We continue our discussion, from Part 14, Part 15, and Part 16, of actions of the Clifford group G on the Clifford geometric algebra of space Cl(V). In Part 14 we have listed four different actions:
1). g·u = g u τ(g),
2). g·u = g u ν(g),
3). g·u = π(g) u τ(g),
4). g·u = π(g) u ν(g).
We have already seen that 1) leads to Lorentz transformations of coordinates in a complex Minkowski space of Special Relativity. Today we will discuss the second action. Let us recall that G consists of all invertible elements of Cl(V). It is defined by the condition gν(g) = ν(g)g =1. Action 2) can thus be written as
g·u
= g u g-1,
and so it consists of automorphisms of the algebra. In particular g·i
= i. The scalar part of u does not transform at all under this
action, only the vector part transforms! What can be the realtion, if
any, to special relativity in this case? For action 1) the scalar part
behaved as the time component of a Minkowski space event. Now the scalar
part does not transform at all. The mystery will be solved immediately
after, anticipating the result, I introduce the notation for this
discussion. The vector part consists of three real and three imaginary
numbers. What can it be? A physicist guesses immediately: write u as u = E+iB,
and see what happens!
Writing g = exp(tX), we find that X+ν(X)=0.
Again we will consider two cases: X odd and X even. Let us start with X
even, thus π(X) = X. As in the case 1) we set X = in, where n is a real unit vector. In this case, for g(t) = exp(itn) we have ν(g(t))=g(-t)=τ(g(t)). The action is the same as in the case 1) - E and B transform as vectors (or covectors?) under 3D rotations in the plane perpendicular to n. The same formulas as for x in the case 1).
Now we cansider the case of X odd. Thus we set X=n. Then exp(tn) is rather easily calculated:
exp(tn) = cosh(t)1 + sinh(t)n.
Exercise 1. Verify the last formula.
Then calculating
(E'+iB') ≡ exp(tn)(E+iB)exp(-tn) = (cosh(t)1 + sinh(t)n) (E+iB) (cosh(t)1 - sinh(t)n)
is a matter of a straightforward algebra using the bi-quaternion multiplication formula
(p0,p)(q0,q) = (p0q0 + p·q, p0q + q0p + i p⨯q). We obtain
E' = ....
B' = ...
I will not give the results now. I will not take the pleasure from the reader to find it himself/herself. To find and then compare with what can be found on internet.
Anyway, it looks like Cl(V) contains not only spacetime, but also "the field strength". If so, then a field would be a function on Cl(V) (treated as complexified space-time) with values in Cl(V) (treated this time as a receptacle for field strengths). What kind of a function? V.V. Kassandrov suggested the answer: it should be "analytic" function, "analytic" in a special way. Then analyticity would automatically lead to something similar to Maxwell equations. But this is still in the future - I am only learning this stuff. So far we are dealing with kinematics, no dynamics yet. Yet there is something else still lacking in this picture - the "conformal infinity", the "absolutes" of projective geometry. Points where parallel lines meet.
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"Points where parallel lines meet."
ReplyDeleteWhat does "meet" means?
(Different) Parallel lines don't have common points.
The answer to this question requires a different Clifford algebra. We will come to this point soon. For now think that we are dealing with a stereographic projection from S^3 to R^3. The parallel lines in R^3 meet at the projection point on S^3.
Delete"The parallel lines in R^3 meet at the projection point on S^3."
DeleteWhat does "meet" means?
-> mean
DeleteIt means that they have a common "limit point". Their *topological closures* are intersecting circles in S^3.
DeleteThings becomes somewhat more complicated with spacetime instead of space. With spacetime the "infinity" is not just a single point. It has a structure. Light rays meet at different points than timelike lines, and still different from points where spacelike lines meet.
DeleteYeah, in mathematical language "meet" translates to "converge to". Check technical drawing of perspective in architecture for example:
Deletehttps://www.architecturemaker.com/what-is-perspective-drawing-in-architecture/
"It means that they have a common "limit point".
DeleteWhat does "have" mean?
"Their *topological closures* are intersecting circles in S^3."
What does "intersect" mean?
It is like two meridians meet at the North and South poles. They have common points there.
ReplyDelete"It is like two meridians meet at the North and South poles."
DeleteMeridians are not circles. They are semicircles.
"Their *topological closures* are intersecting circles in S^3."
Are they really (in the common sense of the word) intersecting or rather those circles are tangent to each other?
Alright, once you insist on this side point of this note, I will devote to this subject next post. Provided the world will survive till Wednesday.
Delete(E'+iB') ≡ exp(tn)(E+iB)exp(tn) = (cosh(t)1 + sinh(t)n) (E+iB) (cosh(t)1 - sinh(t)n) ->
ReplyDeleteplus or minus?
Fixed. Thanks!
DeleteWhat I get is:
ReplyDeletex' = (n·x)n + cosh(2t)(x-(n·x)n) + i sinh(2t)(n × x)
where x = E +iB
This is a first step. The second step would be splitting it into a real and imaginary parts.
DeleteThis first step was big. The second is small.
DeleteE' = E║ + cosh(2t)E┴ -sinh(2t) (n × B),
B' = B║ + cosh(2t)B┴ +sinh(2t) (n × E).
Good: Now we need to compare it with with Wikipedia:
DeleteTransformation of the fields between inertial frames
E and B fields
https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity
Oh, they are very similar.
DeleteIt looks that not only they are similar, but they look identical if we identify "arctanh(v/c)" with "-2t".
DeleteAnd identifying direction of v with direction of n.
DeleteYes, of course, you are right.
Delete"Transformation of the fields between inertial frames
ReplyDeleteE and B fields
https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity"
One more good place to verify the result is the book by P. Lounesto "Clifford algebras and spinors", section 9.4 for E and B Lorentz transformations.
Every time i'm afraid i will fail, but with a little help of Bjab i've got it, exactly!
Ark, thanks a lot for not taking the pleasure to find it myself.
Thanks for bringing it up to my attention. In fact Lounesto derived the formulas using the same Clifford algebra as in my posts in section 9.7. Very good reference! But the Cambridge University Press Version, not the Springer/Kluwer version.
DeleteThis is the book I'm trying to get through! but alas, I am a simple engineer, the symbols and equations are a bit much for me. Any recommendations on an introduction to this?
DeleteIf you tell me what is your main goal, themore details, the better, I will try to think of some replacement.
DeleteAnother possibility is that you go through the book, and whenever you have a difficulty and want a given piece explained, ask, and I will try to answer your question.
DeleteYes, i have the Cambrige Univ. version. The book is written in pedagogical style and really useful. For our discussion, i like that he adds physical spirit to calculations when shows the place of velocity v/c = tanh(a) in the transformation matrix A (p.124) and separates boosts and rotations.
ReplyDelete"What can be the realtion..."-> relation
ReplyDelete"Now we cansider..."-> consider