AlphaNumeric, I don't want to bash you, I am more interested in intellectual discussion. The subject of General Relativity is at least 80 years old, however it has only come to real favor over the last 15 years. And this mostly due to our advanced telescope technology and our high energy colliders. So although we can have some, as well as a firm understanding of the math and physics of spacetime, we including myself are still interesred in learning new things from people such as yourself about the math and physics of GR, Cosmology, and High Energy Physics. Part of growing and learning is engaging in discussions with others that do not necessarily agree with you, and hearing their ideas, wrestled against the facts; that in my opinion is what posting in these forums is about. Agreed. I dont agree with you here. The Schwarzschild Radius is truly a vector having both a magnitude or scalar quantity as you point out, and it is also a vector quantity. Lets stick with Spherical Coordinates. And call the Schwarzshild Radius a static "black Hole" event horizon, which is an isolated net inertial mass system located in some region of space. The space and time in the local vicinity of the mass creates a gradient field in space, and where the curvature is most maximum in that gradient field, is define by the Spherical body whose vector radius is the Schwarzschild Radius. The closed Geodesic associated with the matter and energy in that location is Source of Curvature. Schwarzschild Radius Vector - Tensor \(r_{S} \^r = \2 (\frac{m_{net}G} {c^2}) \^r = (\frac{\frac{1}{2}\ m_{net}c^2} {\frac{1}{4} \frac{c^4}{G}}) \^r = (\frac{T_{ab}} {\frac{1}{4} \frac{c^4}{G}})= \frac{G_{ab}}{2 \pi} \). Spacetime Metric for Non Expanding Spherical Net Inertial Mass Body at the Schwarzschild radius Event Horizon \(s^2 = \({r_{S}^2} + \ {r_{S}^2}\({a}^2 + \ b^2 \sin^2(a_0))\). Vector for Non Expanding Spherical Net Inertial Mass Body at the Schwarzschild radius Event Horizon \(s \^r = \({r_{S}\^r} + (\ {r_{S}{a})\^a} + ({r_{S}{b}}) \sin(a_0) \^b\). . Schwarzschild Radius - - \(r_{S} -> m \) Net Inertial Mass - - \(m_{net} -> kg\) Rest Energy - - \(E_{Rest} = m_{net}c^2 -> \frac{kgm^2}{s^2}\) Speed of Light - - \(c -> \frac{m}{s}\) Universal Gravitational Constant - - \(G -> \frac{m^3}{kgs^2}\) Source of Curvature Geodesic - - \(G_{ab} -> m\) Riemann/Ricci Maximum Curvature Geodesic - - \(R_{ab} -> m\) Stress Energy Tensor - - \(T_{ab} -> \frac{kgm^2}{s^2}\) Tensor Indices, Latitude & Longitude Direction Angles - - \({(a), (b)} -> radians\) I believe that the units do match! Best.
I don't mean to interject on the current love fest but I have a sincere question for Farsight which is partially related to something I've been thinking about for a while: Would you predict that the tick rate of a clock under extreme pressure would be reduced?
High energy colliders don't test GR. And the 'golden age' of GR was the 60s and 70s. Before the 80s there was plenty of stuff we could test in GR, all of the major predictions Einstein made were tested to some degree. before 1990. The errors I've been pointing out aren't a matter of opinion. You cannot equate a rank 2 tensor to a scalar. The metric is not a volume. The Einstein tensor is not always about minimal length geodesics. The radius is a length. The vector you're referring to is \(r_{S} \mathbf{e}_{r}\), ie a vector in the radial direction whose length is the Schwarzchild radius. \(r_{S}\) appears in the line element expression for the SC metric, it is a component of the metric in the usual SC coordinates and thus it is a scalar. An event horizon is a surface, not a length or vector. For a given moment in time the condition \(r = r_{S}\) defines a 2 dimensional surface which happens to be spherical. No, the black hole system is an isolated mass system ('inertial' doesn't really come into it) which happens to have an event horizon. The event horizon itself isn't the system. Incorrect. The event horizon isn't defined by maximum curvature, that occurs at the singularity as can be seen if you compute \(R^{abcd}R_{abcd}\) in the SC metric to get \(\frac{48}{r^{6}}\). The event horizon is the surface defined by the points where there exists one and only one null geodesic which does not lead to the singularity, that followed by a photon moving directly away from the singularity. This isn't the location of maximum curvature because inside the event horizon the curvature is such that there are no such null geodesics, they all lead to the singularity. The SC coordinates are such that certain quantities become infinite on the event horizon but this is a coordinate singularity, not a physical one (look up the difference). By suitable choice of coordinates the event horizon becomes 'nice', such as Kruskal coordinates or Eddington-Finklestein coordinates. This is meaningless. If I'm wrong about that quantify what you mean. You clearly don't understand the concept of 'tensor rank'. I said you couldn't equate a scalar to a rank 2 tensor and now you've 'corrected' it to equate a vector, which is a rank 1 tensor, to a rank 2 tensor. You're simply proving my criticisms are accurate, you lack the basic essential understanding required to do any actual GR. I'll lay it out for you. See the 'a' and 'b' indices on \(G_{ab}\), there's 2 of them so its rank 2. A vector component would be written as \(v_{a}\), there's 1 of them so rank 1. The Riemann curvature tensor is \(R^{a}_{bcd}\), there's 4 of them so rank 4. See how it works? \(G_{ab}\) is short hand for a matrix, \(G_{11}\) is the first colum-first row entry. \(G_{12}\) is first row second column etc. You're equating a matrix to a vector or scalar, which is wrong. If you don't understand why then you need to learn some linear algebra, the sort taught to 1st years. The whistling sound you're hearing is my point flying over your head. I suggest you slink back off into the depths of the internet before you jam your foot any further down your throat. I pity the people you con into buying your books and I think you should be ashamed of yourself for being such a deceitful person to write such things and charge money for them.
But \(T^{ab}\) has units of kg . m^-1 . s^-2 (energy density) so \(G^{ab}\) has units of m^-2 thus \(R\), \(R^{ab}\) and \(R^{abcd}\) all have units of m^-2 thus \(R^{abcd} R_{abcd}\) is a scalar with units of m^-4. This is indeed the standard definition as even tertiary sources like Wikipedia reveal: http://en.wikipedia.org/wiki/Schwarzschild_metric#Singularities_and_black_holes "One such important quantity is the Kretschmann invariant, ... \(\frac{12 r_s^2}{r^6}\) ... " which clearly has units of m^-4.
He doesn't. And he does't have to. Come on, read it again: This space-time variability of the reciprocal relations of the standards of space and time, or, perhaps, the recognition of the fact that “empty space” in its physical relation is neither homogeneous nor isotropic, compelling us to describe its state by ten functions (the gravitation potentials gμν) has, I think, finally disposed of the view that space is physically empty. Now think it through. Gravity isn't some magical mysterious action-at-a-distance tug. We have no evidence for gravitons flying around. When light curves it's because there's a gμν gradient across the space it's moving through. If the space was homogeneous, if there was no variability, there wouldn't be any gμν gradient. And the light wouldn't curve. You can gauge gravitational potential by measuring gravitational time dilation with optical clocks. See this report. If it's the same at the floor as it is at the ceiling things ain't going to fall down from the ceiling to the floor. And whilst people say "time passes faster at higher elevations" those clocks are optical clocks. They're clocking up the motion of light. And the speed of light through space is determined by c = √(1/ε0μ0). The light goes faster near the ceiling because the space there isn't the same as the space near the floor. That's what the evidence is telling you. The same sort of thing applies to a gravitational field. State the problem correctly and you're asking why light moves faster at higher elevations rather than why time passes faster. The rest is easy. Yes. For light to travel in curved lines through space, that space cannot be homogeneous. It cannot be uniform. Your sphere is merely uniformly non-uniform. Then out goes the baby with the bath-water, and you've just made a mystery out of gravity. It comes back to the way a uniform gravitational field is a contradiction in terms. Again, I would urge you to read Einstein's gravitational field My argument is all about semantics. That's semantics in the sense of meaning, not irrelevant hair-splitting. The FLRW metric starts with the assumption of homogeneity and isotropy of space. And the whole point of my argument is that when you take on board what Einstein said, you appreciate that the space within your sphere exhibits a homogeneous inhomogeneity and thus it is a contradiction in terms. My problem is that Σ ranges over a 3-dimensional space of uniform curvature contradicts The FLRW metric starts with the assumption of homogeneity and isotropy of space. Evidence vindicated general relativity fairly quickly. WMAP told us we could throw away some subset of EFE solutions. It didn't predict that. It gave the equations of motion. It predicted how things move. It predicted the curvilinear motion that is nowadays described as curved spacetime, wherein spacetime is often labelled as a "space". Taking what Einstein said at face value. Taking the evidence at face value. We're going to have to agree to differ on this. You might like to look afresh at thestress energy tensor. And remember what I said about the electromagnetic field being frame-dragged space. Space is not a fluid, the metric is not symmetric, it doesn't boil down to ten components. And in this region things don't fall down, light goes straight, and there is no discernible gravitational field. Yep, and when you set aside the expansion of the universe to map the whole of space as it is now, you find that it doesn't curve round on itself. Then when you let the universe expand, it still doesn't. Your initial sentence was I read this as Einstein saying we'd have to learn to view space as being endowed with physical properties, and accept that it could be inhomogenous and anisotropic in those properties. It paid no attention whatsoever to the fact that Einstein was talking about a gravitational field here. You must surely be aware that you did that unconsciously because it doesn't square with what you've been taught. That's just your excuse to avoid facing up to the fact that you're dismissing the inhomogeneous vacuum paper along with Einstein, Newton, patent scientific evidence, and a straightforward chain of reasoning. I don't know. He talked about space as a something, he talked about spatial stress-energy, he talked about a frictionless adiabatic fluid with a pressure and a density. He knew that gravitational fields were inhomogeneous space, affecting the motion of things through space. But somehow he didn't see that if space itself has an innate pressure and is not infinite, it must perforce expand like a ball of perfect gas with no gas around it. Some say he lost something, like his edge or his hunger or his insight after he found fame. Some say he got lost in maths. But like I said, I don't know. And I don't know that we can get much further with this przyk. It's time consuming, and you're putting up barriers every inch of the way.
No, your problem is you don't understand geometry well enough to realise that its possible for space to take a shape which is homogeneous, isotropic and yet not flat. Spheres are such an example. They have constant curvature and are isotropic and homogeneous*. It isn't a contradiction, your understanding is simply too poor to grasp it. * Please don't make the mistake of viewing those properties in terms of a 2 sphere embedded into 3 space, you should view it in terms of the sphere itself, not the space you might embed it in. Please explain why you think you've got the 'true' understanding of Einstein from reading pop science books yet everyone in the GR community for a century, who read his actual word, some even worked with him and all of whom had a working understanding of GR (which you don't have) yet somehow failed to grasp what you have. You keep on with the "You're just parroting what you've been taught" argument but you never actually demonstrate what we've been taught is wrong, you just assert it, and you never demonstrate that your understanding, in the absence of being taught or learning anything, is somehow superior. Do you really believe what you're spouting? Is your world view so distorted you honestly believe it? You talk about reasoning, evidence, Einstein, Newton etc yet every single person who has devoted their lives to those things who has evaluated your work has found it fails on all of those. You demonstrate you don't even know basic geometry, the foundation of Einstein's work. You admit you don't know any of the quantitative details, thus preventing you from learning even 1st year level physics and maths related to Einstein and Newton but you think you've got the best understanding!? Your "You're just parroting textbooks" attitude shows your true colours. You don't like it that others can grasp what you can't and you try to convince yourself its okay because you're not being 'told what to think' and what we're being 'told to think' is, so you claim, wrong anyway. That way you can tell yourself its alright you failed where others (like us) succeeded. You can't simultaneously say "You're just parroting your books" and "Listen to Einstein". As usual, a crank says "Don't accept things unquestioning...... other than what I say!". You don't like textbooks because you don't understand them so you cannot evaluate their merits or claims thus you think "Anyone using a textbook must be accepting it unquestioningly", as that's what you're forced to do. In reality some people can grasp books you can't and thus don't suffer from this problem. Both Rpenner and I explained why your "Lets narrow the scope of a discussion to mass" is unacceptable and unscientific. I'll give you the benefit of the doubt in regards to not replying to it that it was late and you didn't have the time. Ignore it a second time and it'll seem like you're trying to avoid facing up to the requirement people ask you questions about your work you don't have answers to.
Yes, sorry. It was late. No. Take a look at underwater acoustics. The speed of sound underwater increases with pressure. So if your clock was something like a sonic version of the parallel-mirror light clock, the tick rate would increase. But don’t take that as a definitive answer because pressure can get very tricky. It’s kind of in the space between the particles, and if your force per unit area isn’t making something collapse, something is pushing back. Or pulling. Or there’s some combination of both. Like in a sound wave – it’s a pressure wave that travels very fast through an elastic material such as steel.
Rpenner asserted it with a pile of pompous flannel in an attempt to avoid focussing in on something we can get our teeth into. And yet his list of questions betrays his interest. Hmmn. You give more flannel, plus your customary feather-spitting abuse, and you contradict yourself when you say pick a single real world phenomenon which you believe your work is able to describe and explain. Now, how does that square with me suggesting mass, and you saying discussing mass would be unacceptable and unscientific? It doesn't, now does it? Sigh. You simply cannot conduct a sincere discussion, you will not correct a colleague who makes a blatant error, you dismiss Einstein and the scientific evidence, and you even evade the distinction between physical space and a mathematical space. And as ever you resort to abuse. We can't get off the ground with mass, so thanks but no thanks. I'll not waste my time talking to you, troll. Go away and play with your string-theory pseudoscience.
You have a 'model for mass' then? Mass is something components of systems have. I want you to demonstrate you can model even one system in the real world because if you can't then you cannot demonstrate your work has any connection to reality. You know this because you try to use the same argument against string theory. More hypocrisy and double standards from you. I'm the one whose repeatedly tried to engage you in discussion on the mistake you made in your work where you conflated mathematical axiom and physical postulate, your comment is just nonsense. The fact people say things you don't want to hear doesn't mean they're not engaging in sincere discussion, it means you aren't. I've asked you to justify those statements. Your repeated inability to do so, yet continued use of them as lines of argument speaks volumes about you and your sincerity. I don't work in string theory any more. My job now entails many different areas of physics and maths, such as fluid mechanics and quantum theory, specifically applying them to real world problems. Once again you invent your own little world so you can convince yourself you're not the complete and utter abject failure in science you appear to be to everyone else. You can try to insult me for my PhD research but the fact of the matter is I've managed to be more of a scientist than you ever will be. Please Register or Log in to view the hidden image!
OK. If we account for the distance differential between the clock's components due to the pressure, would you predict that the increase in local tension or stress might produce a reduction in the clock's tick rate? This isn't a trick question, it's related to a description of energy/mass that I believe was written by you in the past which is not dissimilar to my own thoughts on the subject...
Are you even reading anything I say? I never said anything about action-at-a-distance or gravitons. Nobody is claiming classical GR includes action-at-a-distace or gravitons. This seems to be the crux of your misunderstanding of GR and the mainstream view of it. Farsight, the \(g_{\mu\nu}\) aren't associated with space-time. They're associated with a particular coordinate system in space-time. The metric components essentially measure how locally stretched and/or distorted the coordinate system they're associated with is, so you can think of the derivatives of the \(g_{\mu\nu}\) as measuring the inhomogeneity of that coordinate system, and not of space-time. With that fixed, the only correct interpretation of what you said is this: when the \(g_{\mu\nu}\) are inhomogenous (i.e. not constant) in a particular coordinate system, you will obtain curved descriptions of the trajectories of massless and massive (GR isn't just about light) particles in that coordinate system. And this is where the Riemann curvature comes in. When the Riemann curvature is non-zero, any coordinate system you try to map space-time with is forced to be inhomogenous, even if space (and/or space-time) itself is homogenous[sup]*[/sup]. Consequently, if space-time is curved, you are forced to accept curved descriptions of at least some trajectories, no matter how you set up your coordinate system. If you mean that, then I agree. [sup]*[/sup]Example: the sphere is homogenous, but you can't map it with a homogenous coordinate system. You can't map the surface of the Earth with a map that faithfully represents distances everywhere on Earth at a fixed scaling factor. This is circular reasoning. You start with the assumption that light can only curve in inhomogenous spaces (which is specifically what you're supposed to be justifying) and then conclude that the example homogenous but curved space I gave you must in fact be inhomogenous. Yes, and in a large family of solutions, the spatial part of the FLRW metric is a 3-sphere. Friedmann, Lemaître, Robertson, and Walker clearly considered the sphere an example of a homogenous space. No, as stated above, it just means that Friedmann, Lemaître, Robertson, and Walker considered the sphere homogenous. In whatever sense you are using the word "homogenous", it is not in the same sense that Friedmann, Lemaître, Robertson, and Walker used it. The action principle for a given metric predicts that. But general relativity also includes the Einstein field equation which gives the reverse relationship: it predicts how space-time responds to the presence of matter. The FLRW family of metrics are solutions to the Einstein field equation, and not of the action principle. They are specific predictions about how the geometry and evolution of space-time depends on the amount of matter and radiation in it, assuming the matter and radiation distribution is homogenous. What? I didn't deny that Einstein was, generally, talking about gravity. I'm saying that, in your quote, Einstein does not say anything anywhere near as specific about the relation between gravity and space as what you attribute to him. No, it is a reason for dismissing the inhomogenous vacuum paper. You say that GR has always been a theory about how gravity is inhomogenous space, yet apparently the best anyone has been able to do, as of 2008, is show that the idea works in a special case? While the Riemann curvature tensor, and not a "graded refractive index", appears explicitly in Einstein's early papers?
Cor, blimey! Wot, me pompous? Yeh arrogant teller of cobblers, yeh bloody great jam roll. I'm rather inclined to opine that you are projecting, since you didn't manage to pull out even one prevarication, equivocation, or misrepresentation to dissect. http://www.thefreedictionary.com/flannel http://en.wikipedia.org/wiki/Psychological_projection
AlphaNumeric, I was going to ignore this, but I don't want anyone to think that you are somehow proving me incorrect. I am not anywhere nears a know it all; and I don't mind being corrected. But this is definitely not the case or the time for being corrected. First, you are only partially correct here, and it is this sort of mathematic that you are describing here that keeps physicist away from GR. The reason that the Super Principia Mathematica was written was to simplify the rigorous mathematics into something that people could sink their teeth into and understand, this also was Newton's Goal when he wrote his Principia. So let's start with a little history; The concept of the Moment of Inertia for a spherical body was first introduced by Leonhard Euler in his book Theoria motus corporum solidorum seu rigidorum in 1765. In this book, he discussed the moment of inertia and many related concepts, such as the principal axis of inertia. Then some 100 or more years later Einstein working on his Field Equations wanted to design a mathematical system for an isolated Net Inertial Mass body, applied the concepts of Euler's Moment of Inertia for a spherical body to his model and came up with 9 different terms components, Axes, or "Ranks;" and one for time for a total of 10 components or "Ranks" to his Field Equation. Einstein's isolated Net Inetial Mass system would describe the moment of inertia of an object about a given axis, which describes how difficult it is to change its angular motion about that axis. Therefore, it encompasses not just how much mass the object has overall, but how far each bit of mass is from the axis. The farther out the object's mass is, the more rotational inertia the object has, and the more force is required to change its rotation rate. Sorry, I don't know how to build the Matrix using LaTex. When Einstein satisfied himself with his 10 terms to his field equation, he was forced to use the 10 terms because he chose to work with "Cartesian Coordinates" (x, y, and z) as his metrics. However, shortly after Einstein's success, Karl Schwarzschild decided to use Spherical Coordinate and determined that this removed the necessity for having to carry around 10 terms, or "Ranks." And that the angles for latitude and longitude was enough to accomplish what Einstein's 10 terms was trying to accomplish. So what does each of the 10 terms describe? Each term describes one degree of freedom for the body, and is an Axis (principle or parallel) relative to the center of mass of the system. All of these rotations are inherent in choosing a spherical coordinate system over a cartesian coordiate system. Here the difference is clearly seen why choosing a Spherical Coordinate over a Rectangular coordinate works; it is because the Spherical Coordinate has a separate Directional Component that is independent of the radius of the sphere. Spherical Coordinate \(s^2 = \({r_{S}^2} + \ {r_{S}^2}\({a}^2 + \ b^2 \sin^2(a_0) \)\). Cartesian Coordinate \(s^2 = \( {r_{S}^2}{sin^2(a)}{cos^2(b)} + \ {r_{S}^2}{sin^2(a)}}{sin^2(b)} + \ {r_{S}^2}{cos^2(a)} \)\). So when you are mentioning "Ranks" to me when I am using Spherical Coordinates instead of Rectangular Coordinates and you throw a \(G_{ab} = G_{11}\), or \(G_{ab} = G_{12}\) this sounds like you are mixing things up, and I am being nice here!! Now I will give you some credit here because Modern Mathematians are using the shorthand of Riemnann where he uses three dimensions of space and one dimension of time and created a four (4) vector or tensor Riemann curvature tensor is \(R^{a}_{bcd}\). But once again this is using the Einstein concepts of 10 terms in one equation and reducing them down to 4 terms in one equation. This not wrong on your part per se, just not user friendly. You may be right here, there is much speculation going on here! Would you please explain how this term is mechanized into the mathematics of General Relativity; more specifically, what equations is term apart of \(\frac{48}{r^{6}}\)? Przyk, this is a really good statement, and very very important! What I would add to this, is that the FLRW family of metrics predict what is happening inside and outside areas of curvature; whether that curvature be cause by matter or heat radiation. Meaning that the FLRW metric predict both Euclidean and Non-Eucliden Space or Spacetime as one system; or Curve and Flat space together as one system. And if anyone want's a good read, that is user friendly to the Conceptualist or Mathematicians on the subject of General Relativity, I recommend the Super Principia Mathematica, by Robert Louis Kemp.
I've added you to my know-nothing-crank list. AlphaNumeric explained multiple mistakes of yours, including quite basic ones showing that you have no idea about the difference between a scalar, a vector and a tensor. Listening to anything else you have to say on the subject of general relativity is bound to be a complete waste of time. Bye! PS The problem here is not so much that you lack knowledge. It's that you show a complete unwillingness to learn from people who have the knowledge you lack.
That doesn't change the fact you haven't got a clue about relativity and differential geometry. You should have put in some effort to learn them before writing books on the subject and trying to con people out of money. I did a degree in a maths department and a PhD in a physics one. I worked along side people researching GR, so I am perfectly aware of the general views of the communities on such things as GR. If you think the level of mathematics I've given is excessive you're just naive. The sort of stuff I've said to you, about tensor ranks and indices, is the stuff 1st year physics students learn about. It is the basics, required knowledge for everyone doing a physics or maths degree. In order to write a book which simplifies complicated stuff you must first understand the complicated stuff. You don't. Plenty of people already can sink their teeth into this stuff, plenty of people take physics and maths degrees. Sure, it'd be nice if more of the general public understood more physics and maths but your approach is only going to hinder that because its important to understand things correctly. I'd rather 1 person knows something complicated correctly than 10 people think they know something simpler correctly but it actually be wrong. Your attitude is "I'm helping people who don't grasp this" but you don't grasp this. You don't know how the number of components in the EFE arise. Rank and components are not the same. A vector in 3 dimensions is rank 1 but has 3 components. More generally a vector in N dimensions is rank 1 but has N components. A general rank M tensor in N dimensions has \(M^{N}\) components. This is then reduced by symmetries. In the case of a rank 2 tensor in 4 dimensions there are 16 components, ie you have a tensor \(M_{ab}\) where a,b can range over 1,2,3,4. In relativity its generally written as 0,1,2,3 where 0 is the time index. If the tensor is symmetric then \(M_{ab} = M_{ba}\). This reduces the number of independent components from \(N^{2}\) to \(\frac{N(N+1}}{2}\). If N=4 you get 10 independent components. That is now 10 components arise in a symmetric 4 dimensional rank 2 tensor, which \(G_{ab}\), \(g_{ab}\), \(T_{ab}\) and \(R_{ab}\) all are. If you're going to go down the route of "Lets review some history" at least get it right. You're obviously used to dealing with people who don't know any relativity and whom you can bamboozle with bullshit. That isn't going to work here, I (and others here) have a working understanding of this area of physics and I've got no qualms in demonstrating it. No, the (x,y,z) coordinates are coordinates, not metrics. You should look up the role coordinates play in general relativity. You'll find it in the first chapter of a GR book because they are dealt with in the definition of 'manifold', which you obviously don't know. Besides, GR is formulated in a coordinate independent way, that's part of its power. Tensors are used because tensor equations are independent of what (valid) coordinate system you use. \(G_{ab} = 8\pi T_{ab}\) is true in Cartesian coordinates or polar coordinates or Rindler coordinates or Kruskal coordinates or any valid coordinate system you care to make up. The 10 terms are nothing to do with angular momentum axes. Besides, even if they were the angular momentum tensor \(I_{ab}\) is symmetric too, \(I_{ab} = I_{ba}\) so there's only 6 independent components in it, you're still 4 short. Properties of the angular momentum tensor are another thing 1st year physicists have to learn. No, he didn't. Unlike you I actually know the derivation of the SC metric via Birkhoff's theorem's application to \(R_{ab} = 0\). The typical form of the metric is written in spherical-like coordinates but that doesn't mean the metric is the result of that choice. People use Kruskal or Eddington-Finklestein coordinates in the SC metric all the time. There are still 16 components in the SC metric, with 10 of them possibly independent. The use of spherical coordinates helps to highlight an underlying symmetry of the metric, that it is spherically symmetric. As such it has fewer independent components than a general metric but it still have more than just latitude and longitude. In fact the whole point of spherical symmetry is that its independent of latitude and longitude, the system doesn't change. I'd tell you to compute the geodesic arc length joining two points on the sphere and then apply an SO(3) rotation and repeat to get the same result but I know you're incapable of doing such a calculation. I'll add spherical symmetry to the list of things you don't understand. No, you're being ignorant. You don't understand the role of coordinates, the coordinate independent nature of tensors, the notion of tensor rank or the difference between rank and components. I'd expect someone starting their first course in GR to know those things, because they come up in prerequisite courses to study GR. The fact you've written books on this stuff makes your errors inexcusable. Did you even open a book on GR before writing them or did you just 'know' how it all works? You haven't understood the curvature tensor either. The rank of the curvature tensor (ie the number of indices) is always 4, whether you're in 1 dimension or 50. If you knew the definition of the tensor you'd know that. This is something anyone studying GR should know, because there's nice little results like the fact the Riemann curvature tensor has only 1 independent component in 1+1 dimensions, which equates to the Ricci curvature scalar. The curvature tensor can be computed from the metric using this equation and the definition of the connection. The way you describe it is as if its used instead of the metric. You continue to confuse rank, components and independent components. Let's make this even simpler. Let \(M_{ab}\) be a general rank 2 tensor in 4 dimensions. Then you have \(M \sim \left( \begin{array}{cccc} M_{11} & M_{12} & M_{13} & M_{14} \\ M_{21} & M_{22} & M_{23} & M_{24} \\ M_{31} & M_{32} & M_{33} & M_{34} \\ M_{41} & M_{42} & M_{43} & M_{44} \end{array} \right) \) 16 different entries or components. Suppose it is symmetric, so that \(M_{ab} = M_{ba}\), then you get \(M \sim \left( \begin{array}{cccc} M_{11} & M_{12} & M_{13} & M_{14} \\ M_{12} & M_{22} & M_{23} & M_{24} \\ M_{13} & M_{23} & M_{33} & M_{34} \\ M_{14} & M_{24} & M_{34} & M_{44} \end{array} \right) \) Now there's only 10 different components. That's how it works for the metric and the Einstein tensor. What about if its antisymmetric, \(M_{ab} = -M_{ba}\)? \(M \sim \left( \begin{array}{cccc} 0 & M_{12} & M_{13} & M_{14} \\ -M_{12} & 0 & M_{23} & M_{24} \\ -M_{13} & -M_{23} & 0 & M_{34} \\ M_{14} & -M_{24} & -M_{34} & 0 -\end{array} \right) \) Now there's only 6. That's how it works for the Maxwell tensor. Notice how 16 = 10+6? That's because every rank 2 tensor splits into a symmetric part and an antisymmetric part. This is linear algebra 101. You can't even get onto a GR course if you can't do this. No, its right, you're the one utterly wrong. You thought that the event horizon was a real singularity, where curvature is maximum. I explained its only due to the choice of coordinates, different coordinates remove the problem. In order to see this rigorously you have to compute curvature dependent quantities which are scalars. \(R^{abcd}R_{abcd}\) is such a thing, so if you're right then the value of that should be maximum at the event horizon. If you crunch through the algebra for the SC metric you get \(R^{abcd}R_{abcd} = \frac{48M^{2}}{r^{6}}\). Therefore maximum value of this quantity is at r=0, not r=2M, thus your claim is false. If you don't know how to go about working that out for yourself then you illustrate you don't have even a basic grasp of the working of relativity. It's little more than a tedious homework problem. I'd wager you're him (since I've accused you of that several times and you haven't denied it) and even if you aren't your utter ignorance of the workings of GR illustrates you didn't learn anything useful from those books. Besides, commercial advertising of books isn't allowed, particularly when they are full of ignorant crap. You, Farsight, Anita Meyer, all of you plug books which failed to meet even basic scientific standards and hence why you're plugging them on websites (or in Farsight's case batshit crazy TV shows on backwater satellite channels). Short conversations with any of you are enough to expose such profound stupidity and ignorance that its clear that the books aren't as amazing and insightful as you'd like us to believe.
This is a tricky one, because I'd be making a postdiction rather than a prediction, and whilst we talk of stress-energy, words like pressure, tension, and stress apply to materials rather than space. I'd rather say that an increase in vacuum impedance produces a reduction in the clock's tick rate. Vacuum impedance is Z0 = √(μ0/ε0), and c = √(1/ε0μ0), so if impedance increases c decreases, and a parallel-mirror light clock will run slower. Then see the NIST caesium fountain clock. It's an atomic clock, but it employs electromagnetic phenomena, so in the wider sense it's a light clock too. So again the tick rate is reduced. Ditto for a quartz clock, which uses an electronic oscillator. Electromagnetic processes within machines go slower too, so a mechanical clock will also go slower.
Sorry to be slow replying, we're moving house next week and there's a lot to do. I'm reading everything you say. The point is that light curves because of some property of the space it's moving through, not because of some magical mysterious phenomena. And it doesn't move through spacetime because there is no motion through spacetime, so curved spacetime is not the causative agent. This is the crux of your misunderstanding, pryzk. Coordinate systems have no real existence. When you measure time with a clock, what you're actually measuring is cumulative local motion through space. When you use a parallel-mirror light clock held flat to avoid issues of length contraction, at different altitudes you think you're measuring different degrees of gravitational time dilation. But what you're actually measuring is different speeds of light. I do accept curved descriptions of some trajectories - I've talked about the curvilinear motion of light previously. But I don't accept curvature as the cause. I'm saying that spatial inhomogeneity is the cause and curved spacetime is the effect. Just map it on a globe. Sure, you can't map a curved sphere onto a flat rectangle without distorting something, but "mapping spacetime" is getting this badly wrong. What's out there is space, and motion through it. Time is a cumulative measure of motion, nothing more, so map the space. It's robust reasoning. Remember what I said above: light curves because of some property of the space it's moving through. I say it's down to a gradient in vacuum impedance, which causes c to vary, which you can measure via a parallel-mirror light clock. We know about vacuum impedance and c, those clock readings are hard scientific evidence, and you should take it all at face value. The clock clocks up the motion of light, so when the clock goes slower the light goes slower. You seem to be dismissing this in favour of "light curves because the space is curved". That's circular reasoning. Agreed. I'm using it the same way. The issue is whether you allow the curvilinear motion of light in homogeneous space. I say if the space is truly homogeneous there is no Z0 gradient and hence light moves in straight lines. And WMAP is telling us which is correct. All I can suggest is that you read it again. I've been quoting Einstein's variable speed of light from his early papers, he refers to Huygen's Principle in section 22 of The Foundation of the General Theory of Relativity. And he isn't giving the equations of curved spacetime, he's giving the equations of motion. Through space. But for some reason people seem to think that a Riemannian manifold is space. It isn't. As they say, the map is not the territory.
Mathematicians and physicists often use 'space' in a synonym for 'manifold' so often you'll read discussion of 'a Riemannian space' when they are referring to 'space-time with a Riemannian metric defined on it'. Things are further complicated by the case space-time is a pseudo-Riemannian manifold but will often still be referred to as 'a Riemannian space'. This is further complicated by the fact a space-like 3 dimensional submanifold of pseudo-Riemannian space-time is itself specifically a Riemannian space (not pseudo). All of these nuances are things you pick up and can automatically put into context if you work with this stuff and can do the specifics. You've already shown you don't grasp these nuances and you have no working understanding of it so its a bit daft you tell others what is meant in papers when you can't understand the details of said papers,. If you rely on the principle "I'll interpret that using everyday usage" (which you do) then you'll end up misunderstanding and its precisely those kinds of misunderstandings we've been trying to correct in you. How long you planning on keeping this up? Are you planning to forever whine on forums about how you're a world expert in electromagnetism and have world leading understanding in so many areas of physics? Are you planning on submitting more work to journals? Are you going to keep throwing your money down the drain on vanity publishing and magazine adverts? You've failed every single test your work has been subjected to, every person versed in physics whose looked at it has rejected it. Internet hacks usually just melt back into the internet but you can't seem to let go. So where are you going from here?
