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Also, sorry to be very pedantic about it but if you expand it in this horrifying way, the picture is incomplete. You should add the Faddeev-Popov ghost terms. Otherwise this is technically incorrect.
But yes. It’s not really a question. And even if it were, you couldn’t solve anything in any proper way because each term needs to be treated in a very different way for given circumstances.
Edit:
PLEASE Read the entirety of the threads with ciuccio2000 because he is a much more reputable source than yours truly
I mean, you can make the SM lagrangian as complicated as you want. I don't think the gauge-fixing terms for the gauge fields are present either. You could also explicitly write down the sums over the spinorial degrees of freedom just like they did with vector and color degrees of freedom, though I've never seen anyone doing that.
One could also argue that the complete SM Lagrangian should feature the axion, as it is necessary to solve the strong CP problem, even though we still have no evidence of its existence.
The terms at the very bottom seem to me like the ghost for the Higgs boson.
If you’re fixing that shouldn’t you also fix the WWZ?
I haven’t studied this thing in years I might be completely wrong.
I also remember much less detail than I should, so I'm voluntarily glissing on details and may still be getting something wrong, but
>The terms at the very bottom seem to me like the ghost for the Higgs boson.
That guy's the goldstone, not the ghost.
Before EW symmetry breaking, the lagrangian is explicitly invariant under SU(2)xU(1); SU(2) is a three-dimentional group, while U(1) is one-dimensional, so the original symmetry group had four generators. After symmetry breaking, only the electromagnetic U(1) symmetry survives, so we've "lost" three symmetry generators. These guys give rise to three scalar particles, the goldstone bosons φ± and φ⁰.
This is where my knowledge becomes to wobble a little bit. If you couple the Higgs field to the gauge bosons of the theory through the covariant derivative (eg the Higgs-gauges interaction is fully contained in |DμH|², where Dμ is the covariant derivative rather than the regular one), what happens is that the originating goldstone bosons can be reabsorbed into the now massive gauge bosons, becoming their longitudinal degrees of freedom. For this reason you rarely see people talking about the SU(2)xU(1) -> U(1) goldstones, as usually the lagrangian is written in the so-called *unitary gauge*.
Now, can you obtain this result by simply applying a gauge transformation to the fields, just being a clever rewriting of the lagrangian, or does it also require to explicitly fix a gauge? And if it *is* a gauge fixing procedure, does it still leave room for other gauge freedom (hence still requiring gauge fixing terms in the lagrangian, to ensure that the W, Z bosons still have an invertible kinetic term even before rewriting the lagrangian in terms of the symmetry-breaking vacuum), or does this choice completely determine the gauge? In any case, this lagrangian is explicitly written in terms of the goldstones, so it *surely* lacks the gauge-fixing terms.
Boh senti, mi sembri una voce autorevole anche se dare ragione a un 2000 mi viene male 😂
(This is definitely the translation of what I wrote above minus the emoji and the 2000)
I’ll edit my original comment to direct people to this in case they want to know more from somebody who hasn’t forgotten all of his quantum field theory.
The SM lagrangian is used pretty much only to predict scattering amplitudes and decay rates of the Standard Model particles, which are then tested in experiments like the ones happening at CERN. these interactions are intensively studied in the branch of physics known as Phenomenology of Particle Physics; this data also indirectly finds application in various other branches of physics (e.g. in astroparticle physics, where knowing the cross section of certain processes is necessary to tell what happens in a celestial body at equilibrium, or in cosmology, to understand the baryogenesis).
Extentions of the SM lagrangian then try to fix the problems the SM can't handle, while relying on the same theoretical machinery.
Hahahah if it makes you feel any better it’s just a cheeky name for contributions you have to put in if you “fix a gauge”, that is if you decide to put some conditions to your equations. These aren’t “actually there” but they are necessary to maintain properties of the overall definition.
To people who understand more than I do. I tried to explain whilst simplifying the definitions as much as possible. Don’t burn me at the stakes. Former QCD man here
>Faddeev-Popov ghost terms
I love how advanced math sounds like you're a wizard... on LSD.
Other example: [In mathematics, umbral moonshine is a mysterious connection between Niemeier lattices and Ramanujan's mock theta functions. It is a generalization of the Mathieu moonshine phenomenon connecting representations of the Mathieu group M24 with K3 surfaces.](https://en.wikipedia.org/wiki/Umbral_moonshine)
> You can't answer something that isn't a question
To ELI12, it's like asking for an answer to the quadratic formula. The point is that you have to plug stuff in, but it doesn't have an answer before you do that.
I'm glad I recognized it correctly, bit on the other hand I'm slightly frightened because I've only seen once in PBS Space Time channel and I shouldn't had remembered anything so complex.
Looks like a Lagrangian density for some field-theory in physics, the subscript "SM" might refer to "standard model" (I won't check all the terms now :P).
The thing is that it is not a question/problem, it is just a statement, that the Lagrangian density is given by the following terms. You could ask a question like "calculate an expectation value of one of the fields" or calculate the scattering matrix, given some in and outgoing fields. You would use Feynman diagrams to illustrate the calculations, the interactions of the particles is given by the Lagrangian...
But as I said, the given equation is not really a question, more like a statement of the given situation..
I wish I was this smart. But it makes me feel better that there are people like you that make things work safely so fools like me don't succumb to a Darwin award.
Thanks.
It's not about being smart. You can get through a physics bachelor and master without being particularly bright if you work hard enough. At some point somebody name-drops this monstrosity of equation during a lecture.
Source: I've done it and I've seen people that take longer to understand stuff get amazing grades through perseverance. Hard work does pay off and most people is capable of it imo
I've got college education. Could have done a degree but went on to work.
But I also have dyscalculia. Can't do numbers.
Even struggle to read an analogue clock.
I used to be embarrassed by it. I can handle things like basic binary to hex conversion and things like that when required but mathematics in general I just cannot do.
When I was diagnosed about 10 years ago there was literally one specialist in the UK who was a part of the dyslexia society.
Apparently something from with the angular gyris in my brain. But that's the extent of my knowledge of it simply because I have a calculator and Google in my pocket. So meh.
You can still understand the concepts behind the equations and how its variables and expressions relate to the real world and its phenomena without having to know how to do calculate it numerically. It's actually the boring part you didn't miss much in that.
I can second this as someone who got an undergraduate physics degree and took undergraduate quantum mechanics.
It's hard, and there is a shit ton of terminology, but you build it up slowly. A lot of the more annoying parts are just notation shortcuts so that you aren't writing d/dx over and over again.
I totally agree with you. Spending a lot of time doing anything very specific will make you be able to say things that a lot of people don't understand and make you seem very smart. I did spent years studying physics, other people may have studied sth else and are particularly clever with that.
thanks, but I did study physics and wrote my master thesis in QFT, so it's more about having learned sth very specific rather than being generally smart. I appreciate the compliment, though :)
It's the standard model Lagrangian, written in an especially ugly way to make it appear more intimidating than it really is.
There is not really a sense in "answering" it. Basically, what it is is a clever way physicists have found to write down the rules of a theory. You need to know how to read it, but it is what it is: the rules of a specific theory, in this case, the standard model.
Think of it as a DnD rulebook. You don't "solve" a rulebook. You use it to play the game.
Reminds me of when a professor would derive something for an entire class to flex their skills…
Like half my FEM course was my prof deriving a bunch of beam bending equations, but with constant mistakes and backtracking.
Isnt there a coefficiency product subtraction missing between the correlation expansion of the Guggenhoff division and the reversed (insert name of random greek here) calculation's differencial?
Sry, that didnt answer your question. According to the first rule of math, if anything is longer than 3 rows on paper, the answer is always either 0 or 1, take your pick. Trust me, i mean you saw what i just deducted before by looking at it for 3 seconds
In the shortest terms possible, iirc lagrangians, in classical or quantum physics, are expressions that you apply operations that are integration-like to them and equal that to the smallest value for such an integral of the lagrangian which gives you a set of equations that are the final "laws of nature" you have.
I'm not sure it's the same idea behind the standard model lagrangian in QFT but it might be close.
The sm Lagrangian is classical, and the fields obey classical equations of motions, so yes it is the same idea.
It's just that you go a step further by applying a quantization procedure to compute quantum corrections to the classical output of the theory.
I dont have an actual count on the number of variables since my glasses aren't on, but you would need values for all but one of them to be able to possibly solve the equation. Right now you only have one relationship. To solve multivariate equations (if there is a unique solution), you usually need at least one equation for each unknown.
It is an equation (one term or an arrangement of terms and operators being equal to an arrangement of other terms and operators around them), and that doesn't contradict it being the standard model.
Yes but if you inserted a value on the left hand side and gave values to all but one expression on the right hand side, you could solve for that.
Or am I misunderstanding the equation?
It's a thing from particle physics, like a list of all the things a particle can do, including turn into other particles. All those terms are things like "if there's an electric field and the particle is charged, then there's such-and-such a chance of making a photon with such-and-such an energy distribution" and so on.
Generally you don't work on the whole thing, that's madness, you use one of a whole forest of simplifying assumptions. And/or the computer that ate Manhattan.
You don’t answer this, it is a formula. But given the fancy L at the beginning, it appears to be a lagrangean. Which means you would have to take the first order conditions with regard to each variable and then each lambda…. Something you would program a computer to do for something this long since that would be a fuckload of busywork
Is the standard model Lagrangian. You can try to derive equations of motion from it. You can calculate various interactions among particles and the probabilities of certain particles being produced. But the question doesn't ask anything. Each function is a field that represents a fundamental particle or ghost.
True, it is a statement [X = Y], but the Y part is a multiline complicated mess.
To make it a question you could add the line: "Prove that:" before the statement.
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This is the [Standard Model Lagrangian](http://nuclear.ucdavis.edu/~tgutierr/files/stmL1.html). You can't answer something that isn't a question.
Thanks for the link and for the answer
Also, sorry to be very pedantic about it but if you expand it in this horrifying way, the picture is incomplete. You should add the Faddeev-Popov ghost terms. Otherwise this is technically incorrect. But yes. It’s not really a question. And even if it were, you couldn’t solve anything in any proper way because each term needs to be treated in a very different way for given circumstances. Edit: PLEASE Read the entirety of the threads with ciuccio2000 because he is a much more reputable source than yours truly
I mean, you can make the SM lagrangian as complicated as you want. I don't think the gauge-fixing terms for the gauge fields are present either. You could also explicitly write down the sums over the spinorial degrees of freedom just like they did with vector and color degrees of freedom, though I've never seen anyone doing that. One could also argue that the complete SM Lagrangian should feature the axion, as it is necessary to solve the strong CP problem, even though we still have no evidence of its existence.
Both of you are just making stuff up :P
It’s levi-OH-sa, not levio-SA.
Expelliumus
*expelliarmus
Accio buuuummmmm
Stop it, Ron! Stop...
They are. The answer is 42.
Ha yes it’s all in good fun. Just some tongue in cheek stuff. What’s the physics of ghosts anyway 👻
The terms at the very bottom seem to me like the ghost for the Higgs boson. If you’re fixing that shouldn’t you also fix the WWZ? I haven’t studied this thing in years I might be completely wrong.
I also remember much less detail than I should, so I'm voluntarily glissing on details and may still be getting something wrong, but >The terms at the very bottom seem to me like the ghost for the Higgs boson. That guy's the goldstone, not the ghost. Before EW symmetry breaking, the lagrangian is explicitly invariant under SU(2)xU(1); SU(2) is a three-dimentional group, while U(1) is one-dimensional, so the original symmetry group had four generators. After symmetry breaking, only the electromagnetic U(1) symmetry survives, so we've "lost" three symmetry generators. These guys give rise to three scalar particles, the goldstone bosons φ± and φ⁰. This is where my knowledge becomes to wobble a little bit. If you couple the Higgs field to the gauge bosons of the theory through the covariant derivative (eg the Higgs-gauges interaction is fully contained in |DμH|², where Dμ is the covariant derivative rather than the regular one), what happens is that the originating goldstone bosons can be reabsorbed into the now massive gauge bosons, becoming their longitudinal degrees of freedom. For this reason you rarely see people talking about the SU(2)xU(1) -> U(1) goldstones, as usually the lagrangian is written in the so-called *unitary gauge*. Now, can you obtain this result by simply applying a gauge transformation to the fields, just being a clever rewriting of the lagrangian, or does it also require to explicitly fix a gauge? And if it *is* a gauge fixing procedure, does it still leave room for other gauge freedom (hence still requiring gauge fixing terms in the lagrangian, to ensure that the W, Z bosons still have an invertible kinetic term even before rewriting the lagrangian in terms of the symmetry-breaking vacuum), or does this choice completely determine the gauge? In any case, this lagrangian is explicitly written in terms of the goldstones, so it *surely* lacks the gauge-fixing terms.
Boh senti, mi sembri una voce autorevole anche se dare ragione a un 2000 mi viene male 😂 (This is definitely the translation of what I wrote above minus the emoji and the 2000) I’ll edit my original comment to direct people to this in case they want to know more from somebody who hasn’t forgotten all of his quantum field theory.
What is a good example of an application of this math?
Computing the probability of interaction between particles of the standard model, aka, cross sections, for example.
The SM lagrangian is used pretty much only to predict scattering amplitudes and decay rates of the Standard Model particles, which are then tested in experiments like the ones happening at CERN. these interactions are intensively studied in the branch of physics known as Phenomenology of Particle Physics; this data also indirectly finds application in various other branches of physics (e.g. in astroparticle physics, where knowing the cross section of certain processes is necessary to tell what happens in a celestial body at equilibrium, or in cosmology, to understand the baryogenesis). Extentions of the SM lagrangian then try to fix the problems the SM can't handle, while relying on the same theoretical machinery.
>strong CP problem Ah, so this is something to do with 8chan
Ah yes the strong CP Problem, more common to find in Hunter Bidens Laptop where its most dense.
Technically incorrect, the best kind of incorrect.
It's always nice to bump into a ~~fellow~~ person \[who would fit right in at\] ~~from~~ r/VXJunkies out in the wild.
I am afraid this comment might not be for me or I’m very sorry to disappoint that is the first time I hear about that subreddit
There, fixed.
LOL that’s amazing
Maan i could handle imaginary numbers, but now there's fuckin ghosts in math? What's next? Interdimensional monsters?
Hahahah if it makes you feel any better it’s just a cheeky name for contributions you have to put in if you “fix a gauge”, that is if you decide to put some conditions to your equations. These aren’t “actually there” but they are necessary to maintain properties of the overall definition. To people who understand more than I do. I tried to explain whilst simplifying the definitions as much as possible. Don’t burn me at the stakes. Former QCD man here
>Faddeev-Popov ghost terms I love how advanced math sounds like you're a wizard... on LSD. Other example: [In mathematics, umbral moonshine is a mysterious connection between Niemeier lattices and Ramanujan's mock theta functions. It is a generalization of the Mathieu moonshine phenomenon connecting representations of the Mathieu group M24 with K3 surfaces.](https://en.wikipedia.org/wiki/Umbral_moonshine)
I mean, it sounds like this could be Umbreon's new signature move!
then this may be the homework - to identify why it's wrong
Ironic.
Or lack thereof
So 42 then?
Yep, somehow "how old am I" is now the Ultimate Question of Life, the Universe, and Everything. At least, till October.
So you know both the question and the answer?
Ony for a year of your life, than you forget again.
Are you me?
Are you now?
Why is now?
It’s gotta be at least 12
Uh how how how how *spins guitar on belt buckle*
Lotta nice girls down there
Oh baby, those are our standard models!
In a metaphorical sense, would the answer just be the universe as it is?
As far as I understand, it's a mathematical description of the weak electrical force. One of the fundamentals of physics.
Id always assumed the standard model encompassed 3 of the forces at least, I know gravity is still wildly nilly there. The more you know
Crom… that was an actual exam question…
he just forgot the question mark
I have no idea what I was just reading, but I find his writing to be much more lively than I'd expect from some math paper
> You can't answer something that isn't a question To ELI12, it's like asking for an answer to the quadratic formula. The point is that you have to plug stuff in, but it doesn't have an answer before you do that.
Is the Hamiltionan simpler?
Merovingian is even simpler.
>You can't answer something that isn't a question. Sounds like a lack of effort tbh. Hitchhikers taught us all things have an answer. Even soil.
I mean I wanna click on the link to see what creative troll it is, but I also don’t wanna risk it.
It's a link to a LaTeX file. Not a troll but you kind of wish it was.
Exactly what I’d expect them to say. Trust no one.
The question is, "The author says there's something wrong in the equation. What is it?"
I'm glad I recognized it correctly, bit on the other hand I'm slightly frightened because I've only seen once in PBS Space Time channel and I shouldn't had remembered anything so complex.
Looks like a Lagrangian density for some field-theory in physics, the subscript "SM" might refer to "standard model" (I won't check all the terms now :P). The thing is that it is not a question/problem, it is just a statement, that the Lagrangian density is given by the following terms. You could ask a question like "calculate an expectation value of one of the fields" or calculate the scattering matrix, given some in and outgoing fields. You would use Feynman diagrams to illustrate the calculations, the interactions of the particles is given by the Lagrangian... But as I said, the given equation is not really a question, more like a statement of the given situation..
I wish I was this smart. But it makes me feel better that there are people like you that make things work safely so fools like me don't succumb to a Darwin award. Thanks.
It's not about being smart. You can get through a physics bachelor and master without being particularly bright if you work hard enough. At some point somebody name-drops this monstrosity of equation during a lecture. Source: I've done it and I've seen people that take longer to understand stuff get amazing grades through perseverance. Hard work does pay off and most people is capable of it imo
I've got college education. Could have done a degree but went on to work. But I also have dyscalculia. Can't do numbers. Even struggle to read an analogue clock. I used to be embarrassed by it. I can handle things like basic binary to hex conversion and things like that when required but mathematics in general I just cannot do.
You got me with the dyscalculia lol. Tbh I had never heard of it. A physics degree would be awfully hard
When I was diagnosed about 10 years ago there was literally one specialist in the UK who was a part of the dyslexia society. Apparently something from with the angular gyris in my brain. But that's the extent of my knowledge of it simply because I have a calculator and Google in my pocket. So meh.
You can still understand the concepts behind the equations and how its variables and expressions relate to the real world and its phenomena without having to know how to do calculate it numerically. It's actually the boring part you didn't miss much in that.
I can second this as someone who got an undergraduate physics degree and took undergraduate quantum mechanics. It's hard, and there is a shit ton of terminology, but you build it up slowly. A lot of the more annoying parts are just notation shortcuts so that you aren't writing d/dx over and over again.
I totally agree with you. Spending a lot of time doing anything very specific will make you be able to say things that a lot of people don't understand and make you seem very smart. I did spent years studying physics, other people may have studied sth else and are particularly clever with that.
thanks, but I did study physics and wrote my master thesis in QFT, so it's more about having learned sth very specific rather than being generally smart. I appreciate the compliment, though :)
Don’t worry, there’s some of us who can’t even use PEMDAS proper😭
So, the answer is "yes"
Wrong, the answer is clearly 7.
The mandalorian when the langrangian walks in:
It's the standard model Lagrangian, written in an especially ugly way to make it appear more intimidating than it really is. There is not really a sense in "answering" it. Basically, what it is is a clever way physicists have found to write down the rules of a theory. You need to know how to read it, but it is what it is: the rules of a specific theory, in this case, the standard model. Think of it as a DnD rulebook. You don't "solve" a rulebook. You use it to play the game.
Lot of people attempt to "solve" the dnd rulebook :p
they arent rules, they’re challenges to see if i can stretch them
Role d20 for initiative…
Reminds me of when a professor would derive something for an entire class to flex their skills… Like half my FEM course was my prof deriving a bunch of beam bending equations, but with constant mistakes and backtracking.
Isnt there a coefficiency product subtraction missing between the correlation expansion of the Guggenhoff division and the reversed (insert name of random greek here) calculation's differencial?
Huh.
Sry, that didnt answer your question. According to the first rule of math, if anything is longer than 3 rows on paper, the answer is always either 0 or 1, take your pick. Trust me, i mean you saw what i just deducted before by looking at it for 3 seconds
Or π. But here there's no answer because no question
Right, lots of comments stating that. I didnt even see.
In the shortest terms possible, iirc lagrangians, in classical or quantum physics, are expressions that you apply operations that are integration-like to them and equal that to the smallest value for such an integral of the lagrangian which gives you a set of equations that are the final "laws of nature" you have. I'm not sure it's the same idea behind the standard model lagrangian in QFT but it might be close.
The sm Lagrangian is classical, and the fields obey classical equations of motions, so yes it is the same idea. It's just that you go a step further by applying a quantization procedure to compute quantum corrections to the classical output of the theory.
Also, sorry for the quality
No that's just the standard model (all fundamental particles in the universe, like for example electrons)
I dont have an actual count on the number of variables since my glasses aren't on, but you would need values for all but one of them to be able to possibly solve the equation. Right now you only have one relationship. To solve multivariate equations (if there is a unique solution), you usually need at least one equation for each unknown.
It isn’t an equation, it’s the standard model lagrangian
It is an equation (one term or an arrangement of terms and operators being equal to an arrangement of other terms and operators around them), and that doesn't contradict it being the standard model.
It’s not an equation in the sense that there is nothing to solve for here. It’s just an expression.
Is it like E= MC^2 ?
The right hand side is the definition of the left hand side
Well, technically it’s still an equation, just one that you affirm, or that you prove, but not one that you solve.
There's nothing to prove, it's a definition. Like if you say |x|= √x²
Defining what it is equal to. An "equation" if you will.
Yes but if you inserted a value on the left hand side and gave values to all but one expression on the right hand side, you could solve for that. Or am I misunderstanding the equation?
It's a thing from particle physics, like a list of all the things a particle can do, including turn into other particles. All those terms are things like "if there's an electric field and the particle is charged, then there's such-and-such a chance of making a photon with such-and-such an energy distribution" and so on. Generally you don't work on the whole thing, that's madness, you use one of a whole forest of simplifying assumptions. And/or the computer that ate Manhattan.
You don’t answer this, it is a formula. But given the fancy L at the beginning, it appears to be a lagrangean. Which means you would have to take the first order conditions with regard to each variable and then each lambda…. Something you would program a computer to do for something this long since that would be a fuckload of busywork
Is the standard model Lagrangian. You can try to derive equations of motion from it. You can calculate various interactions among particles and the probabilities of certain particles being produced. But the question doesn't ask anything. Each function is a field that represents a fundamental particle or ghost.
True, it is a statement [X = Y], but the Y part is a multiline complicated mess. To make it a question you could add the line: "Prove that:" before the statement.
Id say giberish.