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In the complex notation there is a clever device for finding the time average of a product. Suppose f(r,t)=Acos(k×r-ωt+δa)and g(r,t)=Bcos(k×r-ωt+δb). Show that <fg>=(1/2)Re(fg~), where the star denotes complex conjugation. [Note that this only works if the two waves have the same k andω, but they need not have the same amplitude or phase.] For example,

<u>=14Re(ε0E~×E~+1μ0B~×B~)and<S>=12μ0Re(E~×B).~ and .

Short Answer

Expert verified

It is proved that fg=1/2Ref~g~.

Step by step solution

01

Expression for the f(r,t) and g(r,t):

Write the expression for f (r , t).

f(r,t)=Acos(k.r-ωt+δa) …. (1)

Write the expression for g( r , t).

g(r,t)=Bcos(k.r-ωt+δb) …. (1)

02

Determine the <fg>:

Find the fgas follows.

fg=1TgTAcos(k.r-ωt+δa).Bcos(k.r-ωt+δb)dt=ABTgTcos(2k.r-ωt+δa+δb+cos(δa-δb)dt=ABTcos(δa-δb)T=12ABcos(δa-δb)

03

Determine (12)Re(fg)^:

Write the equation in the complex notation.

f~=A~ei(k.r-ωt)g~=B~e-i(k.r-ωt)

Where, A~=aeiδaandB~=Be-iδb.

Determine12f~g*~ as follows.

12f~g*~=A~ei(k.r-ωt)B~ei(k.r-ωt)=12A~B~=12ABei(δa-δb)=12AB(cos(δa-δb)+isin(δa-δb))

Consider the real term.

Re12(f~g~)=12ABcos(δa-δb)=fg

Therefore, it is proved that fg=(1/2)Re(f~g~).

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