This is just to say that the topics of hyperspace and XDs have inspired generations of physicists. And whoever it was who first did the calculation that shows string theory needs extra dimensions to make sense, it must have been one of the most exciting moments I can imagine for a theoretical physicist.

But XDs have come a long way, and were around long before string theory. People sometimes ask me why my talks never mention the earlier works on the topic. The reason is that the theories with XDs proposed in the 1920ies by Theodor Kaluza and Otto Klein, are in their idea different to the 'modern' XDs. Yet, this usually takes too much time to clarify in a talk, so I rather skip it. However, since you - and yes, I mean YOU who you are just raising your eyebrows - are of course the most attentive reader there is, I want to elaborate somewhat on these 'early' XDs since I noticed very little people actually read the original works by Kaluza and Klein.

**General Relativity**

The first mentioning of adding another dimensions to our three space-like dimensions that we experience every day goes to my knowledge back to Nordström in 1913 [2]. He however did not yet use General Relativity (GR) to build his theory upon. Since we know today that the gravitational potential is not a scalar field, but described by the curvature of space time, let us skip to the next attempt which uses GR as we know it today.

GR couples the metric tensor (

*g*) to a source term of matter fields, whose characteristics are encoded in the stress-energy tensor of the matter. All kind of energy and matter results in such a source term, and hence causes the metric to deviate from flat space. This theory does not say anything about the origin of the source terms. The matter and its properties have to be described by another theory - for example by electrodynamics. Electrodynamics on the other hand has a similar problem. The source for the electromagnetic field (charged particles) is not described by Maxwell's equations [5]. They need to be completed by further equations, e.g. the Dirac equations.

In the beginning of the last century, physicists had just understood gravity as a geometrical effect instead of a field in Minkowski space, so it was only natural to try the same for other fields as well, with the obvious next choice being the electric field. The idea of the early XDs is plain and simple. Einstein's field equations are a set of non-linear differential equations for the metric tensor. They are built up of the Ricci-tensor (two indices) which is a contraction over the full curvature tensor (four indices), and the curvature scalar - a further contraction of the Ricci-tensor. Such a contraction is basically a sum over two indices. The indices on these tensors label space-time directions - that is, in the standard case of GR with three space and one time dimensions, they run from 1 to 4 (or, depending on taste, sometimes from 0 to 3).

Now if one had an additional dimension, then two things happen with Einstein's field equation. First, one has more equations because there are more free indices. Since the Ricci tensor and the metric are symmetric, the number of independent equations is

*D*(

*D*+1)/2, here

*D*is the total number of dimensions. The second thing happening is that the equations with the indices belonging to the 'usual' directions acquire additional terms since the sum runs over the additional indices as well. The trick is then to separate the usual part (sum from 1 to 4) from the additional part (sum over the extra dimension), shift the additional part to the other side of the equations, and read it as a source term. In such a way, one obtains a source term

*even if the higher dimensional field equations were source free.*

**Kaluza and Klein**

The result is that components of the higher-dimensional metric tensor appear as source terms for the four-dimensional sub-sector that we observe. The first such approach was Theodor Kaluza's whose ansatz uses one additional dimension. In the remaining entries of the metric tensor (those with one index being a 5) he put the electromagnetic potential with a coupling constant alpha (since the metric tensor is dimensionless but the electromagnetic potential isn't)

(Here, the large Latin indices run over all dimensions, the small Greek indices over the usual four dimensions). Kaluza apparently sent a draft of his paper to Einstein in 1919, to ask for his opinion. It got published with a delay of two years [3].

Kaluza derived the higher dimensional field equations in the linear approximation. Generically, all the components of the metric tensor will be functions of all coordinates, including the additional one. This however is in conflict with what we observe. Kaluza therefore added what he called the 'cylinder condition' that set derivatives with respect to the additional coordinates to zero. In the linear approximation, he then found the ansatz to reproduce GR plus electrodynamics.

However, the use of this linear approximation is not necessary, as was shown by Oskar Klein five years later [4]. Klein used a different ansatz for the metric which has an additional quadratic term:

(Sorry, coupling constant is missing, my fault not Oskar's) And he assumed the additional coordinate is compactified on a circle. Then, one can expand all components in a Fourier-series and the zero mode will fulfill Kaluza's cylinder condition' that is, it is independent of the fifth coordinate. However, if you compare both ansätze [7], Klein's and Kaluza's, you will notice that Klein set the

*g*

_{55}component to be constant to one. This is an additional constraint that generally will not be fulfilled. In fact, the additional entry behaves like a scalar field and describes something like the radius of the XD. At this time however, people had little for additional scalar fields.

Klein's derivation is simply one of the most beautiful calculations I know. One just writes down the higher dimensional field equations, parametrizes the metric tensor according to Klein's ansatz, decomposes the equations - and what comes out is GR in four dimensions (in the Lagrangian formulation as well as the field equations), plus the free Maxwell equations.

(Here, the supscript (4) and (5) refer to the 4 and 5 dimensional part of the curvature/metric). Further, the geodesic equation gets an additional term which is just the Lorentz force term and thus describes a charged particle moving in a curved space with an electromagnetic field.

In the course of this derivation, one is lead to identify the momentum in the direction of the fifth coordinate as the ratio of charge over mass (*q/m*). It can be shown that this quantity is conserved as it should be. Klein concluded that this charge is quantized in discrete steps (this is a geometrical quantization), the first example of the Kaluza-Klein tower.

**Extensions and Problems**

To understand the excitement this derivation must have caused one has to keep in mind that this was 30 years before Yang and Mills, and the understanding of gauge theory was not on today's status. With today's knowledge, the argumentation appears somewhat trivial. One adds an additional dimension with U(1) symmetry, the compactified dimension. The resulting theory needs to show this symmetry that we know belongs to electrodynamics. From this point of view, it is only consequential to extend the Kaluza-Klein (KK) approach to other gauge symmetries, i.e. non-abelian groups. This was done in 1968 [6].

One has to note however that for non-abelian groups the curvature of the additional dimensions will not vanish, thus flat space is no longer a solution to the field equations. However, it turns out that the number of additional dimensions one needs for the gauge symmetries of the Standard Model U(1)xSU(2)xSU(3) is 1+2+4=7 [10]. Thus, together with our usual four dimensions, the total number of dimensions is

*D*=11. Now exponentiate this finding by the fact that 11 is the favourite number for those working on supergravity, and you'll understand why KK was dealt as a hot canditate for unification.

But there are several problems with the traditional KK approach. First, meanwhile the age of quantum field theory had begun, and all these considerations have been purely classical and unquantized. Even more importantly, there are no fermions is this description - note that we have only talked about the free Maxwell equations. The reason is easy to see: fermions are spin 1/2 fields and unlike vector bosons one can not just write them into the metric tensor. One can of course add additional source terms, but this makes the idea somewhat less appealing [8]. The high hope had been to explain all matter and fields from a purely geometric approach.

If one thinks more about the fermions, one notices another problem: right- and left-handed fermions belong to different electroweak representations, a feature that is hard to include in a geometrical interpretation. Furthermore, there is the problem of stabilization of the compact extra dimensions (the sizes should not or only negligibly depend on the time-like coordinate), and the problem of singularity formation from GR persists in this approach. However. If I consider what landscape of problems other theories suffer from, it makes me wonder why the KK approach was so suddenly given up in the early 70ies. A big part of the reason might simply have been that the quark model got established, and it was the dawn of the particle-physics era.

The 'modern' extra dimensions differ from the KK approach by not attempting to explain the other standard model fields as components of the metric. Instead, fermionic- and gauge-fields are additional fields that are coupled to the metric. They are allowed to propagate into the extra dimensions, but are not themselves geometrical objects. Most features of the KK approach remain, most notably the geometrical quantization of the momenta into the extra dimensions and thus the KK-tower of excitations. So remains the problem of stabilization, singularities and quantization (for higher dimensional quantum field theories the coupling constants become dimensionful). However, for me this 'modern' approach is considerably less appealing as one has lost the possibility to describe gauge symmetries and standard model charges as arising from the same principle as GR.

But obviously, the largest problem with the KK approaches was - and still is - that it is not clear whether it is just a mathematical possibility or indeed a description of reality. As Oskar Klein put it in 1926:

*"Ob hinter diesen Andeutungen von Möglichkeiten etwas Wirkliches besteht, muss natürlich die Zukunft entscheiden."*

*"Whether these indications of possibilities are built on reality has of course to be decided by the future."*[9]

**References:**

[1] And if you read the wikipedia entry on Perry Rhodan you find a use for the word 'Zeitgeist'...

[2] G. Nordström,

*"Zur Theorie der Gravitation vom Standpunkt des Relativitätsprinzips"*Annalen der Physik, vol. 347, Issue 13, pp.533-554 (1913); G. Nordström,

*"Über die Möglichkeit, das elektromagnetische Feld und das Gravitationsfeld zu vereinigen"*(

*"About the possibility to unify the electric field and the gravitational field"*) Physik. Zs. 15, 504-506 (1914) [Abstract]

[3] T. Kaluza,

*"Zum Unitätsproblem der Physik'' ("On the Problem of Unity in Physics*, Sitzungsber. Preuss. Akad. Wiss. Berlin (Math. Phys.) (1921) 966.

[4] O. Klein,

*"Quantentheorie und fünfdimensionale Relativitätstheory"*(

*"Quantum Theory and fivedimensional General Relativity",*Z. Phys. 37, 895 (1926).

[5] Unlike to what the Wikipedia entry states, the Lorentz force law can not be derived from the Maxwell equations without further assumptions (like a Lagrangian for the coupled sources). E.g. Maxwell's equations are perfectly consistent for a

**static**superposition of two negativley charged objects, just that we know the charged particles would repel and the configuration can't be static.

[6] R. Kerner,

*"Generalization of the Kaluza-Klein theory for an arbitrary non-abelian gauge group"*, Ann. Inst. Henri Poincare, 9, 143-152 (1968)

[7] Contrary to the wide spread believe, the plural of the German word

*'Ansatz'*is not

*'Ansatzes'*but

*'Ansätze'*(pronounced 'unsetze').

*'Ansatz'*could be roughly translated as 'a good point to start', or a preparation. E.g. the pre-stage for yeast dough is called

*'Ansatz'*...

[8] Which finally brings us to the topic on which I lost two years during my Ph.D. time, namely the question whether one can built up the metric tensor from spin 1/2 fields. I only learned considerably later that most of this approach had been worked out in the mid 1980ies, see e.g. hep-th/0307109 and references therein.

[9] He indeed writes it has to be decided 'by' the future not 'in' the future. Quotation from Ref. [4]

**Further Litarature**

- Physics Notes on Kaluza Klein theories by Viktor Toth
- Overduin and Wesson,
*Kaluza Klein Theory,*gr-qc/9805018 - Van Dongen,
*Einstein and the Kaluza Klein particle*, gr-qc/0009087 - Wikepedia on Kaluza-Klein theories
- Jim Adson,
*Kaluza, Klein, and the Grand Unified Theory* - Häggblad,
*Kaluza Klein Theory* - Van de Schaar,
*Kaluza Klein Theory* - H. Goenner,
*Kaluza's five dimensional unification*(from this living review)

GR's stress-energy tensor is symmetric. It disallows relativistic spin-orbit coupling. GR may measurably fail in binary pulsar PSR J0737-3039A/B, requiring some 20 years of observation,

ReplyDeleteMatters of Gravity, #29Affine, teleparallel, noncommutative gravitations consistent with spin-orbit coupling have chiral vacuum backgrounds. Tests with achiral mass distributions are inert (socks fitted to a left foot). Two days of calorimetry obtain 3x10^(-18) sensitivity (left and right shoes fitted to a left foot). Which is better, 7300 days or 2 days?

Dear Bee,

ReplyDeletethank you for this beautiful exposition!

Do you know why Kaluza calls this condition on the fifth coordinate "cylinder condition"? After all, in his paper the extra dimension is not yet wrapped to a cylinder?

Klein's derivation is simply one of the most beautiful calculations I know.Hm, I guess I should study that paper some day... By the way, it is available online at DOI: 10.1007/BF01397481 - you probably need some institutional access, but at least in Frankfurt, it works.

For the Kaluza and Nordström papers, you unfortunately have to search for the dusty old volumes in the library...

By the way, this Oskar Klein is the very same guy of the Klein Paradox - sorry, unavoidable self-promotion ;-)

Best, stefan

I've read most of these papers in German which I didn't understand too well - but anyway, the main tool of communication were the equations. ;-)

ReplyDeleteThanks for having chastised Clifford for his PC Nazism, and enjoy Rome etc.

Hey, nice historical introduction to KK theory. I played around with it a bit too -- it really is a beautiful unification idea. At one point, I tried allowing torsion. If you do this, you can get the standard model gauge fields in a cool way. Specifically, if we add the compact symmetric space CP2 = SU(3)/U(2) as four XD, then we get the usual SU(3) KK gauge fields in the metric (or vielbein), and we get U(2)=SU(2)xU(1) gauge fields in the spin connection (as the contorsion). Only problem is we have to add a dynamical torsion term to the Lagrangian. Since the electroweak fields are in the connection, it's much easier to build chiral spinors.

ReplyDeleteThis is good fun, but I've abandoned it in favor of what I'm working on now, which is getting gravity and spinors from gauge theory. :)

Hi Garrett:

ReplyDeleteYeah, to me there has always been the question whether it's more promising to understand gravity as a gauge theory or whether the way to go is to understand the geometry of gauge theories. I still can't make up my mind.

Dear Lubos:

I certainly didn't mean to upset Clifford. I just think one has to keep in mind that we are all human, not everything we state does always make sense, and I don't think its good that the media amplifies mistakes by making them seem more important than they are. What's the outcome of that? At some point everybody will be too scared to just talk without having a carefully prepared script that says nothing than what we are supposed to hear. Anyway, I probably shouldn't have commented on that topic without having heard that specific interview. I just thought things like this generally happen far too often to be healthy for an open exchange of opinions - whether or not that NASA person said something silly.

(and of course PI stands for Politically Incorrect)

Dear Stefan:

No, I don't know. As far as I can recall, he doesn't explain that in the paper. It's always been a mystery to me, I wonder if he already had in mind that it could be compactified. (Btw, sorry for not using more of the references you sent, but since the server was down I couldn't access them.)

Best,

B.

Lubos: Yes, one can probably understand the derivations from the equations, but one misses some of the conclusions the authors made, or sentences like the one I cited above about the 'reality' of these possibilities. I forgot to mention there's a Frontiers of Physics edition 'Modern Kaluza Klein theories' that I think has English translations of all the original papers (and is actually quite interesting). Best,

ReplyDeleteB.

Don't forget the extra-dimensional art !

ReplyDeleteReplied to you, Bee, on asymptotia. This was not some questionable statement, this is a matter of policy.

ReplyDeleteI also suggest you talk to a government scientist or two, in NASA, or EPA. They'll tell you, if they're honest, what they really think of say, Al Gore's "An Inconvenient Truth", and then tell you that as a government scientist they're not supposed to tell you that - they can get into trouble for talking to you, and saying, e.g., global warming is real.

I find it highly ironic that one worries about whether statement of stupid NASA policy is merely a "questionable statement", but not about the very real muzzling of scientists in govt. employ.

Good post, Bee. I did my thesis on classical GR in the early 70s and this brings back memories of long calulations of metrics and Ricci tensors, and also using differential form calculations. It is quite beautiful.

ReplyDeleteBTW I managed a short post on Asymptotia briefly supporting Griffin---we shall see if it survives as the last time I tried to post there, it didnt appear ( could be my fault.)

"Yeah, to me there has always been the question whether it's more promising to understand gravity as a gauge theory or whether the way to go is to understand the geometry of gauge theories. I still can't make up my mind."

ReplyDeleteBoth!

I've recently traveled this same path. Gauge theories involve algebra, which is often kind of messy and seems unnatural at first. GR, in contrast, is about manifolds and vector fields -- pretty geometry. The appeal of KK theory is that it allows the messy algebra of gauge theories to be described by the geometry of specific GR solutions. It's an elegant description. But gauge theory has come a long way, and the modern mathematical description of gauge theory is now purely geometric! In fact, it's MORE purely geometric than GR. GR, you see, requires a metric, which is kind of algebraic. In modern (Ehresmann) gauge theory, a Lie group is described by a manifold with a bunch of vector fields on it, corresponding to the generators. This manifold naturally gets a metric, the Killing form, from the Lie algebra of the generators -- which is just the set of Lie derivatives between vector fields. So the metric is emergent in the theory, rather than put in a priori. (This is why I can say it's more purely geometric than GR.) The gauge theory "entire space" is then locally the product of some base manifold with this Lie group manifold, and the connection is a map between vector fields living on this big manifold -- and the vector fields really just describe diffeomorphisms. And, as you alluded to, it's possible to cook up a gauge theory that describes gravity.

After a long relationship with KK, I currently think this "pure geometry" of gauge theory is the way nature works. Especially since I've been able to cook up a gauge theory that includes all standard model fields and gravity. :)

"Sorry, Bee, I don’t think this is just some one questionable statement. This is like Dick Cheney approving of torture, or Abu Gonzales, the Attorney General saying that there is no right to habeas corpus in the Constitution."-Arun

ReplyDeleteArun: Get a grip! This belongs in "Extraordinary Popular Delusions and the Madness of Crowds".

"However, it turns out that the number of additional dimensions one needs for the gauge symmetries of the Standard Model U(1)xSU(2)xSU(3) is 1+2+4=7 (the dimensionality of the groups). "

ReplyDeleteI'm sorry but:

dim(U(1))=1

dim(SU(2))=4-1=3

dim(SU(3))=9-1=8

dim(U(1)xSU(2)xSU(3))=1+3+8=12

which is equal to the number of gauge bosons in the SM

Best,

K

Yo, anonymoose, give Bee some credit!

ReplyDeleteThe dimensions of the Lie groups are as you list. But for KK theory you don't have to tack on Lie groups as XD (although you can). You only have to tack on symmetric spaces, which are Einstein manifolds having the same symmetries (Killing vector fields) as the Lie groups. The smallest symmetric spaces having U(1), SU(2), and SU(3) as symmetries are indeed of dimension 1,2, and 4.

Kaluza-Klein in old unified field theories

ReplyDeleteHi Garrett, Hi Anonymous K,

ReplyDeleteyes, sorry. When I wrote the sentence I thought its not the dimensionality of the groups, I will have to reformulate that properly but then I forgot. Thanks to Garrett for clarifying this, I've put a footnote in the text.

Best,

B.

Dear Arun:

ReplyDeleteI am afraid I didn't make myself clear with what I meant to express. I don't approve of important people stating stupidities, but I'm saying that echoing it in the media back and forth makes it worse instead of better. Don't you see that many people use exactly this tactic to get attention (that attention not necessarily for higher purposes)? It doesn't matter whether their statements make sense or not or whether they actually mean it - as long as the media picks it up and they get attention. That is what worries me because it makes these voices even more important, and the opinion making process even more a game of vanities and pretences.

They'll tell you, if they're honest, what they really think of say, Al Gore's "An Inconvenient Truth", and then tell you that as a government scientist they're not supposed to tell you that - they can get into trouble for talking to you, and saying, e.g., global warming is real.

I find it highly ironic that one worries about whether statement of stupid NASA policy is merely a "questionable statement", but not about the very real muzzling of scientists in govt. employ.

Yes. That's the problem. See, people who work in an organization always have their own opinions that might or might not agree with the majority. There's nothing wrong with that, unless they have to be afraid of having that opinion. That is when things start going wrong. If I'm in Germany and an American tourist tells me, I'm an US citizen, my country leads a war in Iraq but I don't approve that's fine with me. If the American tourist would say I'm an US citizen, my country leads a war, I have an opinion but telling it would bring me in trouble, then I'd be very very concerned.

I agree with Clifford and you that there is the very realistic possibility that what that guy said does indeed reflect more than his opinion and that it therefore is a reason for concern. But so far this is a mere speculation (and I don't know enough about NASA to comment on that). Maybe this is just a reflection of my way to check the arxiv. I just try to ignore the nonsense.

Best,

B.

"The truthiness of string theory leaves one in a state of permanent confuzzlement" ????

ReplyDeleteBee, is there a neat geometric view of spontaneous symmetry breaking?

ReplyDeleteHey,

ReplyDeleteWasnt Minkowski after all the *first* to introduce 'XDs' in 1907?? (Einstein's teacher :) )

And probably lots of people before him, Riemann and Gauss perhaps had the same idea...

That, to me, is by far the most revolutionary step. Treating time as a dimension, in the same footing as space!!

God, thanks for the "enlightenment" (if you understand what I mean ;) )

...Well, it is true it is not *exactly* in the same footing, but let's not dwell into minus signs, we can always go Euclidean...we like Euler too ;)

ReplyDelete@ntlworld.comHi Bee, if everything under the Sun, and everything under the microscope (even electronmicroscope) is in our 3D+T

ReplyDeletehow far do we have to 'magnify' objects to SEE these xtra dimensions, and when we see them - will they not then be in our 3D+T.

If these xtra dimensions are acting on our world of 3D+T, just like earth's gravity acts on the Moon, and the Moon acts on Earth's tides, whilst they both spin around the Sun, in how many dimensions are we moving? before we even start to include what the planets, the Sun, the Solar system and the Milky Way are moving in.

"One has to note however that for non-abelian groups the curvature of the additional dimensions will not vanish, thus flat space is no longer a solution to the field equations. However, it turns out that the number of additional dimensions one needs for the gauge symmetries of the Standard Model U(1)xSU(2)xSU(3) is 1+2+4=7 [10]. Thus, together with our usual four dimensions, the total number of dimensions is D=11. Now exponentiate this finding by the fact that 11 is the favourite number for those working on supergravity, and you'll understand why KK was dealt as a hot canditate for unification."

ReplyDeleteThanks for a nice brief summary of the mainstream KK and supergravity idea, but that vital reference [10] in the test doesn't occur in your list of references, which only goes to reference [9]. It's maybe interesting that the one successful, peer-reviewed and published alternative to KK was immediately censored off arxiv without explanation.

The usual assumption of only one time-like dimension is a bit crazy! According to spacetime, distances can be described by time.

Hence, the 3 orthagonal spatial dimensions can be represented by 3 orthagonal time dimensions.

Back in 1916 when general relativity was formulated, this was no use because the universe was supposed to be static universe, but with 3 continuoysly expanding dimensions in the big bang (there's no gravitational deceleration observable), it makes sense to relate those expanding dimensions to time, and distinguish them from the 3 spatial dimensions of matter which don't expand because the forces holding them together are very strong. Indeed, matter or energy generally is contracted spatially in gravitation. Feynman calculated that Earth suffers a radial contraction of GM/(3c^2) = 1.5 millimetre. That really shows that spatial dimensions describing matter should be distinguished from those describing the continuous expansion of the universe. There's no overlap, you just have 3 overall dimensions split into two parts: those that describe contractable matter and those which describe expanding space.

The one successful, peer-reviewed and published alternative to KK predicts that the cosmological constant is zero, in agreement with observations that the universe isn't decelerating because at big redshifts (over large distances) the gravitational coupling constant falls: vector boson radiation exchanged between gravitational charges (masses) will be redshifted (depleted in energy when received) so there's no effective gravity at high redshifts. That's why there's no deceleration of the universe:

Nobel Laureate Phil Anderson points out:

‘... the flat universe is just not decelerating, it isn’t really accelerating ...’

- http://cosmicvariance.com/2006/01/03/danger-phil-anderson/#comment-10901

Hence the real cc is zero. Moreover, this scheme predicts G.

Take the expanding dimensions to be time dimensions. Then the Hubble constant is not velocity/distance but velocity/time.

This gives outward acceleration for the matter in the universe, resulting in outward force:

F = ma

= m*dv/dt,

and since dR/dt = v = HR, it follows that dt = dR/(HR), so

F = m*d(HR)/{dR/(HR)}]

= m*(H^2)R.dR/dR

= mRH^2

By Newton's 3rd law you get equal unward force - which is carried by gravitons that cause compression and curvature - and this quantitatively allows you to explain general relativity and to work out the gravitational constant G, which turns out to be correct within the experimental error of the data like Hubble constant and density you put into relationship.

It's kind of funny that falsifiable, easily checkable work based on observed facts and extending the applications of Newton's 2nd and 3rd laws to gauge boson radiation, is so ignored by the mainstream.

Even if my mechanism and predictions are ignorable (because for instance, they were only rough calculations at first and only published in the journal Electronics World), you'd think arXiv would have taken Lunsford's highly technical paper seriously as he was a student of David Finklestein and he published his paper in International Journal of Theoretical Physics, Volume 43, Number 1, January 2004 , pp. 161-177(17).

A quick historical note: you credit Kerner with the non-Abelian generalization of Kaluza-Klein. But this was already known to Bryce DeWitt five years earlier. It appears as problem 77 in his 1963 Les Houches lectures (published in _Relativity, Groups and Topology_) -- with the proof left to the reader -- along with a short discussion of the problem of "rigidity" of the metric in the extra dimensions and a mention of the importance of the topology of those dimensions.

ReplyDeleteHi Steve,

ReplyDeleteThanks for pointing this out!

Hi Nige,

The footnote [10] is a link, you might have heard of these. Bring your cursor over it, and click the left mouse button. It should bring you to Garrett's comment above. I fail to see though why this point is 'vital'. It's just curious that the result is also 11. I admit I can't exactly recall the construction of these spaces but its a group theoretical thing (probably Garrett knows).

I have sympathized with the idea of having 3 time like dimensions as well, but eventually gave up because I couldn't make sense out of it. Does a particle move on a trajectory in all 6 dimension, and if so what happens to two particles that have a distances in time but not in space?

Best,

B.

Dear Quasar,

ReplyDeleteIn the 'modern' XD approach 'we' are moving only in the usual 3 dimension. I presume with 'we' you mean standard stuff like quarks, leptons etc. Size of 'large' extra dimension can potentially be as 'large' as 1/100 mm. In the more general setting (including KK as above), one expects XDs to be much smaller ~ 10^-33 cm and we'd 'move' into all of them but it's hard to notice. Takes an enormous amount of energy to resolve such small distances. Best,

B.

'I have sympathized with the idea of having 3 time like dimensions as well, but eventually gave up because I couldn't make sense out of it. Does a particle move on a trajectory in all 6 dimension, and if so what happens to two particles that have a distances in time but not in space?'

ReplyDeleteBee, thanks for giving me the opportunity to explain a bit more. The particle only moves in 3 dimensions (the curvature that corresponds to an effective extra time dimension could have a physical explanation in quantum gravity; curvature is a mathematical convenience not the whole story).

The particle doesn't 'move' in any time dimension except on a graph of a spatial dimension plotted against time. Since there are 3 spatial dimensions, the simplest model is to have 3 corresponding time dimensions.

The assumption that there's only one time dimension is the same as you would have if everything around you was radially symmetric, like living in the middle of an onion, where only changes as a function of radial distance (one effective dimension) are readily apparent: the Hubble constant, and hence the age of the universe (time)is independent of the direction you are looking in.

If the Hubble constant varied with direction then t = 1/H would vary with direction, and we'd need more than one effective time dimension to describe the universe.

The role of time as due to expansion is proved by the following:

Time requires organized work, e.g., a clock mechanism, or any other regular phenomena.

If the universe was static rather than expanding, then it would reach thermal equilibrium (heat death) so there would be no heat sinks, and no way to do organized work. All energy would be disorganised and useless.

Because it isn’t static but continuously expanding, thermal radiation received is always red-shifted, preventing thermal equilibrium being attained between received and emitted radiation, and thereby ensuring that there is always a heat sink available, so work is possible.

Hence time is dependent upon the expansion of the universe.

You can't really measure 'distance' by using photons because things can move further apart (or together) while the photons are in transit.

All cosmological dimensions should - and usually are - measured as time. That's why Herman Minkowski said in 1908:

‘The views of space and time which I wish to lay before you have sprung from the soil of experimental physics, and therein lies their strength. They are radical. Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality.’

It's remarkable that as soon as you start thinking like this, you see that the recession of matter by Hubble's empirical law

v = HRat distanceRis velocityv = dR/dt,sodt = dR/v,which can be put straight into the definition of acceleration,a = dv/dtgiving usa = dv/dt = dv/[dR/v] = v*dv/dR = [HR]*d[HR]/dR = RH^2.So there

really is an outward acceleration of the matter in the universe, which means we are forced to try to apply Newton's 2nd (F=ma) and 3rd (equal and opposite reaction) laws of motion to this outward acceleration of matter. This process quantitatively predicts gravity as mentioned, and other stuff too. Why is this physics being censored out of arxiv in place of stringy guesswork stuff which isn't based on empirically confirmed laws but on a belief in unobservables, and doesn't predict a thing? It's sad in some ways but really funny in others. ;-)Hi Nige,

ReplyDeleteI am afraid you don't understand my question. I, you, and everybody we know experience only one time direction. If you like you can interpret this as a timelike line element in a geometry with more than one timelike dimensions. How do you reproduce what we observe?

If the Hubble constant varied with direction then t = 1/H would vary with direction, and we'd need more than one effective time dimension to describe the universe.This is exactly the problem, because it's definitely not what we observe. Everyone of us, every measurement makes gives the same H(t).

Plus, I'd like to know how you get stable objects like planets - another problem I stumbled across. You can of course make additional symmetry requirements that essentially are a cylinder symmetry for the time like subspace but that's nothing else than reproducing what we already have. To make that point clear: write down Einstein's field equations in Vacuum, spherical symmetric, solve them. In 3+1 you find Schwarzschild, what do you find? (There was a paper about that recently, will try to see if I find that, I recall they also had an additional symmetry requirement that I disliked).

The assumption that there's only one time dimension is the same as you would have if everything around you was radially symmetric, like living in the middle of an onion, where only changes as a function of radial distance (one effective dimension) are readily apparent: the Hubble constant, and hence the age of the universe (time)is independent of the direction you are looking in.Just not true. Objects like particles have to be stable, i.e. independent over at least one timelike dimensions, otherwise your theory doesn't make sense. You don't need spherical symmetry, arount a 'point' (i.e. one instant) in time, you need at least cylinder symmetry. If you have spherical symmetry what you'd expect for a point charge is that it 'falls off' in time as it does in space (i.e. with 1/|t|) which is in conflict with reality.

Best,

B.

"This is exactly the problem, because it's definitely not what we observe. Everyone of us, every measurement makes gives the same H(t).

ReplyDelete"Plus, I'd like to know how you get stable objects like planets - another problem I stumbled across. ... To make that point clear: write down Einstein's field equations in Vacuum, spherical symmetric, solve them. In 3+1 you find Schwarzschild, what do you find? "

1. Spacetime means that for each observed spatial dimension characterized by length element dx, there's a time-like equivalent, dt = dx/c.

2. We observe 3 spatial dimensions, so we have 3 time-like corresponding dimensions:

dt_x = dx/c

dt_y = dy/c

dt_z = dz/c

The fact that time flows at the same rate is here indicated by the fact that c is a constant and doesn't depend on the direction.

Similarly, for situations of spherical symmetry, where r is radius

dx = dr

dy = dr

dz = dr

So we can represent all three spatial dimensions by the same thing, dr. This greatly simplifies the treatment of spherically symmetric things like stars. It's just the same with time.

3. Einstein's 3+1 d general relativity uses the fact that time dimensions are all similar, so all 3 time dimensions can be represented by 1 time dimension in general relativity.

The Riemann metric of Minkowski spacetime is normally (in 3+1 d):

ds^2 = {Eta}(dx^2) = dx^2 + dy^2 + dz^2 – d(ct)^2

The correct full 3+3 d spacetime Riemann metric will be just the same because this is just a Pythagorean sum of line elements in three orthagonal spatial dimensions with the resultant equal to ct.

4. It's the same for the Schwarzschild metric of general relativity. Two time dimensions are conveniently ignored because they are the same, and it's more economical to treat 3 time dimensions as 1 time dimension.

The splitting of 3 spatial dimensions into 6 dimensions isn't a real increase in the number of dimensions, just in the treatment.

It's just more convenient to consider the parts of the 3 dimensions which are within matter (like a planet or a ruler or measuring rod) as distance-like, contractable and not expanding, while the parts of dimensions where distance is uniformly increasing are time-like due to the expansion of the universe.

Obviously, the framework of general relativity for the Schwarzschild metric isn't affected by this. It's only adding to general relativity a mechanism for unification and for making additional predictions about the strength of gravity, other cosmological implications, etc.

Thanks for your interest.

I like extra dimensions too, but for additional reasons. ;-)

ReplyDeleteIt's kind of weird that the other night I was thinking about extra dimensions and how they have become "real" in my lifetime, but when I was a child I had some kooky aunts who talked about all the different "planes" of existence and other 'spiritual' stuff. I know most scientists cringe when kooks talk about that stuff, but you have to admit that it's interesting that the kooks have been predicting extra dimensions for a long time before the scientists took them seriously.

For footnote [10] on the geometry of XD, a decent link would be to this review paper on the arxiv:

ReplyDeletehttp://xxx.lanl.gov/abs/hep-th/9912277

Castellani's exposition is quite good, and thorough.

Rae Ann:

Having traveled the ethereal social circles of Maui, I can assure you that anything new age philosophy gets right about physics is purely accidental. ;)

Nige:

The main problem I see with extra timelike dimensions is tachyons. When we write down a massless wave equation in "extra spatial" KK theory, and find its solutions, there are some wave solutions that just go around in circles -- acting like a zoo of identical massive particles with large masses. This tower of massive particles is hard enough to explain away, but not completely implausible. But for every KK theory one can build with extra timelike dimensions, such as you are proposing, some wave solutions travel in space with speed faster than c (tachyons) which we just don't see. People (humorously called "two timers") have struggled to make this work, but it's a hard problem to make go away.

Garrett, Lubos Motl previously dismissed Lunsford's SO(3,3) unification on the basis that time-like dimensions might form circles and violate causality. However, the time-like dimensions of spacetime are orthagonal just like the three large spatial dimensions.

ReplyDeleteIt's pretty easy to list failures that people in the past have had, and that simply aren't applicable.

However, giving a list of objections without backing them up is helpful if you or someone reading is biased. There is evidence that all observed fermion masses can be built up from a simple quantized mass of similar to the Z_0 mass of 91 GeV. Whether you actually find experimentally confirmed evidence based on a physical mechanism convincing is your own choice.

How can the spacetime correspondence explained above, where dt = (1/c)dx, etc., give rise to wave solutions that go faster than c? It doesn't look as if that is the case here. Perhaps it occurs with incorrect use of extra time dimensions? Thanks.

Hi Nige,

ReplyDeleteYour point 1 is your hypothesis that should result in predictions that can be studied/falsified. Your point 2 is just wrong, in case of spherical symmetry it is not dx=dy=dz=dr but

dr^2 = dx^2 + dy^2 + dz^2 (provided that your x,y,z are euclidean coordinates). Your point 3 doesn't make sense. If you are claiming your theory with 3 timelike dimensions is set up just such that it reproduces exactly the one with only one timelike dimensions, then there's no reason to introduce it in the first place.

I was about to say the same as Garrett, that's why I was asking in which dimensions your particles are allowed to move. Unfortunately, you haven't answered my question. Your answer to Garrett's comment doesn't make sense either.

However, the time-like dimensions of spacetime are orthagonal just like the three large spatial dimensions.The timelike directions being orthogonal doesn't forbid closed timelike curves - the spacelike dimensions being orthogonal doesn't forbid closed spacelike curves either. Best,

B.

Hi Gordon,

ReplyDeletecouldn't find your comment at Asymptotia, but I can't say I'm too sad about it. I really don't wish to blow this up. I have a general problem with the mass media and authorities, maybe that explains my attitude. It's got nothing to do with the NASA, Griffin, or specifically what he said. Best,

B.

This comment has been removed by the author.

ReplyDelete"Your point 2 is just wrong, in case of spherical symmetry it is not dx=dy=dz=dr but

ReplyDeletedr^2 = dx^2 + dy^2 + dz^2 (provided that your x,y,z are euclidean coordinates)." -Bee

Nope, x, y and z are just orthagonal directions, so the definition of spherical symmetry a variation of any function F with respect to any of them is identical.

(dF_x)/dx = (dF_y)/dy = (dF_z)/dz

here F_x = F_y = F_z and dx = dy = dx = dr, where r is radius.

Radius is the same in directions x, y, and z, because, as stated, we're dealing with spherical symmetry. The expansion rate and Hubble constant don't vary with direction.

As I mentioned before, if you're modelling a star, you can do it by taking spherical symmetry and just considering radial dimension r, which is equal to dimensions x, y and z, because of spherical symmetry.

Suppose the star's radius is a million miles. This will be the same in the x, the y, and the z directions, and all are equal to the radius r.

Similarly by spherical symmetry,

(d^n)(F_x)/dx^n = (d^n)(F_y)/dy^n = (d^n)(F_z)/dz^n

where n is 2 for your example (Pythagorean sum).

Here, F_x = F_y = F_x = F_r

because of radial symmetry, which, because (d^n)(F_x)/dx^n = (d^n)(F_y)/dy^n = (d^n)(F_z)/dz^n, implies:

dx^n = dy^n = dz^n = dr^n

where r is radial dimension.

The dimensions x, y and z and all resulting elements are indistinguishable from the radial dimension r, because of the spherical symmetry.Notice now that dx^n = dy^n = dz^n = dr^n with n = 2 gives not the Pythagorean sum but dx^2 = dy^2 = dz^2 = dr^2, and with n = 1 it gives dx = dy = dz = dr.

So your statement that "it is not dx=dy=dz=dr but dr^2 = dx^2 + dy^2 + dz^2" isn't making any sense, and is just confusing spherican symmetry for the Riemann metric.

Both the results are consequences of the same equation, they're not contradictions. It's down to the spherical symmetry involved.

I'm sure you must know this as it is all elementary stuff, and that your question is a tease.

Never mind, thanks for at least having some kind of discussion.

Best regards.

Dear Nige:

ReplyDeleteI'm sure you must know this as it is all elementary stuff, and that your question is a tease.When I said your point 2 is just wrong this was neither a question nor a tease, but a fact. And it is still wrong. It seems to me you slept through the GR session, you evidently have no idea how to make a coordinate transformation. Please look up every standard textbook on differential geometry to find out that I am correct.

A differential operator (or the 1 form rspt) in 3 dimensions with spherical symmetry is

notidentical to three times the one-dimensional case. If it was as you said, then the Coulomb force law in 3 dimensions wouldn't fall with 1/r^2 as it does, but the force of a point charge would remain constant as it does in 1 dimension. This follows if what you said was correct. If you convert the one-forms appropriately with a coordinate transformation, you'll instead find that the Laplace operator in 3 dimensions is not just d^2/dr^2 but actually 1/r^2 d/dr r^2 d/dr (which correctly gives 0 for r neq 0 when applied to 1/r). For a very brief into, see e.g. Wiki on the Laplace operator.Best,

B.

Dear Bee,

ReplyDeleteRepeat question - is there some neat way of spontaneous symmetry breaking in a KK theory?

Thanks!

-Arun

Hi Arun,

ReplyDeletesorry, I wasn't ignoring your questions, I am just kinda stressed out. I don't know any purely geometric scenario that doesn't need additional input like a potential or the like. I have a way to think about it, not sure if you'll like it, but will give it a try. Forget about semi-simple Lie Algebras and just add extra dimensions. If you give it to them that they are time-dependend as well and (at least in the early universe) will expand, then you'd have some kind of structure formation in a higher dimensional setting. I.e. temperature/density drops and stuff clusters. This can break symmetries (locally) much like Weiss domains in a crystal. Now consider formation of submanifolds (branes) in this process, I'd think they don't need to inherit the maximal symmetry. Of course you'll then have to figure out why the remaining symmetries are those of the SM. That however isn't directly related to the KK approach above.

So, the short answer would have been: sorry, I don't know.

Best,

B.

Bee, in radial symmetry,

ReplyDelete(dF_x)/dx = (dF_y)/dy = (dF_z)/dz

This equation says that the gradient of function F is isotropic, i.e., spherically symmetric.

Think about radial variations in electric field strength, or density, or or recession velocities of stars, in directions x, y and z around us.

Look in x direction from the middle of your sphere, and the electric field or density, or recession velocity varies in the same way with distance that it does in directions z, and y.

All are the same.

In

(dF_x)/dx = (dF_y)/dy = (dF_z)/dz

the numerators are all equal and the denominators are all equal,

dF_x = dF_y = dF_z

and

dx = dy = dz

Because of this latter property, it is usual to simplify spherically symmetric cases to just dr:

dx = dy = dz = dr

Now in your claim:

"it is not dx=dy=dz=dr but dr^2 = dx^2 + dy^2 + dz^2"

you're wrong, because dx=dy=dz=dr

which applies to spherical symmetry, has simply nothing to do withdr^2 = dx^2 + dy^2 + dz^2. I don't know why you claim they are the same, but they aren't.The whole physical point of dr^2 = dx^2 + dy^2 + dz^2 is that the squares of the line elements

are not equalhence, dr^2 = dx^2 + dy^2 + dz^2 doesn't generally get used for spherically symmetric cases.If we do apply spherical symmetry to dr^2 = dx^2 + dy^2 + dz^2, then we get:

dr^2 = dx^2 = dy^2 = dz^2

Hence:

3dr^2 = dx^2 + dy^2 + dz^2

Notice that the number 3 appears, which you don't include!

I thought you were maybe joking before, but now it seems as if you are not:

"When I said your point 2 is just wrong this was neither a question nor a tease, but a fact. And it is still wrong. It seems to me you slept through the GR session, you evidently have no idea how to make a coordinate transformation. Please look up every standard textbook on differential geometry to find out that I am correct.

"A differential operator (or the 1 form rspt) in 3 dimensions with spherical symmetry is not identical to three times the one-dimensional case. If it was as you said, then the Coulomb force law in 3 dimensions wouldn't fall with 1/r^2 as it does, but the force of a point charge would remain constant as it does in 1 dimension. This follows if what you said was correct. If you convert the one-forms appropriately with a coordinate transformation, you'll instead find that the Laplace operator in 3 dimensions is not just d^2/dr^2 but actually 1/r^2 d/dr r^2 d/dr (which correctly gives 0 for r neq 0 when applied to 1/r). For a very brief into, see e.g. Wiki on the Laplace operator."

This is just plain wrong, and if you are going to be that abrasive you really do need to have a just cause for doing so!

You are just confusing the Laplacian operator for the divergence operator.Divergence of electric field is

div.F = [(dF_x)/dx] + [(dF_y)/dy] + [(dF_z)/dz]

for spherical symmetry, this produces Coulomb's law from Gauss's law:

Gauss's law:

div.E = charge density/permittivity

hence:

div.E = [(dE_x)/dx] + [(dE_y)/dy] + [(dE_z)/dz]

because there is spherical symmetry:

[(dE_x)/dx] = [(dE_y)/dy] = [(dE_z)/dz]

= 3(dE_x)/dx

= 3(dE_r)/dr

where r is a radial direction (equal to x, y and z, because of the spherical symmetry of the electric field around a point charge).

Hence:

div.E = charge density /permittivity

= 3(dE_r)/dr

When we insert charge per spherical volume (4/3)*{Pi}*r^3 for the "charge density", and rearrange we get:

E_r = charge/(4*{Pi}*r^2)

This immediately becomes Coulomb's law when we remember: force_r = {charge acted upon} * E_r

So that's why you're wrong for the divergence operator, where dx=dy=dz=dr for spherical symmetry.

For the Laplacian operator in spherical symmetry, we have the same kind of thing:

div^2 F = 3(d^2 F_x)/dx^2

= 3(d^2 F_r)/dr^2

where F_x = F_y = F_z = F_r

and

dx^2 = dy^2 = dz^2 = dr^2

This is all easy stuff, and it's

shockingthat you weren't taught any of it.I forgive you for all the abrasiveness, but insist that you learn the facts. If you can find any book which disproves the facts, let me know please. I think someone other than me fell asleep during the math lectures, unless your lecturer was the one who fell asleep. Thanks a lot.

There are two plus signs missing, corrected below in

ReplyDeletebold:Divergence of electric field is

div.F = [(dF_x)/dx] + [(dF_y)/dy] + [(dF_z)/dz]

for spherical symmetry, this produces Coulomb’s law from Gauss’s law:

Gauss’s law:

div.E = {charge density}/{permittivity}

hence:

div.E = [(dE_x)/dx] + [(dE_y)/dy] + [(dE_z)/dz]

because there is spherical symmetry:

[(dE_x)/dx]

+[(dE_y)/dy]+[(dE_z)/dz]= 3(dE_x)/dx

= 3(dE_r)/dr

where r is a radial direction (equal to x, y and z, because of the spherical symmetry of the electric field around a point charge).

Hence:

div.E = charge density /permittivity

= 3(dE_r)/dr

When we insert charge per spherical volume (4/3)*{Pi}*r^3 for the “charge density”, and rearrange we get:

E_r = charge/(4*{Pi}*r^2)

This immediately becomes Coulomb’s law when we remember: force_r = {charge acted upon} * E_r

So that’s why you’re wrong for the divergence operator, where dx=dy=dz=dr for spherical symmetry.

The cylinder condition.

ReplyDeleteHmmmmm.......:)

Kaluza and Klein?

If one was to say that all this math was to accumulate into some new conceptual idea what would that be?

You do the equations and what does this do for you, as you now look at the world, and from where, do the new maths lead one as you think differently of this new perspective?

The symbol for the Laplacian operator above should be namba squared (it is the same symbol as the divergence operator squared without the dot that follows the divergence operator, shortened above to div).

ReplyDeleteNige: would you please at least consider that what I say might be correct and look up a textbook. Alternatively, go and ask anybody who has learned differential geometry if you don't trust me. Again, your reasoning is just manifestly WRONG. As you know I am currently travelling, so I can't give you a page number, but for beginners you might want to look up Misner, Thorne, Wheeler or Weinberg's textbook in the first some chapters to see how a coordinate transformation acts on one-forms and vectors.

ReplyDeleteI was not confusing the Laplace operator with the divergence, I haven't talked about the divergence, I tried to explain why your reasoning is faulty. I used the example of the Laplace operator trying to show that your argumentation is wrong. If it was as you said, then the Laplace operator was d^2/dr^2, which is definitely not the case. Go on and try to derive the correct form using your argumentation.

Further, if you make a sensible coordinate transformation from Euclidean into spherical coordinates, you'll find that the line element changes as

ds^2 = dx^2 + dy^2 + dz^2 = dr^2 + r^2 (d\theta^2 + \sin^2 \theta d\phi^2)

dx^2 = dy^2 = dz^2 = dr^2

This is all easy stuff, and it's shocking that you weren't taught any of it.

I am wondering where you took your classes.

Hence:

3dr^2 = dx^2 + dy^2 + dz^2

Notice that the number 3 appears, which you don't include!

This is complete bullshit. Would you please to me a favor, write down your transformation, transform your one-forms according to it, and derive the above. Good luck.

Of course in spherical symmetry the gradient is spherical symmetric, this is not the point. What you still don't get is that three dimensions - even with spherical symmetry - are not the same as three times one dimension.

When we insert charge per spherical volume (4/3)*{Pi}*r^3This is not a derivation, but a hand-wavy way of putting together factors. I presume you want to derive the field of a point charge, the source is \delta(0) not ~1/r^3. Write down the equations (Laplace on potential equals charge), solve them.

Best,

B.

"I haven't talked about the divergence," - Bee.

ReplyDeleteBee, you wrote above:

"If it was as you said, then the Coulomb force law in 3 dimensions wouldn't fall with 1/r^2 as it does, "

The Coulomb law in operator symbolism is:

divergence.E = charge density/permittivity

with force F_r = {charge acted upon} * E_r

So you need to learn that in vector calculus, Coulomb's law does involve divergence.

"I tried to explain why your reasoning is faulty. I used the example of the Laplace operator trying to show that your argumentation is wrong. If it was as you said, then the Laplace operator was d^2/dr^2, which is definitely not the case. Go on and try to derive the correct form using your argumentation." - Bee

The Laplacian operator is defined as the sum:

{namba}^2 F = [(d^2 F_x)/dx^2] + [(d^2 F_y)/dy^2] + [(d^2 F_z)/dz^2]

Source: Feynman lectures on physics, section on Maxwell's equations (I'm travelling myself and don't have access to a library at present, so can't give page number, but it's in there).

For spherical symmetry,

[(d^2 F_x)/dx^2] = [(d^2 F_y)/dy^2] = [(d^2 F_z)/dz^2]

= [(d^2 F_r)/dr^2]

Hence,

[(d^2 F_x)/dx^2] + [(d^2 F_y)/dy^2] + [(d^2 F_z)/dz^2]

= [(d^2 F_r)/dr^2] + [(d^2 F_r)/dr^2] + [(d^2 F_r)/dr^2]

= 3[(d^2 F_r)/dr^2]

Notice the factor of 3 appear here too!

Notice that in the derivation of Coulomb's force in simple algebra from Maxwell's div.E equation in vector calculus (which I showed in a comment above), it is

essentialthat the sum of three orthagonal differential terms in vector calculus is equal to three times any one of those terms (or the radial direction term). This factor of three cancels out the 3 in the volume of a sphere (4/3)*Pi*r^3, giving the 4*Pi we find in the denominator of Coulomb's law."dx^2 = dy^2 = dz^2 = dr^2" - Bee

This is correct, and what I was saying!

Earlier you claimed "it is not dx=dy=dz=dr but dr^2 = dx^2 + dy^2 + dz^2".

If you were right that

"dr^2 = dx^2 + dy^2 + dz^2"

This earlier statement of yours (which is wrong) is incompatible with your new correct comment that "dx^2 = dy^2 = dz^2 = dr^2" is wrong.

Because dx^2 = dy^2 = dz^2 = dr^2, we can set them all equal to 1 unit, for convenience.

Then your previous claim that "dr^2 = dx^2 + dy^2 + dz^2" becomes:

1 = 1 + 1 + 1.

This is why you need the factor of three! Hence if dx^2 = dy^2 = dz^2 = dr^2, then it follows

3dr^2 = dx^2 + dy^2 + dz^2

i.e.,

3 = 1 + 1 + 1.

Which is correct.

Your new statement

"ds^2 = dx^2 + dy^2 + dz^2"

is correct and has nothing to do with spherical symmetry whatsoever, because ds^2 has nothing to do with dr^2.

ds^2 = dx^2 + dy^2 + dz^2 is nothing to do with 3dr^2 = dx^2 + dy^2 + dz^2 where r is radial distance from the origin in spherical symmetry (the origin being defined x = 0, y = 0, z = 0).

However you then write:

"

Hence:

3dr^2 = dx^2 + dy^2 + dz^2

Notice that the number 3 appears, which you don't include!

This is complete bullshit. Would you please to me a favor, write down your transformation, transform your one-forms according to it, and derive the above. Good luck." - Bee.

This is just wrong. You are, as shown completely confused between radial distances r and the Pythagorean resultant s.

"Of course in spherical symmetry the gradient is spherical symmetric, this is not the point." - Bee.

You claimed earlier that this was the point, stating:

"Your point 2 is just wrong, in case of spherical symmetry it is not dx=dy=dz=dr but

dr^2 = dx^2 + dy^2 + dz^2 (provided that your x,y,z are euclidean coordinates)." -Bee

You here got confused between

ds^2 = dx^2 + dy^2 + dz^2

and

dr^2 = dx^2 + dy^2 + dz^2 (which you write, and which is wrong).

The ds^2 = dx^2 + dy^2 + dz^2 formula is a metric and has nothing to do with the Laplacian operator!

If you want to write dr^2 equals some sum, the correct formula you should write is

3dr^2 = dx^2 + dy^2 + dz^2

for spherical symmetry, or

dr^2 = (1/3)*(dx^2 + dy^2 + dz^2)

This is the reason why the radial contraction of the earth in general relativity is (1/3)GM/c^2 = 1.5 mm, instead of GM/c^2 = 4.5 mm (see Feynman's lectures).

But even if you write dr^2 = (1/3)*(dx^2 + dy^2 + dz^2), that still has nothing to do with dx=dy=dz=dr.

You next state:

"What you still don't get is that three dimensions - even with spherical symmetry - are not the same as three times one dimension." - Bee

If

dx = dy = dz = dr

as in spherical symmetry, then even a child can write down the sum

dx + dy + dz = 3dr

You simply can't add up 1 + 1 + 1 = 3. You keep falsely claiming that because 1 = 1 = 1 = 1, that means 1 + 1 + 1 = 1! Please go and learn to add up to the number 3.

"This is not a derivation, but a hand-wavy way of putting together factors. I presume you want to derive the field of a point charge, the source is \delta(0) not ~1/r^3. Write down the equations (Laplace on potential equals charge), solve them." - Bee

The Laplace operation on potential isn't a Maxwell equation as noted above: the relevant Maxwell equation is div.E. You can write this in other forms but the most useful form is using the divergence operator (see Feynman lectures on physics for example), which you claim is hand-waving nonsense.

It's not a hand-wavy argument:

1. Newton proved that in any spherically-symmetric distribution of inverse-square law force sources, the resultant is the same as for a point source located in the centre.

2. Maxwell's equation for Gauss's law is div.E = charge density/permittivity of free space

charge density = Q/[(4/3)*Pi*r^3]

Hence

div.E = Q/[{permittivity}*(4/3)*Pi*r^3] [Eq. 1]

Now

div.E = [(dE_x)/dx] + [(dE_y)/dy] + [(dE_z)/dz]

where for spherical symmetry

[(dE_x)/dx] = [(dE_y)/dy] = [(dE_z)/dz] = [(dE_r)/dr]

Hence

div.E =

[(dE_x)/dx] + [(dE_y)/dy] + [(dE_z)/dz]

= 3(dE_r)/dr

Setting this equal to Eq. 1 gives:

div.E = Q/[{permittivity}*(4/3)*Pi*r^3] = 3(dE_r)/dr

Hence:

E_r = Q/[4*Pi*{permittivity}*r^2]

Remembering F = q*E_r (where q is charge acted upon by charge Q),

F = qQ/[4*Pi*{permittivity}*r^2]

Which is Coulomb's law. Where is the hand-waving? It's rigorous. We have to use spherical symmetry in it where

[(dE_x)/dx] + [(dE_y)/dy] + [(dE_z)/dz]

= 3(dE_r)/dr

If the number 3 didn't occur here, we'd have a result three times stronger than Coulomb's law, which is experimentally wrong. So the figure 3 must occur here, even if you don't believe that adding three identical functions leads to a resultant which is three times as big as any one of the three!

You need to know this derivation because otherwise you won't be able to derive Coulomb's law in algebra from the vector calculus form in Maxwell's equations (which are written with divergences and curls, not Laplacian operators). If you don't know this, one day you might be caught out on it.

Best,

nige

Bee,

ReplyDeleteThanks! I'll have to think about what you wrote.

Nige,

You are utterly, butterly wrong. Nothing wrong with ignorance, provided it is accompanied with a willingness to learn. But you are too obtuse to accept a correction. I suggest with scant hope that you will heed it that you cease making a fool of yourself in public.

Arun,

ReplyDeleteSo you believe as Bee claimed that the equality of gradients in all directions in a spherical space somehow disproves - or is replaced by - a metric? From someone who confuses ds^2 with dr^2 and whose arithmetic in claiming that dr^2 = dx^2 + dy^2 + dz^2 is equivalent to claiming that 1 = 1 + 1 + 1 since dr^2, dx^2, dy^2 and dz^2 are all equal for the spherically symmetric case where the variable changing in x, y and z directions is independent of the direction?

Bee says:

ReplyDeleteI want to elaborate somewhat on these 'early' XDs since I noticed very little people actually read the original works by Kaluza and Klein.That's because only very little people can fit inside those extra dimensions.

Nige: I am not confusing anything.

ReplyDeleteIn my first anwer assumed you were talking about spherical curves for which d\theta = d\phi = 0, and then dr^2 = ds^2. I apologize if that was not the case.

I have marked all your quotations in italics. Where you remark

you have wrongly attributed your own sentence to me.This is correct, and what I was saying!

You should know that the electromagnetic fields can be written as a gradient of a potential, in which case the equation becomes what I have written (Laplacian on the potential). Again, your reasoning is just wrong. The equation from Feynman's book is of course correct, this is the expression in eucliedean coordinates, maybe he also has the expression in spherical coordinates, then you'd see that my version is correct.

It seems to me you have no idea what the difference between a partial derivatve and a one-form is. The partial derivative is not just the inverse of the one form. You conclusion that dx=dy=dz is nonsensical because these forms are basis elements and independend of each other - otherwise you've managed to collaps a three dimensional space to a one dimensional one. Do yourself a favor and check your above equation with the simplest case you can think of, like Coulomb force or so. Best,

B.

Dear Arun: thanks for your support... Should I ever be around New Jersey, I'll invite you for a dinner :-)

ReplyDelete-B.

"I have marked all your quotations in italics. Where you remark

ReplyDeleteThis is correct, and what I was saying! you have wrongly attributed your own sentence to me." - Bee

Well in that case you maybe need to do something other than use italics when you write a formula, or it looks as if you are just emphasising it with italics.

"You should know that the electromagnetic fields can be written as a gradient of a potential, in which case the equation becomes what I have written (Laplacian on the potential)." - Bee

I don't know why you continue to attack me by making false claims? I of course know that which is

preciselywhy I wrote in my comment above:"You can write this in other forms but the most useful form is using the divergence operator (see Feynman lectures on physics for example), which you claim is hand-waving nonsense."

So what you are doing now is just ignoring what I'm saying and hurling abuse at me which is all completely wrong? I'm sorry to see you write such abuse of a commentator. All this extra-dimensional KK stuff has proved not even wrong. Maybe your replies prove why it's supporters can't be reasoned with. If you claim that Feynman is "arm-waving" then maybe you need to write your ignorant criticisms of him, not me.

The reason I gave the derivation from Maxwell's divergence equation and not the Laplacian, is - as I explained to you above - because it clearly shows that you that:

[(dE_x)/dx] + [(dE_y)/dy] + [(dE_z)/dz]

= 3(dE_r)/dr

which shows that for equal gradients the sum of three of them is three times the radial gradient, which you don't grasp:

"Your point 2 is just wrong, in case of spherical symmetry it is not dx=dy=dz=dr but

dr^2 = dx^2 + dy^2 + dz^2 (provided that your x,y,z are euclidean coordinates)." -Bee

Here you are claiming dr^2 = dx^2 + dy^2 + dz^2 when in fact the correct formula is

3dr^2 = dx^2 + dy^2 + dz^2

because all three are equal to dr^2. Similarly, for the case dr = dx = dy = dz, we get

3dr = dx + dy + dz

Not what you claimed which was dr = dx + dy + dz.

In addition, there is as I explained, still another error in your claim that you still don't seem to grasp:

"Your point 2 is just wrong, in case of spherical symmetry it is not dx=dy=dz=dr but

dr^2 = dx^2 + dy^2 + dz^2 (provided that your x,y,z are euclidean coordinates)." -Bee

The fact is, dx=dy=dz=dr has

nothing whatsoever to do withdr^2 = dx^2 + dy^2 + dz^2.Even if you had the factor of three in dr^2 = dx^2 + dy^2 + dz^2,

i.e., 3dr^2 = dx^2 + dy^2 + dz^2,

your argument that dx=dy=dz=dr should be replaced with dr^2 = dx^2 + dy^2 + dz^2 is entirely wrong.

You then tried to justify this claim by referring to the Laplacian operator for Coulomb's law, and I proved you don't need it - even if your claim was substantive - because you can use Maxwell's equation for divergence of an electric field instead. You really are wrong, and you still can't see it?

"You conclusion that dx=dy=dz is nonsensical because these forms are basis elements and independend of each other - otherwise you've managed to collaps a three dimensional space to a one dimensional one." - Bee

Bee, you are

still totally confusedso I suggest you read the Feynman lectures in physics! In spherical symmetry, the direction-dependent vectors are all equal in x, y, and z directions.(dE_x)/dx is the same in spherical symmetry as (dE_y)/dy which is the same as (dE_y)/dy, which is equal to the general gradient as a function of radial distance (dE_r)/dr.

Because of the spherical symmetry, dE_x = dE_y = dE_z = dE_r and dx = dy = dz = dr.

This is a fact because of the spherical symmetry! You still don't seem to grasp this fact, and instead of admitting that you don't understand it and trying to learn some physics, you instead attack me?

"You conclusion that dx=dy=dz is nonsensical because these forms are basis elements and independend of each other - otherwise you've managed to collaps a three dimensional space to a one dimensional one. Do yourself a favor and check your above equation with the simplest case you can think of, like Coulomb force or so." - Bee

I've already done that above, where the gradients and line elements are equal to the radial line element in spherical symmetry. This is why I can derive Coulomb's law from the divergence of an electric field. The spherical symmetry means that gradients and line elements are

all equalto the radial gradient.So I've already done you the favor of explaining it to you. Maybe you need to read it to see where your errors are.

Best,

Nigel

Nige: (your comments in italics)

ReplyDeletea) I am not attacking you.

b) I am not making any false claims.

If you claim that Feynman is "arm-waving" then maybe you need to write your ignorant criticisms of him, not me.c) I haven't said anything like Feynman being 'arm waving'.

d) Again I tell you that my argument with the Laplace operator was an example to show you that your reasoning is wrong. Apparently you are not able to grasp it. I'm not in the mood to repeat it again.

Not what you claimed which was dr = dx + dy + dz.e) I never said that.

the correct formula is

3dr^2 = dx^2 + dy^2 + dz^2

f) This is just wrong. I have explained it above. I neither have the time nor the patience to repeat myself.

You still don't seem to grasp this fact, and instead of admitting that you don't understand it and trying to learn some physics, you instead attack me?I am not attacking you, I am trying to explain basics of differential geometry that you happily ignore. Maybe try to derive the volume element from your argumentation. Is it dV=dxdydz=dr^3 - as according to your statement that

dx = dy = dz = dr?So I've already done you the favor of explaining it to you. Maybe you need to read it to see where your errors are.Aha. Unfortunately, it's my calculation that is mathematically consistent where yours is not. So I thank you very much for that 'favor' of yours but your confused argumentation that completely ignores what 'dx' actually means doesn't indicate any error in my argument.

Best,

B.

"I am not attacking you." - Bee

ReplyDeleteYou seem to be because you falsely wrote

"It seems to me you slept through the GR session, you evidently have no idea how to make a coordinate transformation. Please look up every standard textbook on differential geometry to find out that I am correct."

I never mentioned coordinate transformations, which have nothing to do with the existence of spherical symmetry of gradients you object to. It's abusive for you to keep on calling the facts as I state them "bullshit" and making the false claim that because I don't mention something irrelevant, I "evidently have no idea how to make a coordinate transformation."

You would feel abused if someone made false statements about your work being wrong, and then said that they evidently didn't know something which they do know, but which isn't relevant for spherical symmetry and equal gradients! So why do you claim that you were not being abusive to me by making false claims about my work and false sneers about what I know?

I don't understand why you keep repeating this endlessly, after I explained by your dismissal is totally vacuous.

The spherical symmetry of gradients has nothing to do with general relativity or with metrics, so you are falsely attacking me.

Spherical symmetry means that x = y = z = r and anything that varies with one direction varies in the other directions in exactly the same way.

Hence dx = dy = dz = dr has nothing to do with the metric.

You claimed

"Your point 2 is just wrong, in case of spherical symmetry it is not dx=dy=dz=dr but

dr^2 = dx^2 + dy^2 + dz^2 (provided that your x,y,z are euclidean coordinates)."

When in fact dx=dy=dz=dr has nothing to do with dr^2 = dx^2 + dy^2 + dz^2, which in spherical symmetry would be 3dr^2 = dx^2 + dy^2 + dz^2, if dr^2 = dx^2 = dy^2 = dz^2 (definition of spherical symmetry is that anything varying as a function of an element in x, y or z is identical).

You again repeat that 3dr^2 = dx^2 + dy^2 + dz^2 is "just wrong. I have explained it above. I neither have the time nor the patience to repeat myself."

It's abusive of you to keep ignoring the facts and keep insisting that they are wrong:

Fact: in spherical symmetry,

x = y = z = r where r is radial distance from the origin.

dx = dy = dz = dr follows from this because of the spherical symmetry. Similarly,

(dx^n) = (dy^n) = (dz^n) = dr^n

This has nothing to do with general relativity and nothing to do with a metricHence the sum

(dx^n) + (dy^n) + (dz^n)

= (dr^n) + (dr^n) + (dr^n)

= 3dr^n

Hence inserting n = 2 we get

3dr^2 = (dx^2) + (dy^2) + (dz^2)

It's elementary stuff, and has nothing to do with a metric, despite the fact that it looks a bit like one - however you were the one who started talking about dr^2 = (dx^2) + (dy^2) + (dz^2), not me.

I then proved you are wrong for spherical symmetry, which is what I'm talking about.

In general relativity you do get the factor 3 come in for spherical symmetry when dealing with earth's contraction. In the Lorentzian contraction, you have

gamma = [1 - (v^2)/(c^2)]^{1/2}

In general relativity, you have a contraction ratio which is similar except v^2 = 2GM/c^2. For small curvature, this gives by the binomial expansion a reduction in length by the small amount, GM/c^2.

But this contraction for gravity is spread over 3 dimensions, as compared to only being a contraction in the direction of motion for the Lorentz contraction! So we have to correct it.

Using Feynman's spherical symmetry,

{delta}x = {delta}y = {delta}z = {delta}r

where r is radial direction, giving:

{delta}x + {delta}y + {delta}z = 3{delta}r

So:

3{delta}r = GM/c^2

Hence: {delta}r = (1/3)GM/c^2.

This result is given in Feynman's lectures on physics, for the Earth's mass

(1/3)GM/c^2 = 1.5 mm,

and since this is radial only (x,y z directions) and not transverse (the circumference is not contracted) you see the need for non-Euclidean geometry.

So even if you are going to bring general relativity into this discussion of spherical symmetry, the sum of small line elements in three perpendicular dimensions is still three times the line element in any one dimension."I am not attacking you, I am trying to explain basics of differential geometry that you happily ignore. Maybe try to derive the volume element from your argumentation. Is it dV=dxdydz=dr^3 - as according to your statement that dx = dy = dz = dr?" - Bee

In spherical symmetry where everything in direction x is the same as everything in directions y and z and all three directions are the same as the radial parameter r, then the product dxdydz = dr^3.

This only holds for spherical symmetry.

"Aha. Unfortunately, it's my calculation that is mathematically consistent where yours is not." - Bee

I explained where your dismissal of my work is flawed. dx is the element of x, and in spherical symmetry it such elements for any gradient are identical in all directions.

"So I thank you very much for that 'favor' of yours but your confused argumentation that completely ignores what 'dx' actually means doesn't indicate any error in my argument." - Bee

You earlier asked me to do a "favor":

"Do yourself a favor and check your above equation with the simplest case you can think of, like Coulomb force or so." - Bee

I had already done it! It proves you wrong. The fact that you make a false claim and refuse to read the facts does indicate an error in your argument. You need to check what you are claiming before you make assertions like:

"Your point 2 is just wrong, in case of spherical symmetry it is not dx=dy=dz=dr but

dr^2 = dx^2 + dy^2 + dz^2 (provided that your x,y,z are euclidean coordinates)." - Bee

Once again, dx is not the same as dx^2 and a sum of squares of line elements doesn't have anything to do with the equality of line elements in spherical symmetry.

That you should invent such a claim as this is like claiming someone who says oranges are orange is somehow wrong because apples are green. You are inventing a false 'disproof' that is totally wrong, and then you're claiming I don't know general relativity.

Maybe you need to be sure you are correct before making assertions, or - if you don't have the time to check everything at the time of making an assertion - you should at least correct errors when pointed out to you.

Best,

Nigel

Dear Nigel,

ReplyDeleteI never mentioned coordinate transformations, which have nothing to do with the existence of spherical symmetry of gradients you object to. [...]

I don't understand why you keep repeating this endlessly, after I explained by your dismissal is totally vacuous.

The spherical symmetry of gradients has nothing to do with general relativity or with metrics, so you are falsely attacking me.

A coordinate transformation, and the metric has not necessarily something to do with general relativity. If you chose a coordinate system different from Euclidean coordinates in flat space (for example spherical coordinates) then the metric tensor will not just be diag (-1,1,1,1).

Nige: In spherical symmetry where everything in direction x is the same as everything in directions y and z and all three directions are the same as the radial parameter r, then the product dxdydz = dr^3.

This only holds for spherical symmetry.

This is wrong. Correct is dxdydx = r^2 sin \theta dr d\theta d\phi, where x,y,z, are the Euclidean coordinates.

(see e.g. here)

Nige: Fact: in spherical symmetry,

x = y = z = r where r is radial distance from the origin.

Is just wrong. The radial distance from the origin is r=\sqrt(x^2 + y^2 + z^2), where x,y,z, are the Euclidean coordinates.

Spherical symmetry means that x = y = z = r andIs just wrong. Spherical symmetry of a function means that that function only depends on the distance to the origin, that distance being r= \sqrt(x^2 + y^2 + z^2) where x,y,z are the Euclidean coordinates. Obviously x is not equal to y and z everywhere.

I am not in the mood to correct all other points that follow from that misunderstanding. Best,

B.

"This is wrong. Correct is dxdydx = r^2 sin \theta dr d\theta d\phi, where x,y,z, are the Euclidean coordinates."

ReplyDeleteNo, in spherical symmetry r is defined as radial direction

which is equal to x, to y and to z.The page you link to is for spherical coordinates, not spherical symmetry!The equation

dxdydx = r^2 sin \theta dr d\theta d\phi

applies to the

generalcase, not specifically to spherical symmetry.Now we see why you are wrong: you are claiming that a specific solution of a formula (spherical symmetry where dr = dx and is equal to all other perpendicular elements) somehow is disproved by the general formula.

If you look at the page you actually linked to,

http://www.tf.uni-kiel.de/matwis/amat/elmat_en/kap_3/basics/b3_2_2.html

you can see quite clearly that it has nothing to do with spherical symmetry, just with spherical coordinates.

"

Spherical symmetry means that x = y = z = r andIs just wrong. Spherical symmetry of a function means that that function only depends on the distance to the origin, that distance being r= \sqrt(x^2 + y^2 + z^2) where x,y,z are the Euclidean coordinates. Obviously x is not equal to y and z everywhere." - Bee

It's not obvious, it's plain wrong because the whole x axis is symmetric to the whole y axis and to the whole z axis. They're all similar!

Spherical symmetry around a point means that all the parameters varying as a function of distance from that point vary as the same function of distance, i.e., as a function of radial distance.If a star is spherically symmetrical, then any variation in any parameter as a function of distance in x direction is mirrored by y and z directions.

This is why we can avoid talking of three separate directions in spherical symmetry and set them all equal.

Suppose that you want to use calculus to find the volume of a sphere. You integrate the surface area, 4*Pi*r^2 over increasing radius,

volume of sphere =

{integral symbol}(4*Pi*r^2)dr

= (4/3)*Pi*r^3.

There is no inclusion of x, y and z dimensions, because they're all the same as r in spherical symmetry!

This relies on the fact that Spherical symmetry means that x = y = z = r.

Suppose you want to integrate to find the total mass observable in spacetime from us, you would need to use x = y = z = r because the universe looks the same in each direction. The fact that there is spherical symmetry means you can dispense with your formulae like

dxdydx = r^2 sin \theta dr d\theta d\phi,

because you are just dealing with the radial direction, outwards from us. For example, to find the total mass of the universe observable to us you would need to integrate the density over increasing radial distance, allowing for the fact that the density increases with apparent distance because we're looking back to earlier epochs of the big bang (when density was higher).

This is the sort of calculation I've been on the basis of spherical symmetry.

"I am not in the mood to correct all other points that follow from that misunderstanding." - Bee

Well, I suppose that this saves me the time required to point out other inaccuracies, so thank you for that. Good luck with your research.

Best,

Nigel

For example of finding observable mass of universe by integrating over radial distance:

ReplyDeleteM = {integral symbol}(4πr^2){rho} dr

Density, {rho}, increases with distance r in proportion to the factor

[1 – rH/c)^{-3}

this is caused by decreasing observable age of the universe we're seeing at earlier times (great distances).

The integral out to r gives infinity, because density tends towards infinity at great distances (corresponding to small times since the big bang).

This raises quite a lot of questions if gravitons are exchanged between all the masses in the universe. You get a problem with an infinite effective mass of the universe.

So you are forced to take account of other factors such as redshift of exchanged gravitons, which reduces the strength of gravitational interactions when gravitons are exchanged over large distances (large redshifts). There are other factors like inflation, which would affect the result. But there are a lot of simple calculations that can be made like this which give clues about the nature of gravity.

Best,

nige

Nige: one uses spherical coordinates for spherical symmetry because it makes the description much easier. Of course you don't HAVE to use them, but once you do, you have to do it properly - using a sensible coordinate transformation.

ReplyDeleteThis relies on the fact that Spherical symmetry means that x = y = z = r.I will repeat it one last time: spherical symmetry does not mean that x=y=z when x,y,z are Euclidean coordinates. The equation x=y=z is a constraint that singles out a one-dimensional curve in space-time (the one on which x=y=z).

If a star is spherically symmetrical, then any variation in any parameter as a function of distance in x direction is mirrored by y and z directions.

This is why we can avoid talking of three separate directions in spherical symmetry and set them all equal.

(you seem to implicitly assume without stating it that the origin of your coordinates is the center of the symmetry.) In case of spherical symmetry one can reduce the differential equations (depending on r, theta, phi) to a system with only a dependence on the radial coordinate, but this equation will not be the same as in the case where you'd only have a space-time with one dimension (as I've tried to explained above, you put another factor in by hand from the volume element, just that this factor is not in your volume element either). You are taking your words and translating them incorrectly into equations. Best,

B.

Hi Damian,

ReplyDeleteThat's because only very little people can fit inside those extra dimensions.:-) I personally favor infinitely large extra dimensions. I love the KK idea but I don't think it could be fundamental either. We'd again be left with the question why these symmetries... could there be a whole landscape? Best,

B.

"(you seem to implicitly assume without stating it that the origin of your coordinates is the center of the symmetry.)" - Bee

ReplyDeleteI explicitly stated:

"The assumption that there's only one time dimension is the same as you would have if everything around you was radially symmetric, like living in the middle of an onion, where only changes as a function of radial distance (one effective dimension) are readily apparent: the Hubble constant, and hence the age of the universe (time)is independent of the direction you are looking in."

This was the whole point I'm making: that the spherical symmetry applies where the origin of the coordinates is the centre of symmetry.

"I will repeat it one last time: spherical symmetry does not mean that x=y=z when x,y,z are Euclidean coordinates. The equation x=y=z is a constraint that singles out a one-dimensional curve in space-time (the one on which x=y=z)." - Bee

That's what I've been telling you: Euclidean coordinates in general have nothing to do with x=y=z. I'm

onlydealing with a very specific, particular situation in which everything around the observer is extremely similar regardless of direction. This is the situation of cosmology."In case of spherical symmetry one can reduce the differential equations (depending on r, theta, phi) to a system with only a dependence on the radial coordinate, but this equation will not be the same as in the case where you'd only have a space-time with one dimension ..." - Bee

That's

only correct if you are not having the symmetry centred on coordinates (0,0,0).You are apparently thinking about the general case, where the coordinates are not the centre of symmetry. This doesn't have anything to do with any of my calculations, which are simplified by the very fact that the expansion rate of the universe, its density, etc., are very nearly isotropic."You are taking your words and translating them incorrectly into equations." - Bee.

Ouch! I think you are mistaken, and I'd like to know where I'm wrong. Misunderstanding my discussion of being at the middle of a spherically symmetric onion (which I still think is a very clear example of what I mean, and translates perfectly into equations), doesn't indicate that my equations are incorrect. Maybe in future I'll add a more detailed discussion of spherical symmetry when I write about it, and show how it is derived from general spherical coordinates. (Then physicists will have to find something else to misunderstand so they can go on falsely claiming it to be completely nonsense.)

Thank you very much for your time.

Best,

nige

ReplyDeleteMaybe in future I'll add a more detailed discussion of spherical symmetry when I write about it, and show how it is derived from general spherical coordinates.Your welcome - but please, not in the comments of this blog. This is definitly not the place to propagate your own private version of vector analysis.

In fact, I could not follow in detail this thread over the last days, but now that I'm trying to catch up, I am quite annoyed how you've been hijacking the thread, insulting the intelligence of your readers and wasting Bee's time...

Take a breath, think once more about it, and maybe this debate can be moved to your blog...

Best, stefan

Hi Stefan,

ReplyDeleteI'm very sorry that the discussion went the way it did but I did my best to be clear and was still misunderstood.

I don't have a "private version of vector analysis". Maths is maths, it's not my private version.

If Bee didn't understand it she could have said so in the first place instead of insulting my intelligence by claiming - falsely - that I'm somehow wrong when Bee missed the point, probably because she was busy with something more important.

I've been answering claims made against my intelligence when I stated facts, not deliberately "hijacking the thread".

Time has been wasted, mine mainly because it's more work to refute insults to your intelligence than to merely claim someone else is wrong without first checking what they are saying. This is elementary background stuff everyone should know, and diverts attention from the physics which is vital!

I have no interest in responding to any more insults, and hope you have the decency to leave it at that. I won't bring up physics points again on your blog, as it just leads to attacks on me lacking any substance, and wasting time.

Still, thanks for your patience. The physics is here.

Best,

nige

Nige:

ReplyDeleteIf Bee didn't understand it she could have said so in the first place instead of insulting my intelligence by claiming - falsely - that I'm somehow wrong when Bee missed the point, probably because she was busy with something more important.I have explained repeatedly why you are not 'somehow' wrong but just wrong. I've explained more than once why several of your 'equations' are faulty. Each time I do that you say something like that equation doesn't apply here or has to be used such, or only in this case. You have no idea what a total derivative is or how to make a coordinate transformation, neither do you use a sensible definition of spherical symmetry in the first place. You've tried to justify your 'derivation' by quoting one equation from Feynman's lecture notes - that equation is about the only correct one. Try to find the rest of your 'derivation' in some textbook.

Stefan: I am thinking about deleting all that crap. I am afraid somebody might pick up one of Nigel's wrong equation from the text if one doesn't read the whole context.

Best,

B.

Dear Bee & Stefan,

ReplyDeleteYou've been very patient, actually.

Looking forward to your visit to New Jersey :)

Best,

-Arun

Bee:

ReplyDelete"I personally favor infinitely large extra dimensions. I love the KK idea but I don't think it could be fundamental either. We'd again be left with the question why these symmetries... could there be a whole landscape?"

I'm in favour of extra spatial dimensions that are as large as the three in the universe experienced, any way. And I have a quite detailed empirical argument on my blog for such dimensions that starts with the evidence of quantum entanglement with the development of a hypothesis, which is then supported by large scale natural evidence.

But then this is a digrammatic but entirely non-mathematical account that represents details of a cause that acts non-locally in addition to all the forces. And this is an account from an unqualified individual, what's more.

So what physicist would take this theory seriously or, even if they did, touch it with a barge pole?

Hi Merlin:

ReplyDeleteSo what physicist would take this theory seriously or, even if they did, touch it with a barge pole?So far I am not aware of a theory with large uncompactified (flat) extra dimensions that could be taken seriously. I didn't say I have one, but I also don't think it's impossible.

B.

But then it's not really that large scale extra dimensions just as such that need be contentious.

ReplyDeleteBut I think my blog presents a fair, if only non-mathematical, argument for maintaining that an extradimensional non-locally acting cause could both produce quantum wave behaviour and astronomically observable effects. And to the extent that a cosmological theory might be developed without cosmic inflation, non-baryonic dark matter or dark energy.

Then I argue that a cause of quantum entanglement, which would act extra-dimensionally so as to maintain the correlations between entangled quantum objects, could act univrsally so as to maintain the form and organisation for atoms. molecules and living organisms. While I also argue that the digrammatic extra-dimensional representation of such a cause resolves the problems of many uniquely experiencing immaterial minds and the indivisibility of consciousness, which have been outstanding since Descartes arguments in the 17th century.

Merlin: In case you try to point me to your blog, a link would be helpful, I am not in the mood to read through all of your writings. Your idea sounds nice but it seems to me you neglect several crucial problems. I.e. how do you confine gravity to 3+1 dimensions, since this is what we observe? Best, B.

ReplyDeleteBy being a cause that necessarily acts where it surrounds objects in 3+1D spacetime, gravity is dimensionally self-confining.

ReplyDeleteJust positing additdonal dimensions of space from where a further cause acts nonlocally - so that, by definition, this cause does not reduce or cease in its effects with distance around objects - does not imply that gravity (or em or the nuclear forces) should also act extra-dimensionally.

I really don't see how the existence of an immaterial, nonlocally acting cause that relates in its effects upon objects from additinal spatial dimensions with a distinct material form conserving property conflicts in any way with the observable evidence.