Discussion : Magneto

You, and AlphaNumeric, have different goals in math, physics, and science in general than I. You took your math talents and ran to give to the rich, and to the best and the brightest minds which headed to the prestigious universities and colleges.

Me I took my math talents and went to those who were a little slower in leaving the gate than others. I went to help those that truly need the help with mathematical understanding, so that they could move to the prestigious next level. As a professor, I have gotten thank you cards, thank emails, gifts of all sorts thanking me for helping them to understand what they did not get from their earlier math teachers. Do you get such Thank You's in your teaching profession??

There was a time in my life when, I found this type of university work very rewarding. But, now I am ready for something more challenging, and like you I am now ready to go to present my talents to the rich.

This is why I applied for a Faculty Position at Caltech. I could clearly earn my PhD there in three (3) to four (4) years; mainly, because I have my own models and research to pull from, and I am ready to start publishing scientific papers.

Truth be told, I was an ideal candidate for Caltech (20 years engineering, 10 years academic, 3 published books) and one of the books "General Relativity" is so close to Kip Thorne's newly discovered work that it is hilarious. It does not make sense not to hire someone that has the proper qualifications and has written a book in the area where you are claiming new research, just reiks of a bad order to me!!

They never did give me a real clear explanation as to why I was not hired. However, I am now in conversing with another university that is competing with Caltech. So we will see what happens in the future???

I totally disagree. Like I stated previously in this post. The internet has changed the way we communicate. And the internet has been opening the door for new business enterprises to emerge.

First and foremost, I have absolutely no affiliation with Gary Sorkin, or the Pacific Book Review. I found that company on blind internet search for book reviewers, just like you did! They claimed that they could review science books, even one on General Relativity, so I took them up on their challenge.

I am actually a fan of book review services like this one.

Why, because no one has free time these days; nor the energy to fill their own personal time to review your book! Ahh!! But for the right price, you could easily find an expert to write a legitimate review of the work! If I paid you the right price and sent you a book, would you have the courage to pen a review and add your name to it?

Remember a review is someone else’s interpretation, or opinion of your work. This reviewer could be the second coming of Einstein, or a high school student who is in AP courses, and is excited about physics.

In your mind's concept. If if is not reviewed by AlphaNumberic and his Academic Gang, then the work is not valid. This is old school thinking. A peer review today is anyone that can read and has an opinion!!!

Some give their opinions for free, and some charge for their opinion. I don't mind paying for an opinion that is written down and signed by the reviewer. It actually takes courage to do that. That you will review someone's work and sign your name to it, is a courageous act!!

So don't be mad at me because the reviewer (Pacific Book Review Company) really liked the book, and learned something. I actually was not surprised by his review, I expected that his reaction would be such, and yours would be also!

 

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So being involved with 'the best and the brightest minds' is somehow a bad thing? Someone has to teach the best and the brightest. As it happens I didn't do my PhD at the university I did my degree. I did my degree at a place which has more Nobel Prize winners than most countries, so clearly the teaching of the brightest and best pays off quite a lot.

The university I did my PhD at requires lower grades to get onto their undergraduate course than the university I was an undergraduate at.

You don't appear to have any mathematical talents. And regardless of who you teach if you're incompetent you're not doing a good thing.

I taught as part of my PhD, I didn't do it professionally. I got thanked a fair bit though.

Besides, I don't think you can take the moral high ground in this stuff because you're charging people to buy your work which you haven't had reviewed and which you seem increasingly unable to defend. Yes, helping educate people is good, its part of the reason I stand up to cranks on forums. Teaching people incorrect stuff is not good though and charging them for the privilege is very bad.

Your attitude of there being some kind of rich/poor divide in science illustrates how little experience you have with the research community.

So you claim but you can't stand up to scrutiny. Plenty of cranks here think their work is worth PhDs or Nobel Prizes. None of them have managed it though.

Provide evidence that Kip Thorn has stolen your ideas. Hell, provide evidence he even read your book. You don't have any such evidence, you just have what you want to believe and you're feeding your own delusions.

Because you have no experience in reputable research, your only maths experience is at a facility little more than a degree mill, you can't do undergraduate level physics and you've got absolutely nothing to show for your supposed decades in academia. Decades in academia and no research?! Your application likely went straight in the bin. At the company I work for we get hundreds of PhDs applying and if they haven't published a paper they are put in the reject pile.

Yes, let us know.

No, that might be 'peer' for you, having a lay person read it but peer review in science is review by scientists. I know it might be a bit of a struggle to grasp that but the distinction is quite important. The whole point of review is that your work is examined by people whose business it is to understand that sort of stuff and so are best positioned to give criticisms and comments.

This is a problem cranks have, they don't like to hear criticism and they try to avoid/ignore it completely. Sure, no one likes being told they are wrong but you won't find a single successful researcher who hasn't been told they are wrong at some point and every single one of them will be able to tell you what they learnt from that. I have had to bin months of work in the past but I learnt from it. I had to rewrite a section of my thesis after my viva but in doing so I greatly expanded my understanding of a particular area I'd been weak on. This is why its important not to say "You're wrong" but to say "You're wrong because....". That's why I've walked you through some of your mistakes, hoping you'd put pen to paper and see where you were mistaken. Unfortunately you seem unwilling to actively put effort into thinking about what I've said and just quote Wikipedia at me (not realising it backs me up).

 

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AlphaNumeric, you have no Idea, who I am, or what I have done, but what you see on paper, that I have released. Since, RPenner was so kind to post my "LinkedIn" page, you are aware of my background. Which means that my country is safe and protected from "Clowns" such as youself, because my Figher Aircraft, Radar, and Orbital Mechanics work.

So while you were in junior high school wearing the "Dunce Hat" sitting in the corner, I was developing new Orbital Mechanical Techniques for Spacecraft, and helping to develop Ion Propulsion engine algorithms at the Jet Propulsion Laboratory, which is actually Caltech.

It is that work that prepared me for subjects like thermodynamics, special relativity, and general Relativity.

And I want to add another thing. In not one post have you heard me make any statements like, I am smarter than anyone. You have many and countless, references to being more exceptional than anyone in math and physics. I have made no such claims.

The only claim that I have made in that regards to your reasoning is that I have developed a theory within the framework of General Relativity. That theory must pass scrutiny, I am aware of that, and I am looking for that kind of scrutiny. This is why I am in your childish challenge.

I have made no claims to have supersede Einstein, or anything that Cranks say, yet you constantly accuse me of being a crank! What you fail to grasp is that I want those models to be challenged! I believe that they can stand up to the challenge, I am aware that there will be some re-tooling.

I am sure you have, probably countless many!! I am doing the same to you now, helping you to greatly expand your understanding of a particular area that you have been weak on!

Every Hooke, will seek out his Newton!

So, lets leave work and academic history out of this, and focus on your "Conversion!!"

I plan to get to this, and your Stress Energy is proportional to Mass question.

 

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I'm going on what you present here. You make claims, I challenge them, you respond, I find your responses lacking and not just quantitatively. You seem to have a chip on your shoulder about people who can do advanced mathematics, yet you want to be seen to be such a person. You dislike people knowledgable reviewing your work yet you want people to sit up and take notice of it. You claim you don't have the time to submit to peer review yet you've been working on this stuff for 2 decades and found time to write a 3 volume book. Prudence would say to get your work reviewed and then publish it, lest you have to rewrite your work due to mistakes, of which you have many.

I didn't read it.

Firstly those things are not 'your' work, they are the culmination of many people's work and not just in your country. I live in a country with similar technological capabilities and my work now is not entirely unrelated to such things.

Whether or not you have contributed to fighter jet, radar or orbital systems technology is irrelevant to the work you're supposed to be defending here. I have plenty of work in physics but that wouldn't mean any claims I make about chemistry would somehow become more valid. Even within physics expertise in one area doesn't mean even competency in another. I know loads of people with physics and maths PhDs and very few understand each other's work.

At least try to be relevant with your insults. I've demonstrated in this thread a lot more understanding than you. My academic record demonstrates plenty of people have been convinced I'm a competent physicist/mathematician so calling me a 'dunce' is just clutching at straws.

As such unless your work has previously been in general relativity, which I can't believe it has been given your inability to do the simplest things in it, your supposed career in aerospace is irrelevant.

What precisely did you do? Did you help build, physically, the ion engine? That's no relativity. Did you design and code the algorithms to control the engine? That's not relativity. Did you do orbital mechanics? That's not relativity. You use the word 'mechanical' which suggests you were involved in something physical, some hardware. That's not relativity.

It's possible to design a control system for an engine for a space craft without knowing anything about orbital mechanics, as you might only be considering controlling power flows to keep the engine in a particular state. Building the engine which puts a GPS satellite into orbit doesn't mean you know the relativity used in designing and calibrating the satellite's systems.

Please be specific about what you've done which involves you working with relativity. Give examples of calculations you did or systems you modelled where you had to explicitly use a relativity formula.

Please provide a link to a post of mine where I said I'm 'more exception than anyone in maths and physics'. I said I think I'm better than you. I said I think I'm better than most. But I certainly didn't say I'm better than anyone. In fact I've said many times on this forum that compared to the people who did similar research to me during my PhD I am quite mediocre. I think I went over this with you before, cranks make the mistake of thinking that because I think I'm better than them then I must think I'm better than everyone. No, you just set the bar very low. It might also illustrate that you do in fact think a lot of your capabilities since if I think I'm better than you and you think you're the best thing since sliced bread then surely I must think I'm 'exceptional compared to anyone'. Given this behaviour is almost universal in cranks I'd say it does indeed illustrate how you view yourself.

And let's not forget, the thread which sparked my challenge was you claiming Kip Thorn stole your ideas! And you claim I'll be converted into 'Paul' and be preaching your work as gospel soon. Clearly you do think a lot of yourself, just rather than explicitly saying it you claim all kinds of ridiculous things about your work's value.

No, you aren't looking for scrutiny because if you were you'd have looked for it before writing a book and charging people money for it.

How is asking you to engage in an honest informed discussion 'childish'? I laid out the rules at the start so it was clear that the central tenants of the discussion are reason and evidence. Questions should be answered, references should be provided. That sort of discourse is the sort of thing peer review involves. When you don't justify a claim you're asked to do so. When you don't provide a reference, you're asked to do so. When you're wrong it is pointed out and constructive criticism comments made. If you can't handle a single guy asking you about stuff you claim to know well you're going to struggle a great deal with it comes to peer review. In less than 2 pages you've decided you don't like having to defend your work and this is all childish. You should have learnt this lesson during your job but good research involves having your ideas challenged. A good physics paper will open by giving an overview of the assumptions it makes and will close with comments on how the work might be improved, tested, which problems might arise in the foreseeable future.

I didn't say you did. You've claimed to have come up with a much simpler derivation of the SC metric and to have developed a streamlined approach to GR in general. You believe your work is valid, worth publishing and worth people paying money for it. You have no provided any reason to think any of that is true, in fact you've provided plenty of reason to think the opposite.

If you'd submitted your work for review before hand, it passed and then got a lot of interest then you'd have other physicists validating the belief you have that your work is worth paying for. I haven't earnt a penny from my published work but that wasn't the point. Instead my published work got me a job because people say what I am capable of. I indirectly made money from it by convincing other physicists I'm a physicist. You've convinced yourself your work is worth money directly but you don't want to hear anyone else's opinion on it if its negative opinion.

Your qualitative view of science, your naivety, your self belief and your quantitative short comings all fit the profile.

Your confusion between \(d(s^{2})\) and \((ds)^{2}\) is a fatal mistake, it utterly invalidates your claim about the SC metric. The \(\theta_{0}\) issue kills it in a different way.

I want to see a full, clear derivation of the SC metric which has the correct d structure. If you end up with something of the form \(d(s^{2})\) then is wrong because you're not obtaining the SC metric but something which, in particular notation, looks like it. Furthermore you cannot have \(\theta_{0}\) in that, it is not the SC metric if its there. That's simply a matter of derivation. If you claim otherwise (which would be ridiculous to argue against a definition) then you must also demonstrate that your claimed 2-sphere metric component \(d\Omega_{2} = d\theta^{2} + \sin^{2}\theta_{0} \,d\phi^{2}\) has the SO(3) invariance it is required to have due to being defined on a 2-sphere. This pertains to the "Do an SO(3) rotation" comment I made, I want to see you explicitly do that if you believe \(d\Omega_{2} = d\theta^{2} + \sin^{2}\theta_{0} \,d\phi^{2}\) and not the usual SC form of \(d\Omega_{2} = d\theta^{2} + \sin^{2}\theta\,d\phi^{2}\)

 

AlphaNumeric, I have made some copy and paste errors along with a mild form of Dyslexia. I sometimes flip things from left to right. You have to watch driving with me, I might say right, when I really mean left. Not all of the time, but every now and again.

For example when I was trying to describe what I picked up from your work on Vacuum Flux, or Fluxes. I stated,

"fluxes could represent the ultra faster than flow light speed and momentum of the particles in this medium."

What I really meant to say was, fluxes could represent the ultra faster than light-speed flow and momentum of the particles in this medium."

It is for this, shortcoming that you continue to whack me with your physics stick for not adhering to proper nomenclature.


For the record I want to be clear, and after this, please don't accuse me of this again; or this will be "Totally Dishonest," but you are probably used to that!!

Spherical Vector Mechanics - Demonstrating Spherical Line Elements of a Radial Vector in three dimensions, and describing the Geodesics associated with that sphere.

Spherical Coordinate Linear 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} = \({r \mathbf{e}_{r}} + (\ {r {\theta})\mathbf{e}_{\theta}} + ({r} \sin(\theta))({\phi}) \mathbf{e}_{\phi} \)\).\( -> m\)


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

\({r} = \sqrt{{x^2} + {y^2} + {z^2}}\)\( -> m\)

Simplest Mathematical Form of a Latitudinal Geodesic on the surface of the Sphere; without regards to any reference of origin for that curvature

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

Simplest Mathematical Form of a Longitudinal Geodesic on the surface of the Sphere; without regards to any reference of origin for that curvature

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

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.


Spherical Mechanics - Demonstrating The Total Area of a Sphere by square the "Line Elements" of the Spherical Radial Vector.

The Total Area of a Spherical Body is equal to the net sum of the square of each three dimensional "line element" vector components of the Spherical Radial Vector equation, obtained using Pythagorean’s Theorem:

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


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


\(s^2 \) = (Area function of distance + Area function of direction) = Total Area of Sphere.\( -> m^2\)


The above Total Area term is comprised of two terms.

1) A fixed Radial Component - \({r^2}\).\( -> m^2\)

2) A Directional Area component - \(\ {r^2}\({\theta}^2 + \ {\phi}^2 \sin^2(\theta) \)\).\( -> m^2\)

The above direction component of Area predicts a patch of area where an object is located in (\((r, \theta, \phi)\) directions on a sphere with a fixed radius.

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


Now the above equation for Total Area is being the net sum of a fixed area radial component, and a fixed patch area with a constant radial component located on the surface of the sphere, and is directional dependent relative to the center of the system. The directionality of a three dimensional sphere can be described by two components a latitude component (\(\theta\)) and a longitudinal component (\(\phi\)).

This is why I feel that we will get into an endless arguments about 2 tensor Ranks, that you will never let me win. I believe that when I describe the location of any object on the surface of the sphere with two components of direction; namely latitude and longitude; I essentially have fulfilled the same job as a 2 Rank Tensor.

Now, lets look at the concepts of Hyperspheres (1D, 2D, and 3D, ...) very briefly only because you were somehow concluding that I was excluding multi-dimensions from General Relativity. A 2-sphere is an ordinary sphere in three-dimensional space, and it is this is what I have been describing above.

In mathematics, a 3-sphere is a higher-dimensional analogue of a sphere. A 3-sphere consists of the set of points equidistant from a fixed central point in 4-dimensional Euclidean space.

A 3-sphere is a compact, connected, 3-dimensional manifold without boundary. It is also simply-connected. What this means, loosely speaking, is that any loop, or circular path, on the 3-sphere can be continuously shrunk to a point without leaving the 3-sphere.

Just as an ordinary sphere (or 2-sphere) is a two dimensional surface that forms the boundary of a ball in three dimensions, a 3-sphere is an object with three dimensions that forms the boundary of a ball in four dimensions.

A 3-sphere is an example of a 3-manifold. A 3-sphere is also called a hypersphere, although the term hypersphere can in general describe any n-sphere for n ≥ 3.

In mathematics, an n-sphere is a generalization of the surface of an ordinary sphere to arbitrary dimension. For any natural number n, an n-sphere of semi-major radius (\({r}\)) is defined as the set of points in (n + 1)-dimensional Euclidean space which are at distance (\({r}\)) from a central point, where the semi-major radius (\({r}\)) may be any positive real number.

It is an n-dimensional manifold in Euclidean (n + 1)-space. In particular, a 0-sphere is a pair of points on a line, a 1-sphere is a circle in the plane, and a 2-sphere is an ordinary sphere in three-dimensional space.

Spheres of dimension n > 2 are called hyperspheres.

The n-sphere of unit radius centered at the origin is called the unit n-sphere. In mathematics, an n-dimensional space is a topological space whose dimension is n (where n is a fixed natural number). In n-dimensional Euclidean space, spaces with large values of n are sometimes called high-dimensional spaces.

Many familiar geometric objects can be generalized to any number of dimensions. For example, the two-dimensional triangle and the three-dimensional tetrahedron can be seen as specific instances of the n-dimensional spaces.

Also, the circle and the sphere can be seen as specific instances of the n-dimensional hypersphere. More generally, an n-dimensional manifold is a space that locally looks like n-dimensional Euclidean space, but whose global structure may be non-Euclidean.

High-dimensional spaces – spaces with a dimensionality substantially greater than 3 – have properties that are substantially different from normal common-sense intuitions of distance, volume, and shape.


Euclidean Metric of Spherical Object.

This Euclidean Metric is a differential form that represents the net sum of square of a line element change in three directions (radial, latitude, and longitude). The line elements for the Volume and Area of the sphere are radial distance and two geodesic line elements, and together they form a closed system in all three dimensions:

Infinitesimal Radius Change relative to the Mean Center of the Sphere


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


\(ds = d(s \mathbf{e}_{r}) = \(dr{\mathbf{e}_{r}} + (\ {r d{\theta})\mathbf{e}_{\theta}} + ({r} \sin(\theta))(d{\phi}) \mathbf{e}_{\phi} \)\).\( -> m\)

Next, we will generate the Euclidean Metric by squaring the individual differential line elements in the equation above, and using Pythagorean’s Theorem.


\(ds^2 = (d(s \mathbf{e}_{r}))^2 = \((dr{\mathbf{e}_{r}})^2 + ((\ {r d{\theta})\mathbf{e}_{\theta}})^2 + (({r} \sin(\theta))(d{\phi}) \mathbf{e}_{\phi})^2 \)\).\( -> m^2\)


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

or

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

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

Note: The dot product of a linear vector is an Area scalar term (\(\(r \mathbf{e}_{r})(r \mathbf{e}_{r}) = {r^2}\)). ......Or (\(\(dr \mathbf{e}_{r})(dr \mathbf{e}_{r}) = (dr \mathbf{e}_{r})^2 = {dr^2}\))


Where the Total Infinitesimal Squared Directional Spherical Scalar Component is


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


Where the Total Infinitesimal Directional Spherical Component is


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


Once you square a vector line element as described above, you naturally create an Area element. If the line is changing the area is also changing is what the above equation describes.

In Metric Differential Form. This Metric differential form represents a line change and is equivalent to an change in the total Area of the sphere via a radial dimension that represents one dimension of space and one dimension of time. And a directional dimension that represents one latitude dimension and a longitude dimension geodesic making a four element vector; which also represents three dimensions of space and one dimension time.


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


\(ds^2 \) = (Area change function of infinitesimal line distance change + Area change function of infinitesimal direction change) = Total Infinitesimal Area Change of Sphere.\( -> m^2\)

The above Total Area term is comprised of two terms.

1) An Infinitesimal Radial Area Change Component - \({dr^2}\).\( -> m^2\)

2) A Infinitesimal Directional Area Change component - \(\ {r^2}\({d\theta}^2 + \ {d\phi}^2 \sin^2(\theta) \) = {r^2}d{\Omega}^2\).\( -> m^2\)

Note: When discussing Area Change, I am referring to when a line changes and you square that result, you get an equilavent area change assocated with that line change. Likewise when discussing Area Change in regards to direction change, I am referring to a change in the patch area on the surface of a sphere associated with square latitude and square longitude angle changes.


Now, AlphaNumeric, after reading this, don't bring up what I did in the past postings, stating, that is not what you wrote before. Here is where I am clearing things up for you!

If you see the Derivation above, you will see that you described your equation incorrectly, and you are questioning my Dyslectic mistakes

This should be

\(d\Omega^2 = (d\theta^{2} + \sin^{2}\theta\,d\phi^{2}) \Not= d\Omega\), as you quote!!

Oh, and by the way the Scharzschild Metric (SC) that you keep refering to is actually an Euclidean Metric (EC). The difference between the two metrics is that the Schwarzschild Metric has an additional space-time expansion term associated with it, (\(\ S = \frac{r}{sqrt{1 - (\frac{r_{S}}{r})}\)), and the Euclidean Metric does not!

 

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

Let, \(d{r^2} = {c^2_{Light}}d{t^2}\), where the time component is given by, (\(d{t^2}\)),


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


Next, dividing both sides by the square of the speed of light, yields a time domain equation for the Euclidean Metric.


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


Next, if you consider having angular velocity in the latitude and longitude directions, there is time there also.

Let the square of the Angular Velocity in the Latitude Direction be given by,

\(\ {\omega^2_{\theta}} = \frac{d\theta^2}{dt^2}\)\( -> radians^2/s^2\)


And, let the square of the Angular Velocity in the Longitude Direction be given by,

\(\ {\omega^2_{\phi}} = \frac{d\phi^2}{dt^2}\)\( -> radians^2/s^2\)


Where, we can now write

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

Finally, we have an Euclidean Metric that is time dependent.


\(\ {dt^2_{s}} = \frac{ds^2}{c^2_{Light}} = \ d{t^2} \(1 + (\frac{r^2}{c^2_{Light}}) ({\omega^2_{\theta}} + \ {\omega^2_{\phi}} \sin^2(\theta)) \)\).\( -> s^2\)

 

That is just nonsense. Whether or not you have dyslexia has nothing to do with the mistakes you've made, they aren't little slips like swapping the order of words or the like, you've made fundamental errors. For instance, let's consider the example you give :

Your mistake has nothing to do with 'not adhering to proper nomenclature', your definition/description is just wrong.

If someone said to you "I'm an expert in quantum field theory" you'd assume they are referring to the study of subatomic relativistic particle based systems. If they then said "Quantum field theory is the study of farm animals" would you say "That isn't quantum field theory" or would you just shrug and think "Well it's just a matter of proper nomenclature, he's not wrong"? I bet you'd contend their definition of quantum field theory. Your definition of 'fluxes' is just wrong, flat out wrong. You aren't rewording 'proper nomenclature', you're just making stuff up.

Fluxes, in the context of my PhD, are quantised gauge field components associated to branes wrapping the non-trivial cycles of the space-time manifold whose perturbations are described by (initially) massless scalar fields known as moduli. Your definition is wrong and no way of bending it will change that.

You know you don't understand my thesis. It's not shameful to say it, I don't understand many other people's PhDs, its the nature of spending years doing research on a very specific thing. But rather than be intellectually honest and just say "Okay you've linked to your thesis, but I don't understand it" you have to lie, just flat out lie, and try to convince me you understand it. Come on now, are you so desperate to avoid saying "I don't know" that you'll try to BS to me about my own thesis topic? Out of all the topics in physics you could try to bluff knowledge on you couldn't pick a worse one.

Remember one of the rules of the discussion is intellectual honesty? The moderators are the deciders of whether or not you've broken that but IMO you have.

I'll skip the bit of your post where you essentially copy and paste in basic high school geometry for about the 4th time.

You've just shown you don't understand what tensors are again. Describing the location of an object on the surface of a sphere is not a matter of direction but of position. The two are distinct notions and you should have come across this before because its something all GR books talk about because its the definition between a point in a space (location) and an element of the tangent space (direction).

Giving a point on a sphere, or any other manifold, is to give a location. If the manifold is N dimensional then you give N scalar values. In flat space \(\mathbb{R}^{N}\) this is a location vector, \(\mathbf{x} \in \mathbb{R}^{N}\). A vector in this regard is a rank one tensor. On a 2-sphere you're considering a 2d space and so you give 2 coordinates. Again, this is not a rank 2 tensor. It's a 2 component vector. A vector is rank 1 because it has one index, \(\mathbf{x} = (x^{1},x^{2}, \ldots,x^{N})\), so general component is \(x^{a}\). A rank 2 tensor has 2 components like \(g_{ab}\) or \(R_{ab}\). You've just demonstrated you still don't understand something so basic its almost assumed knowledge for GR courses at university.

Now for the distinction between location and direction. Latitude and longitude don't define a direction, they define a location. A direction is then defined at a location by another vector. I won't go into the specific construction of the tangent space (look it up, though you should already know it) but it harkens to the TM and T*M I've mentioned before. At every point on a N-sphere, which is usually denoted as \(S^{N}\) you can construct a N dimensional space equivalent to \(\mathbb{R}^{N}\). Thus at every location you can also pick a direction. Thus defines the tangent bundle. If M is the manifold you pick a location from then TM is the tangent bundle where each point p has a 'fibre' formed from the tangent space, \(T_{p}M = \mathbb{R}^{N}\). Its common for people to confuses the base space and the tangent space because in the case of M being Euclidean space you have both the tangent space and the base space being \(\mathbb{R}^{N}\).

I wouldn't be surprised if you made a comment about me 'whacking' you on the head about nomenclature but the distinction between location and direction is central to the proper mathematical framework of differential geometry and relativity in general and you keep getting it wrong. You've confused tensor components with tensor rank and location with direction. The fact you, somewhat indignantly, say "Look, I've done what a rank 2 tensor does!" when you don't know what 'rank' means is just daft.

Is it really so hard for you to say "I don't know"?

You're just in copy and paste mode now. I know you're used to talking to people who don't know what a hyper-sphere is but considering one of the papers of mine I linked you do involved studying the dynamics of a 7 dimensional object wrapped around the \(S^{3}\) cycle of an \(S^{5}\) hypersphere while interacting with a black hole in 10 dimensional \(AdS_{5} \times S^{5}\) space I'd say its clear I don't need to be told what they are.

Your replies seem to be becoming more and more copy and paste. When I challenge you to provide a derivation or to explain something you don't engage in discussion, you just spit out some trivial mathematical statements I've already pointed out the flaws in.

Again, if you can't engage in discussion you're being intellectually dishonest.

Really? Tell me more, I never learnt anything about higher dimensional notions of volume or area when I did a PhD in 6 dimensional compact spaces.

Come on now Magneto, at least try to think about what you're going to copy and paste. I've brought up generalisations of volumes already and I've given links to large quantities of my own work on precisely that. You claimed you understood some of my work (though we both know you don't) so you should have recognised how I regularly make use of higher dimensional volume integrals. Many times I use expressions like \(vol_{n}\) in my work.

If you're going to mindlessly copy and paste stuff at least have the decency to think whether or not its appropriate.

Yes, the square of something with units of length gives something with units of length squared but it doesn't work like that with infinitesimals in the sense that you can just extract out the d's.

For instance consider an area A of a Cartesian rectangular region R, say between x in [0,X] and y in [0,Y]. Suppose you want to compute this from an infinitesimal area element, ie \(A = \int_{R}dA\). Fine. Now what is dA? Well if you consider that the change in area is due to the changes in thw two Cartesian coordinates, x and y, then by basic geometry (which is outlined in every book on vector calculus) you have that dA = dx dy and the integral is \(A = \int_{0}^{X}\int_{0}^{Y}dx dy = \int_{0}^{X}dx \int_{0}^{Y} dy = X Y\). Despite the fact 'dA' has only 1 'd' in the notation is actually represents two infinitesimals. This is standard notion, where the volume element associated to an N dimensional volume region V is written as dV but actually you have to do N integrals over the N coordinates and thus dV is broken down into expressions like \(dx^{1} dx^{2}\ldots dx^{N}\) (in Cartesians). I cover this in my thesis, in the section on differential forms, you should look it up.

As such if you don't take this into account you might end up labouring under the misapprehension about the order of the d's in expressions, as you've indeed done by thinking \(ds^{2}\) is order 1 in the d's, when its actually order 2, ie \(ds^{2} = (ds)^{2}\). It's easy to see why. In order to compute the hyper-volume of an N dimensional region you have to sweep out the region, which requires you varying N coordinates, ie you have to do N integrals. Each integral 'eats' a 'd', as seen in the simple A example I just gave. Since areas are two dimensional you need 2 integral terms and thus the expression must, when you convert to coordinates, be order 2 in the d's. Your original \(ds^{2} = d(s^{2})\) didn't have this.

As James says, time has nothing to do with what you have said. Time doesn't exist in the space you're considering.

No, you don't. This is basic calculus but clearly you need to be told it.

If you have a line of length s you can compute a quantity with units of area by squaring it to \(s^{2}\). Suppose you change \(s \to s + ds\), how does the area change? \(s^{2} \to (s+ds)^{2} = s^{2} + 2s \,ds + (ds)^{2}\). Thus the area change is actually \(2s \,ds + ds^{2}\).

If you consider things order by order then you can use differentiation directly, \(d(s^{2}) = 2s \, ds \not (ds)^{2}\).

You're still incorrect in much of what you say. What you are correct on is stuff where you've referenced Wikipedia. None of what you've just discussed is new, its basic high school geometry and vectors and even then you managed to bork some of it.

So now your \(\theta_{0}\) has magically vanished. Strange, you defended it a number of times and now you're claiming it was a dyslexic mistake? As for \(d\Omega^2 = (d\theta^{2} + \sin^{2}\theta\,d\phi^{2}) \Not= d\Omega\), I used the notation \(d\Omega_{2}\) to represent the 2-sphere line element. It's standard notation.

This is flat out wrong. The SC metric is a distorted version of the Minkowski metric, not the Euclidean metric.If you put in m=0 the the SC metric you get the Minkowski metric. The Minkowski metric has a minus sign in front of the time term (relative to the others), ie

\(ds^{2} = -dt^{2} + dx^{2} + dy^{2} + dz^{2}\)

A Euclidean metric would have a plus sign,

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

That's what makes relativity different, its fundamentally non-Euclidean. This mistake is the worse one you've made because its not some high level concept hidden in algebra (not that anything you've said is high level), its something fundamental to the entire ethos of relativity. The minus sign is what gives relativity its Lorentz invariance, since the Minkowski metric is Lorentz invariant but the Euclidean metric isn't.

You have failed to provide a derivation for the SC metric. Instead you did little more than a copy and paste on stuff high schoolers know. You failed to understand tensor rank vs component count. You've failed to realise the SC metric and relativity metrics in general are non-Euclidean. You've failed to grasp area/volume element integration. Each time I ask you to provide a justification for something you avoid answering the request directly and instead you come back with either an irrelevant or incorrect sound bite. You can't even admit you don't understand my research area, even after I specifically correct you on basic definitions.

I'll ask again, provide a valid derivation of the SC metric using your methods. If you cannot then have the honesty to admit it. You've been dishonest in pretending to understand what 'fluxes' are in regards to my work and your repeated avoidance of direct requests relevant to your claims is beginning to wear thin.

 

If this is actually true, then you are the worst possible advertisement for the "University of Phoenix", or any institution that would claim to be a place of education. You are undoubtedly a fraud, and the worst type of fraud - you attempt to make monetary gain by deceiving people that you are something you are not.

You should be ashamed of yourself.

 

[AlphaNumeric, Dishonest would be a kind word for you right now. I don't even want to quote you. This is the last time that you will accuse a professional of cutting and pasting. I have a reputation to protect too!.

That quote that you are so "Jealous" of, is found is my own personal writing and understanding, and in whole chapter named: The Geometry of Spacetime & 3 Sphere N-Dimensional Spaces - page 139 of Super Principia Mathematica.

I just don't like talking about those types of things, because they are not practical!!

Not liking is not the same thing as not understanding. I don't like caviar, I understand that they are under developed fish fetus. However, some people absolutely love caviar. I say to each his/her own!!

 

You're cutting and pasting from your book, I know. My point (which you always seem to struggle to get) is that rather than engage in discussion specifically what I said you just copy and paste a talking point you have.

For instance, why did you copy and paste in a load of stuff about hyperspheres and higher dimensional volumes? Did you think I didn't know that? Did you think I was unaware of the notion of hyper-spheres and that geometry in dimensions more than 3 or 4 existed? You have been linked to my papers and thesis, which you claim to understand so you have no excuse to thinking I hadn't heard of them, particularly since I mentioned such things in other posts multiple times.

The section of your post on the spherical metric is redundant, since I'm well aware of it and I've been the one correcting you on it. You have a \(\theta_{0}\) in there and I had to explain why it wasn't there and now, magically, you've dropped the \(\theta_{0}\). Yet for some reason you think you need to tell me something I've been telling you?

I know how to get the metric for a sphere, I know all about the Cartesians to polars transformations. I know how to manipulate a variety of metrics in a multitude of coordinates, more so than you. I say that because of your comments about passing through the event horizon of a black hole, as if the SC metric singularity was a problem. Change into different coordinates (like Kruskal or Eddington-Finklestein) and there's no problem. These are techniques taught at university, that's where I learnt them, so if you're as well read an academic as you want us to believe you should know about them.

I want you to stop posting where are little more than trivial exercises in coordinate transforms for 1st years and I want you to provide your supposed simplified derivation of the SC metric. I know how to do \((x,y,z) \leftrightarrow (r,\theta,phi)\), I know how to do \(ds^{2} = dx^{2} + dy^{2} + dz^{2} \leftrightarrow dr^{2} + r^{2}(d\theta^{2} + \sin^{2}\theta \, d\phi^{2})\). Anyone who hasn't slept through a relativity course knows that. Hell, anyone who hasn't slept through a basic course in vector calculus knows how to do that. You claim to have something novel, I want to see it. Each time you avoid doing that and instead reproduce trivial book work you demonstrate yet more intellectual dishonesty.

 

AlphaNumeric, I will have to admit, this is somewhat fun and I have totally enjoyed this.

If they were handing out "Nobel Prizes in Comedy" you would win, with no competition in your wake!!

It is this kind of stuff!! The above is not true! The point is that

\(ds^{2} \Not= dx^{2} + dy^{2} + dz^{2}\)

but is actually equal to

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

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

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


So, once again you are not correct! When are you going to get it right?

 

If these are indeed extracts from your self-published book, then I hope you have given significant references to Wikipedia.

Examples of your plagiarism are a plenty. I first noticed it here

 

You actually think that Wiki owns the right to basic concepts that is the same understanding for everyone. 100 different physicist would write those same statements if they have understanding!!

This would be like you saying that Wiki owns the statement, "The Enclosed Surface Area of a sphere is defined as (\(A = 4\pi r^2 \))."

 

Just to clarify, you're claiming this:

and other examples are coincidence, and you have not simply copied without citation?

In fact, why on Earth would I expect you to be honest about this? You spend your life trying to deceive.

 

You have not convinced me that statements that are generally true and understood by everyone could be considered unique statements. We are suppose to cite and or quote unique statements made by others.

This is the whole concept about patenting, and song copyrights, in order to own rights, you have to prove that your thing whatever it is, is unique!

This is why the laws of nature are not patentable. No one can own only that which belongs to God!

I have not deceived, or tried to deceive anyone, period!!!

 

I think your denial of plagiarism with regards the 51 word extract I've just presented (with punctuation and spelling in tact) gives a very good means for people to measure your honesty and intellect.

I, and I imagine the public at large, are grateful for this opportunity.

 

AlphaNumeric, I missed this one...

Now, finally you got something right!!

In my humble opinion, proving this, and bringing this result "to light" and in line with the Schwarzschild Metric is at the "forefront" of "Classical General Relativity."

I agree with this statement that you wrote down above.

Suppose you change \(s \to s + ds\), how does the area change? \(s^{2} \to (s+ds)^{2} = s^{2} + 2s \,ds + (ds)^{2}\). Thus the area change is actually \(2s \,ds + ds^{2}\).

So you are assuming that your "moduli" or "Higgs particles" somehow morph into something else, some new kind of spacetime "vacuum" energy? What I am more concerned with, is, are these spacetime interactions with the branes traveling at faster than light speed? I have been seeing Tachyon kind of stuff when it comes to vacuum energy, floating around lately!

 

Well done, you've just shown you don't know how to do a coordinate transform in vector calculus.

When you change coordinates \((x,y,z) \to (r,\theta,\phi)\) you're supposed to change all the x,y,z's to the spherical coordinates or vice versa, they are not to be mixed.

Seeing as you fail on such a basic thing I'll walk you through that. A line element in Cartesian coordinates is :

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

Th relationship between Cartesians and sphericals is the following :

\(x = r \sin \theta \cos \phi\)
\(y = r \sin \theta \sin \phi\)
\(z = r \cos \theta\)

Now you ask "If I change the spherical coordinates a small amount what happens to the Cartesians". In other words you apply 'd' to both sides and use the fact its a differential operator (ie use the Leibnitz rule).

\(x = r \sin \theta \cos \phi\)
becomes
\(dx = d\Big( r \sin \theta \cos \phi\Big) = (dr)\sin \theta \cos \phi + r \, d(\sin \theta) \cos \phi + r \sin \theta d(\cos \phi) = (dr)\sin \theta \cos \phi + (d\theta) r \cos\theta \cos \phi - (d\phi) r \sin\theta \sin \phi\)

Here I used \(d(sin \xi) = d\xi \cos \xi\) and \(d(\cos \chi) = - d\chi \sin \chi\), which is basic calculus. Now we have how much x changes if we vary the spherical coordinates. Now do the same for y and z :

\(dy = d\Big( r \sin \theta \sin \phi\Big) = (dr)\sin \theta \sin \phi + (d\theta) r \cos\theta \sin \phi + (d\phi) r \sin\theta \cos \phi\)

\(dz = d\Big( r \cos \theta \Big) = (dr)\cos \theta - (d\theta) r \sin \theta\)

Notice how in each case the change in the Cartesian coordinate is written only in terms of the change in the spherical coordinates, they aren't mixed together. As a result you can swap the dx, dy, dz terms in the \(ds^{2}\) expression for the spherical case. I'll leave you to do the boring algebra but if you put those in you'll find \(ds^{2} = dx^{2} + dy^{2} + dz^{2} = dr^{2} + r^{2}(d\theta^{2} + \sin^{2} \theta d\phi^{2} )\).

To help you along I'll do the coefficient of \(dr^{2}\) for you, which if this method is right should come out to be 1. It'll be the sum of the squares of the coefficient of \(dr\) in each term, ie \(\sin \theta \cos \phi\) from dx, \(\sin \theta \sin \phi\) from dy and \(\cos \theta\) from dz.

\((\sin \theta \cos \phi)^{2} + (\sin \theta \sin \phi)^{2} + \cos^{2}\theta = \sin^{2}\theta (\cos^{2}\phi + \sin^{2}\phi) + \cos^{2}\theta = \sin^{2}\theta + \cos^{2}\theta = 1\)

There, precisely as I said. Try it with the rest, it's easy algebra. I'll go get you a fork for all that humble pie you'll eat afterwards

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Once again you demonstrate you don't have the level of knowledge and ability you so obviously believe you have. Each time you say "OMG you've got that wrong!!" I've been able to explicitly walk you through the method which shows I'm not wrong, you are.

If you'd read any books on relativity you'd be familiar with this. This stuff isn't even relativity, its basic vector calculus. I'll add basic coordinate transformations to the extensive list of things you're incapable of doing.

You also didn't answer my direct questions. Why did you simply copy and paste a load of stuff about what 'hypersphere' means when you know full well I'm more than familiar with the concept? It seems that you don't have anything to actually say but you feel you must say something so you just do a copy and paste from your book something you feel is relevant. It shows you're intellectually dishonest because you're unable to engage in honest discussion. I reply the points you make with things I come up with on the fly, I don't copy and paste from anywhere. I would expect someone engaging in honest discussion to do likewise.

I also asked you why you feel the need to lie about your understanding of 'fluxes'. It wasn't a matter of nomenclature, your attempt at a definition was simply wrong. Your silence on the matter says a lot.

But your honesty is further demonstrated to be non-existent by Guest, who has caught you red handed copying and pasting from sources without reference. Now not only are you copying and pasting from your own book, which you didn't initially admit was your book, but now you're copying and pasting from Wikipedia and passing it off as your own understanding/thoughts.

I'd say that's enough to call an end to this discussion. Metrics, tensors, volume elements, components, ranks, coordinates, transformations, derivatives, units, metric signatures, all of those things pertaining to your 'work' are things you've shown you don't understand. You won't provide derivations when asked, you ignore corrections, you can't admit mistakes, you won't answer questions. You've been thoroughly dishonest and have provided sufficient evidence that you don't understand this material that in the future when you bring up your work all any of us have to do is just point to this thread to demonstrate your work has been thoroughly discredited. You've had plenty of opportunity to step up and you've either not done so or have fallen flat on your face when you've tried. Your last remark about the various expressions for \(ds^{2}\) in different coordinates and Guest catching you plagiarising seals the deal.

If any moderators are reading this then I request an external evaluation of Magneto's general level of honesty and compliance with answering relevant direct questions and providing references for his claims. I contend he's failed multiple times to uphold any of those requirements and thus the discussion is over and the thread can be locked. If Magneto can provide exceptionally valid reasons to reopen it at a later time it can remain as evidence he's a hack.

 

Once again, you demonstrate that you have filled your brain with so much knowledge that you have forgotten the basics of General Relativity.

The result that you prove above is not a Euclidean Metric, and is the trivial result obtained when converting from Cartesian to Spherical coordinates. This result does not produce a Schwarzschild or Euclidean Metric.

What Karl Schwarzschild discovered, was that the real solution to GR problems is to start with Spherical Coordinates and not, I repeat, not start with the Cartesian Coordinates like you tried to prove.

You have revealed that you need to re-read your "First Years" physics text and maybe we can start a new challenge.

Until then I agree, this challenge is over. We can close/lock this post.

I declare you, AlphaNumeric the winner and heavy weight physics champion of the World!!! :bravo: AlphaNumeric :bravo: