Dark Matter and Dark Energy as Undiscovered Particles of Scalar and Vector Fields.

About Hontas Farmer

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I am currently a graduate student of physics at Depaul University working on a

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Hontas
Farmer

Dark matter, dark energy and scalar tensor vector theories of gravity are widely thought of as at odds. I have shown that if the scalar and vector fields common to alternative gravities are considered as any other quantum fields on a curved space time the consequence is the creation of particles and matter which need not couple into the electromagnetic force. In short if there are "extra" fundamental forces of nature then there will be "extra" dark matter particles related to those forces.

(Some people have commented in a way which indicates that they are here looking for a theory of gravity. This particular blog post is only tangentially related to gravity. This post exposes for the first time a little theory of Vector and scalar fields in curved space-time which could explain dark matter. NOT a theory of gravity per se. In fact for this work I am simply taking GR and a certain static curved background as a given.)

Let me explain. In standard and accepted quantum field theory in curved space time it is known that the local value of the metric will effect the probability of interactions occurring. However this geometric effect is usually so small compared ot the influence of the forces involved that it can be totally neglected. The value of the coupling constant, or charge, on the particles involved plays a much greater role. Never the less it is accepted theory that different observers following different path's through a curved space time will observe different numbers of particles. This is known as the Unruh effect.

What I realized was that the scalar and vector fields in theories of alternative gravity like Jacob Bekenstein's (Tensor Vector Scalar) TeVeS and John Moffat's (Scalar Tensor Vector Gravity) STVG are no different from any other fields. I did the calculations and sure enough they too are subject to differential interaction cross sections due to the geometric effect. This effect, which I detail in the linked PDF, is radially symmetric. The effect diverges at the origin, and has a sharp peak at a distance of one Schwarzchild radius from the center of a galaxy, then drops off as 1/r^4.

The difference between the dark forces and the forces we know and love could be the coupling constants. Which I cannot determine from theory alone. If their strength is comparable to or weaker than the geometric effect that would explain the lack of dark matter in the interior of the galaxy and why we have such trouble detecting any particles of dark matter. In our region of the galaxy due to the geometric effect the probability of interactions between dark matter particles is much greater than it is at cosmological distances from the galaxy. Higher probability of interaction means greater chances of an interaction destroying particles, as well as creating them. In this theory the Dark matter particles at cosmological distances from the galaxy are going to be more stable, and their distribution would be spherically symmetric about any galaxy or other collection of "ordinary matter". This could explain observations that suggest a spherical halo of dark matter surrounding spiral galaxy's.

This result shows that it is not necessary for alternative gravities to be at odds with observations of dark matter.

For the full theory please see.

Dark Matter and Energy as Particles and Fields of Unobserved Scalar and Vector
Fields (PDF) (rev2)..* **As submitted to a few Journals which I will not name.

(*It bears mentioning that just today I did a google search on vector scalar dark matter etc and found that work has been done in that direction. One piece I could find on Vector dark matter "Hidden vector dark matter." by Thomas Hambye. As well as much work on scalar dark matter in which various theories predict scalar particles and fields being the source of dark matter. My theory is unique in that I connect these hypothetical fields to the hypothetical scalar and vector fields of the proposed alternative theories of gravity TeVeS and STVG. I also formulated the theory in a realistically curved background space-time. That is how my theory is different from those previous works.)

**The reviewer at two of the journals say that a paper like this needs more experimental data to back it up. I don't know that there is much I can do about that. One said, interestingly that I should have used the Friedman Robertson Walker metric. This would not allow me to say anything about the dark matter in a specific galaxy, but it would tell me something about the universe as a whole. Perhaps it would tell me if this theory would lead to a distribution of dark matter which would prevent inflation. Or not. This is something I will look at in the coming weeks.

Comments

Your graph doesn't show much effect from your force, outside the schwazchild radius, so it wouldn't make much difference to the gravitation of a galaxy, only a minor modification to the motion of particles near the central black hole, I can't see this effect explain dark energy or dark matter. That said TeVes hasn't been too bad at explaining dark matter.

DrBDOAdams (not verified) | 05/13/09 | 16:06 PM

The geometric effect is not a "force" per se. It really is a effect due to the curvature of space-time itself, it is pure non-euclidian geometry. If you look at the paper what is being graphed is the value of ((√g) R) (g is the scalar of the metric tensor and R is the Ricci curvature scalar this is what I have dubbed the geometric effect.) which would be the leading term in each of the Feynman diagrams for every interaction term in the theory. Basically it would be a weight on the probability amplitudes, collision cross sections, decay rates etc. In general the larger the cross section the more likely it is that a particle will interact by way of a given force. Where an interaction can be a collision particle creation or decay.

So when you see that the geometric effect is small outside the schwarzchild radius of the galaxy what you see is that the interactions of particle creation annihilation are less likely outside the schwarzchild radius of the galaxy. What is also less likley is particle decay . Thus particles of dark matter once created are less likely to simply decay...when they are at the fringes of the galxy. When they are closer to the center of the galxy they are much more likely to spontaneously decay...if my model is correct.

Also when asessing this one has to be mindful of how large the schwarzchild radius of something as massive as a galxy is. It's no small distance. For the mikly way......

R_Schwarzchild=2GM/c^2= 2(6.67*10^-11)(5.8*10^11(1.988*10^30kg))/((2.99*10^8)^2)

An order of magnitude estimate is10^17km or about 10^5 light years or approximately 20% of the radius of the galaxy is where the Schwarzchild radius is. This means the effect is still pretty strong even in the middle of the galxy.

Lastly don't be decieved by the way I graphed the effect. I choose to graph it on that scale because it would show most of the interesting features. Suppose I choose to graph the same exact effect in this manner with the graph starting at one schwarzchild radius and extending beyond the edge of the galaxy...

Hontas Farmer | 05/13/09 | 22:07 PM

Your paper contains an elementary algebraic mistake in the calculation of \sqrt{g}, which is a shorthand \sqrt{|\det(g)|}, the square root of the absolute value of the determinant of the metric tensor taken as a 4x4 matrix. If you don't know how to compute determinants, look it up in a Linear Algebra textbook. Hint: it is obvious that your formula is wrong, because it has a \sin(\theta), and the Schwarzschild metric is spherically symmetric (the \sin(\theta) in the arclength element comes from choosing spherical coordinates, where it also appears for the Minkowski metric, after all!).

Anonymous (not verified) | 05/14/09 | 03:01 AM

I am afraid that you are wrong.
You see the g in the equation is not the determinant of a matrix, it is the trace of a tensor. Please see Einstein's summation convention and Tensor Contraction.

Have a nice day. :-)

Correction you are right about the determinant of g. Looking at the alphabet soup of terms confused me. :-) But the rest of what you wrote is demonstrably wrong.

The elements of the Schwarzchild metric tensor are...
g11=-(1-2GM/r), g22=(1-2GM/r)^-1,g33=r²,g44=r²sin[theta]
The metric tensor is diagonal and like any diagonal matrix the determinant will be the product of the diagonal entries.

-(1-2GM/r)(1-2GM/r)^-1r²(r²sin[theta])=-r⁴sin[theta]. Where in the calculation in question I have already absorbed the negative sign which is shown under the square root in the lagrangian I wrote.

Understand?

Hontas Farmer | 05/17/09 | 11:05 AM

Please see Einstein-Hilbert action for the convention that is relevant in this case.

Anonymous (not verified) | 05/15/09 | 02:40 AM

Not unless you are using a very nonstandard formulation of gravity which does not agree with General Relativity (in which case you can't assume that the Schwartzschild metric is a solution of that theory of gravity). The prefactor that multiplies the Lagrangian density for matter fields on a curved Riemannian manifold is indeed the square root of the matrix determinant of the metric tensor (in the Minkovksian signature case one has to take the absolute value first, because the determinant is negative), because that is simply the volume element on the manifold. See also here and here. Moreover, the metric tensor has two covariant (lower) indices, and its cotensor has two contravariant (upper) indices, so you can't contract it without first raising or lowering one of its indices, but then you necessarily get a Kronecker delta (whose trace is simply the dimensionality of spacetime), because that's how the metric cotensor is defined (as the matrix inverse of the metric tensor). Thus, I'm afraid that the Anonymous is right, and your paper does indeed contain a mistake.

Georg von Hippel | 05/15/09 | 03:07 AM

Ahh Georg! Long time no type... Finally a critique from someone who knows what they are talking about. Such is why I post these things to this blog. :-)

Firstly let me admit where you are right... I will refer you to "An introduction to General Relativity Spacetime and Geometry" by Sean M. Carroll Page 394. Just under equation 9.87 it says and I quote... "Aside from the predictable appearance of the metric g_{/mu/nu} and it's determinant, we have also included a direct coupling to the curvatures scalar R, parameterized by a small constant.." Then it goes on to talk of conformal coupling etc. So there you have it g in this instance is a matrix determinant. You are right I was wrong. What confused me was the two different conventions being used side by side. R is most definately not a determinant (The Ricci tensor it is derived from if a rank four tensor not a matrix at all.) As for that annoymous person they said what they said what they said in a really nasty way. Not trying to explain or to help but merely seeking to offend and degrade. You did not do that Georg so to you I listend.

Now here how you and the anonymous one are wrong... You seem to think that the lagrangian I wrote is meant to be the lagrangian of a theory of gravity....it is not. The "force" being modled here is NOT gravity. It is scalar and vector fields which have eluded our detection, because the matter we are made of does not couple to those fields. I am modeling those fields in fixed curved non-dynamical space time. So there is no need for the Einstein Hilbert action. To a good approximation the curvature due to a quantum particle or field can be completely ignored, when compared to the curvature due to the mass of a galaxy.

Making the correction to the theory and using the determinant instead of the trace of the metric tensor the radial dependence of the geometric effect varies like so

Which has fundamentally the same spherical symmetry and radial dependence as I found when I made my small error.

Thankyou Georg for pointing it out nicely.

Hontas Farmer | 05/15/09 | 11:14 AM

Don't thank me too early. The paper still contains a number of mistakes, the most crucial one being that you seem to think that the volume factor r^2 sin(theta) has some physical meaning that it does not have. Write down any theory you like in flat Minkovski space in polar coordinates; the volume element will still be r^2 sin(theta), and it will multiply every term in the action. That does not mean that the couplings increase quadratically with the distance from the origin; it just means that you have chose polar coordinates, where this extra factor is needed to compensate for the r- and theta-dependence of the curvilinear coordinates themselves.

Georg von Hippel | 05/17/09 | 12:40 PM

Georg.... the theory is not written in flat minkowski space time. It is written in a static curved Schwarzschild space-time. There are fundamental differences a number of things which are assumed in flat space time just aren't true in curved spacetime. In curved space time the invariants are all different. Volume is an invariant you are quite right. However in curved space time it is not the same as it is in flat spacetime.

Hontas Farmer | 05/17/09 | 20:46 PM

I was trying to get you to understand where you go wrong by using the very simplest possible vacuum solution to Einstein's field equations -- which happens to be the Minkovksi spacetime. The role of the factor of r^2 sin(theta) is the same, no matter what further factors may in some case arise from the curvature of spacetime. In the Schwarzschild case, those factors happen to cancel to zero, and the volume form there is the same as for flat space.

Georg von Hippel | 05/17/09 | 21:02 PM

I now see where the problem is. Georg. You are thinking of the ricci curvature. Note that in my paper I never said Ricci curvature scalar (there is more than one).

There is g^{/mu/nu}R_{/mu/nu} which you were thinking of. However there are also many other ways to contract that tensor, the one I used being...

R^{/mu/nu/rho/sigma}R_{/mu/nu/rho/sigma}

Which is NOT zero. If you must have a specific name for this one it's called the "Kretcheman invariant". Think about what you are saying physically man. Curvature of space time is what keeps us from falling off the earth and earth going around the sun in this (almost) Schwarszchild metric we live in and call gravity. If it were zero we'd be in trouble! :-)

R^{/mu/nu/rho/sigma}R_{/mu/nu/rho/sigma}= (48 G^2 M^2) r^-6

Hontas Farmer | 06/12/09 | 01:32 AM

No. I've been talking about the volume element sqrt(det(g)), to which you have been attributing a significance which it does not have. If you read what I wrote above, this should be very clear.
Which curvature invariant you couple your new fields to doesn't matter in this context. The square of the Riemann tensor is indeed nonzero for the Schwarzschild metric, and this is how one can see that the singularity at r=0 is real, as opposed to the coordinate singularity at r=2M, which can be made to go away by a change of coordinate. Nobody calls it "R", however.

Georg von Hippel | 06/12/09 | 05:01 AM

I see that nobody calls it R now. However after I have made that clear in a paper say writing that I am using that invariant it should be clear enough. I thought that was clear enough from the math I was using, because it would be wrong if I had used the Ricci scalar. I guess you would have had to assume that I'm not a total fool.

As for the volume element I have shown you direct calculations of the determinant of the Schwarzschild metric let me show you again

|-(1-Rs)                                                |
|           1/(1-RS)                                |
|                         r^2                          |
|    r^2 Sin^2[th] |
(I so can't wait for Latex)

The determinant is just the product of these diagonal elemetnts. It is clearly not equal to one. In your analysis in which you conclude it is one you have noted that the last two elements are the same as they are in flat space, then you have set them equal to what they would be in rectangular coordinates. I will agree that your thought exercise gives a correct result if you can show that by using coorindate transformations. Meaning applying a correct transformation matrix to the Schwarzchild metric to put it in rectangular coordinates, then taking the determinant of that. You will note that when you do that the first two elements will change giving you the same information in an uglier package. I ask for this because the logic you have used to justify that objection to me just is not convincing. Show me a proof

Hontas Farmer | 06/12/09 | 09:09 AM

Of course the volume element is not one in polar coordinates -- that's the whole point! When changing coordinates, the metric (being a rank-2 tensor) picks up two powers of the Jacobian matrix (one for each index). Thus the square root of its determinant picks up one power of the Jacobi determinant, just as it has to (being the invariant volume element). The r^2 sin\theta is precisely that Jacobi determinant. This is really fairly basic, and the proof should be right in whatever textbook you like best. If you want to see that the determinant changes with a coordinate transformation, just look at http://en.wikipedia.org/wiki/Eddington-Finkelstein_coordinates or http://en.wikipedia.org/wiki/Kruskal-Szekeres_coordinates and calculate the volume element in those.

Georg (not verified) | 06/12/09 | 09:28 AM

Georg, once you have chosen a basis to write your tensor's matrix form in , if that matrix form is diagonal then the determinant is simply the product of the terms on the diagonal. That's it that's all.

If you don't want to believe me please see this written by a faculty member at MIT. "For a diagonal matrix it is the product of the diagonal elements." Which for the Schwarzchild metric is exactly what I said it is.

Another line of evidence for you is that a editor for a journal has not objected on that basis...instead he chooses to say I have not done enough comparison to experimental data? Where is there experimental data on dark matter other than the observations which strongly suggest it exist and tell us nothing more!

Hontas Farmer | 06/13/09 | 09:09 AM

In that basis, of course! But the components of the metric tensor (and hence the determinant of the corresponding matrix) depend on the choice of basis. Physical quantities cannot depend on such an arbitrary choice.

Once again: if you change your coordinate basis, the metric (being a rank-2 covariant tensor) picks up two powers of the corresponding Jacobi matrix. By the determinant multiplication theorem, sqrt(det(g)) therefore picks up one power of the Jacobi determinant. That is exactly what is needed to make sqrt(det(g)) dx^1...dx^n the invariant volume form, because the aforementioned Jacobi determinant cancels the Jacobi determinant picked up by the differential form dx^1...dx^n.

Georg (not verified) | 06/13/09 | 10:42 AM

Now I think I see where your mistake comes from -- you think of the transformation law A' = SAS^{-1} for a matrix under a change of basis in a vector space, which of course leaves the determinant invariant. But the metric isn't really a matrix, it is a rank-2 covariant tensor, whose components transform as
g_{\mu'\nu'} = g_{\mu\nu} (\partial x^{\mu}/\partial x^{\mu'})(\partial x^{\nu}/\partial x^{\nu'}),
and whose matrix representation therefore does not have the same determinant under a coordinate change.

Georg (not verified) | 06/14/09 | 01:08 AM

But if the metric tensor is not a matrix then what does it mean to take a determiant of it.

The solution to this problem is to choose a basis to represent the tensor in. In the basis I choose, the standard one for the Schwarzchild metric simple 2nd or 3rd grade multiplication shows that the determinant is what i wrote that it is. Who cares's what it is in some other coordinate system. If you change coordinates on the metric you need to change all of the coordinates in the paper. Which will preserve the physics seen in the coordintae sytem I used, which again is the standard one for dealing with gravitational physics.

Hontas Farmer | 06/14/09 | 10:51 AM

I've run out of patience with this thread, so don't expect another reply. This stuff has made it into Wolfram's mathworld and into Wikipedia, so you can't claim it is not well-known. Think whatever you want to think, but if you want to learn something about physics, your are going about this the wrong way for sure: Self-doubt, not self-assuredness, is a prerequisite to doing sound science.

Georg (not verified) | 06/14/09 | 11:40 AM

I agree. I am pretty patient if people just want to get smarter but there's a difference between asking questions about things you don't understand (what I would have done if I had made an error and Georg corrected me) and trying to get out of a bad position without looking stupid by jumping down theoretical rabbit holes.

Hank Campbell | 06/14/09 | 12:36 PM

Hank, Georg pointed out one genuine error, and two things he thought were errors but weren't. First I used the trace instead of the determinant. I thanked him and made a change

He insisted that the curvature scalar I used was equal to zero... in spite of my referring him repeatedly to a text book which showed the curvature scalar I used is not zero. He kept insisting that a different curvature scalar than the on I used was zero... He thought I was using the Ricci scalar when it is clear from my math that I wasn't. There is in fact more than one curvature scalar. Instead of thinking "maybe Hontas is talking about something else". Or "Maybe I'm missing something." he only thought "Hontas is an idiot therefore she must be wrong." That's the way I took it.

As for the determinant of a tensor, once one chooses a representation of that tensor in a certain coordinate system it is perfectly valid to treat it as a matrix. The determinant of the Schwarzschild metric plainly is not equal to one.

Georg was right about one thing. But he has been wrong about everything else he has said since then.

Hontas Farmer | 06/14/09 | 19:46 PM

Hontas,

Off topic, but I see from your bio. that you are a Firefox fan. I like the search (Ctrl-F) facility, compared with which the IE version is very tedious in operation. But in using the Fox for Scientific Blogging especially, I have encountered the problem that shortly after opening the history (Ctrl+H) window, so valuable in retrieving that item that I looked at hastily two days ago, the program crashes.

This happens both on my work and home computers. Have you any suggestions as to how to hunt down this bug? It seems to be made of dark matter, since I can't find any reference to it on the Internet.

Robert H Olley | 05/14/09 | 12:19 PM

Well I use Macintosh or Linux for most of my purposes. I also don't use the hotkey's that you do. I would try upgrading to the latest version of firefox. Have you tried that?

Hontas Farmer | 05/14/09 | 16:46 PM

There is a little update in the text of the article. JIC anyone is still watching it.

Hontas Farmer | 07/05/09 | 21:27 PM

I hate to say it, but I agree with the critique about having sin(theta) in a scalar. For a spherically symmetric physical situation, as Schwarzschild describes, there can be no dependence on theta (or phi) in a scalar.

You might try starting with a Cartesian version of the Schwarzschild metric. See the useful review paper Catalogue of Spacetimes. What will the determinant be then? Certainly it won't have sin(theta) in it.

I think the real problem is that GR is written to be independent of coordinate choices. I think that this is not a feature but instead it is a bug. The problem is that to fix it, even in the case of a "trivial" solution to Einstein's field equations like the Schwarzschild metric, one must choose some particular coordinate choice as preferential. My choice is Gulltrand-Painleve as they appear when you rewrite GR in the gamma matrices used for spin-1/2 relativistic particles.

Carl Brannen (not verified) | 07/11/09 | 11:08 AM

I hate to say it, but I agree with the critique about having sin(theta)
in a scalar. For a spherically symmetric physical situation, as
Schwarzschild describes, there can be no dependence on theta (or phi)
in a scalar.

You are equating spherical and azimuthal symmetry. I do not blame you for this, but it is a common error because the difference is not commonly taught at any level. Remember we are talking about a hypothetical galaxy. Let us think about this conceptually. Consider for a moment the metric at a distance from a flat disk of matter. Now consider the metric at a distance from a sphere of matter of the same size. We can treat both of these as if they were the same and use the Schwarzchild metric to a good approximation. However there is a difference. In the case of the disk there is theta and phi dependence due to the way the matter is distributed. The field at the edge of the disk will be different than directly above then disk.

Mathematically, the computation of is simple

Which should lay that to rest.

As for a Cartesian version of Schwarzschilds metric.....

That would or should have the same physical features as it does in spherical porlar coordinates. That is from the reference you gave. The physics, once determined, do not depend on coordinate choice. However the choice of cartesian coodinates for this problem is really ugly and reveals no different physics than I have described.

Just to lay this to rest let me show you the comments from CQG's referee. They weren't anything that anyone here has noticed. Certainly not what you and a couple others have mentioned.

Hontas Farmer | 07/11/09 | 12:30 PM

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