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I understand that mass-energy equivalence is often misinterpreted as saying that mass can be converted into energy and vice versa. The reality is that energy is always manifested as mass in some form, but I struggle with some cases:

Understood Nuclear Decay Example

In the case of a simple nuclear reaction, for instance, the total system mass remains the same since the mass deficit (in rest masses) is accounted for in the greater relativistic masses of the products per $E=\Delta m c^2$. When a neutron decays and you are left with a fast proton and a relativistic electron. If you could weigh those two without slowing them down, you would find it weighed as much as the original neutron.

Light in a Box

This becomes more difficult for me when moving to massless particles like photons. Photons can transmit energy from one heavy particle to another. When a photon is absorbed the relativistic mass (not the rest mass) of the (previously stationary) particle that absorbs it increases. But if my understanding is correct, the energy must still be manifested as mass somehow while the photon is in-flight, in spite of the fact that the photon does not have mass.

So let's consider a box with the interior entirely lined with perfect mirrors. I have the tare weight of the box with no photons in it. When photons are present the box has an additional quantifiable amount of energy (quantified below) due to the in-flight photons. Say there are $N$ photons... obviously assume $N$ is large.

$$\Delta m = \frac{ E }{ c^2 } = \frac{ N h }{ \lambda c}$$

Interactions are limited to reflections with the wall, which manifest as a constant pressure on the walls. If I hold this box in a constant gravitational field (like the surface of Earth) then there will be a gradient in the pressure that pushes down slightly. Is this correct? Wouldn't there still technically be mass as the photons are in-flight, which would cause its own gravitational field just as all matter does? How is this all consistent with the assertion that photons are massless? Is it really correct to say that photons don't have mass? It seems to be a big stretch.

Please offer a more complete and physically accurate account of this mirror-box.

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3 Answers 3

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The statement that photons are massless means that photons do not have rest mass. In particular, this means that, in units where $c=1$, the magnitude of the photon 3-momentum must be equal to the total energy of the photons, rather than the standard relationship where $m^{2} = E^{2}-p^{2}$.

But, you can create multi-photon systems where the net momentum is zero, since momentum adds as a vector. When you do this, however, since the energy of a non-bound state is always non-negative, the energies just add. So, this system looks just like the rest frame of a massive particle, which has energy associated with its mass and nothing else.

The statement about gravity is a little bit more subtle, but all photon states will interact with the gravitational field, thanks to the positive results of the light-bending observations that have been made over the past century. So you don't even need a construction like this to get photons "falling" in a gravitational field.

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Would it then be correct to say that the relativistic mass, not the rest mass, of a photon is $m=\frac{h}{\lambda c}$? Then all statements I would make about the mirror box could be done using this. I think that $(m_0 c^2)^2 = E^2 - (p c)^2$ (this might be what you had in mind) still holds, however, since $m_0=0$. –  Alan Rominger May 31 '11 at 14:46
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@Zassoundtsukushi: YOu can do that, but it's somewhat conceptually complicated to use the term "relativistic mass" eventually--the "relativistic mass" is really just the energy, and has properites closer to an allergy. –  Jerry Schirmer May 31 '11 at 15:00
    
Yes, to the extent that your unit system allows it (like $\frac{MeV}{c^2}$ for mass) I agree that the energy of the photons can be looked at as "just the energy". My question is sufficiently answered by nothing that this photon relativistic mass or energy (which one you use is semantics) exhibits all of the properties expected of that quantity of mass. This means that it is affected by gravitational fields and warps space-time itself. This is more than I previously felt comfortable claiming, but the conceptual picture here seems to be consistent. –  Alan Rominger May 31 '11 at 15:45
    
@Zassoundsukushi: the reason why it won't work out is technical--radiation gravitates, but not in the same way as matter, and rest mass and relativistic energies don't add together in the same way when you're creating systems out of many particles. At a first glance, you're ok using them semi-interchangeably, but realize that it's complication-prone if you plan on going deeper into this stuff. –  Jerry Schirmer May 31 '11 at 16:28
    
@Jerry "not in the same way as matter" -> how so? Gravitation couples to stress energy tensor which doesn't really differ all that much between massless and massive systems... –  Marek May 31 '11 at 19:04

I think that there is some confusion in your understanding of relativistic physics in the statement here:

In the case of a simple nuclear reaction, for instance, the total system mass remains the same since the mass deficit (in rest masses) is accounted for in the greater relativistic masses of the products per E=Δmc2. When a neutron decays and you are left with a fast proton and a relativistic electron. If you could weigh those two without slowing them down, you would find it weighed as much as the original neutron.

The correct statement is that the summed four vector of all the decay products would have the effective mass of a neutron.

Masses are not conserved in relativistic physics, in an analogous way that lengths are not conserved when adding vectors in three dimensions. What is conserved is energy and momentum, a four vector whose measure is m*c, where m is the effective mass of the system, similar to the length of a three-vector after the addition of three-vectors .

When a pi0 goes into two photons, it is true that the available energy for each gamma in the centre of mass system of the pio is half the mass of the pio, and the measure of the invariant mass of those two gammas will be the mass of the pio. When more particles are involved , lets say two pio's then the invariant mass of the four gammas four momenta is not additive to two pio masses. It is better to forget about convoluted arguments with masses and think of four momenta when in the relativistic regime.

Now a box of photons will have a four momentum sum in measure equal to E*2/c*2 - |p|*2=m*2*c**2. (please see the wiki link for clear terminology)

If the three vector momentum sums up to zero, the effective mass of the photons in the box will be E/c**2. Small but there to be weighed.

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There is no confusion, these statements are unambiguously 100% correct, and are the reason physicists used relativistic mass as a concept. –  Ron Maimon Jul 4 '12 at 5:50

There are no confusions in your understanding, everything you said is correct, and it is the nontrivial content of Einstein's E=mc^2 paper. These systems are the reason that "relativistic mass", as introduced by Tolman is pedagogically useful. The concept that we call "mass" in our day-to-day life is the energy of a system (in mass units), and when you only use the word mass to mean "rest-mass", the intuitive concept is changed somewhat.

For the atomic fission, the fast moving fragments have energy which is equal to the initial bomb energy. If you weigh them without slowing them down (for example, if they are charged and you capture them by making them do circles in a magnetic field), the weight you would field on a scale once they are captured would increase by the relativistic mass (the energy over c^2).

The photons in a spherical mirror box weigh the box down exactly as the relativistic mass of the photons inside. The pull of the Earth on these photons is on the relativistic mass. If you replaced the photons with a particle gas at the same pressure, and removed a little mass from the walls to make the total mass be positive, the gravitatonal field outside will be the same, this isn't true replacing the photons with a pressureless block with a weight equal to their relativistic mass, only because the pressure contributes to the gravitational field too.

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protected by Qmechanic Mar 14 at 6:51

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