4 Force

Electromagnetic 4-force Using 4D Vector Product © W. F. Esterhuyse*

October 2008

ABSTRACT We explore a new (simpler) formulation of the electromagnetic 4-force. The result is shown to be expressible using the 4D vector product of ref. [8] (ref. [8] is attached for reference). The electromagnetic 4-force is shown to be expressible as the 4D cross product of the 4-velocity with a 4-vector function Y which combines the E and H field. Maxwell's equations are also expressible in terms of Y.

Keywords:

Electromagnetic 4-Force; Electromagnetism; Lorentz Force;

4D Vector Product; Maxwell’s Equations. *e-mail: [email protected]

CONTENTS: 1. Electromagnetic 4-force 2. Maxwell's Equations in terms of Y Bibliography 1. Electromagnetic 4-force This article uses ref. [5], [6] and [8] for the definition of the vector product of two vectors using the 3x4 determinant developed by row.

Throughout this article v, Y, f will be 4-vectors, u, E, H, J will be 3-vectors. The k'th unit vector will be written as [k]. I also shift the 0'th component of 4-vectors to the 4'th component and start indices at 1. The units are in the cgs system.

γ = 1/ (1 - v^2/c^2)^1/2 throughout the article. From ref. [7] the velocity 4-vector with 0'th component restated as 4'th component is: v = γu(t) + cγ[4] = γc {[1] u1 /c + [2]u2 /c + [3] u3 /c + [4]}

(1)

where uk are functions of t. From the same ref. the electromagnetic 4-force (also swapped like above) is: f = γe {E + u x H/c + E • u [4]/c}

(2)

Carry out the dot and vector product: f = γe {[1] (E1 + u2H3/c - u3H2/c) + [2](E2 + u3H1/c - u1H3/c) + [3](E3 + u1H2/c - u2H1/c) + [4](E1u1 + E2u2 + E3u3)/c }

(3)

We try to write this as: evxY=f c

(4)

using the 4D cross product. It will now be shown that (4) has the same form as RS (3). From ref. [6] or [8] and v as in (1): v x Y = γc | [1] [2] | u1/c u2/c | Y1 Y2

[3] [4] | u3/c 1 | Y3 Y4 |

= γc { [1] | u2/c u3/c 1 | - [2] | u1/c u3/c 1 | + [3] | u1/c u2/c 1 | | Y2 Y3 Y4 | | Y1 Y3| Y4 | | Y1 Y2 Y4 | - [4] | u1/c u2/c u3/c | } | Y1 Y2 Y3 |

(6)

These are 2x3 determinants following by development by row 1 of the 3x4 determinant. With the signature ( 1 1 1 -1 ) imposed on the cross product - [4] changes into [4].

This is equivalent to the ( -1 1 1 1 ) signature when time is in the zerot'h component. The signature is imposed after taking the cross product, just like for the dot product (see definition of dot product ref. [7]). Compute the components of (6) separately: 1 ( v x Y )1 = | u2/c u3/c [] | + | u2/c [] 1 | + | [] u3/c 1

γc

| Y2

Y3 [] |

| Y2 [] Y4 |

| [] Y3

|

Y4 |

= | u2/c u3/c | - | u2/c 1 | + | u3/c 1 | | Y2 Y3 | | Y2 Y4| | | Y3 Y4 | by definition of 2x3 determinants ref. [6]. Continue: ( v x Y )1 = [Y2 - Y3 + u2(Y3 - Y4)/c - u3( Y2 - Y4)/c]γc multiply by e/c to get: e ( v x Y )1 = [Y2 - Y3 + u2(Y3 - Y4)/c - u3( Y2 - Y4)/c]γe c

(9)

Similarly for the other components: e ( v x Y )2 = [Y3 - Y1 + u3(Y1 - Y4)/c - u1( Y3 - Y4)/c]γe c

(14)

e ( v x Y )3 = [Y1 - Y2 + u1(Y2 - Y4)/c - u1( Y1 - Y4)/c]γe c

(19)

e ( v x Y )4 = [u1(Y2 - Y3)/c + u2( Y3 - Y1)/c + u3( Y1 - Y2)/c]γe c

(23)

with the sign change (from signature) on component 4 imposed. Comparing (9) with component 1 of RS (3) we get: E1 = Y2 - Y3 H 3 = Y3 - Y4 H 2 = Y2 - Y4

(24)

Similarly for comparing (14) with component 2 of RS (3): E2 = Y3 - Y1 H 1 = Y1 - Y4 H 3 = Y3 - Y4

(25)

Similarly for comparing (19) with component 3 of RS (3): E3 = Y1 - Y2 H 2 = Y2 - Y4 H 1 = Y1 - Y4

(26)

Similarly for comparing (23) with component 4 of RS (3): E1 = Y2 - Y3 E2 = Y3 - Y1 E3 = Y1 - Y2

(27)

The formulae of (27) are equivalent to: E = | [1] | Y1 |1

[2] Y2 1

[3] | = (Y - [4]Y4)x([1] + [2] + [3]) Y3 | 1 |

(28)

(24), (25), (26) (containing Hk) are equivalent to: 3

H=-

∑

[k] | 1

k=1

1 |

| Yk Y4 |

(29)

If we try to write Y in terms of E and H we get a system that is only consistent if some very special conditions on E and H's components are satisfied. This may be because E and H are interrelated and therefore the appending of the two vectors in RS (30) is just allowed on some conditions). This can be seen by writing (28) and (29) as a matrix equation: 0 1 -1 -1 0 1 1 -1 0 1 0 0 0 1 0 0 0 1

0 Y 0 0 -1 -1 -1

= E1 = E2 = E3 = H1 = H2 = H3

(30)

and use Gauss Elimination on the augmented matrix.

2. Maxwell's Equations in terms of Y Using the relations (28) with the operators in Maxwell's equations we get:

∇• E = | ∂/∂x1 ∂/∂x2 ∂/∂x3 | + 0 | Y1 |1

Y2 1

Y3 1

| |

(31)

∇x E = | [1] | ∂/∂x1

[2]

∂/∂x2

| Y2 - Y3

Y3 - Y1

∇x E = | [1] | ∂/∂x1 | | Y2 Y3 | ||1 1 |

[3] | ∂/∂x3 | Y1 - Y2 |

(32)

[2]

[3]

∂/∂x2

∂/∂x3

| |

| Y3 Y1 | |1 1 |

| Y1 Y2 | | |1 1 | |

(33)

We also have:

∂/∂x4 H = ∂/∂x4 {(Y1 - Y4)[1] + (Y2 - Y4)[2] + (Y3 - Y4)[3] + [some f (x1, x2, x3) + constant][4]}

(35)

where the last term is a possibility since partial differentiation with respect to x4 of the last term is zero. Then:

∇x E + ∂/∂x4 H /c = 0 changes into: OP4 (Y) = 0

(36)

where OP4 is some operation that looks like it can be stated like this because it's result must be: OP4 (Y) = [1][ ∂/∂x2 (Y1 - Y2) - ∂/∂x3 (Y3 - Y1) + ∂/∂x4 (Y1 - Y4)/c] -

[2][∂/∂x1 (Y1 - Y2) - ∂/∂x3 (Y2 - Y3) + ∂/∂x4 (Y2 - Y4)/c]

+ [3][∂/∂x1(Y3 - Y1) - ∂/∂x2 (Y2 - Y3) + ∂/∂x4 (Y3 - Y4)/c] -

[4][something evaluating to zero] (37)

Where the subscript 4 refers to the special variable to the operation. By symmetry the last term may be (logically): -[4][∂/∂x1 f(x2, x3 ,x4 ) - ∂/∂x2 g(x1, x3 ,x4 ) + ∂/∂x3 h(x1, x2 ,x4 )]. There is of course other ways the last term can be zero, but this way it always is. (37) comes from (32) plus (35) times 1/c.

∇• E = 4πρ changes into:

∇• (Y - [4]Y4)x([1] + [2] + [3]) = 4πρ.

(38)

∇• H = 0 changes into:

3

∑ ∂/∂x

| Yk Y4 | = 0

k=1

|1

k

(39)

1 |

We also have:

∇x H = | [1] | ∂/∂x1 | Y1 - Y4

[2]

[3]

∂/∂x2 ∂/∂x3 Y2 - Y4 Y3 - Y4 |

| + 0[4] | (40)

and: - ∂/∂x4 E = - ∂/∂x4 {(Y2 - Y3)[1] + (Y3 - Y1)[2] - (Y1 - Y2)[3] + [ f (unknown)][4]}

(41)

Where f (unknown) must be such that - ∂/∂x4 f (unknown) = 0 Then:

∇x H - ∂/∂x4 E /c = 4πJ/c transforms into: OPC4 (Y) = 4πJ/c and the operation OPC4 is in some sense complementary to that of (36) because we see that (32) is analogous to (40) just with the following replacements: (Y1 - Y4) replaces (Y2 - Y3)

(42)

(Y2 - Y4) replaces (Y3 - Y1) (Y3 - Y4) replaces (Y1 - Y2)

(43)

and (except for opposite sign) (35) is analogous to (41), just with the opposite replacements as in (43). (42) comes from (40) minus (41)*(1/c) and LS (42) reduces to: OPC4 (Y) = [1] {∂/∂x2 (Y3 -Y4) - ∂/∂x3 (Y2 -Y4) - ∂/∂x4 (Y2 - Y3)/c} - [2] {∂/∂x3 (Y1 - Y4) -

∂/∂x1 (Y3 -Y4) - ∂/∂x4 (Y3 - Y1)/c } + [3] {∂/∂x1 (Y2 -Y4) - ∂/∂x2 (Y1 - Y4) ∂/∂x4 (Y1 - Y2)/c} + [4]{ ∂/∂x4 g (unknown variable(s))/c}

(44)

OPC4 (Y) may be written as:

∇4 x (∗4Y)

(45)

where:

∇4 ≡ [1]∂/∂x1 + [2]∂/∂x2 + [3]∂/∂x3 + [4](1/c)∂/∂x4,

(46)

∗4Y = [1](Y1 -Y4) + [2] (Y2 - Y4) + [3](Y3 - Y4) + [4]h(x4)

(47)

and h(x4) may be constant or zero. This may be seen by considering the determinants of [1], as follows: | [1] |X |X

X

∂/∂x2

Y2 - Y4

X

∂/∂x3

X

Y3 - Y4

| (1/c)∂/∂x4 | h(x4) |

(48)

where X represents don't care entries. Then the three determinant terms of [1] are:

| ∂/∂x2 ∂/∂x3 | | Y2 - Y4 Y3 - Y4 | - | ∂/∂x2 (1/c)∂/∂x4 | |Y2 - Y4 h(x4) | | ∂/∂x3 (1/c)∂/∂x4 | |Y3 - Y4 h(x4) |

(49)

and the first one gives the first two terms of (44) component 1 and the other two gives term 3 of (44) component 1. Similar computations are doable for [2] and [3]. This

requires for [4] that: [4]

| = [4] (4πJ/c)4

|0

0

0

| ∂/∂x1 | Y1-Y4

∂/∂x2

∂/∂x3 (1/c)∂/∂x4 |

Y2 - Y4 Y3 - Y4 h(x4)

|

(50)

where this component (4πJ/c)4: is still undefined since J4 is undefined. Note that: | ∂/∂x1 | Y1-Y4

∂/∂x2 ∂/∂x3

| = ∇• E

Y2 - Y4 Y3 - Y4 |

(51)

by definition of 2x3 determinants and therefore (by (50), (51)):

∇• E = 4πρ is already present in the fourth component of:

∇4 x (∗4Y) = (4πK/c)

(52)

if we define the K as a 4-vector such that: K = J + cρ[4].

(53)

(51), (52) and (53) are amazing and proves that the non-square determinant is useful in Physics. Therefore the three (at this stage) Maxwell's equations are in these terms:

∇4 x (∗4Y) = (4πK/c)

(54)

OP4 (Y) = 0

(55)

3

∑ ∂/∂x

|1 1

k=1

|Yk Y4 |

k

| = 0 (56)

Since the coefficient h(x4) in (47) and (48) may be zero we have:

∇4 x (∗4Y) = ∇4 x (Y - 1Y4)

(57)

where 1 is the 4-vector with unit components in every direction. We try to write (55) and (56) as a determinant expression. First define the operator doing the replacements of (43) going from left to right as Λ and the inverse as

going from right to left named: Ρ (capital roh). Using this operator and its inverse we may compute:

∇ x E + ∂/∂x4 H /c, and we prove it exists in: OP4 (Y) = Λ3 ∇4 x (Y - 1Y4) = | [1] [2] [3] | ∂/∂x1 Λ ∂/∂x2 Λ ∂/∂x3 Λ | Y1-Y4 Y2 - Y4 Y3 - Y4

[4] | (1/c)∂/∂x4 Ρ | 0 |.

(58)

Compute this: (Λ3 ∇4 x (Y - 1Y4))1 =

∂/∂x2 Λ (Y3 - Y4 ) - ∂/∂x3 Λ (Y2 - Y4) + (1/c)∂/∂x4 Ρ (-Y3 + Y4 + Y2 - Y4) = ∂/∂x2 Λ (Y3 - Y4 ) - ∂/∂x3 Λ (Y2 - Y4) + (1/c)∂/∂x4 Ρ (Y2 -Y3) do the replacements from the table: (Λ3 ∇4 x (Y - 1Y4))1 =

∂/∂x2 (Y1 - Y2 ) - ∂/∂x3 (Y3 - Y1) + (1/c)∂/∂x4 (Y1 -Y4).

(59)

This is equal to: (∇ x E + ∂/∂x4 H /c )1 by straight comparison. Similarly for components 2 and 3: (Λ3 ∇4 x (Y - 1Y4))2 = - ∂/∂x1 Λ (Y3 - Y4) + ∂/∂x3 Λ (Y1 - Y4) - (1/c)∂/∂x4 Ρ (Y1 - Y3)

(60)

(Λ3 ∇4 x (Y - 1Y4))3 =

∂/∂x1 Λ (Y2 - Y4 ) - ∂/∂x2 Λ (Y1 - Y4) + (1/c)∂/∂x4 Ρ (Y1 - Y2) where the replacement and comparison is left to the reader. Next we prove:∇• H is present in the fourth component: (Λ3 ∇4 x (Y - 1Y4))4 =

∂/∂x1 Λ (Y2 - Y4 - Y3 + Y4) + ∂/∂x2 Λ (-Y1 + Y4 + Y3 - Y4) +

(61)

∂/∂x3 Λ (-Y2 + Y4 + Y1 - Y4) = ∂/∂x1 Λ (Y2 - Y3) + ∂/∂x2 Λ (Y3 - Y1) + ∂/∂x3 Λ (Y1 - Y2) = ∂/∂x1 (Y1 - Y4) + ∂/∂x2 (Y2 - Y4) + ∂/∂x3 (Y3 - Y4)

(62)

and this is equal to ∇• H by comparison. Therefore (55) and (56) changes into:

Λ3 ∇4 x (Y - 1Y4) = 0

(63)

where 0 is the zero 4-vector. Since the two operators are complementary (if we take complement as: ~), we have: ~{∇4 x (Y - 1Y4)} = Λ3 ∇4 x (Y - 1Y4) a strange kind of complement, but it is necessary. The Ρ and Λ inside the determinant is justifiable because it effectively has the meaning that the E and H field intermixes within a system and not just by summation. We therefore have our two Maxwell's Equations as:

∇4 x (Y - 1Y4) = (4πK/c)

(64)

Λ3 ∇4 x (Y - 1Y4) = 0

(65)

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