Congratulations on becoming an aetherist.
Congratulations on becoming an aetherist.
The term "perfect fluid" has a strict definition. The cosmological constant fits that definition. You may not like that definition, but that's just the way it is., then by definition you do not have a fluid.
For example http://ned.ipac.caltech.edu/level5/C...arroll1_3.html
Or any other basic textbook on the matter.The vacuum can therefore be thought of as a perfect fluid
From the same link as above
The cosmological constant turns out to be a measure of the energy density of the vacuum
You also seem to be under the impression that the cosmological constant is well understood to be the cause of the accelerating expansion. There is a Nobel Prize committee that should be notified to retract the awards for Perlmutter, Reiss and Schmidt since one would think that you have evidence that no one else in the history of science ever suspected. I do believe I remember reading an article that one of the scientists offered their Nobel to whomever could explain the accelerating expansion to them. Are you planning on taking him up on that offer with your reliance upon vacuum energy of the cosmological constant?
Well thanks, I actually hadn't seen this quote by him. Although it doesn't surprise me, he would at times seem to be a closet aetherist:The vacuum can therefore be thought of as a perfect fluid
Since Lorentz-violating aether fields have so many fun uses, it's important to verify that the theories are well-behaved. With Tim Dulaney, Moira Gresham, and Heywood Tam, we investigated perturbations in the aether. We found that the results of a naive stability analysis were sensitively dependent on what Lorentz frame you do are looking in -- in a boosted frame, a purportedly stable model begins to look unstable. One exception was what we called "sigma-model aether," so we looked at the empirical constraints on that model. Our stability results have subsequently been challenged by Donnelly and Jacobson, who argue that everything can be fixed if you choose boundary conditions carefully.
S.M. Carroll, T.R. Dulaney, M.I. Gresham, and H. Tam, 2008, "Instabilities in the Aether'', arXiv:0812.1049. [abstract; pdf; SPIRES]
S.M. Carroll, T.R. Dulaney, M.I. Gresham, and H. Tam, 2008, "Sigma-Model Aether'', arXiv:0812.1050. [abstract; pdf; SPIRES]
Besides, you're just taking random potshots now, and purely ad hominem at that.
What is under consideration here is your theory, not the "closet states" of well-known physicists. Since your theory is exactly equivalent to the EFE without cosmological constant it will produce the exact same dynamics. And it will not reproduce accelerating expansion, because if we could get that from the EFE without a cosmological constant term, there wouldn't have been any need to add the term in the first place.
So if there is no further relevant argument on that, would it not be better to simply retract the claims made and let the thread be closed?
Then quote me making the claimNo retraction on the second either because the statement stands as written.
Retract what claims? If you mean whether I have misinterpreted your position about the cosmological constant, in the end my opinion on that will not matter. As to whether my equation will accurately model the accelerating expansion is an unknown. Whether it is mathematically sound however is easily provided in a derivation. Although you do not seem to be interested in this theory, your debate has certainly left me feeling much more confident in how it will perform. Considering you are under no obligation to have ever replied to this thread, I sincerely thank you for your time Caveman.
For exampleRetract what claims?
It will not model any accelerating expansion, that much is known.As to whether my equation will accurately model the accelerating expansion is an unknown.
But seriously, why don't you just solve it for a cosmological vacuum solution (ie do the same as normally done for a (anti) de Sitter solution)? It's about the simplest thing possible to do with the EFE.
No argument there.Whether it is mathematically sound however is easily provided in a derivation.
If you mean that i seem sure that your equation will not produce an accelerating expansion then yes, i am absolutely positive about that. If the EFE, without cosmological constant, for a matter distribution does not produce an accelerating expansion then any equation that is perfectly equivalent will neither. That statement seems so obvious to me as to almost not seem worth stating.
Your main counter argument to my whole theory is
also stated as "because I said so." Or perhaps "if it could have been done that way it would have".It will not model any accelerating expansion, that much is known.
You see, I don't quite have the hubris to demand that someone who has never had the opportunity to be informed of the accelerating expansion stick with their original interpretation. Simply because many people worked on General Relativity, does not lessen their efforts.
Yes, you are looking for Neptune where I instead see epicycles. Good luck on your search.If you mean that i seem sure that your equation will not produce an accelerating expansion then yes, i am absolutely positive about that. If the EFE, without cosmological constant, for a matter distribution does not produce an accelerating expansion then any equation that is perfectly equivalent will neither. That statement seems so obvious to me as to almost not seem worth stating.
Last edited by JMessenger; 2012-Jul-18 at 05:23 AM.
I have had a request to explain the equations in a simpler form, and I think the best way to do so is to tackle the last equation I posted.
Solving for the constant we have a homogeneous and isotropic fluid with no "curvature" (the coordinate system is flat/Euclidean). On the right hand side just by simple analysis these two tensor functions are of opposing derivative nature. The increasing positive curvature as increases is offset by the increasing negative curvature as decreases. For velocities much smaller than the speed of light, the Newtonian field gradients for a spherical mass are equal, even if the magnitudes are vastly different (see gradients plot). Substituting in one gradient for the other, from this analysis, it is reasonable to expect the same low velocity Newtonian gravitational effects for small radii. In the above plot of the upside down sombrero hat, if it were somehow possible to have matter/energy without the corresponding reductions of pressure denoted by then the plot would simply show a dome with a linearly increasing downward pointing gradient. There would be no appearance of "attractive" gravity. Without but including the plot would look no different than the stretched blanket analogy of a dip in the middle of flat space-time. Certainly one can place alone into the equation and obtain exceptional empirical predictions for small radii, but even in spite of the numerous objections to action-at-a-distance etc and the now mainstream accepted view that the quantum vacuum has physical properties, it is when we examine our assumptions of General Relativity at the largest and smallest scales (and not in the easily perceived range of our own senses) do the qualitative assumptions mount up against it.
Another post in how I envisage this theory to help explain before the thread is closed out. As this is a qualitative description I am not ready provide equations.
In order to unify all the forces of nature under one umbrella and to develop an intuitive meaning it may be helpful to have another description of how gravity works through
If all the fundamental particles of nature can be described by symmetric four dimensional wave functions and all "forces" are due to distortions of these wave functions, the pressures within the tensor equation of the perfect fluid act to cause these distortions. At all times this pressure is repulsive between each wave function. For gravity, these waves exist as reductions of density in the fluid and, if alone, cause a symmetric reduction in this pressure within a certain radius about them. This would be the curvature of space-time description. However, if two or more particles are within a certain range of each other, there is a superposition of these reductions of pressure, greatest in the line between them. It is not that there is an attraction between the particles but a reduced repulsion. The reduced pressure between these waves causes a distortion of each wave function which is interpreted as a force tending to move them together. In other words, they aren't "pushed" together but instead each wave function in the fluid seeks symmetry. The caveat with this though, is that it must also mean for the accelerating expansion that these waves are still seeking maximum symmetry. I am not sure what that would portend besides the mainstream conclusion that the Milk Way will end alone.
Caveman, despite regrettably dragging you back into a thread that you no longer want to participate in, I did want to point out a quote of yours. You might not have thought it completely through when posting it and I wanted to give you the opportunity to leave a retraction. In essence, even if we disagree about the accelerating expansion, you are still stating that this aether theory would produce the exact same dynamics as General Relativity. Was that your actual intention?
It's like having and , these are perfectly equivalent. You may say "but g is a square-root type function", it does not matter, the same inputs produce the same outputs, that's all that matters.
Your equation is likewise equivalent to the EFE without cosmological constant, and thus will produce the exact same dynamics. You may say "but mine is an aether-type equation", it also does not matter. If the EFE without cosmological constant cannot produce an accelerating expansion, neither can yours, simply because it is equivalent.
Another way to see this is to think of physical theories as "black boxes" with inputs and outputs. If one black box is equivalent to another then it does not matter which we take, we could in fact not distinguish them observationally.
I'm not saying that aether theories are equivalent to general relativity, i'm just saying that your equation is mathematically equivalent to the EFE without cosmological constant (which should be obvious as it is by design). I have thought this through, so no retraction. The reason i've started asking you to actually solve your equation for cosmological models is because it seems to me that you're not fully grasping the mathematical equivalence thing, and i was hoping that by having you solve your equation and noticing that you'll always get the same results as EFE without cc you'd start to realize the point of the argument.
I am convinced that there exists a mathematically equivalent representation of GR having the density convention you are describing, and that this theory is what you intend to represent. caveman seems more qualified than I in pointing out nuances this change in convention may encounter.
I have long held the ATM idea that all of electromagnetism is flawed, because charge is defined with the wrong sign. I am pretty sure that almost everyone has considered this, since the electron's charge became standard on the quantum level anyways.
Science has a long tradition of maintaining priority of conventions until there is a compelling reason to change, occasionally even if this means adding cumbersome terms. What makes your convention more compelling than the standard density convention with a cosmological constant?
Suppose I had a region of space-time filled with a uniform fluid of constant mass and charge density and a constant electric field. This will produce an electric force on the fluid. Will the fluid accelerate according to or ?
If it is the former, then the wouldn't the equivalence principle suggest that the mainstream convention is superior to your density convention?
Pogono, thanks for the vote of confidence.
I am really not trying to be rude, but you are equivocating. The entire tensor structure of a perfect fluid is based precisely on a physical theory.It doesn't matter whether that equation represents a physical theory or whatever.
One of the main arguments against aether theory are two incompatible physical theories (stiff luminiferous aether and solid matter). The physical theory Lorentz presented had equations that were not covariant, thus it does very much matter.
I think we are getting at the heart of the difference between a mathematician and a physical theorist. This line of thinking would have kept Fourier (a mathematician and physical theorist on heat transfer) from realizing that all periodic functions are sums of an infinite series of simple oscillating functions. It isn't just the output that matters in physical theories.It's like having and , these are perfectly equivalent. You may say "but g is a square-root type function", it does not matter, the same inputs produce the same outputs, that's all that matters.
I actually agree entirely if that is how General Relativity were viewed. But for all the arguments through the years about into Feynman diagrams, it seems very odd to me that General Relativity still allows us the brick wall illusion that we are somehow fundamentally different than a wave of light. That we are "over here" and when we move "over there", that there are actual physical bits and pieces that travel. That there must be a "curvature of space-time" because it is "out there" in the middle of nothing. That is no black box theory. A black box makes no actual assumptions about the processes that occur within to produce an outcome. This is not how the mainstream views GR.Another way to see this is to think of physical theories as "black boxes" with inputs and outputs. If one black box is equivalent to another then it does not matter which we take, we could in fact not distinguish them observationally.
Well, if during my research into Friedmann equations, I am not able to vary spatially to develop a smooth gravitational field to account for Dark Matter nor able to vary during cosmological evolution to account for the extra space being developed then I may have to concede that you have a point.I'm not saying that aether theories are equivalent to general relativity, i'm just saying that your equation is mathematically equivalent to the EFE without cosmological constant (which should be obvious as it is by design). I have thought this through, so no retraction. The reason i've started asking you to actually solve your equation for cosmological models is because it seems to me that you're not fully grasping the mathematical equivalence thing, and i was hoping that by having you solve your equation and noticing that you'll always get the same results as EFE without cc you'd start to realize the point of the argument.