Spontaneous Symmetry Breaking and The Higgs Mechanism

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  • Опубликовано: 29 ноя 2024

Комментарии • 86

  • @corsaircaruso471
    @corsaircaruso471 Год назад +22

    This is such a wonderful combination of accessible but thorough. Certainly one of the best examples I’ve ever seen. Bravo, and thank you.

  • @passingshots
    @passingshots Год назад +18

    This is the closest I've come to grasping the concept after many years of watching layman videos. Well done

  • @nice3294
    @nice3294 2 года назад +33

    This is probably the best explanation I've heard so far! Well done

  • @Asrequired274
    @Asrequired274 8 месяцев назад +2

    This made so much more sense than what I was taught in my advanced qft and susy courses in Oxford. Thank you!

  • @alice.vnssss
    @alice.vnssss 2 года назад +5

    I've been looking at many videos or articles on this topic and this is the one i find explained the topic the most clearly, thank you !

  • @tongyizheng4289
    @tongyizheng4289 7 месяцев назад +1

    I hereby announce that you are my third favorite youtube channel on physics!

  • @ButchNews
    @ButchNews Год назад +1

    In order to be in the same place, at the same time, one must move at 4/3pi (speed of light cubed) for a given time... as long as you remain at that velocity, you will present as an object... you may be moving along a path (inertia) but you will occupy a volume of space as you move. It is the formula for the volume of a sphere where the radius is the speed of light for a given amount of time.

  • @PrticlePhysicsMasterclass
    @PrticlePhysicsMasterclass 7 месяцев назад +3

    Thanks and appreciate your work! Great job and video!

  • @narfwhals7843
    @narfwhals7843 2 года назад +13

    Hi ZAP! Glad to see you uploading :)
    I'm confused about a few points.
    First the zero particle state. Don't we want a particle to be a thing we can measure? Otherwise why talk about particles at all? So shouldn't the zero particle state be a well defined state with a certain energy?
    Then I really don't like the fluid analogy for "giving mass", but perhaps I am misunderstanding this.
    A fluid _slows down_ the thing that moves through it until it is at rest with respect to the fluid. That's just friction. It acts on the the first derivative of position.
    But mass is about inertia, is it not? it resists _acceleration_ so it acts on the second derivative.
    It _defines_ a rest frame for the particle. Or is that not a useful way to think in this context?
    Or is that the same thing as saying it gets "slowed down" from moving at the speed of light? Which is the same thing as having a rest frame.
    That seems like it would lead to problems with special relativity.
    And around 14:00 you talk about the possible ways the field could fluctuate. These seem to be highly dependent on our choice of coordinates. Could the field not fluctuate about any arbitrary direction, leading to various masses? Or are these just the obvious extremes?
    I am also getting a bit wary of the language. Is this a particle physicist's interpretation of the underlying field theory? Or is thinking about particles the easiest way to get a grasp on this? I'm not sure anymore everyone means the same thing when they say "particle" in fundamental physics.
    Lastly I have a question about the W boson mass anomaly. People keep saying that the measured value differs from the standard model prediction for the W mass. But as far as I understand the standard model doesn't predict any of the masses. Is this about the difference between bare and dressed mass? And the discrepancy is in the dressed mass, indicating unknown interactions?

    • @zapphysics
      @zapphysics  2 года назад +12

      @Narf Whals Nice to see you still are asking the big questions :) I will try to answer them as best as I can.
      In regards to the zero-particle state, I want to answer your question with a question: how do we observe particles? We observe them through their interactions with some field or detector and this interaction is described by some theory. Now, if I change the zero-particle state in the theory, this will just change how my theory describes the interactions going on in the detector. So, for example, say I have a detector which is set up to detect particle A where initially, the zero-particle state for A is the same as the vacuum state. Then, it is clear that when an A particle hits the detector, I will see a single "ping" in the detector. Now, say in my theory, I shift my zero-particle state so that A->B where the zero-particle state of B is misaligned with the vacuum state. Based on this theory, I can now calibrate my detector to see B particles (fluctuations around the zero-particle state of B) and I no longer see a single "ping" but instead, a steady background of B particles. The key here is that A and B are very different particles and behave differently in the detector, but it's sort of artificial: I can subtract off the steady background of B particles and look instead for fluctuations around this background, but this is exactly the same as looking for A particles. Obviously, for real-world experiments, this isn't practical and it is much better to just use the theory where the zero-particle state and the vacuum are aligned, but it can actually be used as a helpful mathematical tool in some instances. In fact, in gauge theories, one can extract a lot of information by using such a constant shift and looking at interactions with the background field (this is known as "background field gauge").
      I think that your critique of the fluid analogy is a very valid one and honestly, I don't have too much of a defense for it. Intuitive analogies for this are quite challenging without writing down Lagrangians etc., but maybe a more precise thing to say is that as a psi particle propagates, the phi particles which are popping out of the vacuum are imparting some energy to psi. When we shift our zero-particle phi state to be aligned with the vacuum, this imparted energy sort of averages to a total, intrinsic potential energy for the psi particle, which is of course, another way of saying that psi has a mass.
      It actually turns out that our choice of fluctuations is somewhat unique. By that, I mean if we tried fluctuating in other directions, one would get cross-terms in the Hamiltonian that lead to mixings between different particles as they propagate through space. The fluctuations in the video are the "physical" states in that they are the energy eigenstates (the states which actually propagate through space without mixing). Said in other terms, these are the states which diagonalize the Hamiltonian and are unique up to mass degeneracies.
      I think that your being wary of the particle physics language is actually a good thing. Spontaneous symmetry breaking is definitely not just a particle physics thing, and the whole framework I believe was established in condensed matter physics long before it was used in particle physics (this is actually quite often the case for many particle physics phenomena it turns out). As I said in the video, this is all coming from a particle physics perspective because of my personal bias and everything in the video essentially makes an underlying assumption that we are dealing with weakly-coupled theories where it even makes sense to talk about particles to begin with. Spontaneous symmetry breaking also happens in strongly-coupled theories (i.e. low-energy QCD), but this sort of language fails since particles in such a theory are not well-defined.
      For the W mass anomaly, this is a fantastic question. You are exactly correct that one cannot usually predict the values of parameters in a quantum field theory like the standard model. However, the weak interaction is a bit special because it arises from a spontaneously broken symmetry. One result of this is that weak and electromagnetic parameters are not all actually independent of one another, so if one knows a certain set of electroweak parameters from experiment, one can use these as inputs to fit for the other parameters. This is what is done for the theory prediction of the W boson mass. The basic idea of the CDF measurement "not agreeing with the standard model" is another way of saying that it suggests that this relation between the parameters might not be fully correct. To answer your question of which mass they are talking about, I believe that they use the "on-shell" mass, where the mass used in the Lagrangian is the honest-to-goodness Lorentz-invariant rest-mass of the particle.
      I hope that answered all your questions, but feel free to clarify or ask any more which might come up!

    • @narfwhals7843
      @narfwhals7843 2 года назад +2

      @@zapphysics Thanks! I really appreciate that you actually take the time to answer the "big questions". Being able to get these answers is incredibly helpful!
      Is this arbitrariness of the zero particle state related to Unruh and Hawking radiation? Everybody agrees that they are observing the vacuum state, but one observer ascribes to it some particles while another says there are none. But the accelerated observer could just redefine their zero particle state to see no "real" radiation?
      I don't think I understand that. If my detector gives me a "ping" it took some energy from the particle(perhaps amplified by the apparatus itself, but that energy should be distinct from the vacuum, no?). If I reconfigure it to give me pings from the vacuum that means I have to constantly put energy into it to drive the pings.
      I think what I'm looking for to understand the Higgs mechanism is what it does to the wave equation. I don't think analogies are helpful for me at this point.
      You say it adds an intrinsic potential. So in the wave equation we get a term that constrains the frequency and wavelength of the field excitations, correct?
      (I'm basing this question mostly on this part in Physics Explained's video on what is a particle ruclips.net/video/QPAxzr6ihu8/видео.html )
      So the magnitude of this term determines how much mass we get? And we say that is how strongly a particle couples to the Higgs?
      What does this actually depend on and how does it relate to m=E/c²?
      Is there something about this being a relativistic field theory that then places a constraint on the propagation velocity?
      I'm trying to unify in my head the ideas of mass in a field and the existence of a rest frame and notion of mass in special relativity.
      We didn't have a rest frame(one we could easily lorentz-boost to) before we broke the symmetry but we do once it is broken.
      So we chose the coordinate system that allowed us to be the laziest? ;)
      If we "force" another fluctuation It will not be an energy eigenstate, so it won't persist over time?
      Do you, as a particle physicists, think in terms of particles as most people imagine them? Or is it just convenient language and in your head you always translate back to fields?
      The reason I'm wary is because it seems like some science communicators imply this translation(like Sean Carroll) while others actually encourage thinking of individual objects.
      I'm interested to know if that's a miscommunication or an actual difference in mindset. Or does it depend on what you're working with? (As you say, thinking of particles isn't useful in a strongly coupled theory)
      So we can (very) broadly speaking calculate the mass of the W from the charge of the electron? But even these parameters are always "dressed" in experiments, how can we possibly hope to measure them accurately enough to then get such small uncertainties for the parameter we want to calculate...
      And isn't the honest-to-goodness Lorentz-invariant rest-mass of a particle an expectation value, too?

    • @zapphysics
      @zapphysics  2 года назад +5

      @@narfwhals7843 Excellent, I had a feeling that Unruh/Hawking radiation might come up. The short and important takeaway from this is that no, these are fundamentally different. In the case of just inertial observers in flat spacetime, when we shift a field by a constant amount, we still see the same vacuum as before, it just is now populated by particles since we have changed our definition of the particle. In the case of an accelerated observer (the story is similar, but a bit more complicated for a black hole), we can be measuring the same particles (i.e. I never have to re-calibrate my detector), and we will disagree on the *vacuum state*. In other words, if we take the zero-particle state for both observers to align with the vacuum state, the accelerated observer will still see particles while the inertial observer sees nothing. This bath of particles that the accelerated observer sees is also not uniform: it follows a thermal distribution (like a blackbody spectrum), so that is another way to see that these are different cases.
      In terms of taking energy from the field when the detector sees a ping, this isn't really an issue since, when we shift our field by a constant amount everywhere, we are essentially adding or subtracting an infinite amount of energy to the system.
      An important thing to remember is that the wave equation describes a non-interacting, massless field. I will try to show how this changes on the level of the wave equation, so I apologize for the formatting.
      So, we can start with the wave equation for the field A(x,t): (dtt - dxx)A(x,t) = 0 where dtt and dxx are second derivatives acting on A and I've set the wave speed to 1 for simplicity. Now, if I make the change A -> -A, this equation doesn't change, so the wave equation has the same reflection symmetry as was in the video. Now, say I add in some interaction with another field B(x,t) so that the equation is still symmetric under A, B -> -A, -B. One way to do this is to use the equation (dtt - dxx + a B(x, t)^2)A(x, t) = 0 where a is the coupling between the fields A and B. Now, suppose that B spontaneously breaks this reflection symmetry. Then, the vacuum state of B will no longer be at B(x,t)=0, but instead at some value B(x,t)=v so that in the B vacuum, the equation for A is given by (dtt - dxx + a v^2)A(x, t) = 0. Alternatively, we can redefine our field B(x,t) = v + b(x,t) so that b(x,t)=0 in its vacuum state (a convenient choice), in which case we can describe the back-reaction onto this b-field through the same equation
      (dtt - dxx + a v^2 + 2 a v b(x, t) + a b(x, t)^2)A(x, t = 0. Again, notice the addition of the term a v^2 A(x, t) to the equation which stays around even when b is in its vacuum. However, this term has the same form as the mass of a field in the Klein-Gordon equation (dtt - dxx + m^2)A(x, t) = 0, so you see that adding an interaction to a field which spontaneously breaks the symmetry results in a mass.
      For the fluctuations, it isn't necessarily that the states won't last, they will just oscillate, exactly the same way that neutrinos oscillate between weak-interaction eigenstates because these states are not the same as the energy eigenstates.
      From my experience, it gets hard to not think of things as particles because so much of the language in the field is framed this way. For the most part, it isn't too much of a problem since the particle-view tends to be quite a good approximation to the field dynamics. Even when dealing with a strongly-coupled theory like QCD, one can just trade out talking about the quarks and gluons for the hadrons whose dynamics seem to be particle-like. However, this becomes quite messy on the theoretical side so I have mixed feelings about it. I think the big issue is the fact that the true field dynamics can't be solved for in physical cases, so it is hard to develop an intuition whereas particle interactions and Feynman diagrams are a much easier way (though not necessarily fundamentally correct) of understanding what is going on. Now, there are interacting field theories which can be treated in their full, fieldy glory (such as superconformal field theories like N=4 super Yang-Mills) and there are large communities dedicated to studying such theories in order to develop this language more.
      I think there are a couple of questions about this W mass that I want to tackle one at a time. First and foremost, yes, these relations are all found between "dressed" parameters which all are found using some particular renormalization scheme. The important thing is that one stays consistent with these schemes. As for accuracy, this is one of the reasons to do this fit to begin with: it is typically much easier to measure observables which depend on e.g. the EM coupling, weak decay constant, and Z-mass (again in some particular choice of renormalization scheme that must be specified and kept consistent) than the W-mass. The perturbation series for electroweak observables also tends to converge quite quickly, so these relations tend to be pretty precise. I believe that the least accurate input into this fit is the top-quark mass (again, in a particular choice of scheme which is kept consistent), but this arises only at higher orders in perturbation theory, so the errors stemming from this input are going to be suppressed by a factor of 10 or more due to the small couplings.
      A final comment on the mass: the rest-mass of a particle does *not* fluctuate, even in quantum mechanics. If it did, the theory would not be Lorentz-invariant. However, when dealing with unstable particles that we can't simply weigh, one has to use relations like cross-sections, etc. which depend on the mass and *can* fluctuate due to quantum effects. However, this is not a fluctuation in the rest-mass of the particle. For the same reason, a massless particle can never travel at any velocity besides the speed of light, even in quantum mechanics. This is the whole beauty of using field theory: Lorentz invariance is built into field theories, so even when they are quantized, this invariance stays exact.

    • @narfwhals7843
      @narfwhals7843 2 года назад +1

      @@zapphysics Hi! Sorry for the long delay. I've been trying to gather my thoughts about this and I've come to the conclusion that I simply don't know what mass is.
      I think what I mean when I say mass is inertia. And in Special Relativity this has two meanings.
      First the property of not having to move at the speed of light. This is a binary concept, you either have it or not.
      Second the resistance to acceleration. This is proportional to the energy contained in an object.
      So when we trap light in a perfectly mirrored box, this box contains the energy of the light. The box does not move at the speed of light and the energy of the light contributes to its inertia.
      It is tempting to think of the Higgs mechanism in a similar way(and I think the particle bouncing around analogies contribute to this!). The Higgs interaction somehow confines the energy of the particle in some region and so this region gains inertia. This sounds _very_ fundamentally wrong.
      But you have told me before that it is not necessarily useful to think of "mass" as "confined energy"
      So what does mass mean for a particle physicist? Is it just "this term in the equation"? This term does something to the solutions of the wave equation, namely giving them a minimum frequency. And we call this frequency*h/c² the "mass" of the particle?
      And this frequency depends on the strength of the coupling and this strength is not something the standard model predicts?
      But these minimum frequencies still have the same propagation velocity, which in a relativistic field theory is defined to be the speed of light(aka 1). So where do we get the binary property of inertia?
      Feynman shows how we can get a moving particle from a superposition of planewaves, so the very reason particles move is the uncertainty in momentum. Is it something like this? Do we get an inertial "group velocity" from a superpostion of massive plane waves?
      And how can the rest-mass not have an uncertainty attached to it? It must still be a measure of energy, which must be uncertain.
      Regarding Unruh radiation, so the disagreement here is really what the vacuum is, not what a particle is? The accelerated observer does _not_ observe a vacuum state?

    • @zapphysics
      @zapphysics  2 года назад +4

      @@narfwhals7843 I think that the key thing here is to remember that, at its core, we are really dealing with quantum fields. Perhaps the best way to think of mass in this context is as the absolute minimum amount of energy necessary to fluctuate around the vacuum state. If the energy of fluctuations around the vacuum can be arbitrarily close to zero, then the particles observed from these fluctuations will be massless, but if they have some non-zero minimum energy cost (i.e. if the theory has a "mass gap"), then the resulting particles will be massive. So, for example, if you look at the Millennium Prize involving Yang-Mills theory (this is the theory of only gauge fields with no additional fields), the prize isn't for solving the theory in general, but it is to show that the theory has a non-zero mass gap. In other words, the question is: if we have any Yang-Mills theory, will the minimum-energy particles associated with the theory always be massive?
      So, perhaps one way of seeing how the Higgs mechanism works in this context is that, once the scalar field gets a non-zero vev, any other fields which couple to the scalar will be constantly interacting with this vev which essentially adds an additional, finite energy cost to fluctuate around these other fields' vacua, meaning that the resulting particles will be massive.
      Note that this is completely unambiguous in special relativity: all inertial observers agree on the vacuum and the vacuum is a physical state. So, whether or not the field has a mass gap (whether it is "intrinsic" or if it is generated via interactions) is the field-theory connection with your binary inertia statement: the field either has a mass gap or it doesn't.
      Regarding the uncertainty, I think I understand what you are saying and I think it somewhat comes down to a matter of philosophy (or probably more specifically, the limitations of QFT). In QFT, the mass of a particle is determined by the analytical structure of the theory. Certain quantities (two-point correlation functions) have poles whose locations are given by the masses of the fields. Now, these poles are exactly local: they are unambiguously found at a single point. This is what I mean when I say mass doesn't have uncertainty in QFT. Again, this is consistent with special relativity since the square of the four-momentum of a particle is an unambiguous, frame-independent quantity equal to the square of the rest-mass of the particle, so if our rest mass is allowed to fluctuate, then we are actually breaking the rules of special relativity.
      However, since the QFTs we have don't *predict* this mass, we can never know the location of these poles exactly due to the uncertainty principle since there will always be some uncertainty associated with our measurements. I think this is what you are thinking of when you say that the mass *does* have uncertainty (please correct me if I am wrong here) and I think that this a perfectly correct and valid statement.
      For Unruh radiation, you are exactly right. If all fields are in their vacuum state for inertial observers, an accelerated observer will not see a vacuum state, and vice versa.

  • @julianr1778
    @julianr1778 Год назад +4

    Thank you a ton for this! Really makes this topic understandable

  • @maxmann6933
    @maxmann6933 2 года назад +10

    Hi thank you for another great video!
    The only thing I don't particularly like is explaining the Higgs mechanism in terms of friction. There is a great video out from Lenard susskind where he explains it in terms of dipoles in an constant electronic field. Maybe you can take some inspiration from that. Anyway I really enjoy your content!

    • @zapphysics
      @zapphysics  2 года назад +14

      @Max Mann Thank you for the kind words! Yes, looking back, I think that this analogy is also what I am least happy with in this video...perhaps in the future I will make something which clarifies this with a better (and more accurate) explanation.
      I have not seen Susskind's analogy to dipoles before, I will have to look into it, thank you for the suggestion, and I really appreciate the feedback!

  • @seeker313
    @seeker313 Год назад +5

    Excellent!
    Thank you very much for such a comprehensive explanation.
    I will be grateful if it is possible to make a video on Mermin-Wagner-Hohenberg-Coleman theorem.
    --
    With best regards!

  • @blinded6502
    @blinded6502 2 года назад +3

    In case anyone wonders why interactions create mass, you just need to imagine yourself a photon travelling in a circle. Such photon has rest mass, and so it's effective speed equals 0, until we decide to pump momentum into it that is parallel with the axis of it's orbit.
    By the way, this analogy fully agrees with relativistic speed equation of SR.
    v = c*p0/(p0^2+p1^2)^0.5; where p0 is momentum that we gave to the photon, and p1 is mc.

    • @hooked4215
      @hooked4215 Год назад

      Photons do not travell in a circle, they only travel in a straight line.

    •  Год назад

      I'm too unsmart for tjis

    • @ritvicpaarekh6963
      @ritvicpaarekh6963 11 месяцев назад

      So its momentum and energy conserved can then impact space around it where certain psrticles like higgs can interact the generate particles with different properties.

  • @JasonMercurymessenger
    @JasonMercurymessenger 3 месяца назад

    This is consciousness. :-) Making conscious decisions while in various brain wave states, storing memories, generating ideas, sleeping. Conscious electrical energy.

  • @domenicobarillari2046
    @domenicobarillari2046 Год назад +1

    A clearly presented account...very nice. Would not hesitate to present it as a visual and question answer-er in a field theory course. My only criticism might be your RUclips header slide, as it stands likely leaving the student with the notion that this is where - microscopically - most of the everyday mass we notice comes from ( by neglecting to point out somewheres that most hadron mass arises from the strong interaction, or loosely speaking, "chiral symmetry breaking".) But that would require another presentation I think... can we expect one? I would love to see another presentation as good as this one! best regards, D. Barillari

    • @zapphysics
      @zapphysics  Год назад +1

      Thank you very much for the kind words, I am very pleased to hear you liked the video! I also think your criticism is totally fair, and I am definitely planning on making a video on chiral symmetry breaking/ChPT!

    • @domenicobarillari2046
      @domenicobarillari2046 Год назад

      @@zapphysics I am very happy about that. I never intend to be unkind in any feedback! I would be keen to see how you tackle the presentation of such a potentially complex subject. Productions like these expert ones are a cultural treasure in my mind. best regards once more, Domenico

  • @vjfperez
    @vjfperez 2 года назад +1

    The weinberg chair example is strange. Why the schroedinger equation of the chair is not symmetric to a rotation? All atoms wave functions would be just displaced by a rigid motion, and their coupled hamiltonian is only a function of their relative distances?

  • @rayoflight62
    @rayoflight62 Год назад

    The right question to ask would be, I believe, why matter is subject to inertia. Mass has a connection to Space, a connection that we do not understand, but shapes the entire Universe.
    Behind this lack of understanding, is the nature of Space. One of the first fallacies I identified is the correlation between Space and distance. We feel and measure distances, of which the Universe has no idea. The universe work with Space, while the distances are an human construct. We have problems with particles entanglements because of distance...

  • @jrwarfare
    @jrwarfare 2 года назад +1

    This channel is awesome. Keep it up.

  • @Achrononmaster
    @Achrononmaster 2 месяца назад +1

    @18:20 can we have someone give an outline of spontaneous symmetry breaking and the Goldstone bosons that does not refer to the anthropomorphism of "eating"? ffs Something like: the degrees of freedom associated with the would-be Goldstone bosons are physically manifested as the longitudinal polarization components of the gauge bosons, this is what accounts for the gauge bosons becoming massive while still preserving gauge invariance and renormalizability of the theory. Or... in the Lagrangian, terms arise that couple the would-be Goldstone bosons to the gauge fields (by "arise" we mean we cannot suppress them if the Higgs is a reality, which it is). These (unphysical) terms can be eliminated through a gauge transformation, effectively absorbing the Goldstone bosons into the gauge fields. They are unphysical _precisely because_ they can be eliminated by a gauge transform (meaning they were an unphysical redundancy of the Lagrangian formalism). After this transformation, the gauge bosons have acquired mass terms in the Lagrangian, corresponding to their now physical (irremovable) longitudinal polarization.

  • @hiltonmarquessantana8202
    @hiltonmarquessantana8202 2 года назад +3

    Nice! Any books recomendations?

  • @Arseniy_Arseniy
    @Arseniy_Arseniy 2 года назад +2

    Can you explain, please, why do gauge bosons become massive even BEFORE they eat Goldstone bosons via interracting with scalar fields, that have VEV not zero and m local symmetries are violated ? I thing there's smth deeper than water analogy or frequent collisions explanation)
    And why each of massive gauge bosons has 3 states (polarization type)? And why should there be a longitudinal direction of polarization?Accually, I understand the relationship between changing phase angle and what we call charge, but how are these massive bosons and polarization related ? May be there's smth to do with spin? because Higgs have to respect not only phase angle symmetry U(1) but also SU(2) and SU(3). But i can't structurize all of this in my mind
    UPD Aaaaaa! 3 types of polarisation correspond to 3 spin-stares: S(z) = -1, 0, +1 , because S=1. S(z) is projection, S - spin. And you remember that spin is how the particle behaves itself under the space-time transformations. If particle has s=0 it represents as point in space-time, if s=1 (like gauge bosons) - as vector and if s=1/2 (e.g. electrons, proton etc - matter particles) - as spinor (special object which needs to be wrapped around the axis twice so that it returns to its original state)

  • @karkunow
    @karkunow Год назад +1

    Actually due to the Elitzur's Theorem (en.wikipedia.org/wiki/Elitzur's_theorem), there couldn't be a spontaneous symmetry breaking involved in the Higgs Mechanism. So the story told in the universities is actually a wrong one.

  • @suzy6091
    @suzy6091 Год назад

    Why do you sound like one of my mentor? Voice, the way of teaching using drawings, neutrino expert...so many similarities 🤨🧐 Anyway, I love watching and learning from your videos

  • @barryzeeberg3672
    @barryzeeberg3672 2 года назад +1

    I have a "dumb" question. At the beginning of the video, we see the example of a circle that can rotate about its center, and since we see no difference before and after rotation, we consider this is a symmetry. But in actuality, the circumference of the circle is printed using ink, and the molecules of the ink are distinguishable entities. So after rotation actually is different from before rotation, as the ink molecules are displaced. The symmetry was just an illusion, based on using too coarse a viewing system.

    • @jamsonbatista9456
      @jamsonbatista9456 2 года назад

      In the given example the symmetry is about the shape of the object rather than the intrinsic properties of the particles that create the shape.
      How can you distinguish between homogeneous molecules?

    • @barryzeeberg3672
      @barryzeeberg3672 2 года назад

      @@jamsonbatista9456 Well, we need to distinguish between what we can imagine in our minds, and what "really" exists. We can certainly imagine, say, an idealized non-material circle (that, by the way, is not distorted by the fact that a real circle is within a curved space-time). But, anything other than our imagination must at least be composed of molecules that we can detect, presumably by visible pigmentation. Each individual molecule of the pigment is distinct from any other individual molecule of the pigment. This is true whether we can distinguish them or not. If our visual system were good enough, we could presumably see that molecule 1 was laid down a certain number of nanoseconds before molecule 2, and we could keep our eye focused on molecule 1 without losing sight of it for even an instant. When the circle is finally rotated, we would see that molecule 1 has moved, and the rotated circle is indeed different than the original circle. Our inability to see this difference in the real world does not mean that the difference is not there.
      This same type of reasoning could be applied to the basic examples of entropy, such as mixing of 2 gases on different sides of a partition. Just because we cannot see and track each individual molecule does not mean that in principle these could not be seen and tracked. Two "identical" molecules are not identical. If we designate an arbitrary time as t0, then there is "the molecule that was in position x,y,z at t0" and another molecule that was "in position a,b,c at t0". These are different molecules, and in principle we could follow each one, and name it at any time t based on the position it had at t0. There may be some Heisenberg uncertainty about the exact position at t0, but if the gas is dilute enough, then there will not be a huge amount of overlap in the uncertainties of the positions of the molecules at t0.

    • @markell1172
      @markell1172 Год назад

      @@barryzeeberg3672 you want to be unnecessarily precise over stuff that are actually abstract at the moment of studying it, would be funny how you react that the perfect circle abstract existence is actually a consequence of the symmetry of rotation being imposed to exist.

  • @kyspace1024
    @kyspace1024 Месяц назад

    8:40 But you also don't need a spontaneous symmetry breaking to have this effect, you only need the lowest energy state is not zero-particle. Does symmetry really matter in terms of "giving particles mass"?

  • @spencerwenzel7381
    @spencerwenzel7381 2 года назад +1

    Appreciate your work! Great video.

  • @RyanReece
    @RyanReece 2 года назад +2

    Weinberg's chair breaks the symmetries of the theory, but it is not a good example of "spontaneous" symmetry breaking, which means that you can continuously deform the system from being symmetric to not, e.g. rolling down the wine bottle potential.

  • @suzy6091
    @suzy6091 Год назад

    Thanks Bro for this wonderful explanation😊

  • @zlisiecki
    @zlisiecki Год назад +1

    Yes, this is a nice explanation, but what is on the X-axis ? Is it for shure the space x ? If it is the energy on the Y-axis there must be some areas on the X-axis, not individual points. I'd like to know more exact what is this degeneration of energy vaules and it's cause.
    I am very thankful for this video.

    • @TNaizel
      @TNaizel 5 месяцев назад

      The X-axis is time

  • @chalkchalkson5639
    @chalkchalkson5639 Год назад

    I think the particle physics context is probably one of the most difficult to understand spontaneous symmetry braking and even the higgs mechanism. I find the case of a bose einstein condensate much easier as the symmetry, it's breaking and the interactions leading to masses are all very clear things relating to atoms which we have an intimate intuition for. Going through that route doesn't make a friction analogy necessary to "get" the emergent mass. And you even get a nice mexican hat potential more or less for free. It's also neat from a history of science perspective.
    I fully understand that you didn't want to introduce the concepts necessary to go for this explanation though.
    One thing I find very unclear though is how we assign the emergent-ness of masses and fundamental-ity of particles. Why do we see self interaction as non-fundamental, but the higgs mechanism or massive particles arising from broken symmetries as fundamental? I'm not super deep into standard model physics, but there are plenty of particles in condensed matter physics that arise from broken symmetries in the same way as discusses in this video that few would ever consider "real" particles in that system. Wouldn't it be more "pure" in a way to call the fundamental physics those around the symmetric vacuum state and then call the other particles and masses emergent phenomena?

    • @zapphysics
      @zapphysics  Год назад

      Yes, I think this is a fantastic example of SSB as well (of course, like many of the important results in particle physics, it was stolen from condensed matter)! I have been thinking of doing a follow-up with more examples of SSB in lots of different circumstances outside of particle physics, because it is surprising how often it shows up, sometimes without knowing!
      If I am understanding your question correctly, you are asking why we say that the fundamental particles are the ones that arise in the broken theory (the "emergent theory" after SSB), rather than the particles in the electroweak-symmetric theory? (Please, correct me if I am misunderstanding your question!)
      I think that this is a good point, and really, I think it comes down to personal taste. I think one reason that the fundamental particles are typically talked about in the way they are because these are the particles that we actually observe in experiments because we live in the vacuum state with a broken symmetry. So, in some sense, these are the "physical" particles, and we wouldn't expect to see a left-handed lepton doublet in a detector. However, I don't think anyone would correct you if you said that the electroweak-symmetric particles are the fundamental ones, it just feels a bit removed from the observed physics. I am certainly no expert in condensed matter systems, but I would guess that the mindset is a bit different simply due to the fact that we have less control over our systems: we can't really make a system that lives in the unbroken vacuum state, so we describe things using particles that are physically observable.
      On another note, I could also see an argument to be made for the other way around: even though, say, a phonon is obviously an emergent phenomenon, it behaves exactly like a real particle, so why not call it a real particle? I don't see any real problem with this. Of course, I would agree that this isn't "fundamental," but I think in particle physics, the electroweak-symmetric and broken systems are sort of one-to-one, so even though a lot of the phenomena are "emergent," they really are the same theory, just basically related by field redefinitions, so I don't really see any issues using them as "fundamental" interchangeably.

  • @NoNameAtAll2
    @NoNameAtAll2 2 года назад +1

    "mirror symmetric hill" close to beginning made me think it's sinusoidal instead, so got slightly confused because I expected "different point has same property, but not between them and that other point has different worldview" story, oops

    • @NoNameAtAll2
      @NoNameAtAll2 2 года назад

      6:09 phrase "let's look at some examples" is conditioned in me to expect slide transition, so it being delayed a dozen seconds with no screen activity made me think some error occured

  • @boonedockjourneyman7979
    @boonedockjourneyman7979 Год назад

    You need to start over. Almost every sentence past the 10 minute mark needed a great deal of definitional support.
    Imagine that you made this video as a presentation to your Ph.D. committee. What do you think would happen?

  • @markell1172
    @markell1172 Год назад

    Very well explained, nice.

  • @waynelast1685
    @waynelast1685 Год назад

    3:17 why would you require the solutions for a COLLECTION of atoms to be symmetric?

  • @hyperduality2838
    @hyperduality2838 9 месяцев назад

    Symmetric wave functions (Bosons, waves) are dual to anti-symmetric wave functions (Fermions, particles) -- the spin statistics theorem or quantum duality.
    Bosons are dual to Fermions -- atomic duality.
    Symmetry is dual to anti-symmetry -- symmetry breaking is dual.
    "Always two there are" -- Yoda.
    Energy is dual to mass -- Einstein.
    Mass, Fermions and particles are anti-symmetric and pure energy, Bosons and waves are symmetric -- symmetry breaking is dual.
    The Higgs Boson is dual to the Higgs Fermion -- Duality!

  • @justinpakarno4346
    @justinpakarno4346 2 года назад +1

    My boy's wicked smaht!

  • @das_it_mane
    @das_it_mane 2 года назад +1

    I always see explanations like this but I think I need it dumbed down even more to truly understand why this happens.

  • @waqasriaz4163
    @waqasriaz4163 Год назад

    very well explained

  • @themutatedmutalisk7305
    @themutatedmutalisk7305 Год назад

    at 5:34, where you state that if the zero-particle state is not the same as the lowest energy state in the system, then we can assme the lowest energy state(vacuum) state to have multiple particles. I don't get that. Can't we just remove energy particles so that when we do have 0 particles in the system to get the energy lower?

  • @PrivateEyeYiYi
    @PrivateEyeYiYi Год назад

    What has more Higgs Bosons creating its mass, a pound of feathers or a pound of lead?

    • @nickhowatson4745
      @nickhowatson4745 Год назад

      99.99% of the mass of matter comes from the Binding Energy of the Quarks within Protons and Neutrons. The Higgs Boson is responsible for less than 0.01% of the mass of matter.

  • @transient_moonlight
    @transient_moonlight Год назад

    Thanks a lot for this video (and all your work in general, amazing stuff).
    I have a question: as someone new to this topic, why do we assume that the vev of 2 symmetrical vacuum states are different? If the potential is symmetrical, shouldn't they be the same?

    • @zapphysics
      @zapphysics  Год назад +1

      Thank you for the kind words and good question! The potential can be thought of as a function of the vev, and the whole idea of it having a symmetry is that it the potential looks the same under certain transformations of the vev. In other words, if we relate *different* points in "vev space," then the potential looks the same.
      An easy way to think of this is just to think of a circle in the x-y plane. The circle is defined at many different points in the x-y plane, but it will be symmetric under rotations and reflections which essentially interchange these points.
      The main idea underlying spontaneous symmetry breaking is that the system chooses among multiple different, but physically identical (due to they symmetry) true vacua. If there is only a single true vacuum value of the vev, then we can't have spontaneous symmetry breaking.
      Hopefully that somewhat answered your question. I know it's a bit nuanced and might not be the most clear, so feel free to ask any more questions you may have!

    • @transient_moonlight
      @transient_moonlight Год назад

      ​@@zapphysics So, if I got it right, in math terms the potential for a symmetric field can be seen as a non-injective function of its vev? So we can't find a single vev that determines a given potential? Thanks again!
      Also, sorry if I say or ask trivial stuff, I've only recently picked up math and physics again, after years of not touching them, so I feel very slow in grasping things.

    • @zapphysics
      @zapphysics  Год назад +1

      @Moonsong I try to not make too general of statements with cases like these (since there could always be some niche exception that I don't know about) but in most cases that this comes up, the symmetry will definitely mean that the potential is not an injective function. However, that doesn't necessarily mean that there aren't any values of the vev that uniquely map to the potential.
      For example, if we make things very simple and consider a potential like x^2, then this has a reflection symmetry under x-> -x. It isn't injective, but the point at x=0 does uniquely map onto V=0. The key to spontaneous symmetry breaking is that we have multiple possible values of the vev which produce the same minimum value for the potential.

    • @transient_moonlight
      @transient_moonlight Год назад

      @@zapphysics That does it for me, thanks for your time.

    • @chalkchalkson5639
      @chalkchalkson5639 Год назад +1

      It might be less confusing to think about a heavily related case: a bose einstein condensate. Loosely speaking an individual atom might see a potential very similar to the mexican hat potential shown here, something like -r^2+r^4 with polar coordinates r, phi. But when you have a whole bunch of atoms there are now two distinct cases.
      1: they distribute evenly in phi on the bottom of the well. If you then zoom in on an individual atom again and consider the potential it sees, it still sees a symmetric one.
      2: they clump up at one phi position (as they do in a BEC). Now an individual atom cannot see the symmetry any more because the atom-atom interactions deform the potential around the point where the other atoms clumped up.
      It doesn't just matter what the "raw" potential looks like, but how it is occupied.
      You can even get Higgs boson like quasi-particles in BEC systems called "Higgs amplitude modes" so it's a genuinely strong analogy.

  • @battletwo367
    @battletwo367 Год назад

    I can't keep up with the pace of this video, can you pls recommend me some videos which will help me understand this video.

  • @quitchiboo
    @quitchiboo 2 года назад +1

    Is there a nice modell for why interactions with the higgs slow fermions down? Is it generally true that interactions between fermions do this?

  • @solapowsj25
    @solapowsj25 Год назад

    A wave-loop with extension into 3-D space has mass.

  • @arex3632
    @arex3632 Год назад

    this is very very good!

  • @SafetySkull
    @SafetySkull 4 месяца назад

    The more I learn about Gauge symmetries and symmetry breaking, the less sense it makes. Every explanation is like "so you know what a circle is, right? Well have you ever seen a sombrero? Pretty cool, right? Anyway, now the Schrodinger equation has an 'm' in it. Bye!"

  • @StephenGillie
    @StephenGillie 2 года назад

    When symmetry is broken, who fixes it?

  • @eirninlovesyou6071
    @eirninlovesyou6071 2 года назад

    miss your videos :(

  • @marcux83
    @marcux83 2 года назад

    by wearing an Sombrero?

  • @binishbatool248
    @binishbatool248 5 месяцев назад

    physics relavancy starts at 3:27

  • @5ty717
    @5ty717 6 месяцев назад

    Genius

  • @madvoice3703
    @madvoice3703 2 года назад

    Super 😇🙏 , I KNOW SOMETHING THAT YOU NEED TO KNOW 😇🙏

  • @enochbrown8178
    @enochbrown8178 Год назад

    Excuse me?

  • @StephenGillie
    @StephenGillie 2 года назад

    6:00 Accidental Japanese flag.

  • @ToriKo_
    @ToriKo_ 2 года назад

    I could not keep up with this video from about 3:00 onwards :(

  • @nolanhanna
    @nolanhanna 2 года назад

    Woof

  • @hooked4215
    @hooked4215 Год назад

    All this gobbledygook pretends to fix the wrongness implied in the adoption of probability as a mathematic tool to describe quantic reality.