[yor_on]

No, but it's good fun rereading it, lightarrow is another guy I enjoyed. Damn Bill but we're getting old here, eight years of discussing one topic, physics,, and nowhere closer to the goal

[alancalverd]

Does the circumference of a circle with infinite radius equal a straight line?

No. See Cantor on multiple infinities.

[alancalverd]

Depends on the nature of the interaction. At low photon energies the photon generally disappears into heat, a chemical change, or the movement of charge in an electrical circuit, but at energies above the visible spectrum you can get all sorts of secondary emission including photonuclear reactions.

What is the likelihood that emissions arising from these interactions would influence the original beam?
Would there be any effect other than "attenuated intensity"?

I'm dealing with just such a case right now: the phenomenon of "buildup" as a photon beam passes through a concrete barrier.  Multiple interactions within the barrier means that you can end up with more photons coming out than went in, but at a lower mean energy per photon.

This is a useful phenomenon in industrial radiography where we use a thin sheet of tin or lead o "intensify" the image - more, lower energy, photons are captured by the x-ray film than if you just use the raw 300 kV x-ray beam. It's a pain when designing radiotherapy bunkers because you need to use an iterative process to determine the required thickness of concrete - more doesn't always mean less! And it caused great hilarity when a pompous and ignorant Health and Safety Inspector (he always spoke in Capital Letters) insisted that staff handling high-energy gamma sources should wear lead aprons, which not only slowed them down, but actually increased their instantaneous skin dose.

We use all kinds of filters to remove photons of specific energies from a beam to produce a more monoenergetic  (monochromatic) beam of lower intensity.

In these cases; would a filter equate to a “target”, in the original quote?

[/quote] Indeed. At the low energy end of the business, we use aluminum filters to remove the part of the x-ray spectrum that would cause skin burns without producing an image. Other folk use filters to remove high energy bits from the visible spectrum so that your room lights don't interfere with your TV remote control.

[Bored chemist]

I'm dealing with just such a case right now: the phenomenon of "buildup" as a photon beam passes through a concrete barrier.  Multiple interactions within the barrier means that you can end up with more photons coming out than went in, but at a lower mean energy per photon.

A more commonplace example is where the Sun shines on a rock + warms it up.

[Bill S]

Could be, I’m getting there. I’ll try re-wording my interpretation of:

As consequence of such kind of interactions a photon that interacts with the target is completely removed from the incident beam, in other words a beam of photons that cross a medium is not degraded in energy but only attenuated in intensity.

A beam of photons is travelling through (e.g.) a vacuum.  It hits a transparent/translucent target.  Any photon that interacts with the target is removed from the beam.  If this interaction is such that it does not result in the production of further photons, the continuing beam is not influenced by any energy exchange involving these removed photons, and continues, as before, but with fewer photons, of which the energy is unchanged.  However, if the interactions result in the emission of new photons (possibly more photons, with lower energy), the influence on the emerging beam will not be restricted to attenuation.

[Bill S]

A more commonplace example is where the Sun shines on a rock + warms it up.

If this is an outcrop of rock which is warmed through, what is detected on the far side?  Obviously, this would not be visible light, unless the rock became ridiculously hot, but presumably there would be photons of some wavelength.

[jeffreyH (OP)]

OK I'm going to pose a subsequent question. How does the wavelength of a photon affect its ability to penetrate a material?

[Bill S]

Just a guess, but I would think that the shorter the wavelength, the greater the penetrating power.

[yor_on]

Isn't it more a question of what is 'compatible' when it comes to the absorption of a photon?

[evan_au]

I would think that the shorter the wavelength, the greater the penetrating power.

There is another mechanism at work for wavelengths that are much longer than the object size.

If the wavelength of the radiation is much more than twice the size of the object, the radiation tends to go through and/or around the object as if it wasn't there. This is the ultimate penetrating power!

[PmbPhy]

Since the big bang has the number of photons remained constant? If not then what kind of interactions changes this number?

A simple example is when an electron annihilates a positron producing two photons. My TV produces photons while its on.

[jeffreyH (OP)]

I would think that the shorter the wavelength, the greater the penetrating power.

There is another mechanism at work for wavelengths that are much longer than the object size.

If the wavelength of the radiation is much more than twice the size of the object, the radiation tends to go through and/or around the object as if it wasn't there. This is the ultimate penetrating power!

This is something Leonard Susskind discusses in relation to black holes. If the photon wavelength is twice the diameter of the event horizon then the black hole won't capture it.