I GR as a Graded Time Dilation Field in Euclidean Space?

[rjbeery]

The title says it all, really. Are we able to describe GR in terms of a Graded Time Dilation Field in Euclidean space?

From http://cpl.iphy.ac.cn/EN/Y2008/V25/I5/1571 we can see that light curvature can be analogously described via a material with a graded index refraction, so my question is really whether or not the following is is capable of encompassing GR:

t_0 = t_f \sqrt {1 - \frac{r_0}{r}}<br /> <br /> \frac{t_f}{t_0} = \frac{1}{\sqrt{1 - \frac{r_0}{r}}} =analogy to "n" in optical medium

 

[phyzguy]

The metric tensor is a symmetric tensor with 10 degrees of freedom. Some of these are removed by the physical constraints of the theory, but it is certainly not possible to describe GR with a single scalar field as you propose.

 

[Dale]

my question is really whether or not the following is is capable of encompassing GR:

t_0 = t_f \sqrt {1 - \frac{r_0}{r}}<br /> <br /> \frac{t_f}{t_0} = \frac{1}{\sqrt{1 - \frac{r_0}{r}}} =analogy to "n" in optical medium

As phyzguy said, the answer is no, there are not enough degrees of freedom in a scalar field.

Note that the result from the paper you cited applies specifically to a "static spherically symmetrical gravitational field". This particular spacetime has reduced degrees of freedom due to the assumed symmetry, so it can be represented by a scalar field whereas general spacetimes cannot.

 

[Bill_K]

The title says it all, really. Are we able to describe GR in terms of a Graded Time Dilation Field in Euclidean space?

A few things this "theory" does not describe correctly:

• cosmological solutions
• gravitational waves
• black holes
• perihelion precession

 

[rjbeery]

As phyzguy said, the answer is no, there are not enough degrees of freedom in a scalar field.

Note that the result from the paper you cited applies specifically to a "static spherically symmetrical gravitational field". This particular spacetime has reduced degrees of freedom due to the assumed symmetry, so it can be represented by a scalar field whereas general spacetimes cannot.

Yes! I was formulating a response to phyzguy along these lines before I read your response, DaleSpam. I'd like to understand "where" the analogy breaks down. It's difficult for me to extend the analogy into anything other than a gravitational field (e.g. time dilation caused by relative motion, etc), but I'm left wondering if the pure simplicity of a 3-D Euclidean space could possibly be reduced this far; dimensional symmetry, zero curvature...I thought perhaps the excess degrees of freedom could be eliminated...but I don't know which is of course why I'm asking.

A few things this "theory" does not describe correctly:
cosmological solutions
gravitational waves
black holes
perihelion precession

Hi Bill_K!
I would say that the analogy could be extended to some or all of these in addition to DaleSpam's rotating masses, etc, by replacing the calculated GR curvature with the time dilation field. Note that the "formula" I give for calculating this field is derived from the GR time dilation itself.

 

[WannabeNewton]

How would you predict frame dragging using this theory?

 

[PAllen]

I would say that the analogy could be extended to some or all of these in addition to DaleSpam's rotating masses, etc, by replacing the calculated GR curvature with the time dilation field. Note that the "formula" I give for calculating this field is derived from the GR time dilation itself.

NO, it is derived from one solution of GR that is static and has perfect spherical symmetry. There is no part of the universe that exactly meets these conditions. In any realistic solution, curvature cannot be represented by a single function of position and time. This is a mathematical theorem, not subject to debate or interpretation. Specifically, curvature of 2-surface can be represented by a scalar. For more than 2 dimensions, it cannot.

 

[Dale]

I'd like to understand "where" the analogy breaks down.

In the vacuum region I think it breaks down as soon as you break spherical symmetry. In the matter region it breaks down as soon as you break spherical symmetry, or have a non-static distribution of matter, or have a matter field which is not a perfect dust.

A rotating mass is not spherically symmetric, it is at best axisymmetric.

 

[phyzguy]

Yes! I was formulating a response to phyzguy along these lines before I read your response, DaleSpam. I'd like to understand "where" the analogy breaks down.

For example, consider the case of two orbiting neutron stars, losing energy due to gravitational radiation and spiraling into form a black hole. We know they lose energy due to gravitational radiation (see Hulse-Taylor binary or this paper). There is no way to describe a situation like this with a single scalar field. As Bill_K said, it will not even describe gravitational waves.

 

[rjbeery]

NO, it is derived from one solution of GR that is static and has perfect spherical symmetry. There is no part of the universe that exactly meets these conditions. In any realistic solution, curvature cannot be represented by a single function of position and time. This is a mathematical theorem, not subject to debate or interpretation. Specifically, curvature of 2-surface can be represented by a scalar. For more than 2 dimensions, it cannot.

Hi PAllen! I appreciate what you're saying but the OP calls for Euclidean space with zero curvature. We model EM activity as a field in flat 3-D space, yes? Restricting the conversation solely to time dilation caused by gravity could we do the same thing here?

I not claiming that you or others are wrong on this point, I just want to verify that everyone is stripping time out from the geometry before denouncing the idea.

 

[PeterDonis]

the OP calls for Euclidean space with zero curvature.

No, it doesn't. The OP is asking about "GR", in general.

Restricting the conversation solely to time dilation caused by gravity

This leaves out a lot of "GR", since there are many scenarios covered by "GR" that do not even have a meaningful concept of "time dilation caused by gravity". If you want to restrict consideration only to those scenarios that do, then you need to ask about something more restricted than "GR". (Which then leads to the question, why would you be restricting yourself to just those scenarios?)

 

[rjbeery]

This leaves out a lot of "GR", since there are many scenarios covered by "GR" that do not even have a meaningful concept of "time dilation caused by gravity". If you want to restrict consideration only to those scenarios that do, then you need to ask about something more restricted than "GR". (Which then leads to the question, why would you be restricting yourself to just those scenarios?)

The reason is: baby steps. I'm just exploring the idea. Could you give me some examples of scenarios covered by GR in which "time dilation caused by gravity" does not apply? Are you referring to time dilation caused by relative motion? Or something else?

 

[PAllen]

The reason is: baby steps. I'm just exploring the idea. Could you give me some examples of scenarios covered by GR in which "time dilation caused by gravity" does not apply? Are you referring to time dilation caused by relative motion? Or something else?

Any situation that isn't stationary. That is, anything more complex than an isolated body in an unchanging state. It really is only the case of isolated, unchanging, body for which you could define a time dilation field. In all other cases, GR does not specify gravitational time dilation. That is, the whole notion of gravitational time dilation is not a general feature of GR; it is a specific derived feature applicable only in the most restricted cases.

 

[PeterDonis]

Could you give me some examples of scenarios covered by GR in which "time dilation caused by gravity" does not apply?

Previous posters have already listed a number of them, but to summarize:

(1) The concept of "time dilation due to gravity" only applies if the spacetime is stationary, which basically means that one can find a family of worldlines in the spacetime along which the geometry does not change with time. (The technical way to say this is that the spacetime has a timelike Killing vector field.)

Examples of spacetimes that are not stationary: any spacetime in which objects have orbits that are not perfectly circular (such as the planets in the solar system, including effects like the perihelion precession of Mercury); any spacetime where gravitational waves are emitted (such as binary pulsars); black holes (because the region inside the event horizon is not stationary even if the region outside the horizon is); cosmology (because the universe is expanding).

(2) The concept of "time dilation due to gravity" is only *sufficient* to describe all the effects of gravity if the spacetime is static, which means that there is a family of spacelike hypersurfaces that are orthogonal to the family of worldlines along which the geometry does not change with time. The orthogonality property is necessary for us to be able to view the spacelike hypersurfaces as "space at an instant of time".

The primary example of a spacetime that is stationary but not static is any spacetime in which the central object is rotating. The rotation causes effects such as "frame dragging" that cannot be modeled by treating gravity as a scalar field. So even though there is a meaningful concept of "time dilation due to gravity", that by itself is not sufficient to describe all the effects of gravity for a rotating source.

 

[rjbeery]

Any situation that isn't stationary. That is, anything more complex than an isolated body in an unchanging state. It really is only the case of isolated, unchanging, body for which you could define a time dilation field. In all other cases, GR does not specify gravitational time dilation. That is, the whole notion of gravitational time dilation is not a general feature of GR; it is a specific derived feature applicable only in the most restricted cases.

If we could account for motion-induced time dilation in addition to gravity-induced time dilation would you consider the theory more complete? Or are you suggesting that there are additional elements of GR which would not be accounted for?

 

[PeterDonis]

If we could account for motion-induced time dilation in addition to gravity-induced time dilation would you consider the theory more complete?

This wouldn't change anything. "Motion-induced time dilation" can't be modeled as a scalar field anyway; it's not an invariant, it's a frame-dependent quantity, so it's not even the same kind of thing as "gravity-induced time dilation" to begin with.

 

[WannabeNewton]

If we could account for motion-induced time dilation in addition to gravity-induced time dilation would you consider the theory more complete? Or are you suggesting that there are additional elements of GR which would not be accounted for?

Could you account for all of electromagnetic theory using only a scalar electric potential? Could a scalar potential describe the circulating magnetic fields generated by currents and the circulating electric fields generated by time-varying magnetic fields?

 

[PAllen]

OK, what about two scalar potentials?

I'm just speaking loosely and off-the-cuff here, but perhaps two scalars are also sufficient for moving gravity fields.

How about 6, which is the actual number required. Then just call it the metric tensor after specification of arbitrary coordinate conditions.

 

[rjbeery]

How about 6, which is the actual number required. Then just call it the metric tensor after specification of arbitrary coordinate conditions.

PAllen, you've already made the journey through GR (I presume), but don't you agree that teaching gravity as a time dilation field analogous to an EM field in a flat Euclidean space would be...more aesthetically pleasing if nothing else? Why single out gravity as the only force determining the geometry of our surroundings?

 

[PAllen]

PAllen, you've already made the journey through GR (I presume), but don't you agree that teaching gravity as a time dilation field analogous to an EM field in a flat Euclidean space would be...more aesthetically pleasing if nothing else? Why single out gravity as the only force determining the geometry of our surroundings?

You don't have to call it geometry. You can treat it as fields on an abstract background. Weinberg, among others, has worked on expressing it this way. The point is, that if you want it to match the predictions of GR or spin-2 field theory, you need 6 functions not one or two.

 

[WannabeNewton]

PAllen, you've already made the journey through GR (I presume), but don't you agree that teaching gravity as a time dilation field analogous to an EM field in a flat Euclidean space would be...more aesthetically pleasing if nothing else? Why single out gravity as the only force determining the geometry of our surroundings?

The EM field is an antisymmetric 2-tensor field so that doesn't help your case. Secondly, we can describe EM geometrically in a manner very much like GR. This is (part of) gauge field theory. And in my opinion there is nothing more aesthetically pleasing than a geometrical theory of physics, be it gravity or EM or any other gauge field.

 

[pervect]

I'm not sure I understand the proposed theory, but it seems to me like it isn't even consistent with special relativity, having a Euclidean spatial structure rather than a Lorentzian space-time structure.

If it had a Lorentzian structure, we could use the PPN formalism to show the shortcomings of such a theory, but I won't add more remarks on that until we determine more about the theory.

My impression of the theory is that rather than being based on special relativity and the Lorentz transform, it's based on pre-relativity ideas of absolute time, with the addition of a scalar field that controls the "rate" at which the absolute time flows.

I hope my concerns are clear, I fear there may be a language barrier related to the underlying concepts related to absolute/relative time and the relativity of simultaneity, based on previous discussions with the OP.

 

[rjbeery]

The EM field is an antisymmetric 2-tensor field so that doesn't help your case. Secondly, we can describe EM geometrically in a manner very much like GR. This is (part of) gauge field theory. And in my opinion there is nothing more aesthetically pleasing than a geometrical theory of physics, be it gravity or EM or any other gauge field.

Hi Wannabe!

I'm not trying to make a case, really, I'm just exploring this idea. Can you or PAllen help me appreciate why the EM field can be described as a two scalar potential but gravity requires 6? Does it have something to do with the generality of GR vs a presumption of dimensions in studying EM waves?

 

[WannabeNewton]

I'm not trying to make a case, really, I'm just exploring this idea.

Yes poor choice of words on my part, I apologize.

Can you or PAllen help me appreciate why the EM field can be described as a two scalar potential but gravity requires 6? Does it have something to do with the generality of GR vs a presumption of dimensions in studying EM waves?

I'm not sure what you mean by "why the EM field can be described as a two scalar potential but gravity requires 6"...what is a "two scalar potential" to start with? And in what sense does gravity instead require 6 of them?

The electromagnetic field is entirely specified by a given 4-potential $A_{\mu}$ (equivalent up to a gauge $A_{\mu} \rightarrow A_{\mu} + \partial_{\mu}\varphi$) whereas the space-time geometry associated with the gravitational field is entirely specified by a metric tensor $g_{\mu\nu}$ (equivalent up to a gauge = diffeomorphism). In the case of EM, we define a gauge covariant derivative $D_{\mu} = \partial_{\mu} + ie A_{\mu}$ and the curvature of $D_{\mu}$ is the electromagnetic field $F_{\mu\nu}$. In the case of gravity, we have the covariant derivative $\nabla_{\mu}$ defined in terms of $g_{\mu\nu}$ in the usual way through the Christoffel symbols and the curvature of $\nabla_{\mu}$ gives rise to the various fields describing tidal gravitational effects (expansion, shear, vorticity etc.). So this is essentially how we geometrically describe EM except the geometrical object that $D_{\mu}$ is associated with is more abstract than the physical space-time that $\nabla_{\mu}$ is associated with.

Now, why do we need a 4-tensor theory of gravity whereas we only need a 4-vector theory of EM? Well the 4-vector theory of EM follows straight from Maxwell's equations so that's a done deal. We could certainly try to posit a congruent 4-vector theory of gravity. In fact in a good GR textbook you will find a detailed walk-through of scalar and vector theories of gravity in an attempt to show that they are not adequate for describing gravity (see e.g. Straumann section 2.2.4). The key underlying mechanism that allows for a 4-vector theory of EM but rejects a 4-vector theory of gravity is the fact that like electric charges repel whereas all masses attract-it turns out that a Lorentz covariant i.e. relativistic vector field theory necessarily gives rise to repulsion between like "charges". Such a classical field theory cannot describe gravity. We are thus necessarily led to a 4-tensor theory of gravity (as a theory of a dynamical space-time geometry).

To answer your other question, no it doesn't have to do with the degrees of freedom. Both the gravitational and electromagnetic fields have two degrees of freedom at each point. This is the case for massless fields.

 

[rjbeery]

I'm not sure what you mean by "why the EM field can be described as a two scalar potential but gravity requires 6"...what is a "two scalar potential" to start with? And in what sense does gravity instead require 6 of them?

Please see this post

The electromagnetic field is entirely specified by a given 4-potential Aμ (equivalent up to a gauge Aμ→Aμ+∂μφ) whereas the space-time geometry associated with the gravitational field is entirely specified by a metric tensor gμν (equivalent up to a gauge = diffeomorphism). In the case of EM, we define a gauge covariant derivative Dμ=∂μ+ieAμ and the curvature of Dμ is the electromagnetic field Fμν. In the case of gravity, we have the covariant derivative ∇μ defined in terms of gμν in the usual way through the Christoffel symbols and the curvature of ∇μ gives rise to the various fields describing tidal gravitational effects (expansion, shear, vorticity etc.). So this is essentially how we geometrically describe EM except the geometrical object that Dμ is associated with is more abstract than the physical space-time that ∇μ is associated with.

This is too complex for me to absorb easily.

Now, why do we need a 4-tensor theory of gravity whereas we only need a 4-vector theory of EM? Well the 4-vector theory of EM follows straight from Maxwell's equations so that's a done deal. We could certainly try to posit a congruent 4-vector theory of gravity. In fact in a good GR textbook you will find a detailed walk-through of scalar and vector theories of gravity in an attempt to show that they are not adequate for describing gravity (see e.g. Straumann section 2.2.4). The key underlying mechanism that allows for a 4-vector theory of EM but rejects a 4-vector theory of gravity is the fact that like electric charges repel whereas all masses attract-it turns out that a Lorentz covariant i.e. relativistic vector field theory necessarily gives rise to repulsion between like "charges". Such a classical field theory cannot describe gravity. We are thus necessarily led to a 4-tensor theory of gravity (as a theory of a dynamical space-time geometry).

Great response. Good stuff to think about.

To answer your other question, no it doesn't have to do with the degrees of freedom. Both the gravitational and electromagnetic fields have two degrees of freedom at each point.

This...throws me for a loop. I'd like to consider forces all on equal footing, but this makes it even more perplexing to me why gravity cannot be described as a classical field. I'm reading what you've written but I need to ruminate on it.

 

[PAllen]

The 6 functions was an idea I threw out. To match GR, you need to have something equivalent to the metric tensor, which, in GR, is symmetric. Thus 10 components; but you can specify 4 fairly arbitrary gauge conditions, and have 6 independent functions (without loss of generality). So, I was imagining a possible formulation equivalent to GR, where you have 6 functions on an abstract background. You then have rules for extracting observables from them that are equivalent to applying the metric in coordinates satisfying the gauge conditions. You can pretend you have theory of 6 scalar fields on an abstract 'space' whose geometry is unspecified because it plays no role in extracting observables.

 

[atyy]

@rjbeery, if you would like to see a relativistic theory of gravity that is a scalar field, look up Nordstrom's second theory. It is important as the first relativistic theory of gravity, and was formulated as a field in flat spacetime. Einstein and Fokker reformulated it geometrically, analogously to general relativity. Although it is conceptually important, and is consistent with the strong equivalence principle, its predictions are not verified by experiment. For example, it gives the wrong perihelion precession, whereas general relativity does get it right. So perhaps comparing these two relativistic theories of gravity will help you understand the similarities and differences between a scalar and tensor theory of gravity.

As for why gravity cannot be a vector theory, as WannabeNewton says above, a vector theory doesn't describe attraction between "charges" of the same sign (you can think of mass as "gravitational charge").

An introduction to Nordstrom's theory is found in http://arxiv.org/abs/gr-qc/0405030.

You can also find extensive discussion of the analogies between the gravitational field and quantities that are related to a "refractive index" in http://arxiv.org/abs/gr-qc/0505065.

 

[WannabeNewton]

This...throws me for a loop. I'd like to consider forces all on equal footing, but this makes it even more perplexing to me why gravity cannot be described as a classical field. I'm reading what you've written but I need to ruminate on it.

If it helps, consider the special cases of gravitational waves and electromagnetic waves. The two independent degrees of freedom of the associated fields at each point of space just manifest themselves as the two independent polarization states (classically) of the waves i.e. both gravitational and electromagnetic waves have two independent polarizations. This is a consequence of describing EM and gravity through 4-vector and 4-tensor classical field theories respectively (and the associated gauge symmetries).

By the way I think you might have just misstated it but GR is a classical field theory-it's just not a scalar or vector theory on a flat background.

 

[PeterDonis]

As for why gravity cannot be a vector theory, as WannabeNewton says above, a vector theory doesn't describe attraction between "charges" of the same sign (you can think of mass as "gravitational charge").

There's actually a bit of a subtlety here, which I first encountered in the MTW exercise that asks you to compare the scalar, vector, and tensor theories.

If we write the Lagrangian of the vector theory in the usual way, i.e., the same way we write the EM Lagrangian, then yes, charges (i.e., masses) of the same sign repel rather than attract.

However, MTW throws a curve ball here: they write down a Lagrangian for a vector gravity theory that has a sign reversed on one of its terms, compared to the standard EM Lagrangian. In this theory, because of the flipped sign, like charges attract, so two masses attract rather than repel each other. But that doesn't fix the theory; it just shifts the problem to another part of the theory: gravitational waves in this theory carry negative energy (the last item in this part of the exercise is to show this).

 

[atyy]

Wow, PeterDonis, thanks!

 

[WannabeNewton]

There's actually a bit of a subtlety here, which I first encountered in the MTW exercise that asks you to compare the scalar, vector, and tensor theories...

That's really cool. Do you remember which exercise it is explicitly?

 

[PeterDonis]

Do you remember which exercise it is explicitly?

I just went and looked it up, it's Exercise 7.2. (7.1 deals with the scalar theory, 7.3 covers the tensor theory.)

 

[Dale]

To answer your other question, no it doesn't have to do with the degrees of freedom. Both the gravitational and electromagnetic fields have two degrees of freedom at each point. This is the case for massless fields.

How do you figure that? The EM potentials have four degrees of freedom. I could see using the gauge invariance to reduce one degree of freedom, but that still leaves three. How do you get rid of the last degree of freedom?

 

[WannabeNewton]

How do you figure that? The EM potentials have four degrees of freedom. I could see using the gauge invariance to reduce one degree of freedom, but that still leaves three. How do you get rid of the last degree of freedom?

http://www.damtp.cam.ac.uk/user/tong/qft/six.pdf

I just went and looked it up, it's Exercise 7.2. (7.1 deals with the scalar theory, 7.3 covers the tensor theory.)

Thanks!

 

[PAllen]

If it helps, consider the special cases of gravitational waves and electromagnetic waves. The two independent degrees of freedom of the associated fields at each point of space just manifest themselves as the two independent polarization states (classically) of the waves i.e. both gravitational and electromagnetic waves have two independent polarizations. This is a consequence of describing EM and gravity through 4-vector and 4-tensor classical field theories respectively (and the associated gauge symmetries).

By the way I think you might have just misstated it but GR is a classical field theory-it's just not a scalar or vector theory on a flat background.

Speaking of gravitational phenomena as a whole, the metric has 6 degrees of freedom. In what sense can gravity per GR be described as having fewer than this? Does your statement of two apply only to gravitational waves?

[edit: Note, the addendum to this: http://www.staff.science.uu.nl/~hooft101/gravitating_misconceptions.html
presents the view the EM has 3 degrees of freedom, two of which can propagate; and gravity has 6 degrees of freedom, two of which can propagate. ]

 

[WannabeNewton]

[edit: Note, the addendum to this: http://www.staff.science.uu.nl/~hooft101/gravitating_misconceptions.html
presents the view the EM has 3 degrees of freedom, two of which can propagate; and gravity has 6 degrees of freedom, two of which can propagate. ]

Yes. See also p.266 of Wald.

 

[PAllen]

Yes. See also p.266 of Wald.

Ok, so my point was if you are going to 'get going at all' with an attempt to reproduce all GR predictions with scalar functions, you could not possibly use less than 6, due to the 6 degrees of freedom. Whether it can be done meaningfully at all ultimately depends how loose you want to be on 'meanfingful'.

 

[H_A_Landman]

Let's use the Schwartzschild metric as a simple exact example solution of GR. If you take the low-speed (v << c) and weak-field (r >> R_s) limits, you end up with a metric that is just the flat-space metric plus a time term. In this metric, ALL the curvature is in the time dimension, so it looks exactly like flat Euclidean space plus curved time. You can recover Newtonian gravity from that metric, and it matches what we see in the vicinity of the Earth to better than 1 part per billion.

Approximately, therefore, if you are not moving at relativistic speeds and are not close to a black hole or neutron star, mass causes time dilation and time dilation causes gravity (i.e. curved geodesics). The time dilation gradient due to Earth's mass is what's pressing you into your seat right now. The idea that "gravity causes time dilation" is precisely backwards.

Anyway, in this limit GR gravity really is just a graded time dilation field in Euclidean space, which I think may be the best answer to the original question.

 

[pervect]

Let's use the Schwartzschild metric as a simple exact example solution of GR. If you take the low-speed (v << c) and weak-field (r >> R_s) limits, you end up with a metric that is just the flat-space metric plus a time term. In this metric, ALL the curvature is in the time dimension, so it looks exactly like flat Euclidean space plus curved time. You can recover Newtonian gravity from that metric, and it matches what we see in the vicinity of the Earth to better than 1 part per billion.

Approximately, therefore, if you are not moving at relativistic speeds and are not close to a black hole or neutron star, mass causes time dilation and time dilation causes gravity (i.e. curved geodesics). The time dilation gradient due to Earth's mass is what's pressing you into your seat right now. The idea that "gravity causes time dilation" is precisely backwards.

Anyway, in this limit GR gravity really is just a graded time dilation field in Euclidean space, which I think may be the best answer to the original question.

Necro posts are us. The last post in this thread was March, 2014.

Additionally, I don't think I agree that taking the low-speed, high R limit gives only the time componenents. For instance, take gravitational wave detection ala Ligo - the spatial components of the metric are very important for gravitational waves.

 

[rjbeery]

H_A_Landman: Thanks for the timely response ;) Anyway time dilation "causing" gravity is precisely the idea I was exploring, I just didn't word it in that way...

 

[PeterDonis]

I don't think I agree that taking the low-speed, high R limit gives only the time componenents.

For the large R limit, this is true; there is also an extra factor in front of the space components. This factor shows up in, for example, the bending of light by the Sun; it makes the GR prediction twice the Newtonian prediction.

However, in the slow motion limit (which excludes phenomena like light bending, since light is not "slow moving"--neither are gravitational waves), the extra factor in front of the space components becomes negligible, because we are restricting attention to spacetime intervals for which the spatial differentials $dx$, $dy$, $dz$ are much smaller than the time differential $dt$ (we are using "natural" units here in which $c = 1$). So the effective metric in the weak field, slow motion limit does have only an extra time-time component in addition to the flat spacetime metric.

Carroll discusses this briefly in Chapter 4 of his online lecture notes on GR:

https://arxiv.org/abs/gr-qc/9712019

 

[H_A_Landman]

... take gravitational wave detection ala Ligo - the spatial components of the metric are very important for gravitational waves.

Gravitational waves, like light, travel at the speed of light and hence violate the low-speed assumption. You wouldn't expect them to necessarily be handled correctly by the simplified metric.

 

[H_A_Landman]

Carroll discusses this briefly in Chapter 4 of his online lecture notes on GR:
https://arxiv.org/abs/gr-qc/9712019

Yes he does, specifically around equations 4.10 to 4.22 on pages 105-106. Thanks for the reference.