Can a region of space-time be created with no ER?

[San K]

A photon may be considered as an excitation of the electromagnetic (ER) field.

ER is thought to be omnipresent/ubiquitous in time-space (?)

Is it possible to construct a region in space-time (say a "black" box):1. That contains no ER fields?
2. that contains no photons? i.e. no excitation/energy...just a peaceful region with "calm" ER fields

 

[mfb]

If the field strength is zero, does that count as "no EM field"?
Do you count vacuum fluctuations? If you do, it is impossible (see the Casimir effect).

2. that contains no photons? i.e. no excitation/energy...just a peaceful region with calm ER fields

A small hole in a superconductor, cooled sufficiently... should work.

 

[Naty1]

I think, as mfb implies, a lot depends on just what you mean:

A photon is a quanta of an electromagnetic field, that is, a locally detectable manifestation of ER... a locally observable field quantity, while the field is a mathematical construct, not observable.

I don't think any ER is detectable in space unless from an external source. I suspect that is what mfb's small hole in a superconductor implies??

This seems contradictory:

1. That contains no ER fields?
2. that contains no photons? i.e. no excitation/energy...just a peaceful region with calm ER fields

Is a 'calm ER field' 'no ER field'...'calm' is not a term I have seen in these forums. Do you make a distinction between 'no ER' and 'no photons' and 'calm ER fields' ??

You are perhaps thinking of 'detectable' ER...??

You can get rid of most detectable ER with a Faraday cage, but your question may go beyond that to vacuum energy.

 

[dauto]

A calm field doesn't mean no field just as a calm ocean doesn't mean no water. Anyways, a calm field will happen at zero Kelvin (in classic physics) which is impossible by the 3rd law of Thermodynamics. Clasic physics isn't exact though and in Quantum physics even at zero Kelvin there would be some oscillation left on the fields due to zero point energy

 

[.Scott]

There will always be virtual photons.
The problem is that a "true vacuum" is too specific and would violate Heisenberg Uncertainty.