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!"

An interesting way of looking at the Lorentz Transform

So, this is going to be a rather light posting today.  I was at first going to refer to it as short, but apparently, I can be quite verbose, so, so much for that.  Anyway, this post is really meant as a means for me to get the ball rolling on this blog again after a series of events that lead to me being unwilling to blog.  I actually hope to talk a little about that, at some point, but for now, here is a fun little thing I realized about the Lorentz Transform, which actually stemmed from my desire to find a way of assigning a coordinate to every point in space-time without explicitly relying on velocity.

Let me remind you of the Lorentz Transform, if you don't know what it is:
\begin{equation}
x' = \left (x - vt \right )\gamma
\end{equation}

\begin{equation}
y'=y
\end{equation}

\begin{equation}
z'=z
\end{equation}

\begin{equation}
t' = \left (t -\frac{xv}{c^2}\right )\gamma
\end{equation}

where
\begin{equation}
\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
\end{equation}

If you aren't familiar with what this means, wikipedia has a good article on it.

And of course, the inverse transform is given by:
\begin{equation}
x = \left (x' + vt' \right )\gamma
\end{equation}

\begin{equation}
y=y'
\end{equation}

\begin{equation}
z=z'
\end{equation}

\begin{equation}
t = \left (t' +\frac{x'v}{c^2}\right )\gamma
\end{equation}

So, who cares?

Well, it occurred to me that I was wrong when I said that there was no way of uniquely identifying a space-time point without explicitly referring to velocity.  All it takes is a little adjustment to your point of view, and...well...cheating.

You see, you can actually use the inverse Lorentz Transform to fulfill this exact requirement, and the answer is quite trivial--express it in terms of the origin of the rest frame!  Let me just get into some philosophy before this.  When we describe the universe, at least in the non-general relativistic regime, it is always with respect to some fixed reference point.  Sure, you can express your system in polar coordinates, or spherical, or something completely exotic, but there will always be a point with the label (0,0,0,0).  Often, this point happens to be physically meaningful, and not just convenient.  It might represent some central object like the Sun, or the center of mass.  To show you the significance of this, let's try a thought experiment.

Pick a point--any point in the universe.  The only stipulation is that it must be uniquely defined.  Which did you choose?

I suspect that most people who perform this exercise would pick something that they could easily describe.  They might have chosen "my location", or "the center of Earth", or "the center of mass of Messier 83."  Some particularly creative people might have chosen something like "a point that is located 200 billion km in the direction of where the the geomagnetic field of Earth is pointing at my location."

But, here is what I suspect you didn't do--pick the center of mass of all discrete silicate glass materials in the visible universe with a charge of $-2\pm0.1$ coulombs.  Likewise, I doubt you picked the center of mass of the set of living organisms with the largest volume in the visible universe.

Now, you'll notice the difference between the first set of objects and the second.  In the first, each description, no matter how complicated, relied explicitly on some known physical reference point.  The objects in the second set did not explicitly reference any known physical point.  In fact, it is impossible, given our current knowledge of the universe, to determine what these points are.

We need some physical reference point to anchor our description of physical phenomena.  Even the second set of points, though not explicit, referenced a known location, as the visible universe represents the light cone of Earth, and that's before even talking about the need to express these things against the backdrop of an inertial reference frame to fully define them. (I would like to say, for the record, that this would be a fun, nerd game to play with all your nerd friends.  Specifically, see who can identify the most obscure point in space-time possible, especially with alcohol.  There is no doubt, that eventually, penises would work there way in there some where.  They always do.)

Thus, it seems that only by fixing some origin somewhere that we can begin to describe the universe.

If you think about it, this is exactly what we do with the inverse Galilean transform:
\begin{equation}
x = x' + vt'
\end{equation}

\begin{equation}
y = y'
\end{equation}

\begin{equation}
z = z'
\end{equation}

Namely, as long as everybody agrees where the reference point is, everybody can agree on where the event occurred, and since measurements approximate c as being infinite in speed, then everyone knows when it happened, too.

In fact, we can express equation 10 in another way.  Let $O$ represent the position of the reference point.

Then, we we have
\begin{equation}
x = x' - O
\end{equation}
Of course, this seems obvious.  All we're doing is describing points with respect to $O$.  However, to me, this represents an interesting way of looking at things.  In particular, by rephrasing the Galilean transform in this way, we no longer have to worry about actually measuring relative velocity.  All we need to do is figure out where $O$ was in relation to us, and we would be able to unambiguously identify a point in space that corresponds to every event.

So, the question is, can we use this as inspiration for the Lorentz Transform.  Of course!

So, take equation 6 and replace it with:
\begin{equation}
x = \left (x' - O \right )\gamma
\end{equation}
Now, this makes sense, at least partly.  Somebody moving relative to me at velocity $v$ would say that my trajectory is $-vt'$, assuming our origins coincided at when our clocks both read 0s.  Thus, if they treat me as their reference point, $O$, then this parallels the Galilean so far.
"But hold on!" you might protest, "$\gamma$ isn't just a constant.  It too depends on relative velocity."
Well, algebra's not a problem for us.  $O=-vt'$, so $v=-O/t'$, and this is where things begin to get really kind of interesting.  Let's see what this does to $\gamma$

\begin{equation}
\gamma = \frac{1}{\sqrt{1-\frac{\left( \frac{-O}{t'}\right)^2}{c^2}}}
\end{equation}

\begin{equation}
\gamma = \frac{1}{\sqrt{1-\frac{O^2}{(ct')^2}}}
\end{equation}

Now, notice that $ct'$ would be the position of an electromagnetic wave emitted along the x axis at the time the origins coincided.

So, $\gamma$ becomes an expression involving the position of O and the position of the electromagnetic wave.

We can do something similar with equation 9.
\begin{equation}
t = \left (t' -\frac{x'O}{t'c^2}\right )\gamma
\end{equation}
\begin{equation}
t = \left (\frac{(ct')^2 -x'O}{c(ct')}\right )\gamma
\end{equation}

Again, we see the equation clearly expressed in terms of the positions $O$ and $ct'$.

Now, see what I mean about cheating?  In some sense, this is a mathematical disappointment.  All I did was rewrite the Lorentz transformation, but in a physics sense, it has real ramifications for measurement--perhaps not profound--but real, nonetheless.  Just as with the inverse Galilean transform, we see that there is a way to uniquely assign a space-time coordinate to an event without having to measure velocity, by having everybody agree on a reference point.  In other words, you don't need to differentiate at all.  Just know where the reference point is at the time of the event, and whatever event you encounter, you can say exactly where it would have been for the person at rest with that reference point.  Keep in mind, also, that the fundamental forces are conservative, and therefore rely only position.  This form seems to respect that more than the conventional form.  In all honesty, though, I can't really say this is a useful way of looking at things.  But, then again, I have barely had any time to think about it.

So, what do you think?  Is there any use that can truly come from expressing the transform in this way?  I wonder if this might be useful as a pedagogical tool for talking about closed time-like curves resulting from FTL information transfer.  Anyway, I am signing off, for now.

How to Derive Relativistic Momentum Without Magically Divining the Answer

What's that you say?  "Wow, the Furlong, that was a tasty, if somewhat rushed, derivation that nobody asked for.  Please give me another!"

Ok, I will, though I do apologize if my last derivation may have been difficult to follow.  My time is limited and working with LaTeX in Blogger is about as fun as grinding with a porcupine.  I am still getting used to it, so please, bear with me.

Let's back track somewhat and talk a little about an aspect of Special Relativity, that frankly, I find much cooler than Lorentz Contraction and Time Dilation: Relativistic Momentum and Energy.  In Newtonian Mechanics, for those who don't know, we have momentum, p:

\begin{equation}
p = \int \! F \ \mathrm{d}t = mv
\end{equation}

where F is the net force, m is mass, and v is velocity.

And kinetic energy, T:

\begin{equation}
T = \int \! F \cdot \ \mathrm{d}s = \frac{mv^2}{2}
\end{equation}

where s is the path.

In SR, (1) is replaced by
\begin{equation}
p = \mathrm{MAAAGIC} = mv\gamma
\end{equation}

and (2) is replaced by
\begin{equation}
T^2 = \mathrm{ME \> LUCKY \> CHARMS!} = m^2c^4 + p^2c^2
\end{equation}

where $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ is the Lorentz factor.

You'll notice that the intermediate steps involve mystical invocations and supplications to supernatural beings (the kind you find on cereal boxes).  For all intents and purposes, this is how derivation of relativistic momentum and energy is presented in all explanations I have encountered.

An extremely common method used to describe this, which I confess, I have never taken the time to truly appreciate, is one involving a glancing, elastic, collision between two particles.  First, they show that if we assume $p=mv$ momentum, momentum is not conserved.  Then, they skip the most important step and guess that p satisfies (3) and then show that under that arbitrary assumption, that momentum will be conserved.  That's right!  In order to show people how to derive relativistic momentum, they don't actually derive it, but magically guess that it must take that form.

Please permit me to take a moment to tell you how awful, and completely unhelpful this type of pedagogical strategy is.  As physicists, it is in our best interest to teach upcoming physicists not just how to remember equations, but how to apply sound physical reasoning to arrive at those results, in hopes that some day, they will be able to apply those same vital skills to arrive at novel results.  How is this going to help them do that?

The sad thing is, this kind of strategy is rampant in mathematically oriented subjects.  Too often, especially in upper level courses, new concepts are presented without precedent or even the barest motivation--as if they sprang, fully formed, from the foreheads of their originators.  There is no hint of the hours of reasoning that lead to these results.  Even worse, as an artifact of progress, the original derivation, though correct, is often replaced with a sleeker, more efficient, derivation that turns out to be less instructive than its predecessor.  I want to talk more about this phenomenon--perhaps in the next blog post, but for now, I just want to point out how misguided this particular method is.  If you are going to derive something, then for heaven's sake, derive it!  Don't just invoke Baphomet, and then act as if you have actually explained anything worthwhile!

Anyway, it turns out that there is a really illuminating way to look at this situation and obtain the necessary physical insight for deriving this result.  Let's start with relativistic acceleration.  Again, for pedagogical purposes, it suffices to restrict ourselves to 1+1 dimensions.  Let's assume the canonical set up where B passes A moving at a relative velocity of v.  Suppose that B sees some object, say, a fly, moving with a velocity, $v'_f$, in some event, E.  Then to A, the fly will be moving with velocity,
\begin{equation}
v_f = \frac{v'_f + v}{1+\frac{v'_fv}{c^2}}
\end{equation}
in that same event.

I will not explain why $v_f$ takes this value in this post, though it is straightforward to derive from the Lorentz Transformation.

Now, let's change things up a little.  Let's assume that at time t' on B's clock, B is instantaneously at rest with the fly.  Now, suppose that B measures a force, F', to accelerate the fly at time t' on his clock.  Since B is at rest with the fly in this instant, then B does not measure the fly to be moving at relativistic speeds relative to himself, at least not at first.  In other words, during a small amount of time $\Delta t'$ after $t'$, B would measure the fly to be moving, but not at relativistic speeds. In fact, if $\Delta t'$ is small enough the fly's speed will be nearly 0 m/s.  Now, let's think about the principle of relativity:

The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems of coordinates in uniform translatory motion. OR: The laws of physics are the same in all inertial frames of reference.

Keeping this in mind, let's ask the following question:  What if, instead, A were seeing this fly accelerate from rest, instead of B?  Then, we would reasonably assume that in this small interval of time, the fly should obey regular old Newtonian mechanics--more formally, as $\Delta t'$ approached 0, the laws of physics would approach Newtonian physics.  Namely, we would assume the fly had a mass, m, and that this force F', accelerated this fly in the time interval between $t'$ and $t'+\Delta t'$ by approximately $\frac{F'}{m}\Delta t'$

Well, by the First Principal of Relativity, B would have to measure this exact same scenario.  Now, we're in business!  So, let's see what this implies.  At time $t'$, B sees the fly at some position, $x'$, at rest.  Then, at time $t'+\Delta t'$, B sees the fly at position, $x'$, but moving with velocity, $\frac{F'}{m}\Delta t'$.  The following is a table relating how A and B measure things:

$\left(x',\> t'\right)$	$v'_f$	$(x,t)$	$v_f$
$\left(x',\> t'\right)$	$0 \mathrm{ m/s}$	$\left((x'+vt')\gamma,\>(t'+\frac{x'v}{c^2})\gamma\right)$	$v$
$\left(x',\> t'+\Delta t'\right)$	$\frac{F'}{m}\Delta t'$	$\left(\left(x'+v\left(t'+\Delta t'\right)\right)\gamma,\>\left(\left(t'+\Delta t'\right)+\frac{x'v}{c^2}\right)\gamma\right)$	$\frac{\frac{F'}{m}\Delta t' + v}{1+\frac{\left(\frac{F'}{m}\Delta t'\right) v}{c^2}}$

Boy, that looks beautiful, doesn't it?  You just want to smear chocolate all over that table and rub it on your chest.  Oh yeah...

Oh, where was I?  Oh yes, this horrific table. Now, as much as I enjoy writing in LaTeX (having spent the last hour fighting with that stupid table), I am going to skip a few steps.  They are straightforward, and you should be able to do them yourself.  But, basically, we would find that where B sees the fly accelerate by $\frac{dv'}{dt'}=\frac{F'}{m}$, A sees the fly accelerate by $\frac{dv}{dt} = \frac{F'}{m}\gamma^{-3}$

We can rewrite this as
\begin{equation}
F'=m\frac{dv}{dt}\gamma^3
\end{equation}

Now, what are we to make of this?  Well, intuitively, if B sees the fly accelerate more from his frame, A will see it accelerate less, so we should expect that should somehow translate into the fly being "harder to push" when it is moving.  So, even though we have an expression for acceleration, we need some way to connect this to the actual force that A measures.  Personally, I don't think it is immediately clear how to do this.  The answer, however, lies, in finding a force that both A and B can agree on.

There are a number of ways to find such a force.  Perhaps the most obvious one (at least to me) is a uniform gravitational field, but this is immediately problematic--at least conceptually, because we know that, in a gravitational field, everything accelerates at the same rate.  In this example, we have found that the acceleration should differ.  Of course, we haven't actually found a contradiction.  Rather, all we have found is that, at least algebraically, the expressions for accelerations differ, but have not actually shown that the numerical values should be different.  This, of course, indicates that F', m, and $\gamma$ must take on appropriate values to preserve the acceleration, but, as you can see, things are already getting hairy, so it might help to clear things up by using a different force. [UPDATE: I can't believe that I just realized the error I made in this paragraph.  Everyone in a uniform gravitational field experiences the same acceleration, not the same force.  FAILURE!  Sorry about that.  You know, I am told that a while ago, an actual physics professor visited my blog and concluded that I don't know what I am talking about.  The above paragraph could be what induced him to arrive at this conclusion.  Oh well.]

And for that, we'll choose the good ol' Coloumb force.  Let's imagine for a moment that the fly is a dipteran Piotr Nikolaievitch Rasputin, and has used his mutant powers to turn completely metal, and that dipteran storm charged him with lightning.  Furthermore, suppose that everybody is situated between two enormous, charged, parallel, metal plates, orthogonal to $v$, which generate a uniform electric field, which serves as the force.  At the limit that the plates become infinitely large, not only is this field uniform, but it only depends on their charge and area.  As both A and B agree on these quantities, they would agree on the force between the plates.  Now, invoking the First Postulate of Relativity again, we immediately see that both A and B must agree that the force on the fly is the same.

Hence,
\begin{equation}
F = F' = m\frac{dv}{dt}\gamma^3
\end{equation}

So, what exactly is this telling us?  Well, basically, any time you push on a moving object with a force, F' in the moving frame, an equivalent push in the rest frame will result in a lower accelerations.  Consequently, it should take a stronger force to achieve that same acceleration in the rest frame.  So, it seems that we have arrived at the expression for force, as A measures it, on an accelerating object.

From here on in, things get relatively easy, so to speak (depending on how you look at things).  We have done most of the conceptual work with this thought experiment.  Now, let's invoke the handy First Postulate again to find momentum and energy.

In this case, since B would use $\frac{dp'}{dt'} = F'$, then A should use $\frac{dp}{dt}=F$.  Thus, integrating F, we find that $p=mv\gamma$.
Let's write v in terms of p because that will become useful in a moment:
\begin{equation}
v=\frac{p}{\sqrt{m^2 + \frac{p^2}{c^2}}}
\end{equation}

Likewise, we can calculate the kinetic energy

\begin{equation}
T=\int{F\mathrm{d}x}
\end{equation}

Noting that $F=\frac{dp}{dt}$, let's make this easier by changing the variables:

\begin{equation}
T=\int{\frac{dp}{dt}dx} = \int{\frac{dx}{dt}dp} = \int{v \ dp} = \int{\frac{p}{\sqrt{m^2 + \frac{p^2}{c^2}}} \ dp}
\end{equation}

Therefore,
\begin{equation}
T=c^2\sqrt{m^2 + \frac{p^2}{c^2}} = \sqrt{m^2c^4 + p^2c^2}
\end{equation}

And so, we have just derived equations (3) and (4), as we set out to do.

Introduction and An A Priori Derivation of the Minkowski Space Time Interval

Yes, I am GREAT SCIENTIST!

Ok, not really.  I am actually not a professional scientist at all (though I hope to be some day).  However, I thought I would start a blog to share my exploits in amateur research.  Indeed, I confess as I currently have no access to actual experimental equipment, my ideas are all of the gandenken experiment kind.  Thus, you will find no sigmas here, unless or until I actually enter a research program.  Sorry to disappoint you, but in actuality, the purpose of this blog is mostly to share interesting mathematical derivations and wild speculations, mostly related to whatever it is that I have decided to focus my fickle mind on at the moment.  Often, that's physics, but other times, you will encounter other topics ranging from number theory and abstract algebra, to proof theory.

My goal in writing this blog is to create a delicate balance between overly simplified arguments you will find in most blogs geared toward laymen and overly technical, jargon, laden arguments that you will find in a scientific paper.  In my experience, though it is often true that the devil is in the details, it is also frequently as true that much of the structure of the devil can be found in a straightforward manner by a competent non-expert once some fundamental details have been explained.  In short, most profound truths spring from a handful of key insights.  (I suspect, however, that there are some profound results that don't adhere to this principle.  The Four Color Theorem comes to mind). With this in mind, the purpose of this blog is not to introduce you to fundamental concepts, but, in an entirely self-indulgent way, to provide you, the reader, with navigational charts of territory I have covered, am currently covering, and will cover, though not all in order.  Now, on to the derivation...

The Minkowski Metric

Today, I will start with physics, specifically, Special Relativity, one of my favorite topics.  It is a beautiful theory, which serves as a rather startling example of how profound, foundation shaking, predictions can be derived from simple, though obscure, postulates.  It is also a theory that requires few to no measuring apparatuses to actually derive well-tested experimental predictions--that is to say, it is not a theory of the phenomenological kind.  This makes it ideal for an aspiring physicist, such as me, to use it as a springboard off of which to learn physics while in the transitional, awkward, period of trying to make enough money to go back to school and become a full-time crackpot--err researcher...yeah...that's what you heard me say...researcher...

This derivation developed out of something I consider a rather guilty pleasure: arguing with people online.  I won't get into details, but there is a well known commentator in the Physorg community that has a rather--shall we say--loose idea of physics.  Specifically, I would describe his brain as an empirical demonstration of Murphy's law, where if there is a well established principle that every competent person understands, his brain will find a way to misinterpret it.  But, I digress.  Over the course of arguing with this person, it became quite clear that he was unable to resolve the notion that two observers could measure a physical result differently without crashing all of physics like it was pretend 23:59:59, 12/31/1999 in his mind--even though we do it all the time in regular old Newtonian mechanics when we change coordinate systems.  Understandably, the addition of conflicting temporal measurements does introduce some conceptual difficulties, but the wonderful thing about SR is that we've had 100+ years to develop all kinds of pedagogical strategies for carefully guiding the physics neophyte through the difficult waters of simultaneity.  Anyway, it occurred to me that it might help to develop less ambiguous description of space-time points, which served as triggers for his mind to jump to completely incorrect conclusions.  This lead me to asking if there was a way to uniquely identify a space-time point independent of, say velocity.

In other words, I asked, is there a way to assign a unique coordinate to a spatial point that didn't move to anyone?  Obviously, the inverse Lorentz Transform would not satisfy this, for everyone, except the person in the rest frame.

Mathematically, we would want to find a function, F, such that

\begin{equation}
F(r,t) = F(L_r(r,t),L_t(r,t))
\end{equation}

where r is the spatial position, t is the temporal coordinate, L_r is the Lorentz transform of the spatial position, and L_t is the Lorentz transform of the temporal coordinate.  In English, F of  (r, t) is the same applied to (r', t'), where (r', t') is the transformed coordinate.

This kind of equation, where two sides of the equation involve F applied to different arguments, is known as a functional equation.  Functional equations hold a special place in my heart for a number of reasons I will not get into, at least not now.  They are also notoriously difficult to solve.

Now, one general strategy (if not the general strategy short of guessing) for solving awful equations is to try to convert them to more familiar equations that have already been solved.  In this case, I wondered what the differential properties of this equation would be.  It turns out that this equation is ideal for asking that kind of question.  To see this, it suffices to solve only the 1+1 dimensional version of this equation.

In this case, we have
\begin{equation}
F(x,t) =F((x-vt)\gamma,(t-\frac{xv}{c^2})\gamma)
\end{equation}

Now, it's a well known fact that as v approaches 0, the Lorentz factor approaches 1.  Hence, we can choose v to set the Lorentz transformed point arbitrarily close to the original.  This seems to be what we are looking for.
In particular, let's replace v with a differential.  Let's call it dv.

Now, we have
\begin{equation}
F(x,t) =F((x-dvt)\gamma,(t-\frac{xdv}{c^2})\gamma)
\end{equation}

Using the chain rule, and straightforward, but tedious algebra, we can now turn this into a partial differential equation:
\begin{equation}
F(x,t) =F(x,t) - \frac{\partial F}{\partial x}(x,t)t\mathrm{d}v - \frac{\partial F}{\partial t}(x,t)\frac{x}{c^2}\mathrm{d}v
\end{equation}

or

\begin{equation}
\frac{\partial F}{\partial x}(x,t)t =  -\frac{\partial F}{\partial t}(x,t)x/c^2
\end{equation}

One thing that is immediately clear, if you know a little vector calculus, is that this is saying that

\begin{equation}
\nabla F \cdot \begin{pmatrix} t \\ \frac{x}{c^2} \end{pmatrix} = 0
\end{equation}

However, I won't use this.  Instead, I will find curvilinear coordinates that can be used to separate the variables.  In particular, I will find a parametrized family of curves such that F is constant along each curve.  If I can find this family, I will have solved the equation.  Specifically, we start with F(x,y) and completely differentiate it to get

\begin{equation}
dF(x,t) = \frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial t}dt
\end{equation}

and set dF(x,t) to 0 so that
\begin{equation}
\frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial t}dt = 0
\end{equation}

Plugging (5) into (8), we see that
\begin{equation}
-\frac{\partial F}{\partial t}\frac{x}{tc^2}dx + \frac{\partial F}{\partial t}dt = 0
\end{equation}

So,
\begin{equation}
xdx = c^2tdt
\end{equation}

Integrating both sides,
\begin{equation}
x^2 = c^2t^2
\end{equation}

Look familiar?  We're not quite done yet, though.  What we have found is that F must be constant along the curve defined by x² - (ct)² = 0 .

In fact, I lied.  That's not the only solution from integrating both sides.  The general solution is
\begin{equation}
x^2 = c^2t^2 + k^2
\end{equation}

where k is an arbitrary constant.  Thus, we have found our family of parametrized curves, k being the parameter.  F would be constant along each of these.

Indeed,
\begin{equation}
F(x,t) = F(\sqrt{x^2 - c^2t^2},0 \ \mathrm{s})
\end{equation}

It follows that if we define an arbitrary function, H, such that
Indeed,
\begin{equation}
F(\sqrt{x^2 - c^2t^2},0 \ \mathrm{s}) = H(\sqrt{x^2 - c^2t^2})
\end{equation}

Then, the general solution of (1) is
\begin{equation}
F(x,t) = H(\sqrt{x^2 - c^2t^2})
\end{equation}

--That is assuming that F is differentiable.  Obviously, one function that satisfies this equation is then the Minkowski metric.  Thus, not only have we inadvertently derived the Minkowski metric in 1+1 dimensions, but we have found that every single possible differential function must also essentially be the Minkowski metric, too.

Conclusions

There are some interesting things to note here.  First, I seem to have shown that the answer to my initial question is negative.  We cannot uniquely identify a spatial point without, essentially, fixing our origin on something physical that moves.  We must either make ourselves the origin of the universe, or concede that origin to someone else who is not at rest with us.  You'll notice, also, that I did not do a higher dimensional derivation, for the simple fact that I actually haven't yet.  Based on the method outlined above, it seems like we might be able to get away with deriving results far more interesting than that which I obtained for 1+1 dimensions.  In particular, at least naively, it doesn't seem like there isn't any good reason why we couldn't construct a function that is vector valued, and not just complex valued.  Finally, I wonder if we can extend this method to solving a larger class of functional equations.  Anyway, that's all for now.  Signing off.

[Update:
Some things have been brought to my attention.  First, this probably should read Minkowski Space-Time Interval, not metric.  Second, I didn't actually derive this interval for every single pair of points, but only for pairs of points where one is the origin.  However, I am almost certain a similar argument can be made for those space-time intervals too.  If and when I get around to it, I will post the full argument.  Mea culpa.  I am sure the frequency of me saying that will increase as the number of people who know better than I do who read this blog increases.  Much appreciation to the guy who pointed these things out.  I am not sure if he wishes to be named.]