B Center of momentum and mass energy equivalence

[nitsuj]

told so in another thread that two photons traveling in opposite directions "have mass". I understood the concept of centre of momentum, an inertial frame of reference and rest mass. My cynical reply wasn't conveyed and I got well thought out and accurate replies.

the COM is an idealized point, and this couldn't physically be better demonstrated than with things going c in opposing directions and calculating a com and in turn a "mass"...at most I'd call that a region of space with x amount of energy. However not even remotely something identifiable as a mass.

So how is understanding the two photon system as being massive not equivalent to saying relative motion increases mass...

First thought for me is causally how can this be true in practice. Maybe too poetic but, isn't a photon gone forever...

 

[PAllen]

It is no different for photons than any other particles. The mass of a system of two particles moving in the same direction at the same speed is the sum of their rest masses. However, for two particles moving in opposite directions with equal momentum, the mass of the system is given by the sum of their total energy (rest mass plus kinetic energy) / c². This is not some special rule, it just follows from adding 4-momenta, and taking the norm of the result, noting that the norm of the total 4-momentum of a system is its invariant mass.

 

[Dale]

So how is understanding the two photon system as being massive not equivalent to saying relative motion increases mass...

It is more closely related to the triangle inequality for regular vectors. If you have two vectors, then the magnitude of their sum is always less than the sum of their magnitudes. Do you understand that?

Mass is the magnitude of the energy-momentum vector, so a similar rule applies. The difference is that in this case the magnitude of the sum is more than the sum of the magnitudes.

 

[SiennaTheGr8]

Forget about the center-of-momentum point.

The important concept here is the center-of-momentum frame (aka the rest frame).

There are a few ways to understand what this special frame is:

• It is the inertial frame of reference in which a given system has no momentum.

• It is the inertial frame in which the momentum vectors of the system's constituents sum to the zero vector. (Momentum is a vector, and a system's momentum is equal to the vector sum of the momenta of its constituents.)

• It is the inertial frame in which the system's kinetic energy is zero. (Note: a system's kinetic energy is NOT equal to the sum of the kinetic energies of its constituents, so it is NOT true that the rest frame of a system is the inertial frame in which the kinetic energies of the system's constituents sum to zero; generally no such frame exists.)

Now, consider the following equation, which holds for any system:

$E = E_0 + E_k$

where $E$ is the system's total energy, $E_0$ is the system's rest energy, and $E_k$ is the system's kinetic energy. In the system's rest frame, we obtain:

$E = E_0$.

In words, a system's rest energy is its total energy as measured in its rest frame. It is a quantity that all observers agree on (it's invariant).

Say your system is the gas within a sealed jar. The system's rest frame is clearly the frame in which the jar isn't moving. What is the system's total energy in this frame? (Equivalently: what is the system's rest energy?)

Well, let's think about it for a second. The gas molecules are all moving around, so they have kinetic energy. There's also some small amount of electromagnetic potential energy associated with the relative positions of the gas molecules (likely negligible). But don't forget that each molecule also can be regarded as a (sub-)system in its own right, and each one has its own rest energy that must be included as a contributor to your system's total energy! (In fact, the sum of the molecules' rest energies will almost certainly constitute the vast majority of your system's rest energy.) If we add up all the kinetic energies, potential energies, and rest energies associated with the system's constituents (as measured in the system's rest frame), then we'll find the system's total energy in its rest frame (its rest energy).

What if the jar contained photons instead of gas molecules?

The same basic principle would hold: to find the system's total energy in its rest frame, we'd add up all the kinetic energies, potential energies, and rest energies associated with its constituents (as measured in that frame). It's true that a photon can never be at rest and therefore has zero rest energy, but a photon does have kinetic energy, so we can proceed. If we add up all the energy contributions (as measured in the system's rest frame), then we get the system's rest energy.

You should see by now that the rest energy of a system is NOT equal to the sum of the rest energies of the system's constituents. Instead, it is simply the system's total energy as measured in its rest frame. The rest energies of the system's constituents contribute to the system's rest energy, but their kinetic and potential energies do, too.

If your system was a single photon, it would have no rest energy. Why? Because there's no frame of reference in which the system has no momentum. A photon can never be at rest.

If your system had two photons moving in the same direction, then it also would have no rest energy. Why? Because there's no frame of reference in which the momentum vectors of the system's constituents cancel out. For two non-zero vectors to sum to a zero vector, they have to be pointed in opposite directions. But if two photons are moving in the same direction, then they're moving in the same direction for all observers.

If your system had two photons moving in different directions, then it would have rest energy. Why? Because in this scenario there is a frame of reference in which the photons are moving in exactly opposite directions with exactly the same frequency, meaning that their momentum vectors sum to a zero vector. The rest energy of this system would just be the sum of the (kinetic) energies of the photons (as measured in the system's rest frame), since there are no potential or rest energy contributions to consider.

If you've understood all this, then you're not missing anything essential. For completeness, though, we can add Einstein's famous equation to the mix:

$m = \dfrac{E_0}{c^2}$.

What this equation means is that "mass" is just rest energy expressed in different units.

You asked:

the COM is an idealized point, and this couldn't physically be better demonstrated than with things going c in opposing directions and calculating a com and in turn a "mass"...at most I'd call that a region of space with x amount of energy. However not even remotely something identifiable as a mass.

So how is understanding the two photon system as being massive not equivalent to saying relative motion increases mass...

Be careful: the rest energy (mass) of a system is not the same thing as the rest energies (masses) of the system's constituents.

If there's relative motion among the constituents of a system, then there's kinetic energy that contributes to the system's rest energy (mass). The greater this kinetic energy (as measured in the system's rest frame), the greater the system's rest energy (mass). But the rest energies (masses) of the constituents themselves are unaffected.

A photon's rest energy (mass) is always zero, but its kinetic energy can contribute to the rest energy (mass) of a system.

 

[nitsuj]

Forget about the center-of-momentum point.... the rest energy of a system is NOT equal to the sum of the rest energies of the system's constituents. Instead, it is simply the system's total energy as measured in its rest frame. The rest energies of the system's constituents contribute to the system's rest energy, but their kinetic and potential energies do, too.

$m = \dfrac{E_0}{c^2}$.

What this equation means is that "mass" is just rest energy expressed in different units.

I think i get it with the idea that rest energy is mass.

 

[sweet springs]

Hi.
Rest energy $\sqrt{E^2-p^2c^2}$, divided by $c^2$ is the definition of m.
Though m is still the quantity of inertia, I am not confident for cases of many body system.
Who could pull or push a free many particle e.g. photon gas out of the bottle to put them in ?

 

[nitsuj]

Who could pull or push a free many particle i.e. photon gas out of the bottle to put them in ?

I agree, but my understanding from reading here is that my comment of "region of space with x amount of energy", is the same as saying mass in certain conditions. kinda for silly for a two photon system imo causally whatever of the photons moving away from me.

 

[PAllen]

Hi.
Rest energy $\sqrt{E^2-p^2c^2}$, divided by $c^2$ is the definition of m.
Though m is still the quantity of inertia, I am not confident for cases of many body system.
Who could pull or push a free many particle i.e. photon gas out of the bottle to put them in ?

Just imagine a gas of atoms whose nuclei are in an excited stare, enclosed in a bottle. As they decay emitting gamma or x-rays, don't you think the overall inertia of the bottle remains unchanged? Then, the in flight gamma rays must contribute to the mass of the system exactly as described by the SR formulas.

 

[sweet springs]

Hi. I said

photon gas out of the bottle to put them in

You said

enclosed in a bottle

I agree with you. I can throw the bottle and feel its inertia. But how can I catch and throw free dispersing particles or gamma rays in your case ?

 

[SiennaTheGr8]

I can throw the bottle and feel its inertia. But how can I catch and throw free dispersing particles or gamma rays in your case ?

Perhaps you can't, but whether you can in practice apply a force to a given system is a separate question from how the system would respond if you could, which is:

$\vec a = \dfrac{\vec f - (\vec f \cdot \vec v) \, \vec v}{E}$

(in units where $c = 1$, and assuming that the system's rest energy $E_0$ is constant).

Note that this equation models the system as something that accelerates all at once—an idealized rigid body. But no multi-particle system is truly rigid. When you "throw the bottle," you're really applying a force directly to the molecules of the bottle you're "touching," causing them to "bump into" their neighbors, and so on. The bottle is rigid enough that you perceive yourself "feeling its inertia" in one fell swoop, but really you're "feeling" the cumulative effect of the microscopic reaction forces, over the course of some imperceptibly small (but quite real) interval of time.

My point is that whether you're using the above relativistic equation or the Newtonian approximation $\vec a = \vec f / E_0$ (again, $c = 1$; I like to use $E_0$ rather than $m$), technically you're always describing how the system would respond if you applied a force to every part of it at the same time, so that it would accelerate all at once. To the best of my knowledge this is pretty much always an idealization, unless (maybe?) your system is a single point particle like an electron. (Even in the case of gravity it's an idealization, what with tidal forces.)

BUT: the total energy $E$ of the system (which you might call the "measure of inertia" in relativistic dynamics) is not an idealization. It's "there," regardless of whether the equation above can serve as a useful model for the system in a real-world situation. So an unbottled photon gas would indeed resist acceleration if you applied a force to it, regardless of whether you actually can.

That's how I see it, anyway.

 

[Vitro]

@nitsuj, the mass of a system is defined as $m^2=(\Sigma E)^2-(\Sigma \vec p)^2$ (choosing units where c=1). The term $\Sigma E$ is always > 0 since all particles have positive energy. What about $\Sigma \vec p$? Well, momentums being vectors add as such and can cancel each other out. Most often for a system (assuming a flat spacetime) it's possible to find a frame of reference where the total momentum is zero. In fact there's only one situation when that's not possible, when all the particles in the system are massless and all travel in the exact same direction. So by making the total momentum zero the system mass must be > 0, there's no way around it, it's the definition.

 

[Dale]

The bottle is rigid enough that you perceive yourself "feeling its inertia" in one fell swoop, but really you're "feeling" the cumulative effect of the microscopic reaction forces

Very good point. In this sense inertia is never something that you "feel"

 

[robphy]

Here's another way [a geometrical way] to look at the situation.

My comment builds upon the early comments by @PAllen and @Dale.

First, consider this spacetime diagram of the light-signals in a light-clock
[used in my Insight on the Bondi k-calculus and is the basis of the Relativity on Rotated Graph Paper approach].
Event P lies on the unit-hyperbola centered at event O.
M and N are reflection events on the mirror-worldlines [particular parallels to OP], which are not shown.

Vectorially, we can interpret the situation this way.
So, OP is a unit-timelike 4-velocity vector from O.
ON is a future-forward lightlike vector and OM (and NP) is a future-backward lightlike vector
such that, vectorially, OP=ON+NP.
In other words, the timelike vector OP is expressed in light-cone coordinates [with lightlike components ON and NP].

https://www.geogebra.org/m/bFxv6tjaNow,
let's look at the energy-momentum diagram analogue of the above spacetime diagram,
with Energy running vertically upwards and Momentum running horizontally.
[Taylor and Wheeler would call this a momenergy diagram, by analogy.]

So, OP would represent the 4-momentum of a [timelike] nonzero-rest-mass particle.
The hyperbola is interpreted as the mass-shell (here with magnitude 1).
ON and NP would represent two lightlike 4-momenta, decomposing the timelike 4-momentum OP.
That is, again, vectorially, OP=ON+NP.
ON and NP can be thought of as the 4-momenta of "photons" in our frame:
one is a higher-frequency forward-directed light-signal and
the other is a lower-frequency backward-directed light-signal.
("Higher" and "lower" arise by doppler-shifting the "equal-frequency light-signals found in OP's frame".)

...thus, one can use Bondi k-calculus methods to analyze energy-momentum problems in relativity.

 

[Herman Trivilino]

So how is understanding the two photon system as being massive not equivalent to saying relative motion increases mass...

If you want to view each photon as having a mass equal to its total energy, it is equivalent. Choosing to eliminate relativistic mass from the lexicon of physics is a matter of preference.

Proponents of this preference will often assert that its use leads to learner confusion. One such confusion is a fundamental misunderstanding of the Einstein mass-energy equivalence; how a two-photon system can have a nonzero mass when each photon has a zero mass.

 

[nitsuj]

If you want to view each photon as having a mass equal to its total energy, it is equivalent. Choosing to eliminate relativistic mass from the lexicon of physics is a matter of preference.

Proponents of this preference will often assert that its use leads to learner confusion. One such confusion is a fundamental misunderstanding of the Einstein mass-energy equivalence; how a two-photon system can have a nonzero mass when each photon has a zero mass.

Or like how the use of the term "rest mass" has nothing to do with things being at rest, lots to do with symmetry.

 

[vanhees71]

That's why I insist on using the expression "invariant mass" instead of "rest mass".

 

[Herman Trivilino]

Or like how the use of the term "rest mass" has nothing to do with things being at rest, lots to do with symmetry.

That's why I insist on using the expression "invariant mass" instead of "rest mass".

The primary, if not the only, reason those terms are seen as necessary is to distinguish from "relativistic mass". Remove "relativistic mass" from the lexicon and they are no longer needed. Their continued can confuse learners, making them think that there must be some other kind of mass that necessitates their use.

 

[vanhees71]

Well, since even after 110 years nobody was able to remove "relativistic mass" from being used in socalled modern textbooks, we have to make sure to be understood right and thus unfortunately we have to say "invariant mass", although it should be a tautology for everybody who has read Minkowski's article on space and time! Indeed, when physicists used to relativity talk about "mass" nowadays they always refer to "invariant mass", but in the broader physics community still old-fashioned ideas linger around.

 

[Dale]

That's why I insist on using the expression "invariant mass" instead of "rest mass".

Me too. The phrase "invariant mass" makes sense for photons, rigidly spinning objects, and deformable systems, where "rest mass" seems a little odd in those cases.

 

[robphy]

...read Minkowski's article on space and time! Indeed, when physicists used to relativity talk about "mass" nowadays they always refer to "invariant mass", but in the broader physics community still old-fashioned ideas linger around.

In my opinion, Minkowski [a mathematician] isn't given enough credit by the physics community, particularly physics textbook authors. (Some attribute "spacetime" to Einstein. Maybe for curved spacetime for GR, it's appropriate. But "spacetime", "light cone", "world line", "proper time",... are really Minkowski's. He put Einstein's ideas on a firmer mathematical foundation, which led to wider acceptance of special relativity and allowed general relativity to develop.) Einstein is, of course, a physicist... and it seems his 1905 papers (and his formulation and viewpoint then) are where many physicists stop. Later, Einstein did try to move away from the relativistic mass idea. ( See the quote in https://en.wikipedia.org/wiki/Mass_in_special_relativity#Relativistic_mass )

 

[nitsuj]

That's why I insist on using the expression "invariant mass" instead of "rest mass".

I've always used the idea of mass ultimately being some bit of matter, specifically not a massless particle. Seems contradictory to me, especially in a case where the things are moving away from the COM at c...what is physically meaningful of that space? with respect to the things being used (photons) to define the energy within a defined region of a plausible position of the photons without any causal connection to these three points...all light like.

The mass of sun is decreasing, you're saying fair point is from what distance from the COM are you considering part of the system and that technically all the EM energy that has a symmetric photon (opposite direction, same wavelength/energy) since the defined "start" of the sun is part of it's mass?

One of the most interesting physics things I learn was how significant the energy of gluons is and that energy is a remarkable amount of our (atoms) (invariant lol) mass...VERY different from a "open" and "untouchable" two photon system...

Here is an other way to put it, it's taking the very accurate estimate of energy for two "opposing" bits of energy, "adding" it up and saying it's a mass. Where is this mass now? Forever right there? lol terrible story board :/

 

[PeterDonis]

I've always used the idea of mass ultimately being some bit of matter

This interpretation of "mass" is a holdover from Newtonian physics and is best discarded in relativity. There are other ways of taking "quantity of matter" into account (for example, baryon number/baryon density is often used in astronomy); but the direct link between "mass" and "quantity of matter" simply doesn't work in relativity the way it does in Newtonian physics.

 

[Herman Trivilino]

Here is an other way to put it, it's taking the very accurate estimate of energy for two "opposing" bits of energy, "adding" it up and saying it's a mass.

No, no one is saying it's a mass. What we're saying is that it has mass. You are stuck thinking of an object as being a mass when instead it has mass. Saying that something is a mass is nonsense, even in Newtonian physics. Many authors do it and it leaves deep-seated misconceptions that can be seen not just in students but in educators, too.

 

[vanhees71]

In my opinion, Minkowski [a mathematician] isn't given enough credit by the physics community, particularly physics textbook authors. (Some attribute "spacetime" to Einstein. Maybe for curved spacetime for GR, it's appropriate. But "spacetime", "light cone", "world line", "proper time",... are really Minkowski's. He put Einstein's ideas on a firmer mathematical foundation, which led to wider acceptance of special relativity and allowed general relativity to develop.) Einstein is, of course, a physicist... and it seems his 1905 papers (and his formulation and viewpoint then) are where many physicists stop. Later, Einstein did try to move away from the relativistic mass idea. ( See the quote in https://en.wikipedia.org/wiki/Mass_in_special_relativity#Relativistic_mass )

I cannot agree more. The famous talk on space and time by Minkowski was a breakthrough enabling Einstein to go further towards the General Theory. The most astonishing fact is that Minkowski's talk is completely valid today, including the invention of spacetime diagrams now named after him. Indeed the very idea of spacetime and its geometry is Minkowski's genuine work. Not much later he also provided the first complete and consistent covariant formulation of electrodynamics, including macroscopic electrodynamics with the correct constitutive equations.

 

[Dale]

I've always used the idea of mass ultimately being some bit of matter, specifically not a massless particle

The total four momentum is conserved during particle interactions, so the norm of the total four momentum is also conserved. People find such conserved quantities to be useful, so they give them names. In this case "mass".

Your concept is not conserved, so it is not as useful. It is unlikely that the community will discard the concept of a useful conserved quantity in favor a less useful one.

 

[robphy]

A timely video on relativistic mass

 

[nitsuj]

The total four momentum is conserved during particle interactions, so the norm of the total four momentum is also conserved. People find such conserved quantities to be useful, so they give them names. In this case "mass".

Your concept is not conserved, so it is not as useful. It is unlikely that the community will discard the concept of a useful conserved quantity in favor a less useful one.

lol I agree! how ever this is about the equivalence, in this case energy...photons, so mass is the preferable unit for discussing...in what case is this quantity useful to be thought of as a mass instead of energy of those photons?

I understand the idea of centre of momentum and mass. My point isn't about matter having mass...it's that photons don't...a photon is a force waiting to happen. I see how this is just the idealized case of where the energies of the photons when framed is such that the region has mass and idealizing a shape and com..

I see "..in the case of bound systems simply mass.". Would an "unbound" system be two free flowing photons in opposite directions?

I agree completely with this statement from the same page "For an isolated massive system, the center of mass of the system moves in a straight line with a steady sub-luminal velocity" The last part being my issue with photons being used cause it make the whole thing physically moot.

 

[Dale]

in what case is this quantity useful to be thought of as a mass instead of energy of those photons?

The invariant mass is a fundamentally different concept than energy. The mass is the invariant norm of the four momentum and the energy is the frame variant timelike component of the four momentum.

My point isn't about matter having mass...it's that photons don't.

An individual photon doesn't, but a system of two or more does. This becomes important e.g. In analyzing positron emission tomography.

 

[SiennaTheGr8]

lol I agree! how ever this is about the equivalence, in this case energy...photons, so mass is the preferable unit for discussing...in what case is this quantity useful to be thought of as a mass instead of energy of those photons?

I understand the idea of centre of momentum and mass. My point isn't about matter having mass...it's that photons don't...a photon is a force waiting to happen. I see how this is just the idealized case of where the energies of the photons when framed is such that the region has mass and idealizing a shape and com..

I see "..in the case of bound systems simply mass.". Would an "unbound" system be two free flowing photons in opposite directions?

I agree completely with this statement from the same page "For an isolated massive system, the center of mass of the system moves in a straight line with a steady sub-luminal velocity" The last part being my issue with photons being used cause it make the whole thing physically moot.

It seems like maybe you're too focused on the concept of force. You may need to recalibrate how you understand mass on a fundamental level.

As Dale said, the mass of a system is defined as the norm of its four-momentum $\mathbf{P}$:

$m \equiv |\mathbf{P}| = \sqrt{E^2 - |\mathbf{p}|^2}$

(in units where $c = 1$).

In other words, a system's mass is its total energy as measured in the frame in which it has no momentum. If no such frame exists, then the system's mass is zero (i.e., $E = |\mathbf{p}|$ for all observers).

Now, it's also true that mass is related to a variety of physical phenomena. For example, in Newtonian physics, mass is the "measure of inertia" (the quantity responsible for a system's resistance to acceleration). However, this aspect of mass is secondary to its definition as the norm of the four-momentum.

A system of photons moving in different directions is called a photon gas. And a photon gas has mass. Why? Because there exists a frame of reference in which the momenta of all the photons cancel out. The mass of the system is then the sum of all the energies of the photons as measured in this special frame. This is true regardless of whether the "measure of inertia" concept is useful in practice for this system.

Would you have trouble acknowledging that a solar system has mass? What about a galaxy? What about a cluster of galaxies? We could never apply a contact force to any of these systems in practice, so it's rather meaningless to talk about their resistance to acceleration.* In fact, a contact force can only be applied locally, at a "point of contact"—and even then, what we describe as "touching" is really the result of short-range electromagnetic interactions and the Paul exclusion principle (degeneracy pressure). So the very concept of a contact force is a fiction that happens to be useful in many situations. Sometimes it's not a useful concept, but that doesn't mean that the properties related to that concept (like mass) cease to exist.

*An exception here would be gravitation, but photon gases are affected by gravity too!

 

[sweet springs]

Me too. The phrase "invariant mass" makes sense for photons, rigidly spinning objects, and deformable systems, where "rest mass" seems a little odd in those cases.

I assume that mass belong to system not to each component of system, e.g. not to a photon but to photon gas (ref. my post #6). Mass distribution in a system, e.g. a photon has mass, is an annoying idea. Of course a photon has energy. So do we still need concept of mass in addition to energy ? The concept of mass is convenient in cases that energy concentrates in some region of space and does a collective motion in an approximation. Mass can be applied strictly only to an elementary particle the interior details of which we cannot go into. Even in this case we can say it rest energy or invariant energy. Quantity of material, inertia and gravity source(partially) is energy not mass as used to be called.

Not E=mc^2 mass has energy,
I prefer m=E/c^2 what we have called mass is energy as a special case p=0 of m=\sqrt{E^2-p^2c^2}/c^2.

 

[Dale]

So do we still need concept of mass in addition to energy ?

Yes. One refers to the norm of the four momentum and the other refers to the timelike component. They are separate concepts and both are needed.

Mass can be applied strictly only to an elementary particle

It applies to anything with a four momentum.

 

[sweet springs]

Yes. One refers to the norm of the four momentum and the other refers to the timelike component. They are separate concepts and both are needed.

I totally agree that the norm of the four momentum of a particle or a system is a useful idea whether we call it mass, rest energy or invariant energy.
This is my preference not to call it mass due to the concerns I posted. Again do we still need a concept of mass in addition to energy or four momentum more strictly ? Does mass survive under Ockam's razor ?

 

[Dale]

Again do we still need a concept of mass in addition to energy or four momentum more strictly ? Does mass survive under Ockam's razor ?

We are just going in circles here. Yes both concepts are needed and yes mass survives Occham's razor.

 

[sweet springs]

Um... An example where mass cannot be substituted by energy or four momentum including its norm could help me. Thanks in advance.

 

[vanhees71]

I don't understand your question. Already to know, which elementary particle you are observing you need to know its mass (and other parameters like electric charge etc.).

 

[sweet springs]

I agree with you as for system of single elementary particle where mass M and (proper) rest energy Mc^2 are not distinguishable.

Say two photon system e.g. after electron-positron annihilation, has mass M or rest energy Mc^2. In COM each photon has energy Mc^2/2. I do not think each of the two photons has mass M/2 also. This is an example of my point that and energy or strictly four momentum is enough and we do not need concept of mass in addition.

 

[SiennaTheGr8]

We need the concept of [conserved scalar commonly known as "total energy" or just "energy"], and we need the concept of [total energy as measured in a system's rest frame].

We are free to express these two quantities in whatever units we prefer. For much of the twentieth century, both quantities were variously expressed in units of energy ("total energy" / "rest energy") and units of mass ("relativistic mass" / "rest mass"). This was recognized as redundant and confusing, so eventually "relativistic mass" was mostly phased out, and "rest mass" is now usually just called "mass."

Of course, "rest energy" and "mass" are still redundant. For purely conventional reasons, "mass" is generally used. So now we've got a situation where the concepts [total energy] and [total energy as measured in a system's rest frame] are expressed in different units and with different words ("energy" vs. "mass").

This was confusing to me as a beginner, and anecdotal evidence tells me that I'm not alone. At the very least, I think it's beneficial to see some of the equations in special relativity with $E_0$ at some point. For instance, compare these two equations:

$E = \gamma mc^2$

$E = \gamma E_0$

Which of them immediately brings to mind the following equation for time dilation?

$t = \gamma t_0$

(where $t_0$ is proper time). As soon as you see the parallel here, you understand that kinetic energy is a relativistic effect in exactly the same way that time dilation is. That's a nice insight. Gets you thinking, too: yes, the equations are similar, but we've known about kinetic energy (to lowest order) for centuries, whereas differential aging is imperceptible even at what humans naively consider very fast speeds!

Once you understand the physics, none of this matters. And at that point, you can set $c = 1$ anyway, so energy and mass and momentum are all expressed in the same unit after all. This is about semantics and pedagogy (as was the "relativistic mass" debate).

 

[vanhees71]

I agree with you as for system of single elementary particle where mass M and (proper) rest energy Mc^2 are not distinguishable.

Say two photon system e.g. after electron-positron annihilation, has mass M or rest energy Mc^2. In COM each photon has energy Mc^2/2. I do not think each of the two photons has mass M/2 also. This is an example of my point that and energy or strictly four momentum is enough and we do not need concept of mass in addition.

For two photons, in the center-momentum frame you have (with $c=1$)
$$p_1=(|\vec{p}|,\vec{p}), \quad p_2=(|\vec{p}|,-\vec{p}).$$
Each photon has $m_j^2=p_j^2=0$ and the total energy is $\sqrt{s}=2|\vec{p}|$. To analyze this simple kinematics, we need to know that photons are quanta with mass 0. I don't understand, where your problem is or why you must make up one!

 

[DrGreg]

I totally agree that the norm of the four momentum of a particle or a system is a useful idea whether we call it mass, rest energy or invariant energy.

Well, "mass" has four letters whereas "rest energy" has 11 and "invariant energy" has 16, so that's one practical reason to prefer it as the name for this concept.

Also mass is a property of a particle or system, whereas energy depends on both the system and the observer. Mathematically, mass is $\sqrt{|\mathbf{P} \cdot \mathbf{P}|}$, a property of 4-momentum $\mathbf{P}$ alone, whereas energy is $|\mathbf{P} \cdot \mathbf{U}|$, where $\mathbf{U}$ is the 4-velocity of the observer (under the convention $c=1$). I think mass and energy are sufficiently different to justify different names.

Further justification are equations such as $\mathbf{P} = m \mathbf{V}$ or $\mathbf{F} = m \mathbf{A}$ which, as 4-vector equations, look just like the corresponding Newtonian 3-vector equations.

 

[sweet springs]

Other invariants under Lorentz transformation are proper time and proper length. Why we coin the word mass instead of proper energy ?

 

[DrGreg]

Other invariants under Lorentz transformation are proper time and proper length. Why we coin the word mass instead of proper energy ?

Because the word "mass" already existed from Newtonian physics, whereas there's no obvious Newtonian terminology that could be redefined to distinguish between proper time and coordinate time.

 

[sweet springs]

Proper time is a kind of time.
Proper length is a kind of length.
Proper energy is a kind of energy.
We decided to call proper energy mass in 20th century. The word mass still has its traditional meaning in equation of motion and gravity that would cause some confusion or misunderstanding as discussed in this thread and other. So no use of mass but energy is a simple way though I respect the history of physics. Mass is redundant concept. Energy or strictly four momentum is enough.

 

[Dale]

An example where mass cannot be substituted by energy

For example an electron at rest annhiliting with a moving positron in the frame of a PET system. By measuring the energies of the resulting photons you do not know the initial momentum. You must also know the mass as a separate piece of information.

Mass is redundant concept. Energy or strictly four momentum is enough.

Mass is not redundant with energy. They are separate and distinct concepts.

Of course, they are both parts of the four momentum, but they are different parts. Although the engine and the tires are both parts of a car, they are not redundant and you often must distinguish which one needs to be repaired.

 

[PeterDonis]

Proper time is a kind of time.
Proper length is a kind of length.
Proper energy is a kind of energy.

You're assuming that "time", "length", and "energy" are somehow fundamental concepts. They aren't. That's one of the lessons of relativity.

We decided to call proper energy mass in 20th century.

No, we discovered that the concept of "mass" as it's used in Newtonian physics doesn't work when you take relativity into account. (Newtonian physics doesn't have a concept of "proper energy" at all.) So we had to rearrange our concepts.

The word mass still has its traditional meaning in equation of motion and gravity

No, it doesn't. That's one of the key points that people are trying to communicate to you in this thread.

 

[SiennaTheGr8]

We decided to call proper energy mass in 20th century.

I'd say rather that the (somewhat-mysterious) quantity physicists called "mass" before the 20th century turned out to be nothing but proper energy.

On the other hand, the term "relativistic mass" came into fashion early on, so it seems that some physicists may have looked at the mass–energy equivalence in precisely the opposite way, as if what had always been called "energy" had turned out to be nothing more than a kind of extension of the mass concept. This seems backwards to me, but who am I to judge Lewis and Tolman et al.?

The word mass still has its traditional meaning in equation of motion and gravity that would cause some confusion or misunderstanding as discussed in this thread and other.

Only in the Newtonian limit.

That said, I do prefer "rest energy" (or "proper energy") to "mass," as I've made clear. It's only a preference, though I stand by my contention that many beginners could benefit from using $E_0$ sometimes.

 

[PeterDonis]

I'd say rather that the (somewhat-mysterious) quantity physicists called "mass" before the 20th century turned out to be nothing but proper energy.

I don't think the Newtonian concept of "mass", strictly speaking, corresponds to any concept in relativity. Much of the discussion in this thread is illustrating the problems with trying to draw such a correspondence.

 

[Herman Trivilino]

Say two photon system e.g. after electron-positron annihilation, has mass M or rest energy Mc^2. In COM each photon has energy Mc^2/2. I do not think each of the two photons has mass M/2 also. This is an example of my point that and energy or strictly four momentum is enough and we do not need concept of mass in addition.

All you illustrate with this example is that mass is not additive. That is the lesson of the Einstein mass-energy equivalence. The thing that we measure when we measure mass, whatever you choose to call it, is not additive. That additive property is part of the Newtonian approximation.

Instead of a two-photon system let's look at a two-electron system. The electrons move away from each other and it's possible to find a frame of reference where the electrons have identical speeds. We call this frame the center-of-momentum frame, but do not confuse this with a point in space that lies midway between the electrons. The relationship between the two is simply that this point is at rest in this frame of reference. This point need not be the origin of the spatial axes in this frame, for example.

Now to my point. Each electron has a mass $m$, the two-electron system has a mass $M$. Note that $M \neq m+m$. But they are equal in the Newtonian approximation. So it's not the very useful concept of mass that we need to reject, it's the often very useful, but also often very wrong, notion that mass is additive.

 

[Herman Trivilino]

Say two photon system e.g. after electron-positron annihilation, has mass M or rest energy Mc^2.

The relevant point being that prior to the annihilation the system had the same mass $M$ that it has afterwards.

 

[SiennaTheGr8]

I don't think the Newtonian concept of "mass", strictly speaking, corresponds to any concept in relativity. Much of the discussion in this thread is illustrating the problems with trying to draw such a correspondence.

I think it depends on what you mean by "the Newtonian concept of mass." If you mean a heuristic like "amount of matter" or "the measure of inertia," then I agree with you. These concepts have no straightforward counterparts in relativity. By "measure of inertia," one might reasonably mean rest energy, total energy, or even the matrix that relates the force and acceleration vectors.

But I was simply referring to the quantity $m$ that appears in Newtonian equations. That indeed turns out to be nothing but a measure of how much energy a system has as measured in its rest frame: $m = E_0 / c^2$.

 

[SiennaTheGr8]

Further justification are equations such as $\mathbf{P} = m \mathbf{V}$ or $\mathbf{F} = m \mathbf{A}$ which, as 4-vector equations, look just like the corresponding Newtonian 3-vector equations.

Of course, if you use $E_0$ instead of $m$ in the Newtonian approximation, you can still see a correspondence. Using Newton's overdot notation to mean a $ct$-derivative (so $\dot{\vec{\beta}} = \vec a / c^2$), and a little circle to mean a $ct_0$-derivative (so $\mathring{\vec{B}} = \vec A / c^2$, where $\vec{B} = \vec{V} / c$ is the "normalized" four-velocity):

$\vec{p}c \approx E_0 \vec{\beta} \quad$ (for $\beta \ll 1$)

$\vec{f} \approx E_0 \dot{\vec{\beta}} \quad$ (for $\beta \ll 1$)

$\vec{P} = E_0 \vec{B}$

$\vec{F} = E_0 \mathring{\vec{B}}$

(I'd use boldface for the vectors, but it doesn't work for $\beta$.)