Orthogonal group, indefinite orthogonal group, orthochronous stuff

This post appears in the Linear Algebra and Special Relativity courses.

There are several ways to see that the matrices satisfying $A^*A=I$ are related to rotations in some way, other than just expanding out the components like a dumb pygmy chimp -- no, we are the normal chimp:

1. Write it as $A^TIA=I$ -- i.e. the set of matrices that preserve the identity quadratic form. The identity quadratic form corresponds to the $n$-sphere (e.g. a circle), so we're looking for transformations that preserve the $n$-sphere. A clearer way to see this is that preserving the quadratic form $I$ is equivalent to preserving the valuation $x^TIy$ for all $x, y$, i.e. $(Ax)^TI(Ay)=x^Ty$, so it preserves the value of each contour.
2. With the same logic as above, $(Ax)^T(Ay)=x^Ty$, i.e. the preservation of the Euclidean dot product means that all lengths and angles are preserved. These are called "rigid rotations", and are basically the kind of stuff we can do to a sheet of paper without compressing or stretching it in any way -- i.e. if we nudge a vector by a certain angle, every other vector should also be nudged by the same angle.

What kind of transformations preserve the unit sphere? 

The reason this is a good way of understanding things is that there are plenty of other such "dot products" you can define in mathematics, corresponding to different geometries -- each can be based on the bilinear form it preserves, see this later linear algebra article for more details, relating to isomorphisms of such geometries etc.

As for discriminating between rotations and reflections, suppose we define rotations in a completely geometric way -- for a matrix to be a rotation, all its eigenvalues are either 1 or in pairs of unit complex conjugates.

What do the eigenvalues of orthogonal matrices look like? For each eigenvalue, you need $\overline{\lambda}\lambda=1$, i.e. all the eigenvalues are unit complex numbers. If a complex eigenvalue isn't paired with a corresponding conjugate, you will not get a real-valued transformation on $\mathbb{R}^n$. Meanwhile if an eigenvalue of -1 isn't paired with another -1 -- i.e. if there are an odd number of reflections -- you get a reflection. In this sense, the "conjugate eigenvalues" property of rotations can be seen as a generalisation of the "$s_1s_2=r$" property which you may have learned from plane geometry or dihedral groups. The orthogonal (or rather unitary) transformations that do not behave this way are precisely the rotations.

The similarity between unpaired unit complex eigenvalues and unpaired -1's is interesting, by the way -- when thinking about reflections, you might have gotten the idea that reflections are $\pi$-angle rotations in a higher-dimensional space -- like the vector was rotated through a higher-dimensional space and then landed on its reflection -- like it was a discrete snapshot of a process as smooth as any rotation.

Well, now you know what this higher-dimensional space is -- precisely $\mathbb{C}^n$. And the determinant of a unitary matrix also takes a continuous spectrum -- the entire unit circle. In this sense (among other senses) complex linear algebra is more "complete" than real linear algebra. In fact, you will see in Lie theory that the group $SO(n)$ is connected but $O(n)$ is not, while $SU(n)$ and $U(n)$ are both connected. Can you see why?

(original version of above originally posted to math stackexchange)

Well, here, we benefited from the fact that the product of two reflections is a rotation -- so we could just enforce the "even number of flips", i.e. that $\det A=1$, to specify rotations. But what if we're dealing with one of the "generalised geometries" we discussed? What if instead of preserving $I$, we wanted the group $O(m\mid n)$, i.e. that preserves some $\mathrm{diag}(m\mid n)$ with $m$ 1's and $n$ -1's along the diagonal?

Well, then we don't have rotations between the "1"-labeled (spatial) axes and the "-1"-labeled (temporal) axes, only boosts. But compositions between such reflections form rotations! So simply restricting that $\det A = 1$ will -- while still forming a group $SO(m \mid n)$ -- retain all these rotations which can only be understood as compositions of reflections.

So how do we extract the transformations we want? (What transformations do we want? The ones that correspond to changes of reference frame, in special relativity language -- well, in the sense of Lie theory, this means we're looking for the "component connected to the identity" -- do you see why?)

Let's think about this more clearly. Start by noting that not all reflections in spacetime preserve the Minkowski metric $\mathrm{diag}(m\mid n)$ -- only those that preserve the invariant hyperboloids. In the case of 3+1-spacetime, this means infinite spatial reflections and one time-reversal -- in the case of a general $m+n$-spacetime, this means infinite spatial reflections and infinite temporal reflections (in any $m+n-1$-plane whose normal vector is temporal, not to be confused with time-like). When you multiply an odd temporal reflection with an odd spatial reflection, you get an even time-space rotation, which is in $SO(3\mid 1)$.

$$A = \left[ {\begin{array}{*{20}{c}}{{A_t}}&B^T\\C&{{A_s}}\end{array}} \right]$$
(Note on notation: we'll use ${A_T} = \left[ {\begin{array}{*{20}{c}}{{A_t}}&0\\0&I\end{array}} \right]$ and analogously ${A_S} = \left[ {\begin{array}{*{20}{c}}I&0\\0&{{A_s}}\end{array}} \right]$, where $A_t$ and $A_T$ are "basically the same thing", and analogously for $A_s$ and $A_S$ -- in particular $\det A_t=\det A_T$ and $\det A_s=\det A_S$.)

We see the problem: instead of just mandating $\det A=1$, we must mandate that the temporal minor and the spatial minor of the matrix both have determinant 1, $\det A_t=\det A_s = 1$. But this isn't right -- if you have a boost, i.e. some mixing between the space and time co-ordinates, then $A\ne A_TA_S$ and the component determinants are multiplied by a Lorentz factor (even though still $\det A = 1$). So we mandate instead that $\det A_t>0$, $\det A_s>0$ (equivalently $\ge 1$). Such transformations are called the proper orthochronous Lorentz transformations, because in the context of special relativity they are proper Lorentz transformations that do not flip time:

$$SO^{+}(3 \mid 1)=\{A\in O(3 \mid 1) \mid \det A_t >0, \det A_s >0\}$$
OK, how do we show $SO^{+}(m\mid n)$ is a subgroup? You might get the notion that because of the "two sheets hyperbola" topology of the group, the sheet connected to the identity must be a subgroup (and the other sheet a coset) because moving about on the sheet keeps you on the sheet (and that's what group multiplication is -- moving about on the sheet). The formal way to say this is to say that the map $A\mapsto \mathrm{sgn}(\det A_t )$ is a group homomorphism to the cyclic group $\{1,-1\}$, so its kernel is necessarily a normal subgroup (do you see how these are the same thing?).

So the key is to prove that for two matrices satisfying $\mathrm{sgn}(\det A_t )>0$, their product does too. A proof of the $SO^+(m\mid 1)$ case (relevant for relativity) can be found here -- I'm not sure how that proof can be appropriately generalised to $SO^+(m\mid n)$. I've written out the first few steps here:

1. Multiply the two matrices $A$ and $\tilde{A}$ to show $(A\tilde{A})_t=A_t\tilde{A}_t+B^T\tilde{C}$. We want to show the determinant of this is positive.
2. From multiplying out $A^T\eta A=\eta$ and $A\eta A^T=\eta$, we see that $A_t^2-C^TC=A_t^2-B^TB=I$ and analogous for $\tilde{A}$.
3. So $\det((A\tilde{A})_t-A_t\tilde{A}_t)=\det(B^T\tilde{C})=\sqrt{\det(A_t^2-I)\det(\tilde{A}_t^2-I)}$
4. Well, I'm not sure how to proceed at this point. Does $\det(X-PQ)=\det((P^2-I)(Q^2-I))^{1/2}$ imply that $\det P\ge1\land\det Q\ge1\Rightarrow \det X>0$?

Well, I can't think of a way to continue -- and certainly one can think of a much wider category of problems like this, where we have a much simpler topological picture in our heads than rubbish algebra like the above would betray. So we need a topological way of looking at Lie groups.

You might think of just considering something like the orbit of a vector -- e.g. the unit time vector -- under the group for the topology, but this does not fully describe the topology of the group. As an illustration, in the above example, for $n>1$, the orbit of the time vector under $O^+(m\mid n)$ is actually connected (prove this -- you need to count the number of sheets a general hyperbola has), while the entire topology of the group is actually disconnected, as we will see. A simple way to see that these are two different topologies is that spatial rotations/reflections leave the unit time vector unchanged and therefore all correspond to a single point on the orbit.

This will be our starting point to motivate the study of the topology of a Lie group in the Lie theory articles.

Minkowski everything -- spacetime vectors, rapidity

Four-vectors and energy-momentum analogies

Let's look once more at the equation

$$E=\frac{m}{\sqrt{1-v^2}}$$
This looks an awful lot like the equation for time dilation. $E$ is the mass as measured by someone who sees the object moving at $v$ whereas $m$ is the mass as measured by someone who sees the object at rest, e.g. by the object itself.

Similarly, we have the equation $p=vE$, which looks an awful lot like the equation $x=vt$. It therefore makes sense to wonder how far this analogy goes. We could start with analysing the invariant.

Even if I measure the mass of a 1kg rock as 10kg because of my reference frame, I know that if I brought the bag to rest, I would measure it as 1kg. Much like I can tell people's biological age or look at their clocks to determine their proper time, I can look at the moving thing's mass balance and determine its proper mass $m$.

If we just wanted $m$ in terms of the "co-ordinates" $E$ and $p$,

$$m = E\sqrt {1 - {v^2}}  = \sqrt {{E^2} - {v^2}{E^2}}  = \sqrt {{E^2} - {p^2}}$$
$${m^2} = {E^2} - {p^2}$$
Or in 4 dimensions,

$${m^2} = {E^2} - p_x^2 - p_y^2 - p_z^2$$
We call $m$ the "proper mass". In general, "proper" means "as measured in the rest frame" -- proper time, proper length, proper mass, whatever. This equation is also useful because unlike the previous thing, this also works when $v=1$ (i.e. for light), and reduces to $E=pc$.

But this looks an awful lot like the spacetime interval.

That's not all. Consider an object with mass $E$, momentum $p$ and velocity $w=p/E$ in our reference frame $O$. Now boost to a reference frame $O'$ with relative velocity $v$ to $O$. Then the velocity of the object has transformed from $w$ to $\frac{{w - v}}{{1 - wv}}$. So

$$\begin{array}{c}E' = \frac{m}{{\sqrt {1 - {{\left( {\frac{{w - v}}{{1 - wv}}} \right)}^2}} }}\\ = \frac{{m(1 - vw)}}{{\sqrt {{{(1 - wv)}^2} - {{(w - v)}^2}} }}\\ = \frac{{m(1 - vw)}}{{\sqrt {(1 - {w^2})(1 - {v^2})} }}\\ = \gamma (v)\left( {1 - vw} \right)\gamma (w)m\\ = \gamma \left( {1 - vw} \right)E\\ = \gamma (E - vwE)\\E' = \gamma (E - vp)\end{array}$$
And

$$\begin{array}{c}p' = \frac{{m\left( {\frac{{w - v}}{{1 - wv}}} \right)}}{{\sqrt {1 - {{\left( {\frac{{w - v}}{{1 - wv}}} \right)}^2}} }}\\ = \left( {\frac{{w - v}}{{1 - wv}}} \right)E'\\ = \left( {\frac{{w - v}}{{1 - wv}}} \right)\gamma \left( {1 - wv} \right)E\\ = \gamma (wE - vE)\\p' = \gamma (p - vE)\end{array}$$
Or alternatively

$$\left[ \begin{array}{l}{E'}\\{p'}\end{array} \right] = \gamma \left[ {\begin{array}{*{20}{c}}1&{ - v}\\{ - v}&1\end{array}} \right]\left[ \begin{array}{l}E\\p\end{array} \right]$$
In 4 dimensions,

$$\left[ \begin{array}{l}{E'}\\{{p'}_x}\\{{p'}_y}\\{{p'}_z}\end{array} \right] = \left[ {\begin{array}{*{20}{c}}1&{ - v}&{}&{}\\{ - v}&1&{}&{}\\{}&{}&1&{}\\{}&{}&{}&1\end{array}} \right]\left[ \begin{array}{l}E\\{p_x}\\{p_y}\\{p_z}\end{array} \right]$$
Which is precisely the transformation for time and position.

We call vectors that transform like this spacetime vectors or four-vectors. Four-vectors all share the same algebraic properties -- they transform in the same way, they follow vector addition, their norms and in general their dot products are invariant, etc. -- but not necessarily other properties. E.g. energy and momentum have conservation laws, but position and time do not.

The norm of a spacetime vector is taken as:

$${\left| {\left[ {\begin{array}{*{20}{c}}{{q_0}}\\{{q_1}}\\{{q_2}}\\{{q_3}}\end{array}} \right]} \right|^2} = q_0^2 - q_1^2 - q_2^2 - q_3^2$$
Which is distinct from the Euclidean norm, once again telling us that the geometry of spacetime is not Euclidean.

Four-vectors are perhaps the most beautiful example of the symmetry between space and time. They essentially allow you to replace ordinary pre-relativistic vectors like momentum with vectors that also have a time component alongside three spatial components, because the world is 4-dimensional. You just need to find a quantity that behaves with the vector like time behaves with position -- i.e. you need to show the two quantities transform between each other in a Lorentz transformation sort of way.

You end up with truly mind-boggling results -- we already saw that mass is the time-component of momentum, which explains why mass produces inertia -- an object with mass already devotes a lot of its momentum to moving forward in time, so the more the mass, the more of this momentum you need to transform into the spatial direction. This is really what is meant by the transformation law $p'=\gamma(p-vE)$ for mass $E$, generalising the Galilean $p'=p-vE$ (change $E$ to $M$ if that makes you happy). It also explains why massless (meaning zero rest mass) things can move at the speed of light.

Other such four-vectors include:

• Four-force (time-component: $dE/dt$)
• Four-current (time-component: charge density, space-component: current density)
• Electromagnetic four-potential

Other quantities, like the electric and magnetic fields, even though they follow similar invariants (in the electromagnetic field example $E^2-B^2$), do not combine to form four-vectors, but instead objects called "tensors", which we will eventually talk about.

Note that during this transformation (giving something momentum), both mass and momentum increase. Similarly, time dilates when you move something around. This is again because $E^2-p^2$, not $E^2+p^2$ is invariant. The latter would correspond to a circular rotation, with invariant circles, whereas the former corresponds to a skew (a "hyperbolic rotation"), with invariant hyperbolae.

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Rapidity and hyperbolic rotations

Points $(\cos\theta,\sin\theta)$, $(1,\tan\theta)$, $(\cosh\xi,\sinh\xi)$ and $(1,\tanh\xi)$ plotted for varying $\theta$ and $\xi$. While only $\theta$ can be interpreted as an angle too, both $\theta$ and $\xi$ can be interpreted as areas.

This will be a bit of a DIY section, with some guidance.

QUESTION 1

(a) Consider the equation $v' = \frac{{v - w}}{{1 - vw}}$. What trigonometric identity does this remind you of? Could you resolve the differences somehow? (Hint: $v=\tanh\xi$)

(b) Prove that the Lorentz transformations can be written as

$$\begin{array}{l}t' = t\cosh \xi  - x\sinh \xi \\x' = x\cosh \xi  - t\sinh \xi \end{array}$$
(c) Use the hyperbolic analog of angle-addition formulae to show that this is equivalent to, where $\phi=\mathrm{artanh}(x/t)$ is the rapidity of the point $(t,x)$ in the original reference frame.

$$\begin{array}{l}t' = s\cosh (\phi  - \xi )\\x' = s\sinh (\phi  - \xi )\end{array}$$
(d) The above result means that rapidity transforms as $\phi ' = \phi  - \xi $ (which is itself nice, because it tells you that velocity at low speeds is approximately equal to rapidity by a factor of $c$) and $(t,x) = (s\sinh \phi ,s\cosh \phi )$. Relate the former to the idea of invariant hyperbolae and the interpretation of rapidity as an area (hint, hint: area sweeped out by a conic section... Kepler).

QUESTION 2

(a) Results 1(b) and 1(c) are very similar to the effect of rotations on co-ordinate transformations. Here the linear transformations are skews, not rotations, which is why the formulae are different. Draw as many analogs as you can between rotations and skews in linear algebra. Refer to Article 1103-006. Think about the rotational transformation matrix, etc.

(b) Consider (a) directly in the context of special relativity. Pretending that Lorentz boosts are simply rotations (which would imply a metric signature (+,+,+,+) and treat time exactly like space), explain transformations between time and position, etc. Relate this to the actual, skew-y Lorentz transformations. Describe how relativity would behave in this theory.

(c) Write as many relativistic things as you can in the language of rapidity -- the Lorentz factor, the Doppler factor, components of a four-vector (how do $E$ and $p$ look in terms of rapidity), etc.

(d) Graph the hyperbolic functions and explain why the graphs make the results in 2(b) make sense.

(e) How does rapidity interpretation make certain things, like $c$ being the maximum speed, natural?

QUESTION 3

(a) Consider once again the transformation $\phi ' = \phi  - \xi $. What does this tell you about the relative rapidity $\Delta\phi$? Is this invariant, i.e. do all observers agree on what the relative rapidity between two objects is, like observers did on relative velocity in Galilean relativity?

(b) Explain why it would be foolish to expect the quantity $\arctan{v}$, the Euclidean angle (as opposed to rapidity, which we may call the "Minkowskian angle"), to have any physical significance. Think about the quantity $r\arctan{v}$ where $r^2=\Delta t^2+\Delta x^2$ (no minus sign).

It's therefore reasonable to define the dot product on spacetime as $\vec a \cdot \vec b = |\vec a||\vec b|\cosh \Delta \phi $ where $\Delta\phi$ is the relative rapidity/Minkowskian angle/difference in rapidity. This expression implies that $|\vec a|^2=\vec a\cdot\vec a$is manifestly (i.e. obviously) Lorentz invariant, since both norms and relative rapidity are invariant.

(c) Translate this out of rapidity language, i.e. into a language where rapidity is not used as a parameterisation. You should get $a_0b_0-a_1b_1$ (where 0 and 1 are the temporal and spatial components respectively) in two dimensions.

The fact that this modified dot product is invariant under a skew is analogous to how the standard dot product is invariant under rotations ("complex skews"). Indeed, it turns out see that the 4-dimensional Minkowski dot product

$${a_0}{b_0} - {a_1}{b_1} - {a_2}{b_2} - {a_3}{b_3}$$
Is invariant under skews (between the time axis and some other axis) as well as spatial rotations (and all combinations thereof -- i.e. a general Lorentz transformation), as it contains both a "skew-y" part and a "standard dot product-y" part.

---

Some interesting things regarding 2(b):

A circular Lorentz transformation would transform position and time something similar to this:

$$\begin{array}{l}x' = \eta (x - vt)\\t' = \eta (t + vx)\end{array}$$
One can also talk about transforming the positive and negative sides of the axes separately.

$$\begin{array}{l}x' = \eta (x - vt)\\t' = \eta (t + vx)\\ - x' = \eta ( - x - v( - t))\\ - t' = \eta ( - t - v( - x))\end{array}$$

Whereas with hyperbolic functions, there is no sign difference, so you only need to transform twice to return. This is linked to you having to differentiate circular functions four times to return, as opposed to twice for hyperbolic functions, all the sign differences between trigonometric and hyperbolic identities, the whole $ie^{i\theta}$ proof of Euler's formula, etc.

Lorentz transforms lives

Duration

In your years as an infant reading up stuff on wikipedia, you might've seen formulae such as

$$\Delta t = \frac{{t'}}{{\sqrt {1 - {v^2}} }}$$
Or simply $t=\gamma t$. From our knowledge of the Lorentz transformations, we certainly know that the scale on the time axis changes. It would be interesting to find out exactly how this might be observed in real life -- I mean, we know how time as a co-ordinate transforms, but how does duration -- the interval between two points in time -- transform?

You might be tempted to do calculations like ${t'_1} - {t'_2} = \gamma \left( {{t_1} - vx} \right) - \gamma \left( {{t_2} - vx} \right)$, much like people are tempted to sign up for "get rich quick" scams. Doing so woulbe reckless and stupid.

What we need to do is first precisely formulate what we're looking for. We ask:

Suppose there is a clock moving at a constant velocity v relative to me. In my time, how long does it take for the moving clock to tick by 1 second? Assume that we synchronised our clocks in the beginning, i.e. the moving clock and my own clock showed exactly the same time at t = 0 when our positions coincided.

Let's draw a spacetime diagram.

Point A represents the event "moving clock ticks the one second mark". Since lines parallel to the x-axis link points that we (i.e. the stationary observer) consider simultaneous, we draw a horizontal line connecting Point A and the t-axis (remember, we want to find out what tick of our clock is simultaneous, according to us, with 1 second elapsing on the moving clock). Mark this point of intersection B. Then we are interested in finding the duration OB, which we call $t$ in terms of OA, which we call $t'$.

Well, from the Lorentz transformations we know that $t' = \gamma \left( {t - vs} \right)$. We also know, geometrically, that $s = vt$, so we may write $t' = \gamma t \left( {1 - v^2} \right)$, i.e. $t'=t\sqrt(1-v^2)$, or $t=\gamma t'$.

In general, for the duration between two events (where stuff might not pass through the origin at the right time), we may say $\Delta t = \gamma \Delta t'$. This phenomenon is called time dilation.

Distance

We do the same sort of calculation for distances, first operationalising what we mean:

If I hold out a ruler to measure the length of a metre-stick (i.e. something that is 1 metre in its own reference frame) moving at speed v relative to me, what would be the length I measure?

Once again, we draw a spacetime diagram.

This is a little trickier -- when measuring the length of an object, we do so by measuring the two ends of the object simultaneously (or rather, what is simultaneous according to us). However, what is simultaneous for us is not what is simultaneous for the rod. While the rod's reference frame holds O and L as simultaneous, we actually choose another point on the worldline -- K -- as simultaneous with O, because it lies on the x-axis.

Then:

$$x'=\gamma\left(x+SK-vh\right)=x'=\gamma\left(x+vh-vh\right)=\gamma x$$
Hence $x=x'/\gamma$, i.e. length/distance in the direction of motion is contracted under a Lorentz transformation.

Back when I was an infant, I was confused about why it was that time got dilated (multiplied by $\gamma$), while length got contracted (divided by $\gamma$). Well, now you know -- the two phenomena aren't temporal-spatial analogs of each other at all! Length contraction is a result of measuring the two ends of a distance simultaneously

Speed

We have been interested, since the beginning of this series, in finding out how velocities and speeds transform under a Lorentz transformation. Once again, we formulate our question precisely as follows (if you've done DIDYMEUS, you should understand how this forces us to accept logical positivism):

Suppose O' is moving at velocity v with respect to O. In O', the velocity of object K is w. What is the velocity of K in O?

Once again, we draw a spacetime diagram.

So given $x'/t'$, how would we find $x/t$?

Well, here's an idea: we know the Lorentz transformation associated with the velocity $w$. So we just use simple matrix multiplication to find the compound transformation, and figure out what velocity is associated with this transformation.

In other words, we write $L(v)L(w)$ as the co-ordinate system of $K$ with respect to $O$. Performing the matrix product,

$$\begin{array}{c}\gamma (v)\left[ {\begin{array}{*{20}{c}}1&v\\v&1\end{array}} \right]\gamma (w)\left[ {\begin{array}{*{20}{c}}1&w\\w&1\end{array}} \right] = \frac{1}{{\sqrt {\left( {1 - {v^2}} \right)\left( {1 - {w^2}} \right)} }}\left[ {\begin{array}{*{20}{c}}{1 + vw}&{v + w}\\{v + w}&{1 + vw}\end{array}} \right]\\ = \frac{{1 + vw}}{{\sqrt {\left( {1 - {v^2}} \right)\left( {1 - {w^2}} \right)} }}\left[ {\begin{array}{*{20}{c}}1&{\frac{{v + w}}{{1 + vw}}}\\{\frac{{v + w}}{{1 + vw}}}&1\end{array}} \right]\\ = \frac{1}{{\sqrt {\frac{{{v^2}{w^2} + 1 - \left( {{v^2} + {w^2}} \right)}}{{{v^2}{w^2} + 1 + 2vw}}} }}\left[ {\begin{array}{*{20}{c}}1&{\frac{{v + w}}{{1 + vw}}}\\{\frac{{v + w}}{{1 + vw}}}&1\end{array}} \right]\\ = \frac{1}{{\sqrt {\frac{{{v^2}{w^2} + 1 + 2vw - {{\left( {v + w} \right)}^2}}}{{{v^2}{w^2} + 1 + 2vw}}} }}\left[ {\begin{array}{*{20}{c}}1&{\frac{{v + w}}{{1 + vw}}}\\{\frac{{v + w}}{{1 + vw}}}&1\end{array}} \right]\\ = \frac{1}{{\sqrt {1 - {{\left( {\frac{{v + w}}{{1 + vw}}} \right)}^2}} }}\left[ {\begin{array}{*{20}{c}}1&{\frac{{v + w}}{{1 + vw}}}\\{\frac{{v + w}}{{1 + vw}}}&1\end{array}} \right]\\ = \gamma \left( {\frac{{v + w}}{{1 + vw}}} \right)\left[ {\begin{array}{*{20}{c}}1&{\frac{{v + w}}{{1 + vw}}}\\{\frac{{v + w}}{{1 + vw}}}&1\end{array}} \right]\\ = L\left( {\frac{{v + w}}{{1 + vw}}} \right)\end{array}$$
Interestingly, this product is commutative. We may thus write:

$$L\left( v \right)L\left( w \right) = L\left( {\frac{{v + w}}{{1 + vw}}} \right)$$
The reason this is a useful form to write the velocity addition formula is that it conveys the precise positivist sense in which velocity is transformed: as it is observed in the Lorentz transformation of things associated with it.

One may let one of the velocities be $c$ and confirm that $c$ is the same in all reference frames.

What happens when the Lorentz boost is in the direction perpendicular to the direction of motion? Well, distance is not contracted, but time is still dilated, and the velocity is reduced by a factor of $1/\gamma(v)$ where $v$ is the velocity of the observer. This ensures, and you can verify, that the resultant velocity in the new frame doesn't exceed $c$ even by the Pythagorean sum.)

Relativistic doppler shift

This is a surprisingly important lemma to our future derivation of the equation $E=mc^2$, so make sure you're clear with it. Also, it tells you that speeding through a red light might cause it to turn into gamma radiation, if you go fast enough.

We're interested in finding out how the frequency (i.e. colour) of light changes with respect to a moving observer, accounting for all relativistic effects. Frequency is just the inverse of the time period, which is the time interval between two wavefronts.

The red vertical line is the worldline of the source, the blue line is the worldline of a moving observer and the black vertical line is of course the worldline of the observer we consider stationary. The purple lines are the wavefronts emitted by the source. Suppose one wavefront hits the worldlines of both the stationary and moving observers at the origin. Another wavefront hits quite later.

We first find the co-ordinates of the point of intersection between the blue worldline and the worldline of the second wavefront in the stationary co-ordinate system. We simply find the equations of the lines and set: $x=vt$, $t=T-x$ so that:

$$\begin{array}{l}t = T - vt\\(1 + v)t = T\\t = \frac{1}{{1 + v}}T\\x = \frac{v}{{1 + v}}T\end{array}$$
Now we may easily calculate the co-ordinate $t'$, which is the same as $T'$:

$$T' = t' = \gamma \left( {\frac{1}{{1 + v}}T - vx} \right) = \gamma T\left( {\frac{1}{{1 + v}} - \frac{{{v^2}}}{{1 + v}}} \right) = \gamma T\left( {1 - v} \right) = \sqrt {\frac{{1 - v}}{{1 + v}}} T$$
Then

$$f' = \sqrt {\frac{{1 + v}}{{1 - v}}} f$$
This is an important result! It means that even though a photon has the same speed however fast you chase it, you do see it getting less and less energetic.

Sometimes you will see the inverse coefficient $\sqrt{\frac{1-v}{1+v}}$ -- this involves an observer moving away from the source.

How fast would you need to go for a red light to become gamma radiation? Well, it means that $\sqrt {\frac{{1 + v}}{{1 - v}}} =f'/f=10^{19}/(4*10^{14})=2.5*10^4$, i.e. $(1+v)/(1-v)=6.25\times10^8$. Solving for v, one sees that it must be within 1m/s of the speed of light.

(yes, the title is a joke)