- #1
Kara386
- 204
- 2
I'm reading a textbook on electromagnetism. It says that for two vector fields ##\textbf{F}(\textbf{r})##
and ##\textbf{G}(\textbf{r})## their inner product is defined as
##(\textbf{F},\textbf{G}) = \int \textbf{F}^{*}\cdot \textbf{G} \thinspace d^3\textbf{r}##
And that if ##\textbf{F}## is a harmonic mode of an electromagnetic system we can always set ##(\textbf{F},\textbf{F})=1##. To demonstrate this they set
##\textbf{F} = \frac{\textbf{F}'}{\sqrt{(\textbf{F}',\textbf{F}')}}##
Then say this is a scalar multiple of ##\textbf{F}##, so it's really the same solution (which is because you can add two solutions, multiplied by a constant, together to get another solution). Setting ##\textbf{F}## to have this value allows the reader to easily verify that we can always set this inner product to be one. How does it let me easily verify that? That dot product has to be zero for the inner product to be 1, doesn't it?
and ##\textbf{G}(\textbf{r})## their inner product is defined as
##(\textbf{F},\textbf{G}) = \int \textbf{F}^{*}\cdot \textbf{G} \thinspace d^3\textbf{r}##
And that if ##\textbf{F}## is a harmonic mode of an electromagnetic system we can always set ##(\textbf{F},\textbf{F})=1##. To demonstrate this they set
##\textbf{F} = \frac{\textbf{F}'}{\sqrt{(\textbf{F}',\textbf{F}')}}##
Then say this is a scalar multiple of ##\textbf{F}##, so it's really the same solution (which is because you can add two solutions, multiplied by a constant, together to get another solution). Setting ##\textbf{F}## to have this value allows the reader to easily verify that we can always set this inner product to be one. How does it let me easily verify that? That dot product has to be zero for the inner product to be 1, doesn't it?
