exterior algebra and the maxwell-boltzmann equations€¦ · 2) ^ = 0 ^ = ( ^ ) joseph ferrara...

IntroductionThe Space of p-VectorsExterior Differentiation

The Maxwell-Boltzmann EquationsThe Homogeneous Equations

Conclusion

Exterior Algebra and the Maxwell-BoltzmannEquations

Joseph Ferrara

April 15, 2011

Joseph Ferrara Exterior Algebra and the Maxwell-Boltzmann Equations

Conclusion

Overview

1 The Space of p-Vectors

2 Exterior Differentiation

3 The Maxwell-Boltzmann Equations

Conclusion

2-vectorsp-vectors

The Space of 2-Vectors

Let L be an n-dimensional vector space over Rα, β, γ, · · · ,∈ L

a,b,c,· · · ,∈ R

We choose a p ≤ n where each case defines a new space and call∧pL the space of p-vectors on L.

Elements in∧2 L :

∑i ai (αi ∧ βi )

Basic Properties of 2-Vectorsa1(α1 ∧ β) + a2(α2 ∧ β) = (a1α1 + a2α2) ∧ βb1(α ∧ β1) + b2(α ∧ β2) = α ∧ (b1β1 + b2β2)

α ∧ α = 0

α ∧ β = −(β ∧ α)

Conclusion

2-vectorsp-vectors

a,b,c,· · · ,∈ RWe choose a p ≤ n where each case defines a new space and call∧

pL the space of p-vectors on L.

Elements in∧2 L :

α ∧ α = 0

α ∧ β = −(β ∧ α)

Conclusion

2-vectorsp-vectors

a,b,c,· · · ,∈ RWe choose a p ≤ n where each case defines a new space and call∧

pL the space of p-vectors on L.

Elements in∧2 L :

α ∧ α = 0

α ∧ β = −(β ∧ α)

Conclusion

2-vectorsp-vectors

The Space of 2-Vectors continued

Let Ψ ={σ1, σ2, · · · , σn

}be a basis of L.

Consider two vectors, α =∑n

i=1 aiσi and β =

∑nj=1 bjσ

Taking their exterior product yields,

α ∧ β =(∑n

i=1 aiσi)∧(∑n

j=1 bjσj)

∑ni=1 aibj(σ

i ∧ σj)

= a1b1(σ1 ∧ σ1) + a1b2(σ1 ∧ σ2) + · · ·+ a1bn(σ1 ∧ σn)+ a2b1(σ2 ∧ σ1) + · · ·+ anb1(σn ∧ σ1)+ a2b2(σ2 ∧ σ2) + a2b3(σ2 ∧ σ3) + · · ·+ a2bn(σ2 ∧ σn)+ a3b2(σ3 ∧ σ2) + · · ·+ anb2(σn ∧ σ2)+ · · ·=∑n

i<j(aibj − ajbi )(σi ∧ σj).

Conclusion

2-vectorsp-vectors

Let Ψ ={σ1, σ2, · · · , σn

}be a basis of L.

i=1 aiσi and β =

∑nj=1 bjσ

α ∧ β =(∑n

i=1 aiσi)∧(∑n

j=1 bjσj)

∑ni=1 aibj(σ

i ∧ σj)

Conclusion

2-vectorsp-vectors

Let Ψ ={σ1, σ2, · · · , σn

}be a basis of L.

i=1 aiσi and β =

∑nj=1 bjσ

α ∧ β =(∑n

i=1 aiσi)∧(∑n

j=1 bjσj)

∑ni=1 aibj(σ

i ∧ σj)

= a1b1(σ1 ∧ σ1) + a1b2(σ1 ∧ σ2) + · · ·+ a1bn(σ1 ∧ σn)+ a2b1(σ2 ∧ σ1) + · · ·+ anb1(σn ∧ σ1)+ a2b2(σ2 ∧ σ2) + a2b3(σ2 ∧ σ3) + · · ·+ a2bn(σ2 ∧ σn)+ a3b2(σ3 ∧ σ2) + · · ·+ anb2(σn ∧ σ2)+ · · ·

Conclusion

2-vectorsp-vectors

Let Ψ ={σ1, σ2, · · · , σn

}be a basis of L.

i=1 aiσi and β =

∑nj=1 bjσ

α ∧ β =(∑n

i=1 aiσi)∧(∑n

j=1 bjσj)

∑ni=1 aibj(σ

i ∧ σj)

Conclusion

2-vectorsp-vectors

The Space of p-Vectors

The space of∧p L can be derived similarly.

Basic Properties of p-Vectors(aα + bβ) ∧ α2 · · · ∧ αp = a(α ∧ α2 ∧ · · · ∧ αp) + b(β ∧ α2 ∧ · · · ∧ αp)

α1 ∧ · · · ∧ αp = 0 if i 6= j , αi = αj

α1 ∧ · · · ∧ αp = απ(1) ∧ · · · ∧ απ(p) = sgn(π)(α1 ∧ α2 ∧ · · · ∧ αp)

ϕ ∈∧p L⇒ ϕ =

∑H aHσ

H whereH = {h1, h2, · · · , hp} and 1 5 h1 < h2 < · · · < hp 5 n

Conclusion

2-vectorsp-vectors

The Space of p-Vectors

The space of∧p L can be derived similarly.

Basic Properties of p-Vectors(aα + bβ) ∧ α2 · · · ∧ αp = a(α ∧ α2 ∧ · · · ∧ αp) + b(β ∧ α2 ∧ · · · ∧ αp)

α1 ∧ · · · ∧ αp = 0 if i 6= j , αi = αj

α1 ∧ · · · ∧ αp = απ(1) ∧ · · · ∧ απ(p) = sgn(π)(α1 ∧ α2 ∧ · · · ∧ αp)

ϕ ∈∧p L⇒ ϕ =

∑H aHσ

H whereH = {h1, h2, · · · , hp} and 1 5 h1 < h2 < · · · < hp 5 n

Conclusion

Proof of d(dω) = 0

Exterior Differentiation

Properties of the Exterior Derivatived(ω + η) = dω + dη

d(λ ∧ µ) = dλ ∧ µ+ (−1)(degλ)(λ ∧ dµ)

∀ω, d(dω) = 0

∀f , df =∑

i∂f∂x i dx

Conclusion

Proof of d(dω) = 0

∀ω, d(dω) = 0

Proof. Let ω = aHdxH where aH(x1, x2, · · · , xn) is smooth and

differentiable as often as we like. Applying d to ω we obtain,

d(dω) = d(∑

i∂aH∂x i dx

i,j∂2aH∂x i∂x j dx

jdx idxH

∑i,j 2 ∂2aH

∂x i∂x j dxjdx idxH

∑i,j

(∂2aH∂x i∂x j + ∂2aH

∂x i∂x j

)dx jdx idxH

∑i,j

(∂2aH∂x i∂x j − ∂2aH

∂x j∂x i

)dx jdx idxH

Since we assumed that all functions aH are smooth we have equality ofmixed partial derivatives, and thus conclude that ∀ω, d(dω) = 0

Conclusion

Proof of d(dω) = 0

∀ω, d(dω) = 0

differentiable as often as we like. Applying d to ω we obtain,d(dω) = d

(∑i∂aH∂x i dx

jdx idxH

∑i,j 2 ∂2aH

∑i,j

∂x i∂x j

)dx jdx idxH

∑i,j

∂x j∂x i

)dx jdx idxH

Conclusion

Proof of d(dω) = 0

∀ω, d(dω) = 0

(∑i∂aH∂x i dx

jdx idxH

∑i,j 2 ∂2aH

∑i,j

∂x i∂x j

)dx jdx idxH

∑i,j

∂x j∂x i

)dx jdx idxH

Conclusion

Proof of d(dω) = 0

∀ω, d(dω) = 0

(∑i∂aH∂x i dx

jdx idxH

∑i,j 2 ∂2aH

∑i,j

∂x i∂x j

)dx jdx idxH

∑i,j

∂x j∂x i

)dx jdx idxH

Conclusion

Proof of d(dω) = 0

∀ω, d(dω) = 0

(∑i∂aH∂x i dx

jdx idxH

∑i,j 2 ∂2aH

∑i,j

∂x i∂x j

)dx jdx idxH

∑i,j

∂x j∂x i

)dx jdx idxH

Conclusion

Proof of d(dω) = 0

∀ω, d(dω) = 0

(∑i∂aH∂x i dx

jdx idxH

∑i,j 2 ∂2aH

∑i,j

∂x i∂x j

)dx jdx idxH

∑i,j

∂x j∂x i

)dx jdx idxH

Conclusion

Proof of d(dω) = 0

∀ω, d(dω) = 0

(∑i∂aH∂x i dx

jdx idxH

∑i,j 2 ∂2aH

∑i,j

∂x i∂x j

)dx jdx idxH

∑i,j

∂x j∂x i

)dx jdx idxH

Conclusion

EM EquationsGeneral Spacetime Differentials

The Maxwell-Boltzmann Equations

∇ · B = 0∇× E = −∂B

∂t∇ · E = ρ

∇× B− ∂E∂t = J

Conclusion

Magnetic Field

Traditional: B=(Bx ,By ,Bz) = Bx i + By j + BzkExterior : B = Bxdydz + Bydzdx + Bzdxdy

Electric Field

Traditional: E=(Ex ,Ey ,Ez) = Ex i + Ey j + EzkExterior : E = Exdx + Eydy + Ezdz

Unified EMF

α = B + E ∧ dt = B + Edt= Bxdydz + Bydzdx + Bzdxdy + (Exdx + Eydy + Ezdz) dt

Conclusion

Magnetic Field

Electric Field

Unified EMF

Conclusion

Magnetic Field

Electric Field

Unified EMF

Conclusion

We want to find the form of the exterior derivative on a generalspacetime differential, ω.

Let ω = ωHdxH with H = {h1, h2, · · · , hp} where ωH is any function

of spacetime.

Applying d to ω yields,dω = dωHdx

= ∂ωH

∂x dxdxH + ∂ωH

∂y dydxH + ∂ωH

∂z dzdxH + ∂ωH

∂t dtdxH

i=1∂ωH

∂x i dxidxH + ∂ωH

∂t dtdxH

Thus the general form is as follows,

dω = d ′ω + ∂ω∂t dt

We will use H = {dx , dy , dz , dt}.

Conclusion

of spacetime.

= ∂ωH

∂x dxdxH + ∂ωH

∂y dydxH + ∂ωH

∂z dzdxH + ∂ωH

∂t dtdxH

i=1∂ωH

∂t dtdxH

Conclusion

of spacetime.

= ∂ωH

∂x dxdxH + ∂ωH

∂y dydxH + ∂ωH

∂z dzdxH + ∂ωH

∂t dtdxH

i=1∂ωH

∂t dtdxH

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

Traditional vector check

∇ · B = 0

∇× E = −∂B∂t

α = B + Edt

Let’s consider the exterior derivative on α,

dα = d ′α + ∂α∂t dt

= d ′ (B + Edt) + ∂(B+Edt)∂t dt

= d ′B + d ′Edt + ∂B∂t dt + ∂E

∂t dt ∧ dt

= d ′B +[d ′E + ∂B

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

∇ · B = 0

∇× E = −∂B∂t

α = B + Edt

Let’s consider the exterior derivative on α,

dα = d ′α + ∂α∂t dt

= d ′ (B + Edt) + ∂(B+Edt)∂t dt

= d ′B + d ′Edt + ∂B∂t dt + ∂E

∂t dt ∧ dt

= d ′B +[d ′E + ∂B

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

Setting dα = 0 yields the following two equations,{d ′B = 0

d ′E + ∂B∂t = 0

Notice the similarities!

d′B = 0 ⇔ ∇ · B = 0

d′E + ∂B∂t = 0 ⇔ ∇× E + ∂B

∂t = 0

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

Setting dα = 0 yields the following two equations,{d ′B = 0

d ′E + ∂B∂t = 0

Notice the similarities!

d′B = 0 ⇔ ∇ · B = 0

d′E + ∂B∂t = 0 ⇔ ∇× E + ∂B

∂t = 0

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

Let’s first consider d ′B = 0. Then,

d′B = d ′ (Bxdydz + Bydzdx + Bzdxdy)= d ′Bxdydz + d ′Bydzdx + d ′Bzdxdy

= ∂Bx

∂x dxdydz +∂By

∂y dydzdx + ∂Bz

∂z dzdxdy

=(∂Bx

∂x +∂By

∂y + ∂Bz

)dxdydz

0 = ∂Bx

∂x +∂By

∂y + ∂Bz

=(∂∂x ,

∂∂y ,

∂∂z

)· (Bx ,By ,Bz)

= ∇ · B

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

= ∂Bx

∂x dxdydz +∂By

∂y dydzdx + ∂Bz

∂z dzdxdy

=(∂Bx

∂x +∂By

∂y + ∂Bz

)dxdydz

0 = ∂Bx

∂x +∂By

∂y + ∂Bz

=(∂∂x ,

∂∂y ,

∂∂z

)· (Bx ,By ,Bz)

= ∇ · B

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

= ∂Bx

∂x dxdydz +∂By

∂y dydzdx + ∂Bz

∂z dzdxdy

=(∂Bx

∂x +∂By

∂y + ∂Bz

)dxdydz

0 = ∂Bx

∂x +∂By

∂y + ∂Bz

=(∂∂x ,

∂∂y ,

∂∂z

)· (Bx ,By ,Bz)

= ∇ · B

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

= ∂Bx

∂x dxdydz +∂By

∂y dydzdx + ∂Bz

∂z dzdxdy

=(∂Bx

∂x +∂By

∂y + ∂Bz

)dxdydz

0 = ∂Bx

∂x +∂By

∂y + ∂Bz

=(∂∂x ,

∂∂y ,

∂∂z

)· (Bx ,By ,Bz)

= ∇ · B

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

Secondly we consider,d ′E + ∂B

∂t =

d ′ (Exdx + Eydy + Ezdz) + ∂∂t (Bxdydz + Bydzdx + Bzdxdy)

Splitting into two parts for convenience: d ′E and ∂B∂t

First,

d′E = d ′ (Exdx + Eydy + Ezdz)= d ′Exdx + d ′Eydy + d ′Ezdz

= ∂Ex

∂y dydx + ∂Ex

∂z dzdx +∂Ey

∂x dxdy

+∂Ey

∂z dzdy + ∂Ez

∂x dxdz + ∂Ez

∂y dydz

=(∂Ey

∂x −∂Ex

)dxdy +

(∂Ez

∂y −∂Ey

)dydz +

(∂Ex

∂z −∂Ez

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

∂t =

First,

d′E = d ′ (Exdx + Eydy + Ezdz)

= d ′Exdx + d ′Eydy + d ′Ezdz

= ∂Ex

∂y dydx + ∂Ex

∂z dzdx +∂Ey

∂x dxdy

+∂Ey

∂z dzdy + ∂Ez

∂x dxdz + ∂Ez

∂y dydz

=(∂Ey

∂x −∂Ex

)dxdy +

(∂Ez

∂y −∂Ey

)dydz +

(∂Ex

∂z −∂Ez

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

∂t =

First,

= ∂Ex

∂y dydx + ∂Ex

∂z dzdx +∂Ey

∂x dxdy

+∂Ey

∂z dzdy + ∂Ez

∂x dxdz + ∂Ez

∂y dydz

=(∂Ey

∂x −∂Ex

)dxdy +

(∂Ez

∂y −∂Ey

)dydz +

(∂Ex

∂z −∂Ez

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

∂t =

First,

= ∂Ex

∂y dydx + ∂Ex

∂z dzdx +∂Ey

∂x dxdy

+∂Ey

∂z dzdy + ∂Ez

∂x dxdz + ∂Ez

∂y dydz

=(∂Ey

∂x −∂Ex

)dxdy +

(∂Ez

∂y −∂Ey

)dydz +

(∂Ex

∂z −∂Ez

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

∂t =

First,

= ∂Ex

∂y dydx + ∂Ex

∂z dzdx +∂Ey

∂x dxdy

+∂Ey

∂z dzdy + ∂Ez

∂x dxdz + ∂Ez

∂y dydz

=(∂Ey

∂x −∂Ex

)dxdy +

(∂Ez

∂y −∂Ey

)dydz +

(∂Ex

∂z −∂Ez

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

∂B∂t = ∂

∂t (Bxdydz + Bydzdx + Bzdxdy)

=∂Bx

∂t dydz +∂By

∂t dzdx + ∂Bz

∂t dxdy

Summing yields,

d′E + ∂B∂t =

(∂Ey

∂x −∂Ex

∂y + ∂Bz

)dxdy +

(∂Ez

∂y −∂Ey

∂z + ∂Bx

+(∂Ex

∂z −∂Ez

∂x +∂By

We set d ′E + ∂B∂t equal to zero which yields the following three

equations, ∂Ey

∂x −∂Ex

∂y = −∂Bz

∂t∂Ez

∂y −∂Ey

∂z = −∂Bx

∂t∂Ex

∂z −∂Ez

∂x = −∂By

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

∂B∂t = ∂

=∂Bx

∂t dydz +∂By

∂t dzdx + ∂Bz

∂t dxdy

Summing yields,

d′E + ∂B∂t =

(∂Ey

∂x −∂Ex

∂y + ∂Bz

)dxdy +

(∂Ez

∂y −∂Ey

∂z + ∂Bx

+(∂Ex

∂z −∂Ez

∂x +∂By

equations, ∂Ey

∂x −∂Ex

∂y = −∂Bz

∂t∂Ez

∂y −∂Ey

∂z = −∂Bx

∂t∂Ex

∂z −∂Ez

∂x = −∂By

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

∂B∂t = ∂

=∂Bx

∂t dydz +∂By

∂t dzdx + ∂Bz

∂t dxdy

Summing yields,

d′E + ∂B∂t =

(∂Ey

∂x −∂Ex

∂y + ∂Bz

)dxdy +

(∂Ez

∂y −∂Ey

∂z + ∂Bx

+(∂Ex

∂z −∂Ez

∂x +∂By

equations, ∂Ey

∂x −∂Ex

∂y = −∂Bz

∂t∂Ez

∂y −∂Ey

∂z = −∂Bx

∂t∂Ex

∂z −∂Ez

∂x = −∂By

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

These are the equations given by ∇× E. Let’s check it.

∇×E =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

Ex Ey Ez

∣∣∣∣∣∣ = i

∣∣∣∣ ∂∂y

∂∂z

∣∣∣∣− j

∣∣∣∣ ∂∂x

∂∂z

∣∣∣∣+ k

∣∣∣∣ ∂∂x

∂∂y

∣∣∣∣=(∂Ez

∂y −∂Ey

)i +(∂Ex

∂z −∂Ez

)j +(∂Ey

∂x −∂Ex

Furthermore,

−∂Bdt = −∂Bx

∂t i− ∂By

∂t j− ∂Bz

∂t k

Equating the components of these two equations yields∂Ey

∂x −∂Ex

∂y = −∂Bz

∂t∂Ez

∂y −∂Ey

∂z = −∂Bx

∂t∂Ex

∂z −∂Ez

∂x = −∂By

These are exactly the equations we achieved through the exterioralgebra method.

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

∇×E =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

Ex Ey Ez

∣∣∣∣∣∣ = i

∣∣∣∣ ∂∂y

∂∂z

∣∣∣∣− j

∣∣∣∣ ∂∂x

∂∂z

∣∣∣∣+ k

∣∣∣∣ ∂∂x

∂∂y

∣∣∣∣=(∂Ez

∂y −∂Ey

)i +(∂Ex

∂z −∂Ez

)j +(∂Ey

∂x −∂Ex

Furthermore,

∂t i− ∂By

∂t j− ∂Bz

∂t k

∂x −∂Ex

∂y = −∂Bz

∂t∂Ez

∂y −∂Ey

∂z = −∂Bx

∂t∂Ex

∂z −∂Ez

∂x = −∂By

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

∇×E =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

Ex Ey Ez

∣∣∣∣∣∣ = i

∣∣∣∣ ∂∂y

∂∂z

∣∣∣∣− j

∣∣∣∣ ∂∂x

∂∂z

∣∣∣∣+ k

∣∣∣∣ ∂∂x

∂∂y

∣∣∣∣=(∂Ez

∂y −∂Ey

)i +(∂Ex

∂z −∂Ez

)j +(∂Ey

∂x −∂Ex

Furthermore,

∂t i− ∂By

∂t j− ∂Bz

∂t k

∂x −∂Ex

∂y = −∂Bz

∂t∂Ez

∂y −∂Ey

∂z = −∂Bx

∂t∂Ex

∂z −∂Ez

∂x = −∂By

Conclusion

d′B = 0d′E + ∂B

∂t= 0

∂B∂t

We conclude that the two Homogeneous Maxwell Equations,

∇ · B = 0

∇× E = −∂B∂t

are both contained in dα = 0 .

Conclusion

Inhomogeneous ArgumentReferences

Argument for inhomogeneous equations.

∇ · B = 0∇× E = −∂B

∂t∇ · E = ρ

∇× B− ∂E∂t = J

The equation is ?d ? α = 0

dα = 0

?d ? α = 0

Conclusion

∇ · B = 0∇× E = −∂B

∂t∇ · E = ρ

∇× B− ∂E∂t = J

dα = 0

?d ? α = 0

Conclusion

∇ · B = 0∇× E = −∂B

∂t∇ · E = ρ

∇× B− ∂E∂t = J

dα = 0

?d ? α = 0

Conclusion

1 H. Flanders, Differential Forms with Applications to the PhysicalSciences, (1989)

2 J. E. Marsden, and A. J. Tromba, Vector Calculus, Fifth Edition(2003)

3 T. Watson, Introduction to the Hodge Star Operator: DifferentialForms Special Presentation, (2005)

4 S. Owerre, Maxwell’s Equations in Terms of Differential Forms,(2010)

Thank you.

exterior algebra and the maxwell-boltzmann equations€¦ · 2) ^ = 0 ^ = ( ^ ) joseph ferrara...

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