Discussion : Magneto

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

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  1. AlphaNumeric Fully ionized Registered Senior Member

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    What you give there is nothing but a copy and paste of trivial book work. When you're pressed to enter into a discussion you either fail or resort to that.

    I didn't use short cuts, I stated something which you should have known. I didn't go through the explicit steps because I gave you enough credit to know how to do the transformation, you obviously didn't.

    When you go from one set of coordinate to another you should change all the original coordinates into the new ones. Your expression mixes them and is therefore wrong. Not only that it is wrong anyway.

    I walked you through it, I even did the \(dr^{2}\) for you. You either didn't read it or didn't understand it. Either way you haven't bothered to go and find out more. I can give you book references if you want, this is very basic and commonly covered. You aren't 'a little more exacting', you clearly don't give a hoot about being exact or right or accurate else you'd not have made the wealth of mistakes you've made.

    And I responded to it in the very next post. You aren't saying anything I'm unaware of, the definition of manifold means it looks flat close up. But that doesn't mean a space is flat, else ALL manifolds would be flat. A sphere looks approximately flat close up but it is still a sphere and therefore still have constant curvature. You can't get away from that.

    Once again you mouth definitions of basic stuff but you don't show any comprehension of what you say.
     
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  3. Magneto_1 Super Principia Registered Senior Member

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    Finally we can agree. This post is now complete for me!
     
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  5. AlphaNumeric Fully ionized Registered Senior Member

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    Agree? I agree on the definition of a manifold but not in the claim you're making that a sphere is flat because it looks flat close up. All manifolds, no matter how twisted and curved and warped, are locally morphic to \(\mathbb{R}^{N}\) by definition. This is the essence of the construction of atlases from sets of charts. I can very much recommend 'Geometry, Topology and Physics' by Nakahara for a good introduction to this stuff.

    If you take any open set on a N-sphere you can map that into an open set in \(\mathbb{R}^{N}\) but you can't take the entire sphere and map it bijectively into \(\mathbb{R}^{N}\). This is seen in the case of latitude and longitude because the North pole on the sphere maps to a line in Euclidean space, ie \((\theta,\phi) = (0,\phi)\) all represent the North pole and likewise for South with \(\theta = \pi\). Thus they are inequivalent.

    If you claim otherwise then please provide a bijective coordinate transformation between all of an N-sphere to a closed subset of \(\mathbb{R}^{N}\). I'll be nice, you can consider just \(S^{2}\) if you want.

    Except you didn't answer a large number of direct questions I asked you or provide the derivations of results you claimed to have.

    Do you understand your units for the components of \(g_{ab}\) were wrong?
    Do you understand your definition of the SC metric is wrong, in that you don't derive the SC metric?
    Do you understand your definition of 'fluxes' in regards to my thesis was wrong?
    Do you understand the difference between tensor components and tensor rank?
    Do you understand your attempt at a coordinate transform here is wrong, completely and utterly?
    Do you understand a sphere isn't 'flat' just because it looks flat sufficiently close up?
    Do you understand the Minkowski metric is non-Euclidean?
    Do you understand that general relativity, 'complex' or otherwise, does not make predictions about 'multi-dimensions'?
    Do you understand that \(d(s^{2}) \neq (ds)^{2}\), even though the units may match?
    Do you understand you cannot integrate up \((ds)^{2}\) to \(s^{2}\) directly?
    Do you understand how your interpretation, \(ds^{2} = d(s^{2})\), is inconsistent with the arc length formula you've quoted from Wikipedia on a number of occasions?
    Do you understand that coordinates, like latitude and longitude, are not directions but positions and directions are a different concept in differential geometry?
    Do you understand the metric isn't the area or volume of anything, it defines line elements and that volumes are constructed from metrics in a different manner?
    Do you understand that the problem the SC metric has with the event horizon is due to the choice of coordinates, not a physical problem, and that other coordinates allow for the dynamics of an object crossing the event horizon to be computed?
    Do you understand that expressions like \(g_{\theta\phi}\) for metric components doesn't automatically mean that said component is a function of \(\theta\) and \(\phi\)?

    And questions unrelated to your understanding,

    Why did you bother to copy and paste that chunk of text on hyper spheres, despite having seen what my research was on and saying you understood it? Why did you think quoting basic definitions of objects I've made clear I'm familiar with would in any way provide a good response to my questions and comments?

    The original point of this thread wasn't for me to ask a list of questions, you answer and then be done, it was to engage in a back and fore where new questions and points of discussion come up as answers are given. That's how discourse works. I can understand your desire to cut and run, you've had every claim knocked down and you've demonstrated gross misunderstandings and ignorance each time you've attempted to do any kind of mathematics. Continued discussion would only continue to demonstrate your lack of understanding, your intellectual dishonesty and your detachment from the reality of your lack of capabilities. So the question now is whether you've got just enough grip on reality that you know continued discussion would only serve to have you dig your hole deeper or whether your misplaced self confidence is so blinding you think you can turn this discussion around. :shrug:
     
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  7. Me-Ki-Gal Banned Banned

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    Oh you are funny! My hero, Robin Hood
     
  8. Magneto_1 Super Principia Registered Senior Member

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    Thanks.

    I agree, on any curved surface if you describe "flat" or straight line, the "flat" is only approximated because if you took infinitesimal distances of that straight line you would find curves on those local spaces. Curves are Curves period; although they can be approximated by straight lines or "flat spaces."

    See Affine Connection

    Like when playing "racquet ball" or "squash" you hit that "squashable" soft ball in a straight line trajectory towards the wall, it looks like a straight line. But from the consideration of the play room relative to a sphere the size of the earth you see curves and not straight lines.

    Let's, get to this some other time.


    I believe that I derived the Schwarzschild Metric (SC) correctly, in that the (SC) is comprised of three component terms being a net difference and is constructed:

    SC = (Field Term) - [(Expansion Term) + (Patch Area Term)]

    I do know that there is a way within the confines of the field "Complex General Relativity" that derives the SC metric via the Einstein Field Equation.

    From my point of view, I constructed the SC metric it in a way that people can achieve understanding.

    Since you know it so well "professor", why don't you demonstrate for us your exceptional physics and mathematical capabilities, and you derive the SC metric; AlphaNumeric Style!! :shrug:

    Yes. I may have been totally out of left field on that one! My question to you was and still stands. In your concept of "Fluxes" do they have a speed and or velocity component? And if they do, do either of those speed or velocity components travel at speeds much faster than light speed in a vacuum or what ever medium?

    See: Gauge covariant derivative


    Yes. Since you know it so well "professor", why don't you demonstrate for us your exceptional physics and mathematical capabilities, and you explain the difference between tensor components and tensor rank; AlphaNumeric Style!! :shrug:


    No. I do not believe that I have presented physics or mathematics of any sort that has been in your words "wrong, completely and utterly." I in no regards consider myself to be a perfect mathematician, or to have mastered all of physics; physics is vast, and I am still learning new things all of the time. I am sure that I have made some mistakes, hopefully minor, and that which can be corrected with some "reasonable" discussion with a "reasonable," and rationally thinking human being!


    Yes. A curve is a curve, no matter how infinitesimal or infinite.

    Since you know it so well "professor", why don't you demonstrate for us your exceptional physics and mathematical capabilities, and you describe how a sphere isn't 'flat' just because it looks flat sufficiently close up; AlphaNumeric Style!!

    Yes, AlphaNumeric I do understand the Minkowski metric is non-Euclidean

    It probably would have been better to describe the Minkowski/Einstein Metric as such:

    The Minkowski/Einstein Metric - Pseudo-Euclidean:

    The Minkowski/Einstein Metric is also considered Pseudo-Euclidean, is a net difference, and is comprised of two components: 1) a three dimensional space-time radial component and, 2) a two dimensional locational directional component. The Result is a four dimensional element vector (space, time, latitude, longitude).

    The Minkowski/Einstein Metric - Euclidean :

    \(ds^{2} = ({c^2_{Light}}dt^{2}) + r^{2}(d\theta^{2} + \sin(\theta)^{2}d\phi^{2})\) \( -> m^2 \).


    SC = [(Space-Time Term) + (Patch Area Term)]


    The Minkowski/Einstein Metric - Non-Euclidean :

    \(ds^{2} = ({c^2_{Light}}dt^{2}) - r^{2}(d\theta^{2} + \sin(\theta)^{2}d\phi^{2})\) \( -> m^2 \).


    SC = [(Space-Time Term) - (Patch Area Term)]


    The difference between the Euclidean and the Non-Euclidean Metric is the adding or the subtracting of the "Patch Area" term.

    Where the space-time radial component in the Center of Mass frame of reference is given by,

    \(dr^{2} = ({c^2_{Light}}dt^{2})\) \( -> m^2 \).

    This Minkowski/Einstein Metric term fell out of favor, when the Schwarzschild Metric (SC) was introduced; in that there was big contention about the relevancy of Minkowski’s Non-Euclidean Metric.

    Some physicists found the Non-Euclidean Term useful, while others found this term not so useful, hence the contention about this Pseudo-Euclidean term. If Einstein had not jumped on the band wagon, about this concept, it may have ended up in obscurity. Minkowski was one of Einstein's professors.

    I will try to get to the others later; on this condition.

    I don't mind your obnoxious, and sometimes, oh well, a lot of times, abusive humor, I can handle. I want you to be who you are, and you, and you, and that other you that I left out, in my "Quark" joke; I would have had to describe a Neutron!

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    But what I will absolutely not tolerate is "SLANDER." In my country slander is illegal, and you can be sued for that kind of thing! My friend, Slander is akin to bearing false witness, and is even bigger than that, it is a "sin" before God and is listed with one of the ten (10) commandments in the Bible. So whether you can profess to be a "Theist" or "Atheist," we all know what "Slander" is!!

    So, AlphaNumeric, I guess this means that your "Conversion" is Now Complete!! From here on out, "Resistance is Futile!!"
     
    Last edited: Apr 24, 2011
  9. Magneto_1 Super Principia Registered Senior Member

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    Wow!! AlphaNumeric, you are correct I, did do "something wrong"; just not "completely and utterly."

    And the reason that it is not "completely and utterly" wrong, is that I still maintain, that you did not maintain "Frame of Reference Independence" during your "Cartesian/Rectangular" to "Spherical" Coordinates transformation, which I will maintain during the derivation.


    The "Final Result" is correct but, I believe that you wanted to see is the "Total Differential Transform" - where the Chain Rule is applied:

    Scalar Magnitude components (x, y, z) of the three dimensional Cartesian Coordinate Radial Vector as a function of Spherical components of a Sphere described below:

    \(dx_{s} = \ d({r}{sin(\theta)}{cos(\phi)}) = [(d{r}){sin(\theta)}{cos(\phi)}) + \ ({r})\ d({sin(\theta)}{cos(\phi)})] \). \( -> m\)

    \(dy_{s} = \ d({r}{sin(\theta)}{sin(\phi)}) = [(d{r}){sin(\theta)}{sin(\phi)}) + \ ({r})\ d({sin(\theta)}{sin(\phi)})] \).\( -> m\)

    \(dz_{s} = \ d({r}{cos(\theta)}) = [(d{r}){cos(\theta)}) + \ ({r})\ d({cos(\theta)})]\). \( -> m\)

    Remember that the "Dot Product" of two linear vectors is a scalar term.

    And the square of a linear vector is equilavent to the square of the scalar magnitude of that term; independent of whether that line is a fixed term or a differential term ---- \((dx)^2 = dx^2 .\)


    Ok, so the below Metric does transform: :shrug:

    External Observer Frame

    \(ds^{2} = dx^2_{s} + dy^2_{s} + dz^2_{s}\)

    And

    Center of Mass/Momentum Frame - Proper Frame

    \(dr^{2} = dx^2 + dy^2 + dz^2\)

    Magnitude - Center of Mass/Momentum Frame - Proper Frame

    \(r^{2} = x^2 + y^2 + z^2\)


    And is transforms from the Rectangular Cartesian Coordinates above into the Spherical Coordinates. The equation below were obtained using the mathematical method known as the Chain Rule, and playing around with Trigonometric Identities:

    \(ds^{2} = dx^2_{s} + dy^2_{s} + dz^2_{s} \to dr^{2} + r^{2}(d\theta^{2}+\sin^{2}\theta d\phi^{2})\)

    Math transformation process yields


    \(ds^{2} = dr^{2} + r^{2}(d\theta^{2} + \sin^{2}\theta \, d\phi^{2})\)


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


    \(ds^{2} = dx_s^{2} + dy_s^{2} + dz_s^{2} = dx^{2} + dy^{2} + dz^{2} + (x^{2} + y^{2} + z^{2})(d\theta^{2} + \sin^{2}\theta \, d\phi^{2})\)



    Great catch Alphanumeric! So, maybe the statement shoud read that, "Magneto" did "something wrong;" but "AlphaNumeric" you made a mistake "Completely and Utterly" common with that of "First Years!!"

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    Last edited: Apr 23, 2011
  10. Magneto_1 Super Principia Registered Senior Member

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    I kind of "Quasi" agree. I do understand that "Classical General Relativity" does not make predictions when it comes to our modern day terminology for "multi-dimensions;" where things like Branes and Worm Holes are typically discussed. The fields of Super Gravity, Super Strings, and String Theorist alike are typically working.

    However, the "multi-dimensions" that I believe, can be discussed in "Classical General Relativity" is what you describe above with N-Spheres.

    \(\mathbb{R}^{N}\)

    For me the above would completely describe the classical multi-dimensions that I was referring to. Stop trying to be "Dishonest!!"

    Finally, the inverse of the N-Sphere equation is far more useful to me!

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    \(\frac{1}{\mathbb{R}^{N}}\)

    And, you should know this one; will you show me how to express the following using either the "mult-dimension" or "N-Sphere" concept?

    \(ds^{2} = g_{ab}(x^{c})dx^{a}dx^{b}\)


    AlphaNumeric, why do you continue to be so dishonest, we have gone over this a million times already. This that you keep brings up was a typo, when I when I was thinking of the area equation in the above equation where I made the mistake.

    But here is the dishonest part, you keep bringing this up when, It only occurred one time! And the mistake itself is a real mistake in that I left the (d(z)^) off of the equation. So not only did I express the concept incorrectly, I wrote the equation down incomplete also.

    You can't take points off for that. That is totally dishonest. Boy, I would not want you "Grading" my test!!


    Yes. Once again you are bringing up old things that, have already been hashed.

    First to go from (\((ds)^{2}\) ---> \((ds)\)) you have to take the square root

    \(ds = \sqrt{(ds)^{2}} \)

    This you can integrate and square

    \(s^{2} = \(\int_{L}ds\)^2 \)


    Prove it!! ??

    However, while we are here, the above "mistake" that you are erroneously are pointing out, is an "Area Differential" is real and, this is up for further discussion. This kind of "Area Differential" is what Johanes Kepler was doing in his Astronomia Nova in 1609.

    Yes. When you hold your Global Positioning System (GPS) unit in your hand, they are in "cell phones" today, the information is relayed to you in "Geo" location units of "Latitude" and "Longitude."

    However when specifying location relative to an inertial observer at the center of the system, or an external frame observer, they would consider latitude and longitude as being directions relative to the mean center of the system.

    One again, you are having problems with frames of reference; did you remain awake in "First Years - Special Relativity."

    Yes. passing through the event horizon can occur when discussing Heat Radiation Energy, and Vacuum Energy. There is definitely something strange and even an equation that just "Vanishes" at the "Event Horizon."

    \(S = \frac{r}{\sqrt{1 - (\frac{r_{S}}{r})}} \) \( --> m\)


    I am aware that there is a solution known as the Kruskal solution to resolve this. I don't claim to know exactly how it works, but I do know its function. Perhaps, you could explain it?
     
  11. Magneto_1 Super Principia Registered Senior Member

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    Here, is where, I can use your help, my goal here is to reach common ground?

    I do realize that I am defining Volume and Area correctly however a little different than what is currently interpreted by using a Rank 2 Tensor, my use of the indices (\(\theta\) and \(\phi\)), and the Einstein Field Equation.

    I do know that the math work, because I have built some rough "sim" models to validate this. But, now I need a good interpretation or explanation for making the equivalencies, and I could use your help with this matter.:shrug:

    Plus, this is all part of your conversion process anyway!!:bravo:, you've made it!
     
  12. AlphaNumeric Fully ionized Registered Senior Member

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    6,702
    Your result is wrong. The result \( [tex]ds^{2} = dx_s^{2} + dy_s^{2} + dz_s^{2} = dx^{2} + dy^{2} + dz^{2} + (x^{2} + y^{2} + z^{2})(d\theta^{2} + \sin^{2}\theta \, d\phi^{2})\)
    [/tex] is not true. I see why you've done it, you think \(dr^{2} = dx^{2} + dy^{2} + dz^{2}\) because \(r^{2} = x^{2}+y^{2}+z^{2}\).

    I've already explained part of the problem, that if you change coordinates, say from (x,y,z) to \((r,\theta,\phi)\) then the expression for \(ds^{2}\) should change from being in terms of dx, dy, dz to \(dr,d\theta,d\phi\). You can't mix the expressions, it is like saying "I'm going to translate this from English to French" and then end up with a sentence which is a mixture of French and English. It's one or the other, not both.

    You're also making some distinction between x and \(x_{S}\). There is no such thing as \(x_{S}\). What you're referring to is writing Cartesian coordinates in terms of spherical ones.

    So yes, you have got it wrong and in doing so demonstrated you don't know how to do coordinate transformations properly.

    I'm not being dishonest, I'm pointing out mistakes you make. If the mistakes you make are due to you being unable to explain yourself properly then its not my fault. You claim to be well read in this stuff and have been involved in education, so one would expect you to be capable of explaining your thoughts. As such when your explanations are wrong the implication is that your understanding is wrong. At best your "Oh I really meant ...." back peddling explain away one mistake while demonstrating another.

    Let's consider what you said :

    That isn't an N-sphere, it's N dimensional Euclidean space. An N sphere is written as \(S^{N}\) and is topologically distinct from \(\mathbb{R}^{N}\). I even used that notation previously/ Considering such notation is everywhere, EVERYWHERE, in general relativity the fact you don't understand it shows you have no understanding of the terminology you're quoting from Wikipedia. This illustrates that the stuff you quoted to me about hyperspheres was stuff you didn't understand.

    Not only does that have nothing to do with hyper spheres but the notation is meaningless. It is demosntrating that you haven't got a bloody clue what the notation means, much less understand its use within differential geometry and GR. Find me one reputable journal with a paper which uses that notation.

    This demonstrates precisely what I predicted, your continued replies would only serve to dig yourself in deeper.

    Your question illustrates you don't even know what the notation means. Its just the general expression for a line element in Riemannian geometry. It can pertain to any kind of Riemannian space, not just a sphere. It is already a concept which works in any number of dimensions, that is what the indices are for.

    No, it wasn't just a type. You did your derivation via \(s = \) and then \(s^{2} = \) and then you just slapped on 'd' on the side. You didn't do a single typo, you gave an entire derivation based on that misunderstanding!

    The fact you left the dz terms off is irrelevant to your mistakes of understanding (or rather misunderstanding).

    You say that now, after I walked you through it.#

     
  13. AlphaNumeric Fully ionized Registered Senior Member

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    6,702
    Once again you just quote things at me you don't understand. If you did understand you'd have seen how my explanation of such things as the origin of \(ds^{2}\) is completely in agreement with those things.

    Except you haven't given a correct derivation. An incorrect derivation simpler than the correct one is still incorrect.

    Look up Birkhoff's theorem, it's about 2 pages of calculus. Besides, wasn't providing 3 published papers an a 300+ page thesis on string theory enough of a demonstration of my capabilities? What would providing something I literally wrote down in a lecture 5 years ago provide above those?

    Firstly I dont' for a second think you know what a gauge covariant derivative is, you can't even get Euclidean geometry right, never mind gauge theory. Secondly 'fluxes' aren't particles, they are the equivalent of electric charge. Things which have fluxes can move but fluxes aren't specific objects in their own right any more than 'electric charge' is an object. Things can have the property of electric charge, but you can't just have 'electric charge' on its own. None of the objects which carry fluxes move faster than light, faster than light dynamics have nothing to do with this.

    You are simply trying to dig yourself out of a hole. Is it really so hard for you to say "I don't know" when it comes to a tiny area of string theory? I don't know about 98% of string theory and I spent 4~5 years studying it! For god sake have some humility, or are you do desperate not to acknowledge any gap in your knowledge of existence you can't even manage that?

    You say 'professor' as if I'm being all intellectual pointing out the difference between tensor components and tensor rank. I've already been through this with you in this thread, so either you didn't understand or you didn't read it. Either way you're being dishonest in asking again.

    Let's consider a vector in N dimensional space, whose components we define by \(v = v^{a}\partial_{a}\). The components in a given coordinate charge are \(v^{a}\). 'a' is the index which goes from 1 to N (in relativity it is often 0 to N-1 where '0' is the time index). The components are therefore \(v^{1}\), \(v^{2}\), ... \(v^{N}\). There are N of them. N dimensional space, N components. See the connection? Notice how the index notation is such that there's only 1 index on the components, \(v^{a}\). Thus the vector is a rank 1 tensor (given it obeys certain transformation rules). 1 index, rank 1. See the connection?

    Now consider the metric, g. It's components are defined by \(g_{ab} = g(\partial_{a},\partial_{b})\). In N dimensional space both a and b can vary from 1 to N so there are \(N^{2}\) components. A metric is symmetric, \(g_{ab} = g_{ba}\) and thus there's only \(\frac{1}{2}N(N+1)\) independent components. The metric has 2 indices, therefore it is rank 2.

    There, that's two examples of how rank and components are different. In some cases you might get them to be the same, like a vector in 1 dimension will be a rank 1 vector with 1 component but they are not synonymous concepts.

    This is so fundamental to basic vector calculus I can't even begin to describe how much this completely destroys your claims to understand GR. It's a shame other people aren't as familiar with this stuff as the few of us on this forum who study it, as then everyone would see how laughably undermined your claims are.

    Do you think if you simply say that enough times that people won't notice the entirety of this thread has been me pointing out mistake after mistake by you?

    And yet you can't even admit that you don't know what 'fluxes' are in regards to string theory. It's a niche concept in advanced physics, most PhDs in physics would not hear of them, yet you can't bring yourself to admit you haven't, you just keep claiming you somehow were right in your initial assessment/definition.

    Again, you ask a question in a partonising way, as if I'm wrong or being condescending to you yet the question you ask is trivial. A sphere is a manifold, it can be covered in coordinate charts which are homeomorphisms to open sets in Euclidean space. That's a definition (or part of) for 'manifold'. It isn't flat because it has non-zero Ricci curvature. You've got the metric for the sphere of constant radius \(R_{0}\), \(ds^{2} = (R_{0})^{2}(d\theta^{2} + \sin^{2}\theta d\phi^{2}) = g_{ab}dx^{a}dx^{b}\), why don't you work out the explicit expression for \(R = g^{ab}R^{c}_{acb}\). It'll come out to be \(\frac{1}{R_{0}^{2}}\). This is a fundamental result in geometry, something you should be aware of. Something I've already covered in fact.

    The fact you ask questions which you should know the answers to, which you should know are basic results known to all students of the subject, which I've already addressed, shows you don't know basic results and you don't understand things I've already explained.

    Every post of yours uncovers more of your ignorance.

    That isn't the Minkowski metric. Saying "It's the Euclidean version" is just nonsense. You're redefining words and terminology to suit your purposes. It's dishonest and it is not how science is done. The Minkowski metric, in Cartesian coordinates, is of the form \(-ds^{2} = -dt^{2} + d\mathbf{x}\cdot d\mathbf{x}\). There's no two ways about it.

    You are simply making stuff up without reason or justification to suit your purposes. You were wrong but you have such a problem accepting it you just continue lying, hoping you'll either exasperate people enough that they'll stop pointing it out or perhaps you'll bend reality to your whims.

    Provide references for your claim the non-Euclidean version was 'contentious'. The Euclidean form is just the Euclidean metric, it wasn't something also put forward.

    You clearly have no knowledge of how SR came about. Maxwell's equations of Lorentz invariance, which the Minkowski metric also has and which the Euclidean metric doesn't. Lorentz transforms, length contradiction, time dilation and Poincare transforms were all developed towards the end of the 1800s. Einstein then put them all into a single formalism with special relativity. Minkowski then put forth his metric has a way of putting SR into a geometric point of view. Minkowski would not have put forth the Euclidean form because it is not invariant under Lorentz transforms, which was a fundamental component of both special relativity and Maxwell's electromagnetism. The motivation for Minkowski to suggest a metric was based entirely on symmetries which the Euclidean metric doesn't possess.

    The development of the SC metric didn't have any impact on the use of the Minkowski metric. The flat metric is used for special relativity, which doesn't include gravity. The SC metric is a metric derived for a gravitational system, it is not a replacement for the flat metric. Minkowski's metric is the only metric in special relativity and thus is used everywhere special relativity is, including quantum field theory. In quantum field theory Lorentz transformations are used all the time and explicit Lorentz invariance is obtained by using the Minkowski metric. This is because many quantum fields are vector fields, like the photon, and thus transforms under the Lorentz group. In fact all quantum fields are tensor fields, be they scalar (Higgs or Yukawa fields), vector (gluon, photon, weak bosons), tensor (graviton) or even spinorial (leptons and quarks).

    Minkowski's metric has never fallen out of favour or had its relevance questioned, it is the most relevant metric in theoretical physics today.

    I've told you before, you might be used to spouting your made up nonsense to lay persons who don't know any better but that isn't going to cut it here. I actually paid attention in school and unlike you I've actually learnt some maths and physics, including some of the history behind various areas.

    Firstly I'm an atheist and I think the view that Christians have that morals come from the bible and without it you're morally corrupt/empty is laughable (or it would be if it weren't so ignorant). This notion of "OMG you've upset god, you better repent!" is itself morally bankrupt because it leads people to think that if they pray they are absolved of any responsibility for their actions, when instead they should be dealing with the consequences in the real world, addressing the harm/hurt they have done to others. Besides, the bible has some of the most morally reprehensible stuff found anywhere in fiction.

    Secondly I'm not bearing false witness. I've laid out, in detail, your mistakes. When I ask you to provide references you can't. When I ask you to provide derivations you can't. When you do try to respond to a comment I've made you only display more ignorance.

    Of course if you think I'm bearing false witness then submit your work to a reputable journal for review so you can have it assessed by others. I'm certain it'll be rejected. I'm certain you won't get a job doing that stuff at a reputable university.
     
  14. Magneto_1 Super Principia Registered Senior Member

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    You, must be either "really young" or "really crazy", my hunch is Both! Are you on television, this is better than reality TV. What Planet are you from?


    Minkowski, himself claimed that his ideas were radical.

    Riddle me this Batman; Do "Radical" ideas in physics enter into dogma through "smooth sailing" or "contention"??

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    See: Minkowski Space

    Minkowski, aware of the fundamental restatement of the theory which he had made, said:

    "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. –Hermann Minkowski, 1908"

    Look at this statement under the section Structures:

    "Formally, Minkowski space is a four-dimensional real vector space equipped with a nondegenerate, symmetric bilinear form with signature (−,+,+,+) (Some may also prefer the alternative signature (+,−,−,−); in general, mathematicians and general relativists prefer the former while particle physicists tend to use the latter.)"

    I don't know about you, but statements like "in general, mathematicians and general relativists prefer the former while particle physicists tend to use the latter," sound pretty contentious to me.


    I actually agree with this whole heartedly. And you make a valid point the Minkowski Non-Euclidean Metric is isolated to SR, but not limited by SR; and is also a component that can be found in the Schwarzschild Metric (SC).


    I agree 100%.

    I kind of agree here. I agree that the Minkowski metric addressed symmetries; I am not so sure that I would that the Euclidean Metric is not symmetric, in that Newtonian Action/Reaction Forces can also arise from solutions of the Euclidean Metric, which would definitely be symmetric.


    I agree with this statement, "The flat metric is used for special relativity, which doesn't include gravity.

    However, the Minkowski metric is a component that is found in the SC Metric, an demonstrated for you below.

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

    AlphaNumeric, can you spot "Waldo" in the Picture. Do you see the Minkowski "Non Euclidean" Metric there in the equation below?

    \(ds^{2} = ({c^2_{Light}}dt^{2}) - r^{2}(d\theta^{2} + \sin(\theta)^{2}d\phi^{2})\) \( -> m^2 \).


    I am sure that you are quite aware that the (SR) equation above is in the (GR) equation below?


    \(ds^{2} = (1 - (\frac{r_{S}}{r}))({c^2_{Light}}dt^{2}) - \frac{dr^{2}}{(1 - (\frac{r_{S}}{r}))} - r^{2}(d\theta^{2} + \sin(\theta)^{2}d\phi^{2})\) \( -> m^2 \).



    I agree 100%


    Maybe fallen out of favor was a "bad choice of words." Maybe it would have been better to say that it has been contentious in that mathematicians and general relativists prefer to use one form of the equation while particle physicists tend to use another form of the same equation.


    This is what I am actually trying to find out, were you really paying attention; because right now it is Questionable!!

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  15. Magneto_1 Super Principia Registered Senior Member

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    Oh, now, I get it. The reason that you keep harping on this equation, that I continuously, ask you to let go. You are somehow, fascinated, or enamored with!

    Well, since you keep asking, I just did not want to confuse you, I know that is easy to do. But this is such a condition also.

    Area differential based on Area differential

    \(dA_{rea} = d(s^{2}) \)\( ---> m^2\)


    Area differential based on line differential

    \(ds^2_{Line} = (ds)^2 \)\( ---> m^2\)


    So, that is what you have been interested in; you like this "Nomenclature."

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    AlphaNumeric, You, know that is "Check Mate" Right Mate!
     
    Last edited: Apr 24, 2011
  16. Dywyddyr Penguinaciously duckalicious. Valued Senior Member

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    Two minor errors.
    If it's written it's libel.
    And neither is held to be the case (or will stand up as such in court) if the "accusations" are actually true.
     
  17. AlphaNumeric Fully ionized Registered Senior Member

    Messages:
    6,702
    Are you going to provide the reference I asked for, to show that Minkowski put forth a Euclidean metric, or are you going to make vapid irrelevant noise?

    I stand by what I said, he didn't put forth a Euclidean metric because he was specifically trying to put Lorentz transforms on a geometric footing, which the Euclidean metric is not invariant under.

    What you're referring to there is the concept of combining space and time into a single structure, rather than considering just space alone. Your original statement was in reference to 'pseudo-Euclidean' as if there were a choice between Euclidean and non-Euclidean constructions and the choice of non-Euclidean was not favoured initially. A Euclidean formalism was never put forth.

    Yes, special relativity in general took some time to garner support (though nothing like general relativity). However, in this post you try to make the case a Euclidean form is possible and considered. It wasn't. All of your rambling about 'pseudo-Euclidean' etc is utterly baseless, there is only one Minkowski metric and it is, without question or doubt, non-Euclidean.

    No, that isn't contentious because it is purely a matter of convention, like writing 0.5 instead of \(\frac{1}{2}\). In general relativity it is common to write \(-ds^{2} = \eta_{ab}dx^{a}dx^{b}\), not \(ds^{2} = \eta_{ab}dx^{a}dx^{b}\) and thus the different choices have an overall factor of -1 difference in their definitions of the Minkowski metric \(\eta_{ab}\). Some GR researchers prefer the particle physics way and vice versa. I personally prefer the GR way, despite being more of a particle physicist in many way.

    \(\eta\) is not isolated to special relativity. It is found in general relativity too, since SR is a special case of GR. Even in systems with curved space-time local Lorentz invariance exists. I've already been through this with you, remember what I said about veirbeins?

    And you really need to expand your view of GR beyond the SC metric.

    You just demonstrated you don't know what I was referring to. Furthermore you also demonstrated you don't know what I have previously mentioned in regards to SO(3,1). I never said the Euclidean metric doesn't have symmetries, I said it doesn't have Lorentz symmetry.

    In 3 dimensional Euclidean space you have SO(3) symmetry, the rotation group. In 3+1 dimensional Minkowski space-time you have SO(3,1) symmetry, aka Lorentz symmetry. SO(3,1) contains SO(3) as a subgroup but it also includes boosts.

    And before you go Googling to find out about 'symmetric metrics' bear in mind all metrics are symmetric, in that they satisfy \(g_{ab} = g_{ba}\). That isn't the same. If a metric admits a group G symmetry then for all \(T \in G[tex] and [tex]v,w\in TM\) you have that \(g(v,w) = g(T\cdot v , T \cdot w)\). In the case of \(g = \eta\) we have \(G = SO(3,1)\).

    You seem obsessed with viewing the SC metric as just a rescaled version of the Minkowski metric. While that works in spherical coordinates for the SC metric it won't get you very far.

    It would seem you're just clinging to words Wiki or Google find for you without actually comprehending them. If you have any working familiarity with the equations in question you'd know how daft what you just said was.

    Let's consider the energy-momentum equation, \(E^{2} = m^{2} + P\cdot P\) where E is the energy of an object, m is its rest mass and P is its 3-momentum, such that \(P\cdot P = (P^{1})^{2} + (P^{2})^{2} + (P^{3})^{2}\), and c=1. This is constructed within special relativity and can be written as the inner product of a 4-vector with itself using the Minkowski metric. That 4-vector is \(p^{a} = (E,P^{i})\), ie \((p^{0},p^{1},p^{2},p^{3}) = (E,P^{1},P^{2},P^{3})\). It's not hard to see that \(E^{2} = m^{2} + P \cdot P\) can be rewritten as \((p^{0})^{2} = m^{2} + (p^{1})^{2} + (p^{2})^{2} + (p^{3})^{2}\). This can rearranged to \(m^{2}= - (p^{0})^{2} + (p^{1})^{2} + (p^{2})^{2} + (p^{3})^{2}\). If you write \(\eta\) as (-+++) then you have \(m^{2} = p^{a}\eta_{ab}p^{b} = p^{a}p_{a}\). If you write it as (+---) then you have \(m^{2} = p^{a}\eta_{ab}p^{b} = -p^{a}p_{a}\). That is it. That's all the difference. Both equations, when you expand them out, give the same equation, \(E^{2} = m^{2} + P\cdot P\). Therefore there is no contention, it is purely a matter of choice.

    The underlying equation is the same, all that is different is notation. You're reading much more into that quote you provide than is actually the case but then you'd know that if your experience of relativity and particle physics were more than just what you find online and in pop science books.

    Considering you're the guy who thinks \(\frac{1}{R^{N}}\) means something, that \(\mathbb{R}^{N}\) is a hypersphere, who doesn't know the difference between tensor rank and tensor components and who needs to be shown why a sphere isn't flat (and that's just your last few posts I replied to!) I think that's a little rich.

    I just noticed you came out with this gem previously too :

    "The Minkowski/Einstein Metric is also considered Pseudo-Euclidean, is a net difference, and is comprised of two components: 1) a three dimensional space-time radial component and, 2) a two dimensional locational directional component. The Result is a four dimensional element vector (space, time, latitude, longitude)."

    Thus illustrating you still haven't looked up what 'component' means, despite having it explained to you. The 'r' in the SC metric is a single parameter, it is not 3 dimensional. You use 'component' when you should use 'part'. 'Component' has a specific technical meaning, which you should have learnt but obviously haven't. Seeing as I've walked you through the meaning now (on more than one occasion) if you continue to use incorrect terminology you only demonstrate you're either too stupid to understand the correct terminology or are dishonest and want to deliberately confuse people (to hide your mistakes IMO).

    Whenever I ask you to provide a reference for a claim you make you always come back with a Wikipedia link. I've yet to see you mention a book other than your own. I've yet to see you respond to any mathematics I've done with something other than a set talking point.
     
  18. AlphaNumeric Fully ionized Registered Senior Member

    Messages:
    6,702
    I'm not fascinated by it, the problem is you don't seem to understand it.

    I'm the one running rings around you. You think that \(\frac{1}{\mathbb{R}^{N}}\) means something!

    I've already been over this with you here. I'll quote the relevant part in full, so you don't get confused.

    You write \(dA_{area} = d(s^{2})\) but this is a problem in more than one way. Firstly it is not an area element. An area element, as I went through in the above quoted post, has two differential operators, as you have to integrate over two directions. If \(dA\) is the area element for an area then, by definition, it satisfies \(A = \int_{region}dA\). The \(\int_{region}\) becomes two integrals, as done explicitly above, and thus \(dA \to [something] dx^{1} dx^{2}\) where [something] is associated to the particular choice of coordinates, ie the metric and perhaps Jacobian. If you tried to put in \(d(s^{2})\) into the defining formula for the area element then you don't get the right result.

    Secondly, you're considering a single length and squaring it. While this gives the right units, as you're so obsessed with stating, it doesn't give the right qualitative interpretation, never mind quantitative. For instance, consider the area of a rectangle, whose sides are of length X and Y. You can vary those directions independently, varying X doesn't vary Y and vice versa. This is why the area element for such a rectangle is dX dY, not d(XY) or \(d(X^{2})\) or \(d(Y^{2})\) or any other such combination. If you're integrating over an N dimensional region then you need N different terms of the form \(d[something]\) to be multiplied together.

    I went over this with you previously, talking about volume forms. I even commented I covered it in my thesis. By definition the volume of an N dimensional region \(\Omega\) with metric g is \(V = \int_{\Omega}dV\). In differential form notation, picking a section of a frame bundle (the process is explained in detail in my thesis) this becomes \(V = \int_{\Omega} \eta^{12\ldots N}\). Converting into nice coordinates you get \(V = \int \ldots \int \sqrt{|g|}dx^{1}dx^{2}\ldots dx^{N}\). Each coordinate \(x^{j}\) has associated to it an element \(dx^{j}\). Although \(dx^{j}dx^{j}\) has units of length squared it is not a valid area element. This is done in detail here too, where what I've said is backed up entirely. A less formal walk through with pictures is given here, which also includes a short page here on how to use Jacobians to transform from one coordinate system to another, in this case Cartesians to polars. Notice how the result at the end doesn't involve any x or y at all, just r and \(\theta\). That's precisely what I explained to you, that you should convert completely from one to another, not mix them, and there's no \(x[tex] vs [tex]x_{S}\), as you tried to do.

    An alternative approach to see you're incorrect is to consider the arc formula you quoted at me. As I went through here, your \(d(s^{2})\) cannot be put into the formula you quote, as it is meaningless in such a context. Here's what I said again, so you have another chance to think about it :

    Which ever way you come at it your use of \(d(s^{2})\) has problems or is just flat out wrong. Yes, it has units of length squared but there's more to being consistent than that. You need to put your obsession with units aside, they are blinding you to much bigger misunderstandings you have.

    You really do live in your own delusional little world, don't you? At every turn I've demonstrated you have misunderstandings, mistakes and meaningless nonsense in your work. I've walked you through many of them, sometimes more than once. Much of your mistakes are on a level which means its impossible for you to understand general relativity if you read a book on it because you wouldn't understand what the authors mean when they talk about common concepts like tensor rank, tensor components, \(\mathbb{R}^{N}\), directions and coordinates, charts, dimensionality, hyper-spheres, line/area/volume elements, metrics, curvature, tangent space, signatures, Lorentz transforms, symmetries in general, coordinate transforms, Jacobians, parameterisations and more besides.

    Every single one of those you have, at some point in this thread, either explicitly been mistaken about them or have said something which implies you don't understand an important facet of them. Most are crucial to an understanding of relativity, without them you cannot have a working understanding of relativity.

    As such I conclude you do not have a working understanding of general relativity. You have tried the "Play nice, you might be engaging in libel!" (or slander as you mistakenly said). The point of this thread was for you to defend your work. All you've managed is to demonstrate that you don't have any defence for your work. Courts aren't there to assess scientific merit but journals are and I'm certain you'll be rejected from all reputable ones.
     
    Last edited: Apr 24, 2011
  19. Magneto_1 Super Principia Registered Senior Member

    Messages:
    295
    I believe what you say, may be technically true. However, in the Super Principia Mathematica, once I determine that I will never exceed N=4 in my model; It is from this concept that I depart the "Geometry" of "Classical General Relativity" (GR), and head into the description of "Matter in Motion."


    Matter in Motion concepts such as the one that you have described above is described in a much simpler way, that I can guarantee you!!


    Concepts such as the Ideal Gas Nature of Space-time as described by the Stress Energy Tensor (A form of Dark Energy), Expansion, Cosmic Force, Friedmann Cosmological Equations, Classical Vortex Equations; among other various "matter in motion" concepts.

    What we have been discussing mainly in this post is Geometry, and it is this sort of Geometry that makes people run from (GR). But if they are patient after they endure the "bad" medicine of the Geometry, they can get back to Newtonian kinds of mechanics that they are familiar with.

    It gets more fun if you can hang on!!

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    Last edited: Apr 24, 2011
  20. Magneto_1 Super Principia Registered Senior Member

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    I have been walking upon the face of planet earth for 44.5 years and in all of that time, I don't think that I have heard any thing funnier!!:xctd:

    I literally fell out of my chair on this one!

    You and Kruskal must be drinking buddies, or pub mates!

    This was a good one too.

    This is not true I also mentioned, Johannes Kepler, Astronomia Nova (1609).
     
    Last edited: Apr 24, 2011
  21. Magneto_1 Super Principia Registered Senior Member

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    295
    Your, Jealousy reeks of a foul odor!

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  22. Magneto_1 Super Principia Registered Senior Member

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    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 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)
     
  23. Dywyddyr Penguinaciously duckalicious. Valued Senior Member

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    19,252
    Does it?
    Why would I be jealous?
    Is it possibly because I can't "write" a sentence the way you do?
    I would have written (were I given to expressing such sentiments):
    Your (no apostrophe) jealousy (no capital) reeks (since reek in and of itself implies "foul odour"*)!

    * I'll excuse the "spelling" of odor.
     
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