## Magnetism in N-D

Ideas about how a world with more than three spatial dimensions would work - what laws of physics would be needed, how things would be built, how people would do things and so on.

### Magnetism in N-D

A purpose of a magnetic field function is to allow us to calculate the acceleration of a charged particle moving in that field. The two parameters of said function are the location of the particle p and the velocity vector v of the particle. The result is the acceleration vector on the particle a due to magnetic forces.

Instead of having a function with two parameters, there are usually two functions. The first function acts on the location vector, returning a value b. The second function takes this b and v as parameters and returns a.

Traditionally b is a pseudovector called the value of the magnetic field, and the second function is the cross product. This works only in 3D. Instead, we will demonstrate a method that works in any number of dimensions in which there are electric charges and special relativity holds. Special relativity is essentially (1+1)-dimensional -- the other two dimensions are largely irrelevant -- so it is most reasonable to expect relativity to hold in any number of dimensions.

In http://physics.weber.edu/schroeder/mrr/MRRtalk.html it may be seen that magnetism is a direct result of special relativity. It is possible to figure out that magnetism is essentially planar. It is a sort of correction to the electric field due to relativistic effects. The plane of magnetism includes three points: the location of the charge that is making the field, the location where that charge soon will be due to its motion, and the location of the charged particle that is going to be affected by the field.

In geometric algebra that plane is defined by a bivector that we'll call B. A bivector is an ordered pair of vectors that define a plane combined with a scalar. All the detail may be found at http://faculty.luther.edu/~macdonal/GA&GC.pdf or https://archive.org/details/arxiv-1205.5935. We are going to make everything as simple and down-to-earth as possible here, so all we will worry about are a space which includes orthonormal basis vectors e1,e2...eN. Then in 3D the basis of the bivector space may be e1e2, e3e1, e2e3. (We chose e3e1 instead of e1e3 so that the result will resemble the cross product.) Bivectors add together very much as ordinary vectors do, and multiplying by a scalar also works the same. The trickery comes with multiplying an ordinary vector times a bivector. There are three ways to do this, but all we need here is the inner product which we'll denote as |. That is, v|B is geometric algebra's inner product of a vector v with a bivector B. We are going to show that in 3D it can work the same as the cross product.

Let's say that the magnetic field vector we are interested in is b = [ c1e1, c2e2, c3e3 ], and the vector v is [ ae1, be2, ce3 ]. Then the cross product is

v x b =
[
(cc2 - bc3)e1,
(ac3 - cc1)e2,
(bc1 - ac2)e3
]

In the bivector version v is slightly different, it is ae1 + be2 + ce3. The magnetic field bivector is B = c12e1e2+c31e3e1+c23e2e3. We claim that the result is

v|B =
(cc31 - bc12)e1 +
(ac12 - cc23)e2 +
(bc23 - ac31)e3

Aside from superficial notational things, with
c1=c23
c2=c31
c3=c12
we see they are equivalent. So, how was this done? And why use c31 instead of c13? We shall see.

It's too much to show all of geometric algebra, so we shall use only four properties. First is antisymmetry -- reversing the order of the vectors in a bivector reverses the sign.

eiej = -ejei.

Then

eiei = -eiei = 0.

The next is bilinearity of the inner product.

av|cB = cv|aB = acv|B = v|acB etc.

(v1 + v2)|B = v1|B + v2|B

v|(B1+B2) = v|B1 + v|B2

The next two properties are true only for orthogonal vectors, but that is all we are concerned about here. These are:

If all three vectors are distinct then this inner product is zero.

ei|ejek = 0 iff i<>j, i<>k, j<>k.

If the vector ei is the same as the first vector of the bivector then this pair vanishes, leaving the third vector.

ei|eiej = ej

Combining with antisymmetry we then have

ei|ejei = - ej

We shall do the 2D case first as a sort of warmup.

In 2D:

In 2D the bivector field has the basis e1e2. As there is only one plane possible in 2D, the only possible values for the magnetic field bivector are ce1e2 with c a scalar.

v = ae1 + be2
B = ce1e2
v|B = (ae1 + be2)|ce1e2
= c(ae1|e1e2 + be2|e1e2)
= c(ae1|e1e2 - be2|e2e1)
= c(ae2 - be1)

The resulting acceleration vector in 2D is the velocity vector of the charged particle multiplied by c and rotated 90 degrees. Note that this is isomorphic to multiplying a complex vector by ci.

In 3D:

v = ae1 + be2 + ce3
B = c12e1e2+c31e3e1+c23e2e3
v|B = (ae1 + be2 + ce3)|(c12e1e2+c31e3e1+c23e2e3)
=
ae1|c12e1e2 + ae1|c31e3e1 + ae1|c23e2e3 +
be2|c12e1e2 + be2|c31e3e1 + be2|c23e2e3 +
ce3|c12e1e2 + ce3|c31e3e1 + ce3|c23e2e3
=
ac12e1|e1e2 - ac31e1|e1e3 + ac23e1|e2e3 -
bc12e2|e2e1 + bc31e2|e3e1 + bc23e2|e2e3 +
cc12e3|e1e2 + cc31e3|e3e1 - cc23e3|e3e2
=
ac12e2 - ac31e3 -
bc12e1 + bc23e3 +
cc31e1 - cc23e2
=
(cc31 - bc12)e1 +
(ac12 - cc23)e2 +
(bc23 - ac31)e3

This is a thinly disguised version of the traditional result of the cross product which is

v x b =
[
(cc31 - bc12)e1,
(ac12 - cc23)e2,
(bc23 - ac31)e3
]

Q.E.D. In 3D v|B is equivalent to v x b.

Use of e1e3 as a basis vector for the bivectors instead of e3e1 results in a formula that is less familiar but remains correct. As we shall see, this is more natural.

v|B =
(-bc12 - cc13)e1 +
( ac12 - cc23)e2 +
( ac13 + bc23)e3

In 4D:

We may continue as above with six bivectors as a basis for the bivector subfield of the algebra. With

v = ae1 + be2 + ce3 + de4
B = c12e1e2+c13e1e3+c14e1e4+c23e2e3+c24e2e4+c34e3e4

the result of a calculation similar to that of the 3D case is

v|B =
(-bc12 - cc13 - dc14)e1 +
( ac12 - cc23 - dc24)e2 +
( ac13 + bc23 - dc34)e3 +
( ac14 + bc24 + cc34)e4

5D:

v = ae1 + be2 + ce3 + de4 + ee5
v|B =
(-bc12 - cc13 - dc14 - ec15)e1 +
( ac12 - cc23 - dc24 - ec25)e2 +
( ac13 + bc23 - dc34 - ec35)e3 +
( ac14 + bc24 + cc34 - ec45)e4 +
( ac15 + bc25 + cc35 + dc45)e5

and a pattern becomes evident.

v|B =
(0 -bc12 - cc13 - dc14 - ec15)e1 +
( ac12 + 0 - cc23 - dc24 - ec25)e2 +
( ac13 + bc23 + 0 - dc34 - ec35)e3 +
( ac14 + bc24 + cc34 + 0 - ec45)e4 +
( ac15 + bc25 + cc35 + dc45 + 0)e5

Using this pattern, we may return to 3D and rewrite it similarly as

v|B =
(0 - bc12 - cc13)e1 +
(ac12 + 0 - cc23)e2 +
(ac13 + bc23 + 0)e3

A further simplification is possible as the product of a row vector e, a matrix M, and a column vector a.

e = [ e1 e2 e3 ]
M =
[
0 -c12 -c13
c12
0 -c23
c13 c23
0
]
a = [ a b c ]

v|B = eMa

Note that M is an antisymmetric matrix.

This pattern holds in any positive number of dimensions, even 1D where M =  and there is no magnetic field at all.
Last edited by PatrickPowers on Fri Jan 29, 2016 2:13 am, edited 19 times in total.
PatrickPowers
Trionian

Posts: 216
Joined: Wed Dec 02, 2015 1:36 am

### Re: Magnetism in N-D

So what would a drawing of such an N-D magnetic field look like?

Traditionally a 3D magnetic field is drawn as lines with arrows showing the direction of a pseudovector. If the magnetic field is a bivector, then for every dimension greater than two one could instead have little tilted discs that define the plane of the bivector at that point. One could make all of the discs the same diameter and draw them more densely where the field is stronger. Or have a smaller disc represent a stronger magnetic field. Smaller things spin more rapidly.

In two dimensions all of these circles are in the same plane -- the only possible plane -- so a scalar is sufficient to describe the magnetic field.
Last edited by PatrickPowers on Thu Jan 28, 2016 7:03 pm, edited 3 times in total.
PatrickPowers
Trionian

Posts: 216
Joined: Wed Dec 02, 2015 1:36 am

### Re: Magnetism in N-D

The basis of a bivector space in N-D has N!/(2!(N-2)!) elements. That's 45 basis planes in 10D, 4950 in 100D, and 499,500 in 1000 D.

The value of the bivector B may be calculated by summing the bivectors of each charged particle that produces the magnetic field.
PatrickPowers
Trionian

Posts: 216
Joined: Wed Dec 02, 2015 1:36 am

### Re: Magnetism in N-D

The bivector magnetic field B may be calculated by summing over all charges that produce the field. How? This is based on the physics described in http://physics.weber.edu/schroeder/mrr/MRRtalk.html. Essentially it shows that magnetism is a planar phenomenon, and tells how to find the orientation of that plane. The plane is defined by three points. These are the location of the charged particle, the location after dt, and the location of the other charged particle, the one upon which the magnetic field may act.

Let the value of the charge of each particle that generates the field be ci, the position of the particle be pi, and its velocity be vi. This plane is in the span of the vectors vi and p-pi. So to get the bivectorial value B of the magnetic field at point p, sum over civi^(p-pi) where ^ is the geometric algebra outer product. This is defined in http://faculty.luther.edu/~macdonal/GA&GC.pdf. This document shows that curl, divergence, Stokes' theorem, and so forth are easily extendable to N dimensions if geometric algebra is used as a base.

For computational purposes it would seem more convenient to express the bivectors in an orthonormal basis so that their sum may be more readily computed. One would also take advantage of symmetry and use integral calculus rather than a sum.
PatrickPowers
Trionian

Posts: 216
Joined: Wed Dec 02, 2015 1:36 am

### Re: Magnetism in N-D

This is very interesting. I'll have to sit down and work through these equations at some point -- it looks like a promising approach to derive a self-consistent model of 4D electromagnetism that doesn't suffer from the limitations and problems of naive generalizations from 3D.
quickfur
Pentonian

Posts: 2561
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: Magnetism in N-D

We went to the basic physics and concluded that magnetism is naturally represented by a bivector B such that the magnetic force is v | B. However there is a big disadvantage in that the established literature is heavily committed to the cross product. The cross product behaves more like the outer product ^ than the inner product |. So maybe it is better to use ^ instead.

This may be done by using dual spaces. In geometric algebra the dual space of anything is computed by multiplying it by eNeN-1...e1. For example, the dual of aeN is aeNeNeN-1...e1 = aeN-1...e1.

We now define the magnetic field as b as B*. Outside of traditional 3D this is a bit weird in that the magnetic field b is now an (N-2)vector. But now we can define a cross product that works in all numbers of dimensions. To prove this we will use an identity from MacDonald, A Survey of Geometric Algebra and Calculus.

With x and y free variables each of which is a blade (trust me for a moment that we are using only blades)

x | y = (x ^ y*)*

--

We claim that v x b = (v ^ b)*

Proof:

v x b = v | B = (v ^ B*)* = (v ^ b)*

So we have a cross product that works in N D. It isn't defined for general multivectors, just "blades", but since vectors, bivectors, and their duals are all blades, that is OK.

An example of a multivector that isn't a blade would be a vector plus a bivector. It would have elements of grade one AND grade two. Such is excluded from the domain of our new cross product. Blades only. If something even more general is desired then the inner product definition may be used, though I'm not sure whether it would have a physical meaning.

Note: there remains some finicky fiddling with signs. I'm out of time right now.
PatrickPowers
Trionian

Posts: 216
Joined: Wed Dec 02, 2015 1:36 am

### Re: Magnetism in N-D

x x y = -(x ^ y)*

So that is what the sign is. I'm not sure this is true in dimensions other that 3 though.
PatrickPowers
Trionian

Posts: 216
Joined: Wed Dec 02, 2015 1:36 am

### Re: Magnetism in N-D

The electric field is a Vector field. A moving electric field generates a magnetic field. A magnetic field is a bivector field. A moving magnetic field generates what?
granpa
Dionian

Posts: 20
Joined: Sat Nov 10, 2012 12:21 am

### Re: Magnetism in N-D

granpa wrote:The electric field is a Vector field. A moving electric field generates a magnetic field. A magnetic field is a bivector field. A moving magnetic field generates what?

See the Electromagnetism in 4D thread.
PatrickPowers
Trionian

Posts: 216
Joined: Wed Dec 02, 2015 1:36 am

### Re: Magnetism in N-D

PatrickPowers wrote:
granpa wrote:The electric field is a Vector field. A moving electric field generates a magnetic field. A magnetic field is a bivector field. A moving magnetic field generates what?

See the Electromagnetism in 4D thread.

Is it a vector field or a bivector field?
granpa
Dionian

Posts: 20
Joined: Sat Nov 10, 2012 12:21 am

### Re: Magnetism in N-D

PatrickPowers wrote: It's too much to show all of geometric algebra, so we shall use only four properties. First is antisymmetry -- reversing the order of the vectors in a bivector reverses the sign.

eiej = -ejei.

Then

eiei = -eiei = 0.

This is the wedge product, not the geometric product. They coincide if i =/= j, but the geometric product has eiei = 1 =/= 0.

So we have a cross product that works in N D. It isn't defined for general multivectors, just "blades", but since vectors, bivectors, and their duals are all blades, that is OK.

No, a blade is a multivector that is a wedge product of vectors.

e1e2e5 + e3e4e5 = (e1e2+e3e4)^e5 is not a blade. It has grade 3.

It is true that the dual of a blade is also a blade.
mr_e_man
Trionian

Posts: 75
Joined: Tue Sep 18, 2018 4:10 am 