exchemist
Valued Senior Member
This is a day to break out the champagne!!
P.S. Just received a laconic response to my reporting of this morning's efforts by The God, saying "resolved". Indeed so.
Correlating Newtonian Model with Einstein's GR
- Thread starter hansda
- Start date
P.S. Just received a laconic response to my reporting of this morning's efforts by The God, saying "resolved". Indeed so.
In my defense... I had six reports to close all relating to or resulting from this individual
exchemist
Valued Senior Member
Well I guessed I would not be the only one. But I am proud to have added my brick to the wall.In my defense... I had six reports to close all relating to or resulting from this individual
DaveC426913
Valued Senior Member
NotEinstein
Valued Senior Member
Please elaborate; what do I not understand?
Please elaborate; where have I misquoted you?
And Newtonian physics is made up to corroborate to reality. How is that a bad thing? Of course they have to match reality; they would be useless as models otherwise!
From a fundamental perspective.
Absolutely not. Please stop putting words in my mouth.
Since we both agree that Lagrangian mechanics doesn’t introduce any new physics when compared to Newtonian mechanics, there was no need to do so.
I haven’t seen any strong argument from you either. “made up”, “conceptually different”. How is that not sloppy or hand wavy?
I did not withdraw it. I only admitted it doesn't provide a working version Newtonian physics with a limited speed of gravity. It does however show that the limited speed of gravity was something being pondered about long before Einstein. In other words, people were already trying to modify/reject Newtonian physics' infinite speed of gravity.
What? I do not understand what you are referring to?
“Hand waving” doesn’t mean “wrong”. Please look up the meaning of "hand waving" as used in science.
I did not. Please stop misrepresenting my position.
Shall we also count the number of times you backed out of pointing out the mistake(s) in the mathematical derivation of Newtonian physics from GR?
Yes, indeed it is. If I have done so, I apologize about that and will withdraw those statements. I’ve already pointed out so many instances of you doing the same (shall we count those?), I hope you’ll do the same?
NotEinstein
Valued Senior Member
(I wrote the above post before I knew about the permanent banning of The God.)
I guess that leaves just one thing... For hansda to respond to my inquiry about his text not mentioning anything about the correlation between GR and Newtonian physics.
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OK, here's one thing I don't understand.
Newton's playground was flat space. Assume we can extend this to spacetime - that's what the weak field limit seems to mean.
Given the field equations of gravitation, it is easy to see that in the absence of a gravitational source, spacetime is flat. This is because the Ricci curvature tensor everywhere vanishes, which can only happen when the metric field is constant (the Ricci tensor field is a second order differential of the metric field with respect to the coordinates - simple calculus). This is especially pretty because it provides an analogue of the famous Laplace equation - the divergence of the gradient of a scalar field is identically zero in the absence of a source. This is generally written as $$\nabla^2\phi =0$$ where $$\phi$$ is the scalar field.
With this rambling, I get to my question: if GR reduces to the Newtonian law in the weak field limit, and if there exists a source, how can it be that, even so, the curvature tensor vanishes everywhere?
Newton's playground was flat space. Assume we can extend this to spacetime - that's what the weak field limit seems to mean.
Given the field equations of gravitation, it is easy to see that in the absence of a gravitational source, spacetime is flat. This is because the Ricci curvature tensor everywhere vanishes, which can only happen when the metric field is constant (the Ricci tensor field is a second order differential of the metric field with respect to the coordinates - simple calculus). This is especially pretty because it provides an analogue of the famous Laplace equation - the divergence of the gradient of a scalar field is identically zero in the absence of a source. This is generally written as $$\nabla^2\phi =0$$ where $$\phi$$ is the scalar field.
With this rambling, I get to my question: if GR reduces to the Newtonian law in the weak field limit, and if there exists a source, how can it be that, even so, the curvature tensor vanishes everywhere?
NotEinstein
Valued Senior Member
In the weak field limit the metric takes the form of a flat metric plus some small perturbations: $$G_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$$, where the contributions of $$h_{\mu\nu}$$ are small. For more details, see this link I posted somewhere in this thread before: https://www.quora.com/How-can-we-de...avitation-from-Einsteins-theory-of-relativity
I am unconvinced that the term "flat metric" has any mathematical meaning. Moreover, you are just reiterating the assertion for which I was seeking clarification.
Nonetheless, I thank you for your response. However, I have figured it out, so if anyone is interested, here goes........
In general, an $$n$$-manifold is defined as a "jazzed-up" point set that is, at some arbitrary scale, indistinguishable from the plane $$R^n$$. Clearly $$R^n$$ is itself a manifold.
A manifold with a metric is similarly defined, but with $$R^n=E^n$$, by which I mean that the Euclidean metric applies locally. Assume that spacetime is a 4-manifold with a metric. Now because of the fact that one of the coordinates is signed differently from the other 3, at least in the definition of the metric, the Euclidean metric cannot apply, so the Minkowski 4-metric $$\eta_{jk}$$ is used instead. This is, for example, is the metric used in the Special Theory, a theory that ignores forces an acceleration, where this metric is the same at each and every spacetime point for any choice of coordinates. I assume this implies that our 4-manifold is trivial i.e. that the "arbitrary scale" I referred to above is the whole banana.
So if you want the Special Theory to be compatible with the General Theory, it follows that in the absence of a gravitational source ($$T_{jk}=0$$), and for the algebraic reasons I can earlier that the curvature field vanishes ($$R_{jk}=0$$), implying the metric field is constant globally. Again implying (if not down-right proving) that in the absence of a source, spacetime is locally and globally indistinguishable from itself (surprise!!) or flat.
But since the inverse is true - a constant metric field implies the absence of a gravitational source - from the definition of the spacetime 4-manifold, we can take "local" to be a region of weak gravitational source.
Nonetheless, I thank you for your response. However, I have figured it out, so if anyone is interested, here goes........
In general, an $$n$$-manifold is defined as a "jazzed-up" point set that is, at some arbitrary scale, indistinguishable from the plane $$R^n$$. Clearly $$R^n$$ is itself a manifold.
A manifold with a metric is similarly defined, but with $$R^n=E^n$$, by which I mean that the Euclidean metric applies locally. Assume that spacetime is a 4-manifold with a metric. Now because of the fact that one of the coordinates is signed differently from the other 3, at least in the definition of the metric, the Euclidean metric cannot apply, so the Minkowski 4-metric $$\eta_{jk}$$ is used instead. This is, for example, is the metric used in the Special Theory, a theory that ignores forces an acceleration, where this metric is the same at each and every spacetime point for any choice of coordinates. I assume this implies that our 4-manifold is trivial i.e. that the "arbitrary scale" I referred to above is the whole banana.
So if you want the Special Theory to be compatible with the General Theory, it follows that in the absence of a gravitational source ($$T_{jk}=0$$), and for the algebraic reasons I can earlier that the curvature field vanishes ($$R_{jk}=0$$), implying the metric field is constant globally. Again implying (if not down-right proving) that in the absence of a source, spacetime is locally and globally indistinguishable from itself (surprise!!) or flat.
But since the inverse is true - a constant metric field implies the absence of a gravitational source - from the definition of the spacetime 4-manifold, we can take "local" to be a region of weak gravitational source.
The special theory implies Minkowski spacetime (I think Quarkhead has just shown this), i.e. the absence of a gravitational source. But isn't Minkowski spacetime just one solution for "the" vacuum?
And isn't Minkowski spacetime always a local solution (light doesn't bend much over relatively small distances)? Unless of course, you have a universe with no matter in it?
And isn't Minkowski spacetime always a local solution (light doesn't bend much over relatively small distances)? Unless of course, you have a universe with no matter in it?
Schmelzer
Valued Senior Member
The weak field limit would have been better named weak field approximation. It is only an approximation, not really a limit. (The usual sloppy language of physics.)
In the weak field approximation, the clock time is approximately the same as Newtonian absolute time, and the curvature is approximately zero.
If there would be a real limit, it would have to be exactly zero. But this would be only in the case when the gravitational potential is exactly constant.
NotEinstein
Valued Senior Member
First you say that you are unconvinced that "flat metric" means anything, but then you prove that $$\eta_{\mu\nu}$$ results in flat spacetime... Should I instead have said that "in the weak field limit the metric takes the form of the Minkowski metric..." etc., and that that results in a (nearly) flat spacetime?
And I'm not sure the curvature tensor disappears: I don't think it does in the link I posted. The perturbation term still gives a contribution, and it's that which gives rise to Newtonian gravity.
According to Wikipedia:
"
It is a mathematical fact that the Einstein tensor vanishes if and only if the Ricci tensor vanishes. This follows from the fact that these two second rank tensors stand in a kind of dual relationship; they are the trace reverse of each other:
where the traces are:
.
A third equivalent condition follows from the Ricci decomposition of the Riemann curvature tensor as a sum of the Weyl curvature tensor plus terms built out of the Ricci tensor: the Weyl and Riemann tensors agree,
, in some region if and only if it is a vacuum region."
--https://en.wikipedia.org/wiki/Vacuum_solution_(general_relativity)
"
It is a mathematical fact that the Einstein tensor vanishes if and only if the Ricci tensor vanishes. This follows from the fact that these two second rank tensors stand in a kind of dual relationship; they are the trace reverse of each other:
where the traces are:
A third equivalent condition follows from the Ricci decomposition of the Riemann curvature tensor as a sum of the Weyl curvature tensor plus terms built out of the Ricci tensor: the Weyl and Riemann tensors agree,
--https://en.wikipedia.org/wiki/Vacuum_solution_(general_relativity)
Which is OK when physicists talk among themselves, I suppose, but confusing on a forum such as this.
In the present context, I cite 3 more examples.....
The General Theory is a FIELD Theory. This means that every term in the Einstein field equations are themselves fields. So it only makes sense to talk about "the" metric tensor or "the" curvature tensor (say) when you are referring to a specified point in spacetime. And if you are, it makes very little sense geometrically.
Second, the so-called tensor, say, $$A_{jk}$$ is not a tensor at all, neither is it a tensor field - in spacetime coordinates (where it is assumed that all tensor fields are symmetric), it refers rather to the 10 scalar components of a tensor (or outer) product of vector fields.(If you really have no life, read this. I did it here many years ago http://www.sciforums.com/threads/sr-so-whats-a-tensor-for-chrissake.74301/
Third, and annoyingly, the set of all tensor fields on a given manifold is itself a vector space by the usual axioms. Therefore any element in this set i.e. tensor field is by definition a vector.
Confused? you should be!![/quote]
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Schmelzer
Valued Senior Member
That's not the point. The language of physicists is sloppy if considered from point of view of mathematicians. The language in such a forum, full of laymen, is even more sloppy.
This thread is about co-relation of various models of Physics. NM and GR are two models of physics. So they can be correlated. I have explained this in the OP.
This thread is not about derivation.
I think that NotEinstein is pointing out that the only TOE you have are those 10 things in your shoes, but that is just my opinion.
Good joke.
I was invited to check this thread by hansda here: http://www.sciforums.com/threads/universe-expansion.159815/page-7#post-3478871 Since I do not see a "correlating NM with GR through [hansda's] TOE", I assume it must be in the PDF linked in the opening post. So let's take a look at it!
The abstract already makes one thing clear: the word "success" here is not used as it is in statistics. Statistics is the closes one would get to a theory of success in mainstream, but when hansda says there are only qualitative and not-mathematical theories in mainstream science, it cannot be referring to that. This reading is confirmed in the first line of the main text, where "success" is described as "a general desire".
The rest of the introduction is introductory, so let's pick it up again at the start of the next section, called "II. Discussion".
'II.1' through 'II.5' introduce our definitions. Note that action is defined significantly different from the mainstream physics usage of the term. Note also the usage of the word "desire". A "doer" thus must possess a mental process, and cannot be lifeless. So for example, a planet cannot be a "doer".
II.6 starts with a pretty okay description of the Newton's first law. The second paragraph seems to be inconsistent with the previous definitions. It uses the phrasing "duration of the action", however, an action is defined as "any movement or motion of a particle with [sic?] relative to an observer". The duration of a motion is ill-defined, as we can always translate into its rest frame. Additionally, does changing a motion "end" the original motion? But this confusion appears to be irrelevant (for now), so let's move on.
The rest of the page just introduces a (non-standard) notation. What is given here is a partial description of standard classical mechanics. Nothing to comment here.
Page 3 (or actually, equation 2) opens with usage of the term "infinitesimal unit of time". It strongly suggests a mis-use of terminology: "two consecutive instants of time" is NOT the proper way to talk about this. However, this may simply be a wording issue, so let's continue.
The last part of II.6 is weird, as a set of forces is being replaced by its sum. However, let's assume this is not signaling an actual correspondence, but just a lingual nicety. Other than that, no comments as this appears to be quite standard classical mechanics, albeit with an unusual notation.
II.7 introduces another definition. Here we encounter a point of interest: a "technique" is coupled to a (single) action, but it is described by its own C(R)FS. However, these are already summed over all forces that affect a particle, so there can be only one. Due to the uniqueness of the C(R)FS, there can only be one technique (per particle) at a time. Thus only one action at a time. This however is consistent with II.6, and we now find our "duration of a motion": it is as long as a certain technique is applied.
II.8 confirms this reading.
II.9 starts with introducing pretty standard mathematical terms. No comment. Then differentation is defined through a combination of a tangent line and infinitesimals. Not sure if this is mathematically sound, but it wouldn't be far off.
Then this: "We know that a point on a straight line is dimensionless ie its radius is zero of has no length." This is non-sensical. We are talking about calculus here, not geometry. Saying a tangent line has zero radius is just mixing different mathematical domains to the point of meaninglessness. Luckily for us, it appears this "radius" has no real impact on the outcome, so we can safely ignore this confused mess.
Page 4 then throws an infinite set our way by "expanding" the real numbers into intervals of size dx, using "infinite" as if it's a number. This is very improper notation, but it's still clear what is meant, so let's see where this leads.
The values of a function are given the same treatment in equation 5. Such a set can either contain a single value multiple times, or we are expected to filter for uniqueness ourselves; that is not made clear.
Then we connect the notation of this set with the CRFS. This is of course problematic, since one set is defined over an infinite interval, the other over a finite (time-)interval. This subtlety is ignored in the text, but infinities are notoriously dangerous to play such tricks with, so I think the equating of these two sets is not sufficiently supported. But... let's say that is was!
CRFS now thus have a function associated with them. Note that this does not necessarily place any restrictions on these functions: they can in principle be discrete, be non-continious, etc.
II.10 contains no new information: it's just putting several definitions together.
II.11 seems to suggest only one technique can lead to a desired result. This may either be sloppy wording, or a property of techniques that I missed. Other than that, nothing spectacular is written here.
II.12 is just a re-statement of things we saw earlier. It however ends with the line "So, this theory also can be considered as a Theory of Everything". What is meant here is that because all actions and techniques are contained within the framework of this theory, it will contain everything. So this is not a theory of everything in the mainstream scientific usage of the word, but a simple statement that the domain of applicability of this theory is the entire universe.
III is supposed to be the conclusion, but it introduces the idea that one can approach success without reaching it. In other words, there is all of a sudden an ordening in outcomes, not just a "single one succes, rest equal". It also says that doers can perform actions by using techniques leading to success. This isn't an insight of any kind; give something the power of choice and this will follow almost necessarily.
This conclusion does however confirm my reading of "Theory of Everything" as I commented on with II.12.
And there the text ends. This is weird, because it appears to be missing its most essential parts! It does not define "success" outside of "doers", thus it places no restrictions on the motions of (for example) fundamental particles or planets. All non-"success" related parts of the text are standard classical mechanics.
I fail to see where any insight might be gained from this text as it currently stands.
And, I also see no derivation of the Einstein field equations, or anything GR related. hansda, you misled me!
Thanks for reading my paper. Also thanks for your views.