01 January 1970 47 3K Report

Of course, one can use Clifford Algebra Cl(3,1) in electromagnetism. But then, one is not working with Gibbs vectors anymore (NOTE 1.1, 1.2) , but with multivectors (NOTE 2.1, 2.2, 2.3). Better yet, we can try to check with tensors. There is a physics fundamental reason (NOTE 1.31P) why tensors are better than Cl(3,1) when one transforms coordinates.

Also, every element of a geometric algebra, such as Cl(3,1) and spacetime algebra by Hestenes, can be identified with a tensor, but not every tensor can be identified with an element of a geometric algebra. In that sense, tensors are more general than pure grade multivectors. The rank of a tensor is not restricted by the dimension of the base vector space like the grade of a multivector.

This means that there is no way to represent any 2nd-rank tensor by a bivector, vector, and scalar. Or, in particular, no way that any 3×3 matrix could be represented by a 2-vector in 3D!

So, it is not a 3D here and 1D there, but a union of the two, in 4D, that is desirable in electromagnetism spacetime. One can transform length into time, and vice-versa.

Tensors allow that, Cl(3,1) does NOT do that. And electromagnetism may be at least 4D, maybe 6D or more, which tensors can represent, but Cl(3,1) would not.

Similar question: what's the relationship of tensor and multivector . The short answer is that all multivectors are tensors, but not all tensors are multivectors, so geometric algebra is not and cannot be isomorphic to tensor algebra.

Therefore, geometric algebras (Clifford Cl(3,1), Hestenes) seem insufficient for electromagnetism, in all possible generality. What is your view?

NOTES:

1.1. The 3D vector cross product is not a vector, but a tensor. This is well-known and the reason to invalidate its use in Maxwell's equations, and physics equations. This is both a physical and a mathematical reason, commented below. But, would it not serve a restricted purpose adequately? No, it can create mistakes upon reference frame change, for example, simple mirroring, gives us wrong units in the SI MKS, worldwide, and this is all old news that have to be somehow continuously repeated, with a "life of its own," as a misconception in even current college books at competitive US universities.

1. 2. MATHEMATICAL REASON: The 3D vector cross product is not a closed operation in the 3D vector space, it produces a member that does not belong to the same set, the 3D vector space, although it may look like it in some cases.

1.3. PHYSICAL REASON: Both sides of an equation representing a physical relationship, such as A = B, must change equally when the frame of reference changes and the so-called inertial condition is obeyed, as already stated by Galileo, Newton, and Einstein, that the laws of physics are the same for all uniformly moving observers. But if one writes an equation using a 3D vector cross product, such as A= B x C, the left and right side may transform differently if the coordinate frame of reference changes, while still inertial. In the past, this was accommodated, not solved, by considering spurious things such as polar and axial vectors, and pseudoscalars. This is solved using tensors, which maintain the form A = B under inertial reference change.

2.1. Multivectors, in geometric algebras, or Clifford algebra, or Cl(n,1) algebras, or STA Hestenes algebra, solve the mathematical reason (1.2), creating a closed space in all operations. No geometric algebra operation is mathematically unsound.

2.2. Multivectors do not solve the physical reason (1.3), but tensors do. This is another reason, besides lack of isomorphism with tensors and no use of time as a coordinate in spacetime, that invalidates the use of geometric algebras in equations of physics, including electromagnetism.

2.3. With multivectors, If one eliminates the physically wrong results, by requiring an additional step of "filtering" through, let us say, a Hamiltonian, this will not produce those results that are physically valid but were ignored in the first place, using just multivectors. A sequence of filters cannot filter less than the first filter, well-known in physics, math, and engineering. This appears, more easily to see, in non-euclidean spaces, such as anything larger than, let us estimate in general, a few Planck lengths.

3. This thread arrived at a first conclusion, which is stated in the NOTES 1-2 above. There is no room to refuse to notice an obvious thing, or to re-explain here, the reasoning and references are available above, to anyone.

4. We are now moving on, to non-mathematical aspects of using Clifford algebras, even beyond the physical reason given in the NOTES above, where using Clifford algebras would be detrimental to special relativity. We are talking about time.

More Ed Gerck's questions See All
Similar questions and discussions