Rosé and A-M: Geometrization of Quantum Mechanics

[Careful]

in post #127 you stated fiveiticisms:
It seems to me that (a) and (b) are real criticisms, i.e assertions that the research is wrong, while (c), (d), and (e) are like suggestions for future directions. Is this what you are waiting for a response to?

Well for now, I am waiting for one specific thing: there are complex curvatures according to Torsten and Helge, which I disagree with.
If there are no complex numbers coming out, then we have to think again about the link with QM. In the other case, we can proceed towards the construction of the algebra.

Cheers,

Careful

 

[selfAdjoint]

Fair enough. Complex curvatures, yea or nay, is next up then.

 

[torsten]

Well for now, I am waiting for one specific thing: there are complex curvatures according to Torsten and Helge, which I disagree with.
If there are no complex numbers coming out, then we have to think again about the link with QM. In the other case, we can proceed towards the construction of the algebra.
Cheers,
Careful

Dear Careful,
your problems with our approach are grouped into the two classes:
1. the complex curvature
2. dynamical aspects of differential structures including the propagation of induced gravitational waves
ad 1. The usage of the curvature of a complex line bundle instead of using the curvature of a real rank-2-vector bundle is induced by the topological fact, that the embedding of a surface into a 3- or 4-manifold M can be described as the zero set of a section in a complex line bundle.
Now let \Omega be the curvature of a complex line bundle, i.e. a 2-form with values in the Lie algebra i{\mathbb R} of the U(1).
The 2-form \frac{i}{2\pi}\Omega represents te first Chern class of the line bundle, i.e. it is an element of the cohomology H^2(M,{\mathbb R}).
The Poincare dual of that cohomology class represents the homology class of the embedded surface.
Instead to use the complex bundle there is the possibility to use a real rank-2-vector bundle. I think you have that bundle in your mind because the tangent bundle TS of the surface S is such a bundle. Especially the curvature is real and we have the relation \frac{i}{2\pi}\int_S \Omega=\int_S e(TS) with S=PD(\Omega) and e(TS) is the Euler class of the tangent bundle TS. But we want to use the curvature to define the coefficient field of the algebra. The real and the complex bundle are equivalent according to the representation of the structure group. But the set of the complex curvatures forms a number field and that's why we can use it for the coefficients of the algebra (the real case forms only a ring).
More in detail: We use the definition Tr(\Omega)=\int_S \Omega and get a pure imaginary number. Then the square root of this number produces the complex number.
The real curvature of TS embedded into M is described by a matrix-valued 2-form (the two indices of the curvature tensor).
Our goal is the usage of the values of the curvature as the coefficients in the algebra.
But that swaps out the real bundle because the matrices don't form a field which is necessary to define the algebra.
Thus, the description of embedded surfaces (needed to define the product in the algebra) by a complex line bundle and the relation between the square of the coefficient and the value of pure imaginary curvatue produces the complex numbers.
ad 2. As we state in one of previous postings, we are not the real experts in singularity theory.
Thus, as usual, your *careful* critism was useful.
In our papr we studied implicitly stable, singular mappings f:M->N. As we learned now, the singular set \Sigma of the map f is only determined up to isotopy (=smooth homotopy).
That means, that the two spaces \Sigma and \Sigma\times [0,1] are given by the same stable map.
But that shows: our matter is not a sitting duck, it defines a world tube. The additional parameter of the interval [0,1] must be determined by an additional equation which is the dynamics of the matter.
Furthermore, the size of the singular set \Sigma can also change according to the isotopy. This parameter can be described by Einsteins equation directly.
Thus not everything is completely determined by the singular map but by the solution of Einsteins equation we get the metric where the singular set is the source.
Of course you will also get gravitational waves as usual.
A look in my crystal ball shows me, that you will probably ask about the dynamics of the singular map.
In our next paper we will answer this question because much more mathematics is needed. But we think we are able to derive the dynamics for the DS described via spinor fields.
But we have to stop here because no further explanation is possible without introducing more math.
Maybe you are interested now?...
Best regards
Helge and Torsten

 

[Careful]

Hi,

No, I disagree.


(a) ** The usage of the curvature of a complex line bundle instead of using the curvature of a real rank-2-vector bundle is induced by the topological fact, that the embedding of a surface into a 3- or 4-manifold M can be described as the zero set of a section in a complex line bundle.
Now let \Omega be the curvature of a complex line bundle, i.e. a 2-form with values in the Lie algebra i{\mathbb R} of the U(1).
The 2-form \frac{i}{2\pi}\Omega represents te first Chern class of the line bundle, i.e. it is an element of the cohomology H^2(M,{\mathbb R}).
The Poincare dual of that cohomology class represents the homology class of the embedded surface. **

The Chern form is constructed for gauge connections on a vector bundle over the Riemann surface S (which equals M in this case). Gauge connections in the bundle satify *different* transformation laws than
connections on M : therefore you lose spacetime covariance (I told this already). If you do not believe me or think this comment is irrelevant, start from a mapping f:M -> N, two *spacetime* metrics g and h, their Levi Civita connections and show me how complex curvature arises.

** The real and the complex bundle are equivalent according to the representation of the structure group. **

4-D covariance ??

** 2. As we state in one of previous postings, we are not the real experts in singularity theory.
In our papr we studied implicitly stable, singular mappings f:M->N. As we learned now, the singular set \Sigma of the map f is only determined up to isotopy (=smooth homotopy).
That means, that the two spaces \Sigma and \Sigma\times [0,1] are given by the same stable map.
But that shows: our matter is not a sitting duck, it defines a world tube. **

Ah, I thought you knew this, since the same fact was the reason for me to say to Mike2 that matter is not quantized at all in your framework. But this does not help you at all :
(i) You can obtain a singular world tube, but then you lose information again about the ``space´´ your matter is sitting in.
(ii ) the sitting duck has nothing to do with the issue world tube or not, but refers to the fact that your matter does not produce grav waves in the Einstein dynamics (world tube or not - the metric outside the body remains invariant by construction)

** Of course you will also get gravitational waves as usual. **

Nope, your metric does not change outside the source (the connection remains invariant there) just by *construction* of D = df D_M \beta + D_N(1 - df \beta). Perhaps you should comment accurately on this (I told this many times already but no answer came).

Enough for now.

Cheers,

Careful

 

[selfAdjoint]

Helge and Torsten could you perhaps expand upon this statement a bit:

The real and the complex bundle are equivalent according to the representation of the structure group.

Also it would be helpful if you could illustrate this:

The real curvature of TS embedded into M is described by a matrix-valued 2-form (the two indices of the curvature tensor).

and indicate in what sense the complex line bundle provides/generates/transforms to the Levi-Civita curvature of the tangent bundle on spacetime? This is kind of a key link in your effort to represent genuine matter in spacetime by change of DS, and I'm sure you agree it is important that it be absolutely clear.

 

[Mike2]

Dear Helge and Torsten,

Have you found any connection with the results of black holes? It seems with all this talk of supports, curvature, and mass, that this sounds a lot like issues that concern black holes. Recently Bekenstein derived the mass spectrum of black holes, etc. So I was wondering if that might apply to your effort.

 

[torsten]

Helge and Torsten could you perhaps expand upon this statement a bit:
Also it would be helpful if you could illustrate this:
and indicate in what sense the complex line bundle provides/generates/transforms to the Levi-Civita curvature of the tangent bundle on spacetime? This is kind of a key link in your effort to represent genuine matter in spacetime by change of DS, and I'm sure you agree it is important that it be absolutely clear.

Dear Careful, dear selfadjoint,

Of course we are not wizards, so given a metric g and calculate the curvature R then this curvature is real. But we use another way: Given the curvature two form R^a_b on the the 4-manifold (which is real!). Then we we form the first Pontrjagin class p_1(R)=\frac{1}{8\pi^2}Tr(R^a_b\wedge R^c_a) and via the relation e(R)\wedge e(R)=p_1(R) the Euler form. This Euler form is a 2-form and we construct a surface S_R (unique up to diffeomorphism) via the integral \int_{S_R}e(R)=1(deRham theory). This surface is dual to the Euler form. Now we study the geometry of this embedded surface via the embedding F:S_R\hookrightarrow M. That is the point where the complex structure appears: By general bundle theory one knows that every complex line bundle on M defines an embedded surface in M (take a section in the line bundle, then the zero set of the section is this surface). Thus, let L_S be the complex line bundle which corresponds to the surface S_R having the curvature \Omega. That curvature is a 2-form with values in the Lie algebra of the U(1), i.e. the pure imaginary numbers. Of course i\Omega is the real curvature of the surface w.r.t. the embedding F:S_R\hookrightarrow M. That is, the metrik g_{\mu\nu} on M induces via the embedding F a metric g_{ij}=\frac{\partial F^\mu(x)}{\partial x^i}\frac{\partial F^\nu(x)}{\partial x^j}g_{\mu\nu} on S_R.

Now to the representation question: Because of the isomorphism SO(2)=U(1) there is also the possibility to represent S_R via a rank-2-vector bundle instead of a complex line bundle. Then you will get a real matrix-valued curvature R^a_b. But that approach runs into a problem: the two-dimensional matrices form only a ring but not a number field which is needed for the construction of the algebra. That fact singles out the complex numbers.


Secondly some words about grav. waves: We never say that the change of the DS completely determined the corresponding solution of Einsteins equation (by using the singular connnection). Everything is only determined up to some important constants which have to be determined by solving the equation. But then you have the same effect as putting matter in the space and then turn on gravity. In that approach, matter cannot be a sitting duck and grav. waves propagate as it should be.

Regards
Helge and Torsten

 

[Helge Rosé]

Dear Helge and Torsten,
Have you found any connection with the results of black holes? It seems with all this talk of supports, curvature, and mass, that this sounds a lot like issues that concern black holes. Recently Bekenstein derived the mass spectrum of black holes, etc. So I was wondering if that might apply to your effort.

Dear Mike2,

no, we have not. At this stage the approach can not handel such questions.

(By the way:
My opinion is, that a black hole is a infelicitous defined model (but a solution of the E.eq.) and the epistemological problems are too deep. Thus I am not very interessed in black hole solutions of the E.eq.)

 

[Careful]

Hi,

There was a misunderstanding, I thought you were making use of the fibration of the 4 D manifold, but obviously you are not doing that at all. You are just using the natural tangent space of the two disk, connecting this somehow to three surfaces \Sigma. Relating to my comment in post 98, I was doing myself yesterday some calculataions concerning how a distributional part in the ricci tensor should arise in case the nucleus of df on \Sigma is three dimensional. Actually, the difference connection I defined there runs into trouble (because of the mixed indices); remember : (D(v)w)(x) = df D_M(v) \beta w(f(x)) + D_N (df(v(x))) [(1 - df \beta) w(f(x))]
v(x) \in TM_x and w(f(x)) \in TN_f(x).
The latter fact makes that your Ricci tensor shall have an index in TM and one in TN. To avoid all these problems, it is better to choose v,w \in TN_f(x) :
D(v)w(x) = df D_M [ \beta(v(f(x))) ] \beta w(f(x)) + D_N (v(f(x))) [(1 - df \beta) w(f(x))]
This covariant derivative has no problems at all when df = 0 on \Sigma (it would be just D_N there and D_M outside \Sigma). Now, you can compute the Riemann curvature using orthognal projections on the appriopriate subspaces (I am not going to type out the formulae here, because they are just too lengthy). Given this connection, you can look for a suitable combination of g and h (metric on M and N respectively) which are covariantly constant under the difference connection. I did not work that out but indeed it seems you have freedom of some parameters but this is only relevant on \Sigma no ?? Therefore, my comment towards the grav waves remains.


** Then we we form the first Pontrjagin class p_1(R)=\frac{1}{8\pi^2}Tr(R^a_b\wedge R^c_a) and via the relation e(R)\wedge e(R)=p_1(R) the Euler form. This Euler form is a 2-form and we construct a surface S_R (unique up to diffeomorphism) via the integral \int_{S_R}e(R)=1(deRham theory). This surface is dual to the Euler form. Now we study the geometry of this embedded surface via the embedding F:S_R\hookrightarrow M. **

A small technical question : I am not familiar with the Euler form, but (a) does it always exist and (b) does it determine a unique cohomology class ?

Your further references towards the pontryagin class and chern form are confusing and unneccesary since you are only using the natural tangent bundle of the two surface (you are using elephant terminology to whip a mosquito).

I have still some questions concerning the algebra but this is enough for now.

As you notice: a rigorous treatment does not take more place than what you are using and benefits everyone.

Cheers,

Careful

 

[kneemo]

Dear Mike2,
no, we have not. At this stage the approach can not handel such questions.
(By the way:
My opinion is, that a black hole is a infelicitous defined model (but a solution of the E.eq.) and the epistemological problems are too deep. Thus I am not very interessed in black hole solutions of the E.eq.)

Hi Mike2

In the BPS Black Holes and U-duality thread I mention black holes that are built from the ground up (so to speak), as opposed to strictly hunting down charged black hole solutions of the E. eqs.

These "extreme" black holes have a charge that is equal to their mass, so satisfy a condition for unbroken supersymmetry called the BPS condition. Imagine these black holes as ones that have decayed to their ground state through the emission of Hawking radiation. String theorists, at this point, call them D0-branes.

Now the recipe for constructing BPS black holes that have the E6(6), E7(7), and E8(8) U-duality groups is by starting with a 27-dimensional Jordan algebra. Such a finite dimensional Jordan algebra behaves as a function algebra over a finite set of points--with the number of points usually equal to the dimension of the algebra. However, to find out how many points for sure, we have to solve the eigenvalue problem for the algebra.

The proper way to setup the eigenvalue problem for the 27D Jordan algebra stumped researchers for years (Pascual Jordan and John von Neumann included), until Tevian Dray and Corinne A. Manogue discovered how in 1999 (http://www.arxiv.org/abs/math-ph/9910004" ). They found that when you setup the problem properly, you get 3 real eigenvalues. So the 27D algebra becomes a function algebra over 3 points in the frame work of Jordan quantum mechanics.

The geometric picture emerges from the fact that each eigenvalue has a corresponding "eigenmatrix", and these eigenmatrices are points of a projective space called the Cayley plane. The U-duality groups then emerge as the transformation groups of this projective space (e.g. the collineation, conformal, and quasiconformal groups)--no fine tuning needed.

So to recap, we do some quantum mechanics with a 27D Jordan algebra and find that it behaves as a function algebra over 3 points. These 3 points give a discrete approximation of the Cayley plane, which physically describes a BPS black hole. Even better, string theorists have recently found out how to count microstates of these black holes (see: http://www.arxiv.org/abs/hep-th/0512296" ) by using the norm of the 27D Jordan algebra. How cool is that? ~Mike

p.s. I don't mean to hijack the thread. I just wanted to show there was promise in the geometrization of black hole quantum mechanics.

 

[torsten]

Hi,
There was a misunderstanding, I thought you were making use of the fibration of the 4 D manifold, but obviously you are not doing that at all. You are just using the natural tangent space of the two disk, connecting this somehow to three surfaces \Sigma. Relating to my comment in post 98, I was doing myself yesterday some calculataions concerning how a distributional part in the ricci tensor should arise in case the nucleus of df on \Sigma is three dimensional. Actually, the difference connection I defined there runs into trouble (because of the mixed indices); remember : (D(v)w)(x) = df D_M(v) \beta w(f(x)) + D_N (df(v(x))) [(1 - df \beta) w(f(x))]
v(x) \in TM_x and w(f(x)) \in TN_f(x).
The latter fact makes that your Ricci tensor shall have an index in TM and one in TN. To avoid all these problems, it is better to choose v,w \in TN_f(x) :
D(v)w(x) = df D_M [ \beta(v(f(x))) ] \beta w(f(x)) + D_N (v(f(x))) [(1 - df \beta) w(f(x))]
This covariant derivative has no problems at all when df = 0 on \Sigma (it would be just D_N there and D_M outside \Sigma). Now, you can compute the Riemann curvature using orthognal projections on the appriopriate subspaces (I am not going to type out the formulae here, because they are just too lengthy). Given this connection, you can look for a suitable combination of g and h (metric on M and N respectively) which are covariantly constant under the difference connection. I did not work that out but indeed it seems you have freedom of some parameters but this is only relevant on \Sigma no ?? Therefore, my comment towards the grav waves remains.
** Then we we form the first Pontrjagin class p_1(R)=\frac{1}{8\pi^2}Tr(R^a_b\wedge R^c_a) and via the relation e(R)\wedge e(R)=p_1(R) the Euler form. This Euler form is a 2-form and we construct a surface S_R (unique up to diffeomorphism) via the integral \int_{S_R}e(R)=1(deRham theory). This surface is dual to the Euler form. Now we study the geometry of this embedded surface via the embedding F:S_R\hookrightarrow M. **
A small technical question : I am not familiar with the Euler form, but (a) does it always exist and (b) does it determine a unique cohomology class ?
Your further references towards the pontryagin class and chern form are confusing and unneccesary since you are only using the natural tangent bundle of the two surface (you are using elephant terminology to whip a mosquito).
I have still some questions concerning the algebra but this is enough for now.
As you notice: a rigorous treatment does not take more place than what you are using and benefits everyone.
Cheers,
Careful

only a short answer to your technical questions:
ad (a) the Euler form is always well-defined and is the Pfaffian of the curvature 2-form
ad (b) by a theorem of Thom, the cohomology class corresponding to the Euler form is unique and well-defined

Torsten

 

[Chronos]

Torsten, that is the best answer I have seen. Well done. I am thoroughly impressed.

 

[Mike2]

Dear Mike2,

no, we have not. At this stage the approach can not handel such questions.

(By the way:
My opinion is, that a black hole is a infelicitous defined model (but a solution of the E.eq.) and the epistemological problems are too deep. Thus I am not very interessed in black hole solutions of the E.eq.)

For example, it would seem to me that your efforts are the GR equivalent of the QFT approach to the Unruh effect. What I mean is that it seems that in both case of Diffential Structure and in the Unruh effect particles appear as a result of acceleration, a change in spacetime curvature. In the Unruh effect, radiation is observe for any observer that is accelerating, whether that acceleration is due to linear acceleration, gravitational acceleration, or the acceleration due to the expansion of space. But in the Differential Structure picture, particles also appear as a result of a change in the curvature of spacetime. Isn't this change of curvature also equal to an acceleration? Thanks.

 

[Mike2]

For example, it would seem to me that your efforts are the GR equivalent of the QFT approach to the Unruh effect. What I mean is that it seems that in both case of Diffential Structure and in the Unruh effect particles appear as a result of acceleration, a change in spacetime curvature. In the Unruh effect, radiation is observe for any observer that is accelerating, whether that acceleration is due to linear acceleration, gravitational acceleration, or the acceleration due to the expansion of space. But in the Differential Structure picture, particles also appear as a result of a change in the curvature of spacetime. Isn't this change of curvature also equal to an acceleration? Thanks.

I wonder if your approach is only taking into account spatial curvature? I wonder if bosons could be obtained by your approach if the curvature was obtained from linear acceleration instead. Thanks.

 

[selfAdjoint]

Mike I think their approach curves spacetime, not just space. It's a general GR curvature, identified therefore with a momentum-energy tensor. In general that includes the momentum-energy vector (as first row/column), and it varies from point to point, hence acceleration.

But I don't see the Unruh connection at all. Unruh, at least classically, is defined in flat Minkowski spacetime. And it needs a given quantum vacuum, usually assumed to contain scalar particles.

 

[Mike2]

Mike I think their approach curves spacetime, not just space. It's a general GR curvature, identified therefore with a momentum-energy tensor. In general that includes the momentum-energy vector (as first row/column), and it varies from point to point, hence acceleration.

But I don't see the Unruh connection at all. Unruh, at least classically, is defined in flat Minkowski spacetime. And it needs a given quantum vacuum, usually assumed to contain scalar particles.

I thought, by definition, that acceleration is a curvature of spacetime. If so, then we have two effects that associate particles with curved spacetime, the Unruh effect and the Helge Torsten effect. Am I way off base to wonder if they can be equated?

Not only that, but there seems to be various ways to curve spacetime, gravity, linear acceleration, universal expansion of space, etc. So the natural question is whether these are all the same in the Helge Torsten effect. If not, then perhaps a different kind of curvature might produce the bosons that are missing in their picture so far.

 

[selfAdjoint]

I thought, by definition, that acceleration is a curvature of spacetime. If so, then we have two effects that associate particles with curved spacetime, the Unruh effect and the Helge Torsten effect. Am I way off base to wonder if they can be equated?

Acceleration in spacetime is defined by a curved worldline. This is an intrinsic property of the worldline as a 1-dimensional manifold and does not imply anything about the embedding space. In fact Unruh's mechanism is as I said defined in flat Minkowski spacetime and the curvature of spacetime does not come into it. You seem to be reading more into a broad account of Unruh radiation than is really there.

Not only that, but there seems to be various ways to curve spacetime, gravity, linear acceleration, universal expansion of space, etc. So the natural question is whether these are all the same in the Helge Torsten effect. If not, then perhaps a different kind of curvature might produce the bosons that are missing in their picture so far.

The components of the momentum-energy tensor can be identified with rest-energy, four-momentum, and "stress", or rate of change of four-momentum in the various directions of 3-space (not in the time direction). I don't know where you got the idea that acceleration per se curves spacetime. The expansion of space does curve spacetime over and above this, but on the scale of normal acceleration this is negligable. And what missing bosons are you talking about? This is all classical physics - GR and differentiable structures on manifolds; quantum physics is OT.

 

[Mike2]

Acceleration in spacetime is defined by a curved worldline. This is an intrinsic property of the worldline as a 1-dimensional manifold and does not imply anything about the embedding space. In fact Unruh's mechanism is as I said defined in flat Minkowski spacetime and the curvature of spacetime does not come into it. You seem to be reading more into a broad account of Unruh radiation than is really there.

Perhaps. Most of my discussion is in the form of a question. I haven't mathematically proven anything yet, and I'm trying to develop an intuitive sense of these things. However, according to the equivalence principle, acceleration is equivalent to gravitation which does curve spacetime. So, therefore, a linear accelerated observer must feel himself just as much inside a curved spacetime as much as an observer in a gravitational field, or the equivalence principle does not hold.

And what missing bosons are you talking about? This is all classical physics - GR and differentiable structures on manifolds; quantum physics is OT.

Well, as I recall, Torsten and Helge relate the quantum properties of only fermion to the curvature of spacetime. They admit that bosons are a different issue. Thus my question about different kinds of curvature with different kinds of particles.

 

[selfAdjoint]

Perhaps. Most of my discussion is in the form of a question. I haven't mathematically proven anything yet, and I'm trying to develop an intuitive sense of these things. However, according to the equivalence principle, acceleration is equivalent to gravitation which does curve spacetime. So, therefore, a linear accelerated observer must feel himself just as much inside a curved spacetime as much as an observer in a gravitational field, or the equivalence principle does not hold.

Consider this thought experiment: A rocket is taking off. Let's say it's on the Moon, to avoid atmospheric side issues. The astronoauts are pullling six gees, i.e. they are experiencing an acceleration six times that of Earth's gravity at Earth's surface. There is a little satellite coasting by the rocket as it rises (standing in for a comoving inertial frame). Does the satellite feel a gravitational pull toward the rocket? According to the naive view of the equivalence principle it should, since "acceleration is equivalent to gravitation is equivalent to spacetime curvature" But I don't think you're going to assert that this accelerating frame over here necessarily produces a curvature, and curved geodesics, over there.

Or are you?

 

[Mike2]

Consider this thought experiment: A rocket is taking off. Let's say it's on the Moon, to avoid atmospheric side issues. The astronoauts are pullling six gees, i.e. they are experiencing an acceleration six times that of Earth's gravity at Earth's surface. There is a little satellite coasting by the rocket as it rises (standing in for a comoving inertial frame). Does the satellite feel a gravitational pull toward the rocket? According to the naive view of the equivalence principle it should, since "acceleration is equivalent to gravitation is equivalent to spacetime curvature" But I don't think you're going to assert that this accelerating frame over here necessarily produces a curvature, and curved geodesics, over there.

Or are you?

No, I'm not going to assert that the satalite feels the gravity of the accelerating spaceship 6 times that of earth. As I understand it, and I'm only learning by bits and pieces as they seem to be relevant to me, it would be only those things on the spaceship that are experiencing the acceleration that would precieve themselve to be in a curve spacetime as much as if they were in a gravitational field - that there is something inherent to acceleration that transforms their accelerating frame into the same kind of curvature (from their accelerating perspective) that would be observed for an observer in a gravitational field. Is this much right?

 

[selfAdjoint]

Yes I think that the astronauts (or cosmonauts if you prefer) as long as they ignore second order effects like tides, can regard themselves as being in a curved geometry. Of course since they're sitting on the engine, and worrying if it might fail, the curvature assumption might seem a little, mmmm, purist!

 

[Mike2]

Yes I think that the astronauts (or cosmonauts if you prefer) as long as they ignore second order effects like tides, can regard themselves as being in a curved geometry. Of course since they're sitting on the engine, and worrying if it might fail, the curvature assumption might seem a little, mmmm, purist!

The question is whether this kind of spacetime curvature is the same as the curvature in the Helge-Torsten scheme of things. If not, then does a different kind of particle (a boson?) relate to this different kind of curvatue.

 

[selfAdjoint]

The question is whether this kind of spacetime curvature is the same as the curvature in the Helge-Torsten scheme of things. If not, then does a different kind of particle (a boson?) relate to this different kind of curvatue.

There is only one kind of curvature in general relativity: the kind expressed by the curvature tensor. The Helge-Torsten mechanism is to produce that kind of curvature in the spacetime manifold, which is by the Einstein equations the equivalent of momentum-energy, and hence of matter. The fact that the cosmonauts are entitled to treat their acceleration as due to curvature doesn't mean the curvature is really there in the manifold.

 

[Helge Rosé]

Dear Mike2, dear selfAdjoint

this was an interesting discussion! At the first sight it seems like an easy question about the "Fahrstuhl" experiment and we can give a quick answer. But we discuss about your problem a long time and realizing: it is not straight forward. Ok, what is the problem? There are two reference frames:

RF1 the accelerated rocket (no observations to outside allowed)
RF2 the satellite

In RF1 we feel a gravitational force. In RF2 we see a accelerated body (the rocket) and feel no gravitational force. By the equivalence principle both descriptions are equal: a homogeneous gravitation can be transformed to zero if we change to a accelerated reference frame (RF1 -> RF2).

In RF1 is gravitation and gravitation warps the spacetime - does it mean we get a curved spacetime? And in RF2 is no gravitation - no curved spacetime? Do we have two spacetimes - curved and flat?

The problems comes from the meaning of "curved". If "curved" means curvature - the Riemann tensor - is not zero we get the problem: The Riemann tensor is invariant by diffeomorphisms. How could we change the Riemann tensor to non-zero by the transition RF1->RF2:
1) RF1 -> RF2 is no diffeomorphism
2) the curvature is no tensor

The solution is: RF1 -> RF2 is a diffeomorphism, the curvature is a tensor (the Riemann or Ricci tensor). But "curved" does not mean Riemann tensor non-zero.

In RF2 we have a flat spacetime (Riemann zero). In RF1 Riemann is also zero but we have a homogeneous gravitational force! Here we have to take care what we mean by "gravitation" and "curved":

metric - tensor - gravitational potential
connection - no tensor - gravitational field = force
curvature - tensor - change of the gravitational field

By the diffeomorphic transition RF1 -> RF2 we can not change the invariant curvature (Riemann) of the spacetime, but we change the connection - and this is the force in the homogeneous case. This point is different to gauge theories: here the connection is the potential and the curvature is the force.

In ART we can change the force (connection). We can transform away the gravitation in RF1 by a transition to RF2 - the curvature (Riemann tensor) remains invariant.Now, what does it mean for our approach? If we change the differential structure we do not make a diffeomorphic transition like RF1 -> RF2 - were the Ricci tensor is invariant - we do a non-diffeomorphic transition were the curvature changes - e.g. changing a homogeneous gravitational field to a non-homogeneous one. Such a transition can only be caused by matter or verse visa: such a transition - between two reference frames which are belong to two differential structures - is matter.

the best

Torsten and Helge

 

[Mike2]

Now, what does it mean for our approach? If we change the differential structure we do not make a diffeomorphic transition like RF1 -> RF2 - were the Ricci tensor is invariant - we do a non-diffeomorphic transition were the curvature changes - e.g. changing a homogeneous gravitational field to a non-homogeneous one. Such a transition can only be caused by matter or verse visa: such a transition - between two reference frames which are belong to two differential structures - is matter.

the best

Torsten and Helge

OK... so... where there is acceleration, there is an Unruh temperature (for the accelerated observer) and with it an energy density and thus a mass density (for the accelerated observer) and thus a curvature according to your approach, right?

P.S. It occurs to me that you might object with the claim that the Unruh effect is a different effort than yours. But it seems your entire point is that you can get the Algebra of QM through differential structures, and the Unruh effect is a use of that algebra in accelerating reference frames. So I do think my above syllogism has merit.

 

[Mike2]

OK... so... where there is acceleration, there is an Unruh temperature (for the accelerated observer) and with it an energy density and thus a mass density (for the accelerated observer) and thus a curvature according to your approach, right?

As I understand it, Torsten's efforts only apply to fermions, particles of mass. But I don't understand why it can not apply to bosons as well. Since photons have energy and energy can be equated to mass, what's the difference? What is the key point that restricts Torsten's efforts to fermions? Is it the "support" of a singularity? Thanks.

 

[torsten]

As I understand it, Torsten's efforts only apply to fermions, particles of mass. But I don't understand why it can not apply to bosons as well. Since photons have energy and energy can be equated to mass, what's the difference? What is the key point that restricts Torsten's efforts to fermions? Is it the "support" of a singularity? Thanks.

Dear Mike,

sorry for the long time to answer your question but I was ill again.
Your question is correct and in the current version of the paper we can only state that we get fermions. But more is true...
Consieder a 3-manifold with boundary. If the boundary consists of a single compact surface we get the properties of a fermion. But if the boundary is the connected sum of two surfaces or more then we will get the properties of the bosons. Thus our approach is similar to the Crane approach via conical singularities (see gr-qc/0110060 and gr-qc/0306079).

I hope that helps.

Torsten