arcnets
Dec6-03, 01:27 PM
This thread is the sequel to my other thread, 'Quaternions and SR'.
My goal is to write some physical equations with quaternions.
Because I think quaternions represent the 4-dimensionality and metric used in special relativity. See:
A quaternion is a generalized complex number:
A = a_t + ia_x + ja_y + ka_z
with the fundamental equation
i^2 = j^2 = k^2 = ijk = -1.
Quaternions are not commutative, for instance
ij = -ji = k.
Let's define
A_3 = ia_x + ja_y + ka_z
and the dot and cross products as usual for 3-vectors, then the product of two quaternions is
AB = a_tb_t + a_tB_3 + A_3b_t - A_3 \cdot B_3 + A_3 \times B_3.
Thus,
\frac{1}{2}(AB - BA) = A_3 \times B_3
and
\frac{1}{2}(AB + BA) = a_tb_t + a_tB_3 + A_3b_t - A_3 \cdot B_3.
Let's define the commutator
\left[A,B\right] = \frac{1}{2}(AB - BA)
and the anticommutator
\left<A,B\right> = \frac{1}{2}(AB + BA).
Now for physics. Let's define the differential operator
\nabla = \frac{\partial}{\partial t} + i \frac{\partial}{\partial x} + j \frac{\partial}{\partial y} + k \frac{\partial}{\partial z}.
Then Maxwell's equations can be written
1. Coulomb's law:
\left<\nabla, E\right> = \nabla_tE - 4\pi J_0
2. Ampere's law:
\left[\nabla, B\right] = \nabla_tE + 4\pi J_3
3. Faraday's law:
\left[\nabla, E\right] = -\nabla_tB
4. No magnetic monopoles:
\left<\nabla, B\right> = \nabla_tB.
Now if we use a vector potental written as a quaternion A, which satisfies Lorentz's condition
\nabla_t a_t - \nabla_3 \cdot A_3 = 0
and let
E = -\frac{1}{2}\left<\nabla,A\right>
B = \frac{1}{2}\left[\nabla,A\right]
then Maxwell's equations reduce nicely to two wave equations:
4 \pi J = \frac{1}{2}\left<\nabla^2,A\right>
\nabla_t B = \frac{1}{2}\left[\nabla^2,A\right].
That's my result so far. Any comments?
My goal is to write some physical equations with quaternions.
Because I think quaternions represent the 4-dimensionality and metric used in special relativity. See:
A quaternion is a generalized complex number:
A = a_t + ia_x + ja_y + ka_z
with the fundamental equation
i^2 = j^2 = k^2 = ijk = -1.
Quaternions are not commutative, for instance
ij = -ji = k.
Let's define
A_3 = ia_x + ja_y + ka_z
and the dot and cross products as usual for 3-vectors, then the product of two quaternions is
AB = a_tb_t + a_tB_3 + A_3b_t - A_3 \cdot B_3 + A_3 \times B_3.
Thus,
\frac{1}{2}(AB - BA) = A_3 \times B_3
and
\frac{1}{2}(AB + BA) = a_tb_t + a_tB_3 + A_3b_t - A_3 \cdot B_3.
Let's define the commutator
\left[A,B\right] = \frac{1}{2}(AB - BA)
and the anticommutator
\left<A,B\right> = \frac{1}{2}(AB + BA).
Now for physics. Let's define the differential operator
\nabla = \frac{\partial}{\partial t} + i \frac{\partial}{\partial x} + j \frac{\partial}{\partial y} + k \frac{\partial}{\partial z}.
Then Maxwell's equations can be written
1. Coulomb's law:
\left<\nabla, E\right> = \nabla_tE - 4\pi J_0
2. Ampere's law:
\left[\nabla, B\right] = \nabla_tE + 4\pi J_3
3. Faraday's law:
\left[\nabla, E\right] = -\nabla_tB
4. No magnetic monopoles:
\left<\nabla, B\right> = \nabla_tB.
Now if we use a vector potental written as a quaternion A, which satisfies Lorentz's condition
\nabla_t a_t - \nabla_3 \cdot A_3 = 0
and let
E = -\frac{1}{2}\left<\nabla,A\right>
B = \frac{1}{2}\left[\nabla,A\right]
then Maxwell's equations reduce nicely to two wave equations:
4 \pi J = \frac{1}{2}\left<\nabla^2,A\right>
\nabla_t B = \frac{1}{2}\left[\nabla^2,A\right].
That's my result so far. Any comments?