Quaternions and Lorentz gauge transformations

09 Feb 2010

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Tue, 02/09/2010 - 23:03

doug

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Joined: 03/13/2009

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Hello:

A paper of mine was rejected because the editor couldn't imagine how quaternions could be used for something that had to be Lorentz invariant. Quaternions are a double cover for the groups SO(3) [3D spatial rotations] via the group SU(2). There is also a connection between SO(4) [4D spatial rotations] and quaternions, two elements of SU(2) that share the same real value.

There is not a well know connection between quaternions and the Lorentz group, O(1, 3), rotations that preserve the scalar t^2 - x^2 - y^2 - z^2 . This info can be gleaned from a small bit of time on wikipedia.

Being well-educated in group theory can cause problems. The entire reason I began working with quaternions was due to a book by Artmann, "The Concept of Number: From Quaternions to Monads and Topological Fields". This was one of only a dozen books in the Harvard University library that had "Quaternion" in the title and was published after 1950. This was one of the first calculations he did:

$(t, x, y, z)^2 = (t^2 - x^2 - y^2 - z^2, 2 t x, 2 t y, 2 t z) \quad eq.~1$

Having taken 3 classes on special relativity, my heart stopped - that is the Lorentz invariant interval, the target of the group O(1, 3), sitting at the pole position, for free. Even today, I don't know how to describe this using group theory lingo. If you take two elements of SO(4), and multiply them together, then you get another element of SO(4) whose scalar happens to have O(1, 3)? Sounds like gibberish.

Since I cannot show the connection between quaternions and Lorentz transformations with my limited group theory skills, I will use examples, eq. 1 being the prime example. One I had all wrong and confused was this one:

$(\frac{\partial}{\partial t}, \vec{\nabla})(\phi, \vec{A}) = (\frac{\partial \phi}{\partial t} - \vec{\nabla} \cdot \vec{A}, \frac{\partial A}{\partial t} + \vec{\nabla} \phi + \vec{\nabla} \times \vec{A})\quad eq.~2$

This has a gauge term, and -E and +B, nice. Never having done a proof or serious calculation, I thought the scalar would be a Lorentz invariant quantity. Wrongo. Tensors show what combo of derivatives of 4-potentials make a Lorentz invariant scalar:

$\nabla^{\mu} A_{\mu} = (\frac{\partial}{\partial t}, -\vec{\nabla})(\phi, -A) = \frac{\partial \phi}{\partial t} + \vec{\nabla} \cdot \vec{A} \quad eq.~3$

People will add this to the EM action and say they have chosen the Lorenz gauge (and usually toss in an errant "t" before the z to repeat a mistake of an American textbook writer).

There is a plus in there because of the tricky accounting system of tensors. Up is contra is positive, down is out is negative - at least for 4-vectors. The rule is opposite for the derivative. I confess, I never embraced that although it must be true given how man grad student centuries have gone over those definitions. I asked myself this very morning, "Is there any way to get and E field, a B field, and a plus sign in that scalar?" The answer is a rock solid "No". So I went out and walked the dog since she had to pee whether I figured out the relationship between quaternions and Lorentz transformations or not. Walking the dog over the same path I have done for years does not require much effort. I know the E and B fields transform like rank 2 tensors, so the gauge term must absolutely not transform like a Lorentz invariant. The gauge term need to transform like part of a rank 2 tensor. That funky sign flipping rule works in favor of the quaternion differential operator. Luck sign flip? I don't think so. That was a good dog walk.