The Lorentz Transform in modified complex algebra
Author 
Carl Borrell 

carlborrell@btinternet.com 
The Lorentz transform in one spatial dimension:
w' = w.cosh(a)  x.sinh(a)
x' = x.cosh(a)  w.sinh(a)
e = w + ix
e' = w' + ix'
e' = L(e) = [w.cosh(a)  x.sinh(a)] + i[x.cosh(a)  w.sinh(a)]
[Where tanh(a) = V/c; sinh(a) = 1/(1  V^{2}/c^{2})^{1/2} ; w = ct; c = speed of light ; V = relative velocity]
cannot be expressed as a function of a complex variable because the CauchyReimann conditions are not satisfied. However if we change the product rule for complex multiplication:
a = a_{0 } + ia_{1}
b = b_{0 } + ib_{1}
a.b = (a_{0}.b_{0}  a_{1}b_{1}) + i(a_{0}b_{1}  b_{0}a_{1})
We can still regard i as (1)^{1/2} because,
(0 + i).(0 + i) = (0.0  i.i) + i(0.i  i.0) = 1
The algebra is noncommutative:
a.b = (b.a)*
The square of any complex variable is always a purely real number in this algebra:
a^{2} = (a_{0}a_{0}  a_{1}a_{1}) + i(a_{0}a_{1}  a_{0}a_{1}) = a_{0}^{2}  a_{1}^{2}
So the square is defined unambiguously even though the algebra is noncommutative.
This leads to a definition of the reciprocal function as:
1/a = a/a^{2} = (a_{0} + ia_{1}) / (a_{0}^{2}  a_{1}^{2})
^{ }
The Lorentz transform can be expressed as multiplication by a complex constant:
L = cosh(a) + isinh(a)
e' = L.e = [w.cosh(a)  x.sinh(a)] + i[x.cosh(a)  w.sinh(a)]
This suggests the following proof of Lorentz invariance:
e'^{2} = (L.e)^{2}
= L^{2}.e^{2}
w'^{2}  x'^{2} = (cosh^{2}(a)  sinh^{2}(a)).(w^{2}  x^{2})
= w^{2}  x^{2 } = e^{2}
[Because cosh^{2}(a)  sinh^{2}(a) = 1 for all a].
This shows the invariance of the interval w^{2}  x^{2} under this transform. In fact all purely real numbers have an infinite number of square roots corresponding to the spacetime coordinates of the same event viewed from each of the infinite number of inertial frames in standard configuration.
In terms of the velocity vector v = (c + iV) the Lorentz factor can be expressed as follows:
L = 1/(1  V^{2}/c^{2})^{1/2} + i(V/c(1  V^{2}/c^{2})^{1/2})
= c/(c^{2}  V^{2})^{1/2} + i(V/(c^{2}  V^{2})^{1/2})
= (c + iV)/(c^{2}  V^{2})^{1/2}
= v/(v^{2})^{1/2}
Now v^{2} is purely real therefore (v^{2})^{1/2} has an infinite number of solutions. In each case L is analogous to unity and L^{2} = 1.
Consequences for the algebra
Some other consequences for the algebra are as follows:
Theorem: Zero factorisation:
0 = (a + ia).(b + ib)
In terms of special relativity where ct = x the interval corresponds to two events joined by a light pulse.
Theorem: All even powers are purely real:
a = w + ix
a^{2} = (w^{2}  x^{2})
a^{2n} = (w^{2}  x^{2})^{n}
Theorem: Two definitions for odd powers:
a = w + ix
a^{(2n + 1) }= (w^{2}  x^{2})^{n} . a or
a^{(2n + 1) }= a.(w^{2}  x^{2})^{n}
These two possibilities are complex conjugates of each other, the time component is the same but the space component is reversed. Perhaps this is related to chirality?
Binomial
(a + b)^{2} = a^{2} + a.b + b.a + b^{2}
We cannot collect the cross products because the algebra is noncommutative, therefore
(a + b)^{2} = a^{2} + a.b + (a.b)* + b^{2}
= a^{2} + 2Re(a.b) + b^{2}
So (a + b)^{2 } is purely real as required.
Derivatives and partial derivatives
Let x = x_{0} + ix_{1 }; f(x) = f_{0} (x_{0},x_{1}) + if_{1}(x_{0},x_{1})
What is the relationship between the derivative df(x)/dx
and the partial derivatives d
[i.e CauchyReimann conditions do not seem to be satisfied there appears to be a relationship with the curl and divergence operators]
Path Integrals
What are the consequences of the zero factorisation?
Taylor Series
Let s = x^{2}
Then for infinitely differentiable function p(x),
p(x) = a_{0} + a_{1}.x + a_{2}.x^{2}/2! + a_{3}.x^{3 }/3! ...
= f(s) + x.g(s)
Quaternions
A similar change can be made to the fourdimensional algebra defined by Sir W.R. Hamilton in the early 1840's.
[Hamilton's original papers are available at the history of mathematics web site, Trinity College, Dublin ref.
www.maths.tcd.ie/pub/HistMath/People/Hamilton/Quatern1/Quatern1.html
www.maths.tcd.ie/pub/HistMath/People/Hamilton/Quatern2/Quatern2.html ]
In Hamilton's quaternion algebra the product is defined as
i^{2} = j^{2} = k^{2} = 1
i.j = k
j.k = i
k.i = j
j.i = k
k.j = i
i.k = j
a = a_{0} + a_{1}i + a_{2}j + a_{3}k
b = b_{0} + b_{1}i + b_{2}j + b_{3}k
a.b = (a_{0}b_{0} + a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3}) +
(a_{0}b_{1} + a_{1}b_{0} + a_{2}b_{3}  a_{3}b_{2})i +
(a_{0}b_{2} + a_{2}b_{0} + a_{3}b_{1}  a_{1}b_{3})j +
(a_{0}b_{3} + a_{3}b_{0} + a_{1}b_{2}  a_{2}b_{1})k
Which is the origin of the more familiar vector algebra due to Gibbs.
The quaternion product can be expressed in vector algebra, if we regard a quaternion as the sum of a vector and a scalar quantity
a = a_{0} + v ; where v = a_{1}i + a_{2}j + a_{3}k
b = b_{0} + u ; where u = b_{1}i + b_{2}j + b_{3}k
Then a.b = a_{0}b_{0} + v.u + a_{0}u + b_{0}v + v X u
The quaternion algebra can be modified in the same way as the complex algebra as follows:
a.b = (a_{0}b_{0} + a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3}) +
(a_{0}b_{1}  a_{1}b_{0} + a_{2}b_{3}  a_{3}b_{2})i +
(a_{0}b_{2}  a_{2}b_{0} + a_{3}b_{1}  a_{1}b_{3})j +
(a_{0}b_{3}  a_{3}b_{0} + a_{1}b_{2}  a_{2}b_{1})k
[In vector algebra, if
a = a_{0} + v ; where v = a_{1}i + a_{2}j + a_{3}k
b = b_{0} + u ; where u = b_{1}i + b_{2}j + b_{3}k
Then a.b = a_{0}b_{0} + v.u + a_{0}u  b_{0}v + v X u]
which gives
a^{2} = (a_{0}^{2}  a_{1}^{2}  a_{2}^{2}  a_{3}^{2})
i.e. the Lorentz invariant interval in four dimensions.
Field equations
Is it possible to express the laws of electromagnetism in terms of field equations in this 4d algebra?
What is the relationship between the derivative and the curl and divergence operators?
Can the equations be transformed as follows?
f(s) = [p(x)]^{2} ; s = x^{2}
This give a scalar function of a scalar variable in terms of Lorentz invariant quantities.