# QCD Mass Versus Higgs Mass

inflector
Quite frankly, I think this overrepeated statement is a conscious fraud. The Higgs (if it exists) gives less than 5% of the mass we know about, the ordinary mass around us, our inertial and gravitational masses. More than 95% comes directly from glue. This is clearly stated in Wilczek's lightness of being.

I'm trying my best to learn particle physics and cosmology for the last year or so. One of the concepts I'm trying to understand is how the 95% figure humanino quotes above is arrived at. I've seen this figure (or something close to it) a few times here.

I read the paper http://arxiv.org/abs/0906.3599" [Broken], and it indicates that they were able to compute the masses of all hadrons using Lattice QCD. It seems that they (S.Durr et al.) were able to develop a table of mass ratios and then by plugging in one experimental value, they were able to compute the masses for the other particles. This implies that physicists must understand a lot about how the glue mechanism contributes to mass if they can compute mass using QCD theory. Or am I missing something?

How do scientists know that 95% of the mass is attributable to glue while 5% is attributable to the Higgs mechanism?

What it is about the glue that gives particles mass? I've seen descriptions of how the Higgs field is supposed to do this but I haven't seen a discussion of how the glue does this, at least not in terms I could understand. Does anyone know of an accessible text or article on this?

I've also seen some posts from some of the forums better posters (from my personal observation) that say we don't really know what mass is? Is this a widely held opinion? How can we compute it if we don't know what it is?

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ansgar
We can calculate the mass of the proton, neutron with assumption that quarks are massless :) And yes, we know "a lot" about QCD, or rather, we know how to calculate things in QCD and we know that all observables we have calculated so far is in agreement with the experiments testing those observables. So what do you think? did you think scientists GUESS? No, we calculate things.

The issue is that E = m in special relativity, binding energy influences the mass of a bound system!

This is also applicable to non-quantum systems.

For instance the mass of the sun + Earth is larger than the mass of the sun + mass of the Earth due to the binding energy between the sun and the earth!

Another famous example is of course the atomic nucleus, there the binding energy is of the same order as the mass of the constituents themselves!

Finally in the proton and neutron themselves, the binding energy of the gluons is MUCH larger then the masses of the (valence)quarks! In fact, one can show and calculate (PCAC-theorem etc.) that the protons and neutrons have mass even though the quarks are massless.

But to mathematically PROVE that (calculations are not proofs) is difficult, in fact you'll get 1 000 000 \$ from claymath instutite if you can prove that the ground state in Yang Mills theory is not massless ;)

We know what math IS, but we don't know WHY the particles in the standard models HAVE mass. According to the symmetries of the Standard Model, the elementary particles should be massless! So it depends on what one MEANS with "we don't know what mass is".

This is where the Higgs field comes in, the Higgs field BREAKS the symmetry which forbids the elementary fermions and bosons in the standard model for having mass. Now they can have mass, by coupling to this higgs field.

So there are 3 ways a particle can have mass:

1) by binding energy (applies to composite particles)

2) simply by "having" a mass, it just HAS mass in the same way it has charge, spin etc. (this would be true in the REAL world if the standard model were not Chiral, i.e. distinguish between left and right handed particles).

If there is a symmetry such that the particles CAN'T have mass, then they will be massless, and - opposite - if there us a symmetry which FORBIDS the particles to have mass (as the symmetry in the standard model) then they MUST have mass - everything that is allowed can happen in quantum mechanics.

3) by "interacting" with the field which breaks the symmetry which forbids the particles to have mass (see #2 above)

Let me know if you need material to study which is suitable for your academic background.

the_house
Just a couple of quick corrections. Because of the binding energy, the mass of a nucleus is less than the sum of the masses of the individual protons and neutrons. Also, the binding energy is only around 1% of the total mass. Similarly with bound systems such as your example of the earth/sun system, at least in principle (although in contrast this is not a measurable effect.) Even though the sign is different, it's similar in that the main point is the equivalence between energy and mass.

The quick answer to the original question is that we can easily imagine a hypothetical standard model with all massless fundamental particles (quarks). If we calculate the mass of a proton or neutron in this hypothetical theory, it turns out to be around 95% of the actual mass of the real particle in the real world with massive quarks. In turn, protons and neutrons account for essentially all of the mass of the world around us (neglecting dark matter, which we don't really know much about). Thus, it can be argued as somewhat misleading to talk about the bare masses of the fundamental particles in the standard model (Higgs mechanism, etc.) as the source of mass in the universe, since even without these fundamental mass terms, there would still be plenty of mass in the (highly relativistic) bound state systems of the massless constituent particles.

ansgar
yeah, the binding energy of a proton RAISES the energy. Do this comparison then, Imagine that you have a box with walls which are 100% opaque and massless, in that box you put a lot of photons. Now put this box on a scale, it will have mass due to the energy content of the photons...

Have a look at ordinary matter. Approx 2/3 of the particles are nucleons, 1/3 are electrons. As electrons are rather light (511 keV) let's forget about them; protons and neutrons are much heavier: 938 and 940 MeV respectively.

Now we have to compare the mass ratios of the nucleons a) from the Yukawa coupling to the Higgs and b) from the strong interaction. According to the particle data group the masses of "free" u- and d-quarks (as measured in the asymptotic limit of deep inelastic scattering) are ~ 2-3 MeV and 3-6 MeV, respectively. So the contribution of the quark masses to the nucleon mass is below 2%.

diggy
I'm a little confused on the binding energy argument. Typically binding energies lower the composite objects mass, the argument above implies that they raise the quarks mass to the much larger say proton mass.

Also in the paper above (I didn't carefully read through it, running on short time here), but I think they used pions and kaons as parameters with O(200) MeV masses, compared to bare quark masses that are O(5) MeV. So there is a missing factor of ~40 for the u/d -> proton left unexplained (for strange its only an order 5). -- I could be wrong on this last paragraph -- I will try and give the paper a read later.

ansgar
I'm a little confused on the binding energy argument. Typically binding energies lower the composite objects mass, the argument above implies that they raise the quarks mass to the much larger say proton mass.

Also in the paper above (I didn't carefully read through it, running on short time here), but I think they used pions and kaons as parameters with O(200) MeV masses, compared to bare quark masses that are O(5) MeV. So there is a missing factor of ~40 for the u/d -> proton left unexplained (for strange its only an order 5). -- I could be wrong on this last paragraph -- I will try and give the paper a read later.

What do you mean by "typically"?

the_house
Intuitively one expects that the bound state should have a lower energy than the individual free constituents, otherwise it will be unstable and decay. In that sense you could say it's more "typical".

Nucleons are a bit special because the individual quarks and gluons are not valid asymptotic states--a proton has nothing to decay into. (A free neutron will eventually decay into a slightly lower mass proton, but not nearly as fast as you would expect if it could decay into a few ~5 MeV free quarks.) They are highly energetic, relativistic systems, and all this localized energy translates into a mass.

This can be contrasted with what you're calling a "typical" bound state, such as the earth-sun gravitational system, where the kinetic energy involved is smaller than the negative potential energy (otherwise it wouldn't be a bound system at all and the Earth would go flying off into space.)

You must not think about a quantum state as a collection of massive classical particles minus the binding energy.

Think about QCD with massless quarks: the nucleon energy (three quarks in the naive picture) would change only slightly, whereas the pion (one quark, one antiquark) would be exactly massless; this is due to the fact that the pion is the Goldstone boson of the (broken) chiral symmetry. Naively one expects

$$m_\pi = \frac{2}{3}m_{p,n}$$

but in reality one finds (with massless quarks)

$$m_\pi = 0$$

That means that the mass of the nucleon is due to quantum effects and is not (nearly) related to "bare" and unobservable masses of "free" quarks.

ansgar
Intuitively one expects that the bound state should have a lower energy than the individual free constituents, otherwise it will be unstable and decay. In that sense you could say it's more "typical".

I would not draw an equivalence between intuition and classical mechanics though

humanino
I should comment on my wording "fraud". It has a twofold meaning.

First, least importantly, it is wrong to say that the Higgs accounts for our mass. As was described above, most of our mass is in the glue and/or virtual sea of partons of our protons and neutrons.

Second, and most importantly, there is in this claim a denial of importance in this phenomenon. I do agree that the Higgs sector(s) of the standard model is quite important and potentially underlies news physics, such as supersymmetry. But I refuse to compare this importance to the one I think the QCD strong sector deserves.

The confinement of quark and dynamical chiral symmetry breaking are cornerstones of the standard model. They should be logically studied first before embarking on the speculations of new physics beyond the SM in the electroweak sector. Among the many reasons :
• Quantum corrections down to the nuclear structure are small. Starting from the nucleon structure (light quarks cannot be non-relativistic), they become essential. At this scale, our intuition fails altogether, much more badly than anywhere else we can access experimentally (including heavy quark physics). This observation is at the heart of all the difficulties.
• The standard treatment in Feynman diagrams is a perturbation around the vacuum. We do not know the vacuum of QCD. There are many different QCD-inspired models, and it is not always clear how they manage to predict comparable observables with different underlying physics ingredients.
• This non-perturbative sector requires the concept of effective field theory, which is mandatory to go beyond the standard model (indeed, beyond the SM, the SM itself becomes effective). That is beautifully described by Weinberg.
• QCD is today the only unbroken non-abelian gauge theory we have to study experimentally. Possibly, there are other ones (dynamical electroweak symmetry breaking, compositeness, extra-dimensions, and modern technicolor).

Thanks to humanino for mentioning the most relevant topics like "light quarks cannot be non-relativistic", "The standard treatment in Feynman diagrams is a perturbation around the vacuum", ...

The difficulty with QCD is that vacuum and bound states (nucleons) are both "non - non-relativistic" due to the light quarks and non-perturbative due to the complex vacuum structure. This is to be contrasted with the Higgs effect which can be understood based on purely classical reasoning w/o any quantum effect.

diggy
What do you mean by "typically"?

I just meant any "classical-quantum-relativistic" decay where a heavier nucleus (for example) decays into lighter nucleons plus energy. I'd consider that the typical interpretation.

Is the basic argument that the proton mass is larger than its constituent parts on account of the effective energy/mass of the QCD field or some such?

Just for the record I don't have a preference for the outcome of this -- its just news to me (that the Higgs only accounts for 5% of nuclear matter), and I'd like to understand the skeleton of the argument.

inflector
So, it seems to me that the case of quarks binding energy and the nucleon binding energy is different and opposite in nature.

Something about putting protons and neutrons together into a nucleus lowers the masses of the combination. That's why He has a much lower mass-per-nucleon than H, or D. In this case, the nucleons want to be together because they get lower energy by the combination.

In the case of the nucleons themselves, being composed of quarks, their constituents have less mass, so the combination is higher energy. This is, I suppose, what makes them so counter-intuitive. It doesn't seem to our intuition that the quarks should stay together since they'd have less energy apart. But then quarks don't exist apart so that's not a meaningful comparison. Also, I don't see how they can measure the mass of a quark since it doesn't exist as a separate particle (but that's a topic for another thread).

In the case of a nucleus, since most of the mass of the nucleons comes from the glue, and not the quarks, it seems that there must be something about the combination of nucleons in close proximity that reduces the effect of the mass generation from the glue in protons and neutrons that makes the overall combination in a nucleus have less energy when combined than when together.

So, for quarks combining you need more energy to keep them together, but you can't separate them. While for protons and neutrons combining, you need less energy to keep them together.

It's a very strange world indeed inside the atom.

inflector
Thanks everyone for the answers so far. It's helped my understanding.

ansgar
I just meant any "classical-quantum-relativistic" decay where a heavier nucleus (for example) decays into lighter nucleons plus energy. I'd consider that the typical interpretation.

Is the basic argument that the proton mass is larger than its constituent parts on account of the effective energy/mass of the QCD field or some such?

Just for the record I don't have a preference for the outcome of this -- its just news to me (that the Higgs only accounts for 5% of nuclear matter), and I'd like to understand the skeleton of the argument.

Well the colour force is a very "special" force so it should have very special results assigned to it :)

Like massless mediators but still finite range due to asymptotic freedom, that's why I think the analogy with the massless box filled with photons is quite good one actually!

ansgar
So, it seems to me that the case of quarks binding energy and the nucleon binding energy is different and opposite in nature.

Something about putting protons and neutrons together into a nucleus lowers the masses of the combination. That's why He has a much lower mass-per-nucleon than H, or D. In this case, the nucleons want to be together because they get lower energy by the combination.

In the case of the nucleons themselves, being composed of quarks, their constituents have less mass, so the combination is higher energy. This is, I suppose, what makes them so counter-intuitive. It doesn't seem to our intuition that the quarks should stay together since they'd have less energy apart. But then quarks don't exist apart so that's not a meaningful comparison. Also, I don't see how they can measure the mass of a quark since it doesn't exist as a separate particle (but that's a topic for another thread).

In the case of a nucleus, since most of the mass of the nucleons comes from the glue, and not the quarks, it seems that there must be something about the combination of nucleons in close proximity that reduces the effect of the mass generation from the glue in protons and neutrons that makes the overall combination in a nucleus have less energy when combined than when together.

So, for quarks combining you need more energy to keep them together, but you can't separate them. While for protons and neutrons combining, you need less energy to keep them together.

It's a very strange world indeed inside the atom.

The forces between nucleons are short ranged and mediated by colourless- massive particles (pions, rho meson etc.)

Now, to measure the mass of quarks, there are three ways:

1) invariant mass method. Assume you produce a b bbar pair, this you can be sure of due to something called "b-tagging". Measure the invariant mass of it's hadronization and decay products -> "mass" of the b-quark

2) Model tuning. MS-bar Masses of quarks comes in many formulas for models describing observables. Make predictions for such observables and compare with experiment. So now one can have "model-mass" of quarks, which in general will differ from model to model, but one can get a feeling for it atleast.

3) low energy QCD, Chiral Perturbation Theory - also one of those "models" but on a more solid physics ground than those.

What is important to take with you I think is that there are different kinds of masses in QFT, mass is always equivalent to "gravitational inertia" as it is in classical mechanics :)

Good point.

In low energy effective models the constituent quarks masses are approx. 1/3 of the nucleon mass. But here the constituent quark is an effective degree of freedom. If you look closer (as you do in deep inelastic scattering) you see the stream quarks masses with the very small masses I mentioned before.

As you are interested in the origin of the nucleon masses I would use the "model quarks mass" of lattice QCD.

genneth
In the case of the nucleons themselves, being composed of quarks, their constituents have less mass, so the combination is higher energy. This is, I suppose, what makes them so counter-intuitive. It doesn't seem to our intuition that the quarks should stay together since they'd have less energy apart. But then quarks don't exist apart so that's not a meaningful comparison.

A good "classical model" is that you can't get free quarks, so they exist only as 2 or 3 particles connected by elastic bands. (Effectively, renormalising out the possibility of pair creation.) They are also constrained by Heisenberg uncertainty, so are in a state of constant agitation. This stretches the elastic bands, and also gives the quarks kinetic energy, which all contribute to the mass.

As Wilczek fancifully puts it: "quark are born free, but everywhere in chains".

cbd1
Imagine that you have a box with walls which are 100% opaque and massless, in that box you put a lot of photons. Now put this box on a scale, it will have mass due to the energy content of the photons...

Is this serious? Adding photons into a container adds no gravitational weight to the container; the photons have no weight and no inertia. Sure, there will be more mass equivalence in the box, box it will never gain mass of the kind that you can weigh on a scale.

humanino
Is this serious? Adding photons into a container adds no gravitational weight to the container; the photons have no weight and no inertia. Sure, there will be more mass equivalence in the box, box it will never gain mass of the kind that you can weigh on a scale.

The equation E=mc^2 applies whenever the momentum is zero. Your box has energy without total momentum in its rest frame.

cbd1
The equation E=mc^2 applies whenever the momentum is zero. Your box has energy without total momentum in its rest frame.

Yes, but the photons are never at rest unless absorbed. The only time photons can add inertial mass to a system is if they are absorbed by the matter of that system. I do not believe your box will ever show weight on a scale, no matter how many photons you add to it.

Again, the mass equivalence, by E=mc^2, will increase. But, as there is no gravitational force downward on photons, they will not contribute force downward on the scale.

Perhaps you have some literature on this to support your claim?

humanino
I can point you out to a very important historical reference (Bohr-Einstein's debates) which addresses just your thoughts. It's called Einstein's box, and it's so well known one can find it on wikipedia.
(My personal reference at home is "Quantum Theory and Measurement" (Princeton Series in Physics) by John Archibald Wheeler and Wojciech Hubert Zurek)

wikipedia said:
At the sixth Congress of Solvay in 1930, the indeterminacy relation just discussed was Einstein's target of criticism. His idea contemplates the existence of an experimental apparatus which was subsequently designed by Bohr in such a way as to emphasize the essential elements and the key points which he would use in his response.

Einstein considers a box (called Einstein's box; see figure) containing electromagnetic radiation and a clock which controls the opening of a shutter which covers a hole made in one of the walls of the box. The shutter uncovers the hole for a time Δt which can be chosen arbitrarily. During the opening, we are to suppose that a photon, from among those inside the box, escapes through the hole. In this way a wave of limited spatial extension has been created, following the explanation given above. In order to challenge the indeterminacy relation between time and energy, it is necessary to find a way to determine with an adequate precision the energy that the photon has brought with it. At this point, Einstein turns to his celebrated relation between mass and energy of special relativity: $E=mc^2$. From this it follows that knowledge of the mass of an object provides a precise indication about its energy. The argument is therefore very simple: if one weighs the box before and after the opening of the shutter and if a certain amount of energy has escaped from the box, the box will be lighter. The variation in mass multiplied by $c^2$ will provide precise knowledge of the energy emitted. Moreover, the clock will indicate the precise time at which the event of the particle’s emission took place. Since, in principle, the mass of the box can be determined to an arbitrary degree of accuracy, the energy emitted can be determined with a precision ΔE as accurate as one desires. Therefore, the product ΔEΔt can be rendered less than what is implied by the principle of indeterminacy.

The idea is particularly acute and the argument seemed unassailable. It's important to consider the impact of all of these exchanges on the people involved at the time. Leon Rosenfeld, a scientist who had participated in the Congress, described the event several years later:
Rosenfeld said:
It was a real shock for Bohr...who, at first, could not think of a solution. For the entire evening he was extremely agitated, and he continued passing from one scientist to another, seeking to persuade them that it could not be the case, that it would have been the end of physics if Einstein were right; but he couldn't come up with any way to resolve the paradox. I will never forget the image of the two antagonists as they left the club: Einstein, with his tall and commanding figure, who walked tranquilly, with a mildly ironic smile, and Bohr who trotted along beside him, full of excitement...The morning after saw the triumph of Bohr.
The "triumph of Bohr" consisted in his demonstrating, once again, that Einstein's subtle argument was not conclusive, but even more so in the way that he arrived at this conclusion by appealing precisely to one of the great ideas of Einstein: the principle of equivalence between gravitational mass and inertial mass. Bohr showed that, in order for Einstein's experiment to function, the box would have to be suspended on a spring in the middle of a gravitational field. In order to obtain a measurement of weight, a pointer would have to be attached to the box which corresponded with the index on a scale. After the release of a photon, weights could be added to the box to restore it to its original position and this would allow us to determine the weight. But in order to return the box to its original position, the box itself would have to be measured. The inevitable uncertainty of the position of the box translates into an uncertainty in the position of the pointer and of the determination of weight and therefore of energy. On the other hand, since the system is immersed in a gravitational field which varies with the position, according to the principle of equivalence the uncertainty in the position of the clock implies an uncertainty with respect to its measurement of time and therefore of the value of the interval Δt. A precise evaluation of this effect leads to the conclusion that the relation $\Delta E\Delta t\geq h$, cannot be violated.

cbd1
Thanks!

I always go by what seems logical to me, and the quantum mechanics proves to be the opposite.

I like PF for you guys getting me set straight.

cbd1
So, if I am to take your argument as correct, could I go on to say that the following is true:

Photons, when free and not contained in a box, have no gravitational or inertial mass, but only mass equivalence. While, if you place photons in a box at rest, making the photons part of a system which can be considered at rest, then the photons will contribute to gravitational and inertial mass of the system.

So, again it is the preparation of the system that gives the desired result, while if you take a photon not prepared in such a system, it does not contain the property of mass.

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Parlyne
So, if I am to take your argument as correct, could I go on to say that the following is true:

Photons, when free and not contained in a box, have no gravitational or inertial mass, but only mass equivalence. While, if you place photons in a box at rest, making the photons part of a system which can be considered at rest, then the photons will contribute to gravitational and inertial mass of the system.

So, again it is the preparation of the system that gives the desired result, while if you take a photon not prepared in such a system, it does not contain the property of mass.

In neither situation does the photon itself have mass. What has mass is the multi-object system that includes the photons. In fact, you can generically show that in relativity, the mass of a multi-object system not the sum of the masses of its constituents (except in the case that relative velocities among the objects are all 0). If we consider the energies and momenta of the constituents in the rest frame of the system (i.e., the frame where the momenta sum to 0), we find (presuming that all particles are free) that the mass of the system is
$$m = \sum_i \sqrt{m_i^{\phantom{i}2}+|\vec{p}_i|^2}$$.
This is larger than the sum of the masses unless all momenta are 0.

The preparation of the system is only significant in that it makes it natural to consider the system as a system. We could just as easily talk about the mass of a system of photons which are flying apart from each other into empty space. There just wouldn't be a natural way to consider measuring that mass.

humanino
Also, think of a neutral pion for instance. Put your neutral pion at rest in a box and close the box. You never know from outside if the neutral pion (which has a mass) has decayed into two photons (each without mass, but the system of two photon having the invariant mass of the pi0)

cbd1
I am just having great difficulty and confusion trying to figure how it is (by what mechanism) it is that gravity causes the photons added to the box to make force downward on the box.

I can see how if they were absorbed it would add weight to what absorbed them, but just their being in the box doesn't make sense to me how they would cause net force downward when in a gravitational field. In general relativity, photons are only affected by the curvature of spacetime in their paths, not actually pulled or accelerated downward.

Edit: If you think about it, neither QCD mass or Higgs mass would account for this. So there must be a different theory that says how the mass would be accounted for in the weight.

Edit 2: Or could it be accounted for somehow by radiation pressure?

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Antiphon
Photons, when free and not contained in a box, have no gravitational or inertial mass, but only mass equivalence.

I believe they have gravitational and inertial effects which may as well be called mass.

If you shine a beam of light near a star and the beam gets deflected, then the star is also displaced. This is required by conservation of momentum. There is no other way to interpret this than light exerting gravity on a star.

In general relativity, photons are only affected by the curvature of spacetime in their paths, not actually pulled or accelerated downward.
Is there any difference? I am by no means expert on GR, but as far as I understand the curvature of spacetime IS the mechanism by which an object is pulled downward.

humanino
Is there any difference? I am by no means expert on GR, but as far as I understand the curvature of spacetime IS the mechanism by which an object is pulled downward.
Yes, that is the principle of equivalence : an observer in a freely falling box can not perform an experiment(1) to decide whether there is a gravitational field around. The observer follows a geodesic in spacetime (not in space). So, that is all stated locally in an infinitesimal box, but remember that globally, we are NOT saying that observers do not fall !

Light carries energy and momentum, and energy and momentum are the source of gravitational field, not mass. Photons do couple directly to gravitons in perturbative general relativity.

This is a fundamental fact which cannot be explained satisfactorily(2) by any mechanism
wMFPe-DwULM[/youtube] (1) Obviousl...g" in the sense Feynman conveys in the video.

humanino
This is a fundamental fact which cannot be explained satisfactorily(2) by any mechanism
To follow up on that, as Feynman says above, although there is not satisfactory explanation to "why", we do have equations to describe what happens. In the case at hand, the equation is just $E^2=\vec{p}^2c^2+m^2c^4$ where in a given referential, energies and momenta do add up linearly (conservation of energy momentum) so we can infer that mass does not ! (because the equation is not linear)

In particle physics, mass is part of what defines a particle as a representation of the Poincare (restricted Lorentz) group (the other Lorentz scalar (or Casimir operator in terms of group theory) defining the particle being angular momentum). Mass is also the energy in the frame where the momentum is zero. A particle may be defined with a zero mass, but the system of several such particles with zero masses in general will not have a vanishing mass. For the total system in its rest frame, by definition of the rest frame you will get the sum of momenta being zero $\sum_i\vec{p_i}=\vec{0}$, and the sum of energies will give you the mass of the system, which will not equal the sum of the masses :
$$m=\frac{1}{c^4}\sqrt{ \left(\sum_i E_i\right)^2-\left(\sum_i\vec{p_i}\right)^2c^2} =\frac{1}{c^4}\sqrt{ \left(\sum_i E_i\right)^2-\vec{0}^2c^2} =\sqrt{\left(\sum_i\sqrt{ \frac{\vec{p}_i^2}{c^2}+m_i^2} } \right)^2}=\cdots \neq \sum_i m_i$$

Orion1

Wiki said:
Since energy is dependent on reference frame (upon the observer) it is convenient to formulate the equations of physics in a way such that mass values are invariant (do not change) between observers, and so the equations are independent of the observer. For a single particle, this quantity is the rest mass; for a system of bound or unbound particles, this quantity is the invariant mass. The invariant mass m of a body is related to its energy E and the magnitude of its momentum p by:
$$mc^2 = \sqrt{E^2 - (pc)^2}$$

Invariant mass (rest mass):
$$m = \frac{\sqrt{E^2 - (pc)^2}}{c^2}$$

Invariant mass of the system:
$$m=\frac{1}{c^2}\sqrt{ \left(\sum_i E_i\right)^2-\left(\sum_i\vec{p_i}\right)^2c^2} =\frac{1}{c^2}\sqrt{ \left(\sum_i E_i\right)^2-\vec{0}^2c^2} =\sqrt{\left(\sum_i\sqrt{ \frac{\vec{p}_i^2}{c^2}+m_i^2} } \right)^2}=\cdots \neq \sum_i m_i$$

Quantum mass manifests itself as a difference between an object’s quantum frequency and its wave number:
$$m = \frac{\hbar}{\overline{\lambda} c} = \frac{\sqrt{E^2 - (pc)^2}}{c^2}$$

The result is that all energy in any system is quantized by its wave number:
$$\hbar c = \overline{\lambda} \sqrt{E^2 - (pc)^2}$$

$$\hbar c = \sum_i \overline{\lambda}_i \sqrt{ \left(\sum_i E_i\right)^2 - \left(\sum_i\vec{p_i} \right)^2 c^2} = \sum_i \overline{\lambda}_i \sqrt{ \left(\sum_i E_i \right)^2 - \vec{0}^2 c^2} = \sum_i \overline{\lambda}_i \sqrt{\left(\sum_i \sqrt{ \vec{p}_i^2 c^2 + m_i^2 c^4} \right)^2}$$

$$\boxed{\hbar c = \sum_i \overline{\lambda}_i \sqrt{\left(\sum_i \sqrt{ \vec{p}_i^2 c^2 + m_i^2 c^4} \right)^2}}$$

Reference:
http://en.wikipedia.org/wiki/Mass#Summary_of_mass_concepts_and_formalisms"
http://en.wikipedia.org/wiki/Wave_number" [Broken]

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The_Duck
I am just having great difficulty and confusion trying to figure how it is (by what mechanism) it is that gravity causes the photons added to the box to make force downward on the box.

I can see how if they were absorbed it would add weight to what absorbed them, but just their being in the box doesn't make sense to me how they would cause net force downward when in a gravitational field.