math for physics/vector calculus

Vector Calculus (10) - Vector Product

이너피스! 2021. 7. 23. 18:09

23. Vector Product(Cross Product)

C=A×BC=A×B

Ci=j,kϵijkAjBkCi=j,kϵijkAjBk

 

Levi-Civita Density(Permutation Symbol): ϵijk={0+11ϵijk=0+11

0: two or three indexes are same (ϵ112,ϵ113,ϵ121,ϵ131,ϵ211,ϵ311ϵ112,ϵ113,ϵ121,ϵ131,ϵ211,ϵ311)

                                                                      (ϵ221,ϵ223,ϵ212,ϵ232,ϵ122,ϵ322ϵ221,ϵ223,ϵ212,ϵ232,ϵ122,ϵ322)

                                                                      (ϵ331,ϵ332,ϵ313,ϵ323,ϵ133,ϵ233ϵ331,ϵ332,ϵ313,ϵ323,ϵ133,ϵ233)

                                                                      (ϵ111,ϵ222,ϵ333ϵ111,ϵ222,ϵ333)

+1: i,j,k even permutation (ϵ123,ϵ231,ϵ312ϵ123,ϵ231,ϵ312)

-1: i,j,k odd permutation (ϵ321,ϵ213,ϵ132ϵ321,ϵ213,ϵ132)

 

C1=j,kϵ1jkAjBk=ϵ123A2B3+ϵ132A3B2=A2B3A3B2C1=j,kϵ1jkAjBk=ϵ123A2B3+ϵ132A3B2=A2B3A3B2

C2=j,kϵ2jkAjBk=ϵ213A1B3+ϵ231A3B1=A1B3+A3B1C2=j,kϵ2jkAjBk=ϵ213A1B3+ϵ231A3B1=A1B3+A3B1

C3=j,kϵ3jkAjBk=ϵ312A1B2+ϵ321A2B1=A1B2A2B1C3=j,kϵ3jkAjBk=ϵ312A1B2+ϵ321A2B1=A1B2A2B1

 

kϵijkϵlmk=δilδjmδimδjlkϵijkϵlmk=δilδjmδimδjl

 

(ABsinΘ)2=A2B2(1cos2Θ)=(iA2i)(iB2i)(iAiBi)2(ABsinΘ)2=A2B2(1cos2Θ)=(iA2i)(iB2i)(iAiBi)2

    =(A2B3A3B2)2+(A3B1A1B3)2+(A1B2A2B1)2=(A2B3A3B2)2+(A3B1A1B3)2+(A1B2A2B1)2

(ABsinΘ)2=C21+C22+C23=|C|2=C2(ABsinΘ)2=C21+C22+C23=|C|2=C2

C=ABsinΘC=ABsinΘ

 

 

24. Result of Vector Product is Vector

 

A(B×D)=i,j,kϵijkAiBjDk

D(A×B)=i,j,kϵijkDiAjBk=i,j,kϵjkiDiAjBk

A(B×D)=D(A×B) (i,j,k are dummy and can be renamed)

 

if A=B then A(A×D)=D(A×A)=D0=0

A,A×D are perpendicular

C=A×B is perpendicular to A,B(plane defined by A,B)

 

Direction of C: Right Hand Screw from A to B

 

 

25. Vector Product of Unit Vector

^ei×^ej=^ek=k^ekϵijk (i,j,k in cyclic order)

C=A×B=i,j,k^eiϵijkAjBk

    =|^e1^e2^e3A1A2A3B1B2B3|=^e1|A2A3B2B3|^e2|A1A3B1B3|+^e3|A1A2B1B2|

 

 

26. Properties of Vector Product

 

A×B=B×A

A×(B×C)(A×B)×C in general

 

A(B×C)=B(C×A)=C(A×B)ABC

A×(B×C)=(AC)B(AB)C

(A×B)(C×D)=A[B×(C×D)]=A[(BD)C(BC)D]=(AC)(BD)(AD)(BC)

(A×B)×(C×D)=[(A×B)D)]C[(A×B)C)]D=(ABD)C(ABC)D=(ACD)B(BCD)A