# LOG#020. e=mc². Notions of mass.

My article today is dedicated to the most celebrated equation in Physics. Strictly speaking, it is not ONE single equation, but 3 or 4 different equations, despite the fact the the concept and physical idea behind its simple looking ARE “the same”. Thus, the popular and biased teaching of physical concepts by some authors, and the iconic figure of Einstein himself has driven to some very common unlucky misconceptions about what the equation means in several contexts. Of course, generally graduated students of Physics, physicists and experts in the theory of relativity are usually aware of these subtle issues but not always. They do know generally what they are doing but even Einstein himself wondered about these concepts, so sometimes you can even feel strange when you don’t know about what kind of mass people is talking about. Take care and think about it: the idea of mass is not a completely understood concept even in the 21st century!

My second purpose will be to explain some of the different notions of mass in classical physics and to introduce the issue of mass, its concept, as the essence of every current theory, either classical or quantum. I will not talk about the whole problem of mass in quantum theories, but I am trying to provide a broad perspective about one of the deepest concepts in Physics since the emergence of modern Physics: mass and inertia.

Let me be back to previous lessons I gave here. Special relativity, in the framework of Geometry, via spacetime vectors, merge the concepts of space and time into spacetime, the concepts of momentum and energy into momenergy. Indeed, it is just a way of rethinking classical concepts of space or time, momentum and energy, into a larger formalism. It does not say, a priori, anything new. It only explains that what we thought they were independent objects are really related ideas. What did we learn? Well, we did learn that

$\mathbb{P}=(E/c,\mathbf{P})$ and $\mathbb{P}\cdot\mathbb{P}=-(mc)^2$

Moreover, we found that this last equation says that:

$E^2=(mc^2)^2+(\mathbf{P}c)^2$

and that the total relativistic energy and the relativistic rest mass are given by:

$E=Mc^2$

and

$e=mc^2$

where the last equation is sometimes written as $E_0=m_0c^2$. However, since the squared rest mass IS the truly invariant quantity up to a multiplicative constant, Okun and other people have remarked that the most correct equation relatinc mass and energy is $E_0=mc^2$! The equation $E=mc^2$ is “confusing”, misleading, and not completely correct from the relativistic viewpoint. So, be aware with the propaganda of the media! Motto: the “invariant mass” IS the fundamental and right notion of “mass” in special relativity. In spite of all you have read about this topic out there. It is NOT right that mass is velocity-dependent. That is wrong. Relativistic momentum IS NOT the rest mass times velocity in special relativity, it is something more complicated. However, special relativity says that there is a well defined notion of mass, an invariant quantity that, unlike classical mechanics, is not conserved in general.

Finally, we also learnt that the relativistic kinetic energy, defined as the difference between the total energy and the rest mass is given by

$K=\Delta E=\delta mc^2=(M-m)c^2$

So we can write these 3/4 equations to compare them a bit better in a single line:

$\boxed{E=Mc^2}$ $\boxed{e=mc^2}$ $\boxed{K=\Delta E=\delta mc^2=(M-m)c^2}$ $\boxed{E^2=(mc^2)^2+(\mathbf{p}c)^2}$

Note: Once again, the only two invariant relationships you really need in special relativity are

1st. The constitutive-like dispersion relationships $E=E(\mathbf{P},m)$:

$(\mathbf{P}c)^2-E^2=-m^2c^4$

Some theories beyond SR could change this equation.

2nd. The constitutive relationship momentum-velocity $\mathbf{P}=\mathbf{P}(\mathbf{v}, E)$:

$\mathbf{P}=m\gamma \mathbf{v}\leftrightarrow \dfrac{\mathbf{P}c^2}{E}=\mathbf{v}\leftrightarrow \mathbf{P}=\mathbf{v}\dfrac{E}{c^2}$

Again, it could be possible for some kind of theories change this last equation or its equivalent expressions.

A caution and comentary deserves special attention: in many textbooks, specially some old-fashioned books and papers, and a really large amount of popular books about physics and relativity, you find the controversial concept of “relativistic mass”:

$M=\dfrac{m}{\sqrt{1-\dfrac{v^2}{c^2}}}$

You observe that M is bigger as you approach to the speed of light. A massive particle in motion, as seen from an inertial frame then, is increased. Be aware a was deliberately “clever” with my notation, since some people use “m” instead other different name for M. And if you are not careful, you can think and confuse M with m. Some books do in fact make the distinction more clear writing $M=M(v,m)$ instead of M or m. Anyway, the important keypoint is that $M(v,m)=M\neq m$!

It is essential to understand that and what this stuff really means. Perhaps, the first thing we should recall is what it was called “rest mass” and “rest energy”. Indeed, I think it is the easiest concept to teach everyone when he has in mind the Einstein’s mass-energy equation, and even when they think about the concept of mass-energy equivalence that Special Relativity introduces, and it is simple. Simply speaking, energy and mass are two related entities, and the proportion is the square of the speed of light. Equivalently, any particle with a non-zero mass has an energy given by:

$e=mc^2 \leftrightarrow E_0=mc^2 \leftrightarrow \mbox{Rest mass-Rest energy equivalence/proportion}$

Recall that the square of the speed of light is a big number. $c^2\approx 9\cdot 10^{16}m^2/s^2$. Therefore, a single kilogram has about $9\cdot 10^{16}J$ of energy!

By the other hand, remember that the rest mass is itself and invariant as well (it is frame independent) since $\mathbb{P}\cdot\mathbb{P}=-(mc)^2$. Let us continue. What happens in other inertial frames? I mean, how are mass and energy observed from a S-frame if you have a massive particle moving with constant speed in a S’-frame. Well, as we studied before, the relativistic momentum or the total energy are NOT independently invariant. They are boosted, i.e., they are transformed under a Lorentz transformation! In this way, the relativistic mass $M=m\gamma$ means that, from the S-frame viewpoint, masses (or energies) in motion are “bigger” than the rest mass. Please, note that from the point of view of S’, its mass is m, not M, accordingly to the relativity principle as we would expect. Even more, if S’ would measure other mass in motion relative to S’, from the S’ viewpoint, the S-frame mass would be larger. There is no contradiction between these views since the objects are different and they are placed in different frames. Therefore, the known equation as the relativistic mass expresses only the fact that the measures of mass are not frame independent. Indeed, that is the reason because the idea of invariant mass as the square of the momenergy is very important. The invariant mass IS invariant in any frame. The rest mass IS not. It is frame dependent!  However, the total energy can be calculated in the SR framework and, perhaps surprinsingly, it is equal to:

$E=Mc^2=m\gamma c^2=\dfrac{mc^2}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Note, that, this equation has a different physical meaning than that of the equivalence rest mass-rest energy. This equation means that, in Special Relativity, the total energy, including likely potential energies and the rest energy $mc^2$ equals the relativistic mass times the square of the speed of light. It is highly non-trivial! Beyond that, the striking similarity with the previous rest mass-rest energy equivalence equation use to confuse people that are not familiar with these details. Subtle and important keypoints that due to the popular spreading of the theory of relativiy has caused lots of confusions between common people. Let me explain this issue with more detail. Imagine a system of two bodies, A and B, which interact via the electromagnetic field only. Their rest energies are $E_A=m_Ac^2$ and $E_B=m_Bc^2$. If the particles are in relative motion with respect other inertial frame S, and their relative speeds with respect to that frame are $u_A, u_B$, their energies will be:

$E_A=M_Ac^2=m_A\gamma c^2=\dfrac{m_Ac^2}{\sqrt{1-\dfrac{u_A^2}{c^2}}}$

$E_B=M_Ac^2=m_B\gamma c^2=\dfrac{m_Bc^2}{\sqrt{1-\dfrac{u_B^2}{c^2}}}$

When the potential energy is added, the system total relativistic energy reads:

$E=E_A+E_B+E_{pot}=\dfrac{m_Ac^2}{\sqrt{1-\dfrac{u_A^2}{c^2}}}+\dfrac{m_Bc^2}{\sqrt{1-\dfrac{u_B^2}{c^2}}}+E_{pot}$

Take as an easy and natural example the hydrogen atom. It is made from a single proton and an electron (in its simpler isotope, of course). We could separate the energy into several terms

$E=m_pc^2+E_{kin,p}+m_ec^2+E_{kin,e}+E_{pot}$

or, rearranging terms,

$E=(m_p+m _e)c^2+(E_{kin,p}+E_{kin,e}+E_{pot})=(\mbox{Total rest mass}+\mbox{Binding energy})$

or

$E=E_0^{total}+U_{bind}$

In this case, the ground state of atomic hydrogen has the well-known value of $-13.6eV$ as binding energy. The proton’s rest mass and the electron’s rest mass are, respectively, $938 MeV$ and $511keV$. Then, in total, the hydrogen’s atom total energy (equivalenty, its total mass multiplying by the square of the speed of light) is slightly smaller than the sum of the rest energies (respectively, rest masses) of its two components! This phenomenon is called, quite conveniently, mass defect. And it explains the third equation

$\Delta E= \delta mc^2$

In our example, for the hydrogen isotope $_1^1H$, it amounts to about

$\dfrac{13.6}{938\cdot 10^6+511\cdot 10^3}=1.45\cdot 10^{-6}\%$

This tiny fraction is equilibrated or compensated with the emission of a photon, when the hydrogen atom is formed by the recombination of a single proton and an electron. This phenomenon is larger when heavy nuclei are formed from protons and neutrons. The reason is that not only electromagnetism matters. The nuclear strong interaction is important as well. And it is stronger and more important than the electromagnetic interaction. Indeed, consider the $_2^4He$ nucleus. It is the simplest helium atom. The rest mass is:

$2(m_p+m_e)=2\cdot (1.6726+1.6749)\cdot 10^{-27}kg=6.6950\cdot 10^{-27}kg$

However, the true rest mass of the helium atom is lower: $6.6467\cdot 10^{-27}kg$

The mass defect in this case is

$\delta m=\dfrac{6.6950-6.6467}{6.6950}=0.72\%$

In general, the process of FUSSION, the formation of light nuclei from protons, electrons and neutrons, and/or the formation of heavier nucleir from lighter nuclei is associated the energy release. The new nuclear species created by this mechanism has a lower total energy (or total mass) than those of their single constituents. Moreover, exothermic nuclear reactions can be found in many stars like our sun (turning hydrogen into helium, helium into carbon,…), in hydrogen bombs, tokamaks, and it will be of course the goal of the forthcoming International Thermonuclear Experimental Reactor (ITER).

The composition of heavy nuclei through endothermic nuclear reactions happens in Nature only when supernovae explode or in the radioactive elements. Nuclei formed in nuclear reactions by the Nature (inside the stars) can be decomposed into lighter nuclei, a process associated with energy release. This is the known process called FISSION. It also happens in radioactive substances that are unstable and decay into more stable and lighter nuclei. Thus, the equivalence mass-energy helped to explain the mysterious behaviour of radioactive elements/substances like those that were discovered by Pierre and Madame Curie. The nuclear fission of uranium and plutonium is used today in nuclear reactors all over the world, and it is also a power supply for spatial probes like the New Horizons probe, or powerful explorer robotic labs like the Curiosity (to land in Mars in August, 2012). As an additional curiosity, the energy released in the Hiroshima bomb was equivalent to a mas defect of as little as 1 single gram!

In summary, there are three concepts behind the 3 equations above. Of course, experts know about this fact. Physics is about concepts, not about Mathematics even when we used advanced high level Mathematics, the 3 equatios mean different things. However, generally speaking, in spite of representing different ideas, the hidden concept is the same: the equivalence mass-energy in Special Relativity.

$\boxed{E=Mc^2\leftrightarrow \mbox{Total relativistic energy}}$

$\boxed{e=mc^2\leftrightarrow \mbox{Relativistic rest energy}}$

$\boxed{K=\Delta E=\delta mc^2=(M-m)c^2\leftrightarrow \mbox{Mass defect}}$

$\boxed{U=mc^2+K+E_p=mc^2+B.E.\leftrightarrow \mbox{Internal energy=rest mass+binding energy }}$

with

$B.E.=K+E_p=(\mbox{Kinetic Energy+Potential Energy})$

and

$\boxed{E^2=(mc^2)^2+(\mathbf{P}c)^2\leftrightarrow \mbox{Total energy-Rest energy-Momentum equivalence}}$

These equations represent one of the most impressive results of contemporary physics. Don’t forget that if you use cell phones, bateries, electricity, part of the energy you consume every day is due to commercial nuclear reactors. It has its dangers, but mastering the energy production has caused a technological advance in our society. And it has not finished yet.

Well, now we will face a harder problem and issue in Modern Physics. What is the definition of mass? Relativity says only that it is equivalent to energy. But, what is energy? What is mass? Galileo and Newton wondered about this deep issue. Let me review the different notions of mass in classical physics and relativity:

$\boxed{\mbox{Classical mass}\begin{cases}\mbox{Inertial mass:} \;\; \mathbf{F}=m_i\mathbf{a} \\ \mbox{Momentum:}\;\; \mathbf{p}=m_i\mathbf{v}\;\;\rightarrow \mathbf{F}=\dfrac{d\mathbf{p}}{dt}\\ \mbox{Gravitational mass}\begin{cases} \mbox{Active:}\;\; \phi=\int \mathbf{g}\cdot d\mathbf{r}=-4\pi G_NM_g^{active}\\ \mbox{Passive:}\;\; \mathbf{F}_g=m_g^{passive}\mathbf{g}\end{cases}\end{cases}}$

$\boxed{\mbox{Mass and relativity}\begin{cases}\mbox{Mass-Energy equivalence:}\;\; E=Mc^2\;\; \\ \mbox{Generalized Newton's law:}\;\; \mathbb{F}=m\mathbb{A}\;\\ \mbox{Momentum:}\;\; \mathbf{P}=M\mathbf{v}=m\mathbf{p}=m\gamma \mathbf{v}\;\;\rightarrow\mathbb{F}=\dfrac{d\mathbb{P}}{d\tau} \\ \mbox{Invariant mass:}\; m \rightarrow \;\;\mathbb{P}^2=-m^2c^2\;\; \rightarrow e=mc^2\\ \mbox{Relativistic mass:}\;\; M \begin{cases}\mbox{Transversal:}\; M_T=m\gamma \\ \mbox{Longitudinal:}\; M_L=\gamma^2M_T=\gamma^3m \end{cases}\end{cases}}$

Mass in galilean and newtonian physics is an interesting concept. Galileo’s inertia law is indeed one the celebrated Newton’s laws of Dynamics. Inertia is caused by mass, and hence it is called inertial mass $m_i$. This scalar magnitude or number is the multiplicative constant before the velocity in the defintion of classical (linear) momentum. Therefore, according to the fundamental law of Dynamics, inertial mass is a measure of an object’s resistance to changing its state of motion when a external force is applied. This mass can be experimentally determined by applying some force to an object and measuring the acceleration from that force.  For instance, an object with small inertial mass will accelerate more than an object with large inertial mass when acted upon by the same force. Greater the (inertial) mass, greater the inertia. Mathematically speaking, this is expressed as the well known equation

$F=m_ia$

Active gravitational mass $M_g^{active}$ is a measure of the strength of an object’s gravitational flux ( the gravitational flux is equal to the surface of gravitationl field over an enclosing surface, i.e., the number of field lines that pass through a given test surface). The gravitational field can be measured by allowin a small test object/probe to freely fall and measuring its free-fall acceleration. For example, an object in free-fall near the Moon (or Mars) will experience less gravitational field, and hence, it accelerates slower than the same object would if it were in free-fall near the Earth. The gravitational fied near the Moon (Mars) has less active gravitational mass.

$\phi=\int g=\int \mathbf{g}\cdot d\mathbf{r}=-4\pi G_NM_g^{active}$

Passive gravitational mass is a measure of the strength of an object’s interaction with the gravitational field. It is something different from the previous concept. Passive gravitational mass is the proportionality constant between the object’s weight (i.e. the gravitational force) and its free-fall acceleration/gravitational field. Mathematically speaking,

$\mbox{Weight}=\mathbf{F}_g=\mathbf{P}=m_g^{passive}\mathbf{g}$

Two objects within the same gravitational field will experience the same acceleration. We will discuss this empirical fact when we study General Relativity in forthcoming articles. However, the object with a smaller passive gravitational mass will experience a smaller force (less weight) than the object with a larger passive gravitational mass. Please, do not confuse the weight’s symbol $\mathbf{P}=\mathbf{F}_g$ with the momentum. Latin (or even greek) letters are finite, so we are into trouble with the notation sometimes. Fortunately, knowing what symbols are representing, generally the context is enough to avoid this kind of mistakes. Be aware anyway.

There are three additional notions of mass from the previous tables. In Special Relativity, we found the equivalence of mass and energy. Moreover we distinguished three extra new notions of mass: invariant mass, transversal mass (or relativistic mass) and longitudinal mass.

The invariant mass $m$ is related to the squared momenergy vector. The transversal mass $M=m\gamma$ is the mass a body in rest measures when he observes a massive particle (with rest mass m) in motion at constant relative constant respect to him in the perpendicular direction of motion. When the motion (i.e. the velocity) is not parallel to the rest frame in S, the observer measures an additional longitudinal mass $M_L=\gamma M=\gamma^3 m$ in the direction parallel to him. Furthermore, due to the equivalence mass-energy in SR, there are a lot of physical processes (like pair production, nuclear fusion, nuclear fision or the gravitational bending of light) showing how mass and energy can be exchanged or released in high energy (high velocities) experiments. Finally, in spite we have not studied the phenomenon yet, photons and other pure energy particles (with rest mass equal to zero) are shown to exhibit a behaviou similar to passive gravitational mass, i.e., photons are affected by gravitational fields as well.

Let me put an example about “invariant mass” to clarify this important mass concept in Special Relativity.

Suppose a very heavy and massive particle A that decays to two very lightweight (likely massless) particles B and C. What you do in order to find the mass of A is to measure the energy $E_B$ and $E_C$, and the angle $\theta_{BC}$ between the directions of motion of the two particles. Then, you can calculate this quantity

$2 E_B E_C (1 -\cos \theta_{BC})$

and then you take the square root, and divide by the speed of light squared

$m_A=\dfrac{\sqrt{2 E_B E_C (1 -\cos \theta_{BC})}}{c^2}$

This answer is called the invariant mass of particles B and C, and it equals the mass of particle A. Note that it is not a trivial sum of the rest masses of the particles B and C, since B and C are created with some momentum (velocity).  Therefore, you can calculate the mass of particle A simply by knowing $E_B, E_C,\theta_{BC}$, and plugging their values into the above formula!

If particles B and C aren’t that lightweight (e.g.neutrinos) or massless, a more precise answer is to replace the previous formulae with these:

$(m_B)^2 c^4 + (m_C)^2 c^4 + 2 E_B E_C \left(1 - \dfrac{v_B v_C}{c^2} \cos \theta_{BC}\right)$

where $m_B$ and $v_B$ are the mass and velocity of particle B (and similarly for C), and moreover c is the speed of light.  Again, taking the square root of this quantity, divided by $c^2$, is the invariant mass, and this will be the mass of the original/mother particle A. Mathematically speaking:

$m_A=\sqrt{(m_B)^2+(m_C)^2+2\dfrac{E_B E_C}{c^2}\left( 1-\dfrac{v_Bv_C}{c^2} \cos\theta_{BC}\right)}$

Finally, in order to make the ultimate claims about mass, momentum and energy in special relativity, I am going to borrow a nice example from physicist Matt Strassler. Suppose a particle 1 decay into two particles, called 2 and 3 (note that 1,2,and 3 are mere labels to identify the three particles). Suppose we have 3 different observers seeing that decay: Amy, Bob, and Cid. Amy is moving down with respect to the particle 1, Bob is moving to the left and Cid is stationary. What do they observe/measure? Moreover, we suppose the following conditions:

1st. Particle 1 has a mass (measured from its own reference frame) about 126 $GeV/c^2$. Particle 2 and 3 are photons and their rest masses are zero.

2nd. Amy moves with $v=4/5c$. Then $\gamma = 5/3$.

3rd. Bob moves to the left with constant $v=4/5c$ as well.

First Reference Frame. Cid’s frame, particle 1 is stationary in this frame.

Cid observes two photons with $E_2=E_3=63GeV=E_1/2$.

Cid observes particle 1 in rest, i.e., particle 1 is stationary, so p_1=0 and then the photon momenta are $p_2=p_3=63GeV/c$, and $p_2$ and $p_3$ have moved in opposite directions (sometimes is said they are back-to-back in this case), i.e., as vectors, $p_2=-p_3$.

Energy and momentum are conserved, but mass of the objects is not, since the photons are massless and the Higgs was not.  What about the mass of the system?  What is the mass of the system of two photons?  It isn’t zero.  In fact it is obvious what it is. Just as for the particle 1 itself (which initially made up the entire system), the system of two photons has the same energy and momentum as the particle 1 did to start with,

$E(total)=E_2+E_3 = 63 GeV + 63 GeV = 126 GeV$

$p(total)=p_2(up)+p_3(down) = 63 GeV/c-63 GeV/c = 0$

And since $p(total) = 0$ for Cid,

$m(total)=E(total)/c^2 =126GeV/c^2$

which is the particle 1 mass. The total (or system’s) mass did not change during the decay, as we would expect with our physical intuition.

Second Reference Frame. Amy’s frame, particle 1 is not stationary in this frame.

Cid says particle 1 has $p_1=0$, $E_1=126$ GeV. What about Amy?  She says: particle 1  has

$p_1=\gamma v E_1= \left(\dfrac{5}{3}\right)\left(\dfrac{4}{5}\right)E_1=168 GeV/c$, upward.

$E'_1=\gamma E_1=\dfrac{5}{3} E_1=210 GeV$

Moreover, Amy observes in the decay that the photons have:

$E'_2=\gamma (1+v/c)E_2=189 GeV$ and $p_2=E_2/c$ moving upward.

$E'_3=\gamma (1-v/c)E_3=21 GeV$ and $p_3=E_3/c$ moving downward.

The conservations of energy and momentum works again, since, for Amy:

$E_1=210 GeV$ and $E_1=E_2+E_3=189+21=210 GeV$.

$p_1=p_2+p_3=189+(-21)=168GeV$. Note that the vector character of the momentum is important in order to get the right answer for the momentum conservation!

And the mass of the system is equal to the particle 1 mass both before and after the decay, because both before and after the decay we get

$E(total)=210 GeV$ and $p(total)=168GeV/c$, moving upward.

For Amy obtaining the particle 1 invariant mass, she has to use the squared of her 4-momentum:

$210^2-168^2=126^2=\mbox{invariant mass squared}$

Remark: interestingly, the so called rest mass IS NOT conserved between different observers, since for Cid or Amy rest masses before and after the decay change.

Third Reference Frame. Bob’s frame, particle 1 is not stationary in this frame.

Now the same calculation that we did for Marie tells us that the particle 1 energy is $E_1= 210 GeV$ and $p_1= 168 GeV$, but unlike for Amy, for whom the particle 1 is moving upward, for Bob the particle 1’s momentum is to the right. We have to work in “components” (i.e., we have to use vector calculus more directly, although I will keep things simple):

i) up-down part of $p_1$ = up-down part of $p'_1$

ii) $\mbox{right-left part of}\,p_1$= $\gamma\left(\left(\mbox{right-left part of}\, p'_1\right) + v E'_1/c^2\right)$

iii)$E_1=\gamma (E'_1+v(\mbox{right-left part of}\,p'_1))$

And these equations are going to be simpler than they look, because from Cid’s point of view, p has no right-left part; all the momentum is either up or down. So Bob sees the particle 1 having the following components:

i) up-down part of $p_1$=up-down part of $p'_1=0$

ii) right-left part of $p_1=\gamma v E_1/c^2=\left(\dfrac{5}{3}\right)\left(\dfrac{4}{5}\right)126GeV/c=168GeV$, rightward.

iii) $E'_1=\gamma E_1=(5/3)126=210GeV$, as we mentioned before.

In the same fashion, Bob calculates the upward-going photon is having

i) up-down part of $p_1$=up-down part of $p'_2=63GeV$ upward.

ii) right-left part of $p_1=\gamma v E_2=\left(\dfrac{5}{3}\right)\left(\dfrac{4}{5}\right)63GeV/c=84GeV/c$ rightward.

iii) $E_1'=\gamma E_1=\dfrac{5}{3}63GeV=105GeV$.

with the formulas for the second photon being the same except that its up-down part points downward. Notice that $E = p c$ for massless photons, and that if we use the Pythagorean theorem for the size p of each photon’s momentum, we obtain the consistent result

$p_2^2=(\mbox{upward part of}\,p'_2)^2+(\mbox{rightward part of}p'_2)^2$

Plugging the numbers: $(105GeV/c)^2=(63 GeV/c)^2+(84 GeV/c)^2$. Please, note these results are indeed the known pythagorean rule to add components of a vector in order to get the modulus of the given vector.

So again, Bob observes completely different energies and momenta than Amy and Cid. But Bob still observes that energy and momentum are conserved.  Bob  also sees that the system of two photons has a mass equal to the mass of the particle 1 mass.  Why?  The total up-down part of the system’s momentum is zero; it cancels between the two photons.  The left-right part of the system’s momentum is 168 GeV/c; the total energy of the system is 210 GeV; and that’s just what Amy saw, with the only difference that she had the system’s momentum going up instead of to the right.   So, like Amy, Bob also sees that the mass of the system of two photons is $126GeV/c^2$, the invariant mass of the original particle.

Summary

It is important to write some conclusions. The 3 observers, Amy, Bob and Cid:

1. Disagree about how much energy and momentum the particle 1 has and they also disagree about how much energy and momentum each of the two photons has.

2. Agree that: energy and momentum are conserved in the decay, the mass of the system is conserved in the decay. They also agree that the invariant mass of the 2 photons system is $126 GeV/c^2$.

3. Interestingly, they agree moreover that the sum of the masses of the objects in the system was not conserved; it has decreased to zero from $126 GeV/c^2$.

There is no accident.  Einstein knew that energy and momentum were conserved according to previous experiments, so he sought (and found) equations that would preserve this feature of the world.  What is behind this stuff: Poincaré’s group symmetry. This topic is to be discussed in the near future in my blog too.

There are even two more notions of “mass” we have not discussed or sketched in previous schemes:

1st. Curvature of spacetime.

The spacetime has an intrinsic geometry and thus, it has curvature. Curvature itself is a general relativistic manifestation of the existence of mass. Spacetime curvature is extremely weak and hard to measure. It was not discovered after it was predicted by Einstein himself with his theory or general relativity, a relativistic theory of gravitation indeed. Extremely precise atomic clocks on the surface of the Earth planet are found to measure less time (run slower) than similar clocks in space (where clocks run faster). The elapsed time is a form of curvature called gravitational time dilation. Other forms of curvature have been measure using some experiments (like the Gravity Probe B satellite). Even worst, Einstein himself invented something called the cosmological constant. Grossly speaking, the cosmological constant can be understood as the energy/mass of “vacuum” spacetime. Imagine a 1 x 1 x 1 meter cube. Erase with some clever procedure every mass/energy from that cube. Then, the density of energy of what remains there is not zero but equal to:

$\rho_\Lambda=\dfrac{\Lambda c^4}{8\pi G}$

The vacuum “weights” something!

2nd. Quantum mass.

Finally, we want to explain briefly how mass arises in quantum (likely relatistic too) theories. There are two contemporary notions on “quantum mass”. The first idea uses the Compton effect (to be explained here in the future) and the second explains mass as interaction with the so-called Higgs field. In the first case, without entering into further details at this moment, quantum mass appears itself like a difference between an object’s quantum frequency and its wave number. In this fashion, the quantum mass of an elemantary particle, say an electron for instance, it is related to the Compton wavelength and it can be determined through various forms of spectroscopy and it is also connected to the the Rydberg constant, the Böhr radius and the classical electron radius. The quantum mass of larger objects can be directly measured using a watt balance and, in relativistic Quantum Mechanics, mass is one of the irreducible representation labels (quantum numbers) of the Poincaré group.
The second notion of mass in Quantum theories arises in (relativistic) Quantum Field Theories. There is a “bare” mass and a “dressed mass” as well. We won’t discuss these two masses here. We are going deeper. Mass terms coupled to fields are not naturally gauge invariant, and we are forced to introduce the so-called Higgs mechanism in order to give a non-zero mass to the gauge bosons $W^+, W^-,Z$ and the fundamental fermions (leptons and quarks) keeping safe gauge invariance. In the framework of Higgs mechanism one gets:

$M_W=\dfrac{1}{2}gv\approx g\cdot 123 GeV$

$M_Z=\dfrac{1}{2}g_Zv\approx g_Z\cdot 123 GeV$

$m_f=vg_f\dfrac{\sqrt{2}}{2}\approx g_f \cdot 174 GeV$

$M_H=v\sqrt{2}\lambda\approx \lambda \cdot 348GeV$

where

$v=\dfrac{1}{\sqrt{\sqrt{2}G_F}}\approx 246GeV$

is the so called v.e.v (vacuum expectation value) of the Higgs field (v is related to the Higgs boson mass) and $g, g_Z, g_f$ are the coupling constants of the gauge bosons W, Z and the fermions to the Higgs boson, respectively. $\lambda$ is the Higgs boson self-coupling. However, the gluons and the photon remain massless. Higgs boson itself gets a mass due to non-linear self-interaction in the Standard Model framework. Therefore, there is no explanation of the Higgs boson mass in the SM, the current accepted theory of fundamental electroweak interactions and strong interactions. Therefore, the SM is an incomplete (or effective) theory and we do not understand the ultimate origin of mass. We only shifted their final understanding to the yet mysterious origin of the Higgs field (and its Higgs particle constituents).

This is not the whole story, since, for instance, protons (or neutrons) get their masses non only through the constituent mass of its quarks. Quantum Chromodynamics (QCD) is the gauge theory of interacting quarks and gluons, and it is a non-abelian gauge theory. Thus, some part of the proton mass is due to nonlinear QCD interactions, not exclusively (indeed not mainly) to the interactions of quarks and Higg bosons! We leave this discussion here, including a final graph with some concepts of mass we discussed here:

PS: A final test to see if you understood and learnt what the mass-energy equivalence in Special Relativity is. Okun himself created a test about what is the truly fundamental mass-energy relationship in special relativity, and he gave 4 options ( note Okun uses other notation for M and m, but there is only one invariant mass!):

a) $E=mc^2$

b)$E_0=m_0c^2$

c)$E_0=mc^2$

d)$E=m_0c^2$

Hint: Being purist and consistent with the full Lorentz invariance of special relativity, there is only one expression right. It is the one that allows to recover the correct newtonian limit and to preserve the notion of “invariant mass” without using the confusing, obsolete and misleading ideas like “relativistic mass” or “velocity dependent mass”. Indeed, the answer was my original motivation in the selection of the first image of this article.

The answer is c, $E_0=e=mc^2$. Note that the subscript zero is irrelevant and it only confuses things such as the two notions of mass do if you are not careful enough.