". As Brown noted, Farsight's highlighted sentence disagrees with "Einstein's vision of general relativity, as defined in The Foundation of the General Theory of Relativity and other publications", which Brown refers to as EGR (Einstein's General Relativity).

There's nothing wrong with either view of relativity, but it's unusual to see Farsight taking the side of MGR against EGR. For several years now, Farsight has been saying MGR is "wrong", while casting himself as a defender of Einstein's original vision.

A uniform gravitational field would look like an inverted cone. It has no Riemann curvature, it doesn't diminish with distance, and isn't a real gravitational field. The tidal force you're talking about is pretty irrelevant.

If we interpret Farsight's first sentence as a claim that all uniform gravitational fields look like inverted cones, then Farsight's first sentence is flatly incorrect. On pages 16 through 19 of his paper, Peter M Brown gave three examples of "gravitational fields that have zero space-time curvature". Only one of those three examples looks like a cone.

The first of those examples is the uniform gravitational field discussed by edd in the discussion edd (properly) chose to continue here instead of there. Farsight, of course, did not understand edd's uniform gravitational field, even though it was the only example of a "uniform gravitational field" given by Farsight's own proof text. (Brown does not actually refer to either of his next two examples as a "uniform gravitational field".)

Brown's third example of a "space-time with zero curvature is that of a straight cosmic string." As Brown states, that example's spacetime looks like a cone. As Brown states, the cone is flat (has zero curvature) except for a singularity at the apex, where the curvature is undefined. (The flatness of a cone goes against the intuition of people who think of cones as embedded within three-dimensional space, but we're talking about intrinsic curvature here. You can create a cone from a perfectly flat sheet of paper without stretching or tearing the paper, which proves there is no intrinsic curvature on the cone itself (away from its apex).) There are no tidal forces (except at the apex).

Although the cone is flat, closed curves of constant curvature k are shorter than circles of curvature k would be in Euclidean geometry. Brown's third example therefore provides an example of non-Euclidean spacetime without curvature (away from the singularity).

Ctamblyn is trying to bamboozle you I'm afraid.

ctamblyn wasn't trying to bamboozle anybody. ctamblyn's explanation was correct.

 

[Farsight]

Clinger: Yes he was, and no it wasn't. And you're defending him? Shame on you. My reference was to the Ray quote, and we were discussing the "uniform gravitational field" here with respect to the principle of equivalence. Not EGR v MGR. And certainly not cosmic strings.

"look like" in what sense? In the rubber sheet analogy, the analogy which is used to attempt to explain gravity to 12-year-olds in science museums? It's better than nothing, but it's just an analogy, and contains numerous explicitly-misleading features which lead amateurs to incorrect predictions in anything other than trivial cases.

It isn't misleading when you understand it.

In the correct rubber-sheet analogy, you can make a sheet shaped like a cone (there is not "inverted" or "right side up" in the more accurate version of the analogy---your use of the word "inverted" is what clued me into the fact that you're thinking of the misleading kiddie version). In this cone, there is no gravitational force except at a singularity in the center. An object left at rest on the "cone" will not accelerate towards the center. There are no geodesics that orbit the center. You are correct that such a cone has no Riemann curvature, but you're incorrect to say that a cone has a "uniform gravitational field". A conical sheet has the same Newtonian gravitational fields as flat sheet does---none at all.

OK, so what sort of object has a gravitational field like this?

Or perhaps you meant that the gravitational potential energy is a cone, with constant slope and therefore constant gravitational acceleration? Draw the GR version of that and you'll see that it DOES have Riemann curvature. Go ahead, try it.

If it's a cone, it doesn't. See above where even Clinger had to confess that I was right:

Riemann curvature is "the defining feature" of a gravitational field because without it your plot can't get off the flat and level.

Farsight's right about that.

Gravitational potentials do not represent the shape of the underlying space.

Of course they don't. And as I was saying, the principle of equivalence was an "enabling principle". It applies to an infinitesimal region. Which means it doesn't apply at all.

 

[edd]

Clinger: Yes he was, and no it wasn't. And you're defending him? Shame on you. My reference was to the Ray quote, and we were discussing the "uniform gravitational field" here with respect to the principle of equivalence. Not EGR v MGR. And certainly not cosmic strings.

It isn't misleading when you understand it.

OK, so what sort of object has a gravitational field like this?

A... ummm... cosmic string.

 

[Farsight]

The essence of the Newtonian intuition... carries over into GTR: if you have two geodesics with four-velocity u and a nearby geodesic connected with n connecting points of equal affine parameter/proper time, then the geodesic deviation^WP between them is related to the Riemann tensor through
[1] D²n/dλ² = R(u,n)u = -E,
where E is called the tidal tensor or the gravitoelectric tensor (cf. Bel decomposition^WP or the Riemann tensor). A vanishing tidal tensor does not necessarily imply that the spacetime is flat, though that's not directly relevant here.

Huh? Vorpal, it's the crux of it. If there's no Riemann curvature you haven't got a gravitational field. Things don't fall down.

Computing geodesic deviation, in, say, the Schwarzschild spacetime makes it very clear that a radially freely infalling body is both stretched in the radial direction and squeezed in the transverse directions. Thus, spaghettification is not just to radial stretching. Moreover, ctamblyn is also correct that a radially directed constant-magnitude acceleration would give a pure transverse squeeze and would not be a "uniform gravitational field", regardless of whether one considers Ray to be is correct or incorrect about uniform gravitational fields in GTR.

Vorpal, let me spell it out: if the force of gravity does not diminish with distance, there's no radial stretch, no Riemann curvature, and therefore no gravitational field. The uniform gravitational field that Ray was referring to is a contradiction in terms. It's on page 20 of http://arxiv.org/abs/physics/0204044 :

"It is very important to notice that in a freely falling frame we have not
transformed away the gravitational field since the Riemann tensor
(gravitation ←→ Riemann tensor) will not vanish and we will still measure
relative acceleration ….. The first thing to note about the 1911 version of
the principle of equivalence is that what in 1911 is called a uniform
gravitational field ends up in general relativity not to be a gravitational field
at all – The Riemann tensor is here identically zero. Real gravitational fields
are not uniform since they must fall off as once recedes from gravitating
matter".

And by the way, I take exception to you guys crawling out of the woodwork with smoke and mirrors to make things complicated and attempt to say "Farsight is wrong" when I'm not.

 

[Farsight]

A... ummm... cosmic string.

Sorry edd, but cosmic strings are a 40-year old hypothesis with no evidential support.

 

[Perpetual Student]

<...>
And as I was saying, the principle of equivalence was an "enabling principle". It applies to an infinitesimal region. Which means it doesn't apply at all.

There's a man who does not understand calculus!

 

[drelda]

And by the way, I take exception to you guys crawling out of the woodwork with smoke and mirrors to make things complicated and attempt to say "Farsight is wrong" when I'm not.

How can you be so sure?

I noticed you also answered 'No' to the question "Openly accepts ideas could be wrong" in the crackpot thread.

In my experience - those people who most adamantly insist that they are not wrong - are the ones most likely to actually be wrong. Therefore I will have more respect for you if you admit that its possible you are mistaken - either about some details - or even your whole set of ideas. i.e.:

I beseech you, in the bowels of Christ, think it possible that you may be mistaken.

- Drelda

 

[edd]

Sorry edd, but cosmic strings are a 40-year old hypothesis with no evidential support.

That's not the point. Purely theoretical constructs can be useful in gaining understanding of a theory and for demonstrating where someone's understanding might be lacking.

 

[lpetrich]

Sorry edd, but cosmic strings are a 40-year old hypothesis with no evidential support.

It's a *speculation*. Do you think that cosmic strings have been ruled out by anything? If so, what has ruled them out?

It is always possible that cosmic strings exist, but cannot be detected or else distinguished from other effects.


Farsight, imagine that you had lived in some past decade or some past century. What would you have dismissed because of "lack of evidence"?

• Before 1850: chemical elements predicted by Dmitri Mendeleev
• Before 1800: atoms
• Before 1800: visible light's wave nature
• Before 1650: terrestrial gravity extends to the Moon and beyond
• Before 1650: white light contains colors

Etc.

 

[Farsight]

How can you be so sure?

Because I understand general relativity and gravity. because I've read a lot of material, including the Einstein material.

I noticed you also answered 'No' to the question "Openly accepts ideas could be wrong" in the crackpot thread.

They aren't my ideas, drelda. We're talking about the equivalence principle and Riemann curvature here, not something I've made up.

In my experience - those people who most adamantly insist that they are not wrong - are the ones most likely to actually be wrong. Therefore I will have more respect for you if you admit that its possible you are mistaken - either about some details - or even your whole set of ideas. i.e.:

I beseech you, in the bowels of Christ, think it possible that you may be mistaken.

I've admitted to being wrong before now. But I'm not wrong about this. See what Clinger said.