# B Is there a minimum wavelength for electromagnetic radiation?

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1. Oct 5, 2016

### afcsimoes

The wavelength is inversely proportional to the foton energy. So, the limit can be stated by the mass of the full universe.
But how much near that limit can the light be?

2. Oct 5, 2016

### DaveC426913

In theory, or in practice?

In theory, there is no upper limit. In practice, since we observe that the universe is not one very high energy photon, the actual existence of such a thing has not happened.

3. Oct 5, 2016

### afcsimoes

When our universe has came into existence the only "thing" was energy. As radiation, i think.

4. Oct 5, 2016

### Staff: Mentor

But that energy is frame dependent. We can make the energy arbitrarily large and the wavelength arbitrarily small just by choosing a frame in which the light is blue-shifted by an arbitrarily large amount.

5. Oct 6, 2016

### afcsimoes

Yes. I forgot the frames question.
So, theoretically there are no lower limit for the foton wavelength and there are no upper limit also.
I realize now that i was thinking of the observer at the same frame of the local of foton's production.
Fotons are produced when a charged particule is decelerated. No massive particule can reach c. In pratice only near a hyper-super-massive BH exist some possibility of such near-zero-wavelength fotons be produced, rigth?

6. Oct 6, 2016

### Staff: Mentor

Any interaction that transfers energy to the electromagnetic field is going to create photons. In principle we can produce arbitrarily energetic interactions by colliding arbitrarily energetic particles. We can describe these interactions using any frame we choose.

BTW, it is spelled "photon", not "foton". I can't justify English spelling, all I can do is apologize for it.

7. Oct 21, 2016

### Staff: Mentor

Not according to our best current models. The earliest phase of the universe's history that our models give us reasonable confidence for is inflation, and during inflation, the energy of the universe was all in the inflaton field, the field driving inflation. At the end of inflation, that energy was transferred to the Standard Model fields, what we usually think of as "matter" and "radiation".

8. Oct 22, 2016

### Simon Phoenix

Personally I think you should apologise for it instead

9. Oct 22, 2016

### vanhees71

There's no need to apologize in this case. English is better than German in this respect, because it transcribes the greek $\phi$ with ph rather than putting f's instead. In English you still write "photography" while in German it's "Fotografie" :-(.

10. Oct 22, 2016

### afcsimoes

I have learned that in physics the generic definition of field is that it is a region of space where we can assign a value to each and any point.
A gravitational field is generated by one mass. A static electrical field is generated by an electric charge with no motion (at the observer's referential). An electromagnetic field is generated by a moving electrical charge. And so on (for other kind of fields).
What kind of "thing" generates the inflation field? Can it be the reverse of a BH?
Thank you, again.
P.S.: I am sorry by my poor english
Best regards to all.

11. Oct 22, 2016

### Staff: Mentor

Not quite. The field is not the region of space; it's the mathematical object that assigns a value to each point in the region.

All this is not necessarily true. An electromagnetic field can exist in a vacuum, with no sources. A gravitational field, in the modern sense of spacetime curvature, can exist in a vacuum, with no stress-energy. It's true that in our ordinary experience, there are always sources somewhere associated with these fields; but that doesn't mean it must always be true everywhere for all fields.

Nothing you're used to. The usual model for the inflaton field (note that it's "inflaton", not "inflation", when referring to the field) is a scalar field, which is just an assignment of a simple number to every point in some region of space (more precisely of spacetime in this case since we're talking about a quantum field on spacetime). Such a field has no "source" in the usual sense--or, heuristically, you could say its source is itself, since there is energy associated with the field.

12. Oct 22, 2016

### afcsimoes

OK.Now I need some time to think about all that. these are very hard concepts.
Thank you, once more.
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