In Response to A Post on phys.org

So, this blog posting is not for everyone.  I wrote it in response to a comment made here by bschott. Initially, I was just going to post a comment, but it rapidly became too large to be a comment.  Thus, in the interest of not spamming phys.org, I will post it here, and link to it with a comment.  Here it is:

For some context, here is a posting by bschott elsewhere on the site:

Time dilation is a human perceptual artifact, not a physical reality. Assume the entire universe is your reference frame and you can watch anything happening in it at any time. The amount of "time" a series of events requires to occur, if they are the exact same events, will be the same no matter what your location or velocity is in the universe. Dilation only occurs between separate IRF's when comparing events in one to events in another. The effect disappears if you view everything simultaneously as opposed to each group in each RF measuring the ticking of a stopwatch and comparing it only to the other group.

Now, admittedly, there are far crazier things on the site, but it is certainly a demonstration of this poster's failure to understand fundamental concepts of Special Relativity.  In particular, saying "the entire universe is your reference frame" is meaningless.  A reference frame is a coordinate system.  The universe is not a coordinate system.

I mean, what does he mean?  Does he mean the center of gravity of the entire universe?  Well, we don't know that one exists, because we don't know how much universe there is.  Does he mean there is some preferred frame of reference in which everything "plays out"?  In that case, he's begging the question.  He would have to demonstrate that such a preferred frame exists in the first place, but we have no evidence of this.

Another problem is his assumption that even if you could find this frame of reference, causality implies that time ticks the same way everywhere in the universe, but this is just not correct.  First, in terms of SR, not every event is causally related.  Events outside the light cone are not causally related.  That means, depending on how you are moving such events can happen in different order. Furthermore, even for causally related events, time necessarily does not tick in the same way.  The whole point of General Relativity is it does away with the notion that there can be a universal rate of time in the presence of gravity.  Indeed, there are only very specific instances where this can be true: static space-time.  But I digress.  This guy doesn't know what he's talking about, and yet, saw fit to say the following:

Funny, I became a poster at this site due to the amount of crackpottery that is entrenched in mainstream theoretical physics and how militant it's supporters are here.

Concerning Non-Mainstream Theories

Now, there are three main kinds of non-mainstream physics, which are not mutually exclusive:

1. a rigorously sound theory, at least based off of acceptable first principles, which can be falsified, or at least seems likely to be falsifiable in a few years
2. a rigorously sound theory, which has been ruled out by conclusively by experiments
3. a poorly defined theory that is not even wrong.  Either there is little or no evidence for it, it is based off of a faulty misunderstanding of establish physics principles, or it is simply too vague to be falsified.

An example of 1) would be Brans-Dicke theory
An example of 2) would be aether theory
An example of 3) would be intelligent design, and most personal hypotheses peddled by people on phys.org.

You will never find me vehemently denying a theory of type 1).  You might find me clarifying that it hasn't yet been demonstrated conclusively, but, if someone wants to provide a cogent, well research treatise on why they think Brans-Dicke theory explains a particular effect, go right ahead.

When it comes to 2 and 3, however, all bets are off.  At least 2, though, has the benefit of being formally defined, but we know better, and that means you need to get with the program.  It has nothing to do with militarism, and everything to do with not beating a stinking equestrian corpse splayed out on the ground with its tongue hanging out of its mouth.

...and if your theory is 3, well, what can I say?  I work on personal hypotheses all the time, and contrary to what posters of phys.org think, I certainly indulge in bucking the trend.  Indeed, I have even indulged in 2, when I have found pedagogical explanations unsatisfactory.  The difference is that, most of these hypotheses don't ever see the light of day because formalizing them usually reveals a fatal flaw in my thinking.  Even then, I would be very careful with revealing my hypothesis until I can find evidence for it that can't be attributed to something else, and this is something crackpots don't do.

Concerning Crackpot Posters

There are generally four types of posters I engage with on phys.org.

1. is a poster who demonstrates general knowledge of science and peer review, and who tend to toe the line of mainstream science.  This poster typically provides well researched arguments whose pedigree can be easily traced to their original peer-reviewed source.  Posters like Thermodynamics, and Stumpy fall into these category.
2. is a poster who demonstrates knowledge of science and peer review, but who does not toe the line of mainstream science.  In general, his/her arguments are also well researched.  I consider Noumenon to be an example of such a poster.
3. is a poster who doesn't know much, but asks questions, and might submit personal hypotheses, but acknowledges that his/her hypotheses are possibly wrong.  This poster learns from mistakes, and sometimes can even offer insightful comments.
4. is a poster who doesn't know much, and does not demonstrate competent understanding of any particular field, but regardless of that, insists that his personal hypothesis should magically trump all of science, evidence, and cogent arguments be damned.  They are resistant to learning, and react to any criticism with hostility.

These categories are not mutually exclusive either.

I have always been very respectful of any poster who falls into categories 1 to 3.  While they might not actually accept mainstream science, they all have one thing in common: they can be reasoned with.

OTOH, people who belong to 4 are impossible to reason with, inconsistent in their arguments, and never, EVER, willing to acknowledge that another argument has even been made, let alone how correct it is.  I liken arguing with them to catching fish with your bare hands.  You can never pin them down.  Their method of reasoning is so slippery and incoherent, it is impossible to get anywhere with them.  Show them evidence, and it is never evidence.  Provide them with a theoretical argument, and it's "just math."  Every argument you make with them always brings you back to square 1.  Ironically, these same people are also extremely predictable.  They invariably accuse mainstream science of rampant dogmatism, but are too incompetent to even recognize that the only reason for their own beliefs is dogmatism.

This brings us to people like bschott and Benni.  You are of type 4.  Hence, you believe that people who correct you only do so because they are militant, just as you are.  It doesn't matter that you can't tell me what Ricci curvature is, or what it means for gravity to be locally equivalent to acceleration.  It doesn't matter how many experiments confirm gravitational time dilation.  It doesn't matter, because you already decided, a-priori, that your vague understanding of SR and GR is incompatible with your world view.

You are the dangerous ones, because you act like authorities on subjects you don't understand, and your pseudoscientific prattling is just more evidence to sane people who fall into category 3) that scientists have their heads in the clouds.  So, if we seem militant to you, remember, it isn't because we are dogmatic.  It is because you are.

More Garbage

In response to my quote,

if it were up to me, I'd get rid of the entire voting system, but then again, if it were up to me, I would implement a system closer to Stack Exchange's, where you aren't allowed to post freely until you've demonstrated some kind of competence.

bschott said

LMAO...holy shit on a stick! Someone who is trained in mainstream physics is therefore competent...and is also(this is the best part) part of a group of people who math has convinced that 4/5 of the matter in the universe exists outside the EM spectrum. 

Now, I hate the term straw-man, mostly because it gets bandied about by people who don't understand what the term means (hint: it doesn't mean paraphrasing what you actually said or believe), but this is a bonafide straw-man.  We are not convinced by math.  We are convinced by demonstrated theory and evidence.  That means we model our conclusions on two things:

1. Is there empirical evidence that a claim is true or false?
2. In lieu of evidence, is there theoretical justification that a claim is true or false grounded in firmly established principles?

So, math has not convinced us.  Evidence, and theoretical arguments grounded in evidence have convinced us.

One thing I love, in particular, is when crackpots make assumptions about my dogmatism.  For example, they assume that I hold String Theory to same level of esteem that I do, say General Relativity.  Of course, nothing could be further from the truth.  String theory, at this point, is not even wrong in a variety of ways.  To me, it does not qualify for 1 or 2 because it is not grounded in firmly established principles, and certainly is not supported by empirical evidence.

What crackpots like bschott don't understand is that unlike String Theory, General Relativity is grounded in firmly established principles.  It's just that it is all too often popularly presented as if Einstein simply cut the notion that gravity is indistinguishable from curvature in space-time from whole cloth.  This is simply not true.  On the contrary, Einstein's insight was developed over several years, and it wasn't really until he started thinking about how we would measure the value of $\pi$ in rotating reference frames, that he came to this conclusion.  He didn't start with this hypothesis.  He reached it by doing lots and lots of work, and finding a pattern.  And even then, he still needed the expertise of people like Levi-Civita to help him formalize his ideas.

bschott also asked me,

Define competent please, because in the real world it means you didn't fuck up this badly.

It depends on the context.  In the case of Stack Exchange, competence is determined by the value of the questions you initially post.  If your questions don't have value, then you aren't likely to give valuable answers.

In the case of phys.org, ideally, I would want at least general familiarity with the concepts of peer review and the evidence available for hot-button topics.  Understandably, this would be difficult to enforce, so I would at least require an example of the candidate's prose in response to a popular hot-button issue to determine if they really understand what an acceptable comment should be.

A comment should really be about the article itself.  It should be specific, and well researched.  In writing your initial salvo, you should ask if it actually pertains to the content of the article, rather than a general ideological stance.

Therefore, if the article is about the effects of global warming on a certain species of Coral, your comment should actually refer to the effect, and not to whether global warming is a hoax and how much Al Gore is being paid.

If you cannot even write a well-researched response to such an issue, there's no reason to expect you to offer comments of any value in a forum that is historically unmoderated.

I might also require a candidate to provide a sample response to a deliberately inflammatory comment, and check to see if the response is measured and offers anything but more flaming.  Ironically, I might actually not allow myself to comment because I have occasionally offered only comments with the intent to wound rather than to contribute meaningfully to the conversation.  In fairness, however, sometimes this has been in response to people who can't otherwise take a hint that they are being pricks.


Well, that's all for now.

A Quick Start Guide to the Geometric Product - A Beginner's Perspective Part I

I am just learning about a phenomenal way of manipulating vectors and transforms called Clifford Algebra (or Geometric Algebra).  To give you an idea of how cool this, I first encountered this in reading a paper by David Hestenes on the Dirac Equation found here: (http://aflb.ensmp.fr/AFLB-283/aflb283p367.pdf), a paper I found by searching for the phrase "Dirac equation insight".  You see, the Dirac Equation is one of the closest things to black magic you will ever find in all of physics literature.  Now, as I am aware that some of the imaginary people who read this blog probably aren't well versed in Quantum Mechanics, I won't get into it, but I'll just say that he took an equation that uses bewildering abstract concepts, (which, in turn came from a formalism, that, though, esoteric, starts making sense once you strike your head forcefully, and in a disciplined manner, against hard surfaces enough) and reformulated it in a more concrete geometrical way.  In fact, one thing he was able to derive from his new reformulation was an interesting kinematic interpretation of electron Spin.  Now, don't be fooled by it being it being concrete and geometrical.  Indeed, to work his magic, Hestenes decided to rederive the equation using Geometric Algebra, which I will briefly describe in a little while.  As a caveat, I am a (possibly perpetual) theoretical math n00b (as you probably already realized), so if you are looking for a rigorous formulation of Geometric Algebra replete with axioms, and whatever the hell things like
 
are, you'll have to look elsewhere.

Cramer's Rule: An Example to Whet Your Appetite

Now, unlike mathematicians, who almost universally seem to subscribe to the idea that gavaging students with definitions and theorems as an acceptable form of pedagogy, I will introduce you to example of how one might start thinking in a way that naturally leads to geometric calculus.  This section is actually quite long (as I need to account for deficits in understanding from my imaginary readership) so if you want to get to the actual quick start guide, skip it.

Let's start with a 2D system of equations.

$a_{00}x_0 + a_{01}x_1 = b_0\\
a_{10}x_0+a_{11}x_1=b_1$

Those of you familiar with linear algebra will think "no sweat", and and do something like treat the $a_{ij}$s as a matrix, M, treat this as a matrix equation, and then multiply both sides by the matrix inverse $M^{-1}$, which, of course, requires knowing what its components are.  Note, however, the similarity with the 1D case.  Ultimately, you end up multiplying both sides by an inverse.  This is a very consistent way to think about things, and with all linear equations, we can always do this, as long as we actually know how to find the inverse, but therein lies the rub.  It can be calculated, but doing so is laborious, and the end result makes little sense if you don't understand geometrically what's going on.

Now, there are other ways of solving  this system that are more geometrical.  I am going to use cross products.

First, we'll express this equation as 3D vectors:
$\begin{pmatrix}
a_{00}
\\
a_{10}
\\
0
\end{pmatrix}x_0 +
\begin{pmatrix}
a_{01}
\\
a_{11}
\\
0
\end{pmatrix}x_1 =
\begin{pmatrix}
b_0
\\
b_1
\\
0
\end{pmatrix}
$

Then, we'll just dispense with components, by writing

$\mathbf{A_0}x_0 + \mathbf{A_1}x_1 = \mathbf{B}$

Now, cross both sides by $A_1$ to solve for $x_0$

$\mathbf{A_1}\times\mathbf{A_0}x_0 = \mathbf{A_1}\times\mathbf{B}$

Then, we recognize that both sides point along the z axis, and so, we can dot with $\hat{z}$, and then divide by $x_0$'s coefficient on the LHS to get

$x_0 = \frac{(\mathbf{A_1}\times\mathbf{B} )\cdot \mathbf{\hat{z}}}{(\mathbf{A_1}\times\mathbf{A_0} )\cdot \mathbf{\hat{z}}}$

Likewise,

$x_0 = \frac{(\mathbf{A_0}\times\mathbf{B} )\cdot \mathbf{\hat{z}}}{(\mathbf{A_0}\times\mathbf{A_1} )\cdot \mathbf{\hat{z}}}$

Writing them out explicitly,

$x_0 = \frac{-a_{01}b_0 + a_{11}b_1}{-a_{01}a_{10} + a_{11}a_{00}}$
$x_1 = \frac{-a_{10}b_0 + a_{00}b_1}{-a_{10}a_{01} + a_{00}a_{11}}$

Now, if you know linear algebra, you will see that the denominators of $x_0$ and $x_1$ are just the determinant of the matrix

$\begin{pmatrix}
a_{00} & a_{01} \\
a_{10} & a_{11}
\end{pmatrix}$

Moreover, if you replace the first column of this matrix with the "b" vector,
$\begin{pmatrix}
b_0
\\
b_1
\end{pmatrix}$

You get the numerator of $x_0$.

Likewise, if you replace the second column of this matrix with the "b" vector, you get $x_1$'s numerator.

This is called Cramer's Rule, and is one of those things you can't believe nobody told you before.  It actually applies to linear systems of any dimension, and makes I (and many others) prefer it to the more laborious, and less illuminating, process of Gaussian Elimination, which is basically just the high school method of solving linear equations in slightly more respectable mathematical clothing.  Furthermore, Cramer's Rule has a geometric interpretation we can see, here.  Remember that the determinant is just a signed volume.  Therefore, Cramer's rule tells us that the solution to a linearly independent N degree linear equation is just the ratio of two different signed volumes.

The Geometric Product Between 2 Vectors

It turns out that the methods we have been using are precisely the methods that inspired the development of Geometric Algebra.  Now, dot and exterior products are both forms of multiplication, in that they distribute over addition, but they have some undesirable algebraic properties.  For example, the dot product takes two things (vectors), and returns an entirely different thing (a scalar).  In other words, it isn't closed.  Cross Product, while closed, is neither associative, nor commutative.  Neither operation has a useful concept of inverse.

William Kingdon Clifford had the brilliant idea to take the cross product and the dot product and combine them into one product that, in a sense, behaves better than either of them.  What I mean by this is that theoretically, it is more well behaved than either dot or cross product.  It is closed, unlike the dot product, and it is associative.  Furthermore, it allows for many elements to have inverses, and those inverses make geometric sense.  The price for this unification, is that the elements are no longer the familiar vectors and scalars of linear algebra, but are a collection of elements known as "multivectors", or "cliffs", depending on the author, but have no fear.  These elements can be represented as linear combinations of geometrically meaningful elements called blades.

Unfortunately, the Geometric Product is not that simple to present all at once, at least not if you don't want to assault the student with a barrage of new concepts, so we will start small.  So, without further ado, here is the geometric product between two vectors:

$UV = U\cdot V + U\wedge V$

It contains the $\wedge$ operator, which we have not defined, so let's do so.

As I said, the idea is to combine the dot product, and the cross product into a single multiplication.  However, there is are problems with doing that, the largest of which is that the cross-product is not associative.  In order for UV to be associative, $U\wedge V$ whatever it is, also must be associative.

To account for this, we introduce an entirely new operation that is both associative, and encodes the cross-product (or exterior product, for higher dimensions).

$U\wedge V$ is known as the wedge product, and its result is neither a vector, nor a scalar, but something entirely different.  There are two ways to think about it.  First, you can think about it as the parallelogram formed by U and V, with what's called an orientation.  An orientation can be clockwise or counter-clockwise.  Intuitively, imagine drawing out the parallelogram without lifting your pencil from the paper, using U as the starting edge, and V, as the edge.  This forces you to draw clockwise, or counter-clockwise.  If you draw counter-clockwise, your orientation is positive, and if you draw clockwise, your orientation is negative.  This is actually just the right hand rule.  It also means that the wedge product between two vectors encodes the cross product.

There is another way of speaking about this orientation, alegbraically, and that is that the wedge product between two vectors is anti-commutative.

Thus,

$U\wedge V = -(V \wedge U)$

The second way to think about the wedge product is as the subspace spanned by U and V.

The wedge product between two vectors is known as a 2-blade.

Now, some of you might be puzzled by the presence of this wedge, which is not a real number, and the dot, which is a real number, so we are adding un-like objects.  In fact, in geometric algebra, a general object can contain vectors, too!  However, we've added un-like objects to each other for years with complex numbers, so as long as this sum is irreducible to a real or bivector, there is no reason to complain.

So, what happens when we wedge $U\wedge V$ with a third vector, W?  Unsurprisingly, you can think of

$U\wedge V \wedge W$

as the parellelpiped formed by U, V, and W, again with an orientation, which is either -1, or 1, depending on orientations of the faces of the parallelogram.  The way to envision this orientation geometrically, this time, by drawing the outline of the parallelpiped, by drawing U as the first edge, V as the second edge, and W as the third edge, as shown in Fig. 1 (set U = P, V = Q, and W = R).  Unsurprisingly, this is called a 3-blade.  Likewise, wedging 4 times creates a 4-blade, and wedging k times, creates a k-blade.  Not everyone uses this terminology.  It is very common, also, to refer to a k-blade as a multivector, with a 2-blade known as a bivector, and a 3-blade known as a trivector.  I prefer the former terminology because it is less ambiguous, as using "vector" might suggest that the elements in question are types of vectors, which, they aren't at least not in a pedagogical sense.  Again, the other way to think about a k-blade is as the subspace formed by the k-wedged vectors.  The dimension of a k-blade is known as its grade.

Figure 1.  Source: https://slehar.wordpress.com/2014/03/18/clifford-algebra-a-visual-introduction/

Now, in introducing the concept of a k-blade, I lied a little.  It isn't always true that wedging with a vector with a k-blade produces a (k+1)-blade.  Geometrically, this is easy to see.  If the vector is a member of the subspace spanned by k former vectors, then you won't increase the dimension.  Another way to think about this is that the new parellelpiped will have a k-dimensional volume of zero--in other words, it vanishes!  Furthermore, we want to relate this to the cross-product, which we definitely want to associate with the 2-blade.  Therefore, if $U_0$, $U_1$, ..., $U_{k-1}$ are a set of linearly dependent vectors,

$U_0 \wedge U_1 \wedge \text{...} \wedge U_{k-1} = 0$

Let's return to the geometric product.

Now, one cool thing about this definition is that we can then recast dot and wedge in terms of this new multiplication.  Namely, dot product is

$\mathbf{U}\cdot\mathbf{V} = \frac{\mathbf{U}\mathbf{V} + \mathbf{V}\mathbf{U}}{2}$

and wedge is

$\mathbf{U}\wedge\mathbf{V} = \frac{\mathbf{U}\mathbf{V} - \mathbf{V}\mathbf{U}}{2}$

The geometric product encodes a lot of operations we perform with matrices.  As such, it simplifies thinking about things in several dimensions.  If you know it, it makes things like deriving Cramer's Rule trival.

I will give you a taste of this, now.  Take each side of your 2D system of equations and wedge it with $A_1$, to get,
$(\mathbf{A_1} \wedge \mathbf{A_0})x_0 = \mathbf{A_1} \wedge \mathbf{B}$

It turns out that  $(\mathbf{A_1} \wedge \mathbf{A_0})$ has a multiplicative inverse, which we will see in my next posting, so, we get

$x_0 = \frac{\mathbf{A_1} \wedge \mathbf{B}}{\mathbf{A_1} \wedge \mathbf{A_0}}$

Once you know how to find the multiplicative inverse, then all you need to do is plug it in, and you will arrive at your answer.

What's the *%@!!#!!! Definition of Geometric Product Already?

[Update: I think I now have a short definition of the Geometric Product, though, it requires knowing about, and understanding the extension of the inner product to multiple vectors.  Watch for the next posting for this definition.  I must say that after working with Geometric Algebra more intensely over the last week or so, I am strongly suspecting that, contrary to what I have been reading, the geometric product is a fairly simple concept to understand, and does not require lists and lists of definitions and theorems just to establish the key concepts.  On the contrary, I am almost certain that, as usual, the requisite key concepts are few, and easy to understand.]  

Unfortunately, there are several short answers, each with its own advantages and disadvantages.

1. The geometric product is the associative multiplication you must get when you have a vector space, and your only further assumption (plus some minor technical assumptions for consistency's sake) is that a vector times itself is its squared magnitude.
2. The geometric product is a product between linear combinations of blades and returns another linear combination of blades.  If you multiply two blades, the the resulting linear combination can be organized into geometrically meaningful groups, all sharing the same grade.
3. The geometric product is the multiplicative operation of the ring generated by scalars and vectors, where a vector times itself is its magnitude squared.
4. The geometric product is given by the following formula:

Source: http://arxiv.org/pdf/1205.5935v1.pdf

These are probably not very enlightening.  In fact, definition 1 was downright shocking to me.  I actually spent a few days convinced that it was an incomplete definition, and that you needed one other axiom concerning the definition of wedge products to have a complete, theory--but nope! [Update: I had said that you don't need an axiom concerning the definition of wedge products, and depending on your pedagogical needs, this isn't necessarily correct.  Personally, I do think that this axiom is required.  I made my initial claim based on a mistake I made in deriving the general wedge product ] Definition 2 is not really a fundamental definition, but it is at least descriptive, and prepares us for how to start thinking about the geometric product.  Definition 3 is the closest that I will come to giving a formal definition of the geometric product.  As for definition 4...no...just...no.

...

Ok, fine, but don't say I didn't warn you.  Definition 4 is the geometric product between two blades in all all its glory. You can understand it this way: Definition 4 gives a precise equation describing Definition 2.  It introduces the unfamiliar angle-bracket notation, where

$<A_rB_s>_k$ refers to generalizations of the wedge and dot product with grade k.  Specifically,

$<A_rB_s>_0 = A_r \cdot B_s$, and yes you can take the dot product between general blades.

and

$<A_rB_s>_2 = A_r \wedge B_s$

However, at this point, it would be silly to say more about them, and I only included it to show that there is a formula for the geometric product in terms of geometrically meaningful operations.

Unfortunately, none of these definitions are clear enough for my tastes.  It wish I had something as precise and illuminating as "multiplication between two numbers, M and N, is M added to itself N times", but alas, at this point of my understanding, I cannot.  At some point, in the future, when I have a fuller understanding of this subject, I hope that I can, despite Alan Macdonald's frustrating insistence to the contrary.  However, that doesn't mean I can't give a straightforward, albeit, long, description of it by building up concepts.

In fact, whether you realize it, or not, we have already defined the geometric product.  No other geometric product is possible, given the rules we have already established.  It naturally falls out.  The reason for this is our requirement of associativity of the product.  However, I won't leave you high and dry, or this wouldn't be called a "quick start guide."

The Geometric Product for Vectors

Now, I remarked that the geometric product relates two blades to each other in geometrically meaningful ways.  Superficially, you should already have seen this with my definition of the geometric product between two vectors being the sum of the dot and wedge product between them.  But to show you that I am not out of my gourd, and there really is computational power to this product, we'll explore it in more detail.  Let's consider a really concrete example: vectors in 3D space.

The unit vectors are $e_0$, $e_1$, and $e_2$.

Then, the geometric product for two vectors is pretty easy to see.

Namely,

$e_ie_i = {e_i}^2 = 1$

and

$e_0e_1 = -e_1e_0 = e_0\wedge e_1 = -e_1\wedge e_0$

$e_1e_2 = -e_2e_1 = e_1\wedge e_2 = -e_2\wedge e_1$

$e_2e_0 = -e_0e_2 = e_2\wedge e_0 = -e_0\wedge e_2$

In fact, if any two vectors, U, and V are orthogonal, then $UV = U\wedge V$.  This is easy to see, since we have already defined what the geometric product is for two vectors.

So, immediately, from this, we can see that the geometric product of two vectors

$a_0\mathbf{e_0} + a_1\mathbf{e_1} + a_2\mathbf{e_2}$

and

$b_0\mathbf{e_0} + b_1\mathbf{e_1} + b_2\mathbf{e_2}$

is

$a_0b_0 + a_1b_1 + a_2b_2 + (a_0b_1-a_1b_0)\mathbf{e_0\wedge e_1} +  (a_2b_0-a_0b_2)\mathbf{e_2\wedge e_0} + (a_1b_2-a_2b_1)\mathbf{e_1\wedge e_2}$

One important thing to realize about this expression is that the wedges are interchangeable with the geometric product, as long as we are wedging orthogonal vectors.  So, we can replace this expression with,

$a_0b_0 + a_1b_1 + a_2b_2 + (a_0b_1-a_1b_0)\mathbf{e_0e_1} +  (a_2b_0-a_0b_2)\mathbf{e_2e_0} + (a_1b_2-a_2b_1)\mathbf{e_1e_2}$

The coefficients of the $e_ie_j, i \neq j$ blades should look familiar.  They are precisely the components of the cross-product!  In other words, the cross product is actually encoded in the geometric product between two vectors.  This is no accident; it is a result of how we have defined the wedge product. [Update: Holy ^!%#%! and I just figured out how to recover the cross product (and the generalization of this, the exterior product) from this--and it is exceedingly simple.  I will cover this in  a blog posting, or two]

Breaking vectors into orthogonal components sheds light on what the product is, and is the first strategy I used in attempting to understand this algebra.  However, there is another way to understand it, using permutations, that is remarkably more effective for establishing all of its identities.  However, let's just stick to orthogonal vectors for now.  Sets of orthogonal vectors hold a privileged position in geometric algebra, as proven in our first theorem:

Anticommutivity Theorem:

Suppose we have n distinct, orthogonal, vectors, and you multiply them: $v_0v_1v_2...v_{n-1}$.
Well, it turns out that this product must be anti-commutative.

Proof:

To see this, we use associativity, which shouldn't be controversial, since we are assuming our product is a form of multiplication.

Now, without loss of generality, pick $v_0$ and $v_1$ and swap them.
This yields
$(v_1v_0)v_2...v_{n-1}$


Furthermore, we know that the geometric product of two orthogonal vectors is interchangeable with the wedge product, so, we get
$(v_1\wedge v_0)v_2...v_{n-1} = -(v_0\wedge v_1)v_2...v_{n-1} = -v_0v_1v_2...v_{n-1}$

In fact, the geometric product of orthogonal vectors takes a particularly simple form--a k-blade.  Now, initially, when I wrote this posting, I had thought to make this an axiom, and then, I made a mistake, and reneged on it.  However, I see, now, that my initial suspicion seems to be correct--in a sense.  You see, there are two ways of defining the wedge product in association with the geometric product.  The first is to define it explicitly in terms of the geometric product, which, despite being an awful pedagogical approach, is favored by pretty much everyone.  The second is the way I use, which is to define it, and then base the geometric product off of it.    Admittedly, the first approach is more elegant, in that you need fewer axioms, but I don't know why mathematicians (and many physicists) think that elegance is a good substitute for understanding.  It doesn't matter a whit how concise and direct your argument is if you can't get your point across to all but a subset of people who enjoy the gavage method of learning concepts${}^{\text{TM}}$.  Seriously, dudes, I know I harp on this a lot, but guys, how about you tone down the rigor when presenting new concepts?  You might get more people to like your subject, if you do.

Anyway, I keep meaning to make a blog post about my love/hate relationship with theoreticians, but that's for another day.  For now, let's just add an axiom, which I will call the

Fundamental Axiom

If $v_0$, $v_1$, ..., $v_{k-1}$ are all orthogonal vectors, then,

$v_0v_1...v_{k-1} = v_0 \wedge v1 \wedge ... \wedge v_{k-1} $

The motivation for this is consistency (as far as I can tell.  I am not certain what the original motivation was).  For two orthogonal vectors, this is true, so why not have it be true for more than two?  Indeed, there is a more formal reason why, based off of redundancy, but that is too complicated to get into, now.

Anyway, once we have this axiom, the geometric product can be completely defined.  By that, I mean you could write a straightforward, albeit, inefficient algorithm for calculating the geometric product using wedge products and scalars.

To see this, first, I need to show that if you have a product of multiple vectors, its value will depend fundamentally on how we define the product of orthogonal vectors.
Let's say you have a set of orthogonal vectors, O of size m, and you have a product that contains N vectors from this set O, where n > m.  Then, by the pigeonhole principle, this product contains at least one copy.

Let's construct two sets: $C_{odd}$ and $C_{even}$.

$C_{odd}$ is the set of all vectors that occur an odd number of times in the product.
$C_{even}$, likewise, is the set of all vectors that occur an even number of times in the product.

Furthermore, let $k = |C_{odd}|$

Orthogonal Product Theorem

Let $v_0$, $v_1$, ..., $v_{k - 1}$ be the set of vectors in $C_{odd}$.  Then, this product will be equivalent to $s\mathbf{v_0v_1}...\mathbf{v_{k-1}}$, where s is a scalar.

Proof:

I will start with an example.  Let O be {A,B,C,D}.  Suppose you have the product, $AAABBBCD$.

Rearrange this so you have $ABCD$ on the far right:

$AAABBBCD = AABBABCD$.

Now, $AA = A^2 = a$ and $BB = B^2 = b$, where a and b are scalars.

So, the result is:

$abABCD$

So, for the general proof, just as we did for the  we are going to reorganize the product in the following manner, multiplying it by the appropriate sign:
Let the right-most k terms be $v_0v_1...v_{k-1}$.
In the remaining product, there will be vectors in $C_{even}$, and vectors in $C_{odd}$.  Take each set of copies and place them to the left of $v_0v_1...v_{k-1}$ so that all copies of the same vector are adjacent to each other.  Now, the adjacent "runs" of $C_{even}$ will each yield scalars.  Likewise, the remaining adjacent "runs" in $C_{odd}$ will also produce scalars, since they will be even in length.  Then s is the sign multiplied by all the scalars to the left of $v_0v_1...v_{k-1}$.  We have thus proven our theorem.

Therefore, if we are only using orthogonal vectors, the result will simply be a scaled version of whatever the geometric product is for a set of distinct orthogonal vectors.''

Before I go on, let me introduce an abuse of terminology to make my life easier.  Call, a scalar, a single vector, or a product of multiple, distinct, orthogonal vectors a blade.  If it has k elements, then, we can also call it a k-blade. Furthermore, call a product of orthogonal vectors, which might include copies an orthogonal product.

Now, let's use Theorem 2 to prove its general version:

Geometric Product Theorem

Every geometric product of vectors is a linear combination of distinct blades.

Proof:

The proof for this is straightforward, but tedious if you write it out with actual algebra.  For the sake of brevity, I will merely give a prescription for how to do this.  Every vector is a linear combination of basis vectors.  Now, distribute the product over all linear combinations.  Then, you will get a sum of orthogonal products.  By theorem 2, each of these orthogonal products are then simply a blade multiplied by some scalar.

Proof that Objects Don't Always Return Under Gravity

So, I told this crackpot over at phys.org that I would be proving what is written in the title.  It is now late, and the proof is turning out to be really long, really because he's a bit of a weasel, and thinks he can get away with saying vague, hand-wavy, things like, "Heisenberg kicks in when velocity gets low enough."  Here is the proof, so far, so that Mr. Crackpot doesn't think I am sitting on my hands.  NOTE: THIS PROOF IS NOT FINISHED, YET: [UPDATE:  Here's some background for this blog posting.  Over as phys.org, I have a long history of arguing with a crackpot I call Clever Hans.  That isn't his screen name, but a nickname that I gave him after realizing that his approach to understanding math and physics was about as deep and methodical as the approach taken by his namesake.  This is not the first time I have attempted to prove this particularly well known property of gravity to him.  My first attempt, however, involved solving differential equations and using limits, something he summarily dismissed as "only a mathematical model," adding that at some point, "Heisenberg would kick in," thereby demonstrating not just a poor understanding of quantum mechanics, but an even poorer understanding of potentials.  Anyway, to make a long story short, I decided to attempt this task once again, this time with just high school algebra, conservation of energy, and conservation of momentum.  I don't know why I bother, though.  I am sure that his response will be something profound, like, "you forgot a comma, therefore, your proof is invalid."  ]

All right, Mr. Crackpot.  Thank you for your patience. Here is my proof.

Before I start, let me establish something about gravitational potential.

Now, the gravitational potential is given by -GmM/r, according to http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html, where r is the distance between the gravitating objects, and m and M are the masses.

But, you have insisted, often, that the laws of gravity as presented by textbooks, are merely approximations.  Therefore, I will allow for the more general potential, $\frac{-GmM}{r} + s\left(r\right)$, where s(r) is an error correcting term.  Now, at normal distances, s(r) must be approximately 0 because, otherwise, we would have detected it by now.  However, we are concerned with what will happen at large distances, and so, s(r) may come to dominate the potential.  More on that in a little bit.

Now, you argue that, in a universe with only two gravitating objects, they will eventually return to each other, regardless of their initial velocities.  Normally, I would say this means that at least one of the objects will slow to 0m/s, and then change direction.  However, I anticipate that you would counter that some kind of hypothetical quantum effects might allow the object to "jump" between moving forward and moving backward without completely slowing to a halt, and so, I'll grant you that, for argument's sake.  Let's say that whenever an object slows, and turns around relative to another object, there must be some lowest speed threshold, $S_q$, such that AT SOME POINT BEFORE TURNING AROUND, the relative speed will be less than or equal to $S_q$.

For example, let's say $S_q = 10^{-12}m/s$.  Then the relative speed between the two objects must at least slow down to $10^{-12}m/s$, if not less, before at least one of the objects change direction.  However, as with s(r), let's not assign it a value just yet.

With this in mind, all I need to do to prove you wrong is find a counter example to your claim.  So, here is what I will do. I will do a proof by contradiction, and assume that you are correct, and then show that it implies physically unreasonable conditions for a specific case.

For this, I have chosen a very simple set up.  We choose two identical, spherical, objects, A and B, of mass, 10^13kg, about the mass of the comet, 67P/Churyumov-Gerasimenko.  Object A starts out at $-10^{6} $ m from the origin.  B starts out at $10^{6}$ m from the origin.  Their initial velocities are equal in magnitude and in opposite directions, so that A is travelling, initially at $-v_0$, and B is travelling initially at $v_0$.  I'll get to the value of $v_0$ in a little bit.  For now, let's at least require that $|v_0|$ is well below the speed of light so that relativistic effects are negligible.  With this in mind, let's calculate the gravitational potential of A and B for this initial condition.

Well,

\begin{equation}-\frac{Gm^2}{2\times 10^6 \text{meters}} = -3369200000 \text{J}\end{equation},

 so, our potential is $-3369200000 \text{J} + s(2\times 10^6 \text{meters})$.

Now, let's see what conservation of momentum tells us.  Well, the initial total momentum is $mv_0 - mv_0 = 0 \textrm{kg m/s}$.

That means that if vA and vB are the velocities of A and B respectively in the future, m*vA + m*vB = 0, so that vA = -vB.  So, again, vA, and vB will be equal in magnitude an opposite in direction.  Therefore, we can dispense with this notation, and instead assume that at some point in the future, A is travelling at $-v_1$, and B is traveling at $v_1$.  Currently, we don't know what $v_1$ is.  It could be negative or positive or 0.

Suppose that A is at position xA and B is at position xB.  Since A and B are always moving at the same speed, we can also assume that xA = -xB.  As above, I can dispense with the "A" and "B" notation, and instead say that A is at position $-x_1$, and B is at position $x_1$.  Again, we DON'T YET KNOW what $x_1$ is.

Without loss of generality, assume that B turns around at some point in the future.  Now, since the initial conditions are that A and B start out moving away from each other at the same speed, we can assume that A and B will have moved even further away from each other by the time that their relative speed, $2x_1$, slows down to Sq or lower.  So, we can DEFINITELY assume that this happens when B's position, $x1$ is $> 10^6 \text{meters}$.

Now, let's use conservation of energy to get an idea of how large $x_1$ needs to be so that its magnitude is between 0meters/s and $S_q/2$.

The initial total energy is

$mv_0^2 - 3369200000 \text{J} + s(2\times10^6 \text{meters}) = (10^{13 }\text{kg})\times v_0^2 - 3369200000 \text{J} + s(2\times 10^6 \text{meters})$

Well, $\frac{-Gm^2}{2x1} = -\frac{3.3692\times10^{15} \textrm{kg meters}^3/\text{s}^2}{x_1}$

So, the total energy is $(10^{13}kg)\times v_1^2 -(3.3692\times 10^{15} \textrm{kg meters}^3/ \text{s}^2)/x1 + s(2x1)$

and conservation of energy tells us that

$S_q^2 = v_1^2 = v_0^2 - 0.00033692 \text{J\kg} + \frac{s(2\times 10^6 \text{meters}) }{10^{13}\text{kg}}+ \frac{ 3.3692\times10^{15} \textrm{kg meters}^3/\text{s}^2}{x_1\times10^{13}\text{kg}} - \frac{s(2x_1)}{10^{13} kg} = v_0^2 - 0.00033692 \text{J/kg} + \frac{s(2 \times 10^6 \text{meters})}{10^{13}\text{kg}} + \frac{336.92 \text{meters}^3/\text{s}^2}{x1} - \frac{s(2x_1)}{10^{13}\text{kg}}$

Or more succinctly,
$S_q^2 = v_0^2 - 3.3692\times 10^{-4} \text{J/kg} + \frac{s(2\times 10^6 \text{meters})}{10^{13}\text{kg}} + \frac{ 336.92 \text{meters}^3/s^2}{x_1} - \frac{s(2x_1)}{10^{13}\text{kg}}$

Now, there are two cases to consider: that s(r) is bounded as r increases, or that s(r) is unbounded as r increases.  I assert that it is physically unrealistic if s(r) is unbounded as r increases.  I'll assume that Mr. Crackpot agrees with this since his argument is that as distances grow, quantum effects "kick in", meaning that s(r) becomes quantum energy fluctuations.  In other words, the only two reasonable contributors, in this non-relativistic case, to the gravitational potential energy are the classical gravitational potential energy and quantum energy fluctuations.  Now, I am aware that Mr. Crackpot, being unabashedly bad at physical intuition, might, indeed, protest that I haven't considered the case where s(r) is unbounded.  If he does, I have a proof for that, too, but in the interest of not making this post any more of the novella it is, I will assume that it is acceptable to require that s(r) is maximally bounded.

Thus, at most, s(r) is bounded in magnitude by a particular energy scale.  Let's call this scale $s_{ub}$, so that $-s_{ub} \leq s(r) \leq s_{ub}$.  Note that $s_{ub}$ is nonnegative.

With this in mind, we can say that

$\frac{s(2x_1)}{10^{13}\text{kg}} = v_0^2 - 3.3692\times 10^{-4} \text{J/kg} + \frac{s(2\times 10^6 \text{meters})}{10^{13}\text{kg}} + \frac{ 336.92 \text{meters}^3/s^2}{x_1} - S_q^2$

implies the inequalities,

$-\frac{s_{ub}}{10^{13}\text{kg}} \leq v_0^2 - 3.3692\times 10^{-4} \text{J/kg} + \frac{s(2\times 10^6 \text{meters})}{10^{13}\text{kg}} + \frac{ 336.92 \text{meters}^3/s^2}{x_1} - S_q^2 \leq \frac{s_{ub}}{10^{13}\text{kg}}$

We only need the right inequality,

$v_0^2 - 3.3692\times 10^{-4} \text{J/kg} + \frac{s(2\times 10^6 \text{meters})}{10^{13}\text{kg}} + \frac{ 336.92 \text{meters}^3/s^2}{x_1} - S_q^2 \leq \frac{s_{ub}}{10^{13}\text{kg}}$

This is equivalent to writing

$\frac{s_{ub}}{10^{13}\text{kg}} +  S_q^2 \geq v_0^2 - 3.3692\times 10^{-4} \text{J/kg} + \frac{s(2\times 10^6 \text{meters})}{10^{13}\text{kg}} + \frac{ 336.92 \text{meters}^3/s^2}{x_1}$

However, since $x_1$ is positive, it is also true, if we just simply drop the term containing it.

$\frac{s_{ub}}{10^{13}\text{kg}} +  S_q^2 \geq v_0^2 - 3.3692\times 10^{-4} \text{J/kg} + \frac{s(2\times 10^6 \text{meters})}{10^{13}\text{kg}}$

Now, I will reveal the value of $v_0$.  I have assigned it the hugely nonrelativistic value of 340.29 meters / s.

So, we have

$\frac{s_{ub}}{10^{13}\text{kg}} +  S_q^2 > 115797.2841 \text{meters}^2/\text{s}^2 - 3.3692\times 10^{-4} \text{J/kg}+ \frac{s(2\times 10^6 \text{meters})}{10^{13}\text{kg}}$

which is equivalent to

$\frac{s_{ub}}{10^{13}\text{kg}} +  S_q^2 > 115797.28376308\text{meters}^2/\text{s}^2+ \frac{s(2\times 10^6 \text{meters})}{10^{13}\text{kg}}$

Now, $2\times10^6\text{meters}$ is not very large.  It's in the neighborhood of 1000km. We know, then that $s(2\times 10^6 \text{meters})$ is too small to detect, here at home, so that we can simply say that its value, compared with $115797.28376308 \text{meters}^2/\text{s}^2$, is approximately 0.  It is, therefore, acceptable to write,

$\frac{s_{ub}}{10^{13}\text{kg}} +  S_q^2 > 115797.28376308 \text{meters}^2/\text{s}^2$

Now, the above inequality implies that

\begin{equation}\frac{s_{ub}}{10^{13}\text{kg}} \geq 57898.64188154\text{meters}^2/\text{s}^2\end{equation}

OR

\begin{equation}S_q^2 \geq 57898.64188154 \text{meters}^2/\text{s}^2\end{equation}

(they can't both be less than half of $115797.28376308 \text{meters}^2/\text{s}^2$)

If inequality (2) , then

\begin{equation}s_{ub} \geq 57898.64188154\times 10^{13} \text{J}\end{equation}

This is clearly absurdly large.  To give you an idea of how large this is, this is about 6 times as energetic as the 2004 Indian Ocean Earthquake.  Obviously, $s_{ub}$, and hence s(r) is not just a quantum energy fluctuation.

On the other hand, if (3) is true

$S_q \geq 240.621 \text{meters \ s}$.

In human terms, this is a very large velocity, and is equal to 538.252 miles per hour.

Again, this is unacceptable.  Normal objects like cars and humans go much slower than this before turning around.  So, again, we arrive at an absurd physical requirement.

The only possibility, assuming s(r) is due to quantum fluctuations is that your initial premise, that objects always return under gravity is horribly flawed.

The ball is in your court.  Now, I spent a long time writing this up.  Let's see if you are honorable enough to respond accordingly rather than replying with a terse, irrelevant comment, like "it's just math!"