The Laws Of Cosmology May Need A Re-Write?

Discussion in 'Astronomy, Exobiology, & Cosmology' started by Ultron, Apr 18, 2016.

  1. Farsight Valued Senior Member

    IMHO this is right and wrong. There's no overall gravitational field, but there is an energy density. A gravitational field is a place where the spatial energy-density is not uniform. If you were midway between two stars the spatial energy density is uniform so you don't fall down, but it isn't zero.

    A gravitational field is a place where light curves and matter falls down, but it doesn't suck space in. Like I said to paddoboy, the waterfall analogy is garbage. It started with Gullstrand-Painleve coordinates, which Einstein dismissed for good reason. We do not live in some Chicken-Little world. Gravity doesn't stop space expanding.

    Gravitational field energy is positive. See this: "the energy of the gravitational field shall act gravitatively in the same way as any other kind of energy".

    Space has an energy density. If it's uniform there's no gravitational field. If it isn't, there is.

    So they say. But when it comes to gravitational redshift, the photon is emitted with less energy. It doesn't lose energy as it ascends. See this:

    "An atom absorbs or emits light at a frequency which is dependent on the potential of the gravitational field in which it is situated".

    Negative pressure is tension. Think of the balloon analogy. You can expand a balloon by reducing the tensile strength of the skin. Think bubblegum.
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  3. Schneibster Registered Member

    From Wikipedia:
    If you can't show why it's wrong then this is just another unsupported claim. You couldn't do it with Gibbs either.

    The reason nobody bothered to show it's wrong is that, like your ideas, it ignores the fact that if you stick to one set of coordinates you get consistent answers.
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  5. Schmelzer Valued Senior Member

    Get your quotes right. There was, btw, nothing in that Wiki of interest here. And the problem with your "frames" is not that they are wrong (if they are simply an unfortunate name for whatever, this would not make it wrong) but that you do not provide a clear definition. Some parts of your posting suggests it is simply a system of coordinates, others that it is some vector field.

    I have no problem with Gibbs formula, it has the same problem as the pseudotensors in general - they define energy and momentum depending on a system of coordinates, Gibbs formula in dependence on some vector field. Its fine with me if you want to introduce some such preferred objects, but in this case you have to reject GR spacetime interpretation which forbids such preferred objects as anathema. Or you don't have an energy-momentum tensor of the gravitataional field, but one of the gravitational field and some other unknown vector field with unknown physical meaning.
    In the context of my remark about Einstein's paper your remark is completely off, because Einstein's proof was about comparing the resulting energy for different systems of coordinates. He has shown that different systems of coordinates give the same global energy, and this was the whole point of his proof. Unfortunately, the proof works only if the change of coordinates is restricted.

    In general, I do not ignore this, but fully agree with the point that one can do physics in one system of coordinates. But if you can define energy only in one system of coordinates, or depending on some particular vector field, and you want to have an energy as some object of physics, you have to specify which system of coordinates is that which defines energy, or which vector field is the one which defines it. Else you have a lot very different energies, one for every system of coordinates, or for every vector field.
    Last edited: Apr 24, 2016
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  7. paddoboy Valued Senior Member


    It's your own nonsense that is refuted Farsight and it is certainly you that take Einstein out of context, and also fail to realize that what you have taken out of context is also 100 year old statements. You see unlike yourself, Einstein was a great man who was not afraid to admit error or mistakes in his thinking, and he certainly did do that.
    I suggest you take the rest of your misrepresented claims to your book and/or personal web site and then you won't get so much flack.
  8. paddoboy Valued Senior Member

    Schmelzer is driven by his paper re the ether, and the above statement certainly applies to his thinking and failings.
  9. paddoboy Valued Senior Member

    That's your opinion and your opinion is wrong.
    Let me put it again: Any photon emitted just above the EH and directly radially away, will appear in the local frame to hover there forever, never getting away and never falling in. You see whatever analogy you use, the EH of a BH represents an escape velocity of "c" and light/photons always travel at "c"
    Common sense then follows.

    With regards to your dismissal of the waterfall analogy, you have obviously misunderstood or just simply do not understand the fact of what an analogy is.
    But understandable since you also misunderstand and misrepresent Einstein.

    The river model of black holes
    Andrew J. S. Hamilton, Jason P. Lisle (JILA, U. Colorado)
    (Submitted on 12 Nov 2004 (v1), last revised 31 Aug 2006 (this version, v2))
    This paper presents an under-appreciated way to conceptualize stationary black holes, which we call the river model. The river model is mathematically sound, yet simple enough that the basic picture can be understood by non-experts. In the river model, space itself flows like a river through a flat background, while objects move through the river according to the rules of special relativity. In a spherical black hole, the river of space falls into the black hole at the Newtonian escape velocity, hitting the speed of light at the horizon. Inside the horizon, the river flows inward faster than light, carrying everything with it. We show that the river model works also for rotating (Kerr-Newman) black holes, though with a surprising twist. As in the spherical case, the river of space can be regarded as moving through a flat background. However, the river does not spiral inward, as one might have anticipated, but rather falls inward with no azimuthal swirl at all. Instead, the river has at each point not only a velocity but also a rotation, or twist. That is, the river has a Lorentz structure, characterized by six numbers (velocity and rotation), not just three (velocity). As an object moves through the river, it changes its velocity and rotation in response to tidal changes in the velocity and twist of the river along its path. An explicit expression is given for the river field, a six-component bivector field that encodes the velocity and twist of the river at each point, and that encapsulates all the properties of a stationary rotating black hole.
  10. Schneibster Registered Member

    Somehow the post got munged. I think I forgot a quote or something.

    The underlined bold is my emphasis; the bold is in the original.


    I note with interest that there is no requirement that it be rectilinear, and I also note that you have left out the second part of the definition.

    I'm not sure where you think I defined it as a vector field. I don't think I did. It's probably a misunderstanding; can you tell me where you think I implied this so we can correct it?

    I have the strong feeling that you aren't looking at frames in a GRT way; such frames are not rectilinear because they are defined on a manifold that is not flat. Introducing curvature that varies from place to place in a frame is what GRT is all about.

    When you choose a particular solution for GRT, you have defined a frame on a curved manifold. When you change solutions (i.e. change frames), energy is not conserved between them; this is not unexpected because you have changed frames. You have to apply a transform, and that transform is determined using GRT to define the difference between the two frames at each point within each frame. That is why you are using tensors. But the underlying fact is the same; you can't import a tensor that's valid in one solution of GRT into another solution of GRT without transforming it, it's exactly the same as importing a vector from one inertial frame to another without performing a transform on the vector, which is just as invalid a mathematical procedure as importing a tensor from one solution to another without transforming it. It doesn't matter that in the SRT case it's a vector and in the GRT case it's a tensor; the principle is the same.

    Both Einstein and Gibbs in the references I have linked above are very clear about this. And both are very clear about the fact that the supposed "non-conservation of energy (actually the energy-momentum tensor in the case of GRT)" is not due to a defect in the theory, but a defect in procedure. You are noting that the energy-momentum tensor changes when you change solutions in GRT, and this is not unexpected as you implicitly claim (by claiming "(the) energy(-momentum tensor) is not conserved"); this illusion appears because you have performed an illegal importation of a tensor from one solution of GRT to another without transforming it.

    Admittedly the procedure for making a transform set of simultaneous equations for a pair of curved frames on different manifolds is highly complex, and involves making a separate set of transforms from every point of interest in the first curved frame to every equivalent point in the second frame; it is nowhere near as easy as for a pair of rectilinear (SRT and classical physics) frames. It is computationally intensive to say the least. However, this is how it must be done if the results are to have physical meaning. You can't just insert coordinates from one frame into another without transforming them. And it doesn't matter how many complex equations you try to use if your basic procedure is inherently flawed by this defect in processing. GRT provides the bones for this; it's up to you if you use them right.

    I will remind you of your own words: "I have no problem with Gibbs formula, it has the same problem as the pseudotensors in general - they define energy and momentum depending on a system of coordinates..." Precisely. And this definition only works for one particular inertial frame in SRT, or for one set of solutions in GRT. Trying to drag the definition of momentum from one system of coordinates to another without transforming it, whether the two systems are flat or curved, results in unphysical results. This is not a defect in GRT.

    You'll need to show that. To my mind, I have no expectation that some quantity dependent upon the curvature of the particular frame of interest will have the same value in another frame; but I do have an expectation that it will be the same in the same frame, provided it is converted to agree with the definitions of it at the various curvatures that occur across that frame.

    Yes, I think we are agreeing on this. And it is my point: what you see from different locations that have different stress-energy tensors will be different. That results from the difference in the metric tensor at each point on the manifold.
  11. Schmelzer Valued Senior Member

    The quote "The transport vector is a choice of frame" has created this impression.
    I look at frames in GR as a sloppy and unnecessary name for a system of coordinates.
    And this is where your position looks very unorthodox. Because if you have simply different systems of coordinates, you have no different solutions, but the same solutions. And if you, instead, have different solutions, you have not even a well-defined meaning of what is the same point on the two solutions.

    If you simply consider different systems of coordinates of the same solution, you have the well-defined tensor calculus to transform all physical fields. But there is no such transformation rule for the pseudo-tensors. Because the pseudo-tensor is not a tensor. And, therefore, the pseudo-tensor is nothing you can compute in a physically meaningful way from the physical fields (from point of view of the spacetime interpretation).
    A vector is only a particular example of a tensor. But a pseudo-tensor is not a tensor. Which is the problem of GR gravitational energy.
    Completely wrong. The problem with the Einstein pseudo-tensor is that it changes if you describe the same solution in different coordinates. And the problem with the Gibbs expression is that it depends on some vector field which has no physical interpretation in the spacetime interpretation. So, it also defines different energy -momentum tensors for the same solution.
    which is a completely unjustified invention from your side.
    To transfer solutions from one system of coordinates to another one is a standard mathematical exercise which one learns in differential geometry 101.
    This looks like a cultural misunderstanding. For those with a mathematical background, which have transformed some vectors, tensors and so on from one system of coordinates into another one often enough, this becomes a triviality which is not worth to be mentioned. So one would simply write $ds^2 = dx^2+ dy^2 = dr^2 +r^2 d\varphi^2$ with an = sign between the expressions in rectangular and polar coordinates because this is simply "the same" metric, even if described in different coordinates.
    No. I inform you about the facts, that's all. If you want a course in GR, you will have to pay, this is nothing I'm ready to provide for free.
    That means, you simply have not understood what was the aim of Einstein to prove. He has tried to prove that the integral over his pseudo-tensor - the energy - remains the same if we change coordinates inside. Which he has done.

    Then, forget about curvature. You can use curvilinear coordinates in flat space too, this is done all the time, and the whole mathematical apparatus is the same. Instead, "curvature" is a particular tensor field defined for a metric.
    No, we do not agree on this. This makes clear that you have completely misunderstood the point.

    The rules how to compute a pseudotensor are not the rules of transformation of a tensor. Only in some very special circumstances - for linear transformations of coordinates - the pseudo-tensor behaves like a tensor. In general, not.

    This can be easily shown by considering an elementary property of all tensor fields: If all components of a tensor field in a given point are zero, this remains valid in all systems of coordinates. This does not hold for a pseudotensor. Using a particular system of coordinates, all components of a pseudotensor may be zero at a fixed point, while in another system of coordinates the same pseudotensor has nonzero components.
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  12. Schneibster Registered Member

    The fact that you have chosen a transport vector to use indicates that you have chosen a frame; however, which transport vector you choose determines whether you are looking at energy, momentum, classical angular momentum, SAM, or OAM for that frame. A particular transport vector can be applied to different frames; and a different frame can be interpreted in terms of more than one transport vector. You haven't read the Gibbs paper.
    Yes, I know. And ignore the position of the origin, and the orientation due to the curvature of the manifold. Furthermore you ignore the manifold itself. There is more to a frame than just a system of coordinates, but you don't seem to think so. That's because you don't understand frames of reference even in classical or SRT physics.

    First, it's not just systems of coordinates, as I pointed out above. Second, you have also missed that different frames apply to different points on the manifold implied by a particular solution; remember, we are dealing with a patch, so it can have different frames in different locations in the patch. Third, a particular solution to the EFE is valid for one patch; it defines the stress-energy tensor in that patch. A family of solutions implies a field; it is all the solutions that can be found for different patches in that field, ordered by their positions in the field. None of this is in any way unorthodox. This is how GRT describes spacetime.

    I'm giving up at this point. Your misunderstanding of frames pervades the rest of your post. No one is simply considering different systems of coordinates. You will need to reformulate your thoughts after understanding what a frame of reference is.
  13. Schmelzer Valued Senior Member

    The question if some values of the coordinates are associated with some real points is not relevant for the considerations here. Of course, a map you want to use in reality needs such a connection of points of the map with objects in reality. But for the mathematical apparatus we consider here this is irrelevant.
    No, I do not miss this, I simply name it nonsense.
    No. A solution of the Einstein equations is defined by a manifold, and a metric on this manifold. It is a global object. On each patch, it defines a representation in the coordinates of this patch.

    Indeed, time to give up. Learn GR using a standard textbook, like Wald or MTW.

    Not at all. Here I'm simply defending the standard mathematical apparatus of mainstream GR.
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  14. Schneibster Registered Member

    A patch doesn't have coordinates.
  15. Schneibster Registered Member

    This is incorrect. You are defending the view that there is no conserved quantity in GRT that corresponds to energy.
  16. Schneibster Registered Member

    Let's demonstrate this. It's actually quite easy.

    Define the GRT solution for the gravity of an irregular asteroid. At the surface of the asteroid.
  17. The God Valued Senior Member

    No, in GRT there is something called energy pseudo tensor, which can give local energy density, but the problem is that pseudo tensors can be made coordinate specific unlike its a makes shift arrangement.

    But if we say that due to curved nature of both space and time in GR, the Noether's cannot be works only for flat spacetime that is SR....then we don't need to create pseudos.....if Noether's argument is applied on GR (curved spacetime) then energy conservation is a problem......
  18. Schmelzer Valued Senior Member

    Nonsense. Have I ever claimed to be able to find such GR solutions?

    Then, the solution is nothing defined on its surface, but is defined by the the spacetime metric $ds^2 = g_{\mu\nu}(x)dx^\mu dx^\nu$, not by something "at the surface".

    In fact, all you need is a large enough computer, good software for the solution of the Einstein equations, and data about the particular asteoid, and some information about the laws of the matter of the asteroid. All I have is a cheap computer, hardly strong enough for serious 4D PDE solutions. So, no, it is not quite easy, at least not for me.

    No. I simply defend the mainstream view, which, in this case, is that there is no local energy-momentum tensor of the gravitational field in GR. There are only several proposals for pseudotensors, and these proposals have the problem that they do not allow a physical interpretation as usual in classical theories as well as SR, with some well-defined local energy density.

    This is mainly an interpretational problem. The PPN parameters which define energy and momentum violations are zero for GR. If one accepts a preferred frame, violating the basics of the spacetime interpretation, you can define the pseudo-tensor, as defined in this preferred frame, as the true one and have a conservation law.
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  19. Schneibster Registered Member

    I never said it was defined by something at the surface. You've given up and are logic chopping. This is a waste of time.

    But it's possible. And if you do so, close up on the surface of the asteroid, you will find that the gravity field is just as lumpy and bumpy as the asteroid is. In other words, it's not some smooth manifold; it's a combination of a bunch of smooth manifolds; the question is, how detailed do you want to get? Are you gonna go millimeter by millimeter, or is a patch a kilometer square gonna be good enough for your purposes? That determines how many PDEs you have to solve. And along the way, we've found out what a patch is. Whadda ya know.

    How exact do you want this conservation law to be? Choose an appropriate patch size and start calculating, and as long as you do it all from a single frame it will come out consistent. But you can't switch reference frames in the middle and then claim energy isn't conserved; that's your mistake, not GRT's. And all of the evidence you've offered so far has turned out to involve exactly that: switching frames without doing the math.

    And when relativists disagree among themselves about a matter, I'd say it's pretty difficult to pick some view and call it "mainstream." You might be able to pick some collection of views and call them mainstream; in this case, those will include some who say there is some sort of conserved quantity in GRT that corresponds to the exact conservation laws of classical and SRT physics, and some who say there is not. Quite obviously, since Einstein says there is and other relativists say there is not. I merely touched the surface of the matter and found two views that say there is, one of whose is the view of the original inventor of GRT. There are plenty more.

    It's not a "preferred frame." You have to choose a frame and remain consistent in using it, or perform all the necessary transforms to convert math you got from some other frame into your chosen frame. If you don't you're making a grave basic mathematical error. I can't measure your height with my 12-inch stick and then claim your measurement with your 8-inch stick is "wrong" because I got a different number of sticks than you did. That doesn't mean my 12-inch stick (to put it on a more general footing) is a "preferred" stick. It's the stick I've got and I'm using it, and I'm perfectly comfortable multiplying your measurements by 3/2 to transform them into my frame.
  20. Schmelzer Valued Senior Member

    Sorry, the solution of GR for an asteroid is defined on the smooth manifold $\mathbb{R}^4$. The surface of the asteroid may have some edges or so, thus, not be smooth. But the solution $g_{\mu\nu}(x)$ will be smooth. At least the second derivatives will be well-defined.

    The question how accurately you want to approximate the matter configuration of the asteroid has nothing to do with some number of PDEs one has to solve. There is one 10-component PDE - the Einstein equations. There may be more, depending on the chemistry and the thermodynamics of the asteroid. But simply a rigid piece of matter of a fixed density and temperature may be a sufficiently accurate approximation. In this case, all you need is one PDE, with 10 equations for the components. Which has nothing to do with the form of the surface.

    For $\mathbb{R}^4$, one patch is sufficient.

    All your prosa about the asteroids surface looks like you have no idea at all about the meaning of the Einstein equations, but have, instead, read some horribly bad popularization. Sorry.

    Conservation laws are exact laws of physics, so the very question is nonsensical. If I have an exact solution of a theory with a conservation law, the energy will be conserved exactly. I can have some approximate solutions, then the energy will be conserved only approximately, but the conservation law remains exact.
    There is not much disagreement about this. Some people think the pseudo-tensors are completely useless, some think they are of some use for some computations, say, of the energy taken away by gravitational waves. Roughly, whenever it is fine to consider the metric as asymptotically flat, as reasonable for stars and galaxies, one can use it to compute a global energy. For cosmology it is considered useless. Don't forget that science is not based on authority, and so it does not matter what Einstein thought about his own energy-momentum pseudo-tensor. De facto this is not even teached in many GR courses.

    This is just for information, if you don't want to believe it I couldn't care less.
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  21. Schneibster Registered Member

    Here's a little thought experiment:

    Let's suppose someone claimed that the results from the LHC were "wrong" because they put it in Switzerland, not in China, or on the Moon.

    Or let's suppose someone claimed that Switzerland was a "preferred frame."

    Same idea. You pick your frame and you stick to it. The entire point of physics is that you can choose any frame and get the same results, and you can compare the results in one frame with the results in another to verify that. What you're claiming about GRT is that this is not true, and I am objecting to that; your view that results are not duplicatable from one location to another, after the appropriate transforms, under GRT, is just plain flat wrong. And it's certainly not mainstream physics of any kind.
  22. Schmelzer Valued Senior Member

    Sorry, for you, like for many others here, it is also time to make a clear requirement: If you want to criticize one of my claims, quote it instead of inventing some strawmen.

    Because the funny point here is: I fully agree that the point of physics is that you can choose any frame and get the same results. And that you can compare the results in one system of coordinates with the results in another one. And this is what is very important and holds in GR too.

    And the whole problem with the energy of the gravitational field is that there is no definition of the energy of the gravitational field which fulfills this condition. The pseudo-tensors do not have this property, because they are not tensors. To quote Wald, General Relativity p.85: "... there is no meaningful notion of the local stress-energy of the gravitational field in general relativity."
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  23. Schneibster Registered Member

    We cross-posted.

    So you're claiming that gravity doesn't change from place to place on the surface of an irregular asteroid?

    Now you're being silly.

    Sillyness again. The point is to calculate the gravity at different points on the asteroid, and the fact that we can plot mass concentrations on and in the Earth with satellites proves it's correct procedure. Not only that but those calculations are done with, wait for it, GRT. Just sayin'.

    GRT is for calculating gravity at a given point; that's what the stress-energy tensor represents. If you're leaving irregularities out then you won't get an answer that will reflect the real world except in the limit of distance. The entire point is to be sure that you can get theoretical results at any granularity you care to take the time and effort to obtain, and those results will agree with what you measure if your precision of calculation is equal to or greater than the resolution of your instruments.

    It's complete silliness to claim that there's something wrong with comparing theoretical results with real measurements. To get results that agree with the resolution of current instruments, it is necessary to use PDEs for multiple locations on the asteroid. So you've got one set of PDEs for one location on the asteroid, and a different set for another location, and so forth. If you only use one set of PDEs for the whole asteroid your results will not be of sufficient precision to check them at the resolution of our instruments, and therefore they cannot be confirmed.

    Not to calculate the precise gravity we can measure at different points on the surface of the asteroid, it's not. The patch size is chosen based on the precision that patch size gives you compared with the resolution of your instruments, if you want to check your calculations experimentally. It doesn't make any sense otherwise.

    All your BS and your failure to consider the real world looks like you're the one who doesn't understand GRT. I assume this is because actual real world results conflict with your supposed ether theory so you are biased against them. Unfortunately being biased against experimental results has a really bad history in science.

    You can calculate it as closely as you like; and the experimental results will agree with your calculations. The whole point I'm making is that GRT lets you do that, to the precision you care to calculate to. It's impossible to measure with infinite precision, and it's impossible to calculate an infinite number of PDEs, therefore an approximate solution is all you'll ever get either in measurement or in calculation.

    When relativists claim that energy is not exactly conserved in GRT, it's actually a philosophical position, based on pure mathematics; stating that you can calculate to any desired precision that is not infinite is sufficient given limited observational and calculational capabilities. We don't need to know the exact set of PDEs for every proton in the asteroid, and if we were dumb enough to spend the computer time to calculate all those trillions of PDEs, we couldn't measure it anyway.

    Because they concentrate on the math, not on the reality. And they ignore the fact that any desired precision is available, and the results will give conservation to the limits of measurement.

    Likewise. If you want to argue philosophy do us the favor of doing it on the philosophy forum.

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