4D Magnetism Solution

[PWrong]

Well, even though we don't have a mathematical framework for EM, we can still work things out intuitively.

I think tensors might be easier to visualise than you think. I can't remember where I read this, but apparently a rank-2 tensor (i.e. a matrix), isn't much more than a pair of vectors. You've probably heard that |u x v| gives the area of the parallelogram formed by u and v. Well, a tensor apparently represents the parallelogram itself. I'm not sure if it can be represented simply by a pair of vectors though.

Think of the magnetic field around a wire. It's just a set of concentric circles, but at each point in the circle there's a little vector pointing clockwise. I'm guessing that in 4D, you'd get lots of concentric spheres, and at each point, there's a tensor pointing in two perpendicular directions. They cover the whole sphere, but no point can be singled out as unique. So for instance, you couldn't have them all pointing North and East. Maybe that's what they mean by a form field.

Maybe we should take the concepts from bo198214's link, and generalise them a bit. For instance, we could replace the infinitely long wire with a square loop, then a circular loop, and may even an arbitrary parametric curve. Or we could consider two wires, first parallel, then perpendicular, then at different angles.

If we can do all this in 3D, it shouldn't be hard to do it in 4D. What I'm wondering is, what shape should the wire be? Lines and circles are the obvious shapes in 2D. In 3D, we usually consider coils of wire, i.e. a helix shape. But I can't think of any interesting shapes to make with wire in 4D...

[jinydu]

Actually, a matrix (in general) is made up of an arbitrary number of vectors, not just two. A rank-3 tensor is in turn made up of an arbitrary number of matrices, etc.

But I'm quite sure you're right about the concentric spheres; and I think that your description explains why vector calculus alone is insufficient for 4D magnetism: Every point on a sphere requires two vectors to specify the tangent plane at that point. I think that also ties in nicely with what I read about form fields: For every point in (4D) space around a wire, there is a unique "magnetic sphere" passing through that point; presumably, the form describes the behavior of that sphere infinitessimally close to the point. Furthermore, since we know that the surface area of a sphere is proportional to the distance from its center (i.e. the distance from the wire), we expect the magnetic field will be governed by an inverse-square law. Perhaps we can use this as a starting point to deriving equations for magnetism in 4D; this was the starting point in my physics class (though, of course, they derived equations for 3D).

Actually, there is already a formula to find the magnetic field due to an arbitrary arrangement of current (in 3D, of course). It is called the Biot-Savart Law and, not surprisingly, it involves a cross product.

[thigle]

But I can't think of any interesting shapes to make with wire in 4D...

just an amateurish trial of mine:

taking 2 mobiuStrips, then joining them along their boundaries (1 lemniscate for each), a Klein bottle is born.

now this KleinBottle bears a trace of the process of its formation - a scar along the seam where the 2 strips' edges merged together.

is this an interesting loop in 4d, or not at all ? i mean: the edge of mobius strip. (which is some kind of 'all-round-the-surface-curve' or maybe some geodesic of KleinBottle ? )

[thigle]

also, what about this one: PhysicalSpace as Quaternion Structure: Maxwell equations.?

especially, starting from the paragraph V. Conclusion, one can read:

...as a consequence of this, an important field component went missing in Maxwell's equations, and and all of modern physics have developed from there, perpetuating one of consequences of this oversight, namely, that the electromagnetic field possesses six components, whereas, as we have shown, there should be seven.

[bo198214]

Ok, folks, now I have a more concrete generalization.
Though I do not involve relativistics, they showed me the way.
Result: Each magnetic field in a point can be given by an oriented plane through that point plus the field strength (scalar value) in any dimension (so if you want, you can write it as an nx2 Matrix, congratulations to Jinydu). It resembles also a bit the example with the small plate-compass in each point (by PWrong?) and was the mentioning of the two vectors (by PWrong) already the right intuition? Who knows.
I havent done the computations for dynamicly changing fields, but that wasnt really the aim. Read my

computations/explanations

for the non-dynamic case. For me it has similarity with rotations. In 3d you have a rotation axis and ask yourself how this can be generalized to n dimensions. The answer is: use the perpendicular plane instead of the axis. Same with magnetic field.

[PWrong]

Actually, there is already a formula to find the magnetic field due to an arbitrary arrangement of current (in 3D, of course). It is called the Biot-Savart Law and, not surprisingly, it involves a cross product.

But there's also a formula to find the force vector given the magnetic field and the velocity, which also involves a cross product. So we're really getting one vector from three. I think we should be able to cancel the two products out. Here's the two formulas:

B = Integral[ u/(4pi r²) r cross dI ]
F = E + v cross B

Let's forget about the integral (assume the wire is straight) and ignore the constants and the electric field. So we'll pretend that in 4D:

F = v^B = v^(r^I)
where ^ is the elusive wedge product that we know nothing about.

Now, there is a cross product for three vectors in 4D, but I don't think this is it. The force is perpendicular to the magnetic plane, which is perpendicular to r and I. So the force is not perpendicular to these two vectors. In fact, it has to be in the same plane.
So we know that F is perpendicular to v, and is a linear combination of I and r. That is:

F . v = 0 and
F = a I + b r
for some a & b
We could solve this to find the direction of the force, but not the magnitude.

By the way, have a look at Introduction to Tensors for Students of Physics and Engineering
It's a good explanation of tensors without as much confusing notation, although I don't think it has the wedge product.

[bo198214]

ahm, did you read my previous post, or are you ignoring me too?
I gave an operation there, that does work in any dimension, when regarding the magnetic field as an oriented plane with a scalar value (or as two vectors or as an nx2 matrix).

[PWrong]

I just realised that the wedge product is just a Null-Space! I don't know if either of you know about these, but I learned them in linear algebra last year (don't you hate it when boring subjects turn out to be useful?)

We know that the wedge product of a set of vectors A, is another set of vectors X, so that every vector in A is perpendicular to every vector in X. That means all their dot products equal 0. But in matrix notation, we can write this as:
AX = 0

Now the null space of a matrix A is defined as the general solution to
Ax = 0
It's usually represented by a matrix of its basis vectors, and this matrix is some constant multiple of the wedge product.

Mathematica has a function called NullSpace. Unfortunately, you usually have to multiply by some constant to get the wedge product. I haven't yet worked out what that constant is.

For instance:
NullSpace[{{2, 0, 0, 0}, {0, 1, 0, 0}}]
= {{0, 0, 0, 1}, {0, 0, 1, 0}}

but the wedge product would be
{{0, 0, 0,2}, {0, 0, 2, 0}}

[bo198214]

obviously, yes *speechless*

[PWrong]

Sorry bo198214, you must have posted while I was writing. I only skimmed over your link . I didn't realise you wrote it yourself. I'm not just ignoring you.

Your link is a good idea, but I'm not sure that the identity a^(b ^ c) = (a.c)b - (a.b)c applies in every dimension. But now that we have the wedge product, I'll see if I can prove it.

[PWrong]

It just happened again, . Everything's in the wrong order. I honestly didn't mean any offence .

[bo198214]

No, the idea is:

Every magnetic field B (in 3d) arises by a movement of an electrical field E.
And I now propose that instead of characterizing the magnetic field
by the vector vxE, we characterize it by the pair (v,E). Or more exactly by the oriented plane perpendicular to vxE (that is the plane spanned by v and E) plus the scalar |vxE|=|v||E| (for simplicity I disregard the scalars epsilon0 and mu0).
With this characterization we equally can compute the force of a moving charge q in the magnetic field. By using the new operator between v2 and (v1,E) instead of computing v2 x (v1 x E).
And: this new operator is applicable to arbitrary dimensions (even 2).
Except the correction factor gamma and all improvements over the Maxwell-Theory, the relativistic consideration (in n dimensions) would result in the same operator. You can verify it for example for 2 dimensions.

This operator can also easily imagined: When having a moving test charge in the magnetic field given by the plane X. Then first project the velocity v on the plane, then take the perpendicular vector to v' in X. And multiply it with the scalar value of the magnetic field. This is the force on the test charge. This is of course also true in 3d: When having a coil, the magnetic field in the center is given by the plane that each twist spans. (or traditionally it would be the vector perpendicular to the plane). A test charge that moves perpendicular to that plane receives no force (or tradionally a test charge that moves a along the magnetic field line receives no magnetic force).

But thats only examples to make it more visible. The general operator works in n dimensions, and yields the same results as the Maxwell-Theory in 3 dimensions (and is additioinal even similar to the n-d relativistic generalization).

[PWrong]

And I now propose that instead of characterizing the magnetic field
by the vector vxE, we characterize it by the pair (v,E). Or more exactly by the oriented plane perpendicular to vxE (that is the plane spanned by v and E) plus the scalar |vxE|=|v||E| (for simplicity I disregard the scalars epsilon0 and mu0).
With this characterization we equally can compute the force of a moving charge q in the magnetic field. By using the new operator between v2 and (v1,E) instead of computing v2 x (v1 x E).
And: this new operator is applicable to arbitrary dimensions (even 2).

I agree with everything you've said, but your notation, (v,E) is exactly the same as the wedge product v^E.

By the way, the identity you used does hold after all. That is,
a ^ (b ^ c) = (a.c)b - (a.b)c
in any dimension. It's actually quite easy to prove this using the null space definition, so I'll leave the proof as an exercise . [/b]

[bo198214]

Oh, I didnt realize that the wedge product you mention is a concrete known product. Have never heard about it and thought you mean a to be formulated operation.
I didnt found about it in your given reference (at least not by searching for "wedge"). Do you have a reference?
So we considering the same thing? oO
I also wonder what signature this wedge product has.

[jinydu]

I edited my long post to say something more about divergence.

The wedge product is definitely part of accepted mathematics, although I haven't formally learned it yet. In fact, it seems that forms are based on wedges.

http://mathworld.wolfram.com/WedgeProduct.html

[bo198214]

If the wedge product is something that operates on nd-space and as PWrong explained takes a set of vectors as arguments and yields a set of the same number of vectors, then I dont understand why x^(y^z) is again a vector (as the expression (x.y)z - (x.z)x suggests).

[houserichichi]

In three dimensions you'd have something of the form



where there is no need to use brackets since the wedge product is associative. This entity (notice the final bit on the end) is a trivector and can be interpreted as a directed volume. If we were to wedge just e_1 and e_2 we would have the more simple bivector which can be interpreted, as PWrong said, as the parallelogram itself. A bivector is a subset of a directed plane. They are not vectors, they are not scalars.

[PWrong]

If the wedge product is something that operates on nd-space and as PWrong explained takes a set of vectors as arguments and yields a set of the same number of vectors, then I dont understand why x^(y^z) is again a vector (as the expression (x.y)z - (x.z)x suggests).

Actually, it doesn't give you the same number of vectors. If we're working in nD space, and it takes k vectors as arguments, then it gives (n-k) vectors. It actually gives you the subspace containing all the perpendicular vectors.

So in this case, we're in 4D space. y^z is a 2D subspace (because 4-2 =2).
So x^(y^z) is the wedge product of three vectors (a vector and a 2D space), so we get a 4-3 = 1D subspace, i.e. a vector.

Generally, if we're working in nD space, then y^z is an (n-2)D subspace. x^(y^z) has (n-(1+n-2)) = 1 dimension.


I think Maxwell's equations could work exactly as they are in 4D.
The first two are just Gauss's law, which apply easily.

Del ^ E gives a bivector, which is the same as dB/dt
Del ^ B is one vector, which is the same as J and dE/dt.

So we don't have to change anything .

In practice, I imagine it'll be difficult to solve even simple systems. I think an important first step would be to design a 4D compass. In 3D, a solenoid (coil) can act as a simple compass, but I suppose a single loop of wire would work just as well.

[thigle]

sorry to interrupt this technical intermezzo, but i wanna repeat that (even though i am WAY beyond all you guys in terms of what you discuss technically through the last few posts),
i truly believe that you can not suceed in your attempts at finding 4d magnetism rules, if you do not start to use quaternions, however counterintuitive they might seem.

without maxwell's original quaternions formulations, you're all just re-inventing the wheel. original Maxwell's equations were QUATERNIONIC, and there is NO OTHER way to FULLY grasp the 4-dimensional nature of EM dynamics.

please read & do not ignore: quaternion physics
as well as the story behind anti-quaternionic campaign that transformed physics
this, containing Maxwell's hyperdimensional poetry (dig it !)

i know, the last 2 links are at pages that 'normal' people would avoid. but then, normal people are conformally mad.

[bo198214]

Ok, strictly speaking, the wedge product takes 2 sets of vectors and yields another set of vectors, right? Or more strictly speaking (the order of the vectors may play a role for orientation): The wedge product takes two oriented "scaled" subspaces and returns such an oriented scaled subspace.

Now you simply should post the actual definition (unfortunately I cannot see how the definition of houserichichi/wikipedia/wolframscience applies to the usual n-d space) and then also I can verify that our both definitions coincide (what seems quite evident to me, though I want to see the real thing instead of "so I'll leave the proof as an exercise"). I mean the Nullspace does not say anything about orientation which seems quite necessary. On the other hand it seems to suffice to work with unit vectors because the lengthes of the vectors simply multiply, and the rule given by houserichichi/wikipedia/wolframscience should apply. So what is needed are the wedge products of the e_i.

But quite cool how we came to the same solutions, indeed not the same but yielding the same result.

you can not suceed in your attempts at finding 4d magnetism rules, if you do not start to use quaternions

We already did succeed. Its not always true what sounds most divine.

[thigle]

bo, i cannot resist, but are you sure you know what you dismiss ?: on maxwell again
maxwell original quaternion theory
and most importantly: 34 flaws of classical EM theory

just asking...

[bo198214]

Thigle, the aim was not to find the best EM theory, or to improve the Maxwell-Equations. I think the main curiosity was how a force on a moved charge in a magnetic field in 4d would be directed. Because there would be various orthogonal forces to v and B instead of just vxB in 3d. And we have now found a fine generalization for arbitrary dimension, even independently came to the same result.
Whether the original Maxwell equations are correct or not was not much the point here. At least for some important cases they produce correct predictions.

That they are incorrect shows already the two electron example. The relativistic approach can at least correctly predict the two electron example. But has in no way to do with quaternions either. I dont know the original 20 Maxwell Equations with 20 unknowns. But if they are so much different, where is the (thought) experiment, that is correctly predicted by them and wrongly predicted by the relativistic approach? And why should they yield a more comprehensible generalization to 4 or arbitrary dimensions, especially if nobody here knows about those equations?

[thigle]

well, i just never gonna stop inquiring into the quaternions, even though noone seems to like them and i don§t seem to understand them

Thigle, the aim was not to find the best EM theory, or to improve the Maxwell-Equations.

thanx for clarifying i got lost

the main curiosity was how a force on a moved charge in a magnetic field in 4d would be directed.

wow, now, bearing this in mind, i have to read this whole topic again.

i always keep babbling about quaternions, because they provoke me by my inability to understand them yet, in hope (sometimes fulfilled) that someone will spit off some tidbits of quaternion-knowledge and i somehow get it.

[thigle]

but,

if they are so much different, where is the (thought) experiment, that is correctly predicted by them and wrongly predicted by the relativistic approach? And why should they yield a more comprehensible generalization to 4 or arbitrary dimensions, especially if nobody here knows about those equations?

i think 2 of the links i gave, give some definite answers to your questions, in terms of chirality, lost of topological complexity, etc.

i always keep looking for the obscure & rare.

[PWrong]

i always keep babbling about quaternions, because they provoke me by my inability to understand them yet, in hope (sometimes fulfilled) that someone will spit off some tidbits of quaternion-knowledge and i somehow get it.

Start a new thread on quaternions, and what you want to know about them, and I'll tell you everything I know (which isn't that much). :wink:

Edit. Oh, moonlord already did. I'll just post in there now. :wink:

Now you simply should post the actual definition (unfortunately I cannot see how the definition of houserichichi/wikipedia/wolframscience applies to the usual n-d space) and then also I can verify that our both definitions coincide (what seems quite evident to me, though I want to see the real thing instead of "so I'll leave the proof as an exercise").

The reason I left the proof as an exercise is that it's really hard to write matrices on here. The proof is easy (if you know matrices pretty well), but it takes a while to write out.

I'll show you some examples of finding the wedge product, using systems of equations. Remember, I still don't know how to find the magnitude of the wedge product.

First a very simple example.
(1, 0, 0, 0) ^ (0, 1, 0, 0) =
[ 0 0 1 0 ]
[ 0 0 0 1 ]

That is, the wedge product of a vector along the x-axis and along the y-axis, is the zw plane, with magnitude 1.

Let's find the wedge product of (1, 2, 0, 3) and (2, 0, -1, 2)
We start with the equation A x = 0
[ 1 2 0 3 ] [ x1 ] = [0]
[ 2 0 -1 2 ] [ x2 ] = [0]
...................[ x3 ]
...................[ x4 ]

As a system of equations, this means:
x1 + 2 x2 + 3 x4 = 0
2 x1 - x3 + 2 x4 = 0

We solve this to find that:
x1 = x3 /2 - x4
x2 = -x3 /4 - x4
x3 = x3
x4 = x4

Now x3 and x4 become arbitrary parameters. So any vector x that satisfies the equation must be a linear combination of these two vectors: (1/2, -1/4, 1, 0) and (-1, -1, 0, 1)

These vectors are a basis for the null space, and they are also a basis for the oriented plane that forms the wedge product.

Now I'll try to give a sketch of the proof that a^(b^c) = (a.c)b - (a.b)c in 4D, by showing that it satisfies all the equations.

b^c is a matrix with two vectors. Call them u and v
Let y = a^(b^c)
Also let A be the matrix containing b and c as rows.

By definition, Au = 0 and Av = 0
Therefore:
b.u = c.u = b.v = c.v = 0

now let B be the matrix containing a, u and v as rows. Then by definition,
By = 0
so,
a.y = u.y = v.y =0

Let y = (a.c)b - (a.b)c
Then a.y = (a.c)(a.b) - (a.b)(a.c) = 0
u.y = (a.c)(u.b) - (a.b)(u.c)
v.y = (a.c)(v.b) - (a.b)(v.c)
But b.u = c.u = b.v = c.v = 0, so u.y = v.y =0

So y satisfies the equations. QED.

[bo198214]

Look what I'v found!
Re: Magnetism as relativistic electrostatics. A news group thread including relativistic magnetism and wedge product!

[jinydu]

Del ^ E gives a bivector, which is the same as dB/dt
Del ^ B is one vector, which is the same as J and dE/dt.

So you're thinking about Del as a vector and replacing the curl with wedge multiplication?

[bo198214]

Ok, its time to bring some light in the terminology chaos.
The usual wedge Operator is an operator on the exterior algebra /\V. It takes a k-Vector and an l-Vector and the result is a l+k vector. The 1-vectors are the vectors of the vector space V.

PWrong defined a wedge operator as an operator on the subspaces. It takes an k subspace and an l subspace and yields an n-(k+l) subspace.

Then there is the hodge operator * on the exterior algebra. It takes an k-vector and yields an n-k-vector. Then the wedge operator followed by the hodge ^* takes a k-vector and an l-vector and yields an n-(k+l) vector. It seems again something different from what PWrong (though not completely) defined.

For the operation of PWrong and ^*, for vectors a,b,c is a ^* (b ^* c) a vector again. And I would suppose that
a ^* (b ^* c) = (a.b)c - (a.c)b
though I can not prove it because the hodge operator is not explicitly given in the wiki.

Remember, I still don't know how to find the magnitude of the wedge product.

I would suppose that the absolute value is simply the product of the lenghtes of both arguments. The only difficulty is the sign. Thatswhy I insist on the orientation. And the sign is also the remaining problem in your proof. But PWrong, if you learned about it in linear algebra, there must be a definition somewhere (in your books, or scripts)!

But that the magnetic field is only a helping construction can also be seen by: The cross product is not invariant to changing the orientation of the coordinate system but changes the sign (chirality). This must not happen for physical laws. But it doesnt matter because the double application of the cross product is invariant.
So I would prefer to put the both Maxwell equations (without epsilon0 and mu0):
curl B = j + dE/dt
curl E = - dB/dt
into one via:
curl (dB/dt) = dj/dt + ddE/(dtdt)
- curl (curl E) = dj/dt + ddE/(dtdt)
curl curl E = del x (del x E) = (del.del)E - del(del.E) = lap E - grad(div E)
- grad div E + lap E = dj/dt + ddE/(dtdt)

lap E - ddE/(dtdt) = dj/dt + grad div E

which is then applicable to arbitrary dimensions too (and invariant to orientation change of the coordinate system).

[PWrong]

But PWrong, if you learned about it in linear algebra, there must be a definition somewhere (in your books, or scripts)!

I only learnt the nullspaces in linear algebra, not wedge products. The nullspace doesn't have a magnitude.

I would suppose that the absolute value is simply the product of the lenghtes of both arguments. The only difficulty is the sign.

That's only if the arguments are perpendicular. I think the magnitude is |a||b|sin(theta), where theta is the angle between the vectors. But I also don't know how to find the magnitude of a bivector, or even if they have a magnitude at all:?.

- curl (curl E) = dj/dt + ddE/(dtdt)

curl (curl E) always equals 0, for any E.

There's another way to find the wedge product, but I don't know how to express it as two vectors. You just express the arguments in component form, and use the properties of the wedge product to simplify it.
For instance:
(ai + bj) ^ (ci + dj)
= ac i^i + ad i^j + bc j^i + bd j^j
= (ad - bd) i^j

[houserichichi]

(ad - bc) i^j is a 2-form which is the dual of a bivector. It can be shown that every 2-form in the tangent space of R3 can be expressed as the product of two 1-forms (the dual of contravariant (normal) vectors), but I'm not sure if that follows through to R4. I'll see if I can find that out.