# Photons in a Box Contribute Weight

1. Aug 1, 2010

### cbd1

It is said that if you add photons into a box and close it, the weight of the box will be increased. My question is: How?

Photons are understood to be massless, which is what allows them to travel at c. But, when photons are absorbed by a material body, they do add mass to that body. Now, when photons are in a box, but not absorbed by it, they are still lone photons, which are supposed to be massless. How then do they contribute to a net force downward by the box on the scale?

(My understanding is that photons have mass equivalence in their energy by E=mc^2 [which makes them contribute to spacetime curvature], but that they do not actually individually have inertial mass.)

The only way I can figure it is that the light in the box follows spacetime curvature, making there be statistically more radiation pressure on the bottom of the box than on the other sides -- assuming the insides of the box are 100% reflective.

2. Aug 1, 2010

### cbd1

Is it possible that the box will not actually weigh more in experiment? I suppose "possible" would have to be "yes" as the experiment cannot feasibly be done. So, the equations tell us that the photons will add weight to the box; but, in reality, they possibly could not.

Still, if they do, how?

3. Aug 1, 2010

### bcrowell

Staff Emeritus
No, gravitational and inertial mass are always equal. If they were ever unequal, the equivalence principle would be violated, and the whole structure of GR would be destroyed. When we say that photons have no mass, we're referring only to rest mass, and what we mean is that in the equation $m^2=E^2-p^2$, the constant m is zero; this simply means that a photon's momentum is equal to its energy, p=E (in units where c=1).

Yes, that's right. And the only way you can have radiation pressure is if radiation has inertia.

4. Aug 2, 2010

### humanino

To my mind, as I have said in another thread discussing this very same topic, one should be cautious of "cheating" explanations. The reason there is more pressure at the bottom of the box is because the gas is falling. The fact that there is more pressure at the bottom of the box is not an explanation why the gas is falling. It is interesting to note that consistency holds between the pressure exerted and the effect of gravitation on the gas. But it is important to keep in mind that when there is no box and no concept of pressure, a single photon escaping the gravitational field of a star will also be red-shifted. The point is that gravitation has as a universal source energy-momentum and anything carrying energy-momentum (with or without mass) falls along the 4-geodesics consistently computed from the total energy-momentum around.

Take for instance the gravitational effect on a single hydrogen atom. It will be different from the gravitational effect of a single mass equalling the sum of the masses of a free proton and a free electron, the difference is about 13.6 eV. Right ? What is the explanation in terms of a virtual photon flying around exchanged between the electron and the proton ?

I think bcrowell is aware of those caveats, and that's why he added

5. Aug 2, 2010

### cbd1

So, finally, I should be content with the understanding that photons do not possess the quality of mass, but that they do possess momentum. Is there a way that they could have momentum and not inertia? By my understanding something has to have rest mass to have inertia ?

By the means of the hydrogen atom having more weight than just the proton and the electron; what is your explanation for this? Could it be interpreted that the electron has more energy when bound than when free?

[My apologies for not taking in simply that radiation does have inertia: I am just honestly trying to understand it for myself. I thought I had this understood before, and now it is just not making sense for me.]

6. Aug 2, 2010

### humanino

The hydrogen atom is lighter than the free electron and proton separately, because there is a binding energy of 13.6 eV contributing negatively, that 13.6 eV is the energy it takes to pull out the electron from the proton "electromagnetic potential well". You can picture this energy that you need to put in the system classically, to go from the bound state to the 2 free asymptotic states "at infinity".

I mention this because my understanding is that your initial motivations are related to understanding the proton mass. The proton mass is a fairly more advanced problem, one aspect being that there is no free state of the constituents. In the case of quarks, the energy stored in the system will grow with the distance, and the quarks are permanently confined inside hadrons (also called infrared slavery, the evil brother of asymptotic freedom). So there is a quantum mechanical balance between the light quarks trying to escape out of the region in which they are trapped which is smaller than their Compton radius (Heisenberg inequality), and the growing (positive) energy of the system as they fly apart. In this case, our classical pictures are of much less help.

7. Aug 3, 2010

### starthaus

Not exactly. Photons contribute to the equivalent mass of a system via their momentum. This is a rather involved calculation that you can see https://www.physicsforums.com/blog.php?b=1928 [Broken] (second attachment)

Last edited by a moderator: May 4, 2017
8. Aug 3, 2010

### bcrowell

Staff Emeritus
They don't have *rest* mass.

To me, the definition of having inertia is the ability to exchange momentum.

No, light has inertia but no rest mass.

It has less weight, not more.

Mass and energy are equivalent, so less energy is equivalent to less mass.

No, the binding energy is not just a property of the electron, it's a property of the whole system.

9. Aug 3, 2010

### lightarrow

There is no big difference with the case of photons absorbed by a material body (black body). If a single photon has energy Eph then the box energy is increased of a quantity n*Eph so its mass has increased of a quantity n*Eph/c2, because the box stayed at rest (for objects at rest you can use the relation E = mc2).

Furthermore, the fact photons are massless shouldn't make you think that a system of more than one photon is massless: mass *is not* additive.

Consider a system of two photons travelling in opposite directions and with the same energy Eph. The system's energy is 2Eph (energy *is* additive), but the system's momentum is zero, ok?

Then you apply the relation:

E2 = (mc2)2 + (cp)2

which is valid for every object or system with total energy E, total momentum p and total mass m.

In our case we have:

(2Eph)2 = (mc2)2 + 0

so: m =/= 0. A system of two equal photons travelling in opposite directions *HAS* mass.

If the two photons have not the same energy, or are not travelling in opposite directions but however travels in different directions, or both things, you can always find a frame of reference where they have the same energy and are travelling in opposite directions, so the system of two photons still have mass.

10. Aug 3, 2010

### humanino

I tried to make more or less the same analysis for cbd1, so I would like to thank you. I will repeat this experimental simple fact which I mentioned then : the neutral pion with mass 0.135 GeV/c^2 decays in two photons. We use those two photons to calibrate calorimeters in many experiments with calorimeters and neutral pions.

11. Aug 3, 2010

### Naty1

But the act of closing the lid does not account for the increase.

And if you add the same photons to a smaller box, or compress the size of the box after adding the photons, the weight will increase more....due to an increase in frequency and ultimately electron degeneracy pressure....

12. Aug 3, 2010

### cbd1

Thanks to all for filling me in on these things. This is my understanding at this time:

Photons do not have rest mass. And, because rest mass is required to give resistance in to a change in motion, i.e. inertia, the photon does not technically have inertia. This does not mean, however, that the photons do not have momentum; their momentum is equal to their energy; so they do have momentum, which is equal to their energy's mass equivalence. (By this, the momentum works in conservation of momentum for radiation pressure, even though they do not, by my description, have inertia.) And, to add, even if two photons are traveling in opposite directions and their total momentum equals 0, they will still contribute to the mass of a system.

Now, if photons are considered in a system, their mass equivalence contributes to the rest mass of the system, even though they do not individually have any rest mass. I think this (besides, perhaps the interpretation of "inertia") agrees with what has been said by the more experienced contributors -- I, of course, could be misunderstanding yet again.

As far as the other topic, of the mass of a H atom, the system has less mass than the individual particles. And my next question: Assuming the proton has no change in its mass equivalence when the electron is bound to it, *could the loss in energy (mass equivalence) of the atom be interpreted as a loss in the electron's energy? (Since this amount of energy lost is the amount that would be required to be added to the electron to free it from the system again, giving it back its total free energy, an therefore mass equivalence?) In other words, can the loss in energy of the system (atom's mass) be considered to be due to a loss in the electron's energy, which is lost due it being bound?

Last edited: Aug 3, 2010
13. Aug 3, 2010

### bcrowell

Staff Emeritus
No, rest mass is not required to give resistance to a change in motion. A photon does have inertia. For example, when a photon bounces off of a mirror, the mirror recoils from the collision.

The assumption is wrong, and what's the point in trying to figure out what's true under a false assumption? The binding energy is a property of the whole system, not of one particle or the other.

14. Aug 3, 2010

### cbd1

Could this example not be a result of the conservation of momentum?

Here is a way which shows that photons do not have inertia like other objects: When a photon enters a substance, like the atmosphere, from being in the vacuum of space, it slows from c. This, then, would have to be interpreted as a loss of inertia. Then, if the photon were to exit the atmosphere again and go back out into the vacuum of space, it will go back to traveling at c. It would seem then, that if the photon abides by inertia, it has somehow gained inertia spontaneously from nothing. This is something that objects with inertia do not do; an object with inertia would continue its path at the same speed, due to its resistance to change in motion.

With this, can you give other examples of rest mass not being required to give resistance to change of motion?

You say "the whole system" as if that makes a simple cover to say that neither the proton or the electron lose energy, but that "the system" does. Wouldn't the loss of mass when the proton and electron become bound have to be described as a loss of rest mass in both the proton and the electron, with the degree of which between them being uncertain? The following is my logic.

If you take an unbound proton and an unbound electron in a system by themselves (with only vacuous space otherwise), they have total mass x. Then, upon their binding, they have total mass < x in the H atom. Is there any other thing in the system in which the energy/mass can be lost from? Unless you somehow add negative energy to the system when they bind, you must agree that the proton and the electron cannot still have both of their full beginning masses while in the atom.

15. Aug 3, 2010

### JesseM

I think it would be valid to consider the difference in mass to be equal to the decrease in potential energy of the electron when it is moved from far away (unbound) to close to the proton (unbound). My understanding that the inertial mass of any bound system in SR (its resistance to acceleration in its own rest frame, which could be measured by putting it on a scale) is always equal to (sum of rest masses of all its constituent parts) + (sum of kinetic energies of all its constituent parts) - (amount that the potential energy of the parts is lower than the potential energy of the same parts if they were moved far apart), with kinetic and potential energies calculated in the rest frame of the bound system.

Last edited: Aug 4, 2010
16. Aug 3, 2010

### bkelly

While I cannot follow everything here, this is an interesting thread. It makes me wonder what would be the effect if the inside of the box were perfectly reflective. Put a laser in one corner and aim it so that the photons would bounce around almost forever. Turn on the laser. If you want, get the power for the laser from an external device so we don't have to worry about the energy consumption within a battery having an affect on the mass change. Now what happens to the mass of the box when the laser shines?

17. Aug 3, 2010

### humanino

But you do have to worry about it : your laser plugged in to an outside source of power pumps energy in the box. Fill your box with energy, yes the mass increases.

18. Aug 4, 2010

### lightarrow

No, photons cannot travel at speeds lower than c. The electromagnetic wave can (for what concern phase speed or group speed).

Infact it's exactly this way.

No. Consider, to make an example, a fixed region of space filled with electrostatic field. That region has energy and also has mass. So you don't need to ascribe energy or mass to particles only.

19. Aug 4, 2010

### lightarrow

Where is the difference with the question we are discussing? The box acquires mass as soon as photons enter in it, we don't have to wait for photons to be absorbed by the box' walls (or the OP wouldn't even have made the question )

See up. Consider however that the initial mass of the box shoud be much greater of the order of the laser energy/c2, otherwise the box will recoil, and its increase of mass would be lower than that value.

20. Aug 4, 2010

### bcrowell

Staff Emeritus
This is the same as the example of the light wave bouncing off the mirror. You've just substituted the atmosphere for the mirror.

It exchanged momentum with the atmosphere.

No, this is incorrect. Having "inertia" doesn't mean that an object can never change its motion.

I don't understand what you're asking here, or how it relates to the argument you made above.

No.

The binding energy is negative, and it's equivalent to a certain negative amount of mass.