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174 Responses
After re-reading your chapter on mass, a surprising thought occurred to me:
A stationary, isolated electron wavicle will have a mass that will affect space-time in a negligible way in that frame. Yet if it’s observed by an observer travelling at a high enough velocity approaching the cosmic speed limit c: won’t this lead the moving electron to increasingly warp space-time via its gravitational field?
What you ask is an issue for any object, not just elementary particles. You could ask the same thing about stars.
All objects warp space via their gravitational field. And indeed, the effect of that warping looks different to an observer moving at a high speed relative to that object. The effect is similar to what happens to the electromagnetic field of an electrically charged object: thanks to length contraction along the direction of motion, it becomes planar instead of spherical. https://www.researchgate.net/profile/Eugene-Stefanovich/publication/355442704/figure/fig46/AS:1100331211591681@1639350812903/Electric-field-lines-around-a-charge-at-rest-b-charge-moving-at-a-constant-speed.png
The gravitational field of a high-speed particle similarly becomes a planar gravitational shock wave, known as an “axisymmetric vacuum pp wave” — axisymmetric meaning that it is symmetric around the direction of the particle’s motion, but not spherically symmetric. https://en.wikipedia.org/wiki/Aichelburg%E2%80%93Sexl_ultraboost .
Note this planar structure is completely different from the gravitational field around a black hole, which is not planar, and (if the black hole is not spinning) is spherically symmetric.
Greetings Dr. Strassler! I’m curious what the world would be like if the mass of the Higgs boson were higher, without changing the vacuum expectation value of the Higgs field. Would this have just made the Higgs boson harder to find without really affecting any other processes?
To a point, yes, but only to a point. If its mass were five times or so greater than it is, then things would be more or less the same. But the higher the Higgs boson’s mass, the stronger the Higgs force between W and Z bosons… something that hasn’t been measured yet, because in our world it turns out to be weak. Had the Higgs boson been seven or more times greater than it is, then the Higgs boson really wouldn’t exist as a particle at all, and in its place would be very strong interactions among W and Z bosons. Those interactions wouldn’t have been either unique or easily predictable, and so we would have had to measure them directly in order to start to understand them.
This is why the LHC was a “no-lose” machine — it was guaranteed to make discoveries, if it worked properly. Either one or more Higgs bosons had a low-enough mass to be discovered directly, or the whole Higgs field idea would have been replaced with a sort of composite of other fields, bound together through new strong forces that would have been observed (albeit with difficulty) in the behavior of W and Z bosons. Or quantum field theory as a whole was going to break down. Or something else dramatic. There were a lot of possibilities as to what might happen, but business as usual wasn’t one of them, and it would have been very exciting to figure out what was going on.
In the short term, a single low-mass Higgs is the least exciting possibility. Someday it may turn out to be the most exciting, because of the puzzles it poses https://profmattstrassler.com/articles-and-posts/particle-physics-basics/the-hierarchy-problem/
Waves In An Impossible Sea is, hands down, the best book on physics for the layman that I’ve ever read. After subsequently reading The Elegant Universe (Brian Green), which I also love, I’m now reading Waves a second time.
I have a very basic question regarding e=mc2 , that I haven’t gotten an answer for anywhere else.
What does the speed of light (squared) have to do with the equivalence of mass and energy? I get the idea that mass can be converted into energy and vice versa. I just don’t understand why the speed of light squared is the conversion factor. Is it just because it’s a really big number? If so, could Einstein have chosen some other really big number? I’d be very grateful if you would help me understand. Thank you.
That’s high praise indeed! thank you!
Now, about the speed of light — your question is a common one and a very good one, and it deserves a longish, careful answer. I don’t think I’ve ever answered it on this blog, so I think I’ll try to present the answer in a blog post. To give you a preview, let me quote something from the book’s Chapter 2:
“It’s well-known that light has a characteristic speed, which scientists call c ; this is the speed at which each individual photon travels, too. As scientists discovered centuries ago, c is about 186,000 miles per second. That’s fast, in a way. Our fastest spaceships don’t come anywhere close to that speed. Though my last car was with me for fifteen years, I drove it less than 186,000 miles. At the speed c , you could circle the Earth in a blink of an eye (literally) and travel from my head to my toe in a few billionths of a second.
“And yet c is also slow. It takes light more than one second to travel to the Moon, over eight minutes to reach the Sun, and over four years to reach the next-nearest star. If we sent off a robot spaceship at nearly c to explore the Milky Way, it could visit only a few dozen nearby stars during our lifetimes.
“You and I are small, so we think light runs like a rabbit. But the universe is vast, and from its perspective, light creeps like a turtle.”
What I’m emphasizing in these paragraphs is that light’s speed is only fast from our perspective. Our daily world involves slow speeds. Why? Similarly, the fact that the energy in ordinary objects seems so large to us is also a matter of perspective. And so the real question is not about the conversion factor between matter and energy being a “really big number.” The question is: why does the conversion factor seem to us to be so big? In other words, why are the energies of processes in ordinary life so small compared to the energies stored inside of ordinary objects?
This has a very interesting answer which has everything to do with particle physics and its details, but I’ve never written it down carefully and will have to think about how to do it well.
Am I right in thinking that your conversation with Julian Collins highlights the fact that frequency, like energy, is best thought of as part of a four vector: the four wave vector?
There’s a nice article on mass and energy you wrote that highlights the pitfalls when thinking about energy alone rather than combined with momentum: https://profmattstrassler.com/articles-and-posts/particle-physics-basics/mass-energy-matter-etc/mass-and-energy/
“To understand mass and energy, we need to put momentum into the mix, and discuss the differences and relations among these quantities.”
I find it astonishing that the four wave vector is related to energy-moment via Plancks constant: https://en.wikipedia.org/wiki/Wave_vector#In_special_relativity.
Yes. (energy, momentum) form a four-vector, and so do (frequency, wave-number), where “wave-number” points in the direction of motion and has magnitude equal to 1/wavelength. For a quantum — that is, the wave of smallest amplitude for a quantum field — the two vectors are indeed related by Planck’s constant.
They are both also conjugate to the four-vector (time, space); energy is the quantity conserved because of the time-independence of physical laws, and momentum is the quantity conserved because of the space-independence of physical laws.
Hi Matt. Consider a particle with rest mass. The energy of a particle is given by h x frequency. So in a frame where it is moving it must have a higher frequency than a frame in which it is stationary, as it now has motion energy as well as mass energy. However, time dilation in Special Relativity tells us the period of the vibration as measured by the observer relative to whom the particle is moving will be longer, so the frequency will be smaller. Please can you explain this apparent contradiction.
Didn’t you ask this question already? I was planning to answer it, but it will be a bit long — in fact I’m thinking of answering it in a blog post, because your question is a natural one. But you’re making a very common mistake about how to think about relativity and waves. While you wait for my answer, consider this: suppose you asked your question with the electron stationary and the observer moving. Then it would be the observer’s clock that runs slower, right? So then, as the observer monitors the electron using their clock, the electron would seem to vibrate **more often** than to a stationary observer — in other words, the frequency would be *larger*. Now, the fact that I can run the argument in the opposite order and get the opposite answer is a classic indication that you’re being too naive about relativity — for instance, maybe you’re thinking about time, but not remembering that things happen to space too. More specifically, you know that an electron is a standing wave if it is stationary, but somehow our moving observer must see that standing wave as a traveling wave, with crestst and troughs and a wavelength. So how is that going to work? Think about it, and I’ll try to put an answer together in a few days (I have some travel of my own and a deadline, so we’re probably looking at early next week.)
Thanks Matt. I have considered some of the points you make but just can’t quite come up with a proper solution. In relativity you always have the situation where two inertial observers moving relative to one another both see the other’s clock as ticking slowly, so you have to analyse the problem with respect to the observer you are interested in, and it ends up all being consistent. With the identical twins problem, it is the acceleration when the first twin comes back to the home planet that resolves the problem, explaining why the travelling twin finds the stay at home twin long dead when he returns to the home planet. It all makes sense to me relative to the inertial observer who is stationary wrt the particle, but I can’t understand how to analyse it wrt to the inertial observer moving wrt the particle. As you say, it seems he will see a sort of travelling standing wave. I did ask the question before but did it in the wrong way as a reply to one of your answers to another reader, so I thought I had better do it correctly!
Hi Matt. Just a quick thought I had about the cosmological constant problem. When we consider a stationary particle as a standing wave, it is only the lowest frequency standing wave that represents the stationary particle. Using the guitar string analogy, it is only the fundamental frequency that occurs, which I call the first harmonic. We do not get a stationary particle with double the rest mass, represented by the second harmonic with double the frequency, nor a stationary particle with treble the rest mass, represented by the third harmonic with treble the frequency etc. So it seems that it is a property of relativistic quantum fields that only the fundamental standing wave occurs. So surely it is natural to expect that it is only the fundamental frequency standing wave that will occur with the zero point vacuum vibration, with no higher harmonics occurring. Without the higher energy higher harmonics with their higher frequencies, surely the cosmological constant problem would be largely eliminated, with empty space only obtaining a little energy from the zero point vibration of its quantum fields.
Another great question, and this one may also be deserving of a blog post — but it has a quicker answer than your relativity question, so I’ll give it to you today.
If you pluck a guitar string, you get the reasonance frequency. Yes, you can get a countable set of higher harmonics too. These are the set of resonant frequencies at which the string naturally vibrates. They are a tiny subset of all possible frequencies.
However, if you connect the guitar string to an electronic resonator or some other machine that can push the string back and forth continuously, it can force the string to vibrate at any frequency you like. All possible frequencies can be accessed.
Quantum physics does not merely excite the resonant freuqency of a string, or, in the cosmic context, of a relativistic quantum field. It excites all of them, with all possible frequencies and wavelengths, including crazy unphysical ones that would have gigantic rest mass or imaginary rest mass. All of these frequencies and wavelengths contribute to the cosmological constant, and produce the huge effects in calculations. (They also affect the Higgs boson’s mass and create the hierarchy problem. https://profmattstrassler.com/articles-and-posts/particle-physics-basics/the-hierarchy-problem/ and https://profmattstrassler.com/articles-and-posts/particle-physics-basics/the-hierarchy-problem/naturalness/ )
The resonant frequency of a guitar string or a field is special because it does not need quantum physics (or anything else continuously present, like a resonator) to excite it. It can even be excited classically (i.e. without quantum physics) with a brief disturbance, as with the plucking of a guitar string.
For example, the electron field’s resonant vibrations are electrons; in the rest frame of an electron where the electron is a standing wave, its frequency and rest mass are exactly proportional. If it’s moving and thus a traveling wave, the details of the wave are set by relativity (as you’ll see soon.) If you do something striking to the universe, such as in a proton-proton collision at the Large Hadron Collider, you may make one or more electrons.
But all possible frequencies and wavelengths, including ones whose rest mass is huge, zero, and imaginary, contribute to the cosmological constant (and the Higgs boson’s mass.) That’s a purely quantum effect.
This is in every way linked to my pet peeve about using the word “particle” in the phrase “virtual particle” (https://profmattstrassler.com/articles-and-posts/particle-physics-basics/virtual-particles-what-are-they/). “Virtual particles” refers to all the disturbances of a field, including all possible wavelengths and frequencies. They are not particles. The particle is the one that (when at rest, as a standing wave) would vibrate with the field’s resonant frequency. That’s the only one that can travel across the universe or an accelerator hall as a coherent whole, and arrive with a “ding!” in a particle detector.
Hi Matt. I am familiar with applying quantum mechanics to the simple harmonic oscillator. You use the Schrodinger and it can be solved exactly, with the result that the minimum energy is 1/2 x h x resonant frequency. So in this case, quantum mechanics tells us that zero point energy only involves the vibration with the resonant frequency. Is zero point energy for a relativistic quantum field analysed in the same way, using the Schrodinger equation? Why is it so different from the quantum harmonic oscillator, with non- resonant frequencies contributing to the zero point energy – does this come out of the maths? When you say all possible frequencies contribute to the zero point energy, not just resonant frequencies, do you mean all frequencies from zero to infinity? It seems very different to what we learn from the quantum harmonic oscillator, where only a single resonant frequency produces the zero point energy. Sorry for so many questions but your book was amazing, and has really motivated me to understand this in as much detail as I can.
Okay, at this level of detail, we need formulas, some of which are found on this website.
Even in the quantum mechanics case, you have two choices: you can say the zero-point energy just involves the resonant frequency, or it involves a sum over all frequencies (and the sum just happens to be finite, in which case it has no choice but to be proportional to the resonant frequency since that’s the only quantity in the problem — the only real calculation is the 1/2 .) In the first case, you view it as due to the fact that position and momentum are relatively uncertain (i.e. don’t commute as operators). In the second case, you take a path integral viewpoint and calculate the sum as an integral — but there are a lot of subtleties with that, and I can’t quickly find a website that does things that way.
Instead, I’ll use the first method, which is what you’ve really asked about anyway, and point you to this article
https://profmattstrassler.com/articles-and-posts/particle-physics-basics/quantum-fluctuations-and-their-energy/zero-point-motion/
and the several that follow it. This takes the point of view that you want, in which each harmonic oscillator has a zero point energy proportional to its frequency. But the thing you need to keep track of is that — looking way ahead to a late article in the series — a free field is equivalent to an infinite number of coupled oscillators, and the rest mass is the resonance frequency only of the lowest one… a point that will only become clear in the mathematical asides of the fifth article in the series:
https://profmattstrassler.com/articles-and-posts/particle-physics-basics/quantum-fluctuations-and-their-energy/zero-point-energy-of-fields-part-2/
So be patient and keep going, the answer is there.
Meanwhile, the answer I gave you today refers to the other method, in which we do integrals over all possible frequencies and wavelengths. I probably shouldn’t have given that answer since it is less familiar; it uses Feynman’s path integral and propagator method instead of the standard Schrodinger/Heisenberg methods of quantum mechanics class. At some point I’ll try to find time to show you how the two methods are connected.
Hi Matt. It was a long time ago that I studied physics at university, and I haven’t studied it since but have maintained a great interest, so please excuse if I have got this completely wrong! When we use the Schrodinger equation to analyse the simple harmonic oscillator, the lowest energy energy eigenfunction vibrates with a frequency equal to half the resonant frequency of the equivalent oscillator treated classically, and this is the lowest frequency of vibration of any of the energy eigenfunctions. Let’s call this frequency f0. The zero point energy is then obtained very simply as h x f0 (not 1/2 h f0 as f0 is half the resonant frequency). None of the frequencies of the higher energy energy eigenfunctions are used in calculating this zero point energy. So wouldn’t it be natural in relativistic quantum field theory to assume that it is only the frequency of the lowest frequency vibration of the field that should be used in calculating the zero point energy. So for the electron field, in calculating it’s zero point energy, we would take the field’s lowest possible frequency of vibration, obtained from the electron’s mass, and multiple this by h in order to get the zero point energy. This would avoid the massive energy that results from using the frequencies of all the possible higher frequency vibrations in calculating the zero point energy, and hence could help in solving the cosmological constant problem, where clearly something is massively wrong. Now I wait to hear why what I have suggested can’t possibly be right!
So, in the approach you are taking, the mistake is this: the resonant frequency of the field tells you the rest mass of the particle because it tells you about the field’s standing waves. But there are also traveling waves, corresponding to the moving particle. You’re only accounting for the zero-point energy of the lowest-frequency, standing wave mode; but there are a continuously infinite set of higher-frequency traveling wave modes, each of which has zero-point energy also. That’s where the infinite zero-point energy comes from.
In the set of articles I linked to in my previous reply, I put the field in a box, turning its continuously infinite set of traveling waves into a discretely infinite set of standing waves. So the calculation looks a little different in detail from the one I just described here. But as the box’s size becomes infinite, the two calculations of the zero-point energies become identical.
Hi Matt. Thanks for your help on this. It seems to me that what we would ideally like to do to calculate the zero point energy of a quantum field is exactly what we do with the quantum harmonic oscillator – calculate the Hamiltonian of the quantum field and then use the Hamiltonian to calculate the energy eigenfunctions of the quantum field and their respective energies, using the Schrodinger equation. Is it possible to do this calculation?
This is essentially what I did within the link I sent you:
https://profmattstrassler.com/articles-and-posts/particle-physics-basics/quantum-fluctuations-and-their-energy/zero-point-energy-of-fields-part-2/
But if you want it laid out technically, here it is, done by my friend David Tong, professor at Cambridge and one of the greatest masters of pedagogical presentations in quantum field theory and beyond. https://www.damtp.cam.ac.uk/user/tong/qft/two.pdf
You can skim the first few pages, as they will be familiar. Then you want the derivation of equation (2.26), which starts at (2.18). But I imagine you will want to keep reading into later sections.
Also, note that Tong uses the operator (Heisenberg) formalism, not the Schrodinger equation. Almost never is the Schrodinger equation used in quantum field theory, and certainly not in this context, because the wave function is now a wave functional — essentially, a function of a continuously infinite number of variables — and the technicalities of working with such things is extremely hard. It could be done for a free quantum field, but once fields interact with each other, the wave functional becomes challenging to use.
Thanks very much for that Matt. I’m afraid that the maths is beyond my current level but I am satisfied with knowing that it can be done and that it leads to this huge zero point energy. I have been able to work through all the maths in your derivations of the zero point energy both for massless and massive bosons, and that together with knowing that it also can be calculated using the Schrodinger equation, or an equivalent formulation of quantum mechanics – is good enough for me.
Hi Matt. Just been thinking a bit more about zero point energy! If we consider massive particles, say a Z particle, and consider not the standing wave given by the fundamental resonant frequency where the whole field vibrates with no crests or troughs, but the one with the next highest frequency, the one with exactly one wave crest, then I presume that if we add a quantum of energy h x frequency to the zero point mode of this vibration, then we will produce a stationary particle with energy = h x frequency, with this frequency being higher than the resonant frequency that corresponds to the Z particle. So we will produce a stationary Z particle that is more massive. Is this observed or are the energies required too high to be produced in the LHC? Also I thought that the only way you got a higher frequency was for a travelling wave representing a moving particle, where the higher frequency gives you the motion energy plus the rest mass energy. But now it seems we also have higher frequencies for stationary particles?
If you really did put the universe in a box, then yes, for each type of particle that we have normally, you would have an infinite tower of particles of increasing mass, with the familiar one at the bottom. (These are known as Kaluza-Klein states; something similar, where the waves are standing in some dimensions and traveling others, arise if you make a smaller number of dimensions finite, as we do when considering that the universe might have extra dimensions. https://profmattstrassler.com/articles-and-posts/some-speculative-theoretical-ideas-for-the-lhc/extra-dimensions/how-to-look-for-signs-of-extra-dimensions/kaluza-klein-partners-why-step-1/ ) Of course these particles are not moving in three dimensions; they can’t go anywhere. You’ve traded the freedom of motion of one particle for an infinite tower of motionless particles.
However, as you take the box away, these particles of various masses (and thus energies) merge together and become a continuum of states with a continuous range of energies, rather than an infinite discrete set. The traveling waves of the original particle are in this continuum, as superpositions of the states in the tower (using sums and differences of sin and cosine waves.)
This is a classic example of why quantum field theory is so very tricky. There are continuous infinities everywhere, and it’s easy to misinterpret them. Putting the theory in a finite box, or putting it on a lattice of points instead of a continuous space, or both, are among the classic tools of physicists to remove or control some of those infinities and avoid making mistakes. I used the finite box trick because it makes the zero-point calculation easiest.
I could instead have divided the universe into an infinite set of equally sized boxes and only considered states that look the same in each of the infinite number of boxes — then you get a discrete set of traveling waves, instead of a discrete set of standing waves. That’s harder for non-experts to visualize, but it is just as easy mathematically, and gives the same answer.
Any trick you chose, if you use it correctly, will give the correct answer when you take the box away. But don’t confuse the details that appear in a trick for something physical; this infinite tower of stationary particles is just an artifact of a calculation, not a part of the physical world, which has no such box.
Hi Matt. Just one minor point that arose when I worked through the maths of you zero point energy calculation. In the fourth article in the section titled ” The Effect of a Resonance Frequency : The Details”, when I plug in a sinusoidal solution into the type 1 wave equation, I get the same formula for q except the bracket with the fo has a 2 inside. I think this is because I am using for f in the first equation in this section f = c/2Lu whereas you are using f = c/Lu. Clearly such small numerical differences are irrelevant to the argument given the huge numbers involved, and this is why I think you have used f = c/Lu, but am I right in thinking that the lowest frequency standing wave, with only one wave crest, has a wavelength = 2Lu and hence an exact frequency of c/2Lu. Hence, if we use f= c/2Lu the bracket with the fo should have this 2 inside.
There could certainly be factor of two errors in the derivation; no one has checked them carefully. I’ll take a look.
Hi Matt. Just a general question about how the Schrodinger equation is used in quantum field theory. You explained to me that it is in fact hardly ever used in quantum field theory, and instead operator (Heisenberg) formalism, with which I am not familiar, is instead used. I know that the Schrodinger equation is not Lorentz invariant, so even when it is used in quantum field theory, can it only be used when all wavicles are moving at speeds small compared to the speed of light? Is the operator (Heisenberg) formalism Lorentz invariant, so that it can be used for fields whose wavicles are moving at any speed?
No, it can still be used in a relativistic theory. You can always choose a time coordinate and use the Schrodinger equation relative to that choice.
But the problem is more general. A single particle has a position that consists of three variables, making the wave function a function of time t and the particle’s position x, y, z — four variables. Manageable. N interacting particles would have time t and N particle positions x1, y1, z1, x2, y2, z2, … etc — a function of 3N + 1 variables. Barely manageable when N becomes much bigger than 1. But a field has a configuration F(x,y,z) at each moment in time, and so now your wave function is a function of time t and the function F(x,y,z), which is specified at an infinite number of points. Your Schrodinger equation, therefore, is an continuously-infinite-dimensional differential equation. Not manageable.
So even in a non-Lorentz-invariant quantum field theory (such as you might use to describe a metal near a phase transition), the Schrodinger equation is still generally useless.
Hi Matt. I just want to check that some of my thoughts about the quantum field theory of the Standard Model are correct. In your blogs you discuss equations of motion for the various fields. Am I right in thinking that the Standard Model is completely specified by giving the equation of motion of all the fields contained in the Standard Model, or are other equations also needed? Clearly the equation of motion for a given field will have terms where the given field is multiplied by another field with an interaction parameter in front, with these terms representing all the interactions that the given field has with other fields. I understand that if a field has a wave of frequency f, then the corresponding particle has energy given by E = hf. But you explain that virtual particles are more general “disturbances” in the fields. How is the energy of these more general disturbances calculated ? – do they have a particular frequency and then you apply hf, or is there some other way that their energy is calculated?
In general, quantum theories are not specified by the classical equations of motion. There are subtle effects that these equations do not capture. The Standard Model does have such effects (an example is a tunneling effect, referred to as “instantons”, whose effects depend on a parameter that does not appear in any equation of motion; another is renormalization, whereby an apparent constant parameter in an equation of motion turns out not to be meaningful in the quantum theory, and instead is merely representative of something physically relevant that depends on energy.)
Instead the theory is specified by a Hamiltonian and a vector space of quantum states, called a “Hilbert space”, or in the more common formalism of a path integral, by an “action” (an integral of a Lagrangian over spacetime) and a “measure” (that tells one about a seperate integration in which one integrates over over all the fields). There are many profound subtleties that arise in defining these objects, some of which have observable physical consequences.
So while you are right that the equations of motion can be written and are of the form you described, I’m afraid that the math here is enormous. The equations of motion are now operator equations, i.e. equations for continuously-infinite dimensional matrices in the Hamiltonian formalism, or equations relating infinite sets of integrals in the path integral formalism. After all, while a classical field is just a simple function of space and time (such as the electric field in first-year physics), a quantum field is a vast generalization of this idea. In those infinities, many subtleties can be hidden, these are of relevance to physics.
Finally, the energy of disturbances: the simplest statement (overlooking some subtleties) is that a real particle has E=hf and has E^2 – (pc)^2 = (mc^2)^2, where p is momentum and m is rest mass, while a general disturbance (aka virtual particle aka Green function) has E=hf and E^2 – (pc)^2 not equal to (mc^2)^2. See https://profmattstrassler.com/2022/09/09/protons-and-charm-quarks-a-lesson-from-virtual-particles/ for some additional insight.
Hi Matt. I understood from your website that the quantum fields of quantum field theory satisfied class 0 and class 1 equations of motion, at least in the case when fields are small so that interaction terms involving products of fields can be ignored. It was from the class 1 equation of motion that you calculated the mass of the particle at rest. In these equations, you treated the fields as just normal functions of space and time, and differentiated them with respect to space and time. Are you saying that these equations do not really appear in the standard model?
Hi Matt. Can you just confirm that the Class 1 equation that you give in your section “Waves (Quantum)” is the exact equation of motion for a particle with mass that is not interacting with any other field. Taking the electron as an example, can you let me know how the electron field (Z in your equation) is described mathematically. Is it a complex function of space and time or is it a real function of space and time? Does it have only one component, or does it have several components each of which obeys the Class 1 equation?
The electron field is a little more complicated. It has four complex components. Classically, each obeys the class 1 equation (if it is not interacting with anything else) but the four together satisfy a more complicated constraint, through what is known as the “Dirac equation”.
But that’s still speaking classically. Remember that in quantum physics, fields (and anything else) do not *satisfy* their equation of motion. They only do so approximately, and one must really introduce the full power of quantum math to understand how they behave.
Hi Matt. I’m afraid my precise knowledge of quantum physics only extends as far as the Schrodinger equation for the wave function, which as I understand it is exactly satisfied. I understood from you website that the electron field does exactly satisfy a Class 1 equation, providing it is not interacting with any other fields, but I now understand that this is only an approximation. Could you suggest the best introductory text to the mathematics of quantum field theory so that I can try to understand this precisely. I need something that starts from the Schrodinger equation and takes it on from there.
Okay, I can see what your problem is here. You’re getting confused in a very natural way, common among physics students, so let’s take a step back.
1c) In classical physics, a ball on a spring satisfies a simple ordinary differential equation, m a = – k x (where m is mass, a is acceleration, k is the spring constant and x is position.) Or we can compute this using energy, where the kinetic energy is 1/2 m v^2 = p^2 / 2 m (p being momentum) and the potential energy is 1/2 k x^2
1q) In quantum physics, a ball on a spring satisfies no equation, but is described by a wave function Psi(x,t) which does satisfy an equation: the Schrodinger equation with kinetic term D^2/2 m (where D = d/dx) and the potential energy is 1/2 k x^2 as before.
Notice that in (1c) x(t) satisfies an equation, while in (1q) x(t) does not satisfy any equation; instead the wave function Psi(x,t) satisfies an equation of a more complex sort.
If we had two balls on two springs, with positions x1 and x2 and potential V(x1, x2), then in the classical case x1(t) and x2(t) would satisfy their equations of motion (plural! one for x1 and one for x2), but in the quantum case neither would do so, and instead there would be one wave function Psi(x1, x2, t) satisfying one equation (singular!): the Schrodinger equation containing d/dx1^2, d/dx2^2, and V(x1, x2). Even the number of equations is different now.
(2c) A classical field satisfies a wave equation; you know the example of the electromagnetic wave equation, or of a wave on a string. If the field is F(x,t), then the equation is something like (d/dt^2 – c^2 d/dx^2) F(x,t) = 0 (or maybe the right hand side is more complicated.) Actually, if we think of F(x,t) as F[x](t) as a variable, then this partial differential equation can be viewed as an infinite set of coupled ordinary differential equations acting on the time-dependent variables F[x].
(2q) The classical field F(x,t) satisfies no equation. Instead, there is a wave function Psi(F[x], t), where Psi is now a SINGLE function of an infinite number of variables. This is now an infinite-dimensional partial differential equation. There is no agreed upon notation for such a thing because mathematicians generally do not deal with this at all, and certainly it is not taught in undergraduate classes. It would be something like
\integral dx (dPsi(F[x],t)/dx)^2 + V(F[x])Psi(F[x],t) = i d/dt Psi(F[x],t)
but this is a formal definition which would not actually give finite, useful answers if you tried to make sense of this and solve it naively — except in the case where the field is free.
So what I have done in my notes is describe (2c), not (2q), and then used simple facts about the solutions in (2c) to infer simple solutions of (2q), just as one can do when going from electromagnetic waves to photons, as Einstein did. Einstein did not need to work through the Schrodinger equation of 2q, and neither do we, if we just want to learn something about quanta and how they scatter off one another.
Thanks for the clarification Matt. If the quantum field of say an electron does not obey any equation of motion, then how do we assign a frequency of vibration to it and make statements like the rest mass of the electron is h x minimum frequency of vibration of the field. My whole intuitive understanding of particles as vibrations of fields where the minimum vibration frequency of the field determines the rest mass of the particle seems to be breaking down! Is it best just to concentrate on the maths and say that there are certain mathematical procedures, which are specified as part of the theory, that allow you to extract, for example, the rest mass of the electron. I hope that’s not the case as the vibrating field idea, with its musical analogies, was so appealing!
Here you are expressing mystification about quantum field theory, whereas exactly the same things happen in quantum mechanics.
Consider a ball on a spring (the simple harmonic oscillator.) In classical mechanics, its ground state (or vacuum state) has it stationary. In any excited state, the ball bounces back and forth with the system’s resonant frequency.
In quantum mechanics, the ground state is not stationary, and the first excited state does not have the ball bouncing back and forth with the system’s resonant frequency. Just look at the wave function; it literally shows no sign of back-and-forth swinging. Nevertheless, the frequency of the ball’s motion is visible in the wave function as exp[i(E/hbar)t]. It is also measurable when you interact with the system, e.g. when you send in light of a definite frequency and excite the system resonantly from its ground state to the first excited state.
The issue in classical and quantum field theory is exactly — literally exactly — the same as for the quantum ball on the sprin. The ground state classically is an inert field, while an excited state can have a wave in it that waves up and down with a definite frequency. Quantum mechanically, the vacuum state is not inert, and a first excited state — one electron, or one photon, depending on which field we’re talking about — does not wave back and forth the way the classical field would. Nevertheless, its frequency is observable, just as the frequency of the first excited state of a ball on a spring is observable even though the quantum ball does not swing back and forth.
You just need to put 2 and 2 together here; you already know how this works, you’re just not recognizing it.
Thanks for setting me straight Matt. I have been falling into the trap of thinking classically instead of quantum mechanically! Just as the vibration of a quantum harmonic oscillator is inferred from apply the Schrodinger equation to it, and is very different from a classical harmonic oscillator with a discrete infinity of vibration frequencies rather than a single resonant frequency, and no possibility of no vibration at all, the vibration of a quantum field that is not interacting with any other field is also inferred by applying the Schrodinger equation (or some alternative formulation of quantum mechanics), and gives exactly the same results. And it is the first excited stated of the quantum field that represents a single particle (or should I now say wavicle!), with the ground state representing empty space with no wavicles. One interesting thing is that we derive the rest mass of the wavicle by dividing h x frequency by csquared, whereas the energy of the first excited state is actually 3/2 h x frequency not h x frequency. So the rest mass of the particle comes from its additional energy compared to the energy of empty space, not its total energy.
Sounds like you have the right idea now! And yes, a particle’s energy is the energy difference between the state with the particle and the state without it. The energy of the vacuum is (formally/mathematically) infinite, but the energy difference needed to make a particle in the vacuum is (experimentally) finite.
Hi Matt. Would you be able to recommend a good introductory text book to quantum field theory as I would really like to try to understand more of the details of this fascinating theory. Just hope my maths is still good enough to cope. I would like something that shows the logic of going from the Schrodinger equation, to the Dirac equation to quantum field theory and includes the Higgs mechanism. I’m afraid my knowledge of quantum mechanics is purely based on the Schrodinger equation, so if an alternative formulation is used, the text book would need to clearly explain it.
Hi Matt. I would also like it to give a clear explanation of renormalization and why the fact that it is an infinity that is removed is not fundamental to the process but it is instead just compensating for the fact that the effects on unquantifiable field interactions has to be removed, as you mentioned in a recent podcast.
Hi Matt. Can you help with how I should think about the wavefunction of the quantum field. For a particle using the Schrodinger equation, the wavefunction gives you the probability density, that is it enables you to calculate the probability that the particle will be found in a region of space. However, this interpretation doesn’t seem to make sense with a quantum field – the probability that the field is found in some region of space ? – but the field exists everywhere with equal probability. So is it perhaps telling you the probability density that a field has a value between Z and Z + dZ. I came to this conclusion because with the Schrodinger equation applied to a particle, the wavefunction is a function of position x and is giving the you the probability density of finding the particle between x and x + dx. But when the wavefunction is written for a field F, instead of being a function of x it is a function of F, so I concluded that the wavefunction must now give you the probability density that F is between F and F + dF.
In https://www.physicsforums.com/threads/vacuum-state-of-the-klein-gordon-field.248105/ , reader Avodyne answers the question, noting that it is a problem given in chapter 8 of Srednicki’s quantum field theory book. And yes, Psi[F(x,y,z),t] gives you the probability that the field takes on the value F(x,y,z) at time t.
I assure you that this information is almost useless and that you need to learn the operator formalism for the harmonic oscillator.
Hi Matt. Can you just explain a little why the classical ball on a spring is so different to the quantum harmonic oscillator – for instance it can be stationary with zero motion energy and when it vibrates it only vibrates with its resonant frequency, whereas the quantum harmonic oscillator cannot be stationary with zero energy and vibrates with many different frequencies depending on its energy, none of which is equal to the classical resonant frequency. What causes the classical system to be so different, as naively you would think you could just apply the Schrodinger equation to it and end up with the same results as the quantum harmonic oscillator.
It’s all in Wikipedia (incidentally they do cover the “ladder operator” approach) and in particular you should study the section called “coherent states”. https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator Also notice this figure and its caption: https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator#/media/File:QuantumHarmonicOscillatorAnimation.gif
Hi Matt. The problem I see with the coherent states explanation is that they are not energy eigenfunctions but classically we see that a harmonic oscillator with a definite energy behaves very differently to an energy eigenfunction. And quantum mechanics tells us that independent of the spatial distribution of an energy eigenfunction, its time development has a frequency which depends only on its energy, with higher energies giving a faster frequency of vibration, contradicting the classical result. Where am I going wrong?
Quantum physics is correct, and all possible states, including both energy eigenstates, coherent states, and any linear combination of any such states are perfectly acceptable states. Not all states have definite energy.
Classical physics is an approximation to quantum physics, and is not correct. You may think that a ball swinging back and forth has an exactly specified energy equal to 1/2 m A^2, where A is the amplitude of the oscillation. But this is not true; it is only an approximation. The energy of the swinging ball is not, in fact, precisely specified. That is the lesson of the coherent states, which are the truth, about the classical picture of the swinging ball, which is an approximation to the truth.
So you are going wrong by expecting classical physics and quantum physics to be equal. The problem is to understand how classical physics emerges as an approximation to quantum physics. The answer is that for sufficiently heavy objects vibrating with sufficiently high energy, interactions with the world (e.g. the shining of light on the object) force the object very quickly into a state of approximately known position and approximately known velocity, which is sufficiently close to a coherent state that it behaves similar to the way coherent states do, swinging back and forth with near-definite frequency, near-definite position, near-definite velocity, and near-definite energy… just as we would expect for classical physics as an approximation to quantum physics. The fact that energy eigenstates of the quantum system don’t have these properties is irrelevant, because the classical system, interacting as it does with the outside world, is (clearly, as you see in its behavior) not in an energy eigenstate.
As evidence: A usually-classical system that does not interact with the outside world may not behave classically at all. For instance: https://news.uchicago.edu/story/scientists-use-quantum-spooky-action-entangle-objects-you-can-actually-see
Some discussion here: https://bohr.physics.berkeley.edu/classes/221/1112/notes/harmosc.pdf . Though yet again, along the way the author derives and uses the ladder-operator formalism, because it is so much faster and less awkward than the Schrodinger approach to this problem.
Hi Matt. Can you expand a bit on why the Schrodinger equation can be used in a relativistic theory. When you use the so-called time independent Schrodinger equation the kinetic energy operator is given by acting twice with the momentum operator and dividing by 2m – surely this comes from the non-relativistic result and kinetic energy equals momentum squared/2m, which does not apply to Special Relativity.
The correct definition of the Schrodinger equation is: i d/dt Psi = H Psi , where H is the Hamiltonian, expressed as a differential operator that acts on the state of the system Psi, where H and Psi are expressed in terms of whatever variables characterize the system. So you just find the Hamiltonian for your physical system, no matter whether it be relativistic or not, no matter how many dimensions it is in, no matter if there are any spatial variables at all.
Oh, and the time-independent version, appropriate only if Psi is an eigenstate of the Hamiltonian, is H Psi = E Psi, where H is an operator, Psi is a state, and E is a number that tells you the energy of the state Psi.
Thanks for the clarification Matt. You explain in your blog that for a bosonic field that when the operator formalism is applied to it the field has energies (n+1/2) h x frequency (f). From this I understand that a single photon is the situation when we put n = 1 and a million photons is the situation when we put n = 1000,000. But as these correspond to vibration energies of the whole photon field, which covers the whole universe, how do we distinguish the situation of a laser beam where all the photons are moving together, from the situation where we have a million photons and each photon is in a galaxy millions of light years from any of the other 999,999 galaxies where the other 999,999 photons are? It seems we need to consider the vibration of the universal quantum field as localized in a particular place where the photon is, with the energy of that localized vibration being hf, but the formula (n+1/2) hf applies to the whole field, which exists everywhere in the universe, not to a bit of the field the size of a photon. We see this in your blog when you analyse the cosmological constant problem. The zero point energy of the whole universal field is 1/2 hf, you don’t say that each bit of it the size of a photon has energy 1/2 hf. For fermions where the energy is given by (n-1/2) hf, if we have one stationary electron then we put n=1, and if we have one moving electron we put n=1 but f is higher representing the extra motion energy. So if we had a million electrons would we put n = 1000,000 or do we need to worry about the Pauli exclusion principle? And again, as these vibration energies apply to the whole electron field which exists everywhere in the universe, how do we cope with the idea that an electron is tiny, possibly even a point particle, as on the face of it it seems that rather than being tiny, as it corresponds to the first excited state of the whole universal field, each individual electron should exist everywhere.
Julian, at some point you need to take a on-line course in field theory. I am happy to answer basic questions, but I just do not have time to teach you a personal course in quantum field theory on this blog, nor do I have time to write a full quantum field theory textbook on this blog. For today, I’ll just try to answer two of your questions.
“how do we distinguish the situation of a laser beam where all the photons are moving together, from the situation where we have a million photons and each photon is in a galaxy millions of light years from any of the other 999,999 galaxies where the other 999,999 photons are?”
There is an insanely vast set of field theory states that can describe 1000000 photons, and only one of them has them all in exactly the same energy eigenstate. That’s the a perfect (and unrealistic) laser: all the photons are in lockstep, exist across the entire universe, and have exactly the same energy, each one being in the same energy eigenstate. In any realistic system, the photons are in more localized states, where the number of crests and troughs in the wave is finite, meaning that they are approximately but not exactly in energy eigenstates. Once that door is opened — once the photons are not in exact energy eigenstates, but in superpositions thereof — each of the photons may be localized in a different place. Or localized in multiple places. Or in superpositions of all sorts of combinations of states. Quantum field theory is vastly larger in its possibilities that you are yet dreaming of. There’s plenty of room for all sorts of distinguishable situations, far, far more than you have even begun to list.
“The zero point energy of the whole universal field is 1/2 hf,”
No, that’s the zero point energy of each mode in the field, of which there are an infinite number.
Hi Matt. Point taken. I was about to write to you again but unfortunately you have replied before I had the chance. After I asked the question I realised that if we know the energy of a particle exactly, assuming we know its rest mass we know its momentum exactly, so from Heisenberg’s uncertainty principle we know nothing about its position. So my whole idea of a localized free particle with a definite energy is not possible in quantum physics. I’ve fallen into the trap of thinking classically again!
Hi Matt. When we consider the interaction energy between two particles, say the strong interaction between two quarks inside a proton, as interactions between the two quarks are caused by disturbances in the gluon field which are called virtual gluons, would it be correct to say that the interaction energy is the energy of these disturbances, and hence the energy of the virtual gluons? So if the interaction energy is negative, does this mean that the energy of the virtual gluons is negative?
The notion that interaction energy is calculated between pairs of particles is a notion that makes sense if the interactions are weak. It’s an approximation that won’t work if the interactions are too strong. That happens to be the case for quarks and gluons in the proton; you can’t assemble the energy out of particles and particle pairs, because the very notion you’re basing this on is breaking down. Calculations of the proton mass just go in head-first and do the whole thing at once as a problem of fields, because to break it up into particles, whose number and nature are not fixed in the proton’s quantum state, is to take a hard problem and make it much harder.
So let me answer your question in a simpler setting where the approximation does work: positronium, involving an electron and a positron forming a simple atom and moving at non-relativistic speeds. The virtual photons in question here represent nothing other than the usual Coulomb field between the two charged particles.
But the negative energy of the interaction between electron and positron is not the energy of the virtual photon, and in fact, there is no one virtual photon in this problem but rather a coherent sum over many virtual photons whose relationship to one another is important. You cannot reconstruct the Coulomb field, and prove that its energy is negative, without this sum over many virtual photons. [The very ugliness of this sentence is among the many reason that it is not good to talk of these processes as involving “photons”, as though these are really particles with properties that are sensible to state.]
Instead the fact that the interaction energy is negative sits in the fact that the photon coupling to the electron is -1 while the coupling to the positron is +1. If these had both been electrons, the interaction energy would be positive, but the virtual photon would look the same.
The only thing that can be obtained from the virtual photon is the 1/r dependence of the potential energy between the two charged particles (though even to prove that requires an infinite sum.) To know whether the interaction energy is positive or negative you must know the charges of the charged particles, and that is not written in the virtual photon.
Thanks for the clarification Matt.
Hi Matt. Just trying to get a better understanding of the weak interaction, at least at a qualitative level. I know that the photon will mediate the electromagnetic force for any particle that contains non-zero electric charge and unless a particle contains electric charge it will not interact with the photon. For the Z boson, does it mediate the weak force for any particle that contains non-zero weak hypercharge as well as any particle that contains non-zero third component of weak isospin? How about the W bosons? Are there any simple rules for knowing whether the weak force is attractive or repulsive, in the same way that there is for the electric force? I will understand if it does not work in this way, so that these questions are impossible to answer in this simple form.
The W bosons are simple; all particles with weak isospin interact with them in exactly the same way. Some idea of this is in the 4th section of https://profmattstrassler.com/2022/08/01/celebrating-the-standard-model-how-we-know-quarks-come-in-three-colors/
The Z is complicated but in general all particles with weak isospin or weak hypercharge interact with it.
In general, for the Z-generated force, particles and their anti-particles attract, while particles repel themselves, just as for electromagnetism, and for the same reasons. For two different particles it’s more complicated (just as for electric charges) since it depends on the weak charges of the two different particles.
For the W: there is no simple W-generated force, because the W field (thanks to the “breaking” [or “hiding”] of the weak isospin relations by the Higgs field) converts one type of particle to another.
If the Higgs field weren’t present, then the whole structure would have been different. There would not have been a Z and a photon and two W’s, but rather three W bosons and a hypercharge boson, usually called X. Then there would have been a weak-isospin force from the W fields, and a weak hypercharge force from the X field. So the Higgs field really scrambles things. It’s remarkable how extraordinarily well the Standard Model and its story of the Higgs field explains all the complications from this simple underlying structure.
Thanks so much Matt. There is so much information on the internet about particle physics which turns out to be incorrect, so its fantastic having someone who really knows what they are talking about to provide accurate answers to these questions. It’s such a fascinating subject, but one for which it’s so hard to get accurate information. Other than yourself, no one else explains the physical interpretation of the incredibly complicated mathematics of quantum field theory, and its the physical interpretation that is the really interesting thing for most people.
Thanks, Julian. I do my best to be a beacon of information in a storm of misinfomation, and hopefully my reputation in this regard will grow. I always feel it is shrinking; the number of people who comment and ask questions on my blog is far less than it used to be.
Hi Matt. Just wanted to check my understanding on one point. In your notes it says the Higgs field/particle only interacts via the weak force (plus gravity). But then it is discussed how the Higgs field gives mass to particles be interacting with them and flipping left handed to right handed and also the Higgs force is discussed. I just wanted to check that when the Higgs field interacts with left and right handed particles to convert one to the other, that this is a direct interaction between the Higgs field and the particle, not an interaction that is mediated by the W and Z bosons. Similarly for the Higgs force – I understand this has nothing to do with the weak force, but is a force between two particles that both interact directly with the Higgs field. Can we say that the Higgs interaction is mediated by a virtual Higgs boson, in the same way that we say the electromagnetic interaction is mediated by a virtual photon? Is there any good reason why in the literature the Higgs force is not considered to be a fundamental force but gravity, electromagnetism, the strong and weak forces all are considered to be fundamental – so it seems fundamental has some special definition.
“I just wanted to check that when the Higgs field interacts with left and right handed particles to convert one to the other, that this is a direct interaction between the Higgs field and the particle, not an interaction that is mediated by the W and Z bosons.”
It is indeed a direct interaction. The same is true with the Higgs force; the interaction is direct, and is indeed caused by a “virtual Higgs boson” (a complicated and mystifying term that means “an effect of the Higgs field of the same type as one learns in first-year physics for the electric field”).
I cannot explain why so few people put the Higgs force on the list of fundamental forces. I certainly put it on the list. So did Bjorken, and many others do too. Eventually it will get there. One possible reason for the long delay is that it wasn’t clear until 2012 that the Higgs field would act like a fundamental field, and so books written before 2012 don’t necessarily put it there. (And science writers who grew up before 2012 also seem to perpetuate the “four fundamental forces” trope.)
Hi Matt. Just a bit confused on one point. In your notes you say that the energy levels of a fermion field are (n-1/2) hf where n can only equal zero or 1. I understand that the reason why n can only have these values is the Pauli exclusion principle for fermions. I understood that for moving fermions, we have the same formula but with a higher value of f to account for the motion energy. If we call the frequency for a stationary fermion f0, then this formula becomes (n-1/2) hf0 for stationary fermions. If we say consider the electron, this would seem to imply that there can be only one stationary electron in the universe and similarly there can only be one electron in the universe moving at any particular speed, with this speed determining the value of f. Even if we allow all integer values of n and say that there are different stationary electrons which are different superpositions of the wavefunctions of the field represented by different n, then surely this would imply that when we measure the mass of a stationary electron, we would have various probabilities of obtaining different masses.
If the universe were box with sides of length L, then indeed, there’d be only one true ground state with f = f0 and E = m_electron c^2, and only one electron (well, two, because of spin, but let’s ignore that) could be in that state.
However, the first excited state would have energy
Let’s say the universe were 1 meter across; then (remembering the electron has E=mc^2=511000 electron volts) the difference between these two energies is 10^-(20) electron volts. Unmeasureably small. And that’s for a universe 1 meter wide.
For a universe more like ours, ten billion light years across, this number is 10^{-71} electron volts. You could put a very large number of electrons into this box and never measure any one of their masses to differ from that of any other.
Any realistic electron, when free of an atom, is indeed spread out over some region L, and were it stationary, its E would differ from the mc^2 you’d obtain from f0. But the difference is unmeasurably tiny. Also it’s never exactly stationary, but again, by the same token, you can make it stationary enough that you can’t measure the effect of its motion on E.
Thanks for the clarification Matt. You really are amazing how you are able to show how the maths connects to the real world – nobody else seems to do that and for me its fundamental as the maths is just a tool to understand the real world. Just one point – isn’t there a typo – shouldn’t the first term in the square root be msquared cpower4?
Yes, thanks for pointing out the typo.
Hi Matt. On the Higgs boson entry in Wikipedia near the end it has a diagram showing the interactions of the elementary particles which shows charged leptons and neutrinos directly interacting. I don’t think that is right but please can you confirm. The diagram also shows the Higgs boson interacting with itself, the W bosons interacting with themselves and the Z boson, and the Z boson interacting with the W bosons but not with itself. I think that is all correct, but please can you confirm. I read that in neutrino detectors electron neutrinos only every produce electrons, muon neutrinos only ever produce muons and tau neutrinos only ever produce taus. Is that correct and if so why can’t, for instance, a tau neutrino produce an electron?
I don’t see it, but maybe I’m looking in the wrong entry; send the link for the entry, or even the link for the specific figure, and I’ll check it out.
As for neutrinos, there’s a long story. https://profmattstrassler.com/articles-and-posts/particle-physics-basics/neutrinos/neutrino-types-and-neutrino-oscillations/
Hi Matt. I’ve tried to send a reply with the link to the diagram itself a couple of times but my reply does not appear for some reason. The link to the Wikipedia article is
https://en.wikipedia.org/wiki/Higgs_boson
The diagram appears quite a long way down in the section “Theoretical need for the Higgs”
Sometimes a message will trigger the spam filter on the website, and then I have to approve it manually. That happens when there are too many links. I do see it, you are pointing to this link: https://en.wikipedia.org/wiki/Higgs_boson#/media/File:Elementary_particle_interactions.svg.
In a sense, I would agree that the link between e and nu shouldn’t be there, but the reason the person making the diagram probably put it there is that there is an interaction between W bosons, electrons and neutrinos. I have a different way of diagramming it that I think is a bit more clear and consistent; I’ll point you to it later.
Hi Matt. Yes that’s the link. I know the W boson and Z boson interact directly with all leptons, that’s the weak interaction, but I didn’t think that there was a direct interaction between charged leptons and neutrinos, only an indirect interaction via the W and Z bosons. Is that definitely correct or is it a matter that is uncertain in the Standard Model? Please could you also let me know why an electron neutrino can only ever produce an electron etc.
A W boson decays to an electron and an electron neutrino. That requires an interaction of the form W – e – nu_e . D you consider that a “direct interaction” or an “indirect one”? I would have said it was direct, but I’m not sure how you’re using the words here. See https://commons.wikimedia.org/wiki/File:Standard_Model_Feynman_Diagram_Vertices.png , the diagram at the middle of the grid.
As for your last question, I recommended https://profmattstrassler.com/articles-and-posts/particle-physics-basics/neutrinos/neutrino-types-and-neutrino-oscillations/ and will do so again. The very definition of “electron-neutrino” is “that neutrino, among the three neutrinos, which is created along with an electron when a W boson decays to an electron.”
Hi Matt. Just to confirm that I have it right, the W- particle decays into an electron and an electron antineutrino and a W+ particle decays into a positron and an electron neutrino. I read that this is because lepton number is conserved in the Standard Model.
That’s not quite right; after all, neutrinos have mass, and these masses may violate lepton number, so describing lepton number conservation as the cause, rather than the effect, is backwards. The correct answer is that it is built into the structure of the Standard Model, but the reasons are technical. Again, I recommend my neutrino article. Also read : https://profmattstrassler.com/2022/07/18/celebrating-the-standard-model-why-are-neutrino-masses-so-small/
The only way I can avoid going into a long mathematical discussion of how gauge theory works and why mathematical symmetry structures require that only one neutrino interact with an electron and a W is to give you a simplified version of the story, discussed here: https://profmattstrassler.com/articles-and-posts/particle-physics-basics/how-the-higgs-field-works-with-math/4-why-the-higgs-field-is-necessary/ I simplify from four dimensions to two, and simplify the weak interaction from SU(2) to one based on a simple back-and-forth symmetry.
Thanks for the clarification Matt. Just finished reading the article on why neutrino masses are so small are really enjoyed it. I have read many of the articles on your blog but there always seems to be another gem that I haven’t yet read!
Hi Matt. I had read your article about why the Higgs field is necessary before but have just read it again. It seems from the article that the reason we need left handed and right handed particles is in order to describe the spin of fermions but we don’t need left and right handed particles in order to describe the spin of bosons – have I got that right? In the Wikipedia article on the Higgs mechanism it says that it is technically incorrect to say that the Higgs field spontaneously breaks a gauge symmetry as Elitzur’s theorem says that a gauge symmetry can never be spontaneously broken. Is that right as so many sources refer to the spontaneous symmetry breaking of a gauge symmetry.
Hi Matt. Do you know of a good online source that explains the operator formalism in quantum mechanics as it definitely seems that I need to understand this to progress in quantum field theory?
Sorry, I missed this question for a while. You have to understand the harmonic oscillator in this language. For instance, https://www.physics.udel.edu/~jim/PHYS424_17F/Class%20Notes/Class_5.pdf But any QM book or QM lecture series will cover this.
Oh, and also, see Sean Carroll’s latest book; here’s his book talk about it. https://www.youtube.com/watch?v=cHjoFAOmRbU
Dr. Strassler —
Over at the main “questions about the book” page, a few weeks ago, I left a handful of questions triggered by the first 14 chapters. You replied very generously and clearly. Now that I’ve finished the book, I’ve got one overarching question which doesn’t bear directly on the subject — but isn’t entirely unrelated, either. Let’s see if I can cast this in sensible terms…
This question first occurred to me as I studied Chapter 18’s Table 6 — around page 305 or so — particularly the last column, headed “Long-Lived (> 1 second).” (Each value in that column is a simple binary Yes or No.) The note I made for myself there reads:
“Why is one second a significant duration? Is that like a quantum of time? Are there ‘time WAVES’ too? […] Why is time so… **impenetrable**?”
I kept thinking about all that as I continued to read — was reminded of it every time I encountered the word “second” or other units used to measure it. And it occurred to me that, well, MOST references to time in “science for lay readers” writing leave time pretty much untouched, undefined. It seems to be a given — I guess not worth devoting a lot of explanation to, because it’s simply not amenable to experimentation on its own. Its meaning is like what my old grammar teachers called the “‘You’ understood” in imperative sentences (e.g., “[You] close the door”), invisible but implicit. (In science fiction circles, the use of “time” in science writing might be classified as “hand-waving”: something-something-something-[AND THEN MYSTERIOUS THINGS HAPPENED BUT YOU KNOW WHAT I MEAN]-something-something-something.”)
This is all Very Big Stuff, no doubt, and I can’t reasonably ask you to tackle it here. But could you by any chance recommend a good reference which might poke about in time-as-a-subject in the same way that ‘Waves in an Impossible Sea’ pokes about in particle physics?
Thanks as before. LOVED the book.
Best,
John E. Simpson
Actually, “1 second” was a convenient dividing line between things that last long enough to be useful for humans or the cosmos to do interesting things with them versus those that don’t. There was no deeper reason than that.
Time is especially impenetrable to physicists, and few people feel they have much intelligent to say about it. I certainly don’t. You should read Sean Carroll’s book about it, he’s one of the few who has been willing to take the subject on: https://en.wikipedia.org/wiki/From_Eternity_to_Here . It won’t answer all your questions: no one can.
Gravitational waves can be viewed as waves of space and time or as waves of space across time — two different mathematical/conceptual ways of viewing the same thing, though the latter can fail if spacetime is complicated enough — but there are no waves of time alone, there isn’t any way to define such a thing.
If there is anything like a quantum of time, it will likely be billionths of billionths of billionths of billionths of a second; we know for certain that it can’t be as long as billionths of trillionths of a second because we would already have detected a breakdown in our usual equations, which assume no such quantum of time.
Thank you very much!
I’ve actually found a lecture series — one of the “Great Courses” series — taught by Sean Carroll on the physics of time. (For the benefit of anyone following this thread, it’s at https://www.thegreatcoursesplus.com/mysteries-of-modern-physics-time — probably requires a subscription, though.) But I didn’t know about the book, so thanks again.
He wrote the book first, then gave the lectures, so the former should help you follow the latter.
Hi Matt. Consider a particle with rest mass. The energy of a particle is given by h x frequency. So in a frame where it is moving it must have a higher frequency than a frame in which it is stationary, as it now has motion energy as well as mass energy. However, time dilation in Special Relativity tells us the period of the vibration as measured by the observer relative to whom the particle is moving will be longer, so the frequency will be smaller. Please can you explain this apparent contradiction.
Hi Professor,
I read the book and really enjoyed it. Thank you so much for the effort to make some of the most fundamental concepts of our existence more accessible.
I posted this question somewhere else, but not sure where it would be appropriate:
Do the dynamics that you described for a how a proton gains rest mass scale to larger systems like galaxy clusters bound by gravity if certain criterion are met?
Would that even matter in regards to its gravitational mass, or would this basically just be trying to put the term “rest mass” on a no-more-well-predicted system of mass + energy + dark matter?
The principles seem to be describing a very similar phenomenon to dark matter, just with a more discrete playing field.
Thank you again!
Glad you enjoyed the book!!!
The strong nuclear force is special; it creates the walls that hold the proton together. In large gravitational systems such as galaxy clusters, there are no walls; the system is more like the atom, in that the “hole” or “well” in which the clusters sit dies off with distance. So things do not scale: in one case you have the strong nuclear force, which traps objects on a scale of 10^{-15} meters, and on the other you have gravity, which does not trap things, except in black holes which can have any size but are certainly not in play for loosely bound galaxy clusters.
To use some effect of this sort to account for dark matter (or dark energy) would be akin to drastically modifying Einstein’s gravity — which, of course, people attempt to do to explain dark matter (and/or dark energy). It’s very hard to do it without running afoul of other observations.
The only situation in which gravity behaves a little bit like the strong nuclear force in a proton is in the context of a black hole — which, after all, traps things inside it too. But the details are quite different.
So I’m afraid that dark matter cannot be addressed in this way.
That said, the effects that create dark matter could indeed involve forces that trap the way the strong nuclear force does. My 2006 work on “Hidden Valleys” with Kathryn Zurek (see also “Dark Sectors”) is focused on this issue.
Matt,
I read your great book with interest and got very interesting views and insides, and following your web site for over 10 years. Thank you so much.
I am studying QFT on my own and was trying to see if there is a relationship between the rest mass as a standing wave and the electron’s oscillation between the left and right chirality?
Yes, there is a relation. I’ll write this in 1+1 dimensional form, but the same idea applies in 3+1 dimensions.
The wave equation for a free field X whose particles have mass m takes the form (setting Planck’s constant and the cosmic speed limit to 1, to keep the equations shorter)
which for a plane wave
gives
The standing wave is given for p = 0 — a time-dependent but space-independent wave — with E = m (recall that I set the cosmic speed limit to 1.)
Now for an electron, if you set p = 0, then the equations look like
where L and R subscripts stand for the left- and right-chiralities. If m is non-zero, then you can substitute the second equation into the first to get
from which you conclude E = m, a standing wave. But you see that you cannot take this step if m = 0; instead both equations just say that E = 0, and so there is no wave at all.
Matt,
Two questions:
– How does the Higgs field fit in the L-R chirality oscillations? Just by providing mass to particles, or other way around, as I read somewhere, that the Higgs field induces these oscillations and creates massive particles?
– In your 1+1 example the field X oscillates “into” the imaginary axis ‘i’, right? Can we think of that axis as a spatial degree of freedom orthogonal to other spatial dimensions? Because if yes, the oscillations look like Einstein clocks, and nightmarish Lorentz invariance becomes a purely geometric phenomenon, a consequence of the fact that we’re all composed of Einstein clocks and the only time measure available to us is the proper time of these clocks.
1) Neither, really, because it’s not good to focus on the “Left-Right-chirality oscillations” for too long; it’s a math fact, not a physics fact, that one can think about it that way. Once there is a mass, the notion of chirality is meaningless; it can only be still used, as an approximation, when the particle’s rest mass m is very small compared to the particle’s total energy E (/c^2). For a particle at rest, you can see from the equations I wrote in the previous comment that what’s actually happening is that the left- and right-chiral parts of the electron are being set equal; the mixing between them is 100%.
[Also be very careful not to mix up chirality with helicity, which are the same for particles with m=0 but are otherwise different; this is a common mistake made by students.]
You can think of the Higgs field as inducing this mixing, because if you replace “m” in my equations with “y_e H” (where H is the Higgs field and y_e is the interaction strength between the Higgs field and the electron) you see that there is a very explicit way in which the mixing of the two chirality states only occurs when the Higgs field’s value is non-zero and constant.
2) The asnwer to the “Can we think of…” question is no. The X field’s value, both its real and imaginary parts, has nothing to do with physical space — at least, not the three-dimensions of physical space we know about. [Its value could have to do with motion or distortion in some extra dimensions, but that’s pure speculation; and for a fermionic field like the electron field, that idea is also highly non-intuitive, since all familiar dimensions are bosonic.]
Consider the magnetization field covered in chapter 13-14 of the book. The spin waves that wave the magnetization field do not involve the motion of the field’s medium in space, or motion of any other object. We do know that spin waves only change the orientation of the objects that make up the medium — but that’s something we only know because we understand magnets. In the case of the cosmos, we don’t understand it, and we don’t know the detailed nature of the medium for the electron field — so the best we can do is speculate, as I did in the bracketed comment above. Certainly we cannot conclude, merely from the equations I wrote, that there is anything spatial going on.
Matt,
Your subsequent comment to Serge:
“You can think of the Higgs field as inducing this mixing, because if you replace “m” in my equations with “y_e H” (where H is the Higgs field and y_e is the interaction strength between the Higgs field and the electron) you see that there is a very explicit way in which the mixing of the two chirality states only occurs when the Higgs field’s value is non-zero and constant.”,
makes it very clear.
Thank you.
Not so much beyond the book but complimentary to it: Has your book been entered for the 2024 Royal Society Winton Prize? I hope so.
On your blog page I’d like to start from the beginning with your older articles, gradually working my way through them when I have the time. But when I click the first page 108, I get a “404 Uh-oh! You’ve encountered a quantum superposition!” error.
I do not know of the book being entered for any prizes, but thanks for the thought.
When you say “first page 108”, I’m not following you: which link exactly are you clicking on?
“Previous 1 2 3 … 108 Next” is at the very bottom of your main blog page https://profmattstrassler.com/blog . Page 1 has your most recent articles, I’m expecting page 108 to have your first ones from 2011.
Confirmed for https://profmattstrassler.com/blog/page/66/ and links with larger page numbers.
Thanks. Not sure of the cause, will have to investigate.
Searches for earlier posts still work.
That seems to be a bug. I’ll take it up with my website folks.
In chapter 16, you write that a photon can be spread out across a room and then absorbed by “a single atom, located at one microsopic spot on the room’s walls.” Suppose the wall is a box of gas. Are atoms of the gas wavicles, too. Would such an atom still be located at a small spot when it absorbs the photon, or would it be spread throughout the box? If the former, how big a spot, and what kept it from being spread out more?
So, this is indeed “beyond the book”, but worse, probably beyond what I can explain clearly in a few paragraphs. Really, this needs a whole lot of careful buildup. Quantum physics is very, very strange, and to explain it well takes more pedagogical experience than I currently feel I have. But I’ll try.
Let’s keep it as simple as possible: there’s just one atom in the room, not a wall, and there’s one photon crossing the room. Chances are the photon will miss the atom, but the probability of interaction isn’t zero, so let’s look at it.
An atom is not a simple wavicle, because it is made of many objects, but it does have wavelike behaviors that it inherits from the elementary wavicles that it contains. For this reason it can indeed spread out. [This kind of thing is actually observed, most clearly as reported (poorly) in https://www.space.com/atom-transforms-into-quantum-wave-schrodinger … at some point maybe I’ll have time to explain the paper and the physics behind this.]
How spread out is the atom? This will depend in part on when and how the atom was inserted in the room; the more mass an object has, the longer it takes to spread out across a macroscopic space. (See https://profmattstrassler.com/2024/03/19/yes-standing-waves-can-exist-without-walls/ ) But let’s assume it has had enough time to spread out across the room.
Now we have a photon spread out across the room, and an atom spread out across the room. What happens when they interact? Do they interact everywhere at once?
In a sense yes — they could interact anywhere — and in a sense no — when they interact, the interaction occurs in a small region… atomic sized, because that’s the scale of the electromagnetic dipole moment of the atom, which is what interacts with a photon. The fact that both atom and photon are spread out doesn’t change that.
The fact that they are spread out has only the following effect: the region where they interact could have been anywhere in the room, and could not otherwise have been predicted.
Or at least, that’s one consistent way to interpret the math. It’s not the only one, but I don’t think I could correctly go through all the others.
[If we have multiple atoms, then we have to deal with the fact that the atoms are all correlated with one another, which means that the location of one could in principle depend on all the others, complicating things infinitely. Only in a Bose-Einstein condensate, where the correlation is perfect and simple, would things become easier, but I haven’t thought through how a Bose-Einstein condensate interacts with a photon.]
Obviously this raises all sorts of questions about where information, objects, and relationships among objects are actually stored. Correlations between objects matter; in fact, interactions and measurements always create new correlations between objects or between an object and an observer. Reality contains these correlations, and can’t simply be stored in the way that seems intuitive to us, where there is space that pre-exists, and there are definite objects at definite locations in that space that we can measure without significantly affecting them. Trying to picture this, or follow the logical flow where it needs to be taken, goes beyond what is easy for the human brain, and physicists and philosophers are still struggling for easy-to-explain ways to answer questions like yours.
Thank you for your book. Not only did I enjoy it, as a high school physics and math teacher, it taught me a lot, and has helped me give better answers to my curious students when asked about the Higgs field.
I also appreciate your insightful comments about fields and relativity. I feel I can use it when I talk about relativity.
As a physics teacher, I want to give my students a conceptual model of quantum mechanics, to help introduce it in a way that makes some sense. I think I have succeeded in the following, and since intuitive models of physics seem to be your thing, I am hopeful you will be generous enough to give me your thoughts on it.
My conceptual model is: (1) Quantum objects can be thought of as instantly reformable wave packets, all part of one thing. An electron, say, is an infinite bunch of waves of the electron field, summed up to produce a wave packet. All the constituent waves are part of one thing, the quantum object, the electron. Upon measurement (or a proper interaction with other quantum objects) it instantly reforms (collapses), and you get a new wave packet. (2) These wave packets instantly reform when interacting with other quantum objects, in a random way obeying the Born rule. I tell my students (because it makes more sense) that instant reformation (collapse) probably happens when things interact, but we only know for sure that it happens upon measurement. The size of the new wave packet depends upon the wavelength of the quantum object that caused the collapse. I also go over the case (to illustrate what the Born Rule implies) where half of a quantum wave packet hits a detector and the other half hits nothing. The “particle” is detected half the time, half the time it isn’t.
The reason I like this conceptual model is that enables one to reason out the results of the experiment where you send electrons or photons one at a time though a double slit, even if you’ve never heard of the experiment before. Before it encounters the double slit, the electron is a large wave packet, bigger than the double slit, so you can reason that if it makes it through it will interfere with itself, like any wave will do after passing through a double slit. Most of the time the quantum object collapses on the slit material, because most of it is interacting with the material, following the Born rule. Sometimes it makes it through, because of the holes (not all of it is interacting), and interferes with itself as it travels to the detector. Where will it collapse on the detector? It will never collapse at the points where it is cancelling itself out, and has the greatest probability to collapse in the middle, where the squared amplitude is the greatest (the Born rule). When it instantly reforms, it gives its energy to a detector electron, which is much smaller than it, so upon collapse, it shrinks a great deal, becoming a new wave packet. Thus we see small “particles” on the detector screen.
Now of course, I tell my students about other ways to think about quantum mechanics, and discuss particles and waves and so on, and I tell them that what I have just described is one way to look at what the math tells us, but it just seems like a conceptually easier way to think about it all, seems to give one more insight.
In this model, the only sense that a quantum object is a “particle” is that all the waves that make up the wave packet are part of “one thing”. If you split a quantum object in two, the two parts are still part of “one thing” and if one of the parts instantly reforms upon measurement, the other part disappears. The “whole thing” reacts to part of it successfully interacting, which is a “particle” like trait. However, that seems to be it. There is no particle in this model, as there is no “little ball” anywhere.
This conceptual model is also helpful in talking about electrons about an atom, which are trapped wave packets that form standing waves in a spherical sort of enclosure. The wave packets are not accelerating, but are vibrating in place, so they don’t lose photons. Of course, the wave packets are now of a more uniform frequency, and they still cancel out in certain areas, so you can never detect them there, according to the Born rule. I go over this and show how you can get the shape of electron “orbitals” through the standing waves vibrating in place and having a non zero squared amplitude in certain spots.
It’s also useful for talking about the uncertainty principle, but I’m writing too much…
What I really want to ask you about is the way this model deals with entanglement. The natural model for entanglement with this picture is that if you have two entangled electrons, A and B, the waves of A “occupy” B and the waves of B “occupy” A. Upon measurement of, say A, the waves of B are instantly kicked out of A, and vice versa. You can also think of it as, say, the spin of A exists in B (and A) and the spin of B exists in A (and B). Upon measurement, they instantly reform and kick the other spin out. It makes some sense that whatever you measure in A will be related to what you measure in B (like an opposite spin). This makes sense to me, but I haven’t been able to talk about it much with anyone more knowledgeable, like you. How does this sound to you?
Anyways, I don’t want to make too long of a post. When I have presented this to students (with all the necessary caveats – I hope!) they say they really appreciate it and it makes sense to them, much better than chemistry class. Of course when I first came up with it, I was nervous that I was leading them down a wrong path, so I showed this to a few local physicists, and they gave me some approval, which I hope you will as well.
So my question for now (I hope you will indulge me by answering a few more) is: How does this sound to you? Also, very much hoping to get your thoughts on the entanglement model.
Prof. Strassler:
First and foremost, thank you. Your book and other explanations on this site have been wonderful to my superficial understanding of the concepts behind modern physics. I’ve re-read the book many times, and enjoyed it every iteration.
I hope my question isn’t too naive, I’m a physician and have long been an enthusiast of grasping the Standard Model, though I lack the needed math to go deeper.
My concrete question: Could light speed be infinite, and not c? I mean could the cosmic speed limit be just an artifact that we experience as a consequence of our perspective as beings made from massive particles?
If light, and gravitational waves move at c speed, but no matter how fast we go we can’t overtake them, and even measure their speed as the same no matter what, could it be possible that their speed is actually infinite, and not “just” very fast?
Since from the perspective of a photon (assuming it as a being) it would not experience neither time nor space, because of extreme time slowing and extreme length contraction, could it be possible that our perception of a slower time and a physical space be “artifacts” of us being massive, or made out of massive particles?
Again, hope I’m not speaking absolute nonsense, and thanks again.
Thanks for the question, and I am glad you enjoyed the book.
The answer to your question is no, their speed could not be infinite. We directly can measure the time that it takes for light to go from point A to point B, because we have materials that emit light, and screens that absorb light, whose timing is precise to better than a nanosecond (a billionth of a second), which is the time it takes light to travel just 30 centimeters. The travel time observed is always the same: it is the travel distance divided by the cosmic speed limit c.
Another way we can be sure of this is to study not light or gravitational waves but ordinary matter. If we send an electron from point A to point B, we can measure its time of travel, also. Now the travel time is the travel distance divided by the speed of the electron, v. No matter how much motion-energy we give the electron, we find that its speed v is always less than the cosmic speed limit c — that is, the travel time is never shorter than the travel distance divided by c.
Furthermore, there are many details in the relationship between energy and speed that can be measured, and they are in exact accord with Einstein’s relativity equations, which are essential in preserving Galileo’s relativity principle. If light and gravitational waves traveled at infinite speed, they would violate either Galileo’s relativity principle or Einstein’s relativity principles, in conflict with the results of experiment.
Indeed, the entire Large Hadron Collider — the timing for the accelerator, and the timing for the particle detectors — relies on knowing the travel times of photons and other particles to better than one nanosecond. If our understanding of particle speeds were not precise and reliable, the Large Hadron Collider and its particle detectors ATLAS, CMS, LHCb, and ALICE simply would not work.
Dr.Strassler:
I actually posted this question on a different thread, but realized it is more appropriate here. It is concerning fields, specifically the magnetic field. If I have two magnets, a north pole and a south pole facing each other, and pull them apart, doing work on the system. after pulling them apart, I would say the work I did appears as an increase of mass of the system.
I then, heat up one magnet. Or both, until they lose their magnetization. What happens to the energy I put into the system? Does the collapsing magnetic field transfer the energy to something else?
The first paragraph and second paragraph are logically separate.
First paragraph: the work you did does not appear as an increase of the mass *of the system*, unless you tell me what the system is. In particular, if you are part of the system, the mass of you and the magnets has not changed; in fact you will radiate some heat and the mass of the whole system will slightly decrease. If you are not part of the system, then we can’t measure the mass until you let go of the magnets, at which point the question is incomplete. Is something holding the magnets in place? If there is nothing holding them at the moment you let go, they will move, and so the energetics of the system has to be followed carefully. For instance, if they now slam into each other, this will create some heat and electromagnetic radiation, which will carry energy away. In the end the mass of the two magnets, now stuck together, may in fact be reduced rather than increased; one has to think about it carefully.
Second question: what happens if a magnet is heated and loses its magnetization? Well, heating them puts energy in, and increases the mass. The loss of the magnetic field (and any effects thereof) will be smaller than the heat-energy put into the magnet, which will partially be inside the magnetic (in vibration motion of its atoms) and partly radiated away in photons. The magnetic field energy gets swallowed up in this complicated system in a way that can be tracked if necessary, but it’s certainly not simple.
Of course, if you now allow the magnet to cool again, then the energy will be radiated away, although some of it may be used to remagnetize the magnet, either as a whole or locally in magnetic “domains”.
All of this is to say that when you want to do precise bookkeeping and make sure you know where energy came from and went to, and how much an object’s mass may have changed during a process, it requires careful tracking of all the details, and real effort.
Dear Matt,
Thank you for taking the time and effort to write such a wonderful book. You’ve truly exemplified the famous saying often attributed to Albert Einstein: “If you can’t explain it to a six-year-old, you don’t understand it yourself.” As a musician, I often feel like a six-year-old when it comes to the complex concepts you’ve so skillfully broken down into small, easily grasped notions and metaphors. You’ve opened up a new world of understanding for me.
I have a question: Did you play the clarinet? I inferred this from your remarks about memories of cold mornings. Or was this just a metaphor?
For more than 15 years, I’ve struggled to understand the physics of how a clarinet works. While I can’t claim complete understanding yet, your book and explanations have brought a lot of clarity. I can confidently say I now perceive much more about the phenomena behind it.
Many kind regards and thanks, and I eagerly look forward to your next books.
With fond appreciation,
Alex
Thanks for your kind words about the book! And yes, I played the clarinet for about a decade, and have played the piano for five decades. It is absolutely true that in music camp chamber music rehearsals, I was often out of tune before 9:30 am.
While the following website is about flutes, some (not all) of the issues are exactly the same as for a clarinet. Maybe you’ll find it enlightening.
https://newt.phys.unsw.edu.au/jw/fluteacoustics.html
Matt, your book is super awesome! I’ll try to come to Lenox, if stars aligned, to see you in person.
Book gives many good analogies, and I wonder if you have good analogy for energy and field values itself?
E.g. for example electron has that amount of energy because at certain location in electron field value is vibrating (changing back and forth?) with frequency f. What amplitude of such vibration could represent? I mean, if it’s in fermionic field, then it’s bounded by some value, but this value is not part of E=hf formula. So is this property of the field and not the quant then? How can we think of the value that vibrates?
I don’t think I have a question, but perhaps something worthy of an endnote. (Aside: I much prefer footnotes to endnotes, but, whatever works…)
In figure 27, you lay out the geometric optics of rainbow formation. I recall an article 10-20 years ago that explained a small adjustment to that picture that gives the actually observed rainbow angle. The idea was that the total internal reflection induces a surface wave/evanescent wave/plasmon wave/something (this part of my recollection is fuzzy) that travels for about degree, then internally emits the photons from a slightly different point on the surface. Then the remainder of the path in the diagram is adjusted by about a degree. I think I recall that CAM (complex angular momentum) calculations contributed. More recent papers about Regge pole contributions to atomic rainbows swamp my Google searches, so I’m not able to find a relevant reference. Sorry.
Why this is relevant to the book: My mental model of the fundamental fields frequently leans on my understanding of coupled pendula. (Random example: https://phys.libretexts.org/Bookshelves/University_Physics/Mechanics_and_Relativity_(Idema)/08%3A_Oscillations/8.04%3A_Coupled_Oscillators ) The pedula stick out in various directions (in a more-than-3-dimensional space) and are coupled by springs. (If you try to embed this picture in Euclidean 3-space, there’s nowhere to put all the springs.) This is my go-to get-the-brain-rolling model for QFT, providing a mental model for energy transfer between fields. This seems to be an important idea in the book. It’s also happening in the rainbow water droplet — the external to the droplet photon modes are coupling to the internal to the droplet photon modes are coupling to the surface (something) modes — the energy is sloshing around among these various oscillators.
(I prefer footnotes too — this was my publisher’s decision.)
I haven’t ever looked at this subtlety in rainbows; thanks for pointing it out. If you do find a reference to it, let me know.
Your mental model for fields is indeed the one I use.
Are zero-point fluctuations of a fermionic field subjected in some way to the Pauli exclusion principle? My question is motivated by the following reasoning. Quantum theory predicts immense energy density for the lowest possible scale for these fluctuations. If (1) the exclusion principle is applicable, they must seemingly repel each other and expand as a whole at breakneck speed. Then if (2) this expansion is decoupled from spacetime dynamics, this might explain both cosmological constant and hierarchy problems, in a unified setting.
Similar to the cosmic inflation which produces flat fluctuation spectrum due to superhorizon fluctuation “freezing”, effective (observable) zero-point fluctuations spectrum might be flat as well. The cosmological constant and hierarchy then look like the right picture scaled down by tenths of orders of magnitude. The grand cosmic cycle starts at a hot big bang with the zero average Higgs field value. From then the feedback mechanism works in a quenched regime, i.e. currently Higgs field value just hasn’t had enough time to reach higher values. After eons of eons, the feedback mechanism will have pushed the Higgs field value to such height that wall of doom starts, creating hot and dense plasma while relaxing the field value back to zero in a way similar to the reheating during inflation, and renewing the grand cosmic cycle.
Recently (2019) the quantum optoelectronics group at ETH Zurich has performed some measurement the zero-point fluctuation spectrum, what is interestingly they observed excess at the lower end and deficit at the higher end of the probed frequency range. Could it be a hint to the flatness of the spectrum? Fig.3d of https://www.nature.com/articles/s41586-019-1083-9
The Pauli exclusion principle does apply, but not in the way that you are suggesting. It’s the exclusion principle [or rather, the same relative minus sign that leads to the exclusion principle] that makes the zero-point energies negative. More precisely, it makes the energy density negative and the pressure positive — i.e., a negative cosmological constant. The positive pressure actually causes the universe to collapse, not expand, because of the way that gravity reacts — just as gravity would collapse a gas with relativistically-large positive pressure.
These equations aren’t so hard, you can learn them in places such as Kolb and Turner’s book, or indeed any book on cosmology. Possibly even wikipedia. You’ll see that they don’t accord with your speculations.
Dr.Strassler:
In stars, the positive outward pressure, due to thermal motion, balances gravity trying to collapse the star. However, positive pressure…energy density…contributes to curvature and hence gravity. Obviously, there must be a break even point where the increased outward force, due to thermal motion, outweighs the pressure contribution to curvature due to energy density…..and the star is in equilibrium. Is there an established break even point? It seems that in most “normal” stars the increased outward force…due to pressure….overwhelms the pressure contribution to gravity. In extreme density situations, like neutron stars, would the energy density contribution to gravity start becoming dominant?
Until the system becomes extremely relativistic, with all speeds at or near the cosmic speed limit, the energy density is always larger than the pressure, and that results in usual gravity. In fact, even most relativistic systems still are dominated by energy density. It takes special situations for gravity to be significantly impacted by the effects of pressure, and even more special for those to result in expansionary gravity, such as are seen when one has vacuum energy density and comparable negative pressure.
Dr.Strassler:
I may be convoluting some terms here, hopefully you can clarify for me. I was associating “pressure” with the definition from gases where pressure is described as energy per unit volume, isn’t that the same as energy density? So isn’t pressure a statement of energy density? Or, is pressure being used differently in General Relativity?
Even for gases, this isn’t true; you’re thinking only of the kinetic and associated potential energy of the atoms, and forgetting the energy stored in their masses. Gravity knows all about that energy density; if it didn’t, then Einstein’s laws of gravity wouldn’t agree with Newton’s in the appropriate regimes where they ought to do so.
For example, in an ordinary gas, the ideal gas law does indeed say that p = n k T, where n is the number density of atoms or molecules, and k T is proprtional (not necessarily equal!) to the kinetic energy plus internal potential energy per atom or molecule. Typically that kinetic+potential energy equals x k T, where x might be 3/2, 5/2, or 7/2. If we have a gas of atoms only, then x = 3/2, and 3/2 k T = 1/2 m v^2 = the average kinetic energy per atom, where v is the average speed (really, root-mean-squared speed) of the atoms.
That’s what Newton’s followers thought was the entirety of the energy density. But Einstein discovered that the true energy density is n (m c^2 + x k T), and in any normal gas, m c^2 is far larger than the kinetic + potential energy. That’s what gravity then responds to. And you can see that this is always true for a monatomic gas as long as 3/2 k T = 1/2 m v^2 << m c^2 -- i.e., if the gas is made of atoms moving at non-relativistic speeds, with v << c. Only if the speeds of the atoms are of order c --- relativistic motion, far outside what one learns in first-year physics --- is the pressure per atom is of order m c^2. Moreover, in this case both the energy density and the pressure are positive, so the effect of gravity is still attractive. This is true also for a gas of photons, for which the pressure and total energy density really are equal. But no gas of particles can have negative pressure. That has to come from something more exotic. And for it to have large negative pressure, enough to counter the total energy density, it must be both exotic and relativistic. This cannot happen in Newtonian physics.
Dr.Strassler:
Thank you, I see where my error was made now. So, when they are talking about energy density, in General Relativity, they are talking about how tightly the masses are squeezed together….is that correct?
That’s too limited a vision. A photon gas at high temperature has high energy density, but the photons have no mc^2 rest-mass energy. Zero-point energy of a field leads to huge energy-density (and pressure) but that energy-density has nothing to do with masses of particles, and it could be positive (for bosonic fields) or negative (for fermionic fields).
For a non-relativistic gas of particles with non-zero rest mass, the energy density is potentially dominated by the mc^2 energies of the particles in the gas. But if the gas is sufficiently cold, then various quantum effects can contribute additionally to the energy density; this is critical in neutron stars and in white dwarf stars.
So the answer to your question is that you’ve given one correct example, but there are many others.
Dr.Strassler:
Ahh ok, I see where my conceptual understanding was flawed, basically it’s just “energy” including all forms of energy, contributing to energy density. My background is in Aeronautics, and when we think about energy density, it’s how energy is stored in the degrees of freedom of molecules….but we generally don’t consider the energy locked up in the masses of molecules, and we aren’t generally dealing with a photon gas.
Yes, this is unfamilar territory even for most professional scientists. But it’s crucial. Newton’s law of gravity creates a pull proportional to the masses of objects. Einstein’s, however, creates a pull proportional to the *energies* of objects, if all velocities are slow — and proportional to a more complicated combination of energy and momentum (the latter of which contributes to pressure) when relativistic speeds are involved. It’s only because *all* the energy creates gravity, including the mc^2 energy stored in all particles with rest mass, that Einstein’s gravity law agrees with Newton’s gravity law in the regime of slow velocities (and sufficiently low densities.)
In fact, Newton’s law for the gravity of a proton would be wrong. Most of a proton’s mass comes from the kinetic and potential energy of the particles inside it, not from the mass of the particles inside it. This is discussed in Chapters 6-9 of my book.
Could you please give some ways of thinking about negative energy density of fermionic zero-point fluctuations, as for me it’s really confusing. Is it like “borrowed” energy, or like future potential energy?
You already know that binding energy in atoms or planetary systems can be negative. It is no more confusing than that… It is a statement that if you have a field whose particles have very large mass, but with an interaction that makes them massless inside a box, then if the field is fermionic, you will have to add energy if you want the box to expand, while if the field is bosonic, the box will tend to expand on its own accord. [I should make this more precise; probably there’s a way to turn this into an explicit example…]
The question of why a fermionic harmonic oscillator’s zero-point energy is negative should have an intuitive explanation. (For a bosonic harmonic oscillator, it is easy.) I do not have a good one yet, but it’s on my mind.
Hi Matt,
will your book be translated into French?
There has been some interest; stay tuned.
Hi Matt,
Regarding a first possible wave in that impossible sea (smallest oscillating thing?), you got me wondering! In your book, do you explore how Planck’s quantum of angular momentum might act on that condensate of weak hypercharge (Higgs-type field)?
Nigel
Sorry, but I don’t understand the question. Angular momentum doesn’t “act” on condensates, or more specifically on the Higgs field. Maybe ask again, more precisely?
Hi Matt,
Thanks for the prod! Thinking of that smallest possible wave in your impossible sea, mediated by Planck’s constant of action, implied a quantum of curvature (for those keen to quantize GR).
Being irreducible, such a quantum of angular momentum would persist, a “first measurable thing”; effectively a quantum of localized energy, hence density. Hence curvature.
So what I had in mind was: what if this Planck scale oscillator, this smallest spinning thing, had a rest frame, and it dragged that local frame as it spun? If this quantum of curved action is acting in/on a superfluid tachyonic nonzero scalar field (say, a condensate of weak hypercharge), would we have a quantized vortex of Lenny Susskind’s zilch?
Which suggested a new twist on the old axion idea. Meanwhile, could such a “cold” axion, a classical quantum of curved action, actually define Planck’s constant? Such thoughts sprang to mind when considering minimal waves in your marvelous sea!
Nigel
Thanks a lot for the book, it inspired me and made me think (I’ve recently had several intense half-nighters). I don’t want to spam your blog, so ask for an advice of how and with whom better to share:
– Short fictional story illustrating the equivalence principle of GR (in English, characters excluding spaces – 4438)
– Simple geometric derivation of Lorentz transformations, relativistic energy-momentum relation, invariant mass, for boson-like wavicles.
I’m glad you enjoyed it. Where are you located, Serge?
In Israel, so it was quite easy recently to stay awake in the night.
Then there are many great experts on these subjects within an hour’s drive. Which universities are closest to you?
Weizmann Institute of Science, Tel Aviv University, Bar-Ilan University
Several questions about cosmic walls.
– Does the wall of doom you described in The Wizardry of Quantum Fields run at the cosmic speed limit? I think so as you mentioned the “unsuspecting” universe, and we humans will just cede to exist if the wall reaches us, like unfortunate passengers of the Titan submersible for whom its catastrophic implosion went unnoticed.
– The Hubble constant of 70 km/s/Mpc means that at ~4Gpc the expansion is superluminal and nothing which is father away now, including a wall of doom, could reach us. Therefore, we only need to worry about the local walls which have much lower probability to appear due to the restricted volume, right?
– I’m thinking of a more benign version of the domain wall resembling the iron’s magnetic domains and their walls. Suppose the Higgs field acquired some constant non-zero value but with different potential phase across the universe (considering the Mexican hat shape). If so, are the different vacua “compatible” with each other, i.e. can a wavicle traverse the wall between phase domains? And how such a wall might look like?
Excellent questions.
1) In most models it approaches the cosmic speed limit closely after a short time, and yes, we don’t get much if any warning.
2) Yes, what happens outside our cosmic horizon cannot hurt us, but there are some regions that are currently outside our horizon which might someday come within it. This depends in part on what “dark energy” really is and how it evolves.
3) Domain walls of this sort are not possible with our Higgs field, because the different phases you are imagining are physically equivalent (“gauge equivalent” in the language of physicists.) This is different from a magnet where the different directions of pointing are physically inequivalent, making domains and domain walls possible. [Buzzwords: global symmetries can have inequivalent domains related by a symmetry; gauge symmetries are fakes — they are symmetries that act on the math used but not on the physics — so they cannot.] Other scalars fields could in principle have domains; but indeed, such domain walls have big effects on cosmology and so are sharply constrained by observation.
Matt, the electron field interacts with the electromagnetic/photon field to produce an electron; but also an anti-electron, aka positron, to maintain charge conservation. The fact that the two charges are related by a sign change is something I find fascinating and, to my mind, appears to be saying something about (a) some property of the photon that needs to be conserved in the interaction or (b) to maintain a symmetry such as parity: Is this vaguely correct?
It’s mainly just electric charge that’s at stake. The photon’s interaction with the electron field requires that whatever the former does to the latter, the total electric charge can’t change. Therefore, no matter what happens, a photon can only be transformed into n electrons and n positrons (where n is most commonly 1).
Is there a version of the Standard Model that is mathematically consistent in allowing like charges to attract rather than repel?
If so, how does this influence the need for positrons within it?
I’m thinking that at the point of creation, electron/positron wavicles will attract one another because of quantum gravitational effects from their interacting rest-energy/mass; decreasing with adequate separation. Likewise an additional electric attraction is needed to ensure work is done in separating two charged wavicles created at the same instant.
Here’re a couple of ideas for the page’s name to discuss the book’s pedagogical choices. First is “Final Draft” and inspired by a recent addition to the Alan Wake computer game series. The game protagonist is an eponymous writer drawn to some kind of a parallel world by an obscure, powerful and evil entity called Dark Presence. And by reading your “strange space-suffusing entity, an enigmatic presence known as the Higgs’ field” passage it immediately stroked me how similar they are. Another is “Yoctosecond Edition” which plays with the double meaning of “second” and refers to the smallest timescale currently accessible.
The eponymous game protagonist is a writer …
Hi Matt,
Great book (and blog)! You clarify some major issues that, it is true, are left unanswered, or maybe are taken for granted, in most sources (eg. the fact that “particles” are wavicles, etc)!
My question is this: you say, on the one hand, that mass is intransigence while, on the other, that mass doesn’t slow things down. How can these two statements be compatible with each other? And, in the end, what is it exactly that doesn’t allow objects to reach the cosmic speed limit? Mass or inertia? (I recently heard about an, as yet unverified, idea about “quantized inertia” that attempts to explain this fact)
In 12 What Ears Can’ t Hear and Eyes Can’ t See you discuss the question of why a rainbow is narrow. Your explanation implies that dispersion spreads initially bright narrow arc to the spectrum and we see just a small portion of it. That’s true but not the whole story. The question is why that bright narrow arc appears in the first place. The sun has a finite angular diameter of half a degree, and its rays need to be redirected three times to reach an observer. Any initial misalignment would be amplified during the redirection within the constrained drop’s geometry, and we’ll end up seeing a wide band of light. But a thorough analysis shows that the deflection function has an extremum near the rainbow’s angle which acts like a lens focusing part of incoming light into the narrow band. There’s even another extremum having more complex optical path, which gives the outer, less bright arch with inverted dispersion.
You were such a courageous boy jumping while flying in a jet. You might have expected that if you’d jumped high enough, you would have been tackled by the jet’s wall rushing at 500 mph! Here’s a fun exercise also playing with relativity principle, which I came up with many years ago during my long subway commutes. In a subway car take a seat for you to look sideways. Suppose the train goes to the right of you, and you’re in a long steady hop between two stations. Close your eyes. Without any visual clues, feeling just monotonous bumps and shaking, it’s rather easy to trick you mind thinking that the train goes in opposite direction, to the left of you, because by relativity principle they’re indistinguishable. And when the train starts braking, you’ll feel – quite opposite – rather intense speeding up! The feeling lasts until you realize that instead of intensifying, bumps and shaking subside. *Initially I posted it on “Got a Question” page, but now can’t find it there, so repost here.
Can’t stop thinking about Lorentz-invariance. Here’s another toy/wild idea inspired by your latest posts on zero-point energy and standing waves. Starting points:
– Cosmic fields (vacuum) energy density depends on the smallest scale of its constituent parts
– Non-interacting wave excitations (particles) gradually spread
– We have one example when the cosmic field(s) property – curvature – are influenced by matter-energy
Let’s assume that a) the cosmic fields’ constituent parts are like non-interacting particles. Then in absence of other disturbances they will spread, overlap, and drive the smallest scale up suppressing the zero-point excitations. Then assume that b) in presence of any form of matter-energy these constituent particles interact with it and localize, driving the zero-point excitations to their usual level. Then it’s plausible that localization interaction is Lorentz-invariant because it happens due matter-energy presence, i.e. within its frame of reference. Another advantage that it might alleviate the vacuum energy problem as only induced zero-point excitations count.
Concerning the vacuum energy and induced vs suppressed zero-point excitations. On intergalactic scale the suppressed zero-point excitations might still dominate as the space there is mostly empty. Then this model predicts the vacuum equation of state value as close to but still larger than minus one (due to dilution of the induced excitations during the expansion), which is vaguely consistent with the observed value.
Ok, realized that the assumption for constituent particles to be non-interacting contradicts the need to provide a sort of stiffness in response to gravity.
Dear Matt. I have tried to order your book using the discount code but I live in Spain and it only allows me to put a USA address. How can I get the book shipped to Spain?
Kind regards,
Julian Collins
I’ll have to ask the publisher; the publication and shipping of books remains a black box to me. It may be that you have to go through their UK office and I’m not sure they are offering the discount there.
Indeed, you can’t get the US discount, but the publisher says that Amazon in Spain is offering a comparable discount, and that is probably your best bet for the moment. I hope that works for you!
It appears that the reason that 2 atoms cannot overlap is the same as the reason that we have white dwarves or neutron/quark stars. Could you expand on this topic in a blog post on this topic ?
You are correct, and yes, I agree that is a topic that needs to be added to the website.
Hello Dr. Strassler, I’ve been a reader of your blog for several years (your series of posts culminating in measuring the distance to the Sun via meteor showers is fabulous) and I really enjoyed Waves in an Impossible Sea, so thank you for your writing!
I have a question about confinement. In your post “A Half Century Since the Birth of QCD” from November 2023, you describe how confinement is the result of the dual Meissner effect. My question is, does this effect occur in classical Yang-Mills field theory, or is it strictly a quantum effect?
Thanks for your kind comments about the blog and the book!
Confinement is a quantum effect. In fact, the classical theory has a symmetry, known as a “scaling symmetry”, as does electromagnetism; that symmetry implies directly that the force between two particles at a distance r must be proportional to 1/r^2. Both quantum electromagnetism and quantum Yang-Mills theory violate that symmetry; in the former case the force is a little weaker than 1/r^2 at long distance, while in the latter case it is stronger at long distance. This part is easy to calculate in the quantum theory. What is hard to prove is that in the quantum theory the force eventually becomes constant — independent of r. It is this constancy of the force that represents true confinement. It is not only a quantum effect but a non-perturbative effect (i.e. not calculable using the simple techniques of Feynman diagrams or their equivalents.) It can only be caculated using computers — the numerical methods of lattice gauge theory. If someone can prove it using pure mathematics, they win a million dollars from the Clay foundation. Such motivation has not been enough to generate a solution, however.
Dr.Strassler:
I have a question inspired by one of your posts on one of your other sites, about the two slit experiment
its my understanding, that in order to see this characteristic of wavicles, the two slits have to be very close together…..like less than a millimeter, does that sound correct?
also, in the two slit experiment, say for a single photon, how is momentum conserved? in other words, if I fire my “photon gun” one photon, and the photon gun recoils in the negative X direction, the photon should, have momentum in the positive X direction. After it passes thru both slits, and ends up as a mark on the wall…..absorbed by an atom not directly behind the slits, what did it exchange momentum with? does it exchange momentum with the slits, as it passes thru both of them? If the photon gets absorbed by an atom, way off to the side, not directly behind the slits, it must have had sideways momentum given to it at one point. I watched a lecture by Dr. Susskind on this, and he seemed to indicate momentum was conserved, but didn’t say how…where was the momentum exchange?
There is an interplay between the slit spacing, the wavelength of the light, and the size of the interference pattern, so your question doesn’t have an answer unless you are a lot more specific. Not also the interference pattern itself is not a sign of quantum physics; that happens with classical waves. It’s only when you see the interference pattern built up photon by photon that you are seeing quantum physics. [There are many wrong YouTube videos about this.]
The photon can exchange momentum with the wall, just as a classical light wave can.
Thank you so much for your book, which I just finished reading with great interest. Your use of the wavicle concept makes so much more sense to me than the Copenhagen interpretation. One question (of many):
If a single photon is released that could be seen by two observers an equal distance away from the source, will only one of them see it, or both? Is this hypothetical similar to the two-slit experiment or different? Many thanks.
Great question., I wish I could tell you that thinking in terms of wavicles resolves some of the puzzles in quantum physics, but it does not. It rephrases them, to some degree, and I think it prevents one from getting caught up in wrong ways of thinking (while the Copenhagen interpretation of “particle-wave duality” easily leads to unnecessary confusions.) Nevertheless, the puzzles are all still there. I carefully skirted them in the book because the story of wavicles and their rest masses does not require resolving them, and they would have been an enormous distraction from that story.
As for your question: Only one observer (at most) will see your photon. And yes, it is very similar to the double slit experiment. The photon goes through both slits; it interferes with itself, creating a spread-out, complex pattern. Yet only one atom on the screen will grab it.
However, this is not because a photon is a dot-like object when it is absorbed, as the Copenhagen interpretation would want you to imagine. The photon is still a wave, with a frequency. The absorbing atom, too, is not a dot; it has a radius of 1/3 of a billionth of a meter, which is not small on the scale of subatomic objects. The absorption process is an interaction among waves — or more precisely, an interaction among wavicles — that allow the photon to be absorbed by the electron wavicles that make up the outskirts of the atom.
But the process by which the atom absorbs the spread-out photon, thereby making it impossible for any other atom to absorb it — and how this process is described in terms of probabilities rather than certainties — remains confusing. Either you use the many-world picture, in which the universe proliferates into a gazillion branches of possibilities, or you assert that the equations of quantum field theory leave something out, or… or like me, you sit back and hope someone smarter than you will come up with a better way to think about it.
The many-worlds interpretation has the merit of being consistent with the equations (which the Copenhagen interpretation is not) but that does not make it intellectually or emotionally satisfying to most practitioners. It is popular, though, since it seems to be the best we have for now.
Many thanks. This is very helpful. As between the many-worlds interpretation and confusion, I prefer the latter…
Matt,
Swimming with you through the wavey sea was great! No re-reading, except a paragraph here and there, was needed (although I am). You done good!
As a non-physicist and non-mathematician, I have been reading to understand quantum mechanics for six decades. Ruth Kastner’s Transactional Interpretation finally makes sense to me (after four readings) so I would like to see if it is consistent with yours.
While you offer many demurrals, I understand you to be describing that wavicles traverse trajectories in a pre-existing space-time container to translocate energy. Alternatively, Kastner’s (not Cramers’s) Transactional Interpretation proposes that instead, waves interact outside of space-time to translocate energy thereby creating events and their space-time separation. Note that this interpretation replaces particles, trajectories, and a space-time box with events, pre-space-time waves, and an evolving space-time.
I wonder if your “demurrals” would make yours and Kastner’s interpretations compatible?
Many thanks,
Tony Way
Dallas
I’m afraid I can’t comment on that, since I don’t know Kastner’s interpretation well enough. But I am not taking a position on the interpretation of quantum physics in the book, just trying to give readers a useful picture without claiming that it is complete. I did make the remark that we don’t know if space exists or should be thought of as fundamental, and so for myself, as a scientist, I’m not assuming a space-time container in my research. But the picture I provided to readers does assume it, and might need for that very reason to be replaced someday — a point that I tried to make clear at the end of Chapter 14.
Where can I find the Mathematica programs that you’ve used to make your blog animations, esp. the ones from “Fields and Their Particles: With Math?”
BTW, I’m loving reading your new book. Thank you so much for writing it.
Mathematica is here: https://www.wolfram.com/mathematica/
So glad to hear you are enjoying the book!!
I meant to ask where I can find your animation codes. I would like to try to reproduce you plot animations.
Ah. This is all stuff I write myself and it’s kind of a mess — not commented or anything. And I have to think about what I do and don’t want to release. (Believe it or not, despite 10 years of animations, you’re the first one to ask.) Lemme think about it. There are more animations coming and you can remind me about it.
p.s. I have replied to your other message via email, which may settle the issue you’re really most interested in.
It did. Thanks.
Dr. Strassler:
finished the book, absolutely loved it. I have placed it in my library right next to The Feynman Lectures on Physics. I would like, if possible, sometime in the future, a more detailed discussion of how a wavicle starts to spread out, but then collapses back to “point like” when it collides (exchanges momentum / energy) with another wavicle, and then starts spreading again.
Great to hear you enjoyed the book!! Please leave a review on Amazon or GoodReads if you have the time.
As for how to think about what happens when a wavicle collides with another wavicle — ah yes, I’d like a description of that too. This lies at the heart of perhaps the most difficult conceptual issue in physics. It’s a great question, but I’ll have to build up a whole infrastructure to even define and illustrate the question, so this will be something I probably won’t return to for months or even longer.
I’m eager to get my hand on the book now, blame the recast site – the notion of science as “spectator sports” or since I rarely watch sports I don’t perform myself perhaps “spectator architecture” seems fitting. So seeing we have to wait for months or even longer perhaps some preliminaries to expand on the question or to have preliminary partial answers?
In the following I may make the mistake of confusing the wavicle with the wavefunction but unless given hints I have to assume it is essentially the same interaction collapse. FWIW then here is a related question: Is the work “Answering Mermin’s challenge with conservation per no preferred reference frame” published in the less cited Nature Communications useful (does it survive basic criticism)? [Stuckey, W.M., Silberstein, M., McDevitt, T. et al. Answering Mermin’s challenge with conservation per no preferred reference frame. Sci Rep 10, 15771 (2020). https://doi.org/10.1038/s41598-020-72817-7%5D The conceptual issue does not seem testable but the work reinterprets “wavefunction collapse” as a relativistic effect. My naive view is that it adds to the wavepacket time dilation and length contraction the interaction collapse in order to conserve the wavefunction spin for the observer (here: interaction partner?). Then if we can familiarize ourselves with the first two effects, we could do the same with the potential third candidate.
Well, this is not the type of question the book attempts to address; in fact, I deliberately skirt these issues.
You are indeed at risk of confusing wavicle with wavefunction. But worse, it’s far from clear that wavefunctions collapse, or what it would mean for them to do so in a relativistic theory, or what would cause collapse, or how that collapse could be describable and predictable. This is why many of my colleagues either subscribe to Everett’s many-world view (as do Sean Carroll and Max Tegmark) or throw up their hands in dismay and confusion (as I do.) Someday I will try to explain the problem carefully; I have no solution.
Regarding that specific paper (DOI number is wrong, but it is also here: https://www.nature.com/articles/s41598-020-72817-7 ), I would have to study it. Any serious papers trying to clarify how quantum physics really works have to be studied with care; otherwise one is likely to come away with the wrong impression. But real progress on a problem that has troubled us for a century is likely to require either a new experimental discovery or a profound new theoretical idea. This paper sounds potentially interesting, but not likely to be significant. I would expect bigger ideas to arise potentially from the interplay of quantum computing and quantum gravity research.
Matt, thanks for the response! Yes, the paper discuss a relativistic consistent cause for collapse/Born postulate, which else is seen as a problem in search for bigger (likely testable) ideas.
Sorry, I mean to post this as a question but it has been posted as a reply.
I answered it under “Got a Question” as others may share the puzzlement.
I suspect you may be aware of this, but there’s a significant typo in the Book.
Namely, at the start of the Notes section (on pg 339) the address corresponding to the “asterisks in the endnotes” is given as:
http://www.profmattstrassler.com/WavesInAnImpossibleSea
Unfortunately, this like take you to a “404 page”
Thanks!! The book’s correct, but we have a missing redirect on the site. I’ve put in a temporary redirect and will make it automatic shortly. Try it again!
Thanks. Just tried this and the redirect works great.
[P.S. Just a quick note to tell how much I’m enjoying —and learning from— WiaIS.
I’ve been (highly) recommending it to all my friends who find these things interesting.
Finally, I very much enjoyed your appearance on Sean Carroll’s podcast.
And I want to thank you for one specific point.
That is, I’ve struggled for a long time to understand the precise mechanism as to how “the Higgs mechanism imparts mass to certain elementary particles”.
During the episode you briefly mentioned, almost in passing, that the Higgs field affects (say) the electron field in such a way as to alter the form of the waves in the field, consequently changing its frequency, which in turn “determines the mass of the particle”.
I can only tell you that when I heard this, simple as it may have been, lightbulbs went off all over the place. Thank you once again]
Nichael Cramer
Guilford VT
Great to hear! When you get to chapters 17-20, I hope the lightbulbs will turn into blazing stars.