- #36
gabbagabbahey
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fluidistic said:Now we've showed that the second term of the integrand is worth [tex]2 (\nabla(\partial_t u_3) \cdot \nabla u_3)[/tex].
We did?!... Certainly, if [itex]u_3(0,x)=0[/itex], you can also say [itex]\mathbf{\nabla}u_3(0,x)=0[/itex]...that's all you need in order to show the second term vanishes.
In order to say that the integral will be zero for all [itex]t[/itex], you also need to show that [itex]u_3[/itex] satisfies the wave equation, so that you can apply the proof of [itex]E[/itex] being constant.
Real and positive I'd say. Real, not really sure why (I just seen in Born's book on Optics that some general time-harmonic fields are complex and that the real parts represent the fields), but energy has to be real... although I wouldn't be so surprised if it's a complex number whose real part represent the energy.
The first term is positive if it's a real number.
I'd say it is is safe to assume that [itex]u_3[/itex] is real-valued (and hence so are its derivatives). However, [itex](\partial_t u_3)^2[/itex] doesn't necessarily have to be positive, it could also be zero.
Thus the only way for the E to be 0 is that both terms are actually 0.
Right.
Oh... thus [tex]u_3=0[/tex]
You can't directly conclude that from the fact that [itex](\partial_t u_3)^2[/itex] and[itex]\mathbf{\nabla}u_3\cdot\mathbf{\nabla}u_3[/itex] are both zero. However, the fact that [itex](\partial_t u_3)^2=0[/itex] does allow you to conclude that [itex]u_3[/itex] is constant in time. And you also know that [itex]u_3(t=0,x)=0[/itex] so yes, [itex]u_3=0[/itex] and hence [itex]u_1=u_2[/itex] and thus thereis only one unique solution to the given wave equation.