# Where does the energy lost due to light emission come from?

If an object lose energy when it emits light, does that energy come from the mass of the object?

Examples:
Does a flashlight lose mass when it's turned on?
Does a regular object (like a pencil or a piece of paper) lose mass when the environment is colder than the object?

If there is a loss of mass, what specific part of the atomic structure does that mass come from?

Does it come from electrons (or other particles) losing mass? (in which case it would lead to some electrons having more mass than others?)

Or does it come from whole particles being converted to energy?

Or does it come from the kinetic energy of its molecules?

• "If an object lose mass when it emits light, what specific part of the atomic structure does that mass come from?"...why would it? – ACuriousMind Feb 18 '16 at 1:21
• @ACuriousMind, why would it lose mass? It's my understanding that the light being radiated by an object must come from its mass. Does it not? – GetFree Feb 18 '16 at 1:24
• No, it doesn't. There are various processes for the emission of light (e.g. a light bulb gets the light from just heating it until the thermal radiation is visible, lasers use stimulated emission that's based on the energy difference between certain excitation levels,...), and most of them don't do anything to the mass of the object in question. – ACuriousMind Feb 18 '16 at 1:30
• @ACuriousMind, what about regular objects which are not being supplied with external energy? Like a pencil or a piece of paper. – GetFree Feb 18 '16 at 1:34
• How many light-emitting pencils have you seen? – ACuriousMind Feb 18 '16 at 1:36

You are being misled by some of these answers and responses. It is perfectly fine to think of the internal energy of the flashlight as real mass. Of course the percent difference in mass between a charged and uncharged battery is tiny but in principle that mass works like any other.

If you want a rough classical picture of where this mass is located think about the energy density of the fields. There is more energy in the field when a positive and negative charge are further away, and when they come closer some of this energy moves outward as an electromagnetic wave.

• Is that what happens when electrons jump into a lower orbit? They emit a photon as a consequence of that leap? – GetFree Feb 20 '16 at 19:26
• Yes. Of course an atom is not stable unless we treat it quantum mechanically. But just like the classical picture the extra energy of the higher orbits is due to the electromagnetic field, and the photons emitted are oscillations in the electromagnetic field. – octonion Feb 21 '16 at 16:43
• If we have two atoms of the same element, but one of them has an electron in a higher orbit, would the gravitational field of each atom be exactly the same? or would one of them be stronger? – GetFree Feb 22 '16 at 5:47
• The atom in the higher orbit would have a stronger gravitational field, although we are talking about a very weak effect that can't be measured. The electromagnetic field is a source for gravity in general relativity. What could be measured is that the atom in a higher orbit has more inertial mass. It takes more energy to accelerate it to a given velocity. This is standard in particle physics. – octonion Feb 22 '16 at 18:18
• To be fair when I say could be measured, I still don't think anyone has done that for an atom. The fractional difference in mass for hydrogen between the ground state and a high orbital is only around $1\times 10^-8$. But the difference is treated as a real rest mass difference in theoretical calculations. And differences in rest mass for excited "atoms" of quarks are observed. – octonion Feb 22 '16 at 18:39

When an object radiates heat or energy (without supply from external source), it goes into a lower energy state. There are different way for that to happen. I will take one example - the pencil one.

Think the reverse first - what happens if temperature of pencil is raised say by 10 degrees. The molecules in the pencil vibrate faster and gain kinetic energy.

Same way, when temperature of pencil goes down by 10 degrees, it looses the kinetic energy of its molecules.

If you cool the pencil to close to 0 K, then in addition to kinetic energy of molecules, there may be other changes taking place in its structure, or orbit of electrons themselves. I am not sure, but I guess you get the idea.

A flash light looses energy because its battery chemical goes into a lower energy state by loosing energy.

The loss of energy can be considered a loss of mass per E = M * C * C. But it is not like a part of electron (or something else) break into energy.

Some nuclear decay kind of processes may/do actually emit mass particles though in addition to radiation.

There is no general answer to where the energy for light emission comes from. Where the energy comes from ultimately depends on which physical process is used to produce the light. Blackbody radiation stems from thermal energy. A diode converts electrical energy into light. Electron-positron annihilation would indeed essentially draw from mass. And so on.

• I see. I remember long time ago I heard somebody explaining Einstein's E=m*c^2 as an actual mass loss when an object emits light. – GetFree Feb 18 '16 at 2:07
• @GetFree: Interpreting the mass-equivalent of energy as actual mass is a misleading concept. It's good to keep in mind that all energy has a mass-equivalent per $E=mc^2$, and that e.g. objects with more energy will be heavier because gravity acts on them more strongly (a full battery is very, very slightly heavier than an empty one), but when we talk about the mass of an object, we usually mean its rest mass.. – ACuriousMind Feb 18 '16 at 2:09
• You lost me on that last statement. You say that a full battery is slightly heavier but that its "rest mass" doesn't change? In what way the full battery is heavier then? – GetFree Feb 18 '16 at 2:13
• @GetFree: It's just word games: If you think a "full battery" is a different system than an "empty battery", then you would assign both systems different rest mass (precisely by $m=E/c^2$). However, outside of nuclear particles fusing and perhaps molecular bonds, we usually don't consider such types of energy to contribute to the "rest mass" - no one would say a rotating rod has more mass than the same rod lying still despite the rotating rod having more energy. A full battery is like the rotating rod - it has more energy, but it is somewhat ridiculous to say it has more mass. – ACuriousMind Feb 18 '16 at 2:25
• @GetFree if you would/could measure extremely accurately the gravitational force on some system, then it would reflect its total energy, not the total rest mass of all included particles. A hotter diamond crystal is slightly heavier than a cooler diamond crystal with the same number of atoms; a molecule of oxygen has slightly different weight than two separate atoms of oxygen, when Uranium238 decays, the resulting Th234+He have the same number of neutrons/protons but slightly lower rest mass than the U238 had, etc. – Peteris Feb 18 '16 at 2:26

Optical emission comes from (a) internal energy of the system, such as electronic excitation, or (b) external energy which stimulates the system, providing the energy which will later be emitted.

If the light comes from an otherwise closed system, and it emits total energy E, then yes, the mass of that system is reduced by $m=E/c^2$.