In a perfectly dark room, supposed that you have a completely sealed
box with all internal sides mirrored. There is a light source inside
the box and the light is on. No light can be seen escaping from the
box in the dark room. Now suppose the light in the box is turned off
and the box is opened in the dark room. Why isn't light still visible
when the sealed box is opened up? Where did the photons go? If they
left the box they would have been seen in the dark room?  Were they
all absorbed by the material in the sides of the box? If so, how is
that possible, since when light hits a reflective surface, at least
some of the photons are relected back. It seems like there should be a
few residual photons left when the box is opened.

What a good question. Thanks for asking!

Photons exhibit the behaviors of both particles and waves. Only the
wave behaviors are observable. In your example, the particle behavior
of a photon is focus of your queston. As soon as a photon is reflected
(off the inside side of a box), it becomes an electron. Electrons are
invisible. Once the photon source is removed, the life cycle of the
remaining photons lasts only until they are first reflected. As a
practical matter, since light (or a photon) travels at a speed of
299,792,458 m/s, all the photons would be long gone before the box was
opened.

Mine is a very simplified explanation. I'd recommend an excellent,
illustrated tutorial from the ACEPT W3 Group,  Department of Physics
and Astronomy, Arizona State University, Tempe.

Patterns in Nature - Light and Optics
Life Cycle of a Photon
URL: http://acept.la.asu.edu/beta/test/rdg/photon/photon.shtml

For a more technical mathematical dicussion of photon behavior, try:

Wave Equations - Michael Fowler, University of Virginia
http://www.phys.virginia.edu/CLASSES/252/wave_equations.html

search strategy:
Personal Bookmark
Google Search Terms: photon behavior

If you'd like additional clarification, please feel free to ask. 

~larre-ga

---

Request for Answer Clarification by lexluthor-ga on 19 Jun 2002 22:24 PDT

Thanks for your attempt at an answer but something doesn't add up. In
reading Richard Feynman's book QED( Quantum Electrodynamics) reference
is regularly made to the statistical distribution of photons
reflecting off a glass surface. So clearly, photons do not all change
to electrons upon reflection. If they did there would not be any
visible reflection at all.  I suspect the answer lies in the speed of
the photons and the numbers of reflections that occur in an instant of
time and the fact that only some photons are reflected each time. But
I am looking for confirmation of that or some other factor that may
come into play.

---

Clarification of Answer by larre-ga on 19 Jun 2002 22:35 PDT

There seems to be a bit of difficulty reaching the tutorial directly.
Try the link below, and click on the first search result, Life Cycle
of a Photon.

://www.google.com/search?client=googlet&q=life%20cycle%20photon

~L

---

Clarification of Answer by larre-ga on 19 Jun 2002 22:57 PDT

Before I dig deeper, let me make sure I understand exactly what you're
looking for. You want references which depart from the wave-particle
duality theories and rely upon quantum mechanical explanations of
photon absorption/reflection behaviors?

~L

---

Clarification of Answer by larre-ga on 19 Jun 2002 23:18 PDT

May I suggest discussing this question in further depth in a Physics Newsgroup? 

Usenet Physics FAQ
http://math.ucr.edu/home/baez/physics/

---

Clarification of Answer by larre-ga on 20 Jun 2002 01:15 PDT

The question, as originally stated, asked why the photons do not
persist beyond the time the light source is removed. Once the light
source is removed, the optical properties of a mirror become
irrelevant. At that point, the only factors which affect behavior of
"last" photons are the physical properties of the mirror itself,
glass, and a metallic paint, which would react in a manner identical
to a 'plain' piece of metal, relieving the photon of its energy and
adding spin, an electron.

~L

The answer suggests that all reflected photons become electrons. I
don't believe that is true. Photons reflect and one can determine the
statistical distribution of the possible reflective paths the photons
will take.

Hi larre,
I tried visiting http://acept.la.asu.edu/beta/test/rdg/photon/photon.shtml
and it's asking for a username/pass.  Do you have the login info?
Perhaps it was recently changed.

By the way, great answer :)

I do not agree with larre-ga 's answer.
If photons change into electrons the moment they reflect then how can
u reflect light with glass? Shouldnt glass according to your theory
reflect electrons and not photons?
I guess you are connfusing with the nature of metals to emit elecrtons
when struck by light. That is the photo electric effect you are
talking about and is diffrent.
The answer accordign to what i think is that Total Reflection is 
impossible even the most mirrored or silvered surface will absorb a
significant part of the light. And when the loss is multiplied by the
number fo times light reflects because of its speed, the light gets
absorbed in no time. Also if the light source is nto emitting light it
will also act as a source which absorbs light. Hope this answer is
satisfactory.
-gunsich

I'm sorry about the difficulty you're having andyt, I didn't use a
password to access either of these pages.

The following search link, which I constructed by using the title of
the article, lists the Reading as the first search result. Try direct
clicking from Google, or select the Activity result. From that, scroll
to the bottom of the page, and select the "Life Cycle of a Photon"
link.. If you continue to have difficulty with access, tell me the
complete error message and perhaps between us we can figure out the
cause.

://www.google.com/search?client=googlet&q=life%20cycle%20photon

~L

Larre and Lexluthor,

Clicking directly from the Google result worked fine here, so there
ought not be any more troubles. (Thought you might need a click in
from elsewhere to verify that it wasn't browser trouble).

I'm puzzled by the other commenter's assertion that the answer is
wrong, however.  The question and answer deal specifically with
mirrors, not plain glass.  Or perhaps it's just me.  No matter.  The
illustrated tutorial is really nifty - as is the question and its
Answer.

missy-ga

Like lexluthor-ga and gunsich-ga, I disagree that photons always
change into electrons upon reflection. If you read the "Life Cycle of
a Photon" tutorial carefully, the photon-absorbed-electron-emitted
phenomenon is the photoelectric effect, not reflection. The
photoelectric effect can only occur when the incident photon exceeds a
certain energy level (specific to the surface involved). The frequency
of visible light is too low for the photoelectric effect to occur on
almost all surfaces, at least UV light is needed.

What happens in reflection is that the incoming photon is absorbed by
delivering all its energy to an electron. The electron moves to a
higher energy level but does not physically leave the surface. The
electron quickly relaxes to its original energy state by emitting a
photon. So although it looks like a single photon is "bouncing" off
the surface, it's actually two different photons entering and leaving
the surface.

Anyway, I think that one important factor that answers lexluthor-ga's
question is the way the eye works. Light is seen when photons enter
the eye and are absorbed. Only a small fraction of the photons in the
box will hit the observer's eyes when the box is opened. The other
photons will almost instantly be absorbed by the walls and even the
air in the room. Also, because light travels so quickly, the first and
last photons to hit the eye do so at almost exactly the same time.
This extremely short flash of light might not be registered by the
eye-brain system. After all, a TV image is actually a beam of light
scanning across the screen row by row, so fast that our brains only
perceive the complete picture. My guess is that if you had some kind
of high-tech photon detector instead of an observer in this
hypothetical room, you might be able to detect a few photons leaving
the box.

gunsich is correct in pointing out that larre is confusing reflection
with the photo-electric effect. I would also agree with gunsich's
answer:

Let us assume that the walls of the box have a reflectivity of 99%
(compare: the reflectivity of the mirrored area of a CD-R disc might
be around 70% [ http://www.victoryent.com/cdrspec1.html ]) and that we
are dealing with a cubic box which is 1 meter on a side (pretty large
box). Also assume that the box was opened 1 millisecond (pretty fast)
after the light source was turned off. A beam of light travelling
parallel to any edge of the box would then hit a wall around 299,792
times (assuming reflections took a negligible amount of time) between
the time the source is turned off and the time the box is opened. So
by the time you pop the lid, the light is at (.99)^299792 of its
original intensity. Don't bother trying to calculate that, anything
you use will almost certainly just spit back 0. But to give you an
idea, lets pretend that the light hit the walls only 1,000 times. Then
the light would be at .0043% of its original intensity.

Wow.

The answer given is distorting some pretty simple physics.  There is a
lot of good stuff in some of these comments, but some of them are
paying way too much attention to an answer that is just plain wrong. 
I’m sorry to be so direct, but this is college freshman level physics.
 It is very well understood stuff.

Since I am claiming to be an authority, I guess I should state that I
have a master’s degree in physics.

For purposes of answering your question I will assume that the light
source generates light in a frequency range visible to the human eye. 
(If it were ultraviolet, for example, then you would not be able to
see it because you cannot see ultraviolet light, but that isn’t your
question.)

First, let’s take your question at face value in the simplest case. 
Let us assume when you talk about “mirrored sides” you mean the sides
are perfect mirrors.  (As has been pointed out, in practice this is
impossible, but it serves as a good starting point.)

You ask, “Why isn't light still visible when the sealed box is opened
up?  ...  It seems like there should be a few residual photons left
when the box is opened.”

The answer is that all of the photons are left.  The light will be
visible as a flash when you open the box.  The longer you leave the
light on before turning it off and opening the box, the brighter it
will be.

Now, let’s assume this is a real box and the mirrors are not 100%
reflective.  jdog-ga gives a good analysis of this, so I won’t repeat
it.  However, I will point out that the intensity of the light when
you open the box is generally not zero (as in jdog-ga’s example.) 
There will typically be at least “a few residual photons left.”  If
the light source was bright enough, the mirrors were reflective
enough, and the time between when you turned off the light and when
you opened the box was short enough, you would still see a flash. 
More to the point, in general there will be a flash.  The question is
simply one of whether it will be bright enough for you to observe. 
Potentially, there are also issues with whether or not the light will
be the same frequency as the lamp.  Your eyes might not be able to see
it.

You also ask, “Where did the photons go?”

If a mirror is not perfect, when a photon hits the wall some or all of
the photon’s energy is absorbed by the mirror as heat.  (Some or all
of its energy could also be transmitted through the mirror, but your
question seems to state as one of its assumptions that no light is
transmitted because the box is “completely sealed.”)  If all of the
photon’s energy is absorbed, the photon is gone.  (The number of
photons is not conserved.)  If part of the energy is absorbed, the
photon will have less energy when it is reflected.  That is, it will
be a different color after it is bounced off of the wall.

Photons are particles of energy.  “Where they go” is into the heat of
the box.  Heat is also energy.  Energy is conserved.  Over time, the
box will start to feel warmer when you touched it.

Warm objects radiate.  That is to say, even if the box were sealed, as
it gets warmer it will radiate energy to the room in the form of
photons.  (I will also heat up the air around it through conduction.) 
If the temperature is such that you can stand to put your hand on the
box, you will not be able to see the radiated photons because they
will be mostly infrared.  However, with a bright enough lamp you could
create a situation where the box started to glow because it was so
hot.  (It would also hurt to touch it.)

Ok, that’s my answer.  Here is a list of things in the answer that was
posted that are wrong or irrelevant to the question you asked.

--“Photons exhibit the behaviors of both particles and waves.”  --
True, but not really needed to understand this question.  Your
question is mostly one about conservation of energy.  You are already
thinking about particles.  That's fine.

--“Only the wave behaviors are observable.” -- Just plain false.  Both
behaviors are observable and observed.

--“As soon as a photon is reflected (off the inside side of a box), it
becomes an electron.”  -- Although you can produce situations where
photons become electrons, if you were to put a lamp you can buy at a
typical store into a box with really high-quality mirrors, *very few*
(if any) of the photons would ever become electrons.  Particleman-ga
describes what does happen.  An existing electron can absorb the
photon and gain energy.  (The photon then no longer exists.)

Hope this helps.

8ball-ga

PS. Missy-ga, in response to your comment, “I'm puzzled by the other
commenter's assertion that the answer is wrong, however.”  The answer
*is* wrong.

I was going to comment, but 8ball-ga has just about said it all.

There's quite a nice tutorial at
http://www.siggraph.org/education/materials/HyperGraph/illumin/reflect1.htm
which might help a bit here.

I also found a rather odd-ball analysis from the "Absolute Motion
Institute" (!!) at http://www.circlon.com/HTML/reflection.html which
I'll throw into the mix just for good measure!  :-)

Mark

Hm, maybe I'm stupid, but wouldn't there be some heat generated every
time the beam hit a mirror? And at 186,000 mph that'd add up to a LOT
of impacts. I'd bet the light energy would just be transformed to heat
after a few milliseconds, and radiate out the box like that.

googlebrain-ga

8ball, great comment. Just to clear it up, I wasn't trying to imply
that the intensity would actually be zero, just that (.99)^299792 is
so small that you would get zero if you were to try to calculate it.
But yes, lexluthor, there would most likely be some residual photons,
but not necessarily enough to see.

Particleman also makes a good point about air. Although I don't think
the air itself would absorb visible light (i think the wavelength
would have to be around 250 nm or less), water vapor and pollutants in
the air inside the box would. So even if you had perfectly reflective
mirrors, the light wouldn't survive at its original intensity.