Interference of two plane waves in any point of space

[Nikitin]

Homework Statement

For two plane EM waves with identical amplitude, frequency, polarization and phase traveling in the directions of $\hat{k_1}$ and $\hat{k_2}$, show that their average intensity is given by

$2I_0[1+\cos((\vec{k_2}-\vec{k_1}) \vec{r})]$

The Attempt at a Solution

Ugh, I've been struggling for an hour with this one. What am I to do? I tried adding the ponyting vectors, but I got an average intensity of 2I_0, which is obviously nonsense. What I can't understand, is what $\vec{r}$ actually is? I mean, according to the result I'm supposed to get, the average intensity can get as high as 4I_0 even if $\vec{r}$ is perpendicular to one of the $\hat{k}$s!

Can somebody please show how to solve this problem?

 

[Nikitin]

After returning on this problem with a fresh mind, I solved it by assuming the E-vectors have the same direction,,, but this is kind of retarded as the electric field from an EM wave will always be perpendicular to its respective $k$.

Here is how I solved it, but I really don't understand why it works. Can somebody please explain it to me?

https://fbcdn-sphotos-c-a.akamaihd.net/hphotos-ak-frc3/988360_10201874011028587_384761389_n.jpg

where $\Delta \vec{k} = \vec{k_2}-\vec{k_1}$

 

[haruspex]

I know nothing about the theory here, but I would have thought your first two equations represented waves in the same direction but different phase.

 

[Nikitin]

Ahh, now that you say it indeed it sounds like you are correct. But what I'm confused about, is that the waves have the directions of $\vec{k_1}$ and $\vec{k_2}$, yet the "directional vector" or whatever is $\vec{r}$ which is NOT parallel with the direction of the waves.. Look at the dot-product inside the cosine term. Could you please explain?

PS. I updated the photo in the OP: I corrected the direction of the E-fields of the two waves.

 

[Nikitin]

On a related note: Why can't I find the average intensity by simply superimposing the ponyting vectors and taking their mean over a period? When I tried this I ended up with a wrong answer.

Why do I have to find the total mean intensities by finding the total E-vector squared and finding its average value over an entire wave-period?

 

[haruspex]

I think now I may have misled you, and your original solution is correct.
The field is expressed as a function of $\vec{r}$, position in space, so I'm baffled as to what it means to say the two waves have the same phase. Won't the phase difference depend on spatial position, in fact as $\vec{k}.\vec{r}$? Which is exactly what you wrote. Maybe it should say 'same phase at the origin'?

 

[haruspex]

On a related note: Why can't I find the average intensity by simply superimposing the ponyting vectors and taking their mean over a period? When I tried this I ended up with a wrong answer.

Why do I have to find the total mean intensities by finding the total E-vector squared and finding its average value over an entire wave-period?

Not certain I understand the question, but...
Because the intensity is the mean square value, not the square of the mean value. The mean value of the field, over a whole number of cycles, will always be zero.

 

[Nikitin]

I think now I may have misled you, and your original solution is correct.
The field is expressed as a function of $\vec{r}$, position in space, so I'm baffled as to what it means to say the two waves have the same phase. Won't the phase difference depend on spatial position, in fact as $\vec{k}.\vec{r}$? Which is exactly what you wrote. Maybe it should say 'same phase at the origin'?

I think I understand it all now :) The problem was, my understanding of plane waves, polarization etc. was pretty bad before this problem, but now I think I got it.

Not certain I understand the question, but...
Because the intensity is the mean square value, not the square of the mean value. The mean value of the field, over a whole number of cycles, will always be zero.

I'm not sure you're correct here.. the length of the ponyting vector is the instantaneous intensity, and this intensity does have a time varying, positive value. There is nothing wrong with taking the mean of this.

The problem I met when I added the ponyting vectors of the two waves, was that I got an average intensity of $2 I_0$, which was not correct. I still don't understand this.

 

[haruspex]

I'm not sure you're correct here.. the length of the ponyting vector is the instantaneous intensity, and this intensity does have a time varying, positive value. There is nothing wrong with taking the mean of this.

The problem I met when I added the ponyting vectors of the two waves, was that I got an average intensity of $2 I_0$, which was not correct. I still don't understand this.

Then I have ventured too far beyond my expertise. Good luck.
Btw, it's 'Poynting'.