Completely Positively Ionized Material: Invisible?

[Edi]

A completely positively ionized material would be practically invisible?
Normally it interacts with electrons, witch tell them what to do next (reflect, pass trough..), but a completely positively ionized matter wouldn't have any electrons to interact with. Of course they can still interact with the nuclei, but the chances they actually hit the nuclei would be extremely low, like neutrino (?), because of the sheer size of nuclei and because photon doesn't interact with protons electric field.

 

[Vanadium 50]

I'm not sure what to say other than "no, that's not correct".

 

[Edi]

Ok, than what would make it visible? What would the photon interact with (and reflect, get absorbed)?

 

[Astronuc]

Normally it interacts with electrons, witch tell them what to do next (reflect, pass trough..), but a completely positively ionized matter wouldn't have any electrons to interact with.

A fully ionized material is a plasma, but stable plasma have equal amounts of + nuclei and - electrons in order to maintain charge neutrality. Otherwise, a fully + ionized material would disperse by virtue of the strong Coulomb forces (+ charges would repel). The only way to maintain the mass would be to impose a magnetic field. A large magnetic bottle would be rather visible, so confining a large invisible positively ionized mass would seem rather impractical.

Of course they can still interact with the nuclei, but the chances they actually hit the nuclei would be extremely low, like neutrino (?), because of the sheer size of nuclei and because photon doesn't interact with protons electric field.

What is 'they' in "Of couse they can still interact . . . ."? Photons?

A fully positively (or negatively) charge mass will not remain together. Photons scatter off free (unbound) charges.

How does one conclude "a completely positively ionized matter wouldn't have any electrons to interact with"? This is rather impractical, since 1) the positively charged matter would disperse without off-setting negative charges, and 2) the free electrons would not be far way, and 3) the excessive + charge would attract electrons from neutral atoms.

 

[Bob_for_short]

A completely positively ionized material would be practically invisible?

It depends on the material density, temperature and the photon wave-length.

Even normal material is nearly transparent for X-rays, for example.

Of course they can still interact with the nuclei, but the chances they actually hit the nuclei would be extremely low, like neutrino (?), because of the sheer size of nuclei and because photon doesn't interact with protons electric field.

Photons do not practically interact with electric fields of any nature. They interact with charges. If you compare the cross sections for scattering from charges of different masses, the heavier charges scatter less (in a single photon - single charge interaction). But there are collective effects at low frequencies so a dense and large system can reflect the incident EM wave even though the material consists only of nuclei.

Bob_for_short.

 

[Edi]

..., so confining a large invisible positively ionized mass would seem rather impractical.

So it is invisible.
Se, I'm talking about a perfect situation here - it is well separated from electrons and contained.

 

[Astronuc]

So it is invisible.
Se, I'm talking about a perfect situation here - it is well separated from electrons and contained.

Perfect - but fictional (i.e. ignoring the real-world physics). Besides, if one were to have a huge positive charge, it then affects the behavior of neutral (normal) matter around it.

If one is looking for something that is 'non-detectable' as opposed to 'invisible', then it is impractical to have a positively charged matter.

Without electrons, a positively charged nucleus only scatters photons, and of low energy they would not interact. Nuclear primarily interact with (absorb) gamma rays, i.e. photons of energies in the keV-MeV range. One can look at isomeric transitions, and gammas associated with n-capture to see the typical gamma energies.

 

[Vanadium 50]

Let's ignore the very salient point that the sort of material you are describing does not and cannot exist, and go back to your original message:

Of course they can still interact with the nuclei, but the chances they actually hit the nuclei would be extremely low, like neutrino (?), because of the sheer size of nuclei and because photon doesn't interact with protons electric field.

None of this is correct. Nuclei are larger than electrons, not smaller. Light does interact with protons.

 

[Edi]

Yes, of course, i know that! But electron orbits the nuclei and is wave-like, so the photon interacts, because electron is not really a point particle. wave-particle duality anyone?

The material is not solid, of course, and is confined with electromagnetic fields. The field generators are far enough and strong enough so the large positive mass doesn't steal any electrons.

"Without electrons, a positively charged nucleus only scatters photons, and of low energy they would not interact."

Why? What would photons interact with? The chance of hitting a nuclei is extremely small, why then neutrinos can easily fly trough matter with no or minimal interactions?

 

[Bob S]

The three ways photons interact with ordinary matter are the photoelectric effect (including deep core photoejection), Compton scattering, and pair production. With no electrons, the photoelectric effect cross section would be zero. Photons can compton scatter off of protons, but the cross section is lower by about 1836² (proton-to-electron mass ratio squared). The pair production cross section would be nearly the same. The photon cross section for interacting with nuclei would be unchanged. This includes gamma-neutron interactions, for example O¹⁶(γ,n)O¹⁵, which peaks between 20 and 30 MeV.

α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω

 

[Edi]

So, for photoelectric effect photon needs a electron - we can rule out this in this case.
For pair production photon needs to hit a nuclei (with photon energies >1,22 MeV) - we can rule out this for this case. (Because of low chance that photon will actually hit the nuclei)
For Compton scattering it needs either electron or nuclei - rule out the electron case and the nuclei case for the same reason as pair production, low chance.
Almost every photon would go through unaffected, so the thing would be invisible to EM radiation.

Or explain me this: why neutrino can travel through entire planets without hitting anything? Because of the low chance of hitting nuclei.

Ah, I'm going crazy!

 

[malawi_glenn]

particles do not interact with "collision" as if they were tiny balls, they interact via quantum fields.

The neutrino only interact with the "weak force", which is weaker than the electromagnetic force. The weakest force is gravity.

 

[Bob_for_short]

Almost every photon would go through unaffected, so the thing would be invisible to EM radiation.

As I said, if the material is dense and large, and you photon or EM wave is long, the material can reflect the EM wave. It is a collective effect. You may consider the material as a conductor. Conductors reflect long waves. So your material can be visible.

Or explain me this: why neutrino can travel through entire planets without hitting anything? Because of the low chance of hitting nuclei.

Yes, the interaction cross section for neutrino is very small so it makes a long way between two successive collisions.

Bob_for_short.

 

[Bob S]

So, for photoelectric effect photon needs a electron - we can rule out this in this case.
For pair production photon needs to hit a nuclei (with photon energies >1,22 MeV) - we can rule out this for this case. (Because of low chance that photon will actually hit the nuclei)
For Compton scattering it needs either electron or nuclei - rule out the electron case and the nuclei case for the same reason as pair production, low chance.

For Compton scattering, the photon has to transfer some momentum to the proton, so the equation is the same as for Compton scattering off electrons, so the proton mass has to be substituted for the electron mass in the equation. So instead of a cross section of about 0.66 barns, it is about 0.2 microbarns. So it is very small. But as I said in my earlier post, photo-nuclear reactions can have cross sections of 10's of millibarns, depending on specific nucleus. The photonuclear cross section threshold in beryllium is about 1.67 MeV.

Or explain me this: why neutrino can travel through entire planets without hitting anything? Because of the low chance of hitting nuclei.

Unlike photons, the neutrino can only interact via the weak force, not electromagnetic.

 

[mheslep]

I think the basic Feynman diagram describes the situation. The electron shown is of course charged matter. The photon (green gamma line in the diagram) can either be absorbed or emitted. If a photon detector is inserted so that it views the emitted photon, then the electron is 'seen'. Or conversely if the photon has an external source and is absorbed, the lack of its detection also is evidence of the electron, i.e. it is 'seen'. I suppose the same holds true for nuclei.

http://media-2.web.britannica.com//eb-media/71/571-004-4247634B.gif

 

[Edi]

But, since when photons interact with electromagnetic field? I mean, you can't bend a photon (change direction of motion) using a powerful electro magnet!
Yes, the neutrino interacts with week force and because week force has low range, the neutrino has to come close to the nuclei - the range in witch week force acts. But the photon... does it interact with the electromagnetic field around the nuclei? I don't think so.
If that's the case, why can't one bend a photon with powerful electromagnet? :

 

[malawi_glenn]

one can bend photons with electromagnets for sure, since the photon carries both a magnetic and an electric field:

http://en.wikipedia.org/wiki/Electromagnetic_radiation

 

[Edi]

You serious or you just making fun of me? One can bend photons with powerful enough electro magnet? Of course, to do it on macro scale would require extreme amounts of energy and it would be kinda hard to make the magnet, but still - photons can be bent by by a powerful enough electro magnet??


wow, that's something completely new. I must have died and changed universe i live/d in... :noface:
Of course i knew taht photon has magnetic and electric field, but... yeah :shy:

 

[malawi_glenn]

Well, the photon electric field is oscillating, so it would just be averaged out.

Since you are asking this in the particle physics thread, the EM field's quanta are photons, and photons do not couple directly to photons.

To lowest order, the pair prodcution Feynman diagrams are:
http://musr.physics.ubc.ca/~jess/p200/emc2/img48.gif

(Z = the nucleus with atom number Z)

Now recall that this is not what happens in reality, Feynman diagrams are just mathematical tools and representations of terms in an perturbative expansion of the interaction.

 

[Bob S]

I think the basic Feynman diagram describes the situation. The electron shown is of course charged matter. The photon (green gamma line in the diagram) can either be absorbed or emitted. If a photon detector is inserted so that it views the emitted photon, then the electron is 'seen'. Or conversely if the photon has an external source and is absorbed, the lack of its detection also is evidence of the electron, i.e. it is 'seen'. I suppose the same holds true for nuclei.

http://media-2.web.britannica.com//eb-media/71/571-004-4247634B.gif

We are discussing photon scattering off of matter without electrons. Your diagram is for electron-electron scattering with a photon mediator.

 

[Bob S]

Here is the Feynman diagram for electron pair photoproduction off a nucleus or proton. See
http://www.irs.inms.nrc.ca/EGSnrc/pirs701/node22.html
This is independent on whether or not any electrons are present, because the primary pair production occurs off a nucleus. In aluminum, the pair production cross section is about 350 millibarns per atom at 20 MeV. For pair production off a proton, it is about 2 millibarns.

 

[kanato]

What about something like Bragg diffraction? The picture often given in introductory solid state texts is that xrays reflect off of planes of nuclei and constructively interfere at the correct angles. It seems like that might be relevant to this situation but I am not sure that the simplified picture is correct.

 

[Vanadium 50]

The three ways photons interact with ordinary matter are the photoelectric effect (including deep core photoejection), Compton scattering, and pair production.

True. But irrelevant.

The OP was talking about visible light photons. Your reply is only relevant for photons of much higher energy. And now the thread is hopelessly derailed.

You're a smart guy, Bob. People recognize that. But not every question requires a PhD-level answer.

 

[Bob_for_short]

There are two Bobs here. I, Bob_for_short, say that when the photon wave-length is long, etc., the positive charge matter can be reflective thus visible. No nuclear reactions, no pair production, just a conductor in an external EMW.

Vladimir Kalitvianski.

 

[Bob S]

I am the other Bob. When the photon wavelength is long (e.g., visible light), it can scatter off of free electrons. See the Thomson cross section:
http://en.wikipedia.org/wiki/Thomson_scattering.
When the electons are absent, these visible photons can scatter off of free protons by the same process, but the yield of scattered photons is lower by a factor of the mass-ratio squared (1837² = 3,374,000).

 

[Edi]

So, the photon does interact with the electro magnetic field (around the nuclei).
On macro scale one can't really bend light using a magnet, because photons field oscillates thus averring out and not bending in one direction or another.
On particle scale the frequency of relatively long waves isn't enough to average out, so it interacts via the field around nuclei, but instead of the photon changing its path, the nuclei moves with the changing field, thus receiving energy from the photon and later emitting it via another photon in different direction.

 

[Bob_for_short]

So, the photon does interact with the electro magnetic field.

Wrong, on the contrary. The electromagnetic field appears in the charge equations of motion as a superposition of fields from different sources. Superposition means non interaction.

The field equations themselves contain charges as a source, not the other fields.

Bob_for_short.