OK, not really, but my question is this: If you had a sphere that had an interior of perfectly polished reflective surface and had a vacuum - could you bounce a single photon around in it forever? Just a question, Michael
Any closed volume, say a cube would have same answer, which I think is not forever, but a very long time. (we are assuming surface reflections with never any absorbtion, of course.) I have one, for sure and one possible reason why not forever: (1) Even perfect vacuum occasionally has electron and positron briefly appear. The photon could scatter off one and lose a little energy to its recoil and repeat again. etc. (2) There would be thermal photons inside the sphere, a black body spectrum with temperature of the walls. To first, and possibly some higher orders, they all just pass thru each other with no transfer of energy, but I think there are some higher order interactions possible that could degrade the phonon, which I assume you were thinking of is in the general optical region, down into the black body spectrum range eventually. I am assuming that the walls in that Black body range do not have perfect reflectivity.
To expand on this question... I've always wondered the following... and I am too lazy to work out the mathematics to see what would happen. In said sphere, assume that the photon could reflect forever. Does these exist a point and direction of travel within that sphere such that a given photon could cover the entire interior space of this sphere given "enough time?" I know of at least one scenario in which this would not happen...
It is impossible to make a perfect sphere in this reality so energy would be lost. Rather quickly, I'd say.
I thought that all surfaces absorbed some of the energy from photons reflected by the surface. Perhaps it would be better to say that when many photons are directed toward a reflective surface, some are absorbed by the surface, adding to its energy content. This would mean that for a single photon there is some probability of its being reflected & some probability of its being absorbed by the reflective surface. If the above is a valid description of reality, then an individual photon inside a reflective container would reflect some finite number of times before being absorbed. I doubt that it would continue reflecting for more than a few seconds, if that long. Consider how many times per second it would be reflected inside a sphere with a diameter of one foot or one meter. The velocity of light is approximately 186,000 thousand miles per second (about 300,000 kilometers per second). Astronauts placed some reflectors on the moon 50 or so years ago. They are shaped like the corner of a cube with the inside facing toward the earth. A laser pointed at such a reflector results in beam directed toward the laser. These reflectors enable us to very accurately calculate the Earth/moon distance. I wonder if any articles describing those lunar reflectors mention a decrease in the intensity of the return laser beam due to photons being absorbed rather than reflected. BTW: Interior designers sometimes mount 3 mirrors in the corner of a room (usually a bedroom). No matter where you are in the room, you see your reflection in that corner.
The trouble with scenarios such as this is that if we say: A perfectly spherical mirror with perfect reflection and a loss-less medium within, then yes, the photon would bounce indefinitely. However, even on the quantum level, reality kicks in, and there will be losses, shortening the life-time of the bouncing photon. Hans
Agreed. Classically it works. Quantum mechanically, the photon has some probability to tunnel out of the sphere. To Absane's question, I think the answer is that (again, classically) it depends on the initial conditions. consider the case of shooting a photon from the north pole directly at the south pole. Then, I would expect that the photon just bounce between the poles indefinitely. Given a generic initial condition, though, I would expect the entire inside of the sphere to be covered.
But we are being idealist here with the r = 1 assumption. In QM terms, we say the "walls" of the QM well are infinitely tall (or sphere is infinitely thick walled). With everwhere slight diviation from the sperical surface, then the Ergotic theorm makes the photo pass thru every point within the sphere eventually, but if reflecting surface is a perfect sphere then there are still and infite number of different trjactories that avoid passing thru most of the internal volume. To name just one tiny, still infinte, set: The trajectory is an n-sided regular poligon. The more interest ones (probably* a larger infinity, but not sure of that) are the "Lissajue patterns." (sorry about my poor spelling). I think my post 2 reply more correctly tells why the photon dies, eventually, even with r = 1 exactly and a perfectly spherical surface. ---- *I guess "probably" because if we only consider reflections in a plane containing the diameter, just to illustate my thoughts, the n-side poligons are the size of the infinite integers. Let's consider the n= 6 poligon (hexagon) which touches point "a" on the sphere in the plane we are limiting illustration to, and then ask how many of the Lissajue patterns trajectories also have six wall reflections and also touch point "a" before repeating their six wall contact path? - I think the answer is more than one as the angle at the wall need not be 60 degress but several others rotating Lissajue patterns with different number of "self crossing" inside the sphere exist, I think. (If not true of n =6 then try n = 6000. Please Register or Log in to view the hidden image! ) Thus they cannot, I think, be placed into a one to one correspodence with the intergers.
I would imagine that the photon would bounce around in only one plane in the sphere. The plane defined by the center and the points x0 and x1, the first two points on the sphere edge that the photon hits.
So simplify the problem to a circle. Can you shoot a point-mass into the circumference at such an angle that it is never reflected back to a point it has already visited? Would an angle with irrational trigonometric functions be sufficient?
I think all wall points in that plane will be visited, if trajectory is not exactly repeating, but many interior points will not be. If for example, the path is approximately an equilateral triangle, i.e. the "triangle"does not exactly close (not three 60 degree angles but, for example 60.0001 degrees) will slowly rotate inside the sphere, but the smaller circle inscribed inside a 60 degee triangle will never be penetrated if the reflection angle is greater than 60. When reflection angle is only 59, then some smaler circle will never be visited. etc. When only 1 degree, almost all the plane (except for tiny circle around the diameter) will be visited.
Ahh ok. I had worried about this, and I guess you're right. In hindsight, it makes sense (as things always do). Giving the photon a specific momentum will specify a single pattern, in a single plane, as Human001 said.
This is a fairly straightforward problem: wlog assume the particle is fired from a point on the boundary of the sphere, then the angle made between the initial trajectory and the tangent plane to the sphere at the starting point completely determines the dynamics. I leave you all to distinguish that cases in which the "bounces" occur on a dense subset of S^1, and when not.
I have another rather simple equation that I should probably already know, but, the photon does move doesn't it? And this movement requires no energy? I mean suppose we shot a photon in a straight line, it goes forever in a true vacuum (assuming no electrons pop into existence). It just "seems" like it should require some energy to "move"? The photon travel infinitely? Also, is it possible for photons to "orbit" a black hole and not be sucked into it?
That orbit exist, but is not stable. I.e. either the photon would as you say "get sucked into it" or escape. On photons needing help to move,: no. they follow Newton's first law (body at rest... and body in motion continues in motion unless acted upon by an out side force.) In the photon case, I think the only "out side force" possible is gravity,* but it does not change their speed. If falling down the gravitational hill, they just blue shift in frequency. -------------- *But as I understand the modern POV, gravity is a distortion of space, not a force, so perhaps no force ever acts on photons. I.e. they are "born" traveling at C and since no force ever acts on them, alway traveal at C. Yes I know they go slower in glasses etc. but that is because they make bound electrons "wiggle" at their frequency. These wiggling electrons radiated at that same frequency. The sum of the original photon, traveling at C, plus the out of phase elecron radiation, also traveling at C, is wave with peak traveling at less than C in the glass. I.e. one can support the POV that the photon itself always travels at C, even in glass - but certainly the opposite POV can be suported also.
If you google 'whispering gallery mode' you will find something like this sphere.. except its real and losses are quite a bit greater, because the photon does not travel in vacuum but in glass. The glass bead is close in shape to a sphere, but because its outside contact is air, total internal reflection takes place where the refractive index changes. If the angle of the incoming light is small enough with respect to the inside of the bead total internal reflection takes all along a certain mode on the inside of the sphere. This way photons can make many hundreds of thousands circles in the bead, but they do eventually get absorbed (in a matter of microseconds I belive). This is still very useful for some applications where absorption events are very small, for instance in non-linear optics. This way one can get transformation efficiencies several orders greater than in most resonators.
As that thread is dead for couple of years, I will not revive it but just note that the equations which describe light and have time reversal symmetry. I.e. any window or valve that a particular wavelenth of light can go thru (side A to side B) is also open at that same time for light of same wavelength traveling in the opposite direction (B to A).
he must have used sphere due to its lack of corners, which may have an effect. for a sphere i guess at some point it would go back and forth in a straight line.
