Discussion : Magneto

Discussion in 'Physics & Math' started by AlphaNumeric, Apr 14, 2011.

Thread Status:
Not open for further replies.
  1. AlphaNumeric Fully ionized Moderator

    The book uses phrases like 'quasi-Euclidean', while you use 'pseudo-Euclidean'. It is also clear where you got your \(\frac{1}{S^{2}}\) expression from. It appears on page 177 in the discussion on hyper-spheres. In that case 'S' is being used as the radius of a 3-sphere embedded in 4 dimensional Euclidean space. That page's discussion includes several equations you've posted. Is Google Books where you've been reading that book? Is that why you quoted those formulae out of the blue when they didn't address what I'd said? Is that why you've been obsessed with hyper-spheres, it is what that page talks about?

    The author then does some manipulation of the metric and makes the statement that \(\frac{1}{S^{2}} = 0\) gives the Euclidean metric. This is nothing more than saying if the radius becomes infinite the surface becomes flat. What you have misunderstood is that when I've been saying "the 2-sphere \(S^{2}\)" I have not been referring to a radius but the notation for a 2 dimensional spherical manifold using the notation clearly stated here. I also see why you didn't understand \(\mathbb{R}\), that book uses the notation \(\mathcal{R}\) instead.

    Is that the only book you've ever read on this stuff? It uses non-standard notation, which means if that's all you've ever exposed yourself to you'll be unaware of what other notation is used, or rather what the most common notation is used. Heck, the N-sphere page uses \(S^{N}\) and \(\mathbb{R}\) immediately, both of which are the notation I've been using throughout. This illustrates how unfamiliar you are with basic notation.

    However, if even you were unaware of other notation, if you have understood the material you read you'd have realised what I was referring to, ie \(S^{2}\) was obviously not a reference to a sphere's radius. I explicitly said \(R_{0}\) was that.

    Now that you've said where you've been getting your quotes from its clear why you said some of the things you've said. You've just been mindlessly copying and pasting without understanding or trying to bend the discussion towards something you can parrot.

    Once again your attempt to give an insulting retort (to patronisingly tell me to read that book) has back fired. It's now clear why you made that mistake, you not only didn't understand what I said, you didn't understand that book you're parroting! Unlike you I actually understand the notation in its context, that book is telling me things I'm already somewhat familiar with. It's also telling me you don't understand it but then I was already aware of that too

    Please Register or Log in to view the hidden image!

  2. Google AdSense Guest Advertisement

    to hide all adverts.
  3. AlphaNumeric Fully ionized Moderator

    Thank you for admitting you hadn't read a book on vector calculus, geometry, analysis, tensors, differential geometry, special relativity, general relativity or basically anything of a mathematical physics ilk before 2008-2009.

    Your admission demonstrates how you don't realise just how omnipresent that notation is in mathematics and relativity.

    Spheres are not Euclidean. I've already told you that. I told you to compute the Ricci scalar for a sphere given its metric. In fact its given in that book you just told me to read, on page 177. For a sphere of radius \(R_{0}\) the Ricci scalar curvature is \(\frac{1}{R_{0}^{2}}\). That is why, in that book, when \(\frac{1}{S^{2}} = 0\) it becomes Euclidean, because it becomes flat.

    These results are right in front of you but you don't see them, you don't understand them and when they are pointed out you don't accept them.

    You're now babbling in an attempt to avoid facing up to the fact you've admitted to not reading any material, material you claim you have.

    This isn't just nomenclature, it demonstrates that when I've been saying "The 2-sphere \(S^{2}\)" you've been thinking "That's got radius S". It means that if you've read papers on GR then you've misunderstood them. It means every place in your work where you say \(S^{N}\) or \(\mathbb{R}^{N}\) you aren't referring to what the reader thinks you are referring to. Some people use \(\mathbf{R}^{N}\) instead of \(\mathbb{R}^{N}\). That is just a matter of notation. The choice of sign on the \(\eta\) SR metric, that is a matter of notation. Thinking "That's a length" when in fact it's a manifold is not a matter of notation. This isn't about simple notion, it's about fundamental understanding. And you once again show you have none.

    Anyway, I have work in the morning and I need to sleep.
  4. Google AdSense Guest Advertisement

    to hide all adverts.
  5. Magneto_1 Super Principia Registered Senior Member

    Not true. I understand that you have been describing surfaces. I have been wanting you so solve the variable radius case using your "Complex GR."

    Once, again you just fall in "LOVE" with everything that I present to you.

    You ought to buy the book, just for taking my Class!
  6. Google AdSense Guest Advertisement

    to hide all adverts.
  7. 1100f Banned Registered Senior Member

    It reminds me of that guy that never learned math, and decided to learn alone. He took a book and read it. After he finnished reading the book, a friend of him asked him how did he find the book. So his answer was:" I understood everything in the book except for one thing. Why do they sometime write the number eight horizontally instead of vertically? But appart of this nomenclature problem, I now know mathematics"
  8. Magneto_1 Super Principia Registered Senior Member


    You bring up a good point 1100f, this is something that I encourage my students to do, all of the time, when they are taking a test, or working on tasks that have many parts.

    When taking a test, or tackling a surmountable task, first try and answer everything that you can up front, then save the things that you are no so 100% sure of, for later.

    Why use low efficiency and eat up valuable time working on what you don't know, when energy is spent more efficiently moving forward in what you do know!
  9. Magneto_1 Super Principia Registered Senior Member

    AlphaNumeric, you are so dishonest, if I have presented a question that is too hard, just admit it! It is easy to do, honesty, it feels good, trust me, try it!

    Here you have described the "Map/Patch Area" Metric using "Complex General Relativity" CGR

    \(ds^{2} = (R_{0})^{2}(d\theta^{2} + \sin^{2}\theta d\phi^{2}) = g_{ab}dx^{a}dx^{b}\)

    Will you now also express this Euclidean Metric using "Complex General Relativity" CGR?

    \(ds^{2} = dr^{2} + (R_{0})^{2}(d\theta^{2} + \sin^{2}\theta d\phi^{2}) = \) (Complex General Relativity CGR - Metric)

    I have repeatedly asked you to provide me the answer to #2. And all you keep saying is, "I only know how to solve the Manifold Surface Metric.:bawl:"

    However, I do know, and am well aware, that once you read that D.F. Lawden book that I recommended, watch how magically the answer to my question appears??

    Please Register or Log in to view the hidden image!

    Do you really think that someone could "parrott" and articulate concepts in GR and CGR; and also have you "Converted?" What does that say about you?

    Please Register or Log in to view the hidden image!

    Last edited: Apr 26, 2011
  10. AlphaNumeric Fully ionized Moderator

    Evidence says otherwise. I just walked you through why \(\frac{1}{\mathbb{R}^{N}}\) is meaningless. You claim you're familiar with the notion, since 2008-2009, which means you should have understood what I was saying, you should have been aware that such notation as you use it is never used in the literature. I asked you to provide a single example of it being in the literature and you ignored me. Repeatedly. I also explained it has nothing to do with hyper-spheres.

    Do you really think you can just lie your way out of such a blatant mistake?

    And I asked you to explain what you mean. It isn't my 'complex GR', that phrase is yours. GR is GR is GR, there is no 'complex GR' other than in your head. What you keep quoting is just the metric for a 2-sphere. It isn't complex in the sense of \(\mathbb{C}\) and it isn't complicated, so it isn't 'complex GR'.

    I asked you to explain yourself, I'm still waiting. I cannot read your mind and your inability to explain yourself is the bottleneck here.

    Yet another meaningless comment which you clearly seem to think is somehow an insult or a put down when all it is is you being incoherent. I don't 'love' everything you present, I give lengthy replies because you get so much wrong.

    I just demonstrated I understand it more than you. In 30 seconds of finding it on Google Books I realised where all your errors were from and why you've been posting the things you have.

    Besides, you now mention you have a class yet you've been trying to insult me by calling me 'professor'. Doesn't that strike you as a bit of a stupid way to insult someone?

    I asked you to explain yourself here. You never did. I know you probably skipped the post because it was long and concentrating probably gives you a headache but you are being dishonest in claiming I just skipped it for being 'too hard'. I specifically said it was not 'complex', that I was waiting for you to explain yourself.

    No, you have a metric for a 2 dimensional space written in spherical angles. It isn't a 'complex general relativity metric' or a 'map/patch area' metric. I know you want to label everything with your own terminology to make it seem like you understand it but it only serves to demonstrate you don't.

    There is no 'complex general relativity' except in your head. I asked you before to explain what you meant by that since its a term you made up. I can write it in terms of Cartesian coordinates using basic vector calculus, general relativity doesn't have to come into it because its just a change of coordinates.

    \(ds^{2} = (R_{0})^{2}(d\theta^{2} + \sin^{2}\theta d\phi^{2}) = g_{ab}dx^{a}dx^{b}\)
    This is just the metric for a 2-sphere of radius \(R_{0}\) in terms of polar angles. You can get this from the Cartesian Euclidean metric by the usual Cartesian to spherical transformation but with \(r = R_{0}\). This means that the Cartesian form of the metric is the usual one but with the additional restriction of being a loci with regards to the origin.

    \(ds^{2} = (R_{0})^{2}(d\theta^{2} + \sin^{2}\theta d\phi^{2}) = dx^{2} + dy^{2} + dz^{2}\) such that \(x^{2}+y^{2}+z^{2}=R_{0}^{2}\)

    This is seen from the fact we're having to view the space as an embedding of \(S^{2}\) into \(\mathbb{R}^{3}\) and the embedding is defined by the variety associated to \(x^{2}+y^{2}+z^{2}=R_{0}^{2}\). By restricting our consideration to the space of points which satisfy \(x^{2}+y^{2}+z^{2}=R_{0}^{2}\) we restrict the Euclidean metric of \(\mathbb{R}^{3}\) to the spherical metric of \(S^{2}\).

    Please note how I'm using \(S^{2}\) and \(\mathbb{R}\) to refer to manifolds, not radii.

    I don't see what is hard about that or why you would possibly think that was 'too hard'. That's nothing more than things I've already said to you. I've gone through the coordinate transformation on the metric between Cartesians and sphericals already, in generality. I had to explain to you how to do the transformation, after you thought that the general case ended up like \(ds^{2} = dx^{2} + dy^{2} + dz^{2} + (x^{2}+y^{2}+z^{2})(d\theta^{2} + \sin^{2}\theta d\phi^{2})\), thus proving you don't know how to do the transformation. Now you're asking me a simpler case and trying to pretend I don't know how to do it. All that does it demonstrate you don't understand because you fail to realise how this is a trivial corollary of things I've already said.

    And I repeatedly asked you to explain what you wanted me to do since you didn't make it clear what you wanted. You keep referring to my complex general relativity equations when 'complex general relativity' is your phrase.

    Where did I say that. Provide a link to a post where I said that. I have said no such thing. You're now simply inventing things up wholesale. That's just trolling plain and simple.

    No, I already provided the methods in regards to coordinate transformations long before you mentioned that book. I've also demonstrated I understand the book better than you, I immediately spotted the sources of all your misunderstandings and mistakes.

    I think you're copying and pasting without understanding. You haven't articulated anything, you seem to struggle to even articulate your own thoughts. I repeatedly ask you to explain yourself, as 'complex general relativity' is a phrase you started using, not me, in reference to your own definition of what is or isn't relativity. Thus when you ask me to do something involving my 'complex general relativity equations' I don't know to what you are referring. This is a problem with your articulation, not mine. If anything I explain myself too much and thus have long posts.

    And don't think I didn't notice how you skipped over this post. I bet you didn't realise that admitting you hadn't seen \(\mathbb{R}\) before 2008 would be an admission that you haven't read any standard textbook on anything to do with vectors, calculus, tensors, geometry or relativity. Opps! Really put your foot in it there. If you're a typical example of the teaching (and I use that word in its vaguest sense) staff at the University of Phoenix no wonder it isn't accredited. What courses do you actually teach, seeing as there's no maths or physics department and the only maths or physics courses are short distance learning courses on material which would be considered foundation year? What does it say about a teacher when he hasn't read any of the books relevant to his own course?
  11. Dywyddyr Penguinaciously duckalicious. Valued Senior Member

    That's got to be one of the funniest things I've ever read on the internet.
    Thank you for that Magneto_1.
    I'm going to save that quote to my hard drive as an example of crackpot delusion.
  12. Magneto_1 Super Principia Registered Senior Member


    This is what I was expecting to see:

    The "Map/Patch Area" Metric using "Complex General Relativity" CGR

    \(ds^{2} = (R_{0})^{2}(d\theta^{2} + \sin^{2}\theta d\phi^{2}) = g_{ab}dx^{a}dx^{b}\)

    The above equation describes a Map or Patch area on the Surface Area of a sphere.

    This is also known as a Manifold. Why is this known as a Manifold?

    A manifold is a unique mathematical trick, and it is also a conceptual trick. A Manifold is a curved surface that can be "Mapped" onto a spherical surface. The key phrase is "Mapped" onto a surface. However, it is trickery in that you can't get an actual "area - > m^2" or a geodesic "length -> m" without considering some constant as the radius of the sphere in the metric calculation. And at the same time they say that it is independent of the Sphere range.

    The condition for manifolds to hold regarding the Psuedo (Euclidean/Non-Euclidean) Metric is that the range change is zero (\(dr^2 = 0\)). And this condition (\(dr^2 = 0\)) imposed on the "Metric" is true for a Manifold; whether that Manifold be Euclidean or Non Euclidean.

    And this confusion about a Hyperspheres or N-Spheres (\(S^N_{Sph}\)) and N-Manifolds (\(\mathbb{R}^{(N+1)}\))

    See: N-Spheres

    To be considered a Manifold:

    1) it has to have a "Radius or Range Independent" Surface Area that can be mapped onto a sphere that has a range.

    2) and it has to have a "Geodesic Path Length" that is derived from a "range independent" Manifold, and is a curved path length which approximates a straight line, that can be mapped onto a sphere, that has a range.

    On very large distance or range scales the approximates, become such that the area of a sphere becomes a "flat" surface area manifold, and the Geodesic Path Length becomes a "straight" line.

    See: Asymptotically flat spacetime


    Will you now also express this Euclidean Metric using "Complex General Relativity" CGR? This is what I was asking for.

    \(ds^{2} = dr^{2} + (R_{0})^{2}(d\theta^{2} + \sin^{2}\theta d\phi^{2}) = g_{44}(dx^4)^2 + g_{ab}dx^{a}dx^{b} \)

    \(ds^{2} = -{c^2_{Light}}dt^2_0 + g_{ab}dx^{a}dx^{b} \)

    Where the events of a "Space-time" coordinate clock in the Center of Mass Frame where "Proper" time (\(-dt^2_0\)) and distance is denoted by

    \(g_{44}(dx^4)^2 = -{c^2_{Light}}dt^2_0 \)

    Next doing the transformation into the External Observer Frame and his time measurements (\(dt^2\))

    \(dt = \frac{1}{c_{Light}} \int{\sqrt{-g_{44}}dx^4} \)

    Finally, the "Euclidean Metric using "Complex General Relativity" CGR

    \(ds^{2} = g_{ab}dx^{a}dx^{b} - \(\int{\sqrt{-g_{44}}dx^4}\)^2\)

    \(ds^{2} = g_{ab}dx^{a}dx^{b} - {c^2_{Light}}dt^2\)

    See: A coordinate-dependent definition Note: it is down further in the wiki article.

    "A coordinate-dependent definition is the simplest (and historically the first) way of defining an asymptotically flat spacetime assumes that we have a coordinate chart, with coordinates t,x,y,z, which far from the origin behaves much like a Cartesian chart on Minkowski spacetime, in the following sense. Write the metric tensor as the sum of a (physically unobservable) Minkowski background plus a perturbation tensor, gab = ηab + hab, and set r2 = x2 + y2 + z2. Then we require:

    One reason why we require the partial derivatives of the perturbation to decay so quickly is that these conditions turn out to imply that the gravitational field energy density (to the extent that this somewhat nebulous notion makes sense in a metric theory of gravitation) decays like O(1 /r4), which would be physically sensible. (In classical electromagnetism, the energy of the electromagnetic field of a point charge decays like O(1 /r4).)"

    And just to answer your previous question, I on ran into "your cute" (\(\frac{1}{\mathbb{R}^{N}}\)) in 2008 - 2009. I was done with GR by 2009. There is a lot of non standard symbols sometimes used in GR, I believe that this will work itself out in time.

    Last edited: Apr 26, 2011
  13. Guest254 Valued Senior Member

    This is ridiculous! Your posts fail to communicate even the most rudimentary understanding of the things you're attempting to tell others about. Hell, your posts barely convey any sort of meaning! And this is largely because you're trying to fudge material that you don't understand.

    Come on, be honest with us. You don't have the first clue about geometry, do you? If someone came along and said "hey, is this mathematical object a manifold?", you wouldn't have a clue how to help - would you?
  14. AlphaNumeric Fully ionized Moderator

    Once again you mindlessly copy things you don't understand and then botch it by injecting your own ignorance. Let's consider what you set bit by bit.

    A manifold is not 'a surface surface which can be mapped onto a spherical surface'. It is a set endowed with a topology of open sets \(U_{a}\), each equipped with a map \(\phi_{a} : U_{a} \to \phi_{a}(U_{a}) \subset \mathbb{R}^{N}\). This means each of the open sets is homeomorphic to a subset of \(\mathbb{R}^{N}\). In addition \(\phi^{-1}\) exists. Furthermore if U and V are subsets of M such that \(U \cap V \neq \empty\) and \(\phi(U) \subset \mathbb{R}^{N}\) and \(\psi(V) \subset \mathbb{R}^{N}\) then the map \(\phi(U \cap V) \to \psi(U \cap V)\), ie \(\psi_{\circ}\phi^{-1}\), exists and is invertible.

    Spheres are particular cases of manifolds but they are not in the definition of manifolds. The space which any open subset of M must be homeomorphic to is \(\mathbb{R}^{N}\), not \(S^{N}\).

    You know you don't know what manifold means yet you lie anyway. How on earth you think you'll get away with it I don't know.

    Not all manifolds have metrics. Special kinds of manifolds have metrics, Riemannian manifolds. You have to be able to define tangent spaces and co-tangent bundles etc in order to construct a metric. \(dr^{2}=0\) has nothing to do with manifolds in general. In the example we've been discussing setting \(r = R_{0}\) in spherical coordinates of \(\mathbb{R}^{3}\) defines a 2 dimensional variety, \(x^{2}+y^{2}+z^{2} = R_{0}^{2}\), which happens to be a manifold. Setting \(dr=0\) is a corollary of that. 'r' is a coordinate in spherical coordinates. Not all manifolds admit such coordinates in a nice way, nor do all manifolds define sub manifolds by setting dr=0. For instance, you cannot set r=0 (which gives dr=0) in the space-time manifold defined by the SC metric, as it is a singularity and the metric no longer is valid nor is the space a manifold there. It's actually not a manifold in the full sense, it is a manifold with singularities.

    \(\mathbb{R}^{N+1}\) is an example of a manifold, it is not the N dimensional manifold. It's not even N dimensional, it is N+1 dimensional. The clue is in the 'N+1'. It's subtle but even you should grasp that. But I know why you did it. On the Wiki page you just linked to it defines the N-sphere as a variety within N+1 dimensional Euclidean space. You've demonstrated you don't even understand the definition!!! :lol:

    And we've been over this before. I know what a hyper sphere is. I've got research on them! You've demonstrated you didn't even read that Wikipedia page because it uses the notation \(S^{N}\) and \(\mathbb{R}^{N}\) in the manner I've been using.

    Look up the definition of 'manifold' before being stupid enough to just make up what you think it means and then being even more stupid and trying to BS it to someone who does know what 'manifold' means.

    There is no 'complex general relativity', you just made up a phrase. And what you just gave is wrong.

    \(dr^{2} + (R_{0})^{2}(d\theta^{2} + \sin^{2}\theta d\phi^{2}) \) involves only 3 coordinates, yet you then say \( g_{44}(dx^4)^2 + g_{ab}dx^{a}dx^{b} \), which then includes \(x^4\). Therefore your expressions are not equal.

    That isn't Euclidean, despite you calling it Euclidean. I see you still don't know what that means.

    You don't understand the indices because you keep mixing them up and chopping and changing. Also you again call it Euclidean but it isn't Euclidean, the sign on the \(dt^{2}\) term is opposite to the others, making it non-Euclidean.

    You don't know the meaning of words you use, you don't know what the equations mean, you make up your own definitions and then are stupid enough to try and pretend they are right and you ignore direct questions.

    Again with the mindless quoting of stuff which is irrelevant. The perturbation thing has nothing to do with what I said. The O(1/r^4) thing I've already addressed, \(\mathbb{R}^{N}\) has nothing to do with a radius or length, it is in reference to a manifold defined by a Cartesian product of copies of the Reals.

    You're just trying to throw out anything you can, desperately hoping it sticks. All it does is demonstrate you can't back up your claims, you can only quote mine, desperately hoping I'll see something which I'll accept justifies your claims. The problem for you is I know this area of maths and I know you're wrong.

    I have asked on several occasions, at least 3, for you to provide me with one, just one, example of the specific notation \(\frac{1}{\mathbb{R}^{N}}\) being used in a reputable journal or book. You have ignored the question each and every time. Now you're suddenly claiming that it wasn't \(\mathbb{R}\) you saw for the first time in 2008 (after I pointed out that means you'd never read any relevant book on any of this area of maths or physics if that is the case) but you were referring to \(\frac{1}{\mathbb{R}^{N}}\). Name the book or journal article. Or provide a different book or journal article which uses it. Prove your claim. Several other people have chimed into the discussion to laugh at you for using that notation, at least one of whom has a PhD in mathematics. I mentioned it to several maths PhDs I work with and they all immediately went "What?! He thought that was meaningful?!".

    No doubt you'll just skip over the request for the 4th or 5th time, 'conveniently' forgetting to respond to it. That's how you have behaved the entire way through this thread. You didn't know the definition of fluxes so you just made one up and wouldn't admit your mistake. You don't know the definition of manifolds so you've made one up, botching what you've obviously gone and Googled. You didn't even know that \(\mathbb{R}^{N+1}\) is N+1 dimensional!! :roflmao: And that was because you couldn't understand the definition at the top of the Wiki page you linked to!!

    How much worse can you do?! You don't understand simple definitions you link to. That misunderstanding of the Wiki N-sphere page just proves all your previous comments on hyper spheres were also just copy and pastes. You continually uncover more of your own dishonesty. Why you think its a good idea you continue to reply I honestly don't know. Are you simply a troll or are you so detached from reality you think you're good well in this 'discussion'?
    Last edited: Apr 26, 2011
  15. Magneto_1 Super Principia Registered Senior Member

    I explained this to you before, this is for the N-dimensional guys; of which I am not one yet.

    Inverse N-Dimensional Manifold


    Inverse N-HyperSphere


    Just for kicks!! This really is not my thing!

    The way I would explain the above "Inverse N-Manifold" is first to imagine a surface of a sphere. Next realize that I can mathematically and conceptually divide the surface of a sphere into infinitesimal patches of area which we will call the "Map Area." In essence a sum of Map Areas will equal. (\(4\pi {r}^{2}\)).

    Now, let's consider a single Map/Patch Area of the sphere and this patch or map area will we make equal to a Manifold:

    \(ds^{2} = (R_{0})^{2}(d\theta^{2} + \sin^{2}\theta d\phi^{2}) = g_{ab}dx^{a}dx^{b}\)

    Next, now that we have our (N=2) Sphere and our (n = N+1 = 3) Manifold, I will let the "Great Mathematicans" of the world decide what to do with it from here.


    Inverse N-HyperSphere


    Now how can the above two equation be use in Manifolds? What purpose would they have?

    I don't claim to have all of the answers; and as AlphaNumeric says it is not seen in literature.

    But imagine you have a giant planet that is orbiting one star gravitationally, and that same planet is being pulled on gravitationally by two different stars.

    The surface of the planet is experiencing different gravitational effects on the surface; and likewise is experiencing different heating effects on the surface of the planet.

    Using the "Inverse N-Manifold" concept you will able to predict the different intensities of the different patch areas due to the various gravity and heating of different patch areas on the surface of the planet.

    That is just for starters!
  16. Guest254 Valued Senior Member

    Unsurprisingly, the exact same comments I made to your previous post, apply to this new one (and those before it).

  17. Magneto_1 Super Principia Registered Senior Member


    Thanks, for the clarity, AlphaNumeric, and Dr. George James Weatherill.

    The "Non-Euclidean Map/Patch/Manifold Metric" expressed using "Classical General Relativity" GR and "Complex General Relativity" CGR

    \(ds^{2} = (R_{0})^{2}(d\theta^{2} + \sin^{2}\theta d\phi^{2}) = g_{ab}dx^{a}dx^{b}\)\( ---> m^2\)

    The "Non-Euclidean Space-time Metric expressed using "Classical General Relativity" GR and "Complex General Relativity" CGR

    \(ds^{2} = dr^{2} + (R_{0})^{2}(d\theta^{2} + \sin^{2}\theta d\phi^{2}) = g_{44}(dx^4)^2 + g_{ab}dx^{a}dx^{b} \)\( ---> m^2\)

    Last edited: Apr 27, 2011
  18. rpenner Fully Wired Staff Member

    Does he think a coordinate chart is like a patch over a tire puncture?
  19. AlphaNumeric Fully ionized Moderator

    Simply continuing to make up your own notation which is meaningless and already retorted shows you're dishonest. Plain and simple. Furthermore I've already told you that \(\mathbb{R}^{N+1}\) is N+1 dimensional, not N dimensional.

    That sentence is meaningless. The space is a manifold. By having a definition of coordinates over the space you already have a manifold. Giving it a metric doesn't make it a manifold. I've already explained how metrics are not a necessary part of the definition of a manifold.

    So you've obviously read my request you provide a single example of it being used in the literature yet you can't provide one.

    I've asked 4 or 5 times now. How many times are you going to avoid answering the question?

    Prove it.

    You've just making up words and throwing notation together. You keep using \(g_{ab}dx^{a}dx^{b}\) but you obviously don't understand what it means. I've already commented on this. Consider your last expression :

    The first metic is 3 dimensional yet you then give an expression involving \(g_{44}\). They cannot be equal, an error I previously pointed out to you. You don't understand what you're posting else you'd have understood my correction. Further more you don't understand the index notation, a mistake I skipped over in my previous post :

    For a 4 dimensional space-time metric you have \(ds^{2} = g_{ab}dx^{a}dx^{b} = \sum_{a=1}^{4}\sum_{b=1}^{4}g_{ab}dx^{a}dx^{b} = g_{44}dx^{4}dx^{4} + g_{34}dx^{3}dx^{4} + g_{43}dx^{4}dx^{3} + \ldots\)

    The \(g_{44}\) term is included in the index expression. Furthermore the expression \(ds^{2} = g_{ab}dx^{a}dx^{b} - \(\int{\sqrt{-g_{44}}dx^4}\)^2\) is not true. You don't square root, integrate and thern square each component in turn. To find the length of a path you do it to the entire expression.

    \(s^{2} = \left( \int \sqrt{g_{ab}dx^{a}dx^{b}} \right)^{2} = \left( \int \sqrt{g_{44}dx^{4}dx^{4} + g_{34}dx^{3}dx^{4} + g_{43}dx^{4}dx^{3} + \ldots} \right)^{2}\)

    This is NOT valid : \(s^{2} = \left( \int \sqrt{g_{ab}dx^{a}dx^{b}} \right)^{2} = \left( \int \sqrt{g_{44}dx^{4}dx^{4}} \right)^{2} + \left( \int \sqrt{g_{34}dx^{3}dx^{4}} \right)^{2} + \ldots \)

    You don't understand the index notation, you don't understand how to integrate up a line element expression, you don't understand manifolds. Why do you keep knowingly lying? You admitted to reading my question to show \(\frac{1}{\mathbb{R}^{N}}\) appears in the literature yet you ignored it. You don't know the definition of manifold yet you just make one up. Do you really think you're going to get anywhere by lying? I've slapped you down at every turn and I'm nowhere near as knowledgeable as some of the people who'll review for any reputable journal you submit your work to. Lying and plagiarising (which is what copying and pasting is) earn you an instant rejection.

    Making up quotes is another thing which gets you instantly rejected. Here you claimed I said "I only know how to solve the Manifold Surface Metric.". I'm still waiting for you to provide a link to the post of mine where I said there. Yet another thing you can't provide because it doesn't exist.

    Making things up and then refusing to justify them is trolling. I've asked you 4 or 5 times now to provide a single reference for the notation \(\frac{1}{\mathbb{R}^{N}}\). I've asked you to provide a derivation of the SC metric, you haven't. Now you're claiming the use of an 'inverse N-manifold' (which you haven't properly defined) "you will able to predict the different intensities of the different patch areas due to the various gravity and heating of different patch areas on the surface of the planet." Prove it. For once actually justify a claim you make.

    Each time you avoid the questions, each time you lie, each time you change the subject, you show you are intellectually dishonest. You will never get a job doing research at a reputable university or research institution if this is your approach to science.
    Last edited: Apr 27, 2011
  20. Magneto_1 Super Principia Registered Senior Member

    Why are you so DISHONEST!! That must hurt!!

    You know that the following equation is not valid.

    This is NOT valid : \(s^{2} = \left( \int \sqrt{g_{ab}dx^{a}dx^{b}} \right)^{2} = \left( \int \sqrt{g_{44}dx^{4}dx^{4}} \right)^{2} + \left( \int \sqrt{g_{34}dx^{3}dx^{4}} \right)^{2} + \ldots \)

    However, this is "VALID!"

    \(s^{2} = \left( \int \sqrt{g_{ab}dx^{a}dx^{b}} \right)^{2} = \left( \int \sqrt{g_{44}dx^{4}dx^{4}}\right) \left( \int \sqrt{g_{44}dx^{4}dx^{4}} \right) \)


    \(s^{2} = \left( \int \sqrt{g_{ab}dx^{a}dx^{b}} \right)^{2} = \left( \int \sqrt{g_{44}dx^{4}dx^{4} + g_{34}dx^{3}dx^{4} + g_{43}dx^{4}dx^{3} + \ldots} \right) \left( \int \sqrt{g_{44}dx^{4}dx^{4} + g_{34}dx^{3}dx^{4} + g_{43}dx^{4}dx^{3} + \ldots} \right) \).

    You know if you do one (1) honest thing a day, eventually "It" will go away. What It? Dishonesty that is.....Dishonesty!!

    You, know, you really ought to be "ashamed" at yourself for this one! Oh wait, I forgot you have none!! Shame that is.....Utter Shame!!
  21. Magneto_1 Super Principia Registered Senior Member


    The "Non-Euclidean Map/Patch/Manifold Metric" expressed using "Classical General Relativity" GR and "Complex General Relativity" CGR

    \(ds^{2} = (R_{0})^{2}(d\theta^{2} + \sin^{2}\theta d\phi^{2}) = g_{ab}dx^{a}dx^{b}\)\( ---> m^2\)

    The "Non-Euclidean Space-time Metric expressed using "Classical General Relativity" GR and "Complex General Relativity" CGR

    Center of Mass Frame of Reference

    \(ds^{2} = {c^2_{Light}dt^2_0} - (R_{0})^{2}(d\theta^{2} + \sin^{2}\theta d\phi^{2}) = g_{44}(dx^4)^2 + g_{ab}dx^{a}dx^{b} \)\( ---> m^2\)


    External Observer Frame of Reference

    \(ds^{2} = -{c^2_{Light}dt^2} + (R_{0})^{2}(d\theta^{2} + \sin^{2}\theta d\phi^{2}) = \(\int{\sqrt{-g_{44}}dx^4}\)^2 + g_{ab}dx^{a}dx^{b} \)\( ---> m^2\)

  22. rpenner Fully Wired Staff Member

    Repeating your naked assertions without argument of any kind is not a contribution to a dialogue or discussion; it's authoritarian posturing.

    Richard Hamming "Mathematics on a Distant Planet." American Math Monthly 105 640-650 (1998)
    "In mathematics we do not appeal to authority, but rather you are responsible for what you believe."

    Thus the burden is on you to either use the notation used by the rest of the world or take responsibility and actually explain what your notation means in detail and why you did the idiotic thing like co-opting symbols already in use in places as mainstream as math textbooks and Wikipedia. Oh, wait. You liberally cite from Wikipedia, so you aren't using new notation. Well in that case, you are doing it wrong.
  23. RJBeery Natural Philosopher Valued Senior Member

    Ahh rpenner I really like that quote. May I borrow it?

    Please Register or Log in to view the hidden image!

Thread Status:
Not open for further replies.

Share This Page