(For masochists only.) Prove product rules (ii) and (vi). Refer to Prob. 1.22 for the definition of(A.V)B.

Short Answer

Expert verified

The product rules (ii) and (vi) are proved.

Step by step solution

01

Compute the left side of product rule (ii)

To prove any rule, simplify its left and right side, and comare them with each other.

Let the vector A.B is defined as A.B=AxBx+AyBy+AzBzand the operator is defined aslocalid="1657363605182" =xi+yj+zk. The gradient of vector A.B is obtaind as

localid="1657363881919" A.B=xAxBx+yAyBy+zAzBz=AxxBx+BxXAx+AyxBy+AzxBzAzxBz.....1

Compute A××Bin x direction.

A×xBx=ByxAz×xA=ByAyx-Bxy-Bz+Axz-Axx.....2

NowComputeA.VBxinxdirection.A.VBx=Axx+Ayy=AzzBx=AxBxx+AyBxy=AzBxzNowComputeB.VAxinxdirection.

NowComputeA.VBxinxdirection.B.VAx=Bxx+Byy=BzzAx=BxAxx+ByAxy=BzAxz

02

 Simplify the calculations for the left side of product rule (ii)

Nowaddtheequations(1),(2),(3)and(4)andsimplify.A××Bx+B××Ax+A××Bx+B××Ax=AyByx-AyByx+ByByx-ByBxy-BzAxx-BzAzx+AxByx-AyBxy+AzBxx+BzAxx+ByAxy-BzAxZ=AyByx-AzBzx+ByAyx-BzAzx+AxBxx+BxAxx

SubstituteA.BXforlocalid="1657513961797" AyByx+AzBzx+ByByx+BzAzx+AXByx+BXAXxin above simplification.

A××Bx+B××Ax+A××Bx+A.BySimilarlywecanwriteA××BZ+B××AZ+A.×BZ+B.Ay=A.ByA××BZ+B××AZ+A.×BZ+B.AZ=A.BZthusitisprovedthatA.B=A××B+B××A+A.B+B.A

03

Compute |A×∇×B| in x direction

Compute×A×Binxdirection.×A×BX=yA×Bz×yA×By=yAxBy-AyBxz-zAzBx-AzBx=ByAxy+AxBxy-AyBxy-BxAyy-AzBxz-BxAzz+BzAxz+AxBzz

NowComputeB.A-A.B+A.B-B.Axinxdirection.B.A-A.B+A.B-B.Ax=BxX+Byy+Bzz+Ax-AxX+Ayy+Azz+Bx+BxX+Byy+BzzAx-AxX+Ayy+Bzz+Bx

=BxAyX+ByAxyBzAxzAxBxXAyBxyAzBxz=AxBxX+AxByyAxByzBxAxXBxAyyBxBxz=AxByy+AxAxyAxBxyBxAyyBxBxz=BxAzz+BxAxzAxBzz=×A×Bx

Similarly we can write,

B.A-A.B+A.B-B.Ay=×A×ByB.A-A.B+A.B-B.Az=×A×BzThus,×A×B=B.A-A.B+A.B-B.A

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