Does the Imaginary Part of an EM-Wave Determine Its Phase?

[Niles]

Homework Statement

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?

Thanks in advance.


Niles.

 

[tiny-tim]

Hi Niles!

My lecturer today mentioned today that the imaginary part of an EM-wave gives the phase of it.

Yes, "the imaginary part" and "e^i(phase)" are synonymous.

… the imginary part of an EM-wave is not something we can measure.

Yes we can … think fringes etc.

 

[Niles]

Yes, "the imaginary part" and "e^i(phase)" are synonymous.

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?

 

[turin]

... the imaginary part of an EM-wave gives the phase of it.

"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

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

I can't quite understand this statement, considering the imginary part of an EM-wave is not something we can measure.

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).

 

[tiny-tim]

Hi Niles!

When I hear phase, then I associate it with θ in
$$E = E_0e^{i(\omega t + \theta)}$$

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"

 

[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.

 

[Niles]

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"

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).

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.

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

"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

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).

Thanks for that, that cleared things for me!

 

[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.