## Solution: Hyperbolic Rotations

Advanced Exercise: Interpret the active Lorentz transformation as a hyperbolic rotation of proper velocity.
The rotation angle
a |a| is the arc length (spacetime, not Euclidean) on a unit hyperbola,
as expressed by

$\Large{\begin{array}{l}V\,\,\, = \,\,\,\gamma \,\left( {1\,\, + \,\,\frac{{\bf{v}}}{c}} \right)\,\,\, = \,\,\,{e^{\bf{a}}}\,\,\, = \,\,\,\cosh {\bf{a}}\,\, + \,\,\sinh {\bf{a}}\\\cosh {\bf{a}}\,\,\, = \,\,\,\cosh a\,\,\, = \,\,\,\gamma \,\,\, = \,\,\,\frac{t}{\tau }\,\,,\,\,\,\,\,\,\,\,\,\,\,\,\sinh {\bf{a}}\,\,\, = \,\,\,{\bf{\hat v}}\sinh a\,\,\, = \,\,\,\gamma \,\,\frac{{\bf{v}}}{c}\,\,\, = \,\,\,\frac{{\bf{x}}}{{c\tau }}\end{array}}$

Construct a diagram to express velocity composition as a product of hyperbolic rotations:

$\Large{{e^{\bf{a}}}\,{e^{\bf{b}}}\,\,\, = \,\,\,{e^{{\bf{a}} + {\bf{b}}}}}$

and compare with the product of Euclidean rotations:

$\Large{{e^{{\bf{i}}\theta }}\,{e^{{\bf{i}}\phi }}\,\,\, = \,\,\,{e^{{\bf{i}}(\theta + \phi )}}}$

The hyperbolic parameter a above is not the same as the angle we draw on a spacetime map between the vertical map axis and the history of a free particle moving at relative speed v/c. However, the similarities between, on the one hand, this hyperbolic "angle" parameter (and its relation to the hyperbolic functions) and, on the other hand, the trig functions, are very strong. To be reminded of these similarities and to strengthen intuition about relationships for the unit hyperbola [(ct)2 - x2 = 1] and unit circle (y2 + x2 = 1), examine the interactive diagram below. We can view the circle or hyperbola alone, move the parameter slider, or use the RESET button in the top right corner of the diagram.

First, uncheck "Show hyperbola" (leaving only the unit circle) to check
basic trig function definitions and features of the circle parameter (angle β, the angle of the gray pie sector). Notice:

The directed length of the blue vertical line on the unit circle is the definition of sin(β)

The directed length of the red horizontal line on the unit circle is the definition of cos(β).

The length of the green vertical line on the unit circle is the definition of the magnitude of tan(β).

By obvious similar triangles, abs[sin(β)cos(β)] = abs[directed blue seg / directed red seg] = abs[tan(β) / 1]
and if we remove the absolute value restriction, we can use the ratio to define the sign of the tangent function
so that it becomes positive in the first and third quadrants and negative in the second and fourth quadrants.

The parameter β (in radians) measures the Euclidean arc length around the unit circle (circumference = 2πR = 2π).

The value of parameter β = twice the area of the gray pie sector β sweeps out (circle area = πR2π).

Now, uncheck "Show circle" and check "Show hyperbola" and notice:

The directed length of the blue horizontal line on the unit hyperbola is the definition of sinh(a)

The directed length of the red vertical line on the unit hyperbola is the definition of cosh(a) (always positive).

The length of the green horizontal line on the unit hyperbola is the definition of tanh(a).

By obvious similar triangles, sinh(a) / cosh(a) = directed blue seg / directed red seg = tanh(a) / 1.

The parameter a (−∞ → 0 at the vertex → ∞) measures the spacetime arc length from the vertex along the unit
hyperbola (see proof below).

{Note: we've used a in the formulas above and a in the interactive diagram --- not all programs support the same fonts.}

The value of parameter a = twice the area of the orange shape bounded by the curve, the vertical axis and the
free particle history line for
X1.

The Euclidean angle parameter α represents the bottom angle of the orange shape, in the range (−π/4, π/4).

Notice especially:  tan(α) = tanh(a) = v/c, always!

This fact and the above similarities justify thinking of
parameter a as a hyperbolic angle, which we'll call
a spacetime rotation angle. That is convenient, but beware not to go too far in the analogy:

sin(α) ≠ sinh(a)          and          cos(α) ≠ cosh(a).

Proof that parameter a measures the spacetime arc length:

$\Large{\begin{array}{l}{\rm{Algebraic}}\,\,{\rm{definitions}}:\,\,\,\,\,\,\cosh a\,\,\, \equiv \,\,\,\frac{{{e^a}\,\, + \,\,{e^{ - a}}}}{2}\,\,,\,\,\,\,\,\,\,\,\,\,\,\,\sinh a\,\,\, \equiv \,\,\,\frac{{{e^a}\,\, - \,\,{e^{ - a}}}}{2}\\\\\\\\{\rm{Consequent}}\,\,{\rm{theorems}}:\,\,\,\,\,\,{\cosh ^2}a\,\, - \,\,{\sinh ^2}a\,\,\, = \,\,\,1\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{da}}\,\cosh a\,\,\, = \,\,\,\sinh a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,~~~~~~~~~~~~~~~~~\frac{d}{{da}}\,\sinh a\,\,\, = \,\,\,\cosh a\\\\\\\,\,\,\,\,~~~~~~~~~(x,\,\,y)\,\,\, = \,\,\,\left( {\sinh a,\,\,\cosh a} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,~~~~~~~~~~(x,\,\,ct)\,\,\, = \,\,\,\left( {\sinh a,\,\,\cosh a} \right)\\\\~~~~~~(dx,\,\,dy)\,\,\, = \,\,\,(\cosh a,\,\,\sinh a)\,da\,\,\,\,\,\,\,\,\,\,~~~~~~\,\,(dx,\,\,cdt)\,\,\, = \,\,\,(\cosh a,\,\,\sinh a)\,da\\\\\,\,\,\,\,\,~~~~~~~~~\,d{S_{Eu}}\,\,\, = \,\,\,\sqrt {d{x^2}\,\, + \,\,d{y^2}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,~~~~~~~~~~~~~~d{S_{ST}}\,\,\, = \,\,\,\sqrt {d{x^2}\,\, - \,\,{{\left( {cdt} \right)}^2}} \,\,\,\left\{ {{\rm{spacelike}}} \right\}\\\\\\\\\\\\{\rm{Standard}}\,\,{\rm{Euclidean}}\,\,{\rm{arclength}}\,\,\,\,\,\,\,\,\,\,\,\,~~~~~~~~~~~{\rm{Standard}}\,\,{\rm{spacetime}}\,\,{\rm{arclength}}\\\\\,\,\,\,\,\,\,\,~~~~{\rm{for}}\,\,\,~~~~~~{y^2}\,\, - \,\,{x^2}\,\,\, = \,\,\,1\,:\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,~~~~~~~~~~~~~~~{\rm{for}}\,\,\,~~~~~~{\left( {ct} \right)^2}\,\, - \,\,{x^2}\,\, = \,\,1\,:\\\\\\{S_{Eu}}\,\,\, = \,\,\,\int_0^a {\sqrt {{{\cosh }^2}a\,\, + \,\,{{\sinh }^2}a} \,\,da} \,\,\,\,\,\,\,\,\,\,\,\,\,\,~~~~{S_{ST}}\,\,\, = \,\,\,\int_0^a {\sqrt {{{\cosh }^2}a\,\, - \,\,{{\sinh }^2}a} \,\,da} \\\,\,\,\,\,\,\,\,\,~~~~~ = \,\,\,{\rm{elliptic}}\,\,{\rm{integral}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,~~~~~~~~~~~~~~~~~ = \,\,\,\int_0^a {da\,\,\, = \,\,\,a} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}}$

Of course, the difference between the two calculations comes down to the minus sign in the definition of spacetime

Thus, it makes sense to interpret the alternative notation for the Lorentz transformation as a spacetime rotation. It has a form very similar to Euclidean rotations and leaves the magnitude |X| cτ  of the spacetime vector unchanged:

$\Large{\begin{array}{l}V\,\,\, = \,\,\,\gamma \,\left( {1\,\, + \,\,\frac{{\bf{v}}}{c}} \right)\,\,\, = \,\,\,{e^{\bf{a}}}\,\,\, = \,\,\,\cosh {\bf{a}}\,\, + \,\,\sinh {\bf{a}}\\V\,\widetilde V\,\,\, = \,\,\,1\,\,\, = \,\,\,{e^{\bf{a}}}\,{e^{ - {\bf{a}}}}\,\,\, = \,\,\,{\cosh ^2}{\bf{a}}\,\, - \,\,{\sinh ^2}{\bf{a}}\end{array}}$

Before further examining passive versus active Lorentz transformations, we will review passive versus active 2D Euclidean rotations in the following interactive diagram. We're using unit CCW bivector i = e1 e2, so that multiplying by this on the right side of a vector gives a CCW rotation. You can drag the black dot at the tip of the vector X to examine different initial vectors. You can change the desired rotation angle in the range -90° to +90° (plus = CCW here) using the slider.

Try activating only the
Active interpretation checkbox first and using the angle slider. That's pretty straightforward. As the text box indicates, we're causing the rotation by multiplying by the rotor eiθ. The text box shows the coordinates of the original vector X and the rotated vector X'. Set up X to have coordinates (x,y) = (4,3) and adjust the slider (the arrow keys can be used to fine tune) so that X' = (3,4). Notice that this is the result of a CCW rotation of 16.3°.

Now, uncheck Active interpretation and check the
Passive interpretation checkbox. The idea is easy enough in principle, but more work algebraically. The vector X in space will not move. Instead, we'll rotate the primed coordinate frame (equivalent to rotating its orthogonal frame-defining unit vectors). Move the slider to θ = 0 and set up X = (4,3) again. The primed and unprimed frames are aligned. If we rotate the primed framed CW by 16.3°, the primed coordinates of X will change in such a way that it gets closer to the primed green y-axis. As seen by the primed frame, the vector seems to have rotated CCW. The vector hasn't moved and its coordinates in the unprimed frame are still (4,3), but its coordinates in the rotated primed frame are (3,4), as if it had rotated. When we use the symbol X' in the passive interpretation, we are referring to using primed coordinates to represent the stationary vector.

Let's make sure we understand how calculations get the results in this specific example. We want to rotate the primed axes by 16.3° (0.284 rad) CW, so the primed frame orthogonal unit vectors should get right-multiplied by e-0.284i. Then we get the new primed coordinates by using the orthonormal properties of these unit vectors (we'll use the equivalent spacetime ideas below):

$\Large{\begin{array}{l}{\rm{Active}}\,\,{\rm{CCW}}\,\,{\rm{rotation}}\,\,{\rm{angle}}:\,\,\,\,\,\,\\\\\,\,\,\,~~~~~~\,\,\,\,\,\,\,\,{\bf{\hat a\hat b}}\,\,\, = \,\,\,\left( {\frac{{4{{{\bf{\hat e}}}_1}\,\, + \,\,3{{{\bf{\hat e}}}_2}}}{5}} \right)\,\,\left( {\frac{{3{{{\bf{\hat e}}}_1}\,\, + \,\,4{{{\bf{\hat e}}}_2}}}{5}} \right)\,\,\, = \,\,\,\frac{{24}}{{25}}\,\, + \,\,\frac{7}{{25}}\,{\bf{i}}\,\,\, = \,\,\,{e^{0.284{\bf{i}}}}\,\,\, = \,\,\,{e^{{\bf{i}}\theta }}\\\\\\{\rm{Rotation}}\,\,{\rm{CW}}\,\,{\rm{of}}\,\,{\rm{primed}}\,\,{\rm{axes}}\,\,{\rm{unit}}\,\,{\rm{vectors}}:\\\,\,\,\,\,~~~~~~\,\,\,\,\,\,\,{{{\bf{\hat e}}}_1}^\prime \,\,\, = \,\,\,{{{\bf{\hat e}}}_1}\,{e^{ - \,{\bf{i}}\theta }} = \,\,\,{{{\bf{\hat e}}}_1}\,{e^{ - \,0.284{\bf{i}}}}\,\,\, = \,\,\,{{{\bf{\hat e}}}_1}\,\left( {\frac{{24}}{{25}}\,\, - \,\,\frac{7}{{25}}\,{\bf{i}}} \right)\,\,\, = \,\,\,0.96\,{{{\bf{\hat e}}}_1}\,\, - \,\,0.28\,{{{\bf{\hat e}}}_2}\\\\\,\,\,\,~~~~~~\,\,\,\,\,\,\,\,{{{\bf{\hat e}}}_2}^\prime \,\,\, = \,\,\,{{{\bf{\hat e}}}_2}\,{e^{ - \,{\bf{i}}\theta }} = \,\,\,{{{\bf{\hat e}}}_2}\,{e^{ - \,0.284{\bf{i}}}}\,\,\, = \,\,\,{{{\bf{\hat e}}}_2}\,\left( {\frac{{24}}{{25}}\,\, - \,\,\frac{7}{{25}}\,{\bf{i}}} \right)\,\,\, = \,\,\,0.28\,{{{\bf{\hat e}}}_1}\,\, + \,\,0.96\,{{{\bf{\hat e}}}_2}\\\\{\rm{Orthonormality}}\,\,{\rm{conditions}}:\\\,\,\,\,~~~~~~\,\,\,\,\,\,\,\,{\bf{u}} \cdot {\bf{v}}\,\,\, = \,\,\,{\left\langle {{\bf{u}}\,{\bf{v}}} \right\rangle _0}\,\,\, \equiv \,\,\,\left\langle {{\bf{u}}\,{\bf{v}}} \right\rangle \\\,\,\,\,~~~~~~\,\,\,\,\,\,\,\,{{{\bf{\hat e}}}_1} \cdot {{{\bf{\hat e}}}_1}\,\,\, = \,\,\,{{{\bf{\hat e}}}_2} \cdot {{{\bf{\hat e}}}_2}\,\,\, = \,\,\,{{{\bf{\hat e}}}_1}^\prime \cdot {{{\bf{\hat e}}}_1}^\prime \,\,\, = \,\,\,{{{\bf{\hat e}}}_2}^\prime \cdot {{{\bf{\hat e}}}_2}^\prime \,\,\, = \,\,\,1\\\,\,\,\,~~~~~~\,\,\,\,\,\,\,\,{{{\bf{\hat e}}}_1} \cdot {{{\bf{\hat e}}}_2}\,\,\, = \,\,\,{{{\bf{\hat e}}}_1}^\prime \cdot {{{\bf{\hat e}}}_2}^\prime \,\,\, = \,\,\,0\\\\{\rm{Coordinates}}\,\,{\rm{of}}\,\,{\bf{X}}\,\,{\rm{in}}\,\,{\rm{the}}\,\,{\rm{primed}}\,\,{\rm{frame}}:\\\,\,\,\,~~~~~~\,\,\,\,\,\,\,\,{X_1}^\prime \,\,\, = \,\,\,{\bf{X}} \cdot {{{\bf{\hat e}}}_1}^\prime \,\,\, = \,\,\,\left( {4{{{\bf{\hat e}}}_1}\,\, + \,\,3{{{\bf{\hat e}}}_2}} \right) \cdot \left( {0.96\,{{{\bf{\hat e}}}_1}\,\, - \,\,0.28\,{{{\bf{\hat e}}}_2}} \right)\,\,\, = \,\,\,3\\\,\,~~~~~~\,\,\,\,\,\,\,\,\,\,{X_2}^\prime \,\,\, = \,\,\,{\bf{X}} \cdot {{{\bf{\hat e}}}_2}^\prime \,\,\, = \,\,\,\left( {4{{{\bf{\hat e}}}_1}\,\, + \,\,3{{{\bf{\hat e}}}_2}} \right) \cdot \left( {0.28\,{{{\bf{\hat e}}}_1}\,\, + \,\,0.96\,{{{\bf{\hat e}}}_2}} \right)\,\,\, = \,\,\,4\\\,\,\,\,\,~~~~~~\,\,\,\,\,\,\,{\bf{X'}}\,\,\, \equiv \,\,\,3{{{\bf{\hat e}}}_1}^\prime \,\, + \,\,4{{{\bf{\hat e}}}_2}^\prime \,\,\, = \,\,\,4{{{\bf{\hat e}}}_1}\,\, + \,\,3{{{\bf{\hat e}}}_2}\,\,\, = \,\,\,{\bf{X}}\end{array}}$

In the active interpretation, X' is a different vector, rotated from X. In the passive interpretation, X' is the same vector (written using primed coordinates) as vector X (written using unprimed coordinates). You could activate both checkboxes in the above interactive diagram to notice that the two sets of coordinates are independent of which interpretation we choose.

Now we're ready to investigate active versus passive interpretations for Lorentz transformations in the following interactive diagram. The slider now picks the spacetime rotation angle (often called the rapidity, since tanh(a) = v/c as established above), whose values we've chosen to vary from -2.5 to 2.5. You can drag the black dot at the tip of the event vector X to examine different initial events. Increasing the rotation angle makes the rotated event X' more spacelike than X. This looks like a CW rotation on the spacetime map, but remember, unlike spatial rotations above, for our spacetime maps we are considering one dimensional motion, so what it means to be rotating CW is that the worldline from O to X' is tilted more towards the light cone.

Try activating only the Active interpretation checkbox first and using the ST angle slider. As the text box indicates, we're causing the rotation by multiplying by the spacetime rotor ea (= V). For our one dimensional case, this commutes with all of our events, so multiplying on the left is just a chosen convention. The text box shows the coordinates of the original vector X and the rotated vector X', both with respect to the frame being used to draw the map.

Notice that the changing rotation angle pushes
X' along the hyperbola defined by the proper time of X.
Of course, we wouldn't be justified in using the term
rotation if this weren't the case. We require the magnitude
(proper time) of the rotated event to be the same as for the original event.

Set up
X = 4 + e1 and adjust the slider (the arrow keys can be used to fine tune) so that X' = 5 + e1√10 = 5 + 3.16 e1.
Notice that this is the result of an
ST rotation of a = 0.49. That represents the angle between the X and X' event vectors
(an angle of zero
brings the events back into coincidence). In this case, the angle of event vector X is tanh-1(v/c)
= tanh-1(1/4) = 0.255
and the angle of the event vector X' is tanh-1(3.16/5) = 0.745. Sure enough, the difference
between these values is
0.490

Now, uncheck Active interpretation and check the
Passive interpretation checkbox. If you haven't changed X or the slider from the values in the previous paragraph, you'll notice that, even though the interpretation has changed and a primed spacetime grid is now visible, the values of X and X' haven't changed. The event vector X will not move. Instead, we'll apply the opposite spacetime rotation to the primed coordinate frame (equivalent to rotating its orthogonal frame-defining unit vectors). Move the slider to a = 0 and make sure X = 4 + e1. The primed and unprimed frames are aligned. Now, instead of applying ea to X, we will apply e-a to the primed system's orthonormal frame vectors (below, we show algebraically how to implement this idea to recover the numbers in the previous paragraph, analogous to the algebra summary given above for the passive spatial rotation case). Graphically, as you move the slider to  a = + 0.49, the interactive diagram applies e-0.49e1 to the primed axes, making the primed axes' grid tilt further away from the first quadrant lightline. If you read the coordinates of the event vector on the vertical & horizontal unprimed axes, you get X4 + e1, while if you read the same event vector using the tilted primed grid, you get X' = 5 + 3.16 e1.

However, there are some important differences to keep in mind when dealing with spatial orthonormal vectors versus spacetime orthonormal vectors. First, what notation should be use for a unit spacetime frame vector? For a spatial frame, we use
e1 & e2 for the unprimed frame and e1' & e2' for the primed frame. Since spacetime vectors have scalar and vector parts, the notation needs to be a little different. In the following formulas, we'll use 1 & e1 for the unprimed frame and 1' & e1' for the primed frame. More importantly, we need a workable definition for a spacetime dot product. As proposed in NFCM, section 9-2, under Spacetime Trig, for the dot product between two spacetime vectors we'll define A•B = <AB̃>, that is, the spacetime dot product is the scalar part of one vector times the spacetime conjugate of the other vector (doesn't matter which order). Welcome consequences of this definition are 1•1 = 1e1e1 = - 1 (as we should have for a spacelike unit vector), 1e1 = e11 = 0, and X•X = (cτ)2 = XX̃. Then, to work out numbers for the example above:

$\Large{\begin{array}{l}{\rm{Active}}\,\,{\rm{spacetime}}\,\,{\rm{rotation}}\,\,{\rm{angle}}:\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,X'\,\,\, = \,\,\,{e^{\bf{a}}}\,X\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,{e^{\bf{a}}}\,\,\, = \,\,\,\frac{{X'\,\widetilde X}}{{X\,\widetilde X}}\,\,\, = \,\,\,\frac{{\left( {5\,\, + \,\,\sqrt {10} \,{{{\bf{\hat e}}}_1}} \right)\,\left( {4\,\, - \,\,{{{\bf{\hat e}}}_1}} \right)}}{{15}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,~~~~~~~~~~~~~~~~~~\,\,\,\,\, = \,\,\,1.1225\,\, + \,\,0.5099\,{{{\bf{\hat e}}}_1}\,\,\, = \,\,\,{e^{0.49{{{\bf{\hat e}}}_1}}}\\\\{\rm{Backward}}\,\,{\rm{ST}}\,\,{\rm{rotation}}\,\,{\rm{of}}\,\,{\rm{primed}}\,\,{\rm{axes}}\,\,{\rm{unit}}\,\,{\rm{vectors}}:\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{1^\prime }\,\,\, = \,\,\,{e^{ - \,{\bf{a}}}}\,1\,\,\, = \,\,\,{e^{ - \,0.49{{{\bf{\hat e}}}_1}}}\,\,\, = \,\,\,1.1225\,\, - \,\,0.5099\,{{{\bf{\hat e}}}_1}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{{\bf{\hat e}}}_1}^\prime \,\,\, = \,\,\,{e^{ - \,{\bf{a}}}}\,{{{\bf{\hat e}}}_1}\,\,\, = \,\,\,{e^{ - \,0.49{{{\bf{\hat e}}}_1}}}\,{{{\bf{\hat e}}}_1}\,\,\, = \,\,\, - \,\,0.5099\,\, + \,\,1.1225\,{{{\bf{\hat e}}}_1}\\\\{\rm{Orthonormality}}\,\,{\rm{conditions}}:\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,A \cdot B\,\,\, \equiv \,\,\,{\left\langle {A\,\widetilde B} \right\rangle _0}\,\,\, \equiv \,\,\,\left\langle {A\,\widetilde B} \right\rangle \,\,\, = \,\,\,\left\langle {B\,\widetilde A} \right\rangle \,\,\, = \,\,\,B \cdot A\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \cdot 1\,\,\, = \,\,\,{1^\prime } \cdot {1^\prime }\,\,\, = \,\,\,1\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{{\bf{\hat e}}}_1} \cdot {{{\bf{\hat e}}}_1}\,\,\, = \,\,\,{{{\bf{\hat e}}}_1}^\prime \cdot {{{\bf{\hat e}}}_1}^\prime \,\,\, = \,\,\, - \,1\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \cdot {{{\bf{\hat e}}}_1}\,\,\, = \,\,\,{1^\prime } \cdot {{{\bf{\hat e}}}_1}^\prime \,\,\, = \,\,\,0\\\\{\rm{Coordinates}}\,\,{\rm{of}}\,\,X\,\,{\rm{in}}\,\,{\rm{the}}\,\,{\rm{primed}}\,\,{\rm{frame}}:\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,X'\,\,\, = \,\,\,{X_1}^\prime \,\, + \,\,{X_2}^\prime \,{{{\bf{\hat e}}}_1}^\prime \\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{X_1}^\prime \,\,\, = \,\,\,X \cdot {1^\prime }\,\,\, = \,\,\,\left\langle {\left( {4\,\, + \,\,{{{\bf{\hat e}}}_1}} \right)\,\left( {1.1225\,\, + \,\,0.5099\,{{{\bf{\hat e}}}_1}} \right)} \right\rangle \,\,\, = \,\,\,5\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{X_2}^\prime \,\,\, = \,\,\, - \,\,X \cdot {{{\bf{\hat e}}}_1}^\prime \,\,\, = \,\,\, - \,\,\left\langle {\left( {4\,\, + \,\,{{{\bf{\hat e}}}_1}} \right)\,\left( { - \,\,0.5099\,\, - \,\,1.1225\,{{{\bf{\hat e}}}_1}} \right)} \right\rangle \,\,\, = \,\,\,3.16\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,X'\,\,\, \equiv \,\,\,5\,\left( {{1^\prime }} \right)\,\, + \,\,3.16\,{{{\bf{\hat e}}}_1}^\prime \,\,\, = \,\,\,4\,\left( 1 \right)\,\, + \,\,{{{\bf{\hat e}}}_1}\,\,\, = \,\,\,X\end{array}}$

In the active interpretation,
X' is a different event vector, rotated from X. In the passive interpretation, event X' is the same vector (written using primed coordinates) as event vector X (written using unprimed coordinates). You could activate both checkboxes in the above interactive diagram to notice that the two sets of coordinates are independent of which interpretation we choose.

It's easy enough to visualize a 2D spatial rotation of axes (just twist our heads or twist the paper used to draw them), but rotating spacetime axes makes the time and space axes both twist the same amount towards (or away from) the lightcone. How can we correctly draw the resulting skewed spacetime grid for the rotated primed axes as seen in the interactive diagram above? We can simply
look at how the primed frame unit vectors are transformed. Let's try another example which we can check out using the previous interactive diagram:

$\Large{\begin{array}{l}{\rm{ST}}\,\,{\rm{rotation}}\,\,{\rm{of}}\,\,{\rm{primed}}\,\,{\rm{axes}}\,\,{\rm{unit}}\,\,{\rm{vectors}}\,\,{\rm{for}}\,\,{\rm{v/c}}\,\, = \,\,{\rm{0.6}}:\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\gamma \,\,\, = \,\,\,{\left( {1\,\, - \,\,{v^2}/{c^2}} \right)^{ - 1/2}}\,\,\, = \,\,\,1.25\,\,\, = \,\,\,\cosh \,{\bf{a}}\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{\bf{a}}\,\,\, = \,\,\,0.693\,\,{{{\bf{\hat e}}}_1}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{1^\prime }\,\,\, = \,\,\,{e^{\bf{a}}}\,(1)\,\,\, = \,\,\,{e^{0.693\,{{{\bf{\hat e}}}_1}}}\,\,\, = \,\,\,1.25\,\, + \,\,0.75\,\,{{{\bf{\hat e}}}_1}\,\,\, = \,\,\,\gamma \,\, + \,\,{{{\bf{\hat e}}}_1}\,\gamma v/c\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{{\bf{\hat e}}}_1}^\prime \,\,\, = \,\,\,{e^{\bf{a}}}\,\,{{{\bf{\hat e}}}_1}\,\,\, = \,\,\,{e^{0.693\,{{{\bf{\hat e}}}_1}}}\,\,{{{\bf{\hat e}}}_1}\,\,\, = \,\,\,0.75\,\, + \,\,1.25\,{{{\bf{\hat e}}}_1}\,\,\, = \,\,\,\gamma v/c\,\, + \,\,\gamma \,{{{\bf{\hat e}}}_1}\end{array}}$

To set this example up on the previous interactive diagram,
select the Passive interpretation checkbox and move the slider to −0.69 (remember in this diagram, using the passive interpretation, we rotate the primed axes by minus the ST rotation angle so that the event vector X seems to rotate by a positive ST angle). The blue dashed line is the new primed space axis (with corresponding blue dotted parallel grid lines) and the green dashed line is the new primed time axis (with corresponding green dotted parallel grid lines). On paper, we can just fill in the grid with copies of the unit vectors just found at each tail and tip of other unit vectors. For a geometry program, we simply need sets of two points for line drawing commands. And, given sets of two points, we can also write down equations for families of grid lines:

$\Large{\begin{array}{l}\left\{ {{\rm{Standard}}\,\,{\rm{[x,}}\,\,{\rm{y]}}\,\,{\rm{graphical}}\,\,{\rm{point}}\,\,{\rm{notation,}}\,\,{\rm{or}}\,\,{\rm{[x,}}\,\,{\rm{ct].}}\,\,\,\,i\,\,\, \equiv \,\,\,{\rm{an}}\,\,{\rm{integer.}}} \right\}\\\\\\{\rm{For}}\,\,{\rm{blue}}\,\,{\rm{lines}}\,\,{\rm{parallel}}\,\,{\rm{to}}\,\,{\rm{primed}}\,\,{\rm{space}}\,\,{\rm{axis:}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,{\rm{pts}}\,\,\left( {{\rm{tail}}\,\,{\rm{\& }}\,\,{\rm{tip}}\,\,{\rm{of}}\,\,{{{\bf{\hat e}}}_1}^\prime } \right):\,\,\,\,\,\,\,\,\,\,\,\,\left[ {0,\,\,0} \right]\,\,\,\,\,\,\,\,\& \,\,\,\,\,\,\,\,\gamma \,\,\left[ {1,\,\,v/c} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,{\rm{multiples}}\,\,{\rm{of}}\,\,{\rm{1'}}\,\,{\rm{at}}\,\,{\rm{tail}}\,\,{\rm{\& }}\,\,{\rm{tip:}}\,\,\,\,\,\,\,\,\,\,\,\,{\rm{ + }}\,\,i\,\gamma \,\,\left[ {v/c,\,\,1} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\,\,{\rm{point}}\,\,{\rm{pair}}\,\,{\rm{for}}\,\,{\rm{any}}\,\,{\rm{blue}}\,\,{\rm{grid}}\,\,{\rm{line:}}\,\,\,\,\,\,\,\,\,\,\,\,\gamma \,\,\left[ {i\,\,\frac{v}{c},\,\,\,i} \right]\,\,\,\,\,\,\,\,\& \,\,\,\,\,\,\,\,\gamma \,\,\left[ {1\,\, + \,\,i\,\,\frac{v}{c},\,\,\,i\,\, + \,\,\frac{v}{c}} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{Geometry}}\,\,{\rm{software}}\,\,{\rm{can}}\,\,{\rm{plot}}\,\,{\rm{lines}}\,\,{\rm{through}}\,\,{\rm{these}}\,\,{\rm{sets}}\,\,{\rm{of}}\,\,{\rm{points.}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{Standard}}\,\,{\rm{line}}\,\,{\rm{equations:}}\,\,\,\,\,\,\,\,\,\,\,\,ct\,\,\, = \,\,\,\frac{v}{c}\,x\,\, + \,\,\frac{i}{\gamma }\,\,\, = \,\,\,0.6\,x\,\, + \,\,0.8\,i\,\,\,{\rm{(in}}\,\,{\rm{our}}\,\,{\rm{example)}}\\\\\\{\rm{For}}\,\,{\rm{green}}\,\,{\rm{lines}}\,\,{\rm{parallel}}\,\,{\rm{to}}\,\,{\rm{primed}}\,\,{\rm{time}}\,\,{\rm{axis:}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,{\rm{pts}}\,\,\left( {{\rm{tail}}\,\,{\rm{\& }}\,\,{\rm{tip}}\,\,{\rm{of}}\,\,{1^\prime }} \right):\,\,\,\,\,\,\,\,\,\,\,\,\left[ {0,\,\,0} \right]\,\,\,\,\,\,\,\,\& \,\,\,\,\,\,\,\,\gamma \,\,\left[ {v/c,\,\,1} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,{\rm{multiples}}\,\,{\rm{of}}\,\,{{{\bf{\hat e}}}_1}^\prime \,\,{\rm{at}}\,\,{\rm{tail}}\,\,{\rm{\& }}\,\,{\rm{tip:}}\,\,\,\,\,\,\,\,\,\,\,\,{\rm{ + }}\,\,i\,\gamma \,\,\left[ {1,\,\,v/c} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\,\,{\rm{point}}\,\,{\rm{pair}}\,\,{\rm{for}}\,\,{\rm{any}}\,\,{\rm{green}}\,\,{\rm{grid}}\,\,{\rm{line:}}\,\,\,\,\,\,\,\,\,\,\,\,\gamma \,\,\left[ {i,\,\,\,i\,\,\frac{v}{c}} \right]\,\,\,\,\,\,\,\,\& \,\,\,\,\,\,\,\,\gamma \,\,\left[ {i\,\, + \,\,\frac{v}{c}\,,\,\,\,1\,\, + \,\,i\,\,\frac{v}{c}} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{Geometry}}\,\,{\rm{software}}\,\,{\rm{can}}\,\,{\rm{plot}}\,\,{\rm{lines}}\,\,{\rm{through}}\,\,{\rm{these}}\,\,{\rm{sets}}\,\,{\rm{of}}\,\,{\rm{points.}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{Standard}}\,\,{\rm{line}}\,\,{\rm{equations:}}\,\,\,\,\,\,\,\,\,\,\,\,ct\,\,\, = \,\,\,\frac{c}{v}\,x\,\, - \,\,\frac{c}{v}\,\,\frac{i}{\gamma }\,\,\, = \,\,\,\frac{5}{3}\,\,x\,\, + \,\,\frac{4}{3}\,\,i\,\,\,{\rm{(in}}\,\,{\rm{our}}\,\,{\rm{example)}}\end{array}}$

As we can verify using the last interactive diagram, for this example the blue grid lines have ct-axis interrupts at integer multiples of 0.8 while the green grid lines have ct-axis interrupts at integer multiples of 4/3. The grid lines set up in this last example correspond to the case of the primed frame (perhaps the favorite rocket ship) having a relative velocity of 0.6c relative to the unprimed frame, the kind of situation we've encountered in earlier sections. As we now see, we can think of this as giving the rocket frame an active positive spacetime rotation (often called a boost) of ea, where a = 0.693 e1.

In the preceding interactive diagram and associated paragraphs, we've examined two different interpretations for applying Lorentz transformations to spacetime events. We can choose which interpretation to adopt. In either case we can use alternative notations for proper velocity
V = γ(1 + v/c) = γ + γv/c = cosh a + sinh aea. Our typical choice of interpretation for spacetime events is the passive one, preferring to think of one system as a frame which has been boosted with respect to the unprimed frame. Note that even in this interpretation, we are applying V = ea to the unprimed frame vectors to get the primed frame vectors, which gives the resulting coordinate transformation for the event as X' Ṽ X = e-a X. That is, the passive interpretation of coordinate transformation involves an active transformation of the frame-defining vectors. Also, we've seen in the previous section on velocity composition that this last equation leads to U'Ṽ U, or U = V U'

A good way to interpret this is to say that the particle's worldline is tilted (rotated) by the amount
U' = ea with respect to the primed frame, and the primed frame's axes are tilted by the amount V = eb with respect to the unprimed frame, so that the particle's worldline is tilted by U = ec VUeeb = ea+b with respect to the unprimed frame. For the composition of Euclidean rotations,

$\Large{{e^{{\bf{i}}\theta }}\,{e^{{\bf{i}}\phi }}\,\,\, = \,\,\,{e^{{\bf{i}}(\theta + \phi )}}}$

we can understand this as the addition of arc lengths on a unit circle. For the composition of active Lorentz transformations,

$\Large{{e^{\bf{a}}}\,{e^{\bf{b}}}\,\,\, = \,\,\,{e^{{\bf{a}} + {\bf{b}}}}}$

we can understand this as the addition of arc lengths on a unit hyperbola. We've seen that Galilean velocity composition is additive, while spacetime velocity addition is not, so representation of spacetime velocity composition with this notation is convenient, because rapidities are additive.

Below is one more interactive diagram for more practice with active spacetime rotations. The labels were picked to represent events being actively rotated. Three sets of hyperbolas are drawn. Notice how the rotation angle affects whole worldlines, as expected, so that the spacetime rotor works for events whether they have unit proper length or not. If we focus on the unit hyperbola, the symbols might just as well represent proper velocities. For graphical representation of the previous paragraph, let
X2 represent the particle's proper velocity, let X1 represent the primed system's proper velocity, and the unprimed frame is represented by the untilted axes of the diagram. We have sliders for a, the angle from zero for X1, and b, the angle from zero for X2. Underneath the sliders, we can see δ, the angle between X1 and X2. The symbol correspondence with the previous paragraph becomes  U' = X2 (with respect to X1)eδ
V = X1ea, and  U = X2 (with respect to the vertical) eVU' = ea+δ

Move the top slider to a = -1.1 and estimate the relationship between X1 and X1'''. X1 is riding on the red hyperbola (cτ = 1), while X1''' is riding on the orange hyperbola (cτ = 3). By now, we should be used to differences between spacetime and Euclidean distances. For example, the spacetime length of vector X1 (from the origin to the point) is 1, while its Euclidean length is 4.58. So, it may be comforting to realize that the ratio of lengths of X1''' to X1 to is 3, whether calculated by spacetime or Euclidean length formulas.

Here is a final numerical example to check out with the help of the last two interactive diagrams.

A space shuttle moves with velocity
u'/c = 0.6 with respect to its spaceship. It records an event X'' = 10 - 2e1 = 4√6 e-0.203e1 in its own coordinates. The spaceship moves with velocity v/c = 0.8 with respect to Earth. What are the coordinates of the event in the spaceship frame and in the earth frame? What is the velocity of the shuttle with respect to Earth?

$\Large{\begin{array}{l}{\rm{Shuttle:}}\,\,\,\,\,\,\,\,{\bf{u'}}/c\,\,\, = \,\,\,0.6\,\,{{{\bf{\hat e}}}_1}\,,\,\,\,\,\,\,\,\,{\gamma _{u'}}\,\,\, = \,\,\,5/4\,,\,\,\,\,\,\,\,\,U'\,\,\, = \,\,\,\frac{5}{4}\,\, + \,\,\frac{3}{4}\,\,{{{\bf{\hat e}}}_1}\,\,\, = \,\,\,{e^{0.693\,{{{\bf{\hat e}}}_1}}}\\\,\,\,\,\,~~~~~~~~~~\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,X''\,\,\, = \,\,\,10\,\, - \,\,2\,\,{{{\bf{\hat e}}}_1}\,\,\, = \,\,\,4\sqrt 6 \,\,{e^{ - \,0.203\,{{{\bf{\hat e}}}_1}}}\,\,\, = \,\,\,\widetilde {U'}\,X'\\\\{\rm{Spaceship}}:\,\,\,\,\,\,\,\,{\bf{v}}/c\,\,\, = \,\,\,0.8\,\,{{{\bf{\hat e}}}_1}\,,\,\,\,\,\,\,\,\,{\gamma _v}\,\,\, = \,\,\,5/3\,,\,\,\,\,\,\,\,\,V\,\,\, = \,\,\,\frac{5}{3}\,\, + \,\,\frac{4}{3}\,\,{{{\bf{\hat e}}}_1}\,\,\, = \,\,\,{e^{1.099\,{{{\bf{\hat e}}}_1}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,~~~~~~~~\,X'\,\,\, = \,\,\,U'\,X''\,\,\, = \,\,\,{e^{0.693\,{{{\bf{\hat e}}}_1}}}\,4\sqrt 6 \,\,{e^{ - \,0.203\,{{{\bf{\hat e}}}_1}}}\,\,\, = \,\,\,4\sqrt 6 \,\,{e^{0.490\,{{{\bf{\hat e}}}_1}}}\,\,\, = \,\,\,11\,\, + \,\,5\,\,{{{\bf{\hat e}}}_1}\,\,\, = \,\,\,\widetilde V\,X\\\\\\{\rm{Earth}}:\,\,\,\,\,\,\,\,X\,\,\, = \,\,\,V\,X'\,\,\, = \,\,\,{e^{1.099\,{{{\bf{\hat e}}}_1}}}\,4\sqrt 6 \,\,{e^{0.490\,{{{\bf{\hat e}}}_1}}}\,\,\, = \,\,\,4\sqrt 6 \,\,{e^{1.589\,{{{\bf{\hat e}}}_1}}}\,\,\, = \,\,\,25\,\, + \,\,23\,\,{{{\bf{\hat e}}}_1}\\\\\\{\rm{Shuttle}}\,{\rm{ - }}\,{\rm{Earth}}\,\,{\rm{velocity}}:\,\,\,\,\,\,\,\,X\,\,\, = \,\,\,U\,X''\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{One}}\,\,{\rm{way}}:\,\,\,\,\,\,\,\,U\,\,\, = \,\,\,\frac{{X\,\widetilde {X''}}}{{X''\,\widetilde {X''}}}\,\,\, = \,\,\,\frac{{\left( {25\,\, + \,\,23\,\,{{{\bf{\hat e}}}_1}} \right)\,\left( {10\,\, + \,\,2\,\,{{{\bf{\hat e}}}_1}} \right)}}{{96}}\,\,\, = \,\,\,\frac{{37}}{{12}}\,\, + \,\,\frac{{35}}{{12}}\,\,{{{\bf{\hat e}}}_1}\,\,\, = \,\,\,{e^{1.792\,{{{\bf{\hat e}}}_1}}}\\\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{Easier}}\,\,{\rm{way}}:\,\,\,\,\,\,\,\,4\sqrt 6 \,\,{e^{1.589\,{{{\bf{\hat e}}}_1}}}\,\,\, = \,\,\,U\,\,4\sqrt 6 \,\,{e^{ - \,0.203\,{{{\bf{\hat e}}}_1}}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,U\,\,\, = \,\,\,{e^{1.792\,{{{\bf{\hat e}}}_1}}}\,\,\, = \,\,\,V\,U'\end{array}}$

On the last interactive diagram, you can set
a = 1.1 (spaceship angle with respect to Earth) and adjust b so that δ = 0.69 (shuttle angle with respect to the spaceship). Then you can read off b = 1.79 (shuttle angle with respect to Earth), agreeing with our velocity composition results above.

To check the spaceship-to-shuttle coordinate transformations, go back the diagram before the last and set the checkboxes for passive interpretation. Since
11 + 5 e1 is out of range of the diagram, cut the spaceship's event coordinates in half, that is, move the black event dot to 5.5 + 2.5 e1. Remember, in that mode of that diagram, whatever we set the angle slider to, the frame will be rotated the opposite way, and since the angle of the shuttle frame with respect to the spaceship frame is +0.693, we should move the slider to −0.69. Notice the transformed coordinates are 5 - e1, which corresponds to the shuttle's coordinates of 5 - 2 e1.

Similarly, to check the earth-to-spaceship coordinate transformations, move the black event dot to
2.5 + 2.3 e1 (pretty close to the light cone), set the angle slider to −1.1, and notice the transformed coordinates are 1.1 + 0.5 e1, which corresponds to the spaceship's coordinates of 11 + 5 e1. Finally, to check the earth-to-shuttle coordinate transformations, keep the black event dot at 2.5 + 2.3 e1, set the angle slider to −1.8, and notice the transformed coordinates are 1 - 0.21 e1, which corresponds to the shuttle's coordinates of 10 - 2 e1.