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.]