Discussion : Magneto

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

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  1. AlphaNumeric Fully ionized Moderator


    Following this thread Magneto's been good enough to accept a challenge to discuss his work in a dedicated thread (the same can't be said for Farsight).

    Basic ground rules between him and I :

    * Relevant direct questions should be answered in a timely manner, assuming a barrage of questions aren't asked all in one go.
    * Citations/references to reputable sources must be given when requested. Acceptable sources covers major journals, textbooks published by well known scientific publishers (Cambridge or Oxford presses, Wileys etc), lecture notes for courses at reputable universities (Ivy league or Oxbridge sort of thing). Personal websites may be allowed, if the other person accepts, but should be academic websites, not Geocities nonsense.
    * All quotes of the other person should include a link to the original post being quoted. Any editing of the quote (for brevity say) should be explicitly stated.
    * Intellectual honesty is expected.
    * Less than 100% politeness is allowed but not too much. Basically I'll tone it down a bit.

    The moderators here can decide if one of us has violated any of these. As agreed in the above thread link, should one of us violate this a month long holiday is to be handed out.

    Right, to the issues at hand. I've encountered Magneto in the following threads (first relevant post linked to) :

    Conservation of Angular Momentum
    Inflation and Curvature
    Spacetime Vortex Controversy

    The following issues have come up which I'd like him to address :


    I would like Magneto to provide a reference for the first sentence in that quote.

    The second sentence is false. \(R_{ab}\) is not a distance, it is a measure of curvature at a given point in space-time. A space is Ricci flat if \(R_{ab}=0\) but it isn't necessarily literally flat space. The Schwartzchild metric has \(R_{ab}=0\) but it is not flat. Furthermore the notion of 'any distance beyond this tensor' is meaningless.

    Reference needed for it being a volume. It is not, though it can define a volume.

    The metric defines a line element \(ds\) by \(ds^{2} = g_{ab}(x^{c})dx^{a}dx^{b} = g_{00}dx^{0}dx^{0}+\ldots\). The coordinates are the \(x^{a}\) and this tells you that if you change your position by a little bit, \(x \to x + dx\) then the distance you move through space-time, ds, is determined by that formula. This is as line formula, not a volume. It can be turned into a volume (or area or any other region formula) though and in the case of volume you have \(Vol = \int_{region}\sqrt{|g|}d\mathbf{x}\) where g is the metric determinant.

    Thus your "It's a volume but some use it for other things" is false. I can provide references if needed.

    Metrics and Line Elements

    False. The right hand side is a scalar value, a real valued function. The left hand side is a rank 2 tensor expression. A single component of the metric will be a real valued function but not the tensor as a whole. This is a more complicated version of saying the following :

    \(\left(\begin{array}{c} A \\ B \end{array} \right) = 5\)

    A 2 dimensional vector cannot be equal to a scalar. Components of the vector can, A or B could be equal to 5, but not the vector as a whole. Its equating apples with oranges. To give an example more directly related to what you've said consider the standard 2d spherical metric of radius 1 :

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

    In this case the components are \(g_{\theta\theta} = 1\), \(g_{\theta\phi} = 0\) and \(g_{\phi\phi} = \sin^{2}\theta\). Notice how the indices do not imply coordinate dependency, ie \(g_{\phi\phi}\) is not dependent on \(\phi\) but \(\theta\). Putting all this together we have

    \(ds^{2} = d\Omega_{2} = d\theta^{2} + \sin^{2}\theta} d\phi^{2}\)

    Thus if you want to work out the length of a line you have to compute

    \(s(L) = \int_{L}ds = \int_{L}\sqrt{d\theta^{2} + \sin^{2}\theta d\phi^{2}}\)

    Let the line be parameterised by \(\lambda \in [0,1]\), ie \(\theta = \theta(\lambda)\) and likewise for \(\phi\). Thus we get

    \(s(L) = \int_{0}^{1}\sqrt{\dot{\theta}^{2} + \sin^{2}\theta \dot{\phi}^{2}}d \lambda\)

    If \(\theta(\lambda)\) and \(\phi(\lambda)\) are anything other than extremely trivial this is not analytically integrable. You certainly cannot just do the following :

    \(s(L) = \sqrt{a^{2} + \sin^{2}a b^{2}}\) so that \(s^{2} = a^{2} + \sin^{2}a b^{2}\)

    I bring this up for further reference.

    You have equated a scalar, the Schwartzchild radius, with a rank 2 tensor, the Einstein tensor. As with the metric, you cannot do this. A particular component of G might be equal to the expression you've given but not the tensor itself.

    The indicies of \(G_{ab}\) don't represent what you just said. Firstly what you said isn't even a coherent well formed statement. Secondly \(G_{ab}\) and \(g_{ab}\) are distinct and since you go on to quote something resembling the SC metric you appear to be referring to g, not G. In either case you're incorrect.

    For g they represent scalings associated with a line element displacement in directions defined by the vectors \(\frac{\partial}{\partial x^{a}}\), ie \(g_{ab} \equiv g\left( \frac{\partial}{\partial x^{a}} , \frac{\partial}{\partial x^{b}} \right)\), illustrating the tensor structure of g. \(g_{ab}\) tells you that if you move by \(dx^{a}\) in the \(x^{a}\) coordinate and likewise for \(x^{b}\) then you sweep out a line of length \(g_{ab}dx^{a}dx^{b}\).

    As I demonstrated by considering lengths of paths on a sphere, this is not what you get if you use the SC metric to find path lengths in the SC space-time. The last term you give, \(d^2\({a}^2 + \ b^2 \sin^2(a_0))\) is clearly from the 2 dimensional spherical metric component of the SC metric, as the SC metric includes the term \(ds^{2} = \ldots + r^{2} d\Omega_{2}\). You have just dropped all the differential 'd' terms, as if you can just integrate them out. There's a reason they are included, you can't do that. Not only that but as you comment here you change notation such that now you have d's which represent something else. This demonstrates either deliberate attempts to obscure your work or you don't know what you're doing and have just stumbled into terrible notation.

    Furthermore, if \(s\) is supposed to be the length of a curve then you need to say what the curve is. Instead you've just given a expression and said its to do with latitude and longtitude, the a and b. A path starts at some location and ends at another, hence why when you the integration over the line element you have integration limits and the functions \(\theta(\lambda),\phi(\lambda)\) are needed, they define the path. You haven't given a path, you've just given specific values for the \(\theta,\phi\) in the original line element. This can be seen to be incorrect on a 2-sphere by considering geodesics and using SO(3) invariance.

    I can do into this if required.

    That is absolutely not how you compute lengths on a sphere, SC space-time or any other Riemannian manifold. Even putting aside the fact your d's are now just numbers you cannot drop the differential d's. It's like saying that if y is a function of x then \(\frac{dy}{dx} = \frac{d}{d}\frac{y}{x} = \frac{y}{x}\). That is, in essence, what you have done.

    If you claim you have done it correctly then please provide a detailed step by step walk through of how you went from the SC metric to the expressions above.

    More Tensors

    Previously you have said that I must be misquoting you in regards to the scalar-vector-matrix issue ("I did not even look at the old post, but I am sure that I was mis-quoted.") but in the next quote you make the same mistake again, after I have pointed it out to you.

    Firstly you confuse the notion of a radius, which is a scalar quantity, with a radial vector of a given length. Whole notion of a radius is that it is direction independent, it is the distance from the centre of a ball to its edge.

    Secondly you've now added in a vector term, \(\hat{r}\), which you use to represent radial vector, yet you still equate it to the tensors which appear in the Einstein Field Equations. This is a mistake.

    That is in response to my analogy that mismatching tensor structures is like mismatching units.

    As I've already explained, while its possible that the component of \(T_{ab}\) might be some form of mass that doesn't mean \(T_{ab}\) itself is equal to some mass expression. All entries in \(T_{ab}\) must have the same units, as they can be mixed via coordinate transformations. However, this is different from making the tensor as a whole equal to some scalar expression, as my (A,B)=5 example shows.

    Qualitative Stuff

    That's enough quantitative stuff for now, your later posts tend to cover the same material, we can move onto a qualitative comment which is just a matter of completeness.

    The 'golden age' of relativity was the 60s and 70s. By that point the major predictions of Einstein's initial work had been tested and validated. Yes, work has continued since then but to say its only really come into favour in the last 15 years is just wrong. Nor do colliders test GR, they test SR.

    You can't 'wrestle' against facts, you can only wrestle against people's ideas which they have developed (hopefully) based on those facts.

    I'll skip the rather daft comments here about how if I've done any work in science, why haven't you heard of me. But there is something I asked you which you didn't address and I want you to address it.

    I asked you if you'd got any of your papers published. If so, where? If not, is that because you got rejected or because you haven't submitted them? If its due to rejection which journals rejected them, when and for what reasons? If its because you haven't submitted them why not, since peer review is an essential part of good science.


    That's enough for now. As per the initial rules, for each criticism issue I've raised which you disagree with you should provide a sufficiently detailed walk through derivation of the result to demonstrate how you reached it. Initial equations/assumptions should be stated clearly, notation should not be unnecessarily muddled (like changing m and d around). And results you use should be referenced or at the very least named explicitly.
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  3. Magneto_1 Super Principia Registered Senior Member

    There is so much hypocrisy in your post that it is amazing. You think that bombarding me with lots of your physics BS, makes you look like you have understanding. I agree to your rules, however, I still feel a little uneasy about all this, because I don’t think that you and I have the same goals in physics.

    I spent many, many years figuring out how to make the complex mathematics of General Relativity so that people with a basis understanding of physics could solve the basic problems of GR. Your approach is to keep people in the dark which you pomp around like you have all of the mathematical understanding.

    So if we are to continue, you have to agree:
    1) If there is hard way of doing the math and simple way of doing the same math to get the same result. I want to choose the simpler math.

    2) Please provide the metric units (i.e. distance -> m, time – >s, mass -> kg, force -> kgm/s^2) that describe the quantity or term that you are describing.

    3) All equations presented should allow anyone to be able to plug in values to get a useable output; a lot of the math that you present no one could put in their calculator, program, or simulation tools and get a result. Every equation that I will present any physicist will be able to plug in values and get the results.

    I don't have to answer this question but I will as an author and gentleman. Since you made sure to expose that my name is Robert Louis Kemp, the author of Super Principia Mathematica - The Rage to Master Conceptual and Mathematical Physics. Why have you kept your identity a secret, this is the kind of hypocrisy that I continue to speak of.

    You can do a search on my name you will find that I have papers floating around the internet since 1994. I have never submitted one paper to any reputable or main stream journal. So I have never been rejected by any journal "Yet."

    I never submitted to any journal because, I used that twenty years to work, work, and rework my work. I did not want to be like Einstein, and be forced to rework and rethink everything that you submit; which is what happened to Einstein. Times have changed, but in the past and this still happen with reputable journals; but they force you to sign a rights waiver, so that if they publish your work, you turn over the rights to them. This makes entirely no sense to me. I write a work and submit it to a journal and they own all the rights to the work is not right!

    This new system by http://arxiv.org/ you get to maintain your own rights to your own work. I believe that this is the right way of doing things. This is the main reason that I did not publish in a major journal. And the second main reason is that is took me twenty years to get things right. I did not need journals to beat me up while I worked out the details of a theory.

    I am now ready to publish, but now when I publish, I can quote my own book as a reference, in this way I don't mind them having the rights to my work. This is called thinking outside of the box!!

    Now before I address your math and physics, I presented some math that I want you to address.

    AlphaNumeric let's start with this equation above.

    First I want to know; if I wrote the above equation in this form would it have satisfied you?

    The Spacetime Volume Metric

    \(g_{{\theta}{\phi}} = \frac{V_{ol}}{4 \pi} \({4 \pi}sin^2({\frac{\Psi({\theta},{\phi})}{2}})\) = \frac{V_{ol}}{4 \pi} \({2 \pi}(1 - cos({\Psi({\theta},{\phi})})\) \)\( -> m^3 \).

    \(g_{{\theta}{\phi}} = \frac{r^3}{3} \({4 \pi}sin^2({\frac{\Psi({\theta},{\phi})}{2}})\) = \frac{r^3}{3} \({2 \pi}(1 - cos({\Psi({\theta},{\phi})})\) \)\( -> m^3 \).

    Spherical Latitude Angle: \( {\theta}\)\( -> radians \).

    Spherical Longitude Angle: \( {\phi}\)\( -> radians \).

    The Total Spherical Volume:

    \(V_{ol} = \frac{4\pi}{3} r^3 \) \( -> m^3 \)

    Hence for a unit sphere the solid angle of the spherical cap is given as:

    \( \Omega = {4 \pi}(\frac{g_{{\theta}{\phi}}}{V_{ol}}) = \({4 \pi}sin^2({\frac{\Psi({\theta},{\phi})}{2}})\) = \({2 \pi}(1 - cos({\Psi({\theta},{\phi})})\) \)\( -> radians \).

    \( \Omega = \frac{A_{Area}}{r^2} = \frac{A_{{\theta}{\phi}}}{r^2}\).





    You may be able to make this leap:

    \( \Omega^2 = \frac{A_{{\theta}{\phi}}}{r^2} = (\theta^{2} + \sin(\theta_{0})^{2} \phi^{2})\).

    In differential form; if you integrate the equation below you will get the equation above:

    \( d\Omega^2 = \frac{dA_{{\theta}{\phi}}}{r^2} = (d\theta^{2} + \sin(\theta_{0})^{2} d\phi^{2}) = {3}(\frac{dg_{{\theta}{\phi}}}{r^3})\).


    The Schwarzschild metric:

    \(-ds^{2} = -f(r)dt^{2} + f(r)^{-1}dr^{2} + r^{2}(d\theta^{2} + \sin(\theta_{0})^{2}d\phi^{2})\) \( -> m^2 \).

    AlphaNumeric, do you agree or Disagree with the above equations?
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  5. Me-Ki-Gal Banned Banned

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  7. RJBeery Natural Philosopher Valued Senior Member

    I'll give you this, Magneto, you sure jumped right in the ring..!
  8. Emil Valued Senior Member

    In my opinion your attempt is useless to have a serious, argued discussion with AlphaNumeric.
    I simply ignore him.
  9. James R Just this guy, you know? Staff Member


    One of the rules in the opening post was: Relevant direct questions should be answered in a timely manner.

    You have avoided answering quite a few of AN's direct questions to you, such as the ones regarding your equating of a metric to a volume and the ones concerning your equating of tensor quantities with scalar quantities. Do you plan to answer those questions?
  10. rpenner Fully Wired Staff Member

    I'm also on the sidelines with Me-Ki-Gal. So this is just to give feedback on the temperature of the audience watching your discussion.
    Citation? What definition of hypocrisy and what facts do you use to support this claim?
    You are saying that demonstrating knowledge of physics does not give the appearance of understanding of physics, and that further physics is BS?
    "Their missionScientist goals: to fight injusticereject experimentalist error and poor techniques in favor of the best methods and best data, to right thatthose theories which is wrongare incompatible with repeatable observation, and to serve all mankind to the exclusion of ones ego!" (adapted from Superfriends (1973 - 1975))
    At a minimum, one has to be able to do calculations to find out if existing theories are compatible or incompatible with the experimental record, and as all theories are mathematical and all experimental records are fundamentally numerical, correct usage of math would seem to be required to pursue the goals I have always assumed AlphaNumeric to aspire to. But do you enumerate the goals which you think AlphaNumeric does not share?

    "There is no royal road to geometry." (attributed to Euclid) Your "shortcut math" is of no utility unless it actually is as good or better than the math used in textbooks to apply GR. Is it possible that you are acting out because lots of people did understand AlphaNumeric's exposition. I know I did understand it.

    I would add the qualifier "consistently" in there. Even a stopped clock is right twice a day, but you don't rely on it even it weighs a lot less with all those "extra" parts taken out.

    Units are unphysical choices made by humans. No physical theory should have have a preference for SI units over CGS or Imperial units. It's common in GR to use units where c=1 and G=1, so everything is measured in units of (some power of) length or mass. Plugging the SI units back in is an exercise in arithmetic, but more importantly seems to miss the point that getting units wrong is an analogy for confusing a scalar value with a 2-slot tensor quantity. The tensor (in 4-space) has 6 or 10 or 16 times as much data (depending on its symmetry properties) as a scalar value and transforms differently than a scalar quantity under a change of coordinate systems to describe the same physical truths.

    Not all physics is arithmetic. Since Newton, analysis has been part of the math of physics.

    You misread -- he wasn't asking you to break your pseudonym (although in a previous thread you already did this) he just wanted to know the names of the journals in which you are published. "Omni", "Scientific American", "American Physics Teachers Quarterly", "Physics Letters", "Physical Review", .....

    You never fail if you never try. But failure is what scientists learn from.

    Two points: 1) In addition to other work Einstein published 4 great papers in one year (1905) -- a long series of lengthy explorations of the idea of curved space time (1910-1916), a popular physics book (1920?) and a few other papers still relevant today (1920-1935, including the application of the Bose thermodynamics to photons). The last 20 years of his publication life were largely non-eventful not because he couldn't get past editors with his ideas, but because his ideas were not useful. There was no there there. Moreover, Einstein didn't die an unknown. We have the collected notes and papers of Einstein today, so if there was a editor-buried gem, why hasn't someone found it in the last 55 years?
    2) Under US law, you cannot patent or copyright a law of nature or mathematical method. While journals may overreach in their boilerplate contracts, they do want to make sure that they get paid for the hard work of refereeing, editing, proofing and publishing journals with little-to-no advertising revenue. So, yeah, they want to get paid. How does your desire for intangible control compare with their concrete need to survive month-to-month? Or were you already planning to sell your own bound editions of reprints in competition with the big journals?

    ArXiv (says so on it's own pages) is not peer-reviewed and not suitable for papers in early drafts. It's a pre-print server for professional-to-professional communication of papers that will be appearing in journals. Violation of that operating principle commonly gets people kicked off of ArXiv.

    You don't quote in professional physics papers -- you cite. And you don't frequently cite your own work or non-peer-reviewed material unless you want to stink of ego and self-promotion.

    "wow" -Me-Ki-Gal
    Still wrong with tensor left side and scalar right side.
    \(V_{ol}\) is eccentric at best when \(V_{\textrm{Euclidean}}\) is meant.
    \(d V_{\textrm{Euclidean}} = r^2 \left| \left( \sin \theta \right) dr d\theta d\phi \right|\) (both sides have units of volume, which in SI units is m^3)
    Or you can write it as:
    \(d V_{\textrm{Euclidean}} = \sqrt{ \left| g_{ab} dx^{a} dx^{b} \right| } = \sqrt{ \left| \begin{pmatrix} g_{rr} dr dr & g_{r\theta} dr d\theta & g_{r\phi} dr d\phi \\ g_{\theta r} d\theta dr & g_{\theta\theta} d\theta d\theta & g_{\theta\phi} d\theta d\phi \\ g_{\phi r} dr dr & g_{\phi\theta} dr d\theta & g_{\phi\phi} d\phi d\phi \end{pmatrix} \right| } = \sqrt{ \left| \begin{pmatrix} 1 (dr)^2 & 0 & 0 \\ 0 & r^2 (d\theta)^2 & 0 \\ 0 & 0 & r^2 \sin^2 \theta (d\phi)^2 \end{pmatrix} \right| } \\ = \sqrt{ r^4 \sin^2 \theta (dr)^2 (d\theta)^2 (d\phi)^2 } = r^2 \left| \sin \theta dr d\theta d\phi \right| \) (again with units of volume for the entire subexpressions like the left and right sides).

    As a sidebar, you can consider the units of the components of x and g.
    \(x_r, x^r\) (m), \(x_{\theta}, x^{\theta}\) (radians), \(x_{\phi}, x^{\phi}\) (radians), \(g_{rr}\) (no units), \(g_{r\theta}, g_{r\phi}, g_{\theta r}, g_{\phi r}\) (m/radian) , \(g_{\theta\theta}, g_{\theta\phi}, g_{\phi\theta}, g_{\phi\phi}\) (m^2/radian^2). The mixed units are a result of the choice made in parametrizing the coordinate system, not intrinsic in the geometry of Euclidean 3-space. In the Cartesian coordinate system, g is entirely dimensionless (and trivial).

    Often the radian is considered dimensionless and can be dropped, but is somewhat useful to retain should you need to convert to degrees.

    or with the previous result: \(V = \int \int \int d V_{\textrm{Euclidean}} = \int_{r=0}^{R} r^2 dr \int_{\theta=0}^{\pi} \sin \theta d \theta \int_{\phi=0}^{2 \pi} d \phi = \left(\frac{R^3}{3} - 0 \right)\left( 1 - (-1) \right) \left( 2 \pi - 0 \right) = \frac{4 \pi R^3}{3}\)
    The focus on the limits of integration are critical since no part of that information is contained in g.

    I think you mean shell, not cap.

    \(dr = 0 \Rightarrow dA_{\textrm{Euclidean}} = \sqrt{ \left| g_{ab} dx^{a} dx^{b} \right| } = \sqrt{ \left| \begin{pmatrix} g_{\theta\theta} d\theta d\theta & g_{\theta\phi} d\theta d\phi \\ g_{\phi\theta} dr d\theta & g_{\phi\phi} d\phi d\phi \end{pmatrix} \right| } = \sqrt{ \left| \begin{pmatrix} r^2 (d\theta)^2 & 0 \\ 0 & r^2 \sin^2 \theta (d\phi)^2 \end{pmatrix} \right| } \\ = \sqrt{r^4 \sin^2\theta (d\theta)^2 (d\phi)^2} = r^2 \left| \sin \theta d\theta d\phi \right| \).
    So \(dr = 0 \Rightarrow A(R) = \int \int d A_{\textrm{Euclidean}} = R^2 \int_{\theta=0}^{\pi} \sin \theta d \theta \int_{\phi=0}^{2 \pi} d \phi = R^2 \left( 1 - (-1) \right) \left( 2 \pi - 0 \right) = 4 \pi R^2\)
    So boundary conditions are all-important.

    So you have the result that \(\Omega = 4 \pi = \frac{A(R)}{R^2} = \frac{1}{r^2} \frac{dV}{dr}\) which is just as true in Cartesian coordinates as the coordinates that you expressed g in.

    "I say thee nay" -Thor.
    If \(\Omega\) is the angular measure of the surface of a sphere, it is constant (with units of radians^2). If you have some other shape in mind, you need to specify what it is. Then \(\Omega^2\) is constant, and \(d \Omega^2\) is zero. You can't just invent symbols like \(\theta_0\) and throw them into the right hand side. 2 + 1 = 1 + 2 is an equation; \(2 + 1 = 1 + 2 + 3\) is not.

    The above equation makes no sense unless
    1) dt is measured in meters
    2) f(r) is dimensionless
    3) you replace \(\theta_0\) with just \(\theta\).

    And until you specify what f(r) is, there is nothing to agree or disagree with, even if you did bother to write some exposition on the geometry.

    Remember, Schwarzschild defined his coordinates so that the area of the shell of constant radius would always be \(4 \pi R^2\) -- that was a choice he made. The physical meaning of r and f(r) may not be what you think by copying this equation without understanding more of the math of GR than is displayed here.
    Last edited: Apr 15, 2011
  11. Magneto_1 Super Principia Registered Senior Member

    James R, do you remember this one?

    This sounds like you saying, "Hey, let's play poker, and while we are at it, why don't we let my family members and friends hold the real money while we play."

    This statement sound "foolish" a vector is not a scalar; period. A vector is comprised of two components 1) a scalar or magnitude component, and 2) a directional component.

    Secondly I never mentioned anything about a 2 dimensional vector. All vectors that I have been describing are three (3) dimensional vectors. Once again more "hypocrisy."

    Let me describe the difference between a vector and scalar to you, so that you are not confused.

    Rectangular Coordinate Radial Vector for a Sphere. A sphere is a perfectly round geometrical object in three-dimensional space.

    \(r \mathbf{e}_{r} = \({x \mathbf{e}_{x}} + \ y \mathbf{e}_{y} + \ z \mathbf{e}_{z})\). \( -> m\)

    Scalar Magnitude components (x, y, z) of the three dimensional Cartesian Coordinate Radial Vector ofa Sphere described above:

    \(x = \ {r}{sin(\theta)}{cos(\phi)} \). \( -> m\)

    \(y = \ {r}{sin(\theta)}{sin(\phi)} \).\( -> m\)

    \(z = \ {r}{cos(\theta)} \). \( -> m\)

    Now here is where understanding comes in! Do you know why the above Rectangular or Cartesian Coordinate Radial Vector equation is limited when it comes to General Relativity??

    The above Radial Vector equation is limited when it comes to General Relativity because when we need to discuss the "Area" of the Sphere; or for the PhD crowd the "Metric" of the Sphere, the above equation is "trivial" as shown below.

    The Area and or Metric of a Spherical Body is equal to the net sum of the square of each three dimensional vector component of the Spherical Radial Vector equation:

    \((r \mathbf{e}_{r})^2 = \(({x \mathbf{e}_{x}})^2 + \ (y \mathbf{e}_{y})^2 + \ (z \mathbf{e}_{z})^2 \)\). \( -> m^2\)

    \(r^2 = \( {r^2}{sin^2(\theta)}{cos^2(\phi)} + \ {r^2}{sin^2(\theta)}{sin^2(\phi)} + \ {r^2}{cos^2(\theta)} \)\).\( -> m^2\)

    \(r^2 = r^2 \).\( -> m^2\)

    In Metric Differential Form. This Metric differential form represents a change in Area of the sphere via line elements in all three dimensions.

    \(d(r)^2 = \(d(x)^2 + d(y)^2 + d(x)^2\)\).\( -> m^2\)

    Now the "non-trivial" solution for the Area and or Metric of a Spherical Body is realized when using Spherical Coordinates. This is what Karl Schwarzschild realized in 1915 when he solved the first exact solution to the Einstein field equations of general relativity, for the limited case of a single spherical non-rotating mass, which he accomplished in 1915, the same year that Einstein first introduced general relativity.

    Spherical Coordinate Radial Vector for a Sphere. A sphere is a perfectly round geometrical object in three-dimensional space.

    \(s \mathbf{e}_{r} = \({r \mathbf{e}_{r}} + \ {r_{\theta}\mathbf{e}_{\theta}} + {r_{\phi}} \mathbf{e}_{\phi} \)\).\( -> m\)

    \(s \mathbf{e}_{r} = \({r \mathbf{e}_{r}} + (\ {r {\theta})\mathbf{e}_{\theta}} + ({r} \sin(\theta_{0}))({\phi}) \mathbf{e}_{\phi} \)\).\( -> m\)

    Scalar Magnitude components (r, \( \theta \), \( \phi \)) of the three dimensional Spherical Coordinate Radial Vector of a sphere described above:

    Radius relative to the Mean Center of the Sphere

    \({r} = {r} \)\( -> m\)

    This is a Latitudinal Geodesic on the surface of the Sphere

    \({r_{\theta}} = (\ {r {\theta})} \)\( -> m\)

    This is a Longitudinal Geodesic on the surface of the Sphere

    \({r_{\phi}} = ({r} \sin(\theta_{0}))({\phi}) \).\( -> m\)

    In general a Geodesic is a distance on the surface of a sphere, which is equal to a curve. The shortest distance on a sphere is a geodesic or a curve.

    The Area and or Metric of a Spherical Body is equal to the net sum of the square of each three dimensional vector component of the Spherical Radial Vector equation:

    \(s^2 = \({r^2} + \ (r_{\theta})^2 + \ (r_{\phi})^2 \)\).\( -> m^2\)

    \(s^2 = \({r^2} + \ {r^2}\({\theta}^2 + \ {\phi}^2 \sin^2(\theta_0) \)\).\( -> m^2\)

    Now the above equation for area is not the simple square of the radius area of the sphere; but there is an additional component that is directional. This directional component will allow you to predict a location on a sphere, or location in the curve of the sphere.

    The above equation for area of a sphere is a more complete description for generating a position or location on the sphere.

    The below equation for area of a sphere allow you to predict a location on the area of a sphere.

    \((s^2 - {r^2}) = {r^2}\({\theta}^2 + \ {\phi}^2 \sin^2(\theta_0) \)\).\( -> m^2\)

    In Metric Differential Form. This Metric differential form represents a change in Area of the sphere via geodesic line elements in all three dimensions.

    Taking the differential of both sides yields

    \(\frac{d(s^2 - {r^2})}{{r^2}} = d({\theta}^2 + \ {\phi}^2 \sin^2(\theta_0) \)\).\( -> m^2\)

    \(ds^2 = \(d{r^2} + \ {r^2}\(d{\theta}^2 + \ d{\phi}^2 \sin^2(\theta_0) \)\).\( -> m^2\)

    You have used complex mathematics to describe your metric and only a few have understanding. I have produced what you produced without the complexity and yet have conveyed the same message and solution to a much wider audience.

    I consider (\(s^2\)) and (\(ds^2\)) a Total Area and Total Area change term that includes the isotropic radial term plus the directional term. The square root of this Total Area (\(s\)) could be a curved path line element as you describe. I do not disagree that this is line term. I will disagree that it is a predictable path length, this is a total length like a conserved term.

    Do you agree with the mathematics that I produced or not?
    Last edited: Apr 15, 2011
  12. Magneto_1 Super Principia Registered Senior Member

    RPenner, this is good stuff!!
    But, who will be Robin Hood?
    Last edited: Apr 15, 2011
  13. Magneto_1 Super Principia Registered Senior Member

    Everyone is so negative, as if someone is always up to something devious.

    The "intangible control" that you speak of will allow me to publish the work to another group, organization, technical society, or journal with a wider audience any time I choose. Going about it your way, if I publish with a journal that owns the rights to the published work, and later on a new organization emerges with a wider audience, I can't use that paper without modification, with this new group, because the old group owns the rights!!

    Sorry I left the units off on this one.

    This is conditional:

    \( \Omega^2 = \frac{A_{{\theta}{\phi}}}{r^2} = {3}(\frac{g_{{\theta}{\phi}}}{r^3}) = (\theta^{2} + \sin(\theta_{0})^{2} \phi^{2})\).\( -> radians^2\)

    This is general
    \( \Omega^2 = \frac{A_{{\theta}{\phi}}}{r^2} = {3}(\frac{g_{{\theta}{\phi}}}{r^3}) = (\theta^{2} + \sin(\theta)^{2} \phi^{2})\).\( -> radians^2\)

    In differential form; if you integrate the equation below you will get the equation above:

    This is conditional:
    \( d\Omega^2 = \frac{dA_{{\theta}{\phi}}}{r^2} = (d\theta^{2} + \sin(\theta_{0})^{2} d\phi^{2}) = {3}(\frac{dg_{{\theta}{\phi}}}{r^3})\).\( -> radians^2\)

    This is general:
    \( d\Omega^2 = \frac{dA_{{\theta}{\phi}}}{r^2} = (d\theta^{2} + \sin(\theta)^{2} d\phi^{2}) = {3}(\frac{dg_{{\theta}{\phi}}}{r^3})\).\( -> radians^2\)

    Let's review the conditional

    Let \( d\Omega^2 = 0 \), meaning that \( \Omega^2 = constant \)

    \(d\theta^{2} = - \sin(\theta_{0})^{2} d\phi^{2}\).\( -> radians^2\)

    which becomes

    \(\frac{d\theta^{2}}{d\phi^{2}} = - \sin(\theta_{0})^{2} \).\( -> unit less\)


    \(\sqrt{1 + \frac{d\theta^{2}}{d\phi^{2}}} = \cos(\theta_{0}) \).\( -> unit less\)

    See link for reference: Geodesic - Arc Length

    General Form
    The Schwarzschild metric:

    \(-ds^{2} = -f(r)dt^{2} + f(r)^{-1}dr^{2} + r^{2}(d\theta^{2} + \sin(\theta)^{2}d\phi^{2})\) \( -> m^2 \).

    Conditional Form
    The Schwarzschild metric:

    \(-ds^{2} = -f(r)dt^{2} + f(r)^{-1}dr^{2} + r^{2}(d\theta^{2} + \sin(\theta_{0})^{2}d\phi^{2})\) \( -> m^2 \).

    Not true, this is measured in units of time or seconds not meters as you state.

    The term f(r) is a squared velocity term. Explaining the meaning of this term is on the forefront of GR these days. But if you want a conceptual and slightly mathematical description of this term. This term could be considered a "Field" term. This term behaves like a fluid either moving uphill or down hill. This term describes how matter interacts with the "Speed of Light" isotropy of its surroundings while the potential of the gravitational gradient field associated with the mass decreases with "Distance;" Hence why this term is a function of radius or distance.

    The term f(r) is a squared velocity term, and is equal to zero at the Schwarzschild Radius Event Horizon (f(r) = f(\(r_{Schwarzschild}) = 0)\).

    The term f(r) is a squared velocity term, and is equal to the square of the speed of light (\(c^2_{Light}\)) at an infinite distance away from the center (f(r) = f(\(r_{infinity} )= c^2_{Light}\))).

    Some people view this term and describe this term as if light speed is somehow changing. I don't claim to have an excellent conceptual description for how to explain this term, but for sure I would not say that light speed is changing.

    I would only speculate and in someways "posit" that the term f(r) = (\(c^2_{Light} - 2v^2_{Gravity}\)) is interaction between the square of the speed of light Isotropy and the Gravitational Potential Field Gradient, which causes the "Aether" Ideal Gas fluid "Dark Matter" in that location to either speed up or slow down, which means that there is a quantity of spacetime expansion energy that is either gained or lost it depends on your minus sign.

    Therefore, it can be said that as the Gravitational Potential Energy of an Isolated Net Inertial Mass body system is decreased relative to the Mean Center of the system, and some other form of energy, a Vacuum Energy is Increased and is maximum at the Speed of Light.

    What happens within the Event Horizon, I will not choose to speculate at this time.
    Last edited: Apr 16, 2011
  14. AlphaNumeric Fully ionized Moderator

    Just to make it clear I am aware you've replied. Today has been a busy day at work staring at enough R code to kill an elephant from 40 yards. As such, due to the time it takes to compose a full response, I'll reply tomorrow. I will however comment quickly on one thing :

    I categorically disagreee with that because it is simply not true in Riemannian geometry. To drop the infinetesimal 'd' terms you need to integrate, thus ds does not uniquely define s, you need integration limits. Further more \(ds^{2}\) and \(s^{2}\) do not represent total area and total area change in Riemannian geometry. Please provide a citation/reference which says they do.

    Actually I'll save you the bother of looking. \(ds^{2}\) is actually \((ds)^{2}\), not \(d(s^{2})\), hence why I mentioned the square root. You can't just integrate up from \(ds^{2}\) to \(s^{2}\). If you consider \(ds^{2}\) as the change in \(s^{2}\) then you are viewing \(ds^{2} = d(s^{2})\), which is not the case. \(ds^{2} = (ds)^{2}\) is obtained from its definition in terms of line elements. I can provide references if needs be.
  15. Magneto_1 Super Principia Registered Senior Member

    I concede, and choose to agree with you, symbolism and nomenclature will always have to be discussed and clarified, so that confusion won't remain.

    \((ds)^{2} approx (dx)^{2} + (dy)^{2} + (dz)^{2} \)

    and for Clarity, I think that this is good mathematical form for a PhD.

    \(ds^2 = \(d{r^2} + \ {r^2}\(d{\theta}^2 + \ d{\phi}^2 \sin^2(\theta_0) \)\).\( -> m^2\)

    This is the form that I teach the physicist alike. Or to use your term "First Years!!"

    \((ds)^2 = \((d{r})^2 + \ {r^2}\((d{\theta})^2 + \ (d{\phi})^2 \sin^2(\theta_0) \)\).\( -> m^2\)

    What I am not demonstrating in the above equation is the vector symbol on the isotropic and homogenous radius (dr) term and this "World Line" distance (ds). You do know what the trace of a "World Line" would be, don't you?? A "World Line" would trace from the center of the system along the radius and travel along the the surface of a curved surface and in our case a sphere.

    And the reason that I agree with you is that the mechanical tool that is being used is Pythagorean’s Theorem which states (r^2 = x^2 + y^2 + z^2).
    Alphanumeric, don’t try and confuse things. Is that understanding??

    Actually, AlphaNumeric, it is this type of reasoning that you have come to here, why I did not waste time sending my papers to journals. I can just imagine reviewers like you picking me apart as they decide that they don't like my "Nomenclature." I don't have time for this!!
    Last edited: Apr 15, 2011
  16. Magneto_1 Super Principia Registered Senior Member

    AlphaNumeric, there is only one or two questions remaining to your challenge that I have not addressed. This is also for James R!

    I will address this soon; and then this will conclude my addressing all of your challenge questions!!
  17. prometheus viva voce! Moderator

    I don't think I'm going to have too much time to reply to this thread, but I'll certainly read it as it's almost inevitably going to be jolly funny!

    One comment, on a point of confusion chosen entirely at random

    The number of "scalar components" of a vector is exactly the dimension of the vector you're considering, not magnitude and direction. When you move on from the high school definition of a vector you will learn that a vector is a one dimensional array of numbers (scalars) whose length can be whatever you like. Usually in physics we have vectors whose length is the number of dimensions, so 3 or 4 is very common.

    Your confusion lies in the fact that magnitude is certainly a scalar, but direction is not.
  18. Magneto_1 Super Principia Registered Senior Member

    Yes, I totally agree that General Relativity does predict multi dimensions. I call this field of General Relativity, Complex General Relativity.

    The reality is that we can measure in three (3) and four (4) dimensions of space and time. Trying to measure in any other dimensions works mathematically, but is not very practical. I am limiting this discussion to three (3) and four (4) dimensional space and time.

    This sounds ridiculous; for clarity, a direction (\((\theta) or (d\theta)\)) is not a scalar and is a separate component of a vector, is the point that I was making in itemizing the distinction that a vector has two components: a scalar magnitude component, and a separate directional component. However a square of the directional component (\((\theta^2) or (d\theta)^2\)) is a scalar!

    Boy, AlphaNumeric you are totally loved. You have lots of family and friends that totally love and support you! You are a very Blessed Man!! How can I find such favor??

    Maybe you should change your name from AlphaNumeric to Caesar!

    No, better yet, let's change your name to "Paul" so that once you are converted, you can spread the gospel of the Super Principia Mathematica throughout the the outermost regions of the United Kingdom!

    AlphaNumeric, I know why you have not post, your mind is telling you, I will let him win on the geometry stuff. But there is no way he can get the Stress Energy right, for sure he will mess up there! What "Magic" is he going to pull off? There is no way he is going to prove that mass is directly proportional to the Stress Energy Tensor.

    Your mind is playing tricks on you!!!!

    Remember, you initiated this challenge; I was minding my own business, posting and responding to post just like everyone else. But no, you decide that you want to be big man on campus, and wear your PhD on your chest like Superman's Moniker; choosing to pick on whom you will. Next time you might want to pick and choose your battles a little more carefully.
    Last edited: Apr 16, 2011
  19. temur man of no words Registered Senior Member

    Is this Reiku or Farsight? Or were they the same person?
  20. Magneto_1 Super Principia Registered Senior Member

    Temur, don't let the Cosmic Vacuum "Dark" Force apply lots of "Stress Energy" to your brain because it will expand! This is all "Physics Fun!"

    I am not discounting the skill set of AlphaNumeric, from my reading of his many post, he is one of the "Sharpest Tools in the Shed;" and is an excellent theoretician, mathematician, and physicist. If I were putting a team together to do physics, I would definitely want AlphaNumeric on the team.

    Though, he is definitely a "Pompous A**;" hence why I am in his "challenge."
    Last edited: Apr 16, 2011
  21. przyk squishy Valued Senior Member

    No, they're all different people. Farsight is middle-aged, has a family, successful professional career (from what I gather, anyway), and has appeared on TV and has published a book (which, ironically, we know about thanks to AlphaNumeric). He's obsessed with the idea he has a special understanding of Einstein and other icons in the history of physics - an understanding that the rest of us lowly physics graduates have missed because we merely memorized our courses instead of really thinking about them. His posts are generally well written. His early posts absolutely stank of an attitude that made it clear he was here to lecture rather than listen. He seems to think either he can read people's minds or that he's particularly gifted in psychology (or simply that we're all just automatons with exactly the same "mainstream" thoughts), judging by the frequency with which he assumes he knows all about our views on various topics in physics without us even telling him about them.

    Reiku by contrast was in his early twenties, insecure, constantly whining for attention, likely posted drunk sometimes (his spelling ability varied), and was all-round clearly afflicted with mental issues beyond your usual case of self-delusion. He tended to post about stuff like quantum consciousness. His posts were incoherent and read like a sort of collage of material copied with little modification or understanding behind it. I caught him doing a poor job of [POST=2552811]plagiarising one of my own posts[/POST] once.

    Reiku, like Magneto here, also didn't hesitate to post sometimes completely made up math he obviously didn't understand. Farsight at least has the sense to find excuses for avoiding math when he's confronted with it (usually along the lines that it's unnecessary, irrelevant, or even deliberately obscures the pertinent issues). Magneto's confusion with differential elements (dividing both sides of an equation by the 'd' in ds[sup]2[/sup]) is [POST=2005975]Reiku caliber[/POST].
  22. Magneto_1 Super Principia Registered Senior Member

    Please point out and post where, anywhere in this particular thread post, I have presented some form of mathematics that you disagree with?
  23. Magneto_1 Super Principia Registered Senior Member

    Could you explain why you did this?

    I am assuming \(g_{\theta\theta} = 1 = {r^2}\)
    And, you are choosing \(g_{\theta\phi} = 0 = d{r}\)

    \(ds^2 = \(d{r^2} + \ {r^2}\(d{\theta}^2 + \ d{\phi}^2 \sin^2(\theta_0) \) = \(d{r^2} + \ {r^2}\(d{\Omega}^2 \)\).\( -> m^2\)

    So this becomes according to your condition:

    \(ds^2 = \(\ d{\theta}^2 + \ d{\phi}^2 \sin^2(\theta_0) \) = \(d{\Omega}^2 \)\).\( -> m^2\)

    and likewise

    \(ds = sqrt{\(\ d{\theta}^2 + \ d{\phi}^2 \sin^2(\theta_0) \)} = \ d{\Omega} \).\( -> m\)

    Would this equation be a better choice in this form based on your conditions?

    \(s(L) = \int_{L}ds = \int_{\Omega}d\Omega = \int_{\Omega}\sqrt{d\theta^{2} + \sin^{2}\theta d\phi^{2}}\)

    AlphaNumeric, This is also a solution

    In differential form; if you integrate the equation below you will get the equation above:

    This is conditional:
    \( d\Omega^2 = \frac{dA_{{\theta}{\phi}}}{r^2} = (d\theta^{2} + \sin(\theta_{0})^{2} d\phi^{2}) = {3}(\frac{dg_{{\theta}{\phi}}}{r^3})\).\( -> radians^2\)

    This is general:
    \( d\Omega^2 = \frac{dA_{{\theta}{\phi}}}{r^2} = (d\theta^{2} + \sin(\theta)^{2} d\phi^{2}) = {3}(\frac{dg_{{\theta}{\phi}}}{r^3})\).\( -> radians^2\)

    Let's review the conditional

    Let \( d\Omega^2 = 0 \), meaning that \( \Omega^2 = constant \)

    \(d\theta^{2} = - \sin(\theta_{0})^{2} d\phi^{2}\).\( -> radians^2\)

    which becomes

    \(\frac{d\theta^{2}}{d\phi^{2}} = - \sin(\theta_{0})^{2} \).\( -> unit less\)


    \(\frac{ds}{dr} = \sqrt{1 + \frac{d\theta^{2}}{d\phi^{2}}} = \cos(\theta_{0}) \).\( -> unit less\)

    \(ds = (\sqrt{1 + \frac{d\theta^{2}}{d\phi^{2}}}){dr} \).\( -> m\)

    \(s(L) = \int_{L}ds = \int_{L}(\sqrt{1 + \frac{d\theta^{2}}{d\phi^{2}}}){dr}\).\( -> m\)

    See link for reference: Geodesic - Arc Length
    Last edited: Apr 16, 2011
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