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What Is The Geometry Of Spacetime? — What Kinds Of Inner-Products/Metrics?
We ended the second article in this series with the audacious claim that we have obtained the .& Quite frankly, I expected to hear a howl of protest to my brazen assertion that we had actually gotten so far.& We certainly hadn't obtained anything quite like the usual matrix or tensor form of a metric, or even a "line element", nor had we seen how this has any relationship to the usual "strangeness" of the Special Theory of Relativity.It is the issues/questions of the metric, its relationship to the inner or dot product, and what forms it may take that we shall address in this installment.& As for how the possible forms of the metric relate to the usual "strangeness" of the Special Theory of Relativity, we may be able to address such within the comments section.& If not, we'll have a fourth installment.We Have the Metric, You SayOK.& You will recall that we defined the inner or dot product, on our vector space(s) (the tangent space[s]), thus giving us an inner product space(s).& So, the question, now, is how does the inner or dot product relate to what is called the metric, or the metric matrix (or tensor).Recall that from the definition of the inner or dot product we may take any two vectors, such as x,y ∈ V (our vector space), and obtain a scalar (from the associated mathematical field, F, to our vector space) by way of the operation x·y = y·x = g(x,y) ∈ F (which is the Real numbers, in our case).OK.& Fair enough, but how do we "find" these scalars?& Can we simply pick them "out of a hat", so to speak, or is there some more specific procedure?Recall, again, that we are dealing with a vector space, V (our vector space), and vector spaces have certain features and characteristics.& So, we can use these features to always set up a "basis" for our vector space.A Basis for Our Vector Space, and Expanding the Inner ProductI will not go into the proof here, but suffice it to say that for any finite dimensional vector space, of dimension n, we can obtain n linearly independent vectors, called basis vectors, such that any and all elements of our vector space can be written as a linear combination of these basis vectors.& That is, we can obtain a set of n linearly independent basis vectors, β1,β2,…,βn ∈ V (our vector space), such that for any and all x ∈ V, we can find (unique) scalars, a1,a2,…,an ∈ F (the Real numbers, in our case), so x = a1β1 + a2β2 +…+ anβn = Σi=1n aiβi.Therefore, for all x,y ∈ V (our vector space), we can expand the inner or dot product using the chosen basis (so x = Σi=1n&aiβi, and y = Σi=1n&biβi) in the following manner:& x·y = y·x = g(x,y) = Σi=1n&ai&g(βi,y) = Σi=1n&ai&g(y,βi) = Σi=1n&Σj=1n&ai&bj&g(βj,βi) = Σi=1n&Σj=1n&ai&bj&g(βi,βj).& Since for any choice of basis vectors, β1,β2,…,βn ∈ V, the inner or dot product of any two vectors will always be given in terms of the symmetric set of scalars, g(βi,βj) ∈ F, we may simply codify these scalars as a symmetric matrix gij = g(βi,βj) = βi·βj (a Real Gramian [or Gram] matrix).This should be recognizable as the usual, symmetric matrix (or tensor) representation of the metric.& All it took was expanding the inner or dot product in some (any) chosen basis for the vector space.Other Restrictions on the Inner or Dot Product?OK.& The inner or dot product can be characterized by a symmetric matrix (or tensor) of scalars (from the Real numbers, in our case), but is that it?& Are there no other restrictions?The more observant readers will have noticed that only the first two conditions placed upon an inner or dot product by its definition were used in the expansion, above (namely symmetry, and linearity in the first argument).& However, there is a third condition placed upon inner or dot products, namely positive-definiteness.Since the condition of positive-definiteness is expressed in terms of the inner or dot product of any and all vectors with themselves, we need to expand this requirement in terms of the chosen basis, as well.& Recall that the condition states that for all x ∈ V, x·x ≥ 0 with equality only for x&=&0.Using the expansion in terms of the chosen basis, namely x = Σi=1n&aiβi, we first see that x&=&0 implied that all the coefficients ai&=&0 (since the basis vectors β1,β2,…,βn ∈ V are required to be linearly independent).& However, expanding the first part of the condition yields:& 0 ≤ x·x = Σi=1n&Σj=1n&ai aj&gij, for all a1,a2,…,an ∈ F (the Real numbers, in our case), with equality only for all the coefficients ai&=&0.So, the positive-definiteness requirement on an inner or dot product translates directly to a similar requirement upon the metric matrix (or tensor).& We say that the symmetric matrix, gij, is required to be positive-definite (where that is defined by the condition, above).What Kinds of Symmetric Matrices are Possible?Now that we have the inner or dot product expressed in terms of a symmetric matrix (or tensor), it can be quite useful to ask:& What kinds of symmetric matrices are possible?& Are they all different, or are there similarities?& Can they be classified/categorized?Remember that our choice of basis vectors, for our vector space V, is quite arbitrary.& The one and only condition/requirement is that the basis vectors must be linearly independent (meaning that 0&=&Σi=1n&aiβi if and only if all the coefficients ai&=&0).& So we may choose any other set of basis vectors, such as β'1,β'2,…,β'n ∈ V, to investigate the relationships between two sets of basis vectors.One Basis in Terms of AnotherSince all vectors within the vector space may be written as linear combinations of any set of basis vectors, we may certainly write any set of basis vectors in terms of another set.& For instance, for all i ∈ {1,2,…,n}, β'i = Σj=1n&Λij&βj, and, similarly, βi = Σj=1n&Λ'ij&β'j., where the Λ and Λ' matrices represent their respective change of basis transformations.Applying these transformations in a cycle (from the un-primed to the primed, and back again), we see that for all i ∈ {1,2,…,n}, βi = Σj=1n&Λ'ij&β'j = Σj=1n&Λ'ij&Σk=1n&Λjk&βk = Σk=1n&(Σj=1n&Λ'ij&Λjk)&βk.& Therefore, since the basis vectors are linearly independent, we must have Σj=1n&Λ'ij&Λjk = δik (the identity matrix, with ones [1] along the major diagonal [i=j], and zeros [0] everywhere else).& So Λ and Λ' are (matrix) inverses of one another.& Therefore, the change of basis vectors, between any two sets of basis vectors, may be related by an arbitrary, invertible linear transformation.How do the Metrics Relate?Now that we know how one set of basis vectors relates to another, how will the metric in one basis relate to that within another basis?& Remember that the metric is really just the inner or dot product expressed within some chosen set of basis vectors.& So, regardless of the chosen set of basis vectors, they are still expressing the same inner or dot product.Expressing the same inner or dot product in two different sets of basis vector yields:& gij = g(βi,βj) = g(Σk=1n&Λ'ik&β'k, Σl=1n&Λ'jl&β'l) = Σk=1n&Λ'ik&Σl=1n&Λ'jl&g(β'k,β'l) = Σk=1n&Σl=1n&Λ'ik&Λ'jl&g'kl.& So, the metrics are directly related by the matrix congruence (transformation).& Expressed as matrices, we have g&=&Λ'&g'&Λ'T (where the superscript T stands for the matrix transpose).So, What are the Categories of Symmetric MatricesSo, now that we know how these symmetric matrices, that encode the metric, transform between choices in the set of basis vectors for our vector space, we can finally answer the question of what are the different categories.& In other words, what sets of symmetric matrices do not transform into one another by way of this matrix congruence (transformation)?Proposition:& If two symmetric matrices are related by a matrix congruence (transformation), then we shall consider them to be in the same category.& Corollary:& All matrices related to one another, or, equivalently, to some representative matrix, by matrix congruences (or congruence transformations), are in the same category.& So, if we can determine the complete set of such representative matrices for all such categories, then we will have a complete accounting of all such categories.As it so happens, Real symmetric matrices (as we have) have a special quality:& They can all be diagonalized by orthogonal transformations.& So, for any Real symmetric matrix, g, there exist orthogonal transformations (O, for instance, so OT&=&O-1 [the matrix inverse of O]), such that OT&g&O = λ, a diagonal matrix (so the only non-zero values are on the major diagonal of the matrix).& In fact, since changing the order of the elements along the diagonal of a diagonal matrix requires only another orthogonal transformation, and since the product of orthogonal matrices is also orthogonal, we can always arrange the order of the elements along the diagonal of λ to be any order we may desire.Now, in our case, we have all invertible, Real transformations at our disposal, not just orthogonal transformations, but they include all orthogonal transformations as well.& So, not only can we diagonalize all of the symmetric real matrices we may encode (as metrics), we can actually change the values of the diagonal elements.& At least, up to a point.For instance, if we follow the orthogonal matrix that diagonalizes our symmetric matrix, g, by a diagonal matrix, d, we can rescale the diagonal matrix so produced (let M = d&OT):& M&g&MT = d&OT&g&O&d = d&λ&d, which is also diagonal.& In fact, if we choose the elements along the diagonal matrix d to be just right (though the signs don't make any difference), the reader will notice that we may always "normalize" this diagonal matrix, d&λ&d, to have only elements in the set {-1,&0,&+1} along its diagonal (and we may always group them in any order we may choose, since that can be accomplished as a part of the orthogonal transformation O).So, as candidates for the representative matrices for all of the categories of symmetric matrices (encoding our metric), we have diagonal matrices with n-&-1 elements, n0&0 elements, and n+&+1 elements, such that n-&+&n0&+&n+ = n (the dimensionality of our vector space, of course).& Since all symmetric matrices are related to these matrices by matrix congruences (or congruence transformations) there are no other categories.& All that remains is to prove that there are no fewer categories (we can call them congruence classes), by proving that none of these representative matrices are related to any other representative matrices by any matrix congruences (or congruence transformations).& (This will be an assignment for the reader, if any so choose, but remember that the transformation matrices must be invertible [non-singular].& You may want to check into Sylvester's law of inertia.)But we were Talking About MetricsOK.& So we now have a categorization of symmetric matrices that transform via matrix congruences (or congruence transformations).& However, as noted under the heading "Other Restrictions on the Inner or Dot Product?", above, the positive-definite requirement on the inner or dot product restricts the metric matrix (or tensor) to also be positive-definite.& So, which of the congruence classes are positive-definite?Recall that the positive-definite restriction is expressible as:& 0 ≤ x·x = Σi=1n&Σj=1n&ai aj&gij, for all a1,a2,…,an ∈ F (the Real numbers, in our case), with equality only for all the coefficients ai&=&0.& We can recast this in matrix notation as 0 ≤ x&g&xT, for all x&=&(a1,a2,…,an) (a row vector), with equality only for x&=&0&=&(0,0,…,0) (a row vector of all zeros).Recall that there exists an invertible transformation, M, such that M&g&MT is a diagonal matrix, d, with only elements from the set {-1,&0,&+1} along the diagonal.& So we may rewrite the positive-definite requirement once again as 0 ≤ x&g&xT = x&M-1&M&g&MT&(M-1)T&xT = (x&M-1)&d&(x&M-1)T = x'&d&x'T, for all x', with equality only for x'&=&0.It is easy to recognize that if there are any diagonal elements of d that are zero or negative (-1), this condition will be violated.& So, the only congruence class that is positive-definite is the class congruent with the identity matrix (n- = 0 = n0, and n+ = n).& This eliminates a great many possibilities.& Even for two (2) dimensions we are eliminating the negative of the identity (the negative-definite case, which is, as far as science can determine, physically indistinguishable from the positive-definite case), the Special Relativistic like case with one +1 and one -1 on the diagonal (1+1 spacetime), along with the Newtonian/Galilean like case with one zero (0) and one +1 on the diagonal, and the negative of that (not to mention the trivial zero matrix).Furthermore, what makes anyone think that Nature (not the rag) would be bound by such an arbitrary restriction.& Wouldn't it make far more sense to determine the congruence class of the physical metric—the one that is actually used by, or stems from Nature (not the rag)—from experimental evidence, rather than simply as some einsatz imposed by Mathematicians based upon little real-world experience?Please let me know what you think, in the comments section.This can all be succinctly summarized (for those that are comfortable with the terms) by stating that the dot/inner product is a positive-definite Symmetric bilinear (Hermitian) form.Let F be a (mathematical) field (like the Real or Complex numbers).& A vector space over F (or F-vector space) consists of an abelian group (meaning the elements of the group all commute) V under addition together with an operation of scalar multiplication of each element of V by each element of F on the left, such that for all a,b ∈ F and v,w ∈ V the following conditions are satisfied:V1)& av ∈ V.V2)& a(bv) = (ab)v.V3)& (a+b)v = (av) + (bv).V4)& a(v+w) = (av) + (aw).V5)& 1v = v.The elements of V are called vectors, and the elements of F are called scalars.Articles in this series: (first article) (previous article) (this article)
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I am a Ph.D. Theoretical Physicist.&
Not that that makes me an expert, or anything.
In fact, right now, I am working in the Environmental...
Whoever did the audio for that "Metric" soundbite should do Hawking's audio.Since you are a "Ph.D. Theoretical Physicist", and you mentioned special relativity in the opening paragraph, I expected the kind of metric that is not positive definite. &As you ended your article:Wouldn't it make far more sense to determine the congruence class of the physical metric—the one that is actually used by, or stems from Nature (not the rag)—from experimental evidence, rather than simply as some einsatz&imposed by Mathematicians based upon little real-world experience?That seams to be saying something like, this positive definite metric is a math thing, but sometime later we will get to the metric that experimentalist can show Nature uses. &I would have liked this to be say the second paragraph, and go and explore from there.I did not follow the technical content so well, particularly a line of math like this:x·y&=&y·x&= g(x,y) = Σi=1n&ai&g(βi,y) = Σi=1n&ai&g(y,βi) = Σi=1n&Σj=1n&ai&bj&g(βj,βi) = Σi=1n&Σj=1n&ai&bj&g(βi,βj).Six equal signs in a row is like the third helping of turkey and mash potatoes when you decide to cut down a little on the gravy. &It can be done, but does not feel good afterwards. &Still I think I get the notation because I knew about it before reading the article.This touches on a issue that bothers me. &I am going to use a metric that is not positive definite. &The metric is also not diagonal:I have spent time worrying about the off diagonal guys. &What the guys on the diagonals do is pair up a time with a time, a space in dimension 1 with a space in dimension 1, a space in dimension 2 with a space in dimension 2, and a space in dimension 3 with a space in dimension 3, no matter what the basis is. &Kind of like one of those contra dances.The off diagonals are where time and space appear to mix things up. &That bother me from my quaternion-centric view because the product of time with space from dimension 1 actually goes and lives in the dimension 1 location. &In the quaternion product, all the results go into the time position. &So there was some intellectual tension.What I realized while writing this reply is that the metric m above is diagonalizable. &According to Mathematica, the diagonal matrix of m is:I don't yet understand how to switch between these two metrics, but I feel better about the situation. &A diagonalizable matrix that is not written as a diagonal matrix is just someone trying to make a simple statement complicated. &Yes it can be done, but it does not provide insights into the physical meaning.Glad to read you had a productive holiday, Halliday.
| 11/26/12 | 23:09 PM
LagrangiansForBreakfast (not verified) | 11/27/12 | 05:07 AM
Of course, Doug, along the lines of this article, your non-diagonal matrix (but diagonalizable, because it is Real and symmetric) is not only diagnonalizable to the diagonal matrix you expressed, but, through a congruence (transformation) is exactly transformable into the normalized, diagonal matrix with diagonal [+1, -1, -1, -1].& So it is of the same congruence class as that more usual metric of Special Relativity.While it is true that one may always choose coordinates such that the inner or dot product (metric) is diagonal and normalized (for any flat space, like the vector spaces we are working with), sometimes it is reasonable, or even more enlightening, to go ahead and use another coordinate basis.& If nothing else, this frees up one's thinking by not having to be tied to some special class of coordinates.& ;)After all, the important Physics is not contained within the coordinates, or even the transformations between some certain class of coordinates, but in the geometry that is independent of such things.DavidP.S.& I did the "Metric" "soundbite" myself, using GarageBand on my wife's iPad.& :)
| 11/27/12 | 19:20 PM
Before reading your reply, I did an experiment in Mathematica with the most general, real, symmetric metric I could imagine:From there, one asks for:
DiagonalMatrix[Eigenvalues[mg]]&The result looks like classic Mathematica-too-long-to-be-parsed, the root of a whole bunch of fourth order polynomial terms (starting with: -a b c d - c d e^2 - b d f^2...). &The matrix was diagonal. &I tossed in 10 randomly chosen numbers, and got four terms along the diagonal just to make it concrete.It sounds like there is a theorem out there that says for a real and symmetric matrix, then the matrix can be made diagonal. &Search for a few of these words, and out pops the . &I am fixating on the metric kind of tensor, and not say a rotation matrix which is not symmetric and also cannot be made diagonal. &I am thus looking at your reasonable statement with my real, symmetric matrix glasses on:While it is true that one may always choose coordinates such that the inner or dot product (metric) is diagonal and normalized (for any flat space, like the vector spaces we are working with), sometimes it is reasonable, or even more enlightening, to go ahead and use another coordinate basis.& If nothing else, this frees up one's thinking by not having to be tied to some special class of coordinates.& ;)After all, the important Physics is not contained within the coordinates, or even the transformations between some certain class of coordinates, but in the geometry that is independent of such things.Let me challenge your enlightenment claim in the following way. &Nature always figures out some kind of extremum to use consistently. &What is the minimum number of numbers one can use in a metric for 4-vectors? &The minimum number is 4. &The maximum is 10. &I am attracted by the idea of minimalism in the context of metric tensors. &It feels a little radical to be able to look at a real-valued, symmetric metric and know I can reduce it to a diagonal matrix. &The diagonal metric tensor is now more like the 4-vectors that are used to form the dot product.I thought this line of reasoning - whatever its value - would still apply to slightly curved spacetimes. &One could no longer use constants in the metric, but if the values varied, then the spacetime would be curved, and so long as those &non-constant values were real, the resulting metric for curved spacetime could be made diagonal.Anyway, fun stuff to think about. &Back to programming...Doug
| 11/27/12 | 22:13 PM
LagrangiansForBreakfast (not verified) | 11/28/12 | 01:10 AM
I like to work on the quaternion manifold, call it H1, where one only has automorphisms.& It is similar to working with complex numbers on the complex plane C1.& Then one only needs z and z* to accomplish what one can do with two reals on the R2 manifold.
The real numbers are a subgroup of quaternions.& On the manifold of Q1, they would be represented by (q+q*)/2.& The sum of a quaternion and its conjugate would have every property of real numbers, including being a member of a totally ordered set and commuting under multiplication with any other element on the manifold.& The statement holds on a manifold of Z2 or R4 using different ways of representing quaternions.& The statement holds no matter what the choice of coordinates.
I am not going to engage in a battle between tensors and quaternion products.& My future study of quaternion products will be visual and not viewable here (too much javascript).
| 11/28/12 | 23:12 PM
LagrangiansForBreakfast (not verified) | 11/30/12 | 00:12 AM
Destroying information is a bad thing. &You point it out by an example:Good old a has not changed, while b has lost enough information so it is just like a, no longer like b. &That just looks ugly no matter how far one pushes the case.A more reasonable proposal must avoid destroying information. &Just as one would do in the complex plane, there is the real and imaginary parts, which could be worked with like so:It should be clear that information has just been rearranged. &Add the first two lines together, and you get a, while adding the third and fourth lines ends up at b. &Now we can make a few comparisons:These two statements look consistent with the example you provided: the first terms are the same, while the combination of the second, third, and fourth terms are different. &If one boosts a and b, one would not expect the first terms to remain the same because a and b are different. &You did the calculation.I do confess I have trouble speaking the *morphism lingo. &What I meant was I start with a quaternion and I end with a quaternion, with all the properties that entails (closure, associativity, identity, and inverses). &While we could work with some specific coordinate system in 4D, the way I have written this reply was explicitly to avoid 4D. &A similar thing is done with complex numbers like z and z* on a complex manifold.Quaternions happen to be "messy" as you say from a geometry standpoint. &I learned separately about dot product, and cross products, and multiplying a scalar times a vector. &That's everyone's training. &With quaternions, those three separate players all have specific relationships with each other in the same darn house. &I cannot change that quality of quaternions. &I do accept it as is. &I also understand if you don't. &After all, your side won a long time ago :-)
| 12/01/12 | 00:07 AM
Doug:The "argument"/disagreement is not about the "qualit[ies] of quaternions", but about how you "play" with these qualities in your attempts to "shoehorn" them into 4D, and/or vice versa.I've tried, long ago, to direct you toward systems better suited to your needs (though, apparently, no so suited to your desires).As always, I'm ready to help you make the transition whenever you are ready.David
| 12/01/12 | 13:30 PM
Shoehorn quaternions into 4D? Hamilton wanted to chop off one term and only use three using what he termed pure-quaternions.& The four Roman soldiers to this day strike me as a natural for space-time, particularly since one of the three is different from the other, yet related.
What is a convoluted, Klein-bottle-confusing exercise is attempting to match the history of mathematical tools used by physicists for arbitrary dimensional spaces to quaternions.& I don't care about any dimensional spacetime that has more that one dimension for time and three for space because I don't think it reflects any accounting system used by Nature.& People have put in large efforts to look into the possibilities, no doubt about that.
I can find in the literature folks who say one cannot derive the Maxwell equations using real-valued quaternions.& I can find people who have written that the Lorentz group cannot be written using real-value quaternions.& It is my hobby to show those things are not correct.& I will be restrained in promoting the program since I am not a good spokesman.
And now, back to metrics and congruence classes…a story…
I Googled for "Metric Congruence Geometry".& That turned up a book, "Space, Time, and Geometry" which had a chapter titled: "Reflections on a Relational Theory of Space" by Arthur Fine.& I spent a half hour reading some of the article.& It was not possible on line to read the work as pages where not viewable in spots, but I could buy it for $179.& Ouch.& These issues started back in the days of Newton and Leibniz, so about as old school as one can get and still use the label of physics.& Newton was in the absolute space camp, while Leibniz was more relational.& About the only technical point I took away was that continuity is vital for linking congruence classes to metrics.& And this subject is tricky.
| 12/01/12 | 16:24 PM
Doug:Shoehorning?& Yes.& Absolutely.& ;)Are you not aware that every vector space has other, associated vector spaces?& (Consider the exterior algebra.)& This certainly comes into play in physics, first with the cross product (using a mapping, formed from the metric/inner-product, mapping bi-vectors back to vectors, but transforming differently, hence referred to as pseudo-vectors), then with electromagnetism (bi-vectors within 4D spacetime, which no longer map to anything else).So, an n dimensional vector space also has n&choose&0&(=&1), n&choose&2, ... , n&choose&i, ... , n&choose&n&(=&1) dimensional vector spaces associated with it.& So, in all (counting n&choose&1&=&n), there are 2n dimensions!Your beloved quaternions are isomorphic to the Clifford algebra Cl0,2(R):& The Real algebra generated by a two&(2) dimensional space with a negative-definite (pseudo-)inner product (quadratic form).That's 2D, not 4D.& The 4D only comes about because 22&=&4.& So it is a coincidence, not design.Yes, "Hamilton wanted to chop off one term and only use three using what he termed pure-quaternions."& After all, he was looking for 3D, and was trying to shoehorn what he found into that, just as you are doing with 4D.The only reason why he could have any hope of doing any such thing, with something that is more fundamentally associated with 2D, is the "happy coincidence" that the 2D bi-vector (or pseudo-scalar), in the quaternions, or Clifford algebra Cl0,2(R), behaves in an indistinguishable manner to the regular vectors.Now, the ability to project bi-vectors, in 3D, into vectors (as pseudo-vectors), and tri-vectors into pseudo-scalars, does allow one to go from 23&=&8 total dimensions down to 22&=&4.& Hence the way 3-vectors have been used within physics for oh so many centuries.However, with 4D, one can no longer project bi-vectors into vectors, and tri-vectors are no longer pseudo-scalar like.& Instead, tri-vectors may be projected to vectors (as pseudo-vectors), and quad-vectors projected to pseudo-scalars, but bi-vectors project back to bi-vectors!So, where does this leave your beloved quaternions?This is why you have to go through so many gyrations.& This is why you have a mixture of 4D and Hamiltonian 3D, especially when trying to include electromagnetism.& This is why you have so much trouble getting things to transform correctly, and why things like Maxwell's equations, and Lorentz transformations look so unnatural.Of course, this is most certainly not to say such are "impossible".& But such are a matter of shoehorning.David
| 12/10/12 | 16:59 PM
David:This sounded a dissonant chord in my ear:Your beloved quaternions are isomorphic to the Clifford algebra&Cl0,2(R):& The Real algebra generated by a&two&(2) dimensional space with a negative-definite (pseudo-)inner product (quadratic form).That's 2D, not 4D.& The 4D only comes about because 22&=&4.& So it is a coincidence, not design.!!! my experience programming quaternions. &When I use "D", I think about degrees of freedom. In C, Java, Javascript and Mathematica, I always needed 4 variables.The real numbers are Cl0,0(R). &The complex numbers are&Cl0,1(R). &As you point out, the quaternions are&Cl0,2(R).&&The bi-quaternions are&Cl0,2(C). &The Clifford algebra world is&Clp,q(R).&&One is free to study whatever is convenient for the problem at hand.I don't think Nature has such a choice. &I look to experiments to see what numbers really get used. &The real numbers trump all others. &There is lots of counting that gets done in Nature (the reals). &The second most important numbers are the complex numbers whose properties are at the heart of quantum mechanics. &Both of these are division algebras. &With my physics glasses on, that means that processes that involve addition might be reversed via subtraction, and those that involve multiplication can be reverse via division. &I did fight to add that property to the hypercomplex numbers, but had to concede the point. &As you well know, quaternions coincidentally have division. &Yet quaternions remain stunted. &Real analysis is a rich subject. &Complex analysis is even richer. &Quaternion analysis is stillborn. &Follow the path devised by Fueter, and one cannot show q2 is analytic in q.I do have trouble tracking discussions of bi-vectors, tri-vectors and pseudo-vectors. &I can see why people call this baby a bastard:To be painfully honest, this says that one should not think separately about:This is all part of a matching set. &In the quaternion city of Gomorrah, axial vectors do add with polar vectors. &One must not think about the dot product and its oh-so-negative sign without also considering both the simple first term product and the cross product. &The tradition has been to divide into little pieces, but sometimes a puzzle only works when everything is connected to each other.Doug
| 12/11/12 | 14:20 PM
Doug:Perhaps a more realistic reason why "Real analysis is a rich subject", and "Complex analysis is even richer", but "Quaternion analysis is stillborn", or "remain[s] stunted", may have to do with the first two being mathematical fields, while the latter is not.Myself, I might try to explore larger mathematical fields.Of course, there are those that think that geometry should be non-commuting, so, maybe not.However, I did already try to point out where the four "degrees of freedom" (actual vector space dimensions) come from:& 2&choose&0&(=&1) grade zero (scalar) dimensions + 2&choose&1&(=&2) grade one (vector) dimensions + 2&choose&2&(=&1) grade two (bi-vector, or pseudo-scalar) dimensions = 4 total dimensions.& (The Complex numbers, as Cl0,1(R), are an even simpler graded algebra, but almost too simple.)David
| 12/11/12 | 18:37 PM
I would appeal to group theory for hope. &The complex numbers contain real numbers as a subgroup. &All of real analysis is a special case of complex analysis where the imaginary number is always zero. &The same logic applies to quaternions which have real and complex numbers as subgroups. &For quaternions that point in one direction, calculus needs to behave like complex analysis since it is the same thing. &Note that "the same direction" does not have to be a good old straight line since a different set of coordinates could look quite different.
| 12/11/12 | 21:41 PM
Doug:Now you're thinking like a Mathematician, rather than a Physicist.& ;)Do you think you are up to that?You see, a Physicist tends to think in terms of the physical situation(s) and tries to find/develop (Mathematical) tools to most closely "model" the physical situation(s).& A Mathematician, on the other hand, tends to think of it all as various "games", and asks things like "what kind of game do I have if I change the rules?", or "to what extent can I change the rules but still have the same game?".The latter Mathematician sort of thinking seems more akin to what you have expressed, above.Of course, there is another consideration that Mathematicians tend to use:& How can I get the most "bang for my buck" when playing such "games"?You see, having Real analysis, and then Complex analysis, the "natural" path may not be "quaternion analysis", or even some other form of "analysis" using some larger (mathematical) field.& No, the "natural", and most "efficient" (as in "bang for my buck") may be a form of "analysis" that covers a whole class of "numerical" systems that include the Real and Complex numbers as special cases.& One such (infinite) class that may seem "juicy" enough, to Mathematicians, could be the Clifford Algebras.The nice thing, for you, of course, is that the quaternions are included in this class, along with many other algebras.So, you may find it useful to see what has been done in the area of Clifford "analysis".Just a suggestion.& After all, why reinvent the wheel, if you can find that others have already addressed something that includes what you are interested in.& Besides, you may learn a thing or two beyond what you were expecting.David
| 12/15/12 | 14:10 PM
This is why you have to go through so many gyrations. &This is why you have a mixture of 4D and Hamiltonian 3D, especially when trying to include electromagnetism. &This is why you have so much trouble getting things to transform correctly, and why things like Maxwell's equations, and Lorentz transformations look so unnatural.Let's do a friendly shop and compare on one specific technical issue: the teaching cost to get to the Lagrange density needed to derive the Maxwell source equations. &The starting point is the field strength tensor which needs to be contracted:Don't quote me on the fractions, but I hope they are right. &To teach how this equation works, one needs to explain covariant and contra-variant vectors, a challenge in itself. &Then one has to explain why the derivative has a sign flip. &That can be done with enough time. &Then with those notions in hand, one needs to construct this beast. &I say beast because i have done it by hand and it is darn confusing keeping track of which one has the plus sign at this moment and position to drop into the field strength tensor. &Once one has both the field strength tensors written out, then one needs to contract these. &The results need to then be collected into a concise form. &Not a trivial list of tasks.I am confident I could teach any motivated high school student how to do this if they had learned about the cross product. &Just like my days in high school, I'd start at the back of the book and ask them how to factor the result:Figure out how to write B and E two different ways, and one is cooking with gas. &First thing is to write out one of the pairs. &Take a complete view of all possible changes of a 4-potential:Oops, the term is too simple, it has too much. &Do a subtraction:If the order of the operations is changed, then the only term that flips signs is the one with the cross product:We have one difference, and one negative sum. &It terms out that negative sum is right because the imaginaries toss in a minus sign along the way to the Lorentz invariant first term:Bingo, bingo. &I even got the bonus of the Poynting vector. &I found out recently that one of the crowning achievements of one of the most successful quaternion haters of all time, Heavisides, was a derivation of the Poynting-who-published-first vector.I want to be clear that I am not calling anything in the quaternion derivation anything but a quaternion. &Counting up the quaternion side, there are 5 lines of math. &I am not able to count the work required for the tensor approach. &It takes much effort to convince people that the covariant derivative must have a different sign than the covariant vector :-) &I think the tensor approach starts off scary (look, Greek letters, and then the order of the Greek letters reversed). &The quaternion approach starts with one of the first expressions I ever learned how to factor, the difference of two squares. &I am motivated to figure out all four ways each part of a four potential can change. &To my eyes, I am looking at the forest, not worrying about signs of contravariant vectors.Then there is my Evil Gyration(TM), tossing the first terms of the quaternion derivative of a quaternion potential under the bus. &For EM, that term is not physical. &I thought that was standard operating procedure, some stuff is of this world in this situation, and other stuff is not. &There is a subtraction in the standard tensor approach. &That subtraction gets rid of the same terms in the quaternion approach. &The same sin is done in both methods.I appreciate that those who have invested in learning about vector spaces and dual vector spaces have made their vows to tensors. &We can expect a "no, no, no" from someone who understands the meaning of vector spaces and their duals and their metrics. &What I can promise as a counter offer is that I don't understand quaternions. &There just are not enough tools yet to claim an appreciation for them at this time. &I can still work with them and will continue to do so.
| 12/11/12 | 15:40 PM
Well, Doug, it wasn't my intention to put you so much on the defensive.However, I'll see your quaternion derivation, and raise you a more truly tensorial derivation.We begin with the derivation of the Faraday two-form F = dA (A is the one-form expression of the usual four vector potential, and d is the exterior derivative).& Then the inner product of this two form with itself yields F·F = 2(B2 - E2/c2).& (Yes, the inner product uses the metric, just as in the article.)As it turns out, all Yang-Mills theories have very much this same form, regardless of the symmetry group.Only General Relativity (gravity) doesn't have this form, since, even though it has something quite analogous to the Faraday two-form (the Riemann curvature tensor), it has an invariant that can be formed linearly with it, rather than the quadratic form required for all Yang-Mills theories.& (Generally, linear is considered to be simpler than quadratic, so it seems rather striking that General Relativity works so incredibly well even though it is formed of the simplest possible invariant of the Faraday-like two form.)David
| 12/11/12 | 18:11 PM
If I recall correctly, the exterior derivative does not depend on the metric in any way. &It is the contraction of the field strength tensor where the metric does its job.I saw a covariant derivative for the standard model. &Each of the three forces - EM, the weak, and the strong force - came with its own 4 potential and gauge group (U(1), SU(2), and SU(3) respectively). &What struck me as less than elegant where the internal symmetry indexes needed to link the differential forms to the group symmetry generators. &What is your perspective on the needed link between the potential (pretty much the same) and these particular gauge groups (distinct)?
| 12/11/12 | 21:32 PM
Doug:You are correct that "the exterior derivative does not depend [explicitly] on the metric in any way."& However, the exterior derivative depends exactly upon the very same derivative that gives rise to the "co-variant" and "contra-variant" derivatives.& In fact, while the derivative does naturally give rise to the "connection coefficients", these connection coefficients need not depend upon any metric, and, a fortiori, need not be fully determined by any metric (and its derivatives) even when they are "required" to be "compatible" with the metric.Now, as for the way the "covariant" derivative for the Standard Model is dealt with:& I, likewise, am less than impressed when the various gauge groups (U(1), SU(2), and SU(3)) are treated as fully independent.& (Of course, the first two [U(1) and SU(2)] are treated as more fully coupled within electroweak theory.)Part of what I think is often missing is a determination as to whether one should actually be dealing with representations of a (semi-simple) Lie group (such as the product group U(1)×SU(2)×SU(3)), vs. dealing with a product of representations of the various (simple) Lie groups (U(1), SU(2), and SU(3), separately).There seems to be a lack of rigor, in this regard.I think a great deal of the problem is that there are many instances where the two approaches are equivalent.& However, there are other instances where the representations are quite different!As for the "group symmetry generators" (and the indexes used in "coupling" them to the "potential" functions), this can be seen as merely another case of expanding vector spaces in terms of a set of basis vectors.& (It's just that these basis vectors also have further algebraic properties.)& Of course, if people would more commonly make the expansion in terms of the (Tangent) vector space basis more explicit, the expansion in terms of the basis vectors (generators) of the algebras wouldn't appear to be so different.Anyway, so it goes.& ;)David
| 12/15/12 | 15:56 PM
Doug:The "lingo" means something.& So one should be careful in how one uses such, and, even more-so, in how one "violates" such (like allowing inner or dot products, or metrics to be anything besides positive definite).For your information, what you had was an endomorphism.It may serve you well to read up on the Wikipedia article on .& It even talks somewhat about the relationships between various "*morphism lingo."David
| 12/01/12 | 14:29 PM
Indeed:An&&endomorphism of&X&is called an&.The operation (q+q*)/2 is not invertible as it tosses way information. &
| 12/01/12 | 15:04 PM
LagrangiansForBreakfast (not verified) | 12/01/12 | 18:11 PM
It sounds like you are distorting my statements just a little (and a little can make a big difference):Yet in your (q+q*)/2 example, you were explicitly trying to get something other than a quaternion. Talking about (q-q*)/2, or calling it something other than an automorphism, is forgetting what the original issue was.My errant statement which apparently I cannot correct from above [I will do so here]:I like to work on the quaternion manifold, call it H1, where one only has [endomorphisms].To work on that manifold means to work with endomorphisms that map quaternions to quaternions, wether that map is reversible or not. Both (q+q*)/2 and (q-q*)/2 are quaternions. &I can see absolutely nothing in either operation that would make them no longer be a quaternion.What may be going on is a presumption of the role a quaternion should have, namely that it should not commute as all were taught when first exposed to quaternions. &Let's look at what can be done with those quaternions and any other randomly chosen quaternion, call it R:I have seen in so many places people complain about the indisputable fact that quaternions do not commute with each other. &And that is where they stop. &I prefer to divide and explore. &These statements in no way depend on the dimension of the manifold. &They will be true no matter what the choice of coordinates happens to be.I have this odd fantasy that someone who is really great at math starts to work with q, q*, (iqi)*, and (jqj)* on H1 and starts doing all kinds of quaternion variations on the deeply established subject of complex analysis. &I need to work on my fantasy life.Concerning metrics.To make it clear: A four-vector is a different "beast" than a metric. The metric is indeed a "bigger beast" than a four-vector.It certainly is. &A four-vector stores data. &A metric is about how geometry connects two pieces of data. &The metric takes two four vectors and returns one value. &That is how tensor algebra has been constructed.Any alternative has little chance of going through all necessary eyes of needles. &It has happened a few times in physics where things that were different were linked in an odd way. &People did think electricity and magnetic phenomena should be linked, but it took Olmsted to find the odd 90 degree aspect of their relationship. &I do dream of a day when I have an app that looks like a planet going around the Sun. &There are such apps already, but behind the code is the group Q8 and rules of multiplication just dynamic enough to accomplish something similar but not identical to our current metric theory for gravity. &Will have to work and see if it works out.
| 12/05/12 | 23:20 PM
LagrangiansForBreakfast:You need to be careful with assertions like... it doesn't make sense. It's the same with adding a vector to a tensor, or subtracting a scalar from a vector.While this is completely correct within the context of working with vector spaces in the "usual" way, it is quite incorrect within the context of graded algebras, like Clifford or exterior algebras.So you need to be careful.David
| 11/29/12 | 07:39 AM
LagrangiansForBreakfast (not verified) | 11/30/12 | 02:50 AM
LagrangiansForBreakfast:I understand what you are saying with the example of "new math" and such (though I do remember being taught the "new math", and I don't feel like I had all that much "trouble" with it).& I'm sorry if you feel that my "corrective" ("be careful") comment did you and Doug a disservice.& That certainly wasn't my intention.On the other hand, I have found that addressing concepts in such a way as to suggest that something is "always and forever one way only" does significant long term harm:& always to the student (in at least one of two ways), but often to the "teacher" as well.The principle problem is that any sufficiently good student is almost certain to "go beyond" the limits of applicability of any such direction.& The simplest example of which I had direct experience was the "fictitious forces" of Newtonian non-inertial "forces", another (along the lines of this article) is the positive-definiteness "requirement" imposed upon inner or dot products, and, by extension (as seen in this article), imposed upon metrics.When this happens, the student either has a far more difficult time getting out of their old thinking (a "rut"), or they determine that their "teacher" "lied" to them, so they are far less likely to trust what that "teacher" says, and may even have increased difficulty trusting other "teachers".The "teacher", of course, in the second case, looses credibility.So, what I was trying to help you with, with my "be careful" comment, was to help you to avoid such a trap.It's not that you need to actually present all the alternate possibilities to the student up front.& No, the "lesson" is to simply avoid "always and forever one way only" type language.& Instead, one may simply say things like "within the present context", or other similar provisos.In this way, you will not be found to be "hiding" anything "under the rug", even if you, yourself, are unaware of any such alternatives (within Mathematics, you can just about guarantee that alternatives exist for just about any "rule"), and the student will be more prepared for many such alternatives if and when they run into them.So, like I said, just be careful.David
| 12/01/12 | 13:37 PM
LagrangiansForBreakfast (not verified) | 12/01/12 | 16:26 PM
LagrangiansForBreakfast:From your postscript:I'm not sure what you meant by "einsatz". Do you mean "ansatz"? ...I was looking for the spelling of the word I have heard, and used (orally), in Physics and Mathematics.& I was uncertain of its spelling, and while looking for the right word I hit upon the German word "einsatz" that refers to "the beginning of a musical note", and other things like a stake (as in a wager), bit, input, sortie, insert, onset, etc.& So it looked close to what I was shooting for.However, ansatz looks to be what I was looking for (I thought I had tried that spelling, but I suppose not).& It refers to an approach (such as to a topic or subject), attempt, estimate, approximation, objective, beginning, formation, etc.However, rather than "a mathematical guess that will be checked for correctness later", such as the followingIn physics and mathematics, an ansatz is an educated guess that is verified later by its results. An ansatz is the establishment of the starting equation(s), theorem(s) or value(s) describing a mathematical or physical problem or solution. It can take into consideration boundary conditions. After an ansatz has been established (constituting nothing more than an assumption), the equations are solved for the general function of interest (constituting a confirmation of the assumption).I'm really shooting for something more akin to an unjustified assumption imposed upon a topic or subject. In the case of the positive-definiteness "requirement" placed upon inner or dot products (and, by extension, the metric), it is more of a failed ansatz, since experiments violate the positive-definiteness "educated guess".& (It was a perfectly good "educated guess" so long as we were only working with things in our everyday, usual "space", with everything measured at rest with everything else, or all measurements being done simultaneously, without any temporal aspects whatsoever.)Anyway, "ansatz" will do, I suppose.David
| 12/10/12 | 18:43 PM
LagrangiansForBreakfast:Actually, I am quite pragmatic, on this subject.& It's just that my pragmatism differs from yours (undoubtedly colored by different experiences).My pragmatic approach is summed up in the following statement, taken from my previous message:It's not that you need to actually present all the alternate possibilities to the student up front.& No, the "lesson" is to simply avoid "always and forever one way only" type language.& ...So, to use your "teaching addition to a young nephew" example, this pragmatism simply means that one may state that "we can [or may] write 2+2=4", even without the "in this [or the current] context" proviso.& What one may not say is something to the effect that one must "write 2+2=4", without regard for context.You see, quite pragmatic.DavidP.S.& It's unfortunate that you are not a registered user of this site, or, at least, have not chosen to use such an identity.& For, if you had written under the identity of a registered user of this site, I would have had a number of far more private means, at my disposal, with which to convey my "be careful" message, to you.
| 12/11/12 | 12:48 PM
Let me second the P.S. &You do devote much effort to your replies, and thank you for that (even if we disagree on style and substance, making a stand is far better than silence). &I recall an exchange we had about the conventional understanding of gravitomagnetism. &I was willing to forward correspondence I had with someone who wrote a review article on the subject. &I was not to support the position by a private communication which I think would have been OK with that professor. &As a result, to this day I think you still think I am horribly wrong on that particular subject. &If I cannot share the data I have in my position, I cannot care about your opinion on the subject.
| 12/11/12 | 13:01 PM
LagrangiansForBreakfast:There's another very pragmatic reason why I gave you the "be careful" message/warning, and, as far as I can see, you haven't picked up on it, yet.& You see, when you made your statement about how "adding a vector to a tensor, or subtracting a scalar from a vector" just "doesn't make sense", my warning wasn't just about some esoteric area of Mathematics (graded algebras), but was prompted by a very real, imminent issue closely related to the context in which your comment was made.As I'm sure you are aware, almost any discussion with Doug has the quaternion context at least lurking in the doorway.& In fact, within the very same post where you made the comment that prompted my "be careful" message, you, yourself, actually, explicitly brought in the quaternion context.You see, because the quaternions are isomorphic to the Clifford algebra Cl0,2(R), the quaternions, themselves, admit a graded structure.& So, even Doug's lowly quaternions can be seen to "make sense" of what you claimed "doesn't make sense."Do you now see how pragmatically relevant my "be careful" message was?David
| 12/11/12 | 16:54 PM
Doug:You are correct that you have found a "".& In fact, the "" section is quite clear that "Real
are diagonalizable by ".You "challenge" me with:Let me challenge your enlightenment claim in the following way. &Nature always figures out some kind of extremum to use consistently. &What is the minimum number of numbers one can use in a metric for 4-vectors? &The minimum number is 4. &The maximum is 10. &I am attracted by the idea of minimalism in the context of metric tensors. &It feels a little radical to be able to look at a real-valued, symmetric metric and know I can reduce it to a diagonal matrix.& ...Well, actually, even 4 (Real) numbers is an infinite amount of information (think how many bits it takes to represent an arbitrary Real number).& The actual minimum amount of data, for four dimensional metrics, is an integer from 1 to 15 (or, for physically distinguishable metrics, 1 to 9).& In other words, there are only 15 different congruence classes of four (4) dimensional metrics, or only nine (9) physically distinguishable metrics.& (That is, of course, once we "loose" the non-physical positive-definiteness requirement the Mathematicians like to impose, which picks out one and only one of these possibilities.)However, each congruence class involves an infinite set of metrics (well, other than the congruence class of the zero matrix, which contains only a single element), including both an infinite set of diagonal ones, as well as an infinite number of non-diagonal ones.& However, since they all belong to the same congruence class, they are all equivalent.& In other words, even though it takes an infinite amount of information to specify any one of these, such "information" adds nothing new to the geometry:& They are still representations of the same inner or dot product (as a geometric object).Does this help?& If nothing else, it should provide plenty to think about.David
| 11/28/12 | 19:56 PM
Let me see if I can read any tea leaves. &What is the difference between these 9 physically distinct metrics? &A clue is found here:That is, of course, once we "loose" the non-physical positive-definiteness requirement the Mathematicians like to impose, which picks out one and only one of these possibilities.The positive-definite metric has a signature of +4. &The congruence class that has the Minkowski metric for a member has a signature of -2, unless you are one of those people who prefer the other sign convention, which would make it +2. &Add up the integers between -4 and +4, and even young Euler could see there were nine integers. &So I am speculating that the 9 physically distinct metrics have .As should be obvious, my favorite congruence class has a signature of -2. &I am familiar with the mountains of data that show how this congruence class is used. &Since I don't mind asking nieve questions, are there other congruence classes that are similarly supported by mountains of data?, or in this case, infinite, infinite, infinite. &There is a never ending sea of ways to get to the proverbial geometric object known as the inner product. &I was not thinking of information bits. &Instead I was thinking more along the lines of degrees of freedom. &While I could write a signature -2 metric program to use as many as 10 variables, 4 would be enough. &Note, I am not saying that 4 is better - both the 4 and 10 program are in the same congruence class and will be able to represent the dot product. &The qualitative difference is that the diagonal metric has the same number of numbers (or functions) as the 4-vectors that is combines with to form the geometric object called the inner product. &A similarity like that is one thing I think about since the metric was taught to me a bigger beast than 4-vector.
| 11/28/12 | 22:36 PM
Doug:Unfortunately, you appear to have quite missed the point.However, rather than belaboring the point here, I'll just address some of your misconceptions within your "signature" statements.& (I recommend you take a closer, or more thorough look at the Wikipedia article on .)Your use of a single integer to express "signature" has two (2) flaws:& 1) it can only be used for those subsets of metrics that are non-singular, namely, where n0&=&0 (remember what n0 means); and 2) not all integers between -n and +n (-4 and +4, in your case) are allowed signatures, there is always an unusable value between any two usable values (so, instead of 9 available values between -4 and 4, there are but 5).For n = 4, there are 15 possible congruence classes for the metric (I don't wish to enumerate them here).& These include your "+4", "-2", "+2", and "-4", but also 0 (the case with 2&+1s, and 2&-1s), and a whole host of singular cases where n0&&&0.Now, when I talk of "physically distinguishable metrics", please do not confuse this with physically "viable", as in potential matches for the actual, physical metric of Nature (not the rag).& No, what I am referring to is the apparent fact that if two metrics, from different congruence classes, are the negatives of one another (so -1 times one yields the other), then these two metrics are physically indistinguishable:& There is no physical experiment we can perform that will distinguish between the two possibilities.So, of the 15 possible congruence classes, any two that are related by exchanging n+ and n- (remember what they are) are physically indistinguishable.& This leaves only 9 physically distinguishable congruence classes.In the general case of an n dimensional vector space, the total number of congruence classes is (n+2)(n+1)/2, while the number of physically distinguishable congruence classes is [(n+2)(n+1)/2 + floor((n+2)/2)]/2.& (Note:& floor(x) is the largest integer less than or equal to x.)I hope this helps.David[Edit:& A few ns were not properly italicized.]
| 12/01/12 | 13:14 PM
I never was good at reading tea leaves.In my limited exposure, I had stuck to good old metrics with a signature of -2, which can also be written as (1, 3) [for one positive eigenvalue, three negative eigenvalues, and implicitly zero eigenvalues equal to zero], which can also be written as (1, 3, 0).A pair of physically indistinguishable metrics would be (1, 3, 0) and (3, 1, 0).I have never considered a metric where the last digit was anything greater than zero. &It seams destructive to work with a metric like (1, 1, 2) or (1, 0, 3). &I did pick those two out because they remind me of complex and real numbers respectively (I hear a tree branch breaking).Doug
| 11/30/12 | 23:23 PM
Doug:You are absolutely correct that "[a] pair of physically indistinguishable metrics would be (1, 3, 0) and (3, 1, 0)."& This is also why one finds both "conventions" used within the Physics literature.I understand "never [having] considered a metric where the last digit was anything greater than zero" (that is, a singular, non-invertible metric).& However, if one is to either cover all possibilities, or, perhaps more importantly, to be able to address Newtonian/Galilean relativity (an area so many, out "there" [especially those who are so adamantly opposed to Special Relativity], are still "stuck" in), one must include such "strange" cases.It is true that the singular, non-invertible cases are of "order zero" within the space of all possible metrics, and it is true that they can be extraordinarily difficult to work with.& However, they form the "border" cases between all other possibilities.David[corrected a typo]
| 01/23/13 | 14:31 PM
Readers:Does anyone understand what is being represented within the following diagram?Do you all simply see a "light cone" from Special Relativity (in one dimension of time, and two dimensions of space), here?Hint:& This is not a "light cone".Also, notice the red, green, and blue "dots" within this diagram.& What do you suppose they represent?Hint:& These "dots" (representing single points within the diagram, but enlarged so you can actually see them) are directly related to the representatives of congruence classes talked about within the article.& (These are only three of the six [6] total, or four [4] physically distinguishable, for two [2] dimensional vector spaces.)David
| 12/01/12 | 14:03 PM
Does anyone understand what is being represented within the following diagram?David, is&this just a 4 dimensional &manifold representation of spacetime?Also, notice the red, green, and blue "dots" within this diagram.& What do you suppose they represent?I can see a red dot and a blue dot but no green dot, can you describe where it is on the diagram please? Is the red dot an event in future spacetime that can potentially be observed or reached by an observer and the blue dot is an event in 4D spacetime that can never be observed or reached by the observer in their spacetime? Sorry if I've gone completely down the wrong track but I have just reread all your blogs and it seems like this diagram is just another manifold, like a very complicated signpost and that the blue dot would be the equivalent of a sign pointing to the centre of the earth for this woman. I might have completely lost the plot, if I have then I can reread it all again and next time I won't skim over the formulae, just point me in the right direction please :)Also I found this diagram below&and thought it might also be useful for explaining the coloured dots to a laywoman like me?&The Wiki article says that :-Mathematical spaces often form a hierarchy, i.e., one space may inherit all the characteristics of a parent space. For instance, all&&are also&, because the inner product&induces&a&&on the inner product space
My 5 min film 'Hidden Dangers for ALS' entry in the AAN #2015Neurofilm Festival is listed no. 21 of 65 entries at
| 12/01/12 | 17:05 PM
Helen:Thank you, for the Venn diagram of Mathematical spaces.& Yes, all Inner product spaces are Normed vector spaces, are Metric spaces, are Topological spaces.However, one must be careful to recognize that the use of the term "metric" in "Metric spaces" means something different than the use of the term "metric" within Einstein's Relativity theories, and within the context of this series of articles.& Within these contexts, the term concerns some object that permits one to determine lengths and angles, just as an inner product can.& On the other hand, the use of the term "metric" within the context of "Metric spaces" is a far more general concept that simply means there is some "measure" of "distance"/"closeness" that need not even have all the characteristics of a Norm, let alone an Inner product.& (Hence its relationship to such other spaces.)As for the rest of your post, I'll have to get back to it later.& (Hopefully, later today.)David
| 12/08/12 | 09:19 AM
Helen:Unfortunately, despite my hint (practically a warning)Hint:& This is not a "light cone".you appear to have been lead astray by appearances.& As I expected of many readers (at least prior to my message)Do you all simply see a "light cone" from Special Relativity (in one dimension of time, and two dimensions of space), here?Hint:& This is not a "light cone".Well, I'll give you all another big* hint:& Hover your cursor over the image (whether in the article, or within my message, above) to see what the image "title" is.Now, everyone, just parroting back the image "title" doesn't count.& You must go beyond the image "title" to show that you understand what it represents.David*& At least I think it's a big hint.
| 12/09/12 | 00:36 AM
Helen:You asked:I can see a red dot and a blue dot but no green dot, can you describe where it is on the diagram please? ...Well, the green dot is exactly between the red dot and the blue dot.& It is a bit difficult to see, I suppose, because it actually lies on the (upper) green cone shape.The red dot is completely inside the green cone shape, while the blue dot is quite obviously completely outside of the green cone shape.All three dots are on the plane defined by c'&=&0.& In fact, all three dots are on the line defined by the intersection of that plane (c'&=&0) with the plane defined by b'&=&1.& So the red dot is at a'&=&1, b'&=&1, and c'&=&0.& The green dot is at a'&=&0, b'&=&1, and c'&=&0.& While the blue dot is at a'&=&-1, b'&=&1, and c'&=&0.This knowledge, along with a knowledge of what the entire diagram represents, should make it quite apparent what these three points represent.David
| 12/09/12 | 00:55 AM
Oh well at least I understand where the green dot is now and that the diagram is called 'All possible 2D symmetric matrices'! I also looked up Sylvester's law of inertia&which&is a&&in&&about certain properties of the&&of a&&&that remain&&under a change of&.'&'Namely, if&A&is the&&that defines the quadratic form, and&S&is any invertible