Mathematics 138
Minkowski Geometry

This discussion is inspired by Yaglom, Section 12, pp. 174-194, but it draws heavily on Supplements A and B and the use of vectors to achieve simplicity.
The axioms are the same as the axioms for the other two affine geometries, Euclidean and Galilean, except for Group IV, where we now have Yaglom's IV4(M)
There exists a vector a with <a,a> > 0 and a vector b with <b,b> < 0.
Following the approach on p. 206 of Yaglom (the Gram-Schmidt process),
we take the vector a and divide by its length (the square root of <a,a>) to obtain a "timelike" unit vector i with <i, i> = 1.
Now let <i, b> = g. Then the vector b1 = b - gi has the property that
<i, b1> = <i, b - gi> = <i, b> - g <i, i> = g - g = 0.
Furthermore, 
<b1, b1> = <b - gi, b - gi> = <b, b> - 2g <b, i> + g2 = <b,b> - 2 g2 + g2 = <b, b> - g2
By the axiom <b, b> < 0, so <b1, b1> < 0 also.
Divide b1 by the square root of -<b1, b1> to obtain a "spacelike" unit vector m with <m, m> = -1.
Since <i, b1> = 0, <i, m> = 0.
The vectors i and m form an orthonormal basis, which consists of a timelike unit vector and a spacelike unit vector.

Notice that <i + m, i + m> = <i, i> + <m, m> = 1 -1 = 0
and that <i - m, i - m> = <i, i> + <m, m> = 1 -1 = 0
The vectors i + m and i - m have zero length.  Yaglom calls them "isotropic" but a more common term used by physicists is "lightlike."

Any vector u can now be expressed in the form u = ti + ym.
Another vector u1 can be expressed in the form u1 = t1i + y1m.
Then <u, u1> = tt1 <i, i> + yy1 <m, m> = tt1 - yy1.
This quantity may be positive, negative, or zero. If it is zero, the vectors u and u1 are orthogonal.

A vector for which the norm ||u|| = <u, u> > 0 is called timelike. If the tail of such a vector is the 
event A and the head is the event B, the distance between A and B is the length of u, the square root of its norm, dAB =  = . It is an experimental fact that this equals the time that elapses on a clock that is on hand for events A and B, a quantity that is the same in any frame of reference.

A vector for which the norm ||u|| = <u, u> < 0 is called spacelike. If the tail of such a vector is the 
event A and the head is the event B, the distance between A and B is the length of u, the square root of minus its norm, dAB =  = . It is an experimental fact that this equals the spatial separation between events A and B as measured in a frame of reference where they are simultaneous.  Yaglom calls the distance imaginary in this case, but we will not. Since there is no isometry that carries timelike vectors into spacelike vectors, there is never occasion to compare the length of a spacelike vector with that of a timelike vector, and whether the length of a spacelike vector is called real or imaginary is a matter of taste.

The lines of Minkowski geometry fall into three classes.  The important ones are timelike lines
of the form P + lu . where P is an event ( a point) and u is a timelike vector.  These are like the ordinary lines of Galilean geometry. A timelike line corresponds physically to the set of events in the uniform motion of a particle traveling at a constant speed less than the speed of light.

On occasion we will need to consider lightlike lines of the form P + l u, where P is an event and u is a lightlike vector.  These are like the special lines of Galilean geometry. A lightlike line corresponds physically to the set of events in the motion of a photon or neutrino that moves at the speed of light.

Finally we will need to consider spacelike lines of the form P + l u, where P is an event and u is a spacelike vector.  These are also like the special lines of Galilean geometry. A spacelike line corresponds physically to a set of simultaneous events in some frame of reference.

Following the approach that Yaglom takes in Supplements A and B, we regard timelike lines as the "genuine" lines of Minkowski geometry, with lightlike and spacelike lines used only as aids in proving theorems.  For any pair of events (pair of points of the geometry) we can define a distance, but the two events determine a line of the geometry only if the vector from one to the other is timelike.
Now consider two "forward" timelike vectors u and w.  The term "forward" means that the first component is positive, i.e that <u, i> > 0 and <w, i> > 0.  We will only allow isometries that
carry forward timelike vectors into forward timelike vectors. 

Consider the vector l w - u. As the diagram shows, there are values of l for which l w - u is timelike (including l = 0), there is a range of values of l for which l w - u is spacelike (including one for which the first component is zero), and there are two values of l for which l w - u is lightlike.

Consider <l w - u, l w - u> = <w,w> l2 - 2 <w, u> l + <u, u> = 0.
This quadratic equation in l has two real roots, so
4 <w, u>2 - 4 <w, w><u, u> > 0
and <w, u>2 > <w,w><u,u> .
Since w and u are forward timelike, <w, u> > 0, and we see that
.  Thus, since the hyperbolic cosine function has a range from 1 to infinity, we
can define the angle between the vectors u and w as arc cosh .

Isometries of Minkowski geometry must preserve the scalar product and must carry forward timelike vectors into forward timelike vectors and spacelike vectors into spacelike vectors.

The image of  must be a forward unit timelike vector  
The image of  must be a spacelike unit vector orthogonal to this, i.e. 
This choice leaves out spatial reflections -- that is a matter of taste.
So the 3x3 matrix that represents a Minkowski isometry is

In physical terms, this is a Lorentz transformation followed by a translation.
If we write v = tanh a (Yaglom uses v = - tanh a instead)

then we have cosh a = 
and sinh a = tanh a cosh a =  so that the matrix becomes

Notice that  = 
and  = 
so that the lightlike vectors are eigenvectors of Lorentz transformations.  This is in accordance with Einstein's postulate that the speed of light is the same in every frame of reference.
Notice also that the eigenvalues are cosh a + sinh a and cosh a - sinh a, whose product is 1.

Some Minkowski trigonometry.
The Law of Cosines.
Consider a triangle made from three timelike lines (slope < 1)

Consider the vector w from A to C and the vector u from A to B.
Then the vector from C to B is u -w and
<u - w, u - w> = <u, u> + <w, w> - 2 <u, w>.
From the previous definition of angle,
a2 = b2 + c2 - 2 b c cosh A
This is true even if the vector from C to B is not timelike, but in that case we don't really have a "triangle" and the angles at B and C are not defined.
The triangle inequality:
Suppose that c is the longest side of the triangle.
Since a2 = b2 + c2 - 2 b c cosh A and cosh A > 1
a2 < b2 + c2 - 2 b c
a2 < (c - b)2
a < c - b (since we know that both a and c - b are positive)
or c > a + b.
In words, "a straight line has the longest Minkowski distance between two points."
This is correct physically.  If a clock starts at event A and travels to event B, more time will 
elapse on it if it travels with uniform motion (AB) than if it travels with non-uniform motion (AC followed by CB).

The Minkowski Law of Sines
First we must consider a right triangle with two timelike sides and one spacelike side

If u is the vector AC and w is the vector CB (orthogonal to u) then u+w is the vector AB and 
c2 = <u + w, u + w> = <u, u> + <w, w> = b2 - a2 (the Minkowski Pythagorean theorem)
 By the definition of angle
<u, u + w> =  b c cosh A
Since <u,w> = 0 
<u,u>  = b c cosh A
Now <u,u> = b2 so b2 = b c cosh A   and cosh A =  = 
Then sinh2 A = cosh2 A - 1 =  and sinh A =  = 
Notice also that tanh A = a/b.
Now consider a triangle with three timelike sides.

Construct the altitude on side c, a spacelike line of length h.
h = b sinh A = a sinh B.
It follows that . 
By constructing an altitude orthogonal to side b we get c sinh A = a sinh C. 
It follows that  = . 

Distance from a point P to a line.
As in Euclidean and Galilean geometry, this is the length of the vector u, orthogonal to the line,
that extends to the point P from some point on the line.
This means that for a timelike line, the vector u must be a spacelike vector.  In a Euclidean 
diagram, the slope of the vector u is the reciprocal of the slope of the line.
If the line is horizontal, the vector u is vertical.

The vector u has a negative norm, and <u, u> < <w, w> if w is any other vector from a point on the
line to P.  So we are choosing the "minimum squared distance" as in Euclidean geometry.
If the line is not horizontal, the vector u is not perpendicular to the line in a Euclidean diagram!
Below, the line has slope 3/5 and the vector u has slope 5/3.
If you draw the two special lines through P, with slope +1 and -1, the tail of u is the midpoint of
where these lines intersect the given line.

Angle bisectors.
This is tricky and is not well explained in Yaglom.
Given two intersecting (timelike) lines, the angle bisector is another (timelike) line that makes
equal angles with the two intersecting lines.  There is an isometry (a Lorentz transformation) that
carries one line into the bisector while carrying the bisector into the other line.
Any point on the bisector is equidistant from the two lines.
Here is an example where one line is horizontal.  The other line has slope 3/5, and the bisector
has slope 1/3.
SInce tanh 2a =  we can check this for tanh a = 1/3

Point P. on the bisector, is equidistant from the two lines.  The distance is measured along the two dotted lines.
There are other points P that are also equidistant from the two lines.  These lie on a spacelike
line that is perpendicular to the bisector.  These are the lines that are shown as concurrent in
figure 172c on p. 193 of Yaglom.

For the remainder of these notes, there are excellent diagrams on the Web site.

Circles of Minkowski geometry are Euclidean hyperbolas whose asymptotes are special lines.  There are two types.
A circle of the "first kind" (Yaglom's terminology) has a positive radius.  Every point on such a circle is at the same distance from the center. Here is a diagram.
Any chord through two points on the same branch of one of these hyperbolas is a spacelike line, and so is any tangent to the hyperbola.  Like the circles of Galilean geometry that consist of two
special lines, they are rather uninteresting.
A circle of the second kind (again Yaglom's terminology) is the set of points with a constant
spacelike separation from the center.  We will call the radius of such a circle positive in accordance with our earlier convention about the length of a spacelike vector, but remember that a vector u
from the center of the circle to a point on the circle has the property that <u, u> = -r2.

These circles are more interesting, because any chord joining two points on the same branch is a segment of a timelike line, and so is any tangent.  One branch of such a circle corresponds physically to the motion of a particle that is subject to a constant force (which does not imply constant acceleration!)

Almost every one of Euclid's theorems about circles holds also in Minkowski geometry, but the Euclidean diagrams look strange, because the concepts of "distance" and "orthogonal" are the Minkowski ones.

Some theorems about circles of the second kind.
Preliminary: the Pythagorean theorem for a triangle made from two spacelike vectors and one
timelike vector. 

Vector u is spacelike: <u, u> = -a2
Vector w is timelike: <w, w> = b2
Vectors u and w are orthogonal: <u, w> = 0
b < a, so the hypotenuse u+ w is also spacelike.

<u + w, u + w> = <u,u> + <w,w> = -a2 + b2 < 0
So <u + w, u + w> = -c2 where c2 = a2 - b2
Theorem 1: The vectors joining a point on a circle to the two ends of a diagonal are orthogonal (in Euclidean geometry, this can also be stated as "an angle inscribed in a semicircle is a right angle," but "right angle" is not a Minkowski concept.).
Proof: The vectors from the center of the circle to the ends of a diameter are u and -u, where
<u, u> = -r2 (r is the radius of the circle)
The vector from the center of the circle to a point P on the circle is w, and <w, w> = -r2.
The vectors from the point P to the two ends of the diameter are w - u and w + u
<w - u, w + u> = <w, w> - <u, u> = -r2 + r2 = 0, so the vectors are orthogonal.

Theorem 2: The line tangent to a circle at the point Q is orthogonal to the diameter through that
point.
Proof: Take the limit of the preceding result as P approaches the end of the diameter. The direction of one vector is along the tangent; the direction of the other is along the diameter.

Theorem 3: If two tangents are drawn to a circle from a point M outside a circle to points P and Q on the circle, the triangle MPQ is isosceles.
Proof 1 (like Yaglom figure 169)
Carry out a rotation (a Lorentz transformation) so that the point M is on the y axis. The result is
now obvious by symmetry.

Proof 2:
Let O be the center of the circle.  The spacelike vector OP has length r.  The spacelike vector OM
has length s < r. The timelike vector PM is orthogonal to OP.  Call its length b.  By the Pythagorean theorem, s2 = r2 - b2, so b2 = r2 - s2.

The spacelike vector OQ also has length r.  The timelike vector QM is orthogonal to OQ.  Call its length b'.  By the Pythagorean theorem, s2 = r2 - b'2, so b'2 = r2 - s2.
So b = b' and the triangle is isosceles.

Theorem 4: Let A and B be two fixed points on a circle.  Let C be any other point on the circle.
Then the angle between the lines AC and BC is constant, equal to half the angle between the
tangents to the circle at A and B.
Proof: Construct the tangents to the circle at points A, B, and C.
The tangents at A and at C form an isosceles triangle with the chord AC as its base, so they
make equal angles a with the chord AC.
The tangents at B and at C form an isosceles triangle with the chord BC as its base, so they
make equal angles b with the chord BC.
The angle between chords AC and BC is therefore a + b, while the angle between the tangents
at A and at B is 2a + 2b.

Theorem 5: Construct the line tangent to a circle of radius r at a point P, and choose a point M on this line.
Draw a timelike line through M that intersects the circle at points A and B.  Then the product
of the distances MA and MB equals the square of the distance MP.
Proof: (the same as Euclid's proof of the corresponding theorem of Euclidean geometry)
Let a = the distance MA
Let 2c = the distance AB, so that the distance MB is a + 2c.

Let P be the midpoint of the chord AB.  Denote by u the vector from the center of the circle O to P.
This is a spacelike vector.  Being the median of the isoceles triangle OAB, it is orthogonal to the base.  Let <u, u> = -h2
By the Pythagorean theorem, r2 = h2 - c2

Let d = the length of the spacelike vector OM. By the Pythagorean theorem, d2 = h2 - (a+c)2.
So h2 = r2 + c2 = d2 + (a+c)2.

Let e equal the distance MP. The vector OP is spacelike, with length r, and it is orthogonal to the tangent MP, so by the Pythagorean theorem applied to OMP,
d2 = r2 - e2
Thus r2 + c2 = r2 - e2 + (a+c)2
and e2 = (a+c)2 - c2 = (a+2c)a
or the square of the distance MP equals the product of distances MA and MB
Corollary: From a point M outside a circle, draw two timelike lines, one intersecting the circle at A and B, the other at C and D.  Then (MA)(MB) = (MC)(MD)
Proof: Both products are equal to the square of MP.
In fact, as Yaglom states on page 194, this result is true even if one of the lines is spacelike and
intersects each branch of the hyperbola once.