Why is wave intensity defined as displacement^2 rather than just its magnitude?

jeebs
Messages
314
Reaction score
5
pretty straightforward question. for a long time I've been blindly calculating that intensity is (displacement from zero)^2, but never questioned why.

So why is this? I get that by squaring you get a positive value, which helps in, say, quantum probability calculations, but what's wrong with just taking a magnitude of a wave at a given point and calling that the intensity?
is it just one of our arbitrary conventions in physics or is there some real reason?
 
on Phys.org
Intensity is a classical concept, so let's leave the quantum realm alone for now.
To answer your question, we have to look back at what intensity is defined as: crudely, it is power delivered per unit area (albeit perpendicular to the propagation), which works out to energy per unit time per unit area.

Now, the energy carried by a wave is proportional to the square of its amplitude. This can best be illustrated with mechanical waves, such as a wave on a rope, in which every point is essentially in SHM. Consequently, intensity of a wave is also proportional to (amplitude^2) of the wave.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 36 ·
2
Replies
36
Views
10K
  • · Replies 41 ·
2
Replies
41
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
11
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K