Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: thedoc on 14/07/2015 00:50:01
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Harry Audus asked the Naked Scientists:
On the subject of light, a hypothetical question. Suppose I had a perfectly light-tight box with a lid that was perfectly light-tight when closed. The box is standing in a light environment with its lid open. If I close the lid, will the inside of the box stay forever just as light as it was before I closed the lid?
Now suppose the box has one very very very tiny hole in it. This is a magic hole that allows light to travel through it in one direction only, from the inside to the outside. Is it possible to calculate how long it would take for the box to lose all its light (i.e. become completely dark)? What sort of function would the rate of loss follow?
What do you think?
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Harry Audus asked the Naked Scientists:
On the subject of light, a hypothetical question. Suppose I had a perfectly light-tight box with a lid that was perfectly light-tight when closed. The box is standing in a light environment with its lid open. If I close the lid, will the inside of the box stay forever just as light as it was before I closed the lid?
By definition: you wrote that the box is "perfectly light-tight" and this means that it doesn't absorb light. Of course there isn't only the light that has entered, inside of it, but also the radiation the box emits itself because of its temperature. However I don't grasp the exact meaning of the phrase "will the inside of the box stay forever just as light as it was before I closed the lid?" I hope to have given the answer you were looking for.Now suppose the box has one very very very tiny hole in it. This is a magic hole that allows light to travel through it in one direction only, from the inside to the outside.
And not from outside to inside? Physically impossible.Is it possible to calculate how long it would take for the box to lose all its light (i.e. become completely dark)? What sort of function would the rate of loss follow?
A lot of assumptions are missing.
Assuming we could neglect the radiation emitted thermically by the box and the radiation which is present outside of it when you open the hole, that energy is redistributed instantly inside the box and ... some othe things, the energy escaping the hole in the unit time at a certain instant of time t is proportional to the energy density inside the box at that instant t, to the area s of the hole's surface and to c:
dE/dt = - c*s*E/V where V is the box' volume.
Integrating:
Integral dE/E = - Integral c*s/V dt
E = E0e-c*s*t/V
where E0 is the total energy of light at t = 0.
Example.
V = 1 litre = 10-3 m3
s = 1 mm2 = 10-6 m2
the interval of time t needed to halve the energy inside the box is:
t = ln(2)*V/c*s ≈ 2.3*10-6 s = 2.3 μs.
But this equation would be approximately true only at the beginning, at least because of the assumptions I talked about.
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lightarrow
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Harry Audus asked the Naked Scientists:
..... The box is standing in a light environment with its lid open. If I close the lid, will the inside of the box stay forever just as light as it was before I closed the lid?
This is best answered by a simple experiment.
Place the box so the light shines into it. Pass your hand between the light and the box so it casts a shadow into the box. You will see the inside of the box getting darker as you cut off the flow of light into the box.
Now get a large piece of lightproof material, a coat perhaps, and putting it over your head peer into the box as you cover both your head and the box cutting out any incoming light.
Can you see any light coming out of the box?
When you have done this, report back to us what you saw. I think you will find that it answers your question about the hole in the box and how long it will take the light to 'leak out'.
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The questioner is probably imagining a photon bouncing around forever inside the box, but even if the walls are white or mirrors it will quickly be absorbed and its energy will end up being emitted again as infrared photons. This invisible light will continue to move around inside the box, and be reabsorbed, and be emitted again, and escape from the box by being emitted from the wrong side while other heat energy from outside comes in to be emitted as infrared light inside the box, but once the initial visible light has been absorbed, no more visible light will be emitted (ignoring photons being reflected which could involve a moment of absorbtion without the energy being split up and distributed around).
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Harry Audus asked the Naked Scientists:
On the subject of light, a hypothetical question. Suppose I had a perfectly light-tight box with a lid that was perfectly light-tight when closed. The box is standing in a light environment with its lid open. If I close the lid, will the inside of the box stay forever just as light as it was before I closed the lid?
If the mirrors inside the box were perfect reflectors, i.e. all photons were reflected and none of them were absorbed, then yes. However no such mirror exists or can exist.
Now suppose the box has one very very very tiny hole in it. This is a magic hole that allows light to travel through it in one direction only, from the inside to the outside. Is it possible to calculate how long it would take for the box to lose all its light (i.e. become completely dark)? What sort of function would the rate of loss follow?
Maybe, but it'd be very complicated due to the discrete nature of light. Even then it'd be an approximation.
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Light is a form of motion, you must have something to move to have motion. Complex motion can organize mass as well as disorganize mass. As for light remaining in a container, there will always be light in the container for every thing is in the constant act of expanding and contracting!
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......there will always be light in the container for every thing is in the constant act of expanding and contracting!
Could you please explain that under New Theories?
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If we take light as a general term for electromagnetic radiation we have the example of the resonant cavity used at radio frequencies, these are defined by there Q factors this being the ratio of stored energy verses that which dissipates.
Presumably one with a Q factor of 1000 would have the stored energy reduced to 1/e in one millisecond and pro rata.
Cavities where it is desired to store a great deal of energy such as in LHC are constructed of super conductive materiel.
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The light doesn't last forever in real boxes, but you can learn a lot from how long it does last.
https://en.wikipedia.org/wiki/Cavity_ring-down_spectroscopy
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Presumably one with a Q factor of 1000 would have the stored energy reduced to 1/e in one millisecond and pro rata.
The maths is valid, but I don't think you will get a Q of 1000 with the type of box the OP has in mind. [:)]
However it does answer his question:
.....will the inside of the box stay forever just as light as it was before I closed the lid?
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Thanks all for your contributions. Of course I neglected to say that all internal surfaces are perfectly reflective, but I think most of you realised my intent.
As to the calculation of how long it would take for the box to empty of light via the one-way hole (again, of course, a hypothetical construct), I'm afraid the responses assumed a level of knowledge I don't have. Does anyone have a simple explanation? Or perhaps an explanation of why a simple explanation of the solution isn't possible?
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And the box is in an environment at zero degrees absolute.
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Light is a form of motion, you must have something to move to have motion.
That is incorrect. Light cannot be said to be a form of motion. It's a semantically incorrect sentence. Light moves. It doesn't "have" motion and is not a "form" of motion either.
...every thing is in the constant act of expanding and contracting!
That is quite incorrect. I assume that you're referring to the expansion of the universe. If so then you've made a common mistake, i.e. that just because the universe is expanding that the things inside the universe are expanding, which is wrong.
I see you ignore Colin's question. Why?
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Colin what new theory are you referring to? Pmbphy how do you know that's incorrect, by saying what you have said tells me that you believe that nothing can move faster that the speed of light. Everything is made up of mass including the most powerfullest vacuume man or god can produce! All mass is motion having the function of infinity. Conventional wisdom says this can not happen...why because we relate everthing to space time. Space time (change OF distance) with respect to our own size Space time has a begining and a end.
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Pmbphy how do you know that's incorrect,...
Simple observation. Objects in the universe simply aren't expanding. There's no reason or mechanism for it. However I said that I was assuming that you are referring to the expansion of the universe. Is that correct? If so then that applies only to the expansion of distances between galaxies, not to objects of the size of galaxies and smaller. There's simply no reason for it. Why did you say that they are anyway? Forces like gravity and electromagnet forces counteract the expansion of the universe because those forces at the level of galaxies or smaller are so overwhelming compared to the force of expansion that the expansion forces can be neglected. That's common knowledge among cosmologists.
.. by saying what you have said tells me that you believe that nothing can move faster that the speed of light.
That too is a fact. Its demonstrated every day in particle accelerator labs. There is no force in existence that can accelerate an electron to reach the speed of light. If would take an infinite amount of energy to make that happen and there is only a finite amount of energy that can be used to accelerate an electron.
Everything is made up of mass including the most powerfullest vacuume man or god can produce!
A vacuum doesn't have mass. In fact it's a lack of mass by definition.
All mass is motion having the function of infinity. Conventional wisdom says this can not happen...why because we relate everthing to space time. Space time (change OF distance) with respect to our own size Space time has a begining and a end.
You don't sound like you know what you're talking about. What you just said is nothing more than word salad.
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On the subject of light, a hypothetical question. Suppose I had a perfectly light-tight box with a lid that was perfectly light-tight when closed. The box is standing in a light environment with its lid open. If I close the lid, will the inside of the box stay forever just as light as it was before I closed the lid?
Now suppose the box has one very very very tiny hole in it. This is a magic hole that allows light to travel through it in one direction only, from the inside to the outside. Is it possible to calculate how long it would take for the box to lose all its light (i.e. become completely dark)? What sort of function would the rate of loss follow?
What do you think?
Yes, but the light will come out very slowly. I think it's impossibly to calculate how long it will take for the box to black out.
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Harry Audus asked the Naked Scientists:
Now suppose the box has one very very very tiny hole in it. This is a magic hole that allows light to travel through it in one direction only, from the inside to the outside. Is it possible to calculate how long it would take for the box to lose all its light (i.e. become completely dark)? What sort of function would the rate of loss follow?
What do you think?
I suspect that's one of the easier cases to do the calculation on.
If the hole is very small and the "mirrors" are perfect then the light will mainly miss the hole so it will only come out slowly.
That means that most of the light will be in the box for a fairly long time. Diffraction will therefore have time to spread the light out so it is going essentially in random directions in the box..
You then have something that looks a bit like a "gas" of photons in the box - the bounce round randomly till the hit the hole and escape.
I'm pretty sure that would lead to an exponential decay in the light output. So, it would (theoretically) never reach zero (it would take an infinitely long time).
The decay time constant would be something of the order of the time it takes a photon to bounce across the box multiplied by the area of the box divided by the area of the hole.
With a "box" made from two large, very good, flat mirrors at either end of a tunnel through the middle of the earth you could get the "box" to hold light for long enough that you could watch it fade. (just)
(Big boxes make it last longer because the light spends more of it's time in the space in the box rather than being reflected. It's the reflection that absorbs it)
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Pmbphy; Word Salad I love it! Which universe are you talking about? Our universe or the universe of everything? You can not have contraction with-out expansion or vice-versa! We tend to think of our universe as having a size limit. If you could expand your body large enough, until galaxy's looked as if they were clusters of atom's ( orbiting particle system's )then you would realize that the speed of light in our universe would appear to be much faster. Expansion is the easy part, contraction comes with expansion and as I said before in order to have motion you must have something to move. I know of two forms of expansion "unorganized" and "organized" expansion. Organized expansion is slower than unorganized expansion. The organizing expansion rate is proportional to the feed rate that's feeding the organized expansion. This act of feeding the organized expansion is the contraction part. In order to visualize this think of a sphere that it's surface is covered with hour glasses and the sand is moving into the sphere, the sand is in the act of contracting. Of course our universe is diverse and complex, Large or small everything everywhere is always in the constant act of contracting or expanding. when it's slow and we can not see it does not mean that it's not there. I will also say this, all mass is made up of orbiting particle systems including the orbiting particle systems there self.
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Bored Chemist
I assume your long light storage box would have slightly curved mirrors so that the light beam would stay on centre and not spread out.
Rather like the Spectra Physics lasers with which I used to work, your 12882 Km long device should have a quite high "Q" factor with only 23.28 reflections per second.
Assuming otherwise perfect mirrors the loss at each reflection the would be in proportion of the area of the hole compared to the beam width area
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Even with a 1mm hole and a 1m beam width the loss of energy would be trivial compared with the losses incurred by real world mirrors, it seem that the best available have a 0.1% loss at each reflection so making the box long is the best solution.
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Slightly convex mirrors only help a bit; diffraction still spreads the beam out. I did say "large" flat mirrors, and there was a reason why I said it.
I was assuming something like 1% loss so, with about 30 reflections a second the decay time would be of the order of a few seconds.
If you have deep pockets you can get to 99.999% reflectivity over a narrow bandwidth..
http://www.crd-optics.com/crd-mirrors-vis.html
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I too thought such efficient mirrors were available but could not find a supplier, it would be interesting to set up an experimental system in one of the long tubes used for gravity wave research a possible IG noble project
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Thanks all for your contributions. Of course I neglected to say that all internal surfaces are perfectly reflective, but I think most of you realised my intent.
As to the calculation of how long it would take for the box to empty of light via the one-way hole (again, of course, a hypothetical construct), I'm afraid the responses assumed a level of knowledge I don't have. Does anyone have a simple explanation? Or perhaps an explanation of why a simple explanation of the solution isn't possible?
With perfectly reflecting walls, if you neglect the radiation emitted thermically by the container and you neglect the thermal radiation outside the container, the solution is simple: the time needed to empty the container is infinite.
You can see it from the equation I wrote:
E = E0e-c*s*t/V
Solving from t you get:
t = (V/c*s) log(E0/E)
and if you put E = 0, you get t = oo.
But as you saw from the simple computation I made in my previous thread, in a very short time (of the order of microseconds) the energy inside the container is almost reduced to zero. So if you intended to use that system to store energy...it's not very useful.
In a more realistic situation, essentially because of not perfectly reflecting walls, visible radiation will be lost quicker; furthermore, since the outside environments cannot be at a temperature less than 2.75 K (cmbr radiation) the container will stop losing radiation in a finite time, when it will have reached thermal equilibrium with the environment.
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lightarrow
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If the hole is very small and the "mirrors" are perfect then the light will mainly miss the hole so it will only come out slowly.
That means that most of the light will be in the box for a fairly long time. Diffraction will therefore have time to spread the light out so it is going essentially in random directions in the box..
You then have something that looks a bit like a "gas" of photons in the box - the bounce round randomly till the hit the hole and escape.
And how big are the photons comparing to the little hole? [:)]
Sorry, that "image" is not correct, at least from a didactic point of view because it suggests the idea that photons are corpuscles and that they moves in straight lines, but both things are false.
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lightrarrow
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"it suggests the idea that photons are corpuscles and that they moves in straight lines, but both things are false."
Because particle-wave duality goes both ways, the same is true of molecules.
But the calculation of effusion rates from kinetic theory (not quite the same because molecules have a temperature dependent spread of velocities, rather than all being c, but then I did say it "looks a bit like a photon gas") is pretty reliable.
The answer to the question "And how big are the photons comparing to the little hole? " is simply
"small".
We both agree that the time taken to empty the box is infinite and we both agree that it gets there exponentially.
There's not a lot of mathematical difference between a mirror that absorbs 0.1% of the radiation that hits it and one that loses 0.1% because 99.9% of it is perfect, but it has a hole covering 0.1% of its area (the analogy is better if you have a large number of small holes that add up to 0.1%)
I explicitly pointed out that you needed diffraction on-side which means that the photons don't travel in straight lines. Intermolecular collisions mean that gas particles don't either.
It is, also functionally equivalent to your calculation.
Let's compare them. I will put in some letters to represent the quantities.
I said "The decay time constant would be something of the order of the time [T] it takes a photon to bounce across the box [length L] multiplied by the area of the box [A] divided by the area of the hole ."
I talked of the time taken for light to bounce across the box- that's the distance between the sides divided by the speed of light
T= L/C
And I said the decay time constant [K] is of the order of T A/s
So K=L/C (A/s)
K=LA/CS
Where LA is the volume so
K=V/Cs
OK so the time constant is proportional to the area of the box and to the length of the side- that’s proportional to the volume; good- the consensus is that bigger boxes keep light longer
It’s inversely proportional to the area of the hole- a big hole makes the light decay faster so that makes sense
It’s also inversely proportional to the speed of light – if the box were full of something with a really high refractive index the light would bounce back and too more slowly so everything would slow down and the light output would decay more slowly.
And it’s exponential decay so it’s of the form E(t)=Eo exp-(t /K))
Where K=V/Cs
so
E(t)=Eo exp-(t /(V/Cs))
E(t)=Eo exp-(t *(Cs/v))
E(t)=Eo exp-(t*C*s/v)
You had
E = E0e-c*s*t/V
Same expression
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Thanks all for your contributions. Of course I neglected to say that all internal surfaces are perfectly reflective, but I think most of you realised my intent.
As to the calculation of how long it would take for the box to empty of light via the one-way hole (again, of course, a hypothetical construct), I'm afraid the responses assumed a level of knowledge I don't have. Does anyone have a simple explanation? Or perhaps an explanation of why a simple explanation of the solution isn't possible?
With perfectly reflecting walls, if you neglect the radiation emitted thermically by the container and you neglect the thermal radiation outside the container, the solution is simple: the time needed to empty the container is infinite.
You can see it from the equation I wrote:
E = E0e-c*s*t/V
Solving from t you get:
t = (V/c*s) log(E0/E)
and if you put E = 0, you get t = oo.
But as you saw from the simple computation I made in my previous thread, in a very short time (of the order of microseconds) the energy inside the container is almost reduced to zero. So if you intended to use that system to store energy...it's not very useful.
In a more realistic situation, essentially because of not perfectly reflecting walls, visible radiation will be lost quicker; furthermore, since the outside environments cannot be at a temperature less than 2.75 K (cmbr radiation) the container will stop losing radiation in a finite time, when it will have reached thermal equilibrium with the environment.
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lightarrow
The time is only infinite if photons are infinitely divisible, which they are not. One only has to solve the equation for the time required to get to less than 1 photon to find the answer (assuming a long list of things you have already pointed out...) One thing I do not see in this equation, is a reference to the size of the hole, or the surface area of the box--these will certainly have an effect on the rate of photons leaving.
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Thanks all for your contributions. Of course I neglected to say that all internal surfaces are perfectly reflective, but I think most of you realised my intent.
As to the calculation of how long it would take for the box to empty of light via the one-way hole (again, of course, a hypothetical construct), I'm afraid the responses assumed a level of knowledge I don't have. Does anyone have a simple explanation? Or perhaps an explanation of why a simple explanation of the solution isn't possible?
With perfectly reflecting walls, if you neglect the radiation emitted thermically by the container and you neglect the thermal radiation outside the container, the solution is simple: the time needed to empty the container is infinite.
You can see it from the equation I wrote:
E = E0e-c*s*t/V
Solving from t you get:
t = (V/c*s) log(E0/E)
and if you put E = 0, you get t = oo.
But as you saw from the simple computation I made in my previous thread, in a very short time (of the order of microseconds) the energy inside the container is almost reduced to zero. So if you intended to use that system to store energy...it's not very useful.
In a more realistic situation, essentially because of not perfectly reflecting walls, visible radiation will be lost quicker; furthermore, since the outside environments cannot be at a temperature less than 2.75 K (cmbr radiation) the container will stop losing radiation in a finite time, when it will have reached thermal equilibrium with the environment.
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lightarrow
The time is only infinite if photons are infinitely divisible, which they are not. One only has to solve the equation for the time required to get to less than 1 photon to find the answer (assuming a long list of things you have already pointed out...) One thing I do not see in this equation, is a reference to the size of the hole, or the surface area of the box--these will certainly have an effect on the rate of photons leaving.
I don't understand why you don't see a reference to the size of the hole- it's "s".
The area of the box is a bit more obscure but it's in there. It has been combined with the length of the box (which is also important) to get a volume.
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I don't understand why you don't see a reference to the size of the hole- it's "s".
The area of the box is a bit more obscure but it's in there. It has been combined with the length of the box (which is also important) to get a volume.
My mistake then. It would help though, to have a definition of terms to make it easier to interpret these equations.
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Something like this?
"... the energy escaping the hole in the unit time at a certain instant of time t is proportional to the energy density inside the box at that instant t, to the area s of the hole's surface and to c:"
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Something like this?
"... the energy escaping the hole in the unit time at a certain instant of time t is proportional to the energy density inside the box at that instant t, to the area s of the hole's surface and to c:"
precisely!
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Now all I need to do is perfect my time-travel device...
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Something like this?
"... the energy escaping the hole in the unit time at a certain instant of time t is proportional to the energy density inside the box at that instant t, to the area s of the hole's surface and to c:"
precisely!
What is not clear to me is if you realized that Bored Chemist here quoted my first post of this thread.
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lightarrow
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The time is only infinite if photons are infinitely divisible, which they are not. One only has to solve the equation for the time required to get to less than 1 photon to find the answer (assuming a long list of things you have already pointed out...)
This is true, my was just a classical description. But what you say is not "immediate" as it seems. I'd like to see how you would perform the computation.
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lightarrow
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Something like this?
"... the energy escaping the hole in the unit time at a certain instant of time t is proportional to the energy density inside the box at that instant t, to the area s of the hole's surface and to c:"
precisely!
What is not clear to me is if you realized that Bored Chemist here quoted my first post of this thread.
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lightarrow
Sorry, I guess this is the problem with trying to keep up with posts that are so far apart--by the time I read the last post on your equation, I had forgotten the one with the definitions (posted 9 days later, and 20 posts down). My bad! [:I]
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I strongly suspect that Lightarrow's expression is incorrect since it doesn't depend on the geometry of the box, only the volume.
It's complicated.
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I strongly suspect that Lightarrow's expression is incorrect since it doesn't depend on the geometry of the box, only the volume.
It's complicated.
And why should depend on geometry, given the assumptions I made (e.g. that the energy is redistributed instantly inside the box) and the macroscopic dimensions of the box I used in my example? Do you relate to the cavity geometry when you derive the black body spectrum of emission? Remember that we are talking of light here, that is em radiation with a wavelength enormously shorter than the box dimensions. Furthermore the OP didn't say that a thin ray of light enters the box or something like that, he generically talked of "light" out of the box which entered through a lid, so through a macroscopic aperture. In this conditions the radiation fills the box homogeneously and in the end it's not geometrical optics that you have to use, but em field description.
Anyway, as I wrote when I derived my expression, it was just an approximate computation because of the many assumptions made (implicitly or not).
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lightarrow
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Sorry, I guess this is the problem with trying to keep up with posts that are so far apart--by the time I read the last post on your equation, I had forgotten the one with the definitions (posted 9 days later, and 20 posts down). My bad! [:I]
No problem, sometimes I make the same mistake [:)].
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lightarrow
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I strongly suspect that Lightarrow's expression is incorrect since it doesn't depend on the geometry of the box, only the volume.
It's complicated.
And why should depend on geometry, given the assumptions I made (e.g. that the energy is redistributed instantly inside the box) and the macroscopic dimensions of the box I used in my example? Do you relate to the cavity geometry when you derive the black body spectrum of emission? Remember that we are talking of light here, that is em radiation with a wavelength enormously shorter than the box dimensions. Furthermore the OP didn't say that a thin ray of light enters the box or something like that, he generically talked of "light" out of the box which entered through a lid, so through a macroscopic aperture. In this conditions the radiation fills the box homogeneously and in the end it's not geometrical optics that you have to use, but em field description.
Anyway, as I wrote when I derived my expression, it was just an approximate computation because of the many assumptions made (implicitly or not).
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lightarrow
Granted, you wrote "Assuming ... that energy is redistributed instantly inside the box and ... ".
But I don't think that assumption is valid.
In particular if the box is very long and thin (as CRDS systems generally are) the decay time along the length will be a lot longer than the decay time across the width.
Explicitly including the length and area allows for that factor.
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I strongly suspect that Lightarrow's expression is incorrect since it doesn't depend on the geometry of the box, only the volume.
It's complicated.
And why should depend on geometry, given the assumptions I made (e.g. that the energy is redistributed instantly inside the box) and the macroscopic dimensions of the box I used in my example? Do you relate to the cavity geometry when you derive the black body spectrum of emission? Remember that we are talking of light here, that is em radiation with a wavelength enormously shorter than the box dimensions. Furthermore the OP didn't say that a thin ray of light enters the box or something like that, he generically talked of "light" out of the box which entered through a lid, so through a macroscopic aperture. In this conditions the radiation fills the box homogeneously and in the end it's not geometrical optics that you have to use, but em field description.
Anyway, as I wrote when I derived my expression, it was just an approximate computation because of the many assumptions made (implicitly or not).
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lightarrow
Granted, you wrote "Assuming ... that energy is redistributed instantly inside the box and ... ".
But I don't think that assumption is valid.
In particular if the box is very long and thin (as CRDS systems generally are) the decay time along the length will be a lot longer than the decay time across the width.
Explicitly including the length and area allows for that factor.
Granted, that assumption is wrong in the case you write [:)]
When I made the example of box, I had in mind something compact, like a sphere or a cubic box, but I forgot to write it explicitly.
Thanks for correction.
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lightarrow