Fresh air (vacuum) as media for electromagnetics, can it have losses?

[Fernandopozasaura]

Recently I saw an exercise on electromagnetics (plane waves) in which it was said that a plane wave was traveling through vacuum (ε0 and μ0) BUT with losses, i.e, σ ≠0.
Is that possible?
I think I have read in the past that it's OK in some texbook, but now, I have searched for it in a couple of physics texts without success.
As far as I've learned in my search, σ (conductivity) is due to free charges in volume where you want to get wave electric (or magnetic) field, so in vacuum (or fresh air, which is not exactly vacuum but that is another issue) as long as there is no material, no free charges will exist, so, σ = 0.
Thanks for your help.

 

[Dale]

Summary:: losses in vacuum for electromagnetic `plane waves. Is that possible?

Recently I saw an exercise

Can you please provide an exact reference?

 

[Fernandopozasaura]

Well, I saw it in an exame someone showed to me. I'm sorry I don't have more details.

 

[Dale]

There isn’t much we can do then. We can’t tell if you are misremembering the main point or if you forgot some important subtle point. Sorry.

Classically free space is not a lossy medium. That is about all that we can say generally without any specific details.

 

[Fernandopozasaura]

OK, I understand.
As you, my first answer was what you said: free space is not lossy. But somehow I remembered have read something different, that's why I asked. I hoped someone could eliminate other possibility.
Anyway, thank you very much for your help, Dale.

 

[DaveE]

I suspect that this has more to do with semantics; the definition of a vacuum. As a thought experiment, there just can't be any losses in a perfect vacuum everywhere the EM field is (i.e. everywhere). That does raise the question how did the EM radiation arise in a universe with nothing in it.

 

[vanhees71]

Well, obviously there is something in the universe. An we can't observe the vacuum, because to observe something we need to bring a measurement device in it, and then it's no longer a vacuum. A single photon in an otherwise empty universe is stable, i.e., the vacuum is not lossy at all in standard quantum field theory.