Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: jeffreyH on 26/12/2018 19:11:24
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Since the big bang has the number of photons remained constant? If not then what kind of interactions changes this number?
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The number of photons in the universe is constantly changing. Stars produce them through fusion. radioactive isotopes emit them as gamma rays. Materials can absorb them and then re-emit the energy as a larger number of lower energy photons...
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Related to the above answer is the following page from CERN.
http://meroli.web.cern.ch/Lecture_photon_interaction.html
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Magnetic photon splitting is another mechanism that can change the number of photons in a system: http://iopscience.iop.org/article/10.1086/304152/fulltext/35776.text.html
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I just switched on a light, thus destroying the balance of the universe.
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As soon as I read that I switched off a light. Ha ha!
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http://meroli.web.cern.ch/Lecture_photon_interaction.html
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.
Have I interpreted this correctly?
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, so the continuing beam is not influenced by any energy exchange involving the removed photon. Therefore, the photons in the continuing beam retain their original energy, but there are less of them.
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Just to muddy the waters a bit... if there are an infinite number of photons in the universe, then this would have to be unchanged by such limited transactions as absorption and emission. ???
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Any photon that interacts with the target is removed from the beam, so the continuing beam is not influenced by any energy exchange involving the removed photon.
Yes and no. 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.
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.
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Therefore, the photons in the continuing beam retain their original energy, but there are less of them.
If the beam strikes an object that is above the black body temperature, the object will heat up, and will radiate the increased temperature as low-energy photons.
Anything in deep space that is hotter than the CMBR (2.7K) will, overall, radiate many low energy photons.
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You're thinking of the idea in where there only are transformations, but no 'loss' in a universe Jeffrey? If they are 'excitations in a field' then the number should vary with what interactions presents itself under a arrow of time I think.
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You're thinking of the idea in where there only are transformations, but no 'loss' in a universe Jeffrey? If they are 'excitations in a field' then the number should vary with what interactions presents itself under a arrow of time I think.
The question was intended to provoke answers of interest to readers. However, you are correct in interpreting my line of reasoning. It all stems from symmetry and conservation .
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Yes, those concepts are in some ways exceptionally mindnumbing. Symmetries and conservation laws. Remember when JP first explained to me how he looked at a 'photon recoil' as a 'demand' from conservation laws. It's not that far from being in a cave.
https://en.wikipedia.org/wiki/Allegory_of_the_Cave
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Just to muddy the waters a bit... if there are an infinite number of photons in the universe, then this would have to be unchanged by such limited transactions as absorption and emission.
Large quantity of mud! If infinity is not a number, how can there be an infinite number of anything?
Possibly you added the smiley because you guessed I couldn’t resist the bait. :)
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Does the circumference of a circle with infinite radius equal a straight line? If so no part of the circumference would tell you that it forms a circle. Similarly no one photon would tell you it is part of an infinite set.
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Does the circumference of a circle with infinite radius equal a straight line? If so no part of the circumference would tell you that it forms a circle. Similarly no one photon would tell you it is part of an infinite set.
You should have been a politician :) This doesn't answer the question.
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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"?
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?
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Yes, those concepts are in some ways exceptionally mindnumbing. Symmetries and conservation laws. Remember when JP first explained to me how he looked at a 'photon recoil' as a 'demand' from conservation laws.
Do you have a link to that explanation?
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Hmm maybe? It's here on TNS, and I miss that guy.
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Can't find it Bill, but as far I remember it was 'turning my head'
Reality isn't what I expected then.
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Hmm maybe? It's here on TNS, and I miss that guy.
Don't we all?
Is this the thread it came from? I’ve not had time to look through it yet, but could be, perhaps.
https://www.thenakedscientists.com/forum/index.php?topic=33720.msg326035#msg326035
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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
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Does the circumference of a circle with infinite radius equal a straight line?
No. See Cantor on multiple infinities.
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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.
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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.
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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.
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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.
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OK I'm going to pose a subsequent question. How does the wavelength of a photon affect its ability to penetrate a material?
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Just a guess, but I would think that the shorter the wavelength, the greater the penetrating power.
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Isn't it more a question of what is 'compatible' when it comes to the absorption of a photon?
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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!
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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.
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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.