Quantum gravity by relativization of Quantum Field Theory.

  1. 1.  Richard J.Daley College, City Colleges of Chicago
  2. 2.  College of DuPage

Abstract

The question is how can we make quantum field theory part of General Relativity instead of how we can quantize gravity. Here it will be shown that a Hilbert space can be defined such that the bra-ket is a four vector in \mathcal{M}inkowski space-time, <ψ|ϕ>=va\langle\psi|\phi\rangle=v^{a}\in\mathcal{M}. Similar to a Hilbert space over the field of quarternions. Minkowski space is the tangent space at an arbitrary point on a Riemannian manifold. It will then be shown that the Riemann curvature connects these spaces by operating on the probability density 4-current jaj^{a} of the local QFT. Choosing the Klein-Gordon field as a simple example QFT, the quantized Einstein-Hilbert action will then be derived. From there the expected Feynman diagrams for General Relativity can be read off. In this way one may calculate the gravitational effect due to a quantum field theoretical event.

1 Introduction

The motivating question for this study was can Hilbert space can be defined such that the <ψ|ϕ>=va\langle\psi|\phi\rangle=v^{a}\in\mathcal{M}. What kind of physics can be modeled with such a structure. My thought was that QFT could be incorporated into General Relativity, and by that method a new hypothesis of short length scale gravity or possibly quantum gravity could be proposed. In this model QFT would be exactly correct in the tangent spaces of a Riemannian manifold, while gravity would physically connect these spaces.

Quantum field theory has been the most successful program of theoretical physics since General Relativity. The one failure of Quantum Field theory has been an inability to incorporate General Relativity or handle dynamically curving space time. The idea expressed in this short paper is that one could try incorporating QFT into GR instead of the other way around. Instead of quantizing General Relativity we relativize QFT by demanding strict adherence to the Einstein Equivalence Principle (2).

That for an infinitely small four-dimensional region, the relativity theory is valid in the special sense when the axes are suitably chosen.

Each point on a curved space time has a flat Minkowski space time \mathcal{M}, called a tangent space, in which special relativity is exactly valid. With an appropriate inner product in the Hilbert space, four vectors may act as scalars in the Hilbert space. Thus quantum field theory would automatically obey the Einstein Equivalence Principle.

What is proposed in this paper is not completely without precedent. Standard quantum mechanics is formulated in terms of operators on a Hilbert space over the field of complex scalars. It has been shown that the set of quaternions can also be used as the set of scalars over which the Hilbert space is defined(3). The idea that the key to quantum gravity lies in the formulation of the theory in terms of new variables can be found in the literature (1). There are also many works on the subject of algebraic local quantum field theory(4).

What is new in this paper is the concept of \mathcal{H} over the space \mathcal{M} (which has the Clifford algebra Cl1,3(R)Cl_{1,3}(R)). Treating the tangent space attached to each point on a Riemannian manifold as an algebra over which Hilbert space may be defined enables a new approach to quantum gravity. Instead of assuming QFT is fundamental and incorporating gravity into it, we could incorporate QFT into Relativity.

2 Hilbert space over the space of 4-vectors

Latin indicies will indicate vectors in the tangent space at each point on a Riemannian manifold. Greek indicies will indicate vectors in the curved global space-time of the manifold and ket’s will be vectors in Hilbert space. In the following the vierbein formulation of General Relativity will be used.

Let

\mathcal{H} be a vector space with elements |ψ>|\psi\rangle and |ϕ>|\phi\rangle over the space of 4-vectors with vav^{a}\in\mathcal{M} and the inner product defined as follows.

<ψ|ϕ>=12(ψ¯γaϕ+ϕ¯γaψ)\langle\psi|\phi\rangle=\frac{1}{2}\left(\bar{\psi}\gamma^{a}\phi+\bar{\phi}% \gamma^{a}\psi\right) (1)

\mathcal{H} is a Hilbert space if the inner product satisfies certain conditions. I will now demonstrate that this is a Hilbert space starting with equation 2.

<ψ|ϕ>¯=<ϕ|ψ>\overline{\langle\psi|\phi\rangle}=\langle\phi|\psi\rangle (2)

Taking the hermitian adjoint of equation 1 results in equation 3

<ψ|ϕ>¯=12(ψ¯γaϕ+ϕ¯γaψ)¯=12(ϕ¯γaψ+ψ¯γaϕ)=<ψ|ϕ>\overline{\langle\psi|\phi\rangle}=\overline{\frac{1}{2}\left(\bar{\psi}\gamma% ^{a}\phi+\bar{\phi}\gamma^{a}\psi\right)}=\frac{1}{2}\left(\bar{\phi}\gamma^{a% }\psi+\bar{\psi}\gamma^{a}\phi\right)=\langle\psi|\phi\rangle (3)

Therefore the condition set out in equation 2 will be satisfied. The adjoint of the proposed inner product equals the inner product. Now I will consider the norm of a vector in \mathcal{H}, which has to be finite and non zero in the space \mathcal{M}. Applying equation 1 results in equation 4 which shows that the normalizability criteria is satisfied by equation 1 .

<ψ|ψ>=12(ψ¯γaψ+ψ¯γaψ)=ψ¯γaψ0a\langle\psi|\psi\rangle=\frac{1}{2}\left(\bar{\psi}\gamma^{a}\psi+\bar{\psi}% \gamma^{a}\psi\right)=\bar{\psi}\gamma^{a}\psi\neq 0^{a}\in\mathcal{M} (4)

The last requirement is that the inner product would needs to be linear in at least one of it’s arguments. To show this let |ψ1>,|ψ2>,|ϕ>|\psi_{1}\rangle,|\psi_{2}\rangle,|\phi\rangle\in\mathcal{H}, and let va,yav^{a},y^{a}\in\mathcal{M}.

<vaψ1+yaψ2|ϕ>\displaystyle\langle v^{a}\psi_{1}+y^{a}\psi_{2}|\phi\rangle =12((vaψ1+yaψ2)¯γaϕ+ϕ¯γa(vaψ1+yaψ2))\displaystyle=\frac{1}{2}\left(\overline{\left(v^{a}\psi_{1}+y^{a}\psi_{2}% \right)}\gamma^{a}\phi+\bar{\phi}\gamma^{a}\left(v^{a}\psi_{1}+y^{a}\psi_{2}% \right)\right) (5)
=va<ψ1|ϕ>+ya<ψ2|ϕ>\displaystyle=v^{a}\langle\psi_{1}|\phi\rangle+y^{a}\langle\psi_{2}|\phi\rangle

The major physical advantage of this Hilbert space lay in the norm of a ket in this space results in the conserved Noether current (equation 6).

<ψ|ψ>=12(ψ¯γaψ+ψ¯γaψ)=ψ¯γaψ=ja\langle\psi|\psi\rangle=\frac{1}{2}\left(\bar{\psi}\gamma^{a}\psi+\bar{\psi}% \gamma^{a}\psi\right)=\bar{\psi}\gamma^{a}\psi=j^{a} (6)

3 Riemann curvature operator.

What is gained by this formulation is a hypothesis which is valid in the flat space-time tangent to every point on a Riemannian manifold. Using the vierbein formulation of General Relativity I can start from the Cartan structure equations using the gamma matrices as basis

Rba=dωba+ωcaωbcR_{b}^{a}=d\omega_{b}^{a}+\omega_{c}^{a}\wedge\omega_{b}^{c}
Ta=dγa+ωbaγb.T^{a}=d\gamma^{a}+\omega_{b}^{a}\wedge\gamma^{b}.

Note the Greek indices’s are suppressed but they are still in effect.

Then solve for the Cartan connection.

Ta=dγa+ωbaγb(Ta-dγa)γb=ωbaT^{a}=d\gamma^{a}+\omega_{b}^{a}\wedge\gamma^{b}\Rightarrow\left(T^{a}-d\gamma% ^{a}\right)\wedge\gamma_{b}=\omega_{b}^{a} (7)

Set the torsion equal to the probability current for the Dirac field |ψ>|\psi\rangle, ja=<ψ|ψ>j^{a}=\langle\psi|\psi\rangle, and simplify. This results in the exterior derivative of the Cartan connection.

dωba=djaγbd\omega_{b}^{a}=dj^{a}\wedge\gamma_{b} (8)

Next we can solve for the Riemann curvature in terms of Dirac gamma matrices and probability currents.

Rba=djaγb+jaγcjcγbR_{b}^{a}=dj^{a}\wedge\gamma_{b}+j^{a}\wedge\gamma_{c}\wedge j^{c}\wedge\gamma% _{b} (9)

To get a valid operator on the Hilbert space use ja=<ψ|ψ>j^{a}=\langle\psi|\psi\rangle. Also use the outer product |ψ><ψ||\psi\rangle\langle\psi|. Then I define Rba(<ψ|)R_{b}^{a}\left(\langle\psi|\right).

Rba(<ψ|)=(d<ψ|γb+<ψ|γc<ψ|γb)R_{b}^{a}\left(\langle\psi|\right)=\left(d\langle\psi|\wedge\gamma_{b}+\langle% \psi|\wedge\gamma_{c}\wedge\langle\psi|\wedge\gamma_{b}\right) (10)
Rba^=Rba(<ψ|)|ψ><ψ|=(d<ψ|ψ>γb+<ψ|ψ>γc<ψ|ψ>γb)<ψ|\widehat{R_{b}^{a}}=R_{b}^{a}\left(\langle\psi|\right)|\psi\rangle\langle\psi|% =\left(d\langle\psi|\psi\rangle\wedge\gamma_{b}+\langle\psi|\psi\rangle\wedge% \gamma_{c}\wedge\langle\psi|\psi\rangle\wedge\gamma_{b}\right)\langle\psi| (11)

The Riemann operator, with Greek indices’s suppressed, is Rba^\widehat{R_{b}^{a}} with eigenvalues and eigenstates given by equations 12.

Rba^|Rba>=(Rba(|ψ>)|ψ><ψ|)|Rba>\widehat{R_{b}^{a}}|R_{b}^{a}\rangle=\left(R_{b}^{a}\left(|\psi\rangle\right)|% \psi\rangle\langle\psi|\right)|R_{b}^{a}\rangle (12)

Equation 12 relates the curvature of space time near a point to the probability four currents due to local non-gravitational fields. Equation 12 provides eigenvalues and eigenstates of space-time curvature. This is a quantization of gravity which will be as renormalizeable as the theories which go into computing the state of the system.

4 If all the world was essentially Klein-Gordon, how would gravity behave?

The Klein-Gordon field is the simplest field which could be used in this theory. I will approximate all the matter in space with a Klein-Gordon field. Klein-Gordon is a simple field to work with and will illustrate the basics of this model well. However, a more realistic model may result from use of the Higgs field. The possibility of using the Higgs field will be explored in a future paper.

ψ=eikaxa\psi=e^{ik_{a}x^{a}} (13)

Knowing the curvature operator equation 11, and the solution to the Klein-Gordon equation 13 I can find the curvature eigenvalues.

(γaγcγcγb)e-ikaxaγaeu=Rbaeu\left(\gamma^{a}\wedge\gamma_{c}\wedge\gamma^{c}\wedge\gamma_{b}\right)e^{-ik_% {a}x^{a}}\gamma^{a}e^{u}=R_{b}^{a}e^{u} (14)

Equation 14 can be solved for eue^{u} by examination and yields equation 16 which gives the eigenvalues and eigenstates of RbνaμR_{b\nu}^{a\mu}.

(γaμγcλγcλγbν)=Rbνaμ\left(\gamma^{a\mu}\wedge\gamma_{c\lambda}\wedge\gamma^{c\lambda}\wedge\gamma_% {b\nu}\right)=R_{b\nu}^{a\mu} (15)
eu=eikaxa=|Rbνaμ>e^{u}=e^{ik_{a}x^{a}}=|R_{b\nu}^{a\mu}\rangle (16)

Thus the simplest curvature field is the Klein-Gordon field times the curvature eigenvalue. The resulting equation is the quantized Riemann curvature tensor field.

Rbνaμ(xa)=(γaμγcλγcλγbν)eikaxaR_{b\nu}^{a\mu}\left(x^{a}\right)=\left(\gamma^{a\mu}\wedge\gamma_{c\lambda}% \wedge\gamma^{c\lambda}\wedge\gamma_{b\nu}\right)e^{ik_{a}x^{a}} (17)

Contraction of the appropriate indexes allows me to find the quantized Ricci tensor and quantized Ricci scalar. Then the Einstein-Hilbert action can be written down.

S=12κηabγμaγbνηab(γaμγcλγcλγbν)eikaxa-ηabγμaγbνd4xS=\frac{1}{2\kappa}\int\eta_{a}^{b}\gamma_{\mu}^{a}\gamma_{b}^{\nu}\eta_{a}^{b% }\left(\gamma^{a\mu}\wedge\gamma_{c\lambda}\wedge\gamma^{c\lambda}\wedge\gamma% _{b\nu}\right)e^{ik_{a}x^{a}}\sqrt{-\left\|\eta_{a}^{b}\gamma_{\mu}^{a}\gamma_% {b}^{\nu}\right\|}\,\mathrm{d}^{4}x (18)

Treating the position as an operator requires that I Taylor expand equation 18. This expansion leads to a formula for computing the quantum gravitational action to arbitrary precision using a series that is known to converge uniformly (equation 19).

S=12κηabγμaγbνηab(γaμγcλγcλγbν)(n=0(ikaxa)nn!)-ηabγμaγbνd4xS=\frac{1}{2\kappa}\int\eta_{a}^{b}\gamma_{\mu}^{a}\gamma_{b}^{\nu}\eta_{a}^{b% }\left(\gamma^{a\mu}\wedge\gamma_{c\lambda}\wedge\gamma^{c\lambda}\wedge\gamma% _{b\nu}\right)\left(\sum_{n=0}^{\infty}\frac{\left(ik_{a}x^{a}\right)^{n}}{n!}% \right)\sqrt{-\left\|\eta_{a}^{b}\gamma_{\mu}^{a}\gamma_{b}^{\nu}\right\|}\,% \mathrm{d}^{4}x (19)

To get Feynman diagram style rules for this theory the first few terms in the Taylor series will be useful. Let R0=ηabγμaγbνηab(γaμγcλγcλγbν)R_{0}=\eta_{a}^{b}\gamma_{\mu}^{a}\gamma_{b}^{\nu}\eta_{a}^{b}\left(\gamma^{a% \mu}\wedge\gamma_{c\lambda}\wedge\gamma^{c\lambda}\wedge\gamma_{b\nu}\right) in equation 20. The resulting rules are in figure 1.

S=12κR0(1+ikaxa-k2x22-)-ηabγμaγbνd4xS=\frac{1}{2\kappa}\int R_{0}\left(1+ik_{a}x^{a}-\frac{k^{2}x^{2}}{2}-\dots% \right)\sqrt{-\left\|\eta_{a}^{b}\gamma_{\mu}^{a}\gamma_{b}^{\nu}\right\|}\,% \mathrm{d}^{4}x (20)
Figure 1: Feynman diagrams of gravity in this model. Here wavy lines indicate gravitons. Note these rules result essentially from the choice of the Klein-Gordon field for the local QFT.

5 Conclusions

I have shown that a Hilbert space can be defined such that the <ψ|ϕ>=va\langle\psi|\phi\rangle=v^{a}\in\mathcal{M}. With Hilbert space so defined one can reasonably write of a Riemann curvature operator. Then an eigenvalue equation can be found. I have also shown that the Einstein Hilbert action and well know Feynman Diagram rules for Einstein gravity can be derived from this formulation. That is to say equation 19 allows one to compute the gravitational effect on interacting particles at any quantum mechanically interesting length scale, and to any desired precision. (It is also interesting that this theory resembles an f(R)f(R) model once the Taylor expansion is done.) A more complete treatment of this problem would involve using the full standard model to compute a quantized Einstein Hilbert action, Feynman diagrams etc. However, I believe the essential features of this hypothesis are illustrated well enough by equation 19. A more complete treatment may introduce new problems of calculation but, as long as the model used to compute ψ\psi is at least renormalized none of those problems could be fatal.

I would like to acknowledge Doug Sweetser, and Edward "Henry" Brown for their helpful critical comments on my blog which stimulated me to make an important change in this paper. A PDF of those conversations is attached as an additional asset. Since I could not add their reviews as "reviews" myself, I have printed them out to a PDF file and that file is attached to this paper. I would also acknowledge Hank Campbell of Science 2.0 for providing me with a platform on which to blog about science for many years. I would also like to thank all of my financial backers for their help.

References

  • A. Ashtekar (1986) New variables for classical and quantum gravity. Physical Review Letters 57, pp. 2244–2247. External Links: Document Cited by: 1.
  • A. Einstein (1916) The Foundation of the Generalised Theory of Relativity. Annalen der Physik 7 (354), pp. 769–822. Note: In this Wikisource edition of Bose’s translation, his notation was replaced by Einstein’s original notation. Also some slight inaccuracies were corrected, and the omitted references were included and translated from the German original. External Links: Link Cited by: 1.
  • D. Finkelstein, J. M. Jauch, S. Schiminovich and D. Speiser (1963) Principle of General Q Covariance. Journal of Mathematical Physics 4, pp. 788–796. External Links: Document Cited by: 1.
  • S. Hollands and R. M. Wald (2014) Quantum fields in curved spacetime. ArXiv e-prints. External Links: 1401.2026, Link Cited by: 1.

Additional Assets

Showing 5 Reviews

  • David halliday
    David Halliday
    Originality of work
    Quality of writing
    Quality of figures
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    2

    Hontas:

    Unfortunately, even allowing for your "Hilbert" space to be over an Algebra (actually a "beast" of a different name), you would still violate your very first equation.

    First off, as others have already pointed out, there is no algebra made up solely of elements of your M (a Minkowskian vector space), so one must extend to some appropriate Algebra (Cl_{1,3}(R), in your case).  In any case, this means that any linear combination of vectors in this "Hilbert" space, with arbitrary scalars taken from the Algebra, must also be elements of this "Hilbert" space.  This will, unavoidably cause a violation of this very first equation!

    This most certainly does not bode well for the rest of your paper.

    (I've lumped "use of equations" in my rating of "quality of figures".)

    This review has 2 comments. Click to view.
    • David halliday
      David Halliday

      As a note of clarification: When I referred to Hontas' first equation I was not at all referring to the first numbered equation, but the first equation in the first sentence of the Introduction of the paper. Namely, = v^a is an element of ℳ.

      • Hontas%20 farmer
        Hontas Farmer

        I understood that. No algebra made up solely of elements of M? There is the space-time algebra described here http://www.mrao.cam.ac.uk/~clifford/introduction/intro/node11.html While it is correct not all of it's elements are going to be four vectors, (that algebra includes pseudo vectors, and bivectors). It is possible that a direct product of one vector could be taken with another to create a bivector I'd ask why? When we physicist work with physically interesting vectors, j, P, x, F, etc we never have just jp or jf. There is always a inner product (I know as you pointed out even using that term for minkwoskis's "metric" is not strictly correct but that's the physics language.), cross product, outer product or something going on like that.

      • David halliday
        David Halliday

        I actually worked with the Space-Time Algebra (and Geometric Algebras in general) for a number of years, with Pro. Hestenes, starting in 1994. The fact that "that algebra includes pseudo vectors, and bivectors", etc. is precisely the point! The elements of the Algebra (having 2^4 = 16 basis vectors), includes much more than the elements of ℳ (which only has four [4] basis vectors). While the elements of a basis of ℳ can be used as generators of the Algebra, they, alone, are but a small subset of it, and do not form anything close to a closed subset of the Algebra. Hence the unavoidable violation of your very first equation.

      • David halliday
        David Halliday

        Unfortunately, Edward "Henry" Brown already tried to show you this problem, on your Science 2.0 'blog, early last month (August 10--11). He and John Lee have continued to try, here on the Winnower, to bring this, and other deficiencies, to your attention. (Admittedly, a significant part of the problem in the communication looks like it relates to my first review.) Most unfortunate of all is the way you appear to have tried to "blow them off" by quoting documents you appear to have little understanding of. (As they have tried to point out to you, those documents do not support your assertions.)

      • David halliday
        David Halliday

        By the way, are you aware that the cross product cannot be even close to directly generalized to any other space besides the three (3) space it was created in? If you will but carefully read the first couple of sections of "Imaginary Numbers are not Real", you should gain a much better understanding of how the product (within the Space-Time Algebra) relates to "inner product (...), cross product, outer product or something going on like that."

    • Hontas%20 farmer
      Hontas Farmer

      First of all let me say your two reviews taken together are useful and helpful and if I could I would take care to explain things differently to try and address your points. Maybe revise the hypothesis a bit. As it stands this is how I would've rated the paper. "originality" *** "Confidence" ** Quality of writing ***, quality of figures ** Not very different from you. I don't think I am right, I think I'm at least asking a interesting question.

      • David halliday
        David Halliday

        I, too, think you are asking an interesting question. However, Physics is based upon rigorous mathematical language for good reason. We can and should respect that. (Of course, when I say Physics is based upon such, that, by no means, should be taken to suggest that Physics is, in any way, limited to current mathematical tools, understanding, etc. While, by the vary nature of mathematics, we cannot violate what is already known, within that body of knowledge, we can, and often do, push beyond the frontiers, there.)

  • David halliday
    David Halliday
    2

    Hontas:

    A significant part of the issue you are running into with the two reviewers here, so far, is that your use of terminology simply doesn't jibe with current mathematical usage.  For instance, within your abstract you have the mathematically nonsensical statement concerning "a Hilbert space over the field of quarternions."  First, the quaternions are not a (mathematical) field, yet, second, a Hilbert space is (mathematically) defined as being over a (mathematical) field.

    This is magnified within your arguments over whether the elements of the Clifford algebra Cl_{1,3}(R) are "scalars".  You (rightfully) argue from the standpoint that just because something can be written as a composite (like the Complex numbers) does not imply that such are not scalars.  However, you seem to fail to recognize that this argument also fails to show whether any such is, in actuality, a scalar.

    The truth is that the use of the term "scalar" within the context of a Hilbert space, and other forms of vector spaces, boils right down to whether such form a (mathematical) field.  So, no, the elements of the Clifford algebra Cl_{1,3}(R) are not scalars, at least in this sense.

    Part of the "problem" is that we physicists tend to be a bit cavalier about our use of mathematical terms.  Cases in point are Dirac's delta "function" ("not a function, but a distribution"), the Minkowski "metric" ("not a metric, but a pseudo-metric"), and the "Riemannian" geometry of General Relativity ("not Riemannian, but pseudo-Riemannian").  (If we actually have instances of physicists talking/writing about "a Hilbert space over the field of quarternions", then we have another instance of a very similar issue.)

    So, my advice:  Be careful about the use of mathematical terminology, especially when one is involved in disagreements like the simpler ones we have here.




    This review has 2 comments. Click to view.
    • Hontas%20 farmer
      Hontas Farmer

      Thanks for the advice. There is one other way in which something can be a scalar. We define a tensor of rank N by it's properties under a given transformation. So 4-vectors are such because of their transformation properties under Lorentz-Poincare transformations. The same for 4 tensors and Lorentz scalars. These vectors are scalars in this hilbert space in the sense that operators on the hilbert space do nothing to them. You may be correct in the purest mathematical sense of the term. However in the physics sense I was taught we call something a scalar if in this physical situation (in terms of how operators act on it, it's relationship to vectors). I thank you for your kind and useful review. BEAR IN MIND I CANNOT EDIT THE ARTICLE. So there is nothing I can change now. Article versioning is a planned future feature of the Winnower.

      • David halliday
        David Halliday

        You are correct that transformational qualities are another important criteria for "scalar"-ness. However, there is still the matter of calling something a Hilbert space when it violates the basic definitions of such.

        As you can see in my own writings, I tend to "push back" against certain Mathematical definitions. However, I do not do so "willy-nilly". I only do so when physical motivations, in my opinion, override Mathematical definitions rooted in now discarded, former "physical" motivations.

        So, for instance (recalling my previous list) I don't have a problem with referring to Dirac's "function" as a distribution. I see no harm at all in this change of "name". However, when it comes to a (pseudo-)metric that is a physical, realized metric (from the physical standpoint, even if not by the Mathematical standpoint that is rooted in a now discarded, former concept of "physical reality"), I see absolutely no benefit---nay, even harm---in the "pseudo" moniker. (The same reasoning applies to the "pseudo" moniker on the Riemannian Geometry of General Relativity, especially since it stems from exactly the same root as that of "metric".)

        Now, as for a "Hilbert" space that has as scalars element of an Algebra, rather than a (mathematical) field, I don't see the physical argument for such a different use of such a mathematical term. While, off the top of my head, I cannot recall what the mathematical term is for such a "beast", I just don't see a physical argument for not using the proper mathematical term for such. (The only argument I can see for eschewing the proper mathematical term would be a desire to /appear/ to be using the longstanding mathematical mechanics of QFT.)

    • Hontas%20 farmer
      Hontas Farmer

      Question, I've noticed that no one bothers to use the star rating system on my article, not even for "figure quality". Could you perhaps explain for the Winnower and myself why that might be?

      • David halliday
        David Halliday

        As for myself, I was very glad that I could "geta away with" having no ratings at all, as opposed to appearing to provide ratings of zeros.

        As for a rating of "figure quality", you had no figures, unless one wished to extend the concept to the use of equations.

        As for why I wished to "get away with" having no ratings: I wated to avoid "tainting" my constructive criticism with my opinions on such other matters.

      • Hontas%20 farmer
        Hontas Farmer

        By figures I was referring to the Feynman diagrams. I was just curious why no one seemed to use those at all. Not even half a star as if I wrote about cold fusion or something.

      • David halliday
        David Halliday

        I'm sorry, but to me, the Feynman diagrams looked more like a part of equations. I suppose we, as reviewers, could treat equations and diagrams, etc., all as "figures". On the other hand, perhaps it would be even better for the Winnower to add a rating of something like "use of equations".

  • Placeholder
    John Lee
    Originality of work
    Quality of writing
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    Confidence in paper
    2

    I concur with the other reviewers. The initial concept is math gibberish, making anything that follows incoherent.  I worry for the health of this journal, given that this article was actually solicited by the journal.

    It is clear you wish to treat the space of four-vectors as scalars, but borrowing the algebraic structure of Cl_1,3(R) (sometimes referred to as the "space-time algebra") still does not allow you to achieve this.

    To make this point as simple as possible to follow, consider four-vectors in the basis γ_a.  Any four-vector X can be written with real components X^a in this basis. Written explicitly, X = X^a γ_a = X^0 γ_0 + X^1 γ_1 + x^2 γ_2 + X^3 γ_3

    Now consider the following four-vectors A and B: 
    A = 1 γ_0 + 0 γ_1 + 0 γ_2 + 0 γ_3
    B = 0 γ_0 + 1 γ_1 + 0 γ_2 + 0 γ_3

    Then A*B = (1 γ_0 + 0 γ_1 + 0 γ_2 + 0 γ_3)(0 γ_0 + 1 γ_1 + 0 γ_2 + 0 γ_3) = 1 γ_0 γ_1
    This cannot be written as a four-vector.

    Simply put, the result is not a four-vector, but a bivector in the Clifford algebra.  The vector-space M is not closed under the multiplication operation.  Thus this space of four-vectors cannot be used as a scalar for a Hilbert space.

    I recommend you take the time to seriously consider the issues already pointed out.  You are misunderstanding basic definitions, making mistakes in simple operations, and misreading about geometric algebra enough to think it supports your claims when in does not.

    This review has 3 comments. Click to view.
    • Hontas%20 farmer
      Hontas Farmer

      "The result is not a four vector but a bivector". That is your idea? Check this reference.

      http://www.mrao.cam.ac.uk/~clifford/introduction/intro/intro.html


      http://www.mrao.cam.ac.uk/~clifford/introduction/intro/node5.html#SECTION00023000000000000000

      The product you are having a problem with is given in almost identical form in that publication which was (Foundations of Physics
      September 1993, Volume 23, Issue 9, pp 1175-1201
      Imaginary numbers are not real—The geometric algebra of spacetime
      Stephen Gull, Anthony Lasenby, Chris Doran)

      IT IS CLEARLY SHOWN IN THE REFERENCE PROVIDED THAT SUCH A PRODUCT IS NOT A BIVECTOR. Thanks for the feedback though. :)

      • Hontas%20 farmer
        Hontas Farmer

        The relationship is =x^a . This means given two vectors in Hilbert and the inner product we get one vector in Minkowski...that same object which is a vector in Minkowski with the clifford algebra associated with that space... is at the same time a scalar in Hilbert space. That sounds odd but remember in standard H over complex numbers = c in C a complex number a+bi may be written as a vector as well.

      • Placeholder
        Edward Brown

        John Lee, In case you are unaware of it, please look at the discussion on the author's blog. I tried to break it down into basics like you are doing here, and nothing helped. I wish you luck, but be aware that you are most likely just wasting your time.

      • Placeholder
        John Lee

        Edward, Thank you for the information and concern. Currently Hontas does not know enough to understand the degree of the mistakes being made both in the mathematics and in self-evaluating skill levels relative to others in this discussion. This is the Dunning-Kruger effect, which I believe can be overcome with sufficient teaching effort. Even if my efforts end up wasted, I feel it is better to try than to presume the individual 'unteachable'. I see that you cared enough to try yourself, so hopefully you can understand.

      • Placeholder
        John Lee

        Hontas,
        There is a reading comprehension issue here. The links you provide are just an html version of the paper you mention, which directly contradicts your claim. Let's look at the claims side by side. I state: "the result [of multiplying the four-vectors A and B] is not a four-vector, but a bivector in the Clifford algebra". You state: "IT IS CLEARLY SHOWN IN THE REFERENCE PROVIDED THAT SUCH A PRODUCT IS NOT A BIVECTOR." The paper explains: pg 4 "the product of orthogonal vectors is a bivector" (exactly in agreement of my example with two orthogonal four-vectors). Also table 5.2 on pg 15 shows the vectors and bi-vectors of your algebra, and indeed the bivector σ_1 = γ_1 γ_0 , the product of two four-vectors. Once you understand you have made a mistake here, we can then move on to the other issues if you are still in need of clarification, but I'd like to focus on one point at a time. The point here is: in the algebra Cl_1,3(R) the product of two four-vectors is not a four-vector, and thus the space of four-vectors, M, is not closed under this operation. Do you now agree with that statement?

    • Hontas%20 farmer
      Hontas Farmer

      What you failed to keep straight is that this model contains two very different kinds of space. There is the proposed Hilbert space with ket vectors and then there is the Minkowski space with Clifford algebra of it's vectors. One must not confuse the two. The ket vectors of hilbert space could be expressed over any number of fields, or skew fields or algebras. The mathematical physics literature is full of them. (i.e. Ket vectors expressed over the space of square integrable functions.) Just because you are unfamiliar with one example does not make it nonsense.

      • Hontas%20 farmer
        Hontas Farmer

        Maybe I shouldn't say "you failed to keep it straight"... I failed to write clearly enough that one needs to remember that the ket vectors exist in one space, and the four vectors exist in another space related to each other by =x^a I thought that was plain enough. That was the cause of all this confusion.

    • Hontas%20 farmer
      Hontas Farmer

      Both of you seem to insist that the ket vectors and the minkowski vectors live in the same space. That is the only way for you to reach the conclusion which you are reaching. Again, the inner product on H produces a vector in M which acts as a scalar in H. I have provided a reference for this again and again which you both seem to refuse to read. Specifically yo Edward,...,don't pretend that your objections come from concerns related to geometric algebra. You don't appear to have been aware of it until I told you.

      • Placeholder
        John Lee

        I have read your reference, and even quoted from it to show that it directly contradicted your claim, and agrees with what I have been trying to explain to you. If you really don't understand the relevance of my question, that is disappointing, but we can discuss that error in understanding later. Let's focus on one of your misunderstandings at a time. I have shown with a clear example that with the algebra Cl_1,3(R), the result of multiplying two four-vectors is not a four-vector, and here it was a bi-vector. You didn't discuss what step you felt was incorrect in the example, and instead yelled "SUCH A PRODUCT IS NOT A BIVECTOR". I have further shown that your reference explicitly contradicts your claim that the product is not a bivector, and instead the reference agrees with my statement. Again, if you somehow feel this is irrelevant, we can discuss that issue later, but to make any progress it is important that you accept and learn from your mistake here. The point was and is: in the algebra Cl_1,3(R) the product of two four-vectors is not a four-vector, and thus the space of four-vectors, M, is not closed under this operation. Do you now agree with that statement? Please respond either yes, or no with your explanation including what you believe A*B is in that example.

  • Placeholder
    Edward Brown
    Confidence in paper
    2

    This article is unrecoverable, as the the very starting idea which the article wishes to derive consequences from is incoherent and not mathematically well posed.  The author misuses notation and terminology throughout the article.

    As this is a "post publication review" it is not entirely clear what a review for a paper of this quality should entail.  Is the function to warn others not to waste their time with the paper?  Essentially crowd sourcing the ability to separate the "grain from the chaff"?  I have rated my confidence in this paper accordingly.

    For the author, here are some issues in just the first few equations to make it clear the extent of the problems: 
    (eq 1) This defines an operation taking two elements of H and producing a vector v stated to have real valued components.  The other terms are not clearly defined, and appear to be used inconsistently in the paper.  
    * Four-vectors in the tangent space of the space-time manifold are not elements of a field.  Simply put, four-vectors are not scalars, so this operation cannot be an inner product.  Thus this operation is not part of a Hilbert space.  This comes directly from the definitions of inner product and Hilbert space.
    * The gamma symbols are presumably the Dirac matrices, which means the components of the resulting four-vector are at least complex valued and possibly even still a matrix depending on what phi and psi are.  Thus this is already inconsistent with the statement on what v is.
    (eq 3) It is assumed that the gamma terms are Hermitian, but in the usual representation only one is, while the other three are anti-Hermitian.
    (eq 4) This equation originally tried to check if the operation satisfied one of the requirements of a Hilbert space.  As the space of four-vectors is not totally ordered, the equation made no sense mathematically.  In response to this complaint the equation in the article was changed from a check of positive-definiteness to just checking if the result is not equal to the zero vector. But that misses the original intent, as now this is no longer checking the requirement of a Hilbert space.  The point was that this operation did not satisfy the requirements of a Hilbert space.

    I'll stop here. It is clear the very starting point of this paper is irrecoverably flawed.

    This review has 8 comments. Click to view.
    • Hontas%20 farmer
      Hontas Farmer

      Thank you for posting a thoughtful review! To which I have given a thoughtful response. I apologize if the comments are kinda bunchy but their formatting seems to be limited to one paragraph no HTML. At least for now.

    • Hontas%20 farmer
      Hontas Farmer

      "(eq 1) This defines an operation taking two elements of H and producing a vector v stated to have real valued components. The other terms are not clearly defined, and appear to be used inconsistently in the paper.
      * Four-vectors in the tangent space of the space-time manifold are not elements of a field. Simply put, four-vectors are not scalars, so this operation cannot be an inner product. Thus this operation is not part of a Hilbert space. This comes directly from the definitions of inner product and Hilbert space."

      NOTE THE PAPER NOW SAYS

      Let
      ℋ be a vector space with elements ∣∣ψ> and ∣∣ϕ> over the space of 4-vectors with va∈ℳ and the inner product defined as follows.

      IT NO LONGER SAYS FIELD. YOU POINTED THAT OUT EARLIER AND I FIXED IT BEFORE FINALIZING IT. Please reivew the current version not the preliminary version.

    • Hontas%20 farmer
      Hontas Farmer

      "* The gamma symbols are presumably the Dirac matrices, which means the components of the resulting four-vector are at least complex valued and possibly even still a matrix depending on what phi and psi are. Thus this is already inconsistent with the statement on what v is."

      NO THE GAMMA MATRICES ARE SIMPLY A CONVENIENT BASIS FOR THIS COMPUTATION, there is nothing fundamental to the theory about them. They are just one way to represent the clifford algebra CL_1,3(R).

      What you write is like saying a 2 D vector space would have to be complex because one may write them using a+bi when they can be re written (a,b) etc.

    • Hontas%20 farmer
      Hontas Farmer

      (eq 3) It is assumed that the gamma terms are Hermitian, but in the usual representation only one is, while the other three are anti-Hermitian.

      WRONG, you have missed an important part of the text. "Let
      ℋ be a vector space with elements ∣∣ψ> and ∣∣ϕ> over the space of 4-vectors with va∈ℳ and the inner product defined as follows."

      HERMITIAN CONJUGATION IS WHAT YOU DO TO OBJECTS IN THE HILBERT SPACE (or their representations) NOT THE UNDERLYING (field,space, or algebra)

      In the framework of this theory it would be wrong to hermitian conjugate the gamma matrices. They are just one basis in which the four vectors are represented. They are convenient because those same matrices appear next to the observable four vectors of QFT ie. j P A etc

    • Hontas%20 farmer
      Hontas Farmer

      (eq 4) This equation originally tried to check if the operation satisfied one of the requirements of a Hilbert space. As the space of four-vectors is not totally ordered, the equation made no sense mathematically. In response to this complaint the equation in the article was changed from a check of positive-definiteness to just checking if the result is not equal to the zero vector. But that misses the original intent, as now this is no longer checking the requirement of a Hilbert space. The point was that this operation did not satisfy the requirements of a Hilbert space.

      See the following references http://arxiv.org/abs/1011.4031 http://www.math.columbia.edu/~woit/QM/fermions-clifford.pdf

      Now this is a harder one to respond to! My answer is that since the underlying space is not totally ordered it is not meaningful to speak of a negative or positive vector in any absolute sense. One can always rotate it to be any sign we wish. However THE REASON WE WANT A POSITIVE DEFINITE BRA-KET IS BECAUSE WE NEED A NORMALIZE ABLE BRA-KET. For a Hilbert space over a clifford algebra that is not a problem don't take my word for it.

    • Hontas%20 farmer
      Hontas Farmer

      One more reply to your last point. Please see the following references. See the following references http://arxiv.org/abs/1011.4031 http://www.math.columbia.edu/~woit/QM/fermions-clifford.pdf

      THERE IS EVEN A WHOLE GROUP THAT DOES SIMILAR WORK AT CAMBRIDGE. (Which shows great minds do think alike!)
      http://www.mrao.cam.ac.uk/~clifford/publications/

    • Hontas%20 farmer
      Hontas Farmer

      Here are more references to work along simmilar conceptual and mathematical lines. C. J. L. Doran, A. N. Lasenby and S. F. Gull.
      Grassmann Mechanics, Multivector Derivatives and Geometric Algebra
      In Z. Oziewicz, A. Borowiec and B. Jancewicz, editors, Spinors, Twistors, Clifford Algebras and Quantum Deformations (Kluwer Academic, Dordrecht, 1993), p. 215-226.

      A. N. Lasenby, C. J. L. Doran and S. F. Gull.
      2-spinors, Twistors and Supersymmetry in the Spacetime Algebra
      In Z. Oziewicz, A. Borowiec and B. Jancewicz, editors, Spinors, Twistors, Clifford Algebras and Quantum Deformations (Kluwer Academic, Dordrecht, 1993), p.233-245.

    • Placeholder
      Edward Brown

      Note to site maintainers: I feel it is inappropriate to give articles a DOI number if the author continues to edit it. Especially when those previous versions appear to be gone forever. A system similar to arXiv would be appreciated. Response to author's numerous comments: You do not appear to be understanding the issues here. It one of your edits you now want the extra structure of an algebra instead of just the vector-space of the four-vectors, but this isn't a magic panacea, as problems persist. In the comments you seem to be linking papers which most (frankly I didn't look at all the links) seem to make contact merely by the existence of Clifford algebras. You appear to want reviewers to respond to all this and continue responding until we reach some common consensus or mutual understanding. It would be one thing if there was discussion about misunderstanding or misrepresentation of some minor point in the paper. But here there are serious problems with the paper. As such, it is _extremely_ unreasonable to expect a reviewer to give educational lectures to bring everyone up to speed. This cannot (or at least should not) be the function of a review. If you insist on "learning by committee on the internet", please explore these math concepts on your blog where people can respond with mathematical notation. I consider this review closed.

      • Hontas%20 farmer
        Hontas Farmer

        I didn't edit the article after assigning the DOI. That is not allowed on this website. I edited the article in response to your comments elsewhere and if you look I even mention your comments on science2.0 in the acknowledgements. :)

      • Hontas%20 farmer
        Hontas Farmer

        One last comment, everything you say is fundamentally wrong only proves your own very common misconceptions. Read this paper and be educated Foundations of Physics
        September 1993, Volume 23, Issue 9, pp 1175-1201
        "Imaginary numbers are not real—The geometric algebra of spacetime"
        Stephen Gull, Anthony Lasenby, Chris Doran http://www.mrao.cam.ac.uk/~clifford/introduction/intro/intro.html QUOTE: These considerations all indicate that our present thinking about quantum mechanics is infested with the deepest misconceptions. We believe, with David Hestenes, that geometric algebra is an essential ingredient in unravelling these misconceptions.

        On the constructive side, the geometric algebra is easy to use, and allows us to manipulate geometric quantities in a coordinate-free way. The -vectors, which play an essential role, are thereby removed from the mysteries of quantum mechanics, and used to advantage in physics and engineering. We shall see that a similar fate awaits Dirac's -matrices.UNQUOTE Your review demonstrates all of the misconceptions they speak of.

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