Great video! Can you tell me if the quantum fields are real things that exist, or if they comprise a mathematical model that is more or less a useful abstraction? What programming language did you use for your code in the git repository? I was wondering if a simulation that is based on a finite-size circular quantum field will generalize to an infinite size quantum field on a cube? That is, does the formation of "particles" depend on the disturbances being restricted and bumping into one another within a small region of a special shape? Is there any useful insight to be gained by looking at the Fourier Transform of the quantum field?
spin isn't a hard thing to grasp...... youre make it confusing..... it's the magnetic field that irradiates in and around a particle... the field spins on its axis thus creating angular momentum and spins helically with respect to x as in as it travels along x the field spins in a circle from up to down..... like the earth spinning creating day and night and the angular momentum creating seasons almost but spinnin up and down by a full half of the particle.... wave functions aren't hard to understand..... people just explain this stuff like assholes.....
@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.
@4:00 Mr Higgsino is probably wrong there, imho. Has he considered the Frauchiger-Renner Inconsistency? QM cannot consistently account for QM. It is a convoluted case of Wigner's Friend, a bit like a Cantor-Gödel diagonal argument. Totalizing QM (a fail) is not just about large objects making it trickier to compute the interference, it is about the larger and more exposed a system the more likely entanglement structure breaks down, and so interference does not occur, and that makes a classical account valid. (See also Jacob Barandes, who describes this - without needing an unphysical Hilbert space - as a breakdown of non-Markov indivisibility in the underlying stochastic dynamics transition probabilities). Applied to the universe as a whole you can then see gravity does not need to be (re)quantized, since GR was already a quantum theory (if spacetime is nonclassical, i.e., admits nontrivial topology, i.e., indivisible non-Markov transitions, i.e., entanglement topology).
All due respect, but you are the one with the misconception. Whatever the measurement does, it is causal. (Otherwise... are you kidding me! You would destroy science: principle of sufficient reason. Denial of that amounts to Satanism (of some sort). LOL.) Correlation, as you well know, is not causation. So _whatever it is_ that "collapses" the ψ description, whether ontic, epistemic, or MW branching, that is an *_interaction_* by any other name dude. What you can say is that this is nonclassical. How? No one knows. But there is a nice toy model or metaphor or maybe even exact mode for this interaction, ER=EPR. Classical matter cannot traverse nor probe an ER=EPR wormhole due to topological censorship (qv. Friedman, et al., maybe also Geroch?) but a qubit can (proven by experiments in teleportation). But in all known Bell pair entanglement it is only ever a qubit. Case closed?
This is consciousness. :-) Making conscious decisions while in various brain wave states, storing memories, generating ideas, sleeping. Conscious electrical energy.
@27:35 not necessarily. The left-right symmetric SM is almost all we need. Dark matter = RH neutrino. Abundance of RH ν accounted for by usual letogenesis. Early cosmology = CPT-symmetry ⇒ metric analyticity (no inflaton, thermodynamics accounts for flatness, homogeneity and isotropy), Big Bang goes through a conformal zero, so GR holds if you analytically continue going around the pole at conformal 𝜏=0. SM then only needs one new thing, 36 dimension zero Bogoliubov scalars, which are vacuum transforms curing vacuum energy and Weyl and trace anomalies (not new particles), and this predicts 3 generations of fermions exactly. These scalars probably make a composite Higgs. Predicts no primordial gravity waves (no tensors modes in the CMBR). Predicts lightest neutrino mass _has_ to be exactly zero. See talks & papers by Turok & Boyle.
I wouldn't say that EARTH has orbital angular momentum. The EARTH/SUN SYSTEM has that. Earth actually has rotational angular momentum, all by itself - no reference to other bodies required. But that's not so re: the orbital angular momentum. I mean, you can just arbitrarily associate the orbital part with "Earth," when you name it, but it's not really a good physical picture.
I'm learning QM at my own and i easy understood the concepts and can say that this video is absolutly awesome and beautiful , i think that someone who did not understood it before the video would get enought information to start learning at it's own
The spin of quantum mechanics is affected by electromagnetic waves, and in superposition and entanglement, the entanglement is antimatter and the electric charge can be calculated. An example of seeing particle and wave nature at the same time is the blue glowing phenomenon of the fuel control rod. Western quantum mechanics is based on uncertainty. It is not the quantum mechanics of the invisible calculus like the principle, but the quantum mechanics that is visible starting from the eight trigrams of Eastern philosophy, where non-pole is taegeuk and taegeuk is non-geuk. It shows microscopic quantum mechanics and macroscopic atomic magnetic force. A quantum computer requires knowledge of elements and principles. Even a regular computer can produce the effect of a quantum computer. It is an error to explain the Earth's revolution and rotation in terms of angular motion. When the revolution is at perihelion, the amplitude is small. However, when it is at perihelion, the amplitude is large and numerous. This is why the four seasons are distinct. The rotation period is 230-400 million years, the amplitude is 17-8 years, and the rotation period is the same as the galaxy rotation period. This result is the absence of human science that has made it an unconditional feast of numbers with calculus equations that do not take into account the fact that the sun also orbits and the influence of the rotation of the galaxy on the sun and Earth.
To see why order/disorder is also a good means of understanding entropy one has to deeply/philosophically understand order/disorder: Order ultimately reduces to low information (when with a few parameters in a formula aka low-information, every entity in our system can be described=they are obeying a universal system-wide rule/pattern/law). That commonality/correlation (between system entities' properties) is the source of low information (order). So "one law/pattern to rule them all", "one formula to describe them all" means low information (low entropy). Disorder fundamentally means high amount of information/high-entropy (when every entity in our system is doing its own thing, not ruled by a universal law/rule/pattern or parameter or order and so we need lots of individual pieces of information to capture the whole system). I hole that was helpful.
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!"
I've always assumed there must be a higher dimension that our flat 3D cosmos is expanding into, otherwise our universe would have to curve back onto itself.
Interesting video, as usual. I have to make a picometre correction though: At 1:50, ℏ is the reduced Plack constant, also called the Dirac constant. It is equal to the Planck constant divided by 2π So we have: ℏ = h/2π en.wikipedia.org/wiki/Planck_constant#Reduced_Planck_constant_%E2%84%8F
When you take any "m" value between 0and 1and multiply with "hand write down it using zero and decimal we will reach very two value of classical continuum. And fraction is always an incomplete state What is then expected symmetry??? But question is what is intrinsic. What is the critical set up for intrinsic and extrinsic???
I sort of get it now when people mention it. Incomplete mathematics is the issue of divergence in certain circumstances. One would think that's annoying to live with but with renormalization, you're doing algebra essentially. Add gook, get a better understanding, simplify, logically consistent. The most useful scale is the quickest way to the answer. It's only when you can't be logically consistent that infinities are a real problem. Thanks!
Why is that a strange quark is more likely to decay into a charm quark than an up quark? As fas as I know from weak decays the more energy a system gives to create the W boson the more likely it is for the decay to happen because it needs less energy to borrow, so why is that?
Its way too complicated that just requires the dark portion of the universe which in essence isn't dimensional- just invisible -to propagate matter and quanta energy in waves. Maybe all this is designed to keep us confused?counting dark matter as N05 would satisfy relativity in wave form as gravitational waves are super massive.
This series on the SM is outstanding. It goes into far more depth than most other videos that cover these topics. Thank you for this wonderful group of videos.
I'm having trouble with the relationship between symmetry and charge. If I understand correctly, when we have a local symmetry, we need the gauge field to accommodate for the local transformation. To me, this indicates that _every_ particle that has this *local* symmetry must couple to the gauge field, because that's the point of the gauge field. This would mean that the charge of the theory is the same property as having the local symmetry. In QCD this seems to be the case as SU(3) is just the symmetry of the color space. Can we have particles that have global symmetry in this space but not local? Is that what confinement is? In QED we usually talk of the charge separately. Like we can have particles that have local phase invariance but do not need the gauge field to make up for it. Is that the case? Do neutral particles have local phase invariance? In terms of the covariant derivative, with the gauge as the connection term, we have this term multiplying the charge, which is zero for neutral particles. But that seems like its just a convenient way to write it. In your QED video, (if I understand correctly) you write this term as psibar*psi times the gauge derivative. Will this turn out to be zero, if the field is not locally symmetric?
@narfwhals7843 I think most of your intuition is correct, but I will add a couple of comments to hopefully clarify some things. > If I understand correctly, when we have a local symmetry, we need the gauge field to accommodate for the local transformation. > To me, this indicates that every particle that has this local symmetry must couple to the gauge field, because that's the point of the gauge field. > This would mean that the charge of the theory is the same property as having the local symmetry. You're exactly correct that any field that transforms under the local symmetry must couple to the gauge field. Keeping in mind that the gauge field is really the connection between two points which are now allowed to transform differently under the symmetry transformation, this is very easy to see: suppose I don't couple such a field to the gauge connection. Then, I am not accounting for this extra local transformation as I compare two different points, so two field configurations which are related by the local symmetry transformation will actually "look" different according to the theory and the symmetry is explicitly broken (two fields which are related by a symmetry transformation should always look the same). The charge is essentially how much the field transforms under the symmetry, and is always going to be proportional to the generators of the symmetry group. In QED (and any U(1) theory in general), the generators are any real numbers, so fields can have any charge under a U(1) symmetry. For a more complicated symmetry group like SU(3), fields are restricted to only transform as representations under the group (this is true for U(1) as well, it's just that any real number corresponds to a valid representation). > In QCD this seems to be the case as SU(3) is just the symmetry of the color space. Can we have particles that have global symmetry in this space but not local? Is that what confinement is? Not quite. When we make a symmetry transformation, the transformation has a set number of transformation parameters (the same as the number of generators), i.e. "how much" you are transforming by. These transformation parameters can vary on spacetime, but they will be the same for any field we are transforming. A global symmetry would then need to be completely disconnected from the local one since it would have totally different transformation parameters, so it would probably make more sense to just consider them as separate symmetries. Confinement just has to do with the behavior of couplings/potentials of a theory. However, you can do the opposite, where you have a large global symmetry and only gauge a sub-group of it. For example, I could have a global U(2) symmetry where I gauge the U(1) sub-group of the U(2). > In QED we usually talk of the charge separately. Like we can have particles that have local phase invariance but do not need the gauge field to make up for it. Is that the case? Do neutral particles have local phase invariance? > In terms of the covariant derivative, with the gauge as the connection term, we have this term multiplying the charge, which is zero for neutral particles. But that seems like its just a convenient way to write it. Any time that you have a theory which will depend on *differences* (i.e. derivatives) of fields at spacetime points where the fields transform under a local symmetry, you will need a gauge field. Again, this just has to do with the fact that we are really connecting different points of fields that could be transformed differently, and the gauge field automatically accounts for this. It absolutely isn't wrong to say that a neutral field transforms with "charge" zero. This is just because the "zero" representation is always going to be a representation in any group, since this corresponds to the identity operator and by definition, a group must have an identity element. In this case, yes, the gauge term in the covariant derivative vanishes. I don't think that this is just a convenient way to write it, but it is the correct way to think about it. > In your QED video, (if I understand correctly) you write this term as psibar*psi times the gauge derivative. Will this turn out to be zero, if the field is not locally symmetric? I'm not sure what you mean here by "not locally symmetric." If you mean that the theory truly isn't invariant under local transformations of this field, then the symmetry is explicitly broken, and you can no longer talk about having a local symmetry. If you mean that it doesn't transform under the symmetry (or I guess more properly that it transforms as a 0 representation of the group), then the part of the covariant derivative containing the gauge field will vanish when acting on this field, but the standard derivative will still hang around. However, it will only vanish in the coupling to that field, not to any fields which actually transform non-trivially under the symmetry. Hope that helps!
@@zapphysics Wow, thank you for always giving such extensive answers! So having a non-zero charge tells us how much the gauge transformation affects the field. A zero charge means the field "transforms trivially", so no change at all. I think my confusion was exactly the difference between your last two paragraphs. A trivial transformation is still a local symmetry. So the "amount" of charge is a distinct property of the fields, which is restricted to representations of the gauge group. And that is why we (sometimes) need to talk about it separately. How is that related to the psibar*psi term in the derivative here ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-qtf6U3FfDNQ.html ? Or am I misunderstanding which derivative that is?
@narfwhals7843 ah, this term comes exactly from trying to take the derivative of the gauge-transformed field. I apologize in advance for the formatting, but when I take psi-> exp(i*e*theta(x))*psi and psibar -> psibar*exp(-i*e*theta(x)), where e is the charge of the field and theta(x) is the spacetime-dependent transformation parameter, the impertant terms with the derivative of psi transform as psibar*dpsi/dx -> psibar*(dpsi/dx + i*e*psi*dtheta/dx). The first term isn't a problem, since this is the term we started with, but the second term is an issue, since it adds a term we didn't have before, which is clearly an issue for a symmetry. However, when we have a gauge-covariant derivative, we also have a coupling like psibar*i*e*A*psi where A is the gauge field. Under a gauge transformation, the gauge field also transforms like A -> A - dtheta/dx. So, when we add this all together, the combined transformation of the gauge field as well as the local transformation of the charged fields exactly cancel, leaving the theory invariant under the combined transformation. Note that this is exactly related to the fact that the charge in the transformation is the same as the charge in the gauge-covariant derivative: the transformation of the gauge field makes no reference to the particular charge of any one field, so in order for the cancellation to happen correctly, the gauge field needs to paired with the appropriate charge in the covariant derivative acting on the field. Now, this story gets slightly more complicated for a non-abelian group, but the general idea is the same. Just replace the charge with the appropriate generators of the group corresponding to the representation of the given field.
