Define inner product of vector fields EM

Click For Summary
SUMMARY

The inner product of two vector fields, denoted as ##(\textbf{F},\textbf{G})##, is defined mathematically by the integral ##\int \textbf{F}^{*}\cdot \textbf{G} \thinspace d^3\textbf{r}##. When ##\textbf{F}## is a harmonic mode of an electromagnetic system, it can be normalized such that ##(\textbf{F},\textbf{F})=1##. This normalization is achieved by setting ##\textbf{F} = \frac{\textbf{F}'}{\sqrt{(\textbf{F}',\textbf{F}')}}##, which allows for the verification of the inner product being equal to one. The discussion clarifies that the dot product must be zero for the inner product to equal one, emphasizing the importance of proper substitution in calculations.

PREREQUISITES
  • Understanding of vector calculus
  • Familiarity with electromagnetic theory
  • Knowledge of inner product spaces
  • Proficiency in performing integrals in three-dimensional space
NEXT STEPS
  • Study the properties of inner products in vector spaces
  • Explore harmonic modes in electromagnetic systems
  • Learn about normalization techniques in vector fields
  • Investigate the implications of dot products in vector calculus
USEFUL FOR

Students and professionals in physics, particularly those focused on electromagnetism, vector calculus, and mathematical physics, will benefit from this discussion.

Kara386
Messages
204
Reaction score
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?
 
Physics news on Phys.org
Just substituting in works fine, sorry! Made a mistake!
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 24 ·
Replies
24
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K