Today a reader asked me *“Out of the quantum fields which have mass, do any of them also have weight?”* I thought other readers would be interested in my answer, so I’m putting it here. (Some of what is discussed below is covered in greater detail in my upcoming book.)

Before we start, we need to rephrase the question, because **fields do not have mass**.

### Mass and Weight of Particles and Other Objects

For ordinary objects ranging from particles to planets, the corresponding question is meaningful, but still it is a bit subtle. *“Out of the objects which have mass, do they also have weight?”*

Gravity is a universal force that responds to energy and momentum, and universal means applicable to every object and indeed to anything that has energy. *Because all objects have energy, all objects can have weight , and correspondingly all objects have gravitational mass.*

*NOTE ADDED: as a reader pointed out, the above statement was not written clearly. It should instead read: “Because all objects have energy, all objects have gravitational mass, which means they will have weight*

**if there’s some gravity around that can pull on them!**” An object out in the middle of deep space, in the absence of anything to create a noticeable gravitational force on it, will have no weight, no matter how much gravitational mass it has.(If this universality of gravity weren’t true, one could not think of gravity as a manifestation of curved space and time, as Einstein did. In the presence of gravity, objects without weight would act as though space and time were flat, while everything else around them would act as though space-time is curved. That would ruin Einstein’s whole idea!)

Yet even though all objects have gravitational mass, not all of them have **rest mass**. This leads to confusions that I address in the book, as in this paragraph:

*Here’s another strange thing. If you have read a variety of books about particles and mass, you will probably have noticed that some say that photons have mass and others say that they don’t. It’s hard to believe there could be disagreement about something so fundamental in nature. But the origin of the discrepancy is simple: it depends on which version of mass you’re asking**about.*[From*“Waves in an Impossible Sea”*.]

Photons have gravitational mass, like everything else. But they have zero rest mass, which is why they must always move at the cosmic speed limit *c*, 186,000 miles (300,000 km) per second.

An electron, by contrast, has both gravitational mass and rest mass. But one still has to be careful: an electron’s gravitational mass is usually larger than its rest mass, unless (from your perspective) it is stationary.

### Why Don’t Fields Have Mass?

A photon is a ripple in the electromagnetic field. An electron is a ripple in the electron field. More precisely, each is a quantum — a gentlest possible ripple — of its field. Since electrons have rest mass and photons do not, should we say that the electron field has mass and that the electromagnetic field does not?

No. That would be misguided.

It is true that some physicists will sometimes say, “**the electron field has mass**“. But they are using a potentially confusing shorthand when they do so. What they actually mean is: “**the quanta of the electron field have rest mass**” — i.e., electrons, ripples in the electron field, have rest mass — and that **“the electron field’s equations include a term corresponding to this non-zero mass.” ** The *field’s* rest mass simply cannot be defined; it is meaningless. Here’s why.

Rest mass is a measure of how difficult it is for you to move an object that is currently stationary. But the electromagnetic field, present across the entire cosmos, is not something that can move. It has no such thing as speed, and you can’t move it. It’s part of the cosmos. This is true of the electron field as well. Fields of the universe are not things to which you can attribute motion. Since both rest mass and inertial mass have to do with motion, neither type of mass can be attributed to these fields. Nor can the fields cause gravity the way objects do — they are everywhere, so they can’t pull objects in any direction, as gravity does, or feel weight, which would cause them to move in some direction or other.

Ripples in these fields are a different matter! They do move, and they carry energy and therefore can cause gravity. Quanta of fields definitely can have all possible types of mass, and they have weight. But the fields themselves do not.

So: objects, including all elementary particles (i.e. quanta of the universe’s fields), have weight and gravitational mass, and some have non-zero rest mass. However, the fields of the universe are not objects, and they have neither weight nor mass of any kind.

### Vacuum Energy-Density of fields

Despite this, quantum fields can have gravitational effects and intrinsic energy — more precisely, what is known as **vacuum energy-density**. Even when a field is sitting undisturbed, an effect of quantum physics causes it to be uncertain, creating a constant amount of energy in each spatial volume. This is true whether its quanta have non-zero rest mass or not. Vacuum energy-density contributes to the cosmological constant, and is potentially among the sources of the universe’s widespread “dark energy” *(which, despite the name, is in fact energy-density and negative pressure)*. Ordinary objects, including the elementary particles they’re made from, can’t have vacuum-energy.

Vacuum energy-density has potentially enormous gravitational effects on the cosmos as a whole. But you shouldn’t think of it as directly analogous to weight and mass, which are properties of localized objects made from electrons and other quanta. It is something quite different, with its own distinct effects. *[For instance, you might think positive vacuum energy-density, like the positive energy inside a planet, would cause things to fall inward; but instead it causes the universe’s space to expand. And while the rest mass of ordinary objects in empty space can only be zero or positive, vacuum energy-density can be negative.]*

I hope this somewhat clarifies how the properties of fields differ from the properties of their particles. It’s a very different thing to be spread out across the whole cosmos than to be a localized, movable object.

“Because all objects have energy, all objects have weight, and correspondingly all objects have gravitational mass.”

“(If this weren’t true, one could not think of gravity as a manifestation of curved space and time, as Einstein did. Objects without weight would act as though space and time were flat, while everything else around them would act as though space-time is curved. That would ruin Einstein’s whole idea!)”

1. Energy ~ Weight ~ Mass … = Energy ~ EMR (which includes the gravitational waves)

2. Gravity is “a manifestation of curved space and time”.

3. “Objects without weight would act as though space and time were flat, while everything else around them would act as though space-time is curved.”

4. “That would ruin Einstein’s whole idea!”

I’m a firm proponent of this premise, and submit to you, doesn’t this mean that the universe is NOT open (flat) but, indeed, CLOSED (spherical: bubble)?

I also submit to you that IF ( a big IF, indeed) the universe was a sphericak bubble then we can formulate “a theory of everything” based on “nothing” where this “nothing” is just “cuvred space and time”.

The invisible, yet real, the vacuum, “aether” (curved spacetime … universal bubble)!

PS: Here’s where I may screw my theory all up, :), the “fields” are boundaries of refraction and reflection of the unversal waves, which are created by variable density fluxes based the “bubble universe” hypothisis.

The universe could be flat on average and be curved locally. Indeed, that’s what observational evidence about the universe suggests — it seems to be flat on large scales, but fortunately isn’t flat everywhere since otherwise the Earth wouldn’t orbit the Sun.

If this is the case, then the following statement by Einstein and Infeld can’t be accepted: “Our world is not Euclidean. The geometrical nature of our world is shaped by masses and their velocity” (The Evolution of Physics etc, Touchstone, 2007, p. 235). Now the most recent “observational evidence” suggests that the universe is flat (Euclidean) on large scales and curved (“shaped by masses”) on small scales only. That implies that the Universe appears to be (again) open and infinite, in my opinion.

“Rest mass is a measure of how difficult it is for you to move an object that is currently stationary. But the electromagnetic field, present across the entire cosmos, is not something that can move. It has no such thing as speed, and you can’t move it. It’s part of the cosmos.”

If space can be curved (e.g., Mercury’s orbital precession “error”), and the curvature can change (e.g., 2 neutron stars merging), is there any “resistance” (inertia?) to changing curvature?

If so, does changing curvature “move” the electomagnetic field? Or perhaps this movement of the field is implicit in the geometry, and there is no apparent motion?

“… perhaps this movement of the field is implicit in the geometry, and there is no apparent motion?”

Yes seam to be suggesting what I was asking, the geometry of the spherical universe is one of the vital components to “create” a universe, but yes where does the motion (velocities, quantum oscillations) come from? There needs to be one more requirement the quantum oscilation, to maintain and create the visible universe otherwise it will tend to “rest” and vanish, puff.

But, here’s where the motions/oscillations may arise from, hopefully we can use AI to create a model that could shed more “light”. When the universe reverses and collapses it will do so in such a violent manner that there will be no time to null itself and vanish, instead, it will compress to, yet, another “singularity” and bounce back, another Big Bang. Is this why the speed of light is finite and so HUGE, is the speed of light prevent the universe from vanishing? Maybe?

We didn’t start the fire,

It was always burning,

since the world’s been turning …

– Billy Joel

I was under the impression that changing (gravitational) curvature results in the emission/generation of gravitons (& hence a reduction in the total mass-energy) So this should act as the “resistance”

Is my understanding correct ?

It is true that gravitational waves (presumably made from gravitons) are emitted by certain *changes* of curvature in a system. The Earth-Sun system emits them, for instance. I’m not sure I would want to think about that as resistance. When an atom drops from a highly excited state to its ground state, it emits photons; you could say that the emission of photons resists the electron’s highly excited orbit, but I don’t think that’s a wise use of the word. It is definitely a form of “dissipation”, by which the energy of the system is redistributed out to the wider world. But it’s not resistance in the sense of “inertia” or “mass”; it doesn’t appear in an (F=ma)-type equation.

One has to be very careful not to create confusion by letting multiple meanings of a single word (such as “resistance”) imply relations between topics that are in fact unrelated.

I think you’re imagining the changing of curvature of space as though that is motion… as though space is a slab sitting inside an even larger, fixed space. Space does not change position when it curves — it is that thing which *defines* what position means. And motion means a change in position… so space does not move, because it helps define what it is to move. (It’s like asking “how fast does time change.” Time defines what it means “to change,” so the question has no meaning.)

“Resistance” is the wrong term, but changes in curvature do involve energy. But there is no one answer to this question because changes in curvature can occur for many different reasons — waves in space, the effect of a star on space, the expansion of space — and each one requires looking at Einstein’s equations and doing a calculation.

“Photons have gravitational mass, like everything else. But they have zero rest mass, which is why they must always move at the cosmic speed limit c, 186,000 miles (300,000 km) per second”. (Strassler)

The question may be asked “who or what does the moving” as they “must move” If there is space where there is no gravity, then there is no motion so who or what decides when and where it will move? From a theological aspect, our answer is definitive, but in theory there is debate

The question may be asked. However, philosophy and theology texts are full of debates about questions that may be asked but should not have been. Photons are things that move; they are traveling waves, and traveling is what they do.

This question has been bugging me for a looooong time, can you please shed some light on it!

Space shuttle launch: I assume that a vast amount of fuel must be used up during the initial seconds of lift-off when the entire combo of shuttle, fuel tank & booster is at their heaviest. Why didn’t they put the shuttle on a 747 to give it a good initial velocity to start (around 1000Km/hr), saving some/all of the external tank & boosters?

Thank you professor.

I am no expert on rocket science! However, I would guess that a fully loaded shuttle with sufficient fuel for space flight would have been far too heavy for a 747 to lift; when they carted shuttles around on a 747, they were empty of fuel. Instead, the early designs for a shuttle had a rocket-powered plane to lift it partway instead of the booster rockets. https://en.wikipedia.org/wiki/Space_Shuttle_design_process I don’t remember what the problems were that led to them to use booster rockets and an external fuel tank instead.

Northrop Grumman Pegasus, but only for very small payloads. The space shuttle was intended to be used as a ferryboat, a reusable truck ferrying material and people back and forth. Too too complicated and will never be done again. Worked on it for 20 years.

Human spaceflight is a scam, a white collar welfare program to “pay the mortgages”.

Professor, what is your opinion on the space elevator as a way to get material and people to lower earth orbit?

It’s a balance of ease vs efficiency (and sometimes practicality). For example, a good launch site would be elevated (To avoid the dense lower layer of the atmosphere, a kilometer’s elevation would save a hefty chunk of fuel costs.) and near the equator (so Earth’s rotational speed can be given to the rocket.) Yet a lot of launch sites are where it’s easy to move people and cargo because that’s a big part of any project.

Launches are pretty extreme. An illustrative example of this is SpaceX’s Raptor engines were tested on a launchpad without a flame trench to divert exhaust. The force destroyed the pad, pelting local infrastructure with rocks and chunks of cement. Launching from a 747 presents some very big problems. Modifying the plane to handle such a load is just the start. Even with fuel savings you need a lot of oomph behind the rocket that your plane must work around somehow; a plan to either endure some of the stress. Plus something in case the launch needs to be aborted. With more moving parts involved there’s an increased risk something will go wrong and you lose the stability of the ground which often has millions of dollars’ worth of preparation involved when building a launch pad.

I wouldn’t say the challenges are insurmountable, but they must be balanced against doing something that works in the meantime and which is likely cheaper than the money needed to develop an alternate system.

Very small orbital rockets can be launched from large aircraft (see Pegasus and the former Virgin Orbit’s LauncherOne), but this is infeasible for anything larger.

In practice you need a ΔV of about 10 km/s to get to low Earth orbit (orbital speed is 7.8 km/s, but there are also significant gravity losses on the way there). A 747 can get up to 0.3 km/s. The rocket would still need to provide the remaining 9.7 km/s of ΔV, which wouldn’t let you cut down the size of the external fuel tank or boosters much.

A bigger problem is that neither a 747 nor any other airplane would do. The Space Shuttle stack has a mass of about 2000 tons shortly after liftoff. This is an order of magnitude larger than the amount of cargo any airplane can carry, not to mention the aerodynamic and structural problems of trying to fit a fully-fueled Space Shuttle, boosters, and external fuel tank on an airplane.

In its early days SpaceX was developing an air-launched Falcon 5. They abandoned that idea, finding that building a Falcon 9 and having the first stage return for a landing to be reused was far simpler and more efficient.

Matt,

Intriguing blog. I have to read it again. but a quick question. If you put an electron in a place which has ideal zero gravity, wouldn’t the mass create its own gravitational field and have self energy which can be interpreted as gravitational mass =self energy/c^2?

The gravitational mass of a *stationary* electron is not the energy of its gravitational field (or that of its electromagnetic field, which is much larger), but instead comes from its rest mass, which has a different origin. And an electron does not feel its own effects on the gravitational field; if it were the only source of gravity in its vicinity, it would still have no weight — though it would still have gravitational mass, which reflects its ability to pull on other objects gravitationally, if any such objects were to wander by, and to be pulled by them.

“a *stationary* electron … energy … comes from its rest mass, which has a different origin.”

Interesting, … What is this different origin?

I ask this in the context of how do these fundamental particles get trapped in their own small space? Is there a fifth force of nature that would give a simplier explanation of trapped waves? I presume this fifth force would have an influence well within Planck’s scale and be enormous to keep the electon stable for billions of years.

Is this what this g-2 experiment in Fermi Lab all about?

The mass of a charged particle should include the mass–energy in its electrostatic field (electromagnetic mass). Assume that the particle is a charged spherical shell of radius re. The mass–energy in the field is:

M(em) = ∫1/2 E^2 dV = (q^2)/(8*pi*Re)

which becomes infinite as re → 0. This implies that the point particle would have infinite inertia and thus cannot be accelerated. Incidentally, the value of re that makes M(em) equal to the electron mass is called the classical electron radius, which (setting q=e and restoring factors of c and epsilon(0)) turns out to be:

Re = (e^2) / (4*pi*epsilon*Me*c) = alpha * (ℏ)/(Me*c) = 2.8 x 10^-15 m

where alpha = 1/137 is the fine-structure constant, and (ℏ)/(Me*c) is the reduced Compton wavelength of the electron.

– Wikipedia (sorry, :-))

Here’s my question as a layman, what if, the point particle DOES have infinite inertia and, yes, it cannot be accelerated, it’s “fixed”, stationary? [“fixed” must be thought of in the context of the whole universe, bubble]

What if this fixed particles defining evey point in the universe is DARK MATTER?

What if the equivalent energy of these fixed points of “infinite mass” is DARK ENERGY?

What if these fixed points are the “aether”?

BTW: I asked the question whether the experiment at Fermi Lab, g-2, is basically to find the fifth force of nature that keeps the electon/muon/etc, stable? Could this fifth force be the fundamental force of nature that keeps these “fixed” points (aether) together like Blackholes?

PS: Prof. Strassler, will you be writting article on the recent work that won the 2023 Nobel Prize in physics? The faster stroboscope yet designed to “see” inside an electron?

Matt,

Intriguing blog. I have to read it again. But a quick question. If you put an electron in an ideal place with zero gravity, wouldn’t the mass create its own gravitational field and will have self energy and gravitational mass =E/c^2? Thanks.

Matt, thank you for explaining so clearly the difference between objects and Fields. It emboldens me to ask you a question which has puzzled me since I first read about quantum fields, probably in one of your earliest articles on this site: How can all fields occupy, simultaneously and co-extensively, every location in space? To a naïve mind like my own, that looks like an incredibly tight squeeze, yet that is how I am asked to picture the progress of, say, a free electron, progressing through space, continuously deriving its defining properties from these perpetual fields: its electron ‘nature’ from the electron field, its negative charge from the electromagnetic field, its speed falling short of “c” partially from the Higgs field, etc. (is there an etc?). And as for quarks, quark-field, electromagnetic field, Higgs, PLUS fields for the color charge and the strong force.

I’ve heard some physicists say the question is meaningless because fields are ‘mathematical entities’; but their ‘ripples’ are real enough, and fundamental to all the physical reality we know. Does that mean Pythagoras was right and all things are made of math? Or is it that we shouldn’t take exclusive occupation of local space for granted: is it only fermions that follow the Pauli Exclusion Principle, and fields are akin to bosons, happily cosying up together?

Any clarification would be much appreciated!

I’ll give you a quick answer here, but I cover it much more extensively in the upcoming book’s “Fields” section.

To some degree, the answer for the universe’s fields is that “we don’t know exactly”, since we don’t really understand what those fields are. They could arise in multiple ways.

But it is common for ordinary fields to occupy the same space at the same time. Air has many properties: wind, pressure, density. Each of these three properties is a field — it exists everywhere in the atmosphere and changes over time. Yet these three fields co-exist simultaneously across the atmosphere. How can they do that? They can do it because they are distinct properties of a single substance: air.

Another example: a magnetizable chunk of iron has both density and magnetization. Density tells you how many atoms there are in each little cube of iron. Magnetization tells you the degree to which those atomic spins are aligned. These are two distinct fields of the iron (with two very different types of waves) and they coexist everywhere within the iron.

As for the fields of the cosmos, might they, too, represent properties of some sort of medium? This is not known, though it is certain that if so, that medium cannot be anything like an ordinary medium [a point heavily emphasized in the book.] So we really can’t be sure.

But the idea that multiple fields could coexist in space is not unfamiliar; it happens all the time inside ordinary media. (An example of how electromagnetism and gravity can coexist is Kaluza-Klein theory.)

With this perspective, a lot of your later questions don’t have to be asked. A plane does struggle to fly through air, but it doesn’t have three separate struggles against density, pressure and wind, because those fields simply aren’t separate substances.

As to the right way to think about an electron and where its mass and charge come from and how it moves effortlessly through the fields of the universe, much of that is carefully discussed in the book.

Fields are artefacts that arise when we force a universe of three-dimensional scalar motion into a Cartesian reference system that can only represent a single dimension of vectorial motion. [[Abridged by host.]]

This site is devoted to explaining mainstream physics clearly, and should not be viewed as a venue for alternative approaches to physics. You’re welcome to offer theories like this on your own website, but please do not use this one as a forum. Thanks.

As you wish, Matt, but since when is Einstein’s equivalence principle an alternate approach to physics? I just argued that its field of application is obviously wider than is generally recognized. I hereby leave this forum, whose main purpose seems to be promote your forthcoming publication.,

Many thanks, Matt, for this quick but helpful answer: I’m very glad to hear you will address the issue of coextensive fields in detail in the book. As you tentatively use the term ‘medium’, qualified by ‘that cannot be anything like an ordinary medium’, are we moving towards a new, improved version of the long rejected theory of cosmic ‘ether’?

Matt,

All electrons have the same rest mass, similarly and respectively for all other types of particles.

I understand that the rest mass is determined by the quantum: is that right?

What determines the « size » of the quantum and/or rest mass pf each particles’ type.

Thank you.

About 2/3 of my book is dedicated to answering precisely these questions. If they had really quick answers, no book would have been needed… so I have to ask you to be a little patient until the book comes out in March.

I am still trying to understand your blog. Your statement, “Photons have gravitational mass, like everything else. But they have zero rest mass”. Does it mean that usual statement about equivalence of inertial and gravitational mass is wrong?

In terms of photons there’s an issue, since inertial mass involves acceleration -something that photons can’t do in the way massive particles can. If I recall correctly the gravitational mass of something differs from its rest mass, if it is moving or contains moving parts. The Equivalence Principle (Or more exactly the Universality of Free Fall or UFF) have been tested by numerous groups,coming up with essentially null results each time and I believe that’s where our confidence there stems from. But then I’m an engineer, pi is 3.1. I probably missed something.

Kudzu, inertial mass in general relativity can’t even be separated from gravitational mass; that’s the point of the equivalence principle. You can’t tell the difference between acceleration and uniform gravity. And yes, you can “accelerate” a photon. Acceleration of a near-light-speed particle, such as a proton at the LHC, changes its energy but barely changes its speed. In the limit of something traveling exactly at c, acceleration changes only its energy and leaves the speed unchanged. In short: acceleration of a photon increases its energy and frequency — and that is exactly the same as what happens to a falling photon, as must be the case by the equivalence principle.

No. Photons have inertial and gravitational mass. Rest mass is not the same thing as inertial mass.

Hello Matt,

I have a hard time to understand how the vacuum energy-density of a quantum field can have a negative pressure.

Can this be explained to a person with some basic knowledge in GR and QFT?

Yes, check out https://phys.libretexts.org/Bookshelves/Relativity/Book%3A_Special_Relativity_(Crowell)/09%3A_Flux/9.02%3A_The_Stress-Energy_Tensor — you don’t need every detail but the punchline is in equation 9.2.14, combined with the fact the energy momentum tensor of vacuum energy density is proportional to the metric g^{mu nu}, which has a +1 and three -1’s on the diagonal.

Thank you for the link. It explains that a cosmological constant (lambda-term) is equivalent to a negative pressure in the energy momentum tensor. But how can we understand that the zero point energy of a QFT corresponds to a lambda term with negative pressure? From a naive point of view I would have assumed that such an energy is attractive not repulsive. The textbooks always mention that there is discrepancy between theory and observation in the order of 120 magnitudes but they never give a hint why this energy is repulsive.

It’s not immediately intuitive. But it’s the only possibility allowed by special relativity.

First, forget about gravity and look at what the fields are doing.

A gas has positive pressure. A gas of photons even has p = rho (pressure = energy-density). But a gas of atoms or of photons (like the cosmic microwave background) breaks special relativity’s symmetries; there is a preferred rest frame. Only in that rest frame is the energy-momentum tensor diagonal; only for observers in that frame is there equal pressure in all directions and no overall flow of the gas. A gas is not Lorentz-invariant. You can see this in equations 54-56 of https://www.ruf.rice.edu/~baring/phys532/phys532_2023_lec_030823.pdf .

By contrast, whatever a quantum field does spontaneously within the vacuum must

preservespecial relativity’s symmetries. Otherwise special relativity would be completely broken, even in the absence of gravity, by the fields of the cosmos in their vacuum state. Experimentally, we know that’s not our world.That means the energy momentum tensor created by a quantum field’s fluctuations must be proportional to a matrix (really a “tensor”) that is left invariant under all special relativity transformations. Only the metric tensor, \eta_{\mu\nu} = diagonal{1,-1,-1,-1} has this property. If we set the energy momentum tensor proportional this matrix, it corresponds to positive energy density and equal negative pressure. As you can see from the above link, that’s the unique case where the metric can be diagonal and with the same form

in all inertial frames.Now, let’s add gravity. Einstein’s equations for a uniform gas imply that the curvature scalar R is proportional to the trace of the energy-momentum tensor, namely energy-density + 3*pressure. The three is important. Of course, a gas of ordinary material with positive pressure will gravitate normally: R > 0. So will a photon gas. What about vacuum energy? Now we find R < 0. If the energy-density is positive and the pressure equal in magnitude by negative, then R < 0 and we get completely different gravitational behavior from an ordinary gas. More precisely, in curved spacetime we replace the flat metric \eta_{\mu\nu} with the full metric g_{\mu\nu}, set the energy momentum tensor proportional to the metric, and solve, as in https://ned.ipac.caltech.edu/level5/Carroll2/Carroll1_2.html .

Does that help?

This is why it is so important that dark energy is not energy. Positive energy without pressure, or with positive pressure, is attractive but specifies a preferred frame of reference. Vacuuum-energy-density-and-negative-pressure leaves special relativity unbroken, and is unique.

That helped a lot. The missing piece was that the vacuum energy momentum tensor of a QFT must be proportional to the metric tensor. My textbooks do not mention this fact. Thanks for your explanation.

Matt,

OK! So the mass of an electron field is meaningless concept. Then why everyone puts an m(expt) term in the eq. for electron field. Ideally then one should put 0 in place of m? Why not just let interaction with Higgs field take care of that? But will that not make difference in Feynman diagram calculation unless you always include Higgs in the calculation? That may be enormously complicated. What do you think? Thanks.

Everyone puts “m_electron” into the equation for the electron field because they have to. An electron is a ripple in the electron field. Where else could the electron’s mass come from if it did not appear in the electron field’s equation?

Nevertheless, “m_electron” is the mass of a particle, not the mass of a field. For the field, m_electron changes its “dispersion relation” — the relation between frequency and wavelength for its waves.

An electromagnetic wave — a wave of the electromagnetic field — satisfies

frequency = speed/wavelength,

an equation familiar from beginning physics. But for a wave of the electron field,

frequency = speed * sqrt [1/(wavelength)^2 + constant]

where the constant depends on m_electron. You can verify this yourself, or look at https://profmattstrassler.com/articles-and-posts/particle-physics-basics/fields-and-their-particles-with-math/fields/ ; specifically, at the very end of the article, I discuss what I call Class 0 and Class 1 fields. Class 0 fields are familiar from 1st year physics, but the electron field (and indeed all fields whose particles have mass) are Class 1 fields.

The point is that the electron field *does* have a parameter which changes the field’s frequency-to-wavelength relation AND, in quantum physics, gives its quanta (i.e. electrons) mass. Nevertheless, this parameter

does notgive the field a mass, as such a concept would be meaningless.Some of this is discussed in conceptual detail, without math, in the upcoming book. Many of the technicalities are contained in the series of articles that include the one I cited above, which starts here: https://profmattstrassler.com/articles-and-posts/particle-physics-basics/fields-and-their-particles-with-math/

Evidently I’m a bit rusty after all these years.

I must say though, I’ve not ever heard of a photon being accelerated like that. It makes sense enough, but I’ve never had that perspective outlined before. An interesting little gem of information if ever there was one.

I’m trying to find more precise evidence for these statements in the equations. Acceleration in special relativity is always potentially confusing because there are different ways to ask the question; for example, the acceleration as experienced by an accelerating observer is one thing, while the acceleration of the accelerating observer

as seen by an inertial observeris another thing, and there are both “three-acceleration” and “four-acceleration” versions of the concept, where the latter is a Lorentz vector. But you understand the logic; whatever happens to extremely fast electrons must smoothly connect with what happens to photons of the same energy, and fast electrons definitely have gravitational mass. This is also why photons’ paths, like fast electrons’ paths, must bend when crossing a non-zero gravitational field.That said, I have not taught GR and I am not a leading GR expert, so even though I’m 99.9% confident in my logic, I’d like to see a real GR expert make this statement along with equations to prove it.

Dr. Strassler:

I would like to check my understanding, based on the contents of this blog. I have always understood the following: the quantum fields (the fundamental fields) permeate the entire universe, and coexist together…electron field, quark field…etc. the fields themselves have zero energy & zero mass (or at least the fields hover about at zero energy). When something shoves some energy into a field, let’s say the electron field, that excitation of the field, the electron has mass….not the field. The act of shoving energy into that field creating an electron is what gives the electron its mass. Now, the purpose of the Higgs field is to say that the fields that interact with the Higgs field, require a certain amount of energy to create the “particle” rest energy….and any extra energy manifests as kinetic energy. Photons do not have any rest energy, the electromagnetic field does not interact with the Higgs field, any energy shoved into that field must immediately go into kinetic energy required by moving at the speed of light at some wavelength.

So, any fields that interact with the Higgs field, demand a minimum amount of energy to create the particle, because it has rest energy. Is this somewhat on track? Also, I preordered your book from Amazon.

This is basically right. What you’ll get from the book, I hope, is a more coherent, organized and clear vision of what you’ve just written. But you’ve already understood many of the main points.

There are lots of very useful comments for this article, It is great..