4D Magnetism Solution

[PWrong]

(ad - bc) i^j is a 2-form which is the dual of a bivector.

Oh ok. I didn't realise they were different things :?. So which one is the parallelogram (or in this case, square) formed by i and j? And is the dual of the 2-form perpendicular to the 2-form, or not?

Let's look at the infinitely long straight wire again. I've almost solved it completely. In fact, I think I now understand 4D wires better than 3D wires. The maths is pretty long and difficult, and I haven't explained every step, but I've checked it a few times. If it's at all comprehensible, I wouldn't mind some questions or corrections. If not, you might want to just skip to the end.

Suppose the wire goes up the w-axis. Since we have symmetry along this axis, we'll separate all the vectors into a w component and an x,y,z component. So, R = w + r = (0,0,0,w) + (x,y,z,0)

Clearly, the electric field will fall off with 1/r^2. For the magnetic field, we have:
B= (mu I/2pi²) Integral[ 1/R³ dw^ R^~ ]
R^~ is R-hat, the unit vector in the direction of R

We already worked out that this field is in the form of spherical shells. It's perpendicular to the w-axis and the r vector. So the "direction" of B is w^~^r^~

My physics textbook shows how to find the magnitude of B in the 3D case. The 4D case isn't much more difficult. We draw a diagram and use the fact that |dw^R| = |w||R| sin(theta), and that sin(theta) = sin(pi - theta) = r/R. In n dimensions, we end up with:
B = (mu I/2pi²) Integral[ r / (r²+w²)^(n/2) dw]
In 4D, when we integrate from (-infinity) to infinity, we get
B = mu I / (4pi r²)

Now we can find the force on an arbitrary charge.
F = q (E + v^B)
E = lambda / (epsilon * 4 pi r²) r^~ = a/r² r^~
B = mu I / (4pi r^2) w^~^r^~ = b/r² w^~^r^~

Using the indentity I established before,
F = q/r² (ar^~ + b v^(w^~^r^~
= q/r² (ar^~ + b [ (v.w^~)r^~ - (v.r^~)w^~ ] )

Let's get rid of some of those hats.
F = q/r³ (a r + b [ (v.w^~)r - (v.r)w^~ ] )
Now we separate F into two components and use F = ma. Note that (v.w^~) = w'(t)

m r''(t) = q/r³ (a + b w'(t) ) r
m w''(t) = - b q / r³ * (v.r)
Now we have some relatively simple differential equations that we could solve to find the motion of the charge.

Now it turns out that v.r = r(t) r'(t), so
m w''(t) = - b q r'/ r^2
Integrate both sides:
m w'(t) = b q / r + m w'(0)
Subsitute into the other equation:
m r''(t) = q/r³ (a + w'(0) + q b²/(m r) ) r

Now we can interpret this easily. If b² << a + w'(0), we get the same equation as 3D gravity, so charges move in ellipses around the wire . If q b² /m >> a + w'(0), we get the equation for 4D gravity, and charges go all over the place . It also moves up and down the w-axis a bit (I haven't worked that part out yet).

Unfortunately, we can't compare this to a wire in 3D, because I haven't worked 3D wires yet. I probably should have done that first .

[houserichichi]

Oops, forget what I said about (ad-bc)i^j...it's a bivector since i and j are vectors. I don't know what the heck I was thinking...

[bo198214]

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

And unfortunately your Nullspace-product has not much to do with the wedge product, albeit you can not even compute the magnitude.

A k-vector can be regarded as a (n over k) dimensional vector or as n-dim tensor with k indices. So I dont see how you will identify a k-vector with a k-subspace. If you see, then tell.

To finish the dragging discussion about the "unknown" wedge product I now finally proved that a ^* ( b ^* c ) = (-1)ⁿ ( (a.b)c - (a.c)b ) where ^* is the usual wedge product followed by the hodge operator. For n=3 is already known that a ^* b = a x b and the above formula results in a x ( b x c ) = (a.c)b - (a.b)c. Though of course it makes not much sense (for me) to compute an (n-2)-ary tensor a ^* b instead to directly use the last equation.

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

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

Please inform yourself before stating nonsense. It should already be conspicuous that always dj/dt=-ddE/(dtdt) when assuming your assertion. So for a static electric field it may be true (because there is no magnetic field at all), but of course not for a dynamic field. For more details see also The curl of the curl.

Let's look at the infinitely long straight wire again...I wouldn't mind some questions or corrections

The wire is electrostaticly uncharged, so the force on q is merely
F = q v^B instead of F = q (E + v^B)

The equation for the magnetic force can also derived by computing the electric field of the line charge (by similar integration as you used):
E=(lambda/(4pi eps)) r/r³
then using the already established equation
F=qc^-2 ((v_q.E)v_E-(v_q.v_E)E)
where v_E=wI/lambda is the velocity of the current and v_q the velocity of the test charge. So that we have
F=(mu qI/4pi r³) ((v_q.r)w-(v_q.w)r)
which is the same as your result except the sign.

Now I would assume that two parallel equally directed currents in 4d also have an attractive force as in 3d, i.e. for v_q=w the force for r perpendicular to w should be attractive:
F = (mu qI/4pi r³) ((w.r)w-(w.w)r)=-(mu qI/4pi r³) r

So the sign was correct (qI>0) and we see that we must apply a factor (-1)ⁿ to the above a ^* ( b ^* c ) when computing forces to get the right results in n dimensions.

[PWrong]

Please inform yourself before stating nonsense. It should already be conspicuous that always dj/dt=-ddE/(dtdt) when assuming your assertion. So for a static electric field it may be true (because there is no magnetic field at all), but of course not for a dynamic field. For more details see also The curl of the curl.

Sorry, you're absolutely right. I must have been thinking of the divergence of a curl.

And unfortunately your Nullspace-product has not much to do with the wedge product, albeit you can not even compute the magnitude.

The wedge product gives a plane and a magnitude. The nullspace just gives the plane. Apart from the magnitude, they have the same definition. The magnitude is |a||b|sin(theta), so in most cases it's relatively easy.

A k-vector can be regarded as a (n over k) dimensional vector or as n-dim tensor with k indices. So I dont see how you will identify a k-vector with a k-subspace. If you see, then tell.

I'm still not certain what a k-vector is, technically. But it's obvious that k vectors define a kD subspace (like a line or a plane), but not the other way around. Also, for every k-subspace in nD, there is exactly one (n-k) subspace perpendicular too it.

which is the same as your result except the sign.

I had a feeling I got a sign wrong somewhere. It doesn't really matter. You can always change the sign of the charge or the direction of the current.

The wire is electrostaticly uncharged, so the force on q is merely
F = q v^B instead of F = q (E + v^B)

I never said it was electrostatically uncharged. It's better to consider the effects of both forces. If you want only the magnetic force, let a=0. If you just want an infinite line of charge, let b=0.

I think it's important to note that charges move in elliptical orbits around an infinite line of charge. It suggests that there may be 4D galaxies containing long strands of charged suns, which planets could orbit around. The magnetic field adds an element of 4D instability, and also makes the charges travel up and down the wire. I think charges tend to move in elliptic toroidal orbits in general, and I'm trying to see this in mathematica.

[bo198214]

The wedge product gives a plane and a magnitude.

Not exactly, for example for n=3:
(a₁ e₁ + a₂ e₂ + a₃ e₃) ^ (b₁ e₁ + b₂ e₂ + b₃ e₃ )
= (a₁b₂- a₂b₁) e₁^e₂ +(a₂b₃- a₃b₂) e₂^e₃ + (a₁b₃- a₃b₁) e₁^e₃
So one could regard it as a linear combination of all "elementary" planes of Rⁿ.

The nullspace just gives the plane.

Only in 4D, otherwise n-2-subspace (for two vectors). I dont know why you always intermix the wedge and its dual. The nullspace of k-dim and l-dim subspace is n-(k+l) dim, where the wedge of k-vector and l-vector give k+l vector (and not n-(k+l) vector) but the wedge followed by the hodge star (this is the dual) returns similarely to the null-space an n-(k+l)-vector.

The magnitude is |a||b|sin(theta), so in most cases it's relatively easy.

What is |b|? a is a vector and b is an n-2 subspace.

I'm still not certain what a k-vector is, technically.

Its simply a linear combination of all the elementary k-subspaces. There are (n over k) elementary k-subspaces.

I think it's important to note that charges move in elliptical orbits around an infinite line of charge.

But its the same for an infinite line of mass.
So maybe in a 4d universe the masses have to arrange in threads. But whether this is stable at all ... who knows. I could imagine the threads become longer by masses getting attracted in the direction of the axis (which has the strongest gravity). But then also the protons have to arrange to threads. Would you mind to compute the orbits around a ring?
Though there is the question how to visualize, but perhaps the movement takes only place in a 3d subspace. (But better put in the 4d gravity thread and not here.)

[PWrong]

Only in 4D, otherwise n-2-subspace (for two vectors)

Yes, that's what I meant.

I dont know why you always intermix the wedge and its dual. The nullspace of k-dim and l-dim subspace is n-(k+l) dim, where the wedge of k-vector and l-vector give k+l vector (and not n-(k+l) vector) but the wedge followed by the hodge star (this is the dual) returns similarely to the null-space an n-(k+l)-vector.

Oh ok, I understand all this now. Just read every ^ as ^* in my earlier post.

What is |b|? a is a vector and b is an n-2 subspace.

a and b are the two vectors. Although I guess they could be any kind of subspace.
|a ^* b| = |a||b| sin(theta)

But its the same for an infinite line of mass.

Yes, but the magnetism might make it more interesting. Actually, I've heard that general relativity is a kind of magnetism for gravity, so we might see the same effects with gravity. But GR is obviously a lot more complicated. But if it's concievable that magnetism in 4D could be a very long range force, like gravity. All we need is some system where charges don't get mixed up and cancel each other out.

I could imagine the threads become longer by masses getting attracted in the direction of the axis (which has the strongest gravity). But then also the protons have to arrange to threads.

If you had some kind of giant sticky reaction in which one mass splits into two, but a lot of matter stays connected between them, that would result in lots of straight threads. You'd need an extra force to keep it sticky though.

I'll try to work out a circular current loop tomorrow.

[bo198214]

What is |b|? a is a vector and b is an n-2 subspace.

a and b are the two vectors. Although I guess they could be any kind of subspace.

Yes, you have deal with the case 1 x 1 and with the case 1 x (n - 2) if you want to substitute the cross product in a x ( b x c ).

But its the same for an infinite line of mass.

Yes, but the magnetism might make it more interesting.

More complicated - if you like it ...

Actually, I've heard that general relativity is a kind of magnetism for gravity, so we might see the same effects with gravity.

Already you can compute the "gravito-magnetic" effects with special relativity in the same way as you compute magnetic forces for electrostatic fields (because equation for electric forces is same as for gravitational forces) as in the paper mentioned in the beginning of this thread. Though the result, as I can overview it, is that the increased force cancels out by the increased mass (regarding accelleration).

I'll try to work out a circular current loop tomorrow.

Thanks, for visualization it suffices to do it in 3 dimensions (i.e. let the forth dimension=0).

[jinydu]

Hmmm... I thought I had finally understood a more intuitive way of understanding areas in 4D, but my linear algebra professor pointed out a flaw in my thinking...

Back to the drawing board...

It's also becoming increasingly clear how far I am from fully and correctly understanding Maxwell's theory... I may have to wait for upper division courses.

[PWrong]

It turns out that the magnetic field for a circular current loop is difficult even in 3D. If the loop is in the xy plane, the only simple case is when the charge is on the z axis. I tried to solve the general problem geometrically, but I got some ugly answer. Then I gave up and used mathematica to solve it directly.

Suppose we have a loop of wire in the xy plane, radius R. We have a charge somewhere in the xz plane. I'm using the 4D formulas (i.e. inverse cube laws), but I'm ignoring the w axis. Here's the mathematica code. xs is the position vector of the charge, and ls is the parametric equations for the loop of wire, with s as the parameter. BiotSavartdB gives dB/ds, given xs and ls.

BiotSavartdB[xs_, ls_] := Cross[Derivative[1][ls], xs - ls]/((xs - ls) . (xs - ls))^2

Now we integrate the magnetic field for our circular loop of wire from 0 to 2 Pi.

Integrate[(BiotSavartdB[{x, 0, z}, {Cos[s], Sin[s], 0}]), {s, 0, 2 Pi}, Assumptions -> x > 0 && x != 1 && z > 0] // OutputForm

Code: Select all

               4 Pi x z
{------------------------------------, 0,
   4      2        2          2 2 3/2
(x  + 2 x  (-1 + z ) + (1 + z ) )

                      2    2
           2 Pi (1 - x  + z )
  ------------------------------------}
    4      2        2          2 2 3/2
  (x  + 2 x  (-1 + z ) + (1 + z ) )

This is an algebraic expression in terms of x and z, which is nice. The y component of the field is zero, so the magnetic field is always either inwards, or in the z direction. This makes perfect sense. I graphed a contour plot of the magnitude of this in xz, and it looks pretty good.

All that's left to do is to find the force on a charge, which should be a simple calculation, and then solve the differential equations (numerically, obviously). Then we'll extend to 4D and do it again.

[bo198214]

Now I finally know, what the wedge operator can be good for: for adding fields.

Usually we have the superposition principle, i.e. we can add fields and then expect that the forces also add, for example:
E=E₁+E₂ then F=q eps (E₁+E₂) = q eps E₁ + q eps E₂= F₁ + F₂

My original definition for for the magnetic field to take only the pair (v_E,E) doesnt have this behaviour. F on the test charge q was computed by:

F=(q/c^2)((v_q.E)v_E-(v_q.v_E)E )

where v_E was the velocity of the electrical field E and v_q the velocity of the test charge. If we now for example choose H₁=(e₁,e₂) and H₂=(-e₁,-e₂) i.e. they both lie in the same plane but rotated (for example two opposite points in the computation of the magnetic field of the unit circle). Then we know the magnetic field (in the traditional sense) points in the same direction. So we cannot simply take the component wise sum of H₁+H₂=(e₁,e₂)+ (-v_E,-E)+(v_E,E)=(0,0) for adding up the forces because (q/c^2)((v_q.0)0-(v_q.0)0 ) =0.

And here the wedge can help. Let us call the v_E^E the magnetic 2-vector. Then we already know that the force can be computed by
F = (-1)ⁿ(q/c²) v_q ^* (v_E^E)^*=(q/c^2)((v_q.E)v_E - (v_q.v_E)E ).
I extend my original defintion of |x to 2-vectors as second argument in the following way:
a |x (e_i^e_j) := (-1)ⁿ a ^* ( e_i ^ e_j)^* = (a.e_j)e_i - (a.e_i)e_j
(together with linearity of course)

So for magnetic 2-vectors H we can compute the force as usual (without the detour of going 2 -> n-2 -> n-1 -> 1):
F = (q/c²) v_q |x H
but with the advantage that we now can simply add up the magnetic fields:
F=(q/c²)v_q |x (H₁ + H₂) = (q/c²)v_q |x H₁ + (q/c²)v_q |x H₂ = F₁ + F₂
because wedge and hodge are distributive.

So for 4 dimensions PWrong simply need to exchange the Cross by ^ in his computation of the magnetic field. And later on to compute the force he simply uses |x instead of Cross.

[PWrong]

Now I finally know, what the wedge operator can be good for: for adding fields.

I don't see exactly what you mean. I know the wedge product is distributive over addition; that is, (a+b)^(c+d) = a^c+a^d+b^c+b^d, but I don't see your point.

where vE was the velocity of the electrical field E and vq the velocity of the test charge.

Do you mean v_E is the current I? I use l(s) for the loop of wire. In this case, l(s) = {Cos[s], Sin[s], 0}. The direction of current is the tangent vector dl/ds.
I'm getting confused with your notation. I haven't seen the notations (a,b) or |x used anywhere before.

And here the wedge can help. Let us call the vE^E the magnetic 2-vector.

The magnetic field is already defined, and it has an extension to 4D via the Biot-Savart law.
B = Integrate[ 1/r^4 dl ^* r ],
where r is the vector from the charge to part of the wire, that is, r = x - l

So for magnetic 2-vectors H we can compute the force as usual (without the detour of going 2 -> n-2 -> n-1 -> 1):

You still have to integrate over the wire. Even though it makes little difference, I find it's easier to integrate the magnetic field first, i.e. compute v_q^(Integral[ dB ]) rather than Integral[v_q^dB].

So for 4 dimensions PWrong simply need to exchange the Cross by ^ in his computation of the magnetic field.

I already did :?

By the way, I've discovered that the magnetic field I calculated before is much simpler in 4D than 3D. It's a simple algebraic function, where the 3D magnetic field involves elliptic functions. I'm still working on getting a set of differential equations.

[bo198214]

*Sigh* we really seem to live in different worlds.

I don't see exactly what you mean. I know the wedge product is distributive over addition; that is, (a+b)^(c+d) = a^c+a^d+b^c+b^d, but I don't see your point.

Yes, but what have the wedge product to do with magnetism?

I summarize the results.
1. Magnetic field is only a helping field to compute the force of a moving charge in a moving electric field. The classic equation in 3 dimensions is: F = (q/c²) v_q x ( v_E x E). Where v_E is the velocity of the field and v_q is the velocity of the test charge.
2. The magnetic forces result from relativistic effects, that can be approximated by the above formula.
3. In any dimension this (approximated) force can be computed by F=(q/c²)((v_q.E)v_E-(v_q.v_E)E).
4. We can express this law also with the wedge product and hodge star by F=(q/c²)(-1)ⁿ(v_q ^* (v_E ^* E)) as I showed here.
5. To compute a ^* ( b ^* c ) which is a (1-)vector for a,b,c (1-)vectors one usually have to switch between various dimensions: 1 ^* ( 1 ^* 1) -> 1 ^* ( 2 )^*) -> 1 ^* (n-2) -> (n-1)^* -> 1.One can short this by defining a |x d = a ^* d^* (for simiplicity I let out the factor (-1)ⁿ). Then a ^* ( b ^* c) = a |x ( b^c). We know that a |x d can easily be computed by a ^* ( e_i ^ e_j)^* = (a.e_j)e_i - (a.e_i)e_j which defines it on the 2-basis.
6. We can regard v_E ^ E as the magnetic field in F=(q/c²)v_q |x (v_E ^ E). Though in 3 dimensions this 2-vector is orthogonal (as much as one can say about a relation between a vector and a 2-vector) to the classic magnetic field. Thatswhy I call v_E ^ E the magnetic 2-vector (its possible in every dimension)
7. This new invented magnetic field for arbitrary dimension alias magnetic 2-vector behaves indeed like a field because the forces add up if the fields add up. For only one field with one velocity one dont need 2-vectors. But if we have to add various Fields with velocities then its useful to be able to add up the fields, instead of only be able to add up the forces.

Do you mean v_E is the current I?

I is charge per time (through the regarded point). You can compute the velocity of the charges in a wire by I/lambda where lambda is the charge densitiy. E of a single charge should be clear.

I'm getting confused with your notation. I haven't seen the notations (a,b) or |x used anywhere before.

I used it in my previously posted article.

The magnetic field is already defined, and it has an extension to 4D via the Biot-Savart law.

Yes, the magnetic field is already defined for 3 dimensions, and our task was to generalize it n dimensions. The Biot-Savart Law only tells how to sum up (integrate) magnetic fields of the wire. I.e. mainly by integrating over the moving point charges of the wire.

You still have to integrate over the wire. Even though it makes little difference, I find it's easier to integrate the magnetic field first, i.e. compute v_q^(Integral[ dB ]) rather than Integral[v_q^dB].

Yes, and exactly thatswhy I mentioned that you may do it (if you use the magnetic field as a 2-vector).

So for 4 dimensions PWrong simply need to exchange the Cross by ^ in his computation of the magnetic field.

I already did :?

In your previous post you only regarded 3 dimensions.
In your post about the straight wire, you never computed the magnetic field as a 2-vector. Incidently additionally at that time you still confused ^ with ^*. And you used the law a ^* ( b ^* c ) = (a.b)c - (a.c)b which was not known to us at that time. Except your considerations about the null space, but the null space is only very losely connected with ^*. I only say: if you want to sum up the magnetic fields before computing the force you must use the 2-vector (you can of course also use the n-2-vector but that wouldnt be efficient). Null space is not sufficient and the simple pairing I used is also not sufficient.

[PWrong]

Magnetic field is only a helping field to compute the force of a moving charge in a moving electric field.

This doesn't really matter, but vector fields don't "move", they just change. You can express B in terms of the change in E, but only with Maxwell's equations, which I don't yet understand. The current is what concerns us in these simple situations.

Except your considerations about the null space, but the null space is only very losely connected with ^*.

Well, it seems now that the nullspace is equivalent to the hodge star. That is, X . *X = 0, where "." is ordinary matrix multiplication. Both operations produce a subspace perpendicular to the original.

6. We can regard vE ^ E as the magnetic field in F=(q/c2)vq |x (vE ^ E). Though in 3 dimensions this 2-vector is orthogonal (as much as one can say about a relation between a vector and a 2-vector) to the classic magnetic field. Thatswhy I call vE ^ E the magnetic 2-vector (its possible in every dimension)
7. This new invented magnetic field for arbitrary dimension alias magnetic 2-vector behaves indeed like a field because the forces add up if the fields add up. For only one field with one velocity one dont need 2-vectors. But if we have to add various Fields with velocities then its useful to be able to add up the fields, instead of only be able to add up the forces.

I understand this idea now. I don't think it's a good idea to call it a magnetic field though. Maybe a magnetic dual field, since it seems to be the dual (hodge star) of the classic magnetic field. I can see why it would be useful, especially in 5 or more dimensions, but I think your approach might make the integrals harder.

Quote:
You still have to integrate over the wire. Even though it makes little difference, I find it's easier to integrate the magnetic field first, i.e. compute vq^(Integral[ dB ]) rather than Integral[vq^dB].

Yes, and exactly thatswhy I mentioned that you may do it (if you use the magnetic field as a 2-vector).

I can apply the Biot-Savart law to find the classic magnetic field, but your magnetic dual field is something different altogether. Do you have anything like the Biot-Savart law that we can plug into mathematica and compute the magnetic dual?

In your previous post you only regarded 3 dimensions.
In your post about the straight wire, you never computed the magnetic field as a 2-vector.

No, but I did compute the classic magnetic field. It forms concentric spheres around the wire. Your magnetic dual consists of the w-axis vector, and the radius vector (outwards from the wire). I suppose both kinds of field have the same magnitude, right?

[houserichichi]

Sorry to butt in but I found this little gem of a paper on Two, Three and Four-Dimensional Electromagnetics
Using Differential Forms that you might enjoy browsing through.

Carry on.

[bo198214]

but vector fields don't "move", they just change. You can express B in terms of the change in E, but only with Maxwell's equations,

If you have an infinite line charge moving along the axis the electric field is static, though there is a magnetic field. I think the maxwell equations can handle this by using the current density j. The magnetic field in the whole space is then already determined by the j along the line (something like that). So maxwell equations make this topic a bit more complicated. And Biot-Savart is also nothing else than adding up the magnetic fields of each moving charge in the wire.

Maybe the maxwell-equations are even a bit too general. I would suppose that certain electric fields cannot be achieved by whatever charge distribution you choose. But every electric field is created by charges, so even the electric field is only a helping field. What physically matters is what force have 2 charges (or charge differentials) to each other. And indeed everything can be derived from using Coloumbs law only.

Well, it seems now that the nullspace is equivalent to the hodge star. That is, X . *X = 0, where "." is ordinary matrix multiplication. Both operations produce a subspace perpendicular to the original.

But it is not equivalent. PWrong become finally clear about this! A k-subspace is more coarse than a k-vector. Perhaps you can figure out the concrete correspondence between the two (I dont see it in the moment, its not even clear what would be a normal form of a subspace - because it does not depend on the concrete basis), but its clear that a k-vector has much more information than a k-subspace. I admit that null-space and hodge star look quite similar.

Maybe a magnetic dual field, since it seems to be the dual (hodge star) of the classic magnetic field.

Hm, dual does not contain that it is a 2-vector in every dimension, what seems to me the mainly interesting property. And there is no known magnetic 2-vector field anyway, so it can not become confused. But dual magentic field is a shorter term, so i dont mind to use it.

I can apply the Biot-Savart law to find the classic magnetic field, but your magnetic dual field is something different altogether. Do you have anything like the Biot-Savart law that we can plug into mathematica and compute the magnetic dual?

Yes, thatswhy I said simply exchange the Cross by ^. Dont know whether mathematica can compute with forms. But for integration a k-form is simply a (n over k)-dimensional vector. The wedge of two 1-vectors is easy to compute, i.e.:
(sum_i a_i e_i)^(sum_i b_i e_i) = sum_i<j (a_ib_j-a_jb_i) e_i^e_j

In your previous post you only regarded 3 dimensions.
In your post about the straight wire, you never computed the magnetic field as a 2-vector.

No, but I did compute the classic magnetic field.

How can you compute a classic magnetic field in 4 dimensions?
Your computation in the circle wire play in 3 dimensional subspace, so maybe its even correct in 4d, am not sure about that.

It forms concentric spheres around the wire. Your magnetic dual consists of the w-axis vector, and the radius vector (outwards from the wire). I suppose both kinds of field have the same magnitude, right?

Yes, its the wedge of the velocity v_E, which is in w-direction, and of E, which is directed to the location vector with coordinate w set to 0. I am not sure what the length of a k-vector is.

@houserichichi
I only skimmed over that paper, but it unfortunately regards only 2+1 and 3+1 where +1 is time ...

[PWrong]

And Biot-Savart is also nothing else than adding up the magnetic fields of each moving charge in the wire.

It's more than just adding things up. The magnetic field I found took 11.6 seconds to calculate with mathematica (although there are obviously faster ways to do it). It's a definite integral of a complicated, discontinuous function, with several arbitrary parameters. I don't want to integrate more stuff than I have to.

A k-subspace is more coarse than a k-vector. Perhaps you can figure out the concrete correspondence between the two (I dont see it in the moment, its not even clear what would be a normal form of a subspace - because it does not depend on the concrete basis)

A subspace is a set of infinitely many vectors, like a plane is the set of all the vectors in the plane. A pair of (non-parallel) vectors can define a plane, but not the other way round. The two vectors are a basis for the subspace. A k-vector may have more information than a k-vector, but the information is redundant.

A nullspace is also a set of infinitely many vectors. When we calculate the nullspace as a matrix, we're actually calculating the basis for the nullspace, which is arbitrary. The matrix is not technically the subspace itself. The hodge star does the same thing. It takes a subspace, and gives us a perpendicular subspace, for which we can calculate the basis. The basis for either subspace is just as arbitrary.

The hodge star and the nullspace are the same thing, except that the hodge star also has a magnitude. I don't know how to find this magnitude in general, but we know that |a ^ *b| = |a||b| sin theta. We also know that if a and b are contained in the same realm (for instance, if their w-component is 0), then |a ^ *b| = |a X b|. We can choose one of the basis vectors of a ^ *b to be up the w-axis. Then the other basis vector can be a X b.

Yes, thatswhy I said simply exchange the Cross by ^. Dont know whether mathematica can compute with forms. But for integration a k-form is simply a (n over k)-dimensional vector. The wedge of two 1-vectors is easy to compute, i.e.:
(sumi ai ei)^(sumi bi ei) = sumi<j (aibj-ajbi) ei^ej

That's the usual way to represent a k-form. Unfortunately mathematica doesn't use this notation. But I'm sure you could also represent it just by putting its basis vectors into a matrix. Thus ei^ej = [[1,0,0,0] [0,1,0,0]]
This isn't a unique representation (you could also use [[1,0,2,0] [0,1,2,0]] or any other row operation), but it seems to work.

Your computation in the circle wire play in 3 dimensional subspace, so maybe its even correct in 4d, am not sure about that.

I'm pretty sure it is correct. The B field vector I calculated is one basis vector. The other one has the same magnitude, and points up the w-axis. For calculating the field, we can choose our axes so that the charge's position has no y component and no w component. For finding the forces and equation of motion, we'll have to be more general.

There's something I don't quite get. Does a^*b mean a^ (*b) or *(a^b), or are they the same thing? :?

[bo198214]

And Biot-Savart is also nothing else than adding up the magnetic fields of each moving charge in the wire.

It's more than just adding things up.

Ok, for me integrating is adding up, maybe this was a misunderstanding between us. The Biot-Savart law simply adds up (i.e. integrates) the magnetic fields of the moving dq.

The hodge star does the same thing. It takes a subspace, and gives us a perpendicular subspace, for which we can calculate the basis. The basis for either subspace is just as arbitrary.

PWrong, you are really incorrigable. The k-vector *is not* a k-subspace with some redundant information! (As a matrix can be seen) Ever thought about adding k-vectors? How do you add two k-subspaces? Is the addition of two k-subspaces again a k-subspace? I really dont know why I write a long article about adding fields.
As I see it 2-vectors are just the appropriate extension of "valued subspaces" to be able of adding fields. As I said they are linear combinations of the elementary 2-subspaces (e_i^e_j), so they are ideal for adding subspaces. As I also already said if we only have one electrical field and one velocity we dont need k-vectors. In this case we can simply use the subspace spanned by (or orthogonal to) v and E with an appropriate value. If the velocity is always in the same direction, we can even add up only the E vectors (and vice versa). But we can not add two different (valued) subspaces. The basis addition goes wrong as I showed with the example (-v,-E) and (v,E).

But I'm sure you could also represent it just by putting its basis vectors into a matrix. Thus ei^ej = [[1,0,0,0] [0,1,0,0]]

Thats what I tried first, simply putting (v_E,E) for the magnetic field until I realized, that we can not add the fields then. Thatswhy I wrote the article about adding fields. And that the wedge is indeed useful for that.

There's something I don't quite get. Does a^*b mean a^ (*b) or *(a^b), or are they the same thing? :?

I defined it as wedge followed by hodge, i.e. a ^* b = (a^b)^*. Usually the hodge star is written in front, but I find it more convenient to write like an exponent (avoiding parenthesis). For example like conjugate, or transpose, or inverse. Its not the same as a ^ (b^*).

[PWrong]

Ok, for me integrating is adding up, maybe this was a misunderstanding between us. The Biot-Savart law simply adds up (i.e. integrates) the magnetic fields of the moving dq.

What I meant was, the expression we have to integrate is very complicated. Mathematica is (understandably) very slow at finding a definite integral of several complicated expressions at once.

As I said before, for an arbitrary wire of current, l and a charge at x, we have:
B = Integrate[ 1/|x-l|^4 dl ^* (x-l)]

Correct me if I'm wrong, but your magnetic dual field could be expressed as: dB* = 1/|x-l|^4 dl ^ (x-l)
But do we know that Integral[ dB*] = Integral[ dB ]* ? I guess we could prove this if we knew that (a* + b*) = (a+b)*. Is that what you're trying to demonstrate about adding subspaces? :?

F=(q/c2)((vq.E)vE-(vq.vE)E)

You go straight from the electric field to an expression for the force, without any mention of integration. You ignore the 1/r^4 factor, and the fact that E points from the wire to the particle, neither of which are at the origin, so E is actually a difference of two vectors.

Are we supposed to find this expression for the force from part of the wire, and then integrate the whole thing? Have you tried expanding out this expression, with each vector into its four parts? And by the way, could you use the usual notation for electromagnetism, like I = dl instead of v_E and r = x-l instead of E? It's confusing to use E in an expression for the magnetic field.

Ever thought about adding k-vectors? How do you add two k-subspaces? Is the addition of two k-subspaces again a k-subspace? I really dont know why I write a long article about adding fields.

I don't know why either. Adding subspaces should be a simple matter of adding the components together. I still don't see why that's a problem.:? Obviously (-v,-E) + (v,E) = (0,0).

As I see it 2-vectors are just the appropriate extension of "valued subspaces" to be able of adding fields. As I said they are linear combinations of the elementary 2-subspaces (ei^ej), so they are ideal for adding subspaces.

I'd like to use this notation, because I see why it's potentially easier to use, especially in 5 or more dimensions. But I don't have any formulas to use with it, and the classic magnetic field is still more intuitive. My method seems to work fine for the two cases I've analysed so far. If you could go into more detail, maybe I'd understand.

[bo198214]

Mathematica is (understandably) very slow at finding a definite integral of several complicated expressions at once. But happily it can find the definite Integral.

Maple for example can not. Interestingly its no problem to find the indefinite integral, but there are contained terms like a*arctan(b*tan(ph)) so that tan(0)=tan(2*Pi)=0 and arctan 0 = 0, disregarding that arctan is only definite up to constants k*Pi. So probably arctan(a*tan(2*Pi)) should be regarded as Pi. When assuming this, the result equals yours, but am not in the mood to always substitute in such monster formulas.

But do we know that Integral[ dB*] = Integral[ dB ]* ? I guess we could prove this if we knew that (a* + b*) = (a+b)*. Is that what you're trying to demonstrate about adding subspaces? :?

Its already well known (or even definition :?) that the hodge star is linear, so yes, we know that Integral[ dB*] = Integral[ dB ]*. And no, it mainly was not what I wanted to demonstrate *nudge*.

F=(q/c2)((vq.E)vE-(vq.vE)E)

You go straight from the electric field to an expression for the force, without any mention of integration. You ignore the 1/r^4 factor, and the fact that E points from the wire to the particle, neither of which are at the origin, so E is actually a difference of two vectors.
...
And by the way, could you use the usual notation for electromagnetism, like I = dl instead of v_E and r = x-l instead of E? It's confusing to use E in an expression for the magnetic field.

To be clear: F is the magnetic force caused by a (with velocity v_E) moving electrical field E on a (with velocity v_q) moving test charge q. Its all evaluated at the location of q.

How can this applied to a wire? In the wire are (with v_E) moving charges dQ. Say they have a line charge lambda, i.e. dQ/ds=lambda (where ds is the the considered line segment). Each charge at x_Q creates the electric field (at x)
dE = E' dQ
where E' = (1/4*Pi*eps) ((x-x_Q)/|x-x_Q|ⁿ)
The magnetic force on our testcharge is the sum of the forces from each line segment (like the electric force is the sum of the electric forces from each line segment)
F = integral (1/c²)((v_q.E')v_E-(v_q.v_E)E') dQ
And now is dQ = lambda ds and v_E lambda = I = I I^~ = I (dx_Q/ds)^~, so
F = c^-2 integral ((v_q.E')I-(v_q.I)E') d x_Q
Then you can put in the formula for E' and cancel 1/ (c² * 4*Pi*eps ) = mu / (4*Pi).
This is how you can compute the magnetic force on a moving charge q caused by a wire with current I in any dimension (without wedge, nullspace & co).

If this does not suffice and you must have an intermediate step computing the magnetic (dual) field, then you can integrate over the magnetic (dual) field differentials
B = mu integral v_Q ^(*) E' dQ = mu integral I ^(*) E' ds = mu I integral (dx_Q/ds)^~ ^(*) E' ds
(for integration it doesnt matter that much using ^ or ^*) which results roughly in your equation (let l=x_Q):

B = Integrate[ 1/|x-l|^4 dl ^* (x-l)]

Correct me if I'm wrong, but your magnetic dual field could be expressed as: dB* = 1/|x-l|^4 dl ^ (x-l)

Its right (except for some constants).

Are we supposed to find this expression for the force from part of the wire, and then integrate the whole thing? Have you tried expanding out this expression, with each vector into its four parts?

How to do it with the forces, I showed above, the result will be the same whether first integrate the magnetic fields and then compute the force, or first compute the forces and then integrate them (but only if we really use 2-vectors and not subspaces or vector pairs).

I really dont know why I write a long article about adding fields.

I don't know why either.

*lol*

Adding subspaces should be a simple matter of adding the components together. I still don't see why that's a problem.:? Obviously (-v,-E) + (v,E) = (0,0).

Ok, lets resolve this in 3-space (where we already know that we can add magnetic fields): and let us add
(-v)x(-E)+vxE = vxE+vxE = 2 vxE != 0 = 0 x 0 = (-v+v) x (-E + E)
On the dual hand also
(-v) ^ (-E) + v^E= 2 v ^ E
In this case we added the same (but differently expressed subspace). Adding components lead to the null-space.
For properly behaving fields the addition of two fields must result in the addition of the forces on the test charge/particle/gauge. This is satisfied for distributive operation like x and ^ and * but not for pairing/gluing of matrices/vectors.

But I don't have any formulas to use with it, and the classic magnetic field is still more intuitive.

First of all can you explain, what you mean by classic magnetic field?! I understand it as vxE but this is only useful for 3 dimensions. But I think you mentioned it also in a 4d context.
Second, you didnt accept the formula I gave you. It was only "the usual way to represent a k-form". But despite that it was only a 2-form, this is the formula, that you can use like vectors for adding, deriving and integrating.

My method seems to work fine for the two cases I've analysed so far.

In the first case you analyzed (straight infinite wire 4d) the velocity (and the current) was always the same (direction). So you could add/integrate the magnetic fields simply by adding the electric fields: (v,E₁) + (v,E₂) = (v,E₁+E₂), wedge was not needed (v^E₁ + v^E₂ = v^(E₁+E₂)). The second case (circle wire), where the wedge would be necessary in Biot-Savart, was anyway 3d (subspace), so you used simply the cross product.

Let me try if I can prove that the projection of ^* on a 3d subspace is equal to the x on that subspace... but not now, I am a bit in a hurry

[jinydu]

I took a look at the upper division textbook in electrodynamics used at my university and it appears that special relativity (in particular, Lorentz transformations) can indeed be used to derive some of the formulas in electromagnetism.

I should have a lot to look forward to :wink: .

[PWrong]

Sorry for the late reply. I've just started back at uni this week.

First of all can you explain, what you mean by classic magnetic field?! I understand it as vxE but this is only useful for 3 dimensions. But I think you mentioned it also in a 4d context.

I took it to mean the B field, in any dimension, as opposed to B*, which is the dual field. The classic field is an (n-2)-form in nD.

I'll try to solve the circular loop problem again, this time finding the B* field. We'll end up with 6 expressions to integrate, which I guess is better than 8. They're all pretty similar anyway.

l = {cos s, sin s, 0, 0}
dl ^ (x - l) = {-sin s, cos s, 0, 0} ^ { x - cos s, y - sin s, z, w}

Expanding this out, we get:
[1 - x cos s - y sin s] e₁ ^ e₂
- z sin s e₁ ^ e₃
- w sin s e₁ ^ e₄
+ z cos s e₂ ^ e₃
- w cos s e₂ ^ e₄

Of course, we still have to divide this by |x-l|^4. We can reduce this to three integrals by collecting sin and cos.

[bo198214]

For completeness I give the "classical" magnetic field in 4 dimensions, i.e. v ^* E. The always a bit tedious task with the hodge star is the sign. We know that
e_i₁ ^ ... ^ e_{i_k} = +/- e_j₁ ^ ... ^ e_{j_n-k}
where {j₁,...,j_n-k} is the complement of {i₁,...,i_k}
Then the sign is defined positive when the permutation (i₁,...,i_k,j₁,...,j_n-k) of (1,...,n) is even, and negative if it is odd. So we have for 4 dimensions:
(e₁ ^ e₂)^* = e₃ ^ e₄
(e₁ ^ e₃)^* = - e₂ ^ e₄
(e₁ ^ e₄)^* = e₂ ^ e₃
(e₂ ^ e₃)^* = e₁ ^ e₄
(e₂ ^ e₄)^* = - e₁ ^ e₃
(e₃ ^ e₄)^* = e₁ ^ e₂
So we have generally for n=4:
(sum a_ie_i) ^* (sum b_ie_i) = (a₃b₄-a₄b₃) e₁ ^ e₂ + (a₄b₂-a₂b₄) e₁ ^ e₃ + (a₂b₃-a₃b₂) e₁ ^ e₄ + (a₁b₄-a₄b₁) e₂ ^ e₃ + (a₃b₁-a₁b₃) e₂ ^ e₄ + (a₁b₂-a₂b₁) e₃ ^ e₄
Hm, twice as long as in 3d ....

My original idea, when I asked you to compute the magnetic field of the circle, was to establish an "elementary" magnet when letting the radius R go to 0 (for example the magnet of an hydrogen atom). So that for materials we simply integrate over the elementary magnets and so can see what is in 4d. I am still not completely sure what the elmentary magnet is for 3d. I think the magnetic field is nonzero only on the straight line perpendicular to the circle plane, through the circle center. And reduces by 1/r² there. Letting R to 0 means differentiate with respect to R, i.e. compute dB/dR. Which should also be accomplished by first deriving and then integrating, though havent tried it yet.

For the elementary magnet with R to 0 it should come out the same result regardless whether taking a circle or a square. So maybe it is easier to calculate the square, though I also didnt try it.

[PWrong]

(sum aiei) ^* (sum biei) = (a3b4-a4b3) e1 ^ e2 + (a4b2-a2b4) e1 ^ e3 + (a2b3-a3b2) e1 ^ e4 + (a1b4-a4b1) e2 ^ e3 + (a3b1-a1b3) e2 ^ e4 + (a1b2-a2b1) e3 ^ e4

That has a similar determinant formula to the cross product :
| e₁ e₂ e₃ e₄ |
| e₁ e₂ e₃ e₄ |
| a₁ a₂ a₃ a₄ |
| b₁ b₂ b₃ b₄ |
(Intepret e₁ e₂ as e₁^e₂)

Letting R to 0 means differentiate with respect to R, i.e. compute dB/dR. Which should also be accomplished by first deriving and then integrating, though havent tried it yet.

It might be better to find the taylor series of B about R=0, and look at the first term. I'll try this if I get some spare time.