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Thread: Questions on General relativity

  1. #1 Questions on General relativity 
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    Hello,
    I`ve read Marcus thread on General relativity and I have a few questions concernig the problem as I do not understand GR at all I am merely interested in subject. I am creating new thread as I didn`t want to flood that excellent thread by my stupid questions .

    1) Some physicists today search for gravitational waves. I suppose these are some cases of vacuum solution to EFE same as EM waves are solution to Maxwell equations? If so they would be something like propagating "wave" in metric that would through connection look like propagating wave in curvature? What these waves represent? (change in distribution of matter??? or?) When they should be created and why we didn`t detected them yet? Would they transport energy same as EM waves and if so wouldn`t it mean that they have to have momentum -> act as force?

    2) Marcus said that from infinite possible connections Einstein chose Levi-Civita connection as it has zero stress. What would mean that property of spacetime is stress? Can it be somehow experimentaly verified that our spacetime is stress-free? Or is it possible that some other connection would get better results than Levi-Civita? Or does all connections yield same results and they are just some kind of gauge?

    3) Cosmological constant. From our observation of red shift shouldn`t we be able to quantify it? Does additional term changes solutions to EFE a lot or does it just add some kind of time dependence or ???

    4) The source term of EFE is Energy-momentum tensor. But shouldn`t GR TEM tensor contain energy tensors of all forces basically? Same as for example EM field of electron contribues to it`s inertia. Could deviation from real solution be relevant if one ignores for example EM stress tensor? How about other forces? Is gravity TEM tensor dominant source?

    5) Somewhen someone said that you can formulate GR from Lagrangian. Basicaly get EFEs as Euler-Lagrange equations. What I don`t understand is that Lagrangian is defined as scalar in spacetime (basicaly). How can one make connection between Lagrangian and curvature/metric? Also if such Lagrangian exist shouldn`t it among other symmetries be gauge invariant? Shouldn`t then be there gauge field accompanying?

    6) As EFEs contain only Ricci tensor how is it possible that we doesn`t lose some information about Riemann tensor? Or does Ricci tensor somehow contain all necessary information about curvature? Also are EFE`s solutions unique?

    Thanks in advance for your time


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  3. #2  
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    Quote Originally Posted by Gere View Post
    I suppose these are some cases of vacuum solution to EFE same as EM waves are solution to Maxwell equations? If so they would be something like propagating "wave" in metric that would through connection look like propagating wave in curvature?
    Yes, correct.

    What these waves represent?
    They represent periodic changes in the metric; in other words, they are described by a metric tensor the components of which are explicitely time-dependent.

    Marcus said that from infinite possible connections Einstein chose Levi-Civita connection as it has zero stress.
    No, the Levi-Civita connection has zero torsion. If I said "stress" anywhere, then I made a mistake.

    Or is it possible that some other connection would get better results than Levi-Civita?
    If you allow torsion to be present you get a theory called Einstein-Cartan gravity. The main difference is that ECT does not allow singularities to form, that the Big Bang is replaced by a Big Bounce, and that fermions are spatially extended objects. Whether or not this is better than GR is a matter of debate.

    Cosmological constant. From our observation of red shift shouldn`t we be able to quantify it?
    Yes, it's value can be estimated based on observations : Cosmological constant - Wikipedia, the free encyclopedia

    Does additional term changes solutions to EFE a lot or does it just add some kind of time dependence or ???
    The solutions to the EFEs contain the cosmological constant; depending on its value this may change the properties of the metric so obtained.

    But shouldn`t GR TEM tensor contain energy tensors of all forces basically?
    If you look at a system where the other fundamental forces are important ( like the nucleus of an atom e.g. ), then yes, you have to account for all those forms of energy as well.

    Could deviation from real solution be relevant if one ignores for example EM stress tensor?
    Yes, of course. All forms of energy are relevant so far as gravity is concerned.

    Somewhen someone said that you can formulate GR from Lagrangian.
    Yes, you can - it's called the Einstein-Hilbert Action. If you plug this into the Euler equations you can obtain the EFEs - see the link I gave just now, it demonstrates exactly how it is done.

    Also if such Lagrangian exist shouldn`t it among other symmetries be gauge invariant?
    Yes, GR is in fact gauge invariant; however, unlike most other gauge theories the gauge field of GR is a tensor field, called the Lanczos tensor.

    As EFEs contain only Ricci tensor how is it possible that we doesn`t lose some information about Riemann tensor? Or does Ricci tensor somehow contain all necessary information about curvature?
    Since the source term in the EFEs is a rank-2 tensor ( the energy-momentum tensor ), then the field which couples to it must be rank-2 as well, being the Einstein tensor. Physically, another reason has to do with conformal flatness, but to be honest I am not 100% clear on that particular aspect yet, so I can't elaborate at the moment. Perhaps KJW can explain this particular bit a little further.

    Also are EFE`s solutions unique?
    The theory is gauge invariant, so you can always apply a coordinate transformation to a metric without affecting the physics. The solutions are thus not "unique" in the usual sense of the word.


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  4. #3  
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    Actually, this should clear things up a bit :

    ricci.weyl

    The Ricci tensor is a contraction of the full Riemann tensor, so obviously cannot contain the same information; the bits which are missing are precisely encapsulated in the Weyl tensor. How this tensor behaves depends on the initial conditions chosen for the particular problem at hand - for example, in the Schwarzschild metric we demand the source to be spherically symmetric and static, and to coincide with Newtonian gravity at infinity.
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    Thank you Marcus this is very helpful. I plan to look more closely at GR in future when I have time as I am a bit interested in field theory on curved background. Your primer nicely explained meanings of all those Ricci, Riemann, Einstein-things that I didn`t understood


    Quote Originally Posted by Markus Hanke View Post
    No, the Levi-Civita connection has zero torsion. If I said "stress" anywhere, then I made a mistake.
    Yeah you got torsion in your Primer I just made mistake. I have no idea what consequences connection with torsion would mean. On the other hand I see no math reason why it should be 0 other than pure intuition which is fine in physics

    I glipsed on Einstein-Cartan theory on wiki a bit and field equations seems almost the same. Do they have same solutions eg. Schwardshild metric, FLRW etc. ? I was a bit confused when they said that Dirac equation is not linear there. That would be a bit funny. Too funny for my taste.
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    Quote Originally Posted by Gere View Post
    I plan to look more closely at GR in future when I have time as I am a bit interested in field theory on curved background
    For me it is the other way around - I want to complete my understanding of GR first, and then make a start on quantum theory and QFT, of which I know very little.

    I have no idea what consequences connection with torsion would mean. On the other hand I see no math reason why it should be 0 other than pure intuition which is fine in physics
    Well, it permits an additional degree of freedom into the theory, and increases the number of independent components in the various tensors. Physically, it has consequences such as the ones I listed earlier.

    I glipsed on Einstein-Cartan theory on wiki a bit and field equations seems almost the same.
    Careful - they look almost the same, but really aren't. The Ricci tensor is now no longer symmetric in its indices, and the usual ( symmetric ) energy-momentum tensor has been replaced by its canonical counterpart. This set of equations is mathematically distinct from the usual EFEs.

    Do they have same solutions eg. Schwardshild metric, FLRW etc. ?
    The solutions are the same only in the absence of energy-momentum, i.e. for the vacuum field equations. Therefore you can recover the "normal" exterior Schwarzschild metric also from Einstein-Cartan theory. However, the solutions will differ in the presence of energy-momentum, i.e. all interior and cosmological metrics ( in the presence of matter ) will not be the same as for standard GR. It should also be mentioned that even though the exterior metrics look the same, their physical effects might not be; for example, gravitational light deflection in Einstein-Cartan theory will yield a different value than standard GR, making the theory testable in principle.

    I was a bit confused when they said that Dirac equation is not linear there.
    Yes, that's because torsion couples to fermionic matter in non-linear ways, thereby destroying the usual symmetries of the Dirac equation. This causes problems in QM and QFT, as you can probably appreciate.
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    OK.

    Referring back to the OP, lemme just blurt out a few facts, since "proofs" seem not to be much in fashion here (I do have proofs, but they are hellacious)

    1. An affine connection on an arbitrary manifold (hereafter called a "connection") is symmetric if and only if the torsion tensor is zero.

    2. The converse is true - if a connection is symmetric then the the torsion tensor MUST vanish - just look at the definitions!

    (Why this is a requirement in GR I am not sure, but I suspect it is related to the fact that in general, since torsion is just a generalization of the curl in vector algebra, and if the curl of a vector field is zero, then that field can be interpreted as the gradient of a scalar field- a 1-form, or type (01) tensor. I also suspect this is just Einstein making his life easier than it might otherwise have been - as I say, I am not sure)

    3. Any connection is compatible with the metric if and only if the metric is preserved under parallel transport
    o
    4. For any manifold with a semi-Riemann metric there exists only one unique metric-compatible symmetric connection - this is the Levi-Civita connection

    5. The "full" Riemann 4-index curvature tensor cannot be fully symmetrisized by including in its definition a symmetric connection, whereas the contracted 2-index Ricci curvature tensor can.

    6. No "information is lost" by contracting the Riemann tensor to the Ricci tensor - indeed, it would not be a valid operation if that where so

    I urge you all to do the arithmetic - it is extremely challenging, but well worth it in the long run
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  8. #7  
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    Quote Originally Posted by Guitarist View Post
    The "full" Riemann 4-index curvature tensor cannot be fully symmetrisized by including in its definition a symmetric connection, whereas the contracted 2-index Ricci curvature tensor can.
    Good point.

    No "information is lost" by contracting the Riemann tensor to the Ricci tensor
    This bit I don't quite get - the Riemann tensor has 20 independent components, the Ricci tensor only 10 ( in GR space-times ). How is there no information loss in index contraction, in the sense that a given Ricci tensor does not uniquely imply a specific Riemann tensor ? For example, in a Schwarzschild vacuum the Ricci tensor vanishes everywhere, whereas the Riemann tensor obviously doesn't ( in the general case ); the "difference" between the two is then the Weyl tensor. There is no obvious way to get from Ricci "back" to Riemann, unless further information is supplied in the form of boundary conditions, a metric ansatz etc etc.
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  9. #8  
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    Quote Originally Posted by Markus Hanke View Post
    Quote Originally Posted by Guitarist View Post
    No "information is lost" by contracting the Riemann tensor to the Ricci tensor
    This bit I don't quite get - the Riemann tensor has 20 independent components, the Ricci tensor only 10 ( in GR space-times ). How is there no information loss in index contraction, in the sense that a given Ricci tensor does not uniquely imply a specific Riemann tensor ? For example, in a Schwarzschild vacuum the Ricci tensor vanishes everywhere, whereas the Riemann tensor obviously doesn't ( in the general case ); the "difference" between the two is then the Weyl tensor. There is no obvious way to get from Ricci "back" to Riemann, unless further information is supplied in the form of boundary conditions, a metric ansatz etc etc.
    I thought that, but perhaps Guitarist is talking about the components of the Riemann tensor being functions rather than numbers (considering the field globally rather than locally). While it's true that the Weyl tensor is algebraically independent of the Ricci tensor, the covariant derivatives of the two fields are not fully independent (the covariant divergence of the Weyl tensor can be expressed in terms of the covariant derivatives of the Ricci tensor).
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    There are no paradoxes in relativity, just people's misunderstandings of it.
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    Oh dear, this is soooo disheartening!

    I have given chapter and verse about tensors and their algebra more than once on this site, and I really cannot be bothered to repeat myself.

    I strongly suspect that both Markus and KJW are using a common (and regrettable - in my view) notational shorthand without fully understanding what it is shorthand FOR..

    The shorthand I refer to is firstly, to refer to tensors by their components and secondly, to adopt the so-called summation convention. I say this is all well and good IF AND ONLY IF those that use it have a full understanding of what it is shorthand for.

    To be specific to KJW: of course I am aware the the components of any tensor are scalar. But I equally insist that a tensor is a multilinear function from the Cartesian product of vector spaces onto the Real numbers. But if you write tis function-this tensor - in terms of its scalar components only,you have a dilemma - scalars become functions unless you EXACTLY what you are doing.

    Regarding "loss of information" Makus: consider and , these being mutual dual spaces.

    Then as we know. But if we regard as a type (0,1) tensor, as we may, as a type (1,0) tensor and as a type (0,0) tensor, we would have no difficulty seeing this as a contraction of a type (1,0) tensor by a type (0,1) tensor.

    Note the "=" sign implies identity i.e. no "loss of information" - this generalizes

    So and so that, since we may have that . Since indices are arbitrary set so .

    Note the arbitrary step I took, note also that it is not immediately obvious that

    Reader to supply the proof
    Last edited by Guitarist; September 24th, 2013 at 04:14 PM.
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    Ok, this is funny. I get that you may contract some tensor and not lose information about it by bringing additional boundary conditions, ansatz etc. But you are saying that contracting any tensor does not result in losing information?
    Consider your example. Let be scalar function (field) in space.
    Then I define arbitrary vector field .
    To this arbitrary vector field I can find covector field (linear functional or whatever is correct mathematical term?) that satisfies .
    How is this not a loss of information since I have arbitrariness in choosing vector field and not other way around?
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  12. #11  
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    Quote Originally Posted by Guitarist View Post
    So and so that, since we may have that .
    Ok, I get that.

    Since indices are arbitrary set so .
    That's all good and fine, but arbitrariness of indices is a symmetry choice you introduce yourself. If you are only given the contraction ( a scalar ), then this does not uniquely specify a rank (1,1) tensor from which the contraction resulted, unless more information is given, such as j=k. Otherwise, there are infinitely many vector spaces and their duals which will fit this procedure and yield the same scalar in the end. That is what I meant by loss of information - it is not uniquely reversible without additional information given, in the above case your choice of symmetry j=k. Likewise on manifolds - given only a Ricci tensor there are infinitely many choices of Riemann tensor which will fit that, unless more information is given; in other words Ricci curvature alone does not uniquely determine geodesic deviation on a general Riemann manifold.

    Or am I seeing this wrong ?

    How is this not a loss of information since I have arbitrariness in choosing vector field and not other way around?
    Yes, that's my thinking too.


    ADDENDUM : The Weyl conformal tensor can be considered the trace-free part of the Riemann tensor, whereas the Ricci tensor can be considered the "trace" of Riemann. Hence you need both to uniquely specify Riemann curvature globally in space-time. The same is not true in 1D, 2D or 3D though, because here the Weyl tensor is identically zero, and hence the Ricci tensor is sufficient on its own.
    Last edited by Markus Hanke; September 30th, 2013 at 05:03 AM.
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    Refer also to this document, page 22 :

    http://www.ita.uni-heidelberg.de/~du...sor/tensor.pdf

    It explicitely states that the operation of index contraction leads to a loss of information, since the resulting object has fewer components; this makes perfect sense to me, and it is how I always understood contraction.
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    Sorry to be blunt, but you guys are talking in riddles. (My bluntness comes from the fact that I have literally spent hours trying to explain this stuff on this site)

    The assertion seems to be that, if a function is bijective i.e. invertible it somehow contains "more information" (whatever that means) than one that is not. OK, tough guys, make the case and will accept it.

    Sigh.....

    1. Any tensor of rank 2 or greater can be written as the tensor product of vectors.

    2. Any tensor of any rank has a matrix representation, whose entries are the scalar components of each of the vectors entering into the tensor product, taken in all possible permutations.

    3. The trace of any matrix is the sum of its diagonal elements - no information is thrown away, the off-diagonal elements are simply ignored for the purpose at hand

    4. In the case of a type (1,0) tensor tensored by a type (0,1) tensor, say, equating indices and summing is exactly equivalent to taking the trace of its matrix representation

    5. And using the usual definition for inner products this is exactly equivalent to taking the inner product. Making use of the metric tensor, this can be made true of 2 tensors of the same type and rank.

    6. The trace of the matrix representation of a tensor is invariant under arbitrary coordinate transformations

    7. It follows that this is true of the inner product - though there is a major caveat there; ask me if you dare!

    So that is the mathematics. Why Einstein chose the Ricci tensor over the Riemann one for curvature, I cannot say - I suspect Physics comes into it, though I doubt it has much to do with the hand-waving stuff we have treated to here
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    no information is thrown away, the off-diagonal elements are simply ignored for the purpose at hand
    This is a contradiction in itself - since the off-diagonal elements are ignored, the trace cannot contain the same amount of information as the original matrix, so there is loss of information. Perhaps you are using a different definition of the term "information", but in physics textbooks and in this particular context it is taken to mean the number of independent components of a tensor; it is thus obvious that, if you contract a tensor across two or more of its indices, you reduce that number of components since you are reducing the tensor's rank, and hence what we are referring to as "information". For example, in GR the Riemann tensor has 20 independent components, the Ricci tensor only 10 ( or 256 and 16 respectively, if we ignore the various symmetries ).

    Guitarist, you know very well that we all respect your mathematical knowledge and abilities, but given the definition "information = number of functionally independent components" the situation seems quite clear and straightforward to me. I have a feeling though that you define "information" in a different way, hence the disagreement. Perhaps you can elaborate what your understanding of "information" in this context is.

    The assertion seems to be that, if a function is bijective i.e. invertible it somehow contains "more information" (whatever that means) than one that is not.
    The assertion is quite simply that the operation of index contraction yields a tensor which has fewer functionally independent components than the original tensor, and hence carries less physical information. How this relates to invertibility and bijectiveness, I cannot tell.

    Why Einstein chose the Ricci tensor over the Riemann one for curvature, I cannot say
    Well, the source term of gravity is a rank-2 tensor, so the left hand side had to be a rank-2 tensor as well; furthermore the covariant derivative of the left hand side needs to vanish, in the same way as the covariant derivative of the stress-energy tensor vanishes, in order to guarantee energy-momentum conservation. The Einstein tensor is the only possible choice. To the best of my knowledge there is no rank-4 tensor that one could use instead of the stress-energy tensor ( it's the conserved Noether current associated with translation invariance, after all ), so Riemann cannot appear in the field equations, simply because it doesn't have the correct rank.

    Misner/Thorne/Wheeler ( Gravitation ) put this in a different way ( quote page 343 ) : "The stress-energy tensor provides information only about a certain combination of components of the Riemann curvature tensor, the combination that makes up the Einstein tensor."
    Last edited by Markus Hanke; October 1st, 2013 at 02:22 PM.
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    Quote Originally Posted by Guitarist View Post

    3. The trace of any matrix is the sum of its diagonal elements - no information is thrown away, the off-diagonal elements are simply ignored for the purpose at hand
    This actualy reminds me a story our teacher told us. We had this guy on Classical electrodynamics which is subject that is concerning everything that has some connection to Maxwell equation. Maxwells are simply postulated on the first lecture and then solved for various physical problems. This professor once met with his friend a mathematician. The matematician asked him why is there such a course? From mathematician point of view Maxwell equations are a set of coupled partial differential equations that given a boundary condition have a unique solution so he couldn`t understand what was the purpose of three month course. Well the thing is that we would actualy want to know something about those solutions not that they just exist.
    This case is the same. You say we don`t lose some information we just ignore some off-diagonal elements. But we actually want to know those elements from traced tensor which is in general impossible.
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    OK, I waited a week to see if my anger would cool - it seems it hasn't to any noticeable degree.

    So I am about to make myself extremely unpopular - possibly terminally so

    Quote Originally Posted by Markus Hanke View Post
    Guitarist, you know very well that we all respect your mathematical knowledge and abilities, but given the definition "information = number of functionally independent components"
    I leave aside the patronizing note, and merely ask where this "equality" comes from (and no, I don't care which text or website - what is the argument?)
    I have a feeling though that you define "information" in a different way, hence the disagreement. Perhaps you can elaborate what your understanding of "information" in this context is.
    The introduction of "information" in this context was yours, not mine - I would not use this term.

    Let me try again. Suppose a finite-dimensional vector space with . Then one defines an inner product on by, say, so that

    But we know that, for any vector space there exists a dual space such that for any that for any .

    And iff there is an inner product defined as above we may have some particular such that . Notice the "equals" sign - it matters One assumes thereby that

    Using your stupid notation - , say, and , say.

    Since these are equal, where is the "loss of information given by reduction of functional components"?

    the source term of gravity is a rank-2 tensor, so the left hand side had to be a rank-2 tensor as well;
    If E. had made that argument he would have been a mathematical illiterate - he wasn't. You do NOT force an equality like this - it is circular reasoning

    Markus - and here I make myself persona non grata here - do not lecture us on the E. field equations if you not yourself understand this basic linear algebra
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    Ok, this is getting a bit ridiculous. In short, the trace operator is not injective. Now that this is out of the way can we continue the original discussion?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    Eh, I`m sorry Guitarist but I think that we all mean very different things using same words. I blame this on difference between mathemathicians and physicists notation.

    1. To your argument that Einstein couldn`t force equality like this. Well from what my friend here (astrophysicist) say the reason why he chose Ricci tensor of all possible contractions of Riemann tensor is that all other contractions will vanish trivially or something like that. I didn`t really understood him. He knew that left side of eq. has to do something with curvature therefore with Riemann tensor. Also the source part of equation has rank 2. So he knew that there has to be some contraction of Riemann tensor and Ricci tensor is only nontrivial possible contraction.
    The reason this "derivation" is mathematicaly so displeasing is that this is probaby fundamental physical law. You cannot derive fundamenal law precisely from it`s phenomenological approximations (Newtons law) you just have to have intuition and compare predictions based on that intuitively derivated law with experiments which Einstein did.
    Many physical laws are derived this way. Schrodinger equation, Dirac equation, Higgs coupling etc.

    2.To you assumption that equality sing preserve information. Mathematicians obviously have some different notion of information. If I go to extreme consider this where H is differential operator describing physical system, E is real number and is eigenfuction of H. Now left hand side describes in detail what physical systm are you studying, tells you it`s energies and wavefunctions. Does right hand side tells you the same? Your entire proof (or explanation of basic linear algebra) is based on assumption that equal sign preserve information. This is obviously not true.
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    and no, I don't care which text or website - what is the argument?
    The argument is simply that ( in general ) all components of a tensor have a distinct physical interpretation and meaning; hence, if you reduce the number of components, you reduce the amount of physical information encapsulated in the object. For example, you can form the trace of the stress-energy tensor, but then you loose all information as to the stresses, fluxes and momentum densities of the system in question.

    I leave aside the patronizing note
    Using your stupid notation
    do not lecture us on the E. field equations
    This is definitely not the direction or the tone I want this discussion to go to, so I won't reply to this.

    Ok, this is getting a bit ridiculous.
    I would tend to agree.
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    If E. had made that argument he would have been a mathematical illiterate
    I have just laboured my way through chapter 15 of MTW's Gravitation, where the field equations are derived as a geometric consequence of just three basic principles :

    1. The law of energy-momentum conservation
    2. The identification of energy-momentum density with the Cartan moment of rotation
    3. The principle that "the boundary of a boundary is zero"

    Given these ( and the stress-energy tensor ), the field equations practically come falling into your lap in just a handful of brief steps. However, not wishing to cause any more controversy I will refrain from going into further details - just thought I mention it here as an answer to why we find the Einstein tensor in the EFEs, and not the full Riemann tensor.
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    Quote Originally Posted by Guitarist View Post
    Using your stupid notation
    Ok, I'll bite... What is wrong with the notation? And what makes the notation you are using any better?
    There are no paradoxes in relativity, just people's misunderstandings of it.
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    The simple answer is Gravity is the missing force needed to preexist to compress a singularity enough to create a big bang effect. The so called, gravity particle did not exist pre-big bang. A tensor force is created when the preexisting collapsing force begins to rotate.
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    Quote Originally Posted by YangYin View Post
    The simple answer is Gravity is the missing force needed to preexist to compress a singularity enough to create a big bang effect. The so called, gravity particle did not exist pre-big bang. A tensor force is created when the preexisting collapsing force begins to rotate.
    The OP's question is about GR. Gravity is not a force in GR, so your answer starts off wrong and rapidly gets worse. If you don't have an answer based on science, please don't post it in the general section. Confine your ignorance to Pseudo or Trash, please.
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    Does anyone have a full set of Einstein's equations.As I recall his equations represented a good framework for the observable universe matter and energy interactions.The trouble is that I forgot what his equations were,but I do remember they were good.
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    Quote Originally Posted by Phd. Cubs View Post
    Does anyone have a full set of Einstein's equations.As I recall his equations represented a good framework for the observable universe matter and energy interactions.The trouble is that I forgot what his equations were,but I do remember they were good.
    In brief, the Einstein equations ( without cosmological constant for now ) are



    You will find further information in the stickies in the physics section of the forum.
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    This will be lengthy, as I anticipate it will be last time on this forum, so bear with me
    Quote Originally Posted by Gere View Post
    2.To you assumption that equality sing preserve information. Mathematicians obviously have some different notion of information. If I go to extreme consider this where H is differential operator describing physical system, E is real number and is eigenfuction of H. Now left hand side describes in detail what physical systm are you studying, tells you it`s energies and wavefunctions. Does right hand side tells you the same? .
    This is just silly - the equality you quote is called the "characteristic equation" for the operator and comes quite simply from rearrangement of the "skeletal" Schroedinger eqn And by taking the determinant of the characteristic eqn - yes it IS a matrix - you get the "characteristic polynomial", whose roots are found, as always for a polynomial, by setting to zero and solving. These are of course the eigenvalues for the S. HamilitonianWhere is the loss or gain in "information by doing this?

    To KJW: It is very dangerous to refer to tensors by components unless you know EXACTLY what you are doing. Here's why..

    A tensor is defined as a (multilinear) map from the Catesion product of vector spaces to the Reals, like this

    A type (1.1) tensor is an element in the pace of all mappings . For example I called this a "mapping - strictly it is a "functional" (since the Cartesian product of vector spaces is itself a vector space)

    Write as and and similarly for the other vectors, one has the type (1,1) tensor

    Now assume a manifold such that the type (1,0) tensors, say, entering into the direct product are defined specifically at the point that is In other words, our manifold may not be globally i.e "flat"

    It can be proved - I did it for you elsewhere - that in this circumstance one can make the identification so that where, by the definition of a manifolds the are the coordinate functions from an open neighbourhood of our manifold to the Reals.Also in this circumstance it can be proved that , that is a type (1,0) tensor acrs as a linear operator

    I also showed how these 2 interpretations of a tensor - first as a (multi)linear functional and second as a linear operator on functions can be brought into register.

    All of the above generalizes - it applies to any tensor of any rank and any type (whatever their so-called "physical interpretation")

    BUT all this subtlety is completely lost once you write say which latter is quite simply, in the general case, an array of real numbers. It is logically equivalent to defining the function, say as

    I now wish you all farewell - a sad day for me, but as I seem no longer to be heeded here, I conclude I am no longer needed.

    Good luck
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    Dear Guitarist I have often read your posts in eager anticipation that some of it might slowly start to sink in, this supremely high level at which you seem to be able to express with such consumate ease never fails to amaze me, I'm sure like many others I've been enriched by being able to see your work and to read your posts. I feel that whatever has led you to believe that presence or ability is not needed really does need to be resolved, because your presence here is surely of great benefit to all of us and most definately desirable to all us that hold both your skill and ability in the highest esteem.

    I can only hope to express how much of loss to all of us your leaving would be and just how highly valued your presence actually is. I really wish you well for the future but with the strongest wish that you may yet come to realise the true respect in which you are sincerely held and reconsider, best wishes.
    Everything has its beauty, but not everyone sees it. - confucius
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    Quote Originally Posted by Guitarist View Post
    To KJW: It is very dangerous to refer to tensors by components unless you know EXACTLY what you are doing. Here's why..
    It is very dangerous if you are trying to prove some deep theorem about tensors but completely reasonable to use component notation if you are using tensors as a tool. Some things that are important for the theory are not important for the application - a theme that is repeated over and over again in mathematics (Just look at in any stochastic calculus text, or non-measurable sets in integration theory).

    This whole thread has taken a really weird turn and I'm not really sure why.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    Quote Originally Posted by river_rat View Post
    Some things that are important for the theory are not important for the application
    I would agree with this. There is a place for tensors as abstract geometric objects understood in the context of vector spaces ( with all the finer points, proofs, etc ), and there is a place for tensors as simply an array of indexed components, to be manipulated with the usual rules of calculus and index gymnastics. I am firmly rooted in the physics side of things, and in my mind the index notation is indispensible - it allows me to perform explicit, analytical computations, using tools such as MAPLE and Mathematica, and sometimes pen, paper and whiteboard. I really don't understand why it is such a big deal at all.
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    OK I`m completely lost. Somebody please explain me what is going on. I ask a question about general loss of information about curvature, Markus answers me that I need Weyl conformal tensor and then Guitarist starts acting angrily and begins explaining basic linear algebra and meaning of tensors repeatedly. Did I miss some subtle point? Is there some deep mathematics that I`m missing?
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  33. #32  
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    Quote Originally Posted by Gere View Post
    Did I miss some subtle point? Is there some deep mathematics that I`m missing?
    No, I don't think you really missed anything so far as the physics is concerned. I am equally baffled as to why the thread went the way it did, but I don't think it has anything to do with the maths or physics involved, so I think we should just move on. The physical facts in GR are simple :

    1. In GR the source of gravity is the energy-momentum tensor
    2. The energy-momentum tensor does not in turn specify the full Riemann tensor, but only a certain combination of its components, called the Einstein tensor ( which is itself a linear combination of the contractions of full Riemann )
    3. Therefore, given just the Einstein tensor ( i.e. the Ricci tensor and Ricci scalar ), it is not in general possible to construct a unique Riemann tensor, because G does not contain the same physical information about curvature as the Riemann tensor. The "extra" information that is "missing" is encapsulated in the Weyl tensor
    4. The Weyl tensor ( or, indeed, the full Riemann tensor ) does not appear in the field equations simply because the source term contains information only about the components of the Einstein tensor, nothing more.
    5. As KJW has pointed out in post #8, while the Weyl and Ricci tensors are not algebraically related, their covariant derivatives are not fully independent.

    Does this answer the original question ?
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    Yes, thanks.
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    Quote Originally Posted by Gere View Post
    Yes, thanks.
    No problem, always glad to help when and where I can...you might give me a hand one day with a few questions I have regarding quantum physics, which is totally not my area of expertise. I'm not quite ready yet, but I'll let you know
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