Weight loss of a flashlight due to light emission I watched this video. At one point, the man explains how a flashlight loses some mass due to light emission when turned on. OK, I get that. Then he talks about a different situation, where the flashlight is inside a mirror box. I understand that this is a closed system, no light gets out, so there is no change in mass of the system. 
But there is one thing that I don't get. Suppose you put the mirror box with the flashlight on scales. In that case, after turning the flashlight on, the scales don't move. In the case of flashlight without the box, the scales move a little due to the loss of mass. So far so good. Now, suppose you have the mirror box on the scales and break one wall. It is no longer a closed system, light is emitted and the scales should move. BUT, how do the scales know what I consider to be a closed system? Do they move at a time the light leaves the box through the crack in the wall? What if the box were just larger, would the weight loss happen later? Or, if I had a box full of cracks and sent a beam of light through it, would the scales indicate weight increase at a time the light enters the box? This seems totally wrong, but I can't see a difference, from the point of view of the scales, between a cracked box with beam of light going through it and a sealed box with a beam of light moving inside. I just can't get my head around how do the scales know what to weigh...
 A: The problem here is that the intuition about fields is different than for particles. And light and particles are pretty similar in that they behave like both or either one on occasion.
Regarding the light as a stream of pretty localized photons, I would say the following: The light gives the flashlight a bit of recoil when it leaves. Say the flashlight lies flat, then this recoil might be to the right as the light leaves. In that direction it should not tip the scales at all. The flashlight then weighs a bit less and is pulled down by gravity less.
The light will just hit the mirrors on the left and right side and not contribute to the weight. At some point it might hit the bottom of the box and then contribute to the weight. When it hits the top, it will lessen the weight. This is just pressure, it does not have an effect on the scale on average.
Since the photon has an energy, gravity will attract it downwards. If gravity was super strong, then it should surely make the light go to the bottom of the box eventually. If gravity was so strong that the light would bounce around like a ping-pong ball and never hit the top again, it would contribute to the total weight as there is more pressure downward than upward.
My intuition for this is really sketchy, but I would think that the mass (not weight!) of the closed box would stay constant. Its weight on the scales would be reduced until the photon hit the bottom of the box. Then the scale would have a sudden slight increase. On average (thinking about a ping-pong ball) the light should hit the bottom more often.
You will need lots of photons such that the “photon gas” can actually diffuse down and add some weight to the bottom of the box. Thinking about single photons makes the concept of “weight” void since the scale only shows something that is averaged over a short time. In that short time the photon will have traveled a lot of distance and be reflected a lot of times.
Realistically, the weight of the photon is negligible and it would be absorbed by the imperfections on the mirror before you could measure anything.
A: I think both of the pre-existing answer violate (or at least play with the idea of violating) general relativity. General relativity is in a nuthell that gravity and accerlation cannot be distiguished. Just from that principle, we can solve this entire problem!
I think the situation is quite simple afterall. The photon will bounce up and down, but experiences a gravitational red/blue shift, and thus exerts a different radiation pressure to top and bottom of the vessel. When the light escapes, this excess radiation pressure drops and the reading in the scale will change as soon as the information travels (which is at speed of light). This information is that no photon will bounce from the scale anylonger, since it has escaped. (In other words, one should be very careful with causality violating statements like "The weight reading should drop as soon as the photon leaves the flashlight.").
For completeness, let's derive that: (The derivation ended rather lengthy and inelegant since I did not probably choose the best path, but it should be very easy to follow, since there is only basic algebra involved. There might be some careless errors though.).
So without the light on, the scale reads weight $F=m_0g$.
To make it as simple as possible, imagine a photon with a wave length equal to the height of the box L. The energy of the photon is now $E = hc / L = h / T$, where T is the time it takes for the photon to go from one wall to another. 
Now we make the box to accelerate upwards with acceleration g (remember, this is indistinguishable from gravity). The photon will reach the top of the box from the bottom of the box in time $T_{up} c = L + 0.5 g T_{up}^2$, and to the opposite direction in time
$T_{down} c = L - 0.5 g T_{down}^2$.
These quadratic equations are easily solved:
$$T_{up} = \frac{\sqrt{c^2 + 2gL} - c}{g} \\
T_{down} = \frac{c - \sqrt{c^2 - 2gL}}{g} \\$$
and for photon energies going up and down we get
$$ E_{up} = h / T_{top} = \frac{hg}{\sqrt{c^2 + 2gL} - c}\\
   E_{down} = h / T_{down} = \frac{hg}{c-\sqrt{c^2 - 2gL}}
$$
Since g is actually very small, we can expand to second order
$$ E_{up} = h / T_{top} = \frac{hg}{c \sqrt{1 + 2gL/c^2} - c}\\
   E_{down} = h / T_{down} = \frac{hg}{c-c \sqrt{1 - 2gL/c^2}}
$$
$$ E_{up} = h / T_{top} = \frac{hg}{gL/c + 1/2 (g^2L^2/c^3) }\\
   E_{down} = h / T_{down} = \frac{hg}{gL/c - 1/2 (g^2 L^2/c^3)}
$$
$$ E_{up} = h / T_{top} = \frac{hc/L}{1 + 1/2 (g L/c^2) }\\
   E_{down} = h / T_{down} = \frac{hc/L}{1 - 1/2 (g L/c^2)}
$$
$$ E_{up} \approx hc/L - 1/2 h (g/c) )\\
   E_{down} \approx hc/L + 1/2 h (g/c) \\
$$
We have just derived the gravitational red/blueshift of a photon.
Thus, the momentum difference going up and down is 
$\Delta p = \Delta E / c = h g / c^2$
and this happends roughly in $\Delta T = L/c$. Thus the net momentum transfer rate is
$F = \Delta p / \Delta t = hg/Lc$. Using from the beginning $E = hc/L \rightarrow L = hc/E$, one gets $F = \Delta p / \Delta t = hg/(hc/E)c = g E / c^2$. If we now use the famous $E=mc^2$, we actually get $F = mg$. That is, the reading of the scale due to light will be indistinguisable from rest mass of a massive partlicle.
If the flashlight points sideways, the light still bends more downwards due and similar, but probably more tedious derivation is possible.
Now, if the box is opened, the light will escape and will no longer cause radiation pressure, and hence the scale will show that the light has escaped. The scale will show a difference reading, within bounds of causality, since no information can travel faster than light. This is in perfect agreement with the weight caused by the radiation pressure argument.
A: This is one of those cases where you need to distinguish mass and weight.
Mass (equivalently energy, by the famous $E=mc^2$) is defined in two ways. There's inertial mass, which relates the acceleration of a body to the force applied ($F=ma$), and there's the gravitational mass, which scales the gravitational force exerted by a body ($F=GMm/r^2$). The equivalence principle states that these two masses are in fact identical. If you want to talk about measuring mass directly you need to either apply a known force to the system whose mass you want to measure and then measure its acceleration, or measure it's gravitational force "directly" (a good way, at least schematically, to do this would be to stick the box in a large empty piece of space, put another test particle a little way away and measure the time it takes for them to collide). With a proper measurement of the mass of the system would correctly count all the mass-energy of everything - box, flashlight, photons - in the system, provided you design the measurement appropriately for the system of interest.
The trouble with a scale is that it measures the weight of the box, which is a bit different. This is the force, presumed due to gravity, transmitted down on the scale. But there are a few issues. The force needs to be transmitted, so a photon flying freely won't register. The weight reading should drop as soon as the photon leaves the flashlight. The scale is also sensitive to other forces applied. Photons bouncing around in the box have transfer momentum - a photon reflecting off a mirror on the inside top of the box applied a small "kick" upward, which would momentarily reduce the weight reading of the box.
