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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?

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?

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?

......there will always be light in the container for every thing is in the constant act of expanding and contracting!

Presumably one with a Q factor of 1000 would have the stored energy reduced to 1/e in one millisecond and pro rata.

.....will the inside of the box stay forever just as light as it was before I closed the lid?

Light is a form of motion, you must have something to move to have motion.

...every thing is in the constant act of expanding and contracting!

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.

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?

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?

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?

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.

Quote from: Harry Audus on 18/07/2015 11:23:12Thanks 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 = E_{0}e^{-c*s*t/V}Solving from t you get: t = (V/c*s) log(E_{0}/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. --lightarrow

Quote from: lightarrow on 23/07/2015 18:47:59Quote from: Harry Audus on 18/07/2015 11:23:12Thanks 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 = E_{0}e^{-c*s*t/V}Solving from t you get: t = (V/c*s) log(E_{0}/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. --lightarrowThe 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.

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:"

Quote from: Bored chemist on 24/07/2015 18:35:13Something 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!

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...)

Quote from: chiralSPO on 24/07/2015 19:13:29Quote from: Bored chemist on 24/07/2015 18:35:13Something 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.--lightarrow

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.

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]

Quote from: Bored chemist on 25/07/2015 14:16:27I 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).--lightarrow

Quote from: lightarrow on 25/07/2015 21:56:30Quote from: Bored chemist on 25/07/2015 14:16:27I 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).--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.