# Light in a box.

1. Oct 24, 2008

### mrpeach32

Lets say you had a box who's inside walls were perfectly reflective. In addition to this, because you are just so very awesome, you are somehow able to put a lamp inside that box that doesn't interfere with and of the light rays in the box except to produce more. Now since this is a silly and impossible set up, I'm going to ask some silly and ridiculous questions about it.

1) Would the box get more massive?

2) Would the box eventually have to "pop?"

3) If it was broken open, would there be a flash of light before it dissipated?

2. Oct 24, 2008

### DaveC426913

There's really no such thing as perfectly reflective. Eventually the walls will absorb photons and heat up.

How is your lamp powered? If there's no way in for electricity, then the light will eventually stop. If it's battery-powered then you're putting all the "ingredients" for the photons in the box before you close it.

3. Oct 24, 2008

### mrpeach32

The entire idea was based out of it being completely impossible, really. I was just wondering whether it was possible to "fill" something up with light if certain conditions were met. So at the risk of sounding, well, stupid, I guess I don't really know what the lamp is powered by. Maybe this question is a bit too hypothetical for realistic consideration :-D

4. Oct 25, 2008

### pallidin

The following is from: http://en.wikipedia.org/wiki/Total_internal_reflection

"Total internal reflection is an optical phenomenon that occurs when a ray of light strikes a medium boundary at an angle larger than the critical angle with respect to the normal to the surface. If the refractive index is lower on the other side of the boundary no light can pass through, so effectively all of the light is reflected. The critical angle is the angle of incidence above which the total internal reflection occurs."

----------------------- And further in the article:

"Total internal reflections can be demonstrated using a semi-circular glass block. A "ray box" shines a narrow beam of light (a "ray") onto the glass. The semi-circular shape ensures that a ray pointing towards the centre of the flat face will hit the curved surface at a right angle, this will prevent refraction at the air/glass boundary of the curved surface. At the glass/air boundary of the flat surface, what happens will depend on the angle. Where θc is the critical angle (measured normal to the surface):

If θ < θc, as with the red ray in the above figure, the ray will split. Some of the ray will reflect off the boundary, and some will refract as it passes through.
If θ > θc, as with the blue ray, the entire ray reflects from the boundary. None passes through. This is called total internal reflection."

--------------------

Just something to offer your question.

5. Oct 25, 2008

### Phrak

1)yes
2)what does 'pop' mean?
3)yes

A laser cavity is light-in-a-box, when supplied with both mirrors fully reflective.

Last edited: Oct 25, 2008
6. Oct 25, 2008

### pallidin

It is also important to note that your type of question/scenario is one that has been and continues to be of considerable interest.

For example, at 50% efficiency, a 1 second pulse of photons contained as such and then released at one time(in-phase or not) in one direction, will effect a photonic impact pulse 90,000 times that of the initial input.

Of course, the pulse will necessarily be 90,000 times shorter in pulse-length.

Basically, this is photon density amplification, and should not be confused with a laser, though lasers are often used for the initial pulse.

7. Oct 25, 2008

### pallidin

On a related note, have you ever had one of those plastic "neon" hand-held clipboards, which emit a very bright light from its edges even in low-light surroundings? Doesn't work in total darkness of course.

This is due to internal reflectance from ambient light on the 2 flat surfaces, then exiting the edges.

Not "perfect" internal reflectance or efficiency, but quite remarkable.

I have one here as we speak.

8. Oct 25, 2008

### Nick89

1) How does it get more massive, because the energy of the photons can be seen as mass in a relativistic view or something?

2) I think by 'pop' he means that the box shatters or breaks. I would have to say yes, because photons do carry momentum. If you have a large number of photons bouncing off the walls of a box the walls will be pushed outwards by a tiny amount. If you just put in enough photons they would eventually break, depending of course on how much they can take. Note that the momentum of photons is very small and you would need a hell of alot of photons to achieve this...

9. Oct 25, 2008

### DaveC426913

I wouldn't break, it would just start radiating heat as the impacts started heating the walls. If it couldn't radiate fast enough, it would eventually break down whatever the reflective surface is made of.

10. Oct 25, 2008

### Phrak

In this way: m=F/a. Put it on a scale, it weights more. Try to push on the box, it has more inertia. And in this way: m^2=p^2 for a standing wave. (odd, that it wouldn't work for massless fermions)

11. Oct 25, 2008

### Gear300

I wouldn't have to say its completely pointless...its sort of like shifting the open end of the box so that the Sun is contained in it, then closing the open part...

12. Oct 25, 2008

### Nick89

Well ok, but I guess you could see that last thing as the box breaking no?

Weight is not the same as mass though... If you put the box on a scale and press down on it that wouldn't make the box more massive, but it would make the scale read a larger value.

Why?

I'm not trying to say it's not going to be more massive (for all I know it could), I just don't completely understand how.

13. Oct 25, 2008

### DaveC426913

Yeah, I'm not exactly convinced of this.

14. Oct 25, 2008

### Phrak

Convinced of what?

15. Oct 25, 2008

### DaveC426913

You can't apply classical Newtonian physics to photons and get sensical results.

16. Oct 25, 2008

### Phrak

By the way, I can't believe I screwed that up. p=0, so that E^2=m^2

17. Oct 26, 2008

### HooDude

Just my 2 cents, but if photons are massless particles, how can an increase in their concentration cause an increase in mass?

I would think the box would see an increase in pressure and temperature but not an increase in mass?

18. Oct 26, 2008

### Phrak

What, if not mass, is responsible for the deflection of the scale?

You have a 2 kilo, ideally reflective box, containing 10 kilo of matter and 10 kilo of antimatter. It is suspended by a ballon capable of lifting 7 kilo. Placed on a scale, the scale reads 15 kilo. You enable the contents to annihilate. The scale still reads 15 kilo.

Let's imagine for a moment that the mass of the contents vanishes to zero upon annihilation. The ballon lifts the box and an additional weight to some astonishing altitude. Some means is provided to return the box to it's initial state. It decends, and the process is repeated, where another weight is lifted.

Now we can object, and say that the annihilation process is thermodyamically irreversible and the box will not decend. This places us in the position of claiming that conservation of energy is a result of the second law; we would making the interesting claiming that energy is only statistically conserved.

Last edited: Oct 26, 2008
19. Oct 26, 2008

### Nick89

The force that the photons exert on the walls of the box.
(But now I think about it, on average that force would be zero so the weight of the box would not increase).

Anyway as I've said, weight and mass are two different things. Your weight can increase while your mass does not. The topic starter (and me) asked about the mass, not the weight.

20. Oct 26, 2008

### HooDude

Mass and weight are two different things, this is correct. Weight is relative. For example, something with a certain mass will have a a different weight on the moon than on the earth. Although the mass does not change, the weight does. In this case, weight is relative to gravity.