# Homework Help: Phase of a complex EM-wave

1. Mar 9, 2012

### Niles

1. The problem statement, all variables and given/known data
Hi

My lecturer today mentioned today that the imaginary part of an EM-wave gives the phase of it. I can't quite understand this statement, considering the imginary part of an EM-wave is not something we can measure. Was he right?

Best regards,
Niles.

2. Mar 9, 2012

### tiny-tim

Hi Niles!
Yes, "the imaginary part" and "ei(phase)" are synonymous.
Yes we can … think fringes etc.

3. Mar 9, 2012

### Niles

Hi, thanks for answering. When I hear phase, then I associate it with θ in
$$E = E_0e^{i(\omega t + \theta)}$$
With this definition of the phase, then I guess the imaginary part of E cannot be regarded as the phase?

4. Mar 9, 2012

### turin

"gives" is ambiguous in this context. I would give your instructor the benefit of the doubt. Perhaps a more thorough/complete statement would be:
The comparison of the imaginary part to the real part gives the phase as
Code (Text):

phase = arctan( Im{amplitude} / Re{amplitude})

The imaginary part can't be measured directly in the sense of comparison to a real-valued standard, of course, but it does have physical consequences. It is not so much measureable as formal: something that allows you to frame the wave propagation in a beautifully elegant formalism (that has more profound consequences in quantum theory).

5. Mar 10, 2012

### tiny-tim

Hi Niles!
yes, that's really the phase difference

the difference between the phase of that and of $E = E_0e^{i\omega t}$ …

but we shorten "phase difference" to "phase" in the same way as eg we often shorten "height difference" to "height"

6. Mar 12, 2012

### sunjin09

To be more precise, the imaginary part is uniquely determined by the real part through a Hilbert transform, and they form something called a complex analytic signal that is equivalent but easier to work with. So yes you can measure the imag part, but only indirectly.

7. Mar 12, 2012

### Niles

Thanks! So I guess my professor wasn't 100% right when saying that hte phase difference θ is merely the imaginary part of the EM-wave (see Turin's answer).

Great, thanks. I guess you are referring to the Kramers-Kronig relations?

Thanks for that, that cleared things for me!

8. Mar 12, 2012

### sunjin09

I was convinced that Kramers-Kronig relations are closely related, but are not the same as Hilbert transforms, but I don't know what the subtle difference is. In this situation, I think Hilbert transform is more appropriate, since we're only looking at the real t axis, not the complex z=iwt plane.