Higgs Boson Discovered?!

[theprestige]

Once again, for the fourth time, and for the removal of doubt, I was NOT questioning the physics; I'm not even in a position to question the physics. I was questioning Clinger's assertion that Farside's algebraic transposition was incorrect and a fail at eighth grade level. It was not, and Clinger has now acknowledged that in post 627.

Actually, Clinger asserted that Farsight's interpretation failed at the eighth grade level (by misapplying the algebraic transposition). Post 627 reaffirms this assertion, and expands on the justification for this assertion first made in post 590.

The correctness of the algebraic transposition itself doesn't really matter much to the discussion, which is actually about how Farsight applied it.

 

[sol invictus]

I have no doubt that if Einstein were alive today, he would be deeply immersed in the research of the Higgs field and all its implications.

It's a matter of historical record that Einstein near the end of his life was trying to formulate a unified field theory that would explain both gravity and matter. One of his primary goals? Understanding the origin of mass, so as to account for the electron/proton mass ratio (the other particles were just beginning to be discovered in the 20s and 30s when he started on this).

Did Einstein's own ideas about the origin of mass also violate E=mc^2, Farsight?

 

[phunk]

Noted, lenny.

No. When we fire the cannonball straight up, it's slowing down due to gravity. When it reaches its maximum height it's momentarily motionless. At that moment it isn't moving. So it has zero kinetic energy and zero momentum.

Conservation of energy means that the kinetic energy hasn't mysteriously vanished, it's now potential energy, which is in the cannonball. In previous posts I've referred to this as "hidden kinetic energy", but it's hidden momentum too. The thing that's hiding is energy-momentum, and it makes the cannonball's mass increase a little. In similar vein its mass increases a little when you heat it up.

Haven't caught up on the whole thread yet so forgive me if this is covered already. Momentum is conserved in the motion of the cannonball and the earth at all times. At no point is any of it "hidden". At the time it's fired it's momentum is X, at the top of the trajectory it's 0, and when it gets back to the ground it's -X. And Earth's momentum (relative to the cannonball) starts at -X (recoil from the cannon), hits 0 along at the same time as the cannonball, and is X when the ball hits the ground. At all times, they add up to 0, and none of the momentum is "hidden" in the cannonball.

 

[W.D.Clinger]

a personal progress report

Disclaimer: I am not a physicist, and I know nothing about particle physics.

In an earlier post, I quoted the first sentence of the Higgs paper and asked for help in understanding it.

Lorentz-covariant field theories, symmetry, Lie groups? I was okay with that.

The Goldstone theorem, spontaneous breakdown of symmetry, gauge fields? I hadn't a clue.

Perpetual Student and edd offered some helpful suggestions, which I read. Those readings reminded me of a book ben m mentioned in another thread, which has been sitting on my bookshelf for over a year, unread:

Francis Halzen and Alan D Martin. Quarks & Leptons: An Introductory Course in Modern Particle Physics. John Wiley and Sons, 1984.​

The Higgs mechanism is covered in chapter 14, and the Higgs particle in chapter 15. Sections 14.6 (on spontaneous symmetry breaking) and 14.7 (spontaneous breaking of a global gauge symmetry) were most helpful to me.

Since we've been speaking of algebra, here's a personal anecdote. Section 14.7 starts with the following Lagrangian:

ℒ = (∂_μφ)*(∂^μφ) - μ²φ*φ - λ(φ*φ)²​

That's equation (14.48). If you write the complex scalar field as φ = (φ₁ + iφ₂)/(√2), with φ₁ and φ₂ real, then the potential energy part of that Lagrangian is minimal on the circle with

(φ₁² + φ₂²) = v² = - μ²/λ.​

That's equation (14.49). Picking φ₁ = v and φ₂ = 0 as the values of those real fields at some convenient representative point on that circle, we can examine the Lagrangian in the neighborhood of that point by substituting

φ(x) = (1/√2) (v + η(x) + iξ(x))​

into the Lagrangian, where η(x) and ξ(x) are infinitesimal real fields that model the variation in φ as you move away from the representative point. The result of that substitution is equation (14.51):

ℒ' = ½(∂_μξ)² + ½(∂_μη)² + μ²η² + constant + cubic and quartic terms in η, ξ​

That's what they claim, anyway. Ignoring the use of μ to mean two distinct things in that equation, my eyeball substitution said there should be terms linear in η and quadratic in ξ. When I worked through the algebra, however, those terms cancelled.

The authors immediately explain the geometric reason, shown in Figure 14.5, but I hadn't read that far when I did the algebra.

Section 14.8 (the Higgs mechanism) is basically the same calculation for a local gauge symmetry, so the partial derivatives of the Lagrangian in section 14.7 are replaced by covariant derivatives, which introduces a vector gauge field. To keep the algebraic manipulations from creating the appearance of an unphysical real field, a different substitution is used. We end up with a massive vector boson and a Higgs particle.

That, at least, is what I understood from skimming chapter 14 last night. I'm going to have to read most of the book before I can do justice to chapters 14 and 15. That will take me a while. (The authors' preface suggests chapters 3 through 6 could be the basis for an undergraduate course on QED.)

On the other hand, I am no longer stuck on Higgs's first sentence. I can now read the entire paper, noting the details I still don't understand.

 

[sol invictus]

Disclaimer: I am not a physicist, and I know nothing about particle physics.

Judging by the rest of your post, the second statement at least is false.

The Goldstone theorem, spontaneous breakdown of symmetry, gauge fields? I hadn't a clue.

The Goldstone theorem has a fairly simple mathematical proof that will probably make sense to you. But let me try to give an intuitive explanation.

Suppose you have a symmetry of the laws of physics, like translation invariance. Translation invariance means the laws of physics are the same everywhere in space.

Now suppose we consider a state containing some object that is localized at some position in space. Physicists say that such a state "spontaneously breaks" translation invariance, because the state itself is not invariant under translations even though the laws of physics are. But translation invariance still tells us something - it tells us that the energy (and charge and momentum and every other property) of that object cannot depend on where we put it.

Consider the a degree of freedom that corresponds to the location of that object, let's call it x. What is the energy of associated with x? Clearly, if the object is in motion, there will be extra energy (the kinetic energy of the object, proportional to the time derivative of x, squared). But there cannot be any energy if the object is at rest, no matter where it is. Therefore, there cannot be a term like x^2 or x^4 in the Lagrangian or Hamiltonian that describes this system. Therefore, x is "massless" - in field theory, x would be a field, the symmetry would be a symmetry of the field space, and the absence of x^2 terms would mean x is a massless field.

And that's the basic idea behind Goldstone's theorem - broken symmetries give rise to massless particles, or so-called Goldstone bosons.

ℒ' = ½(∂_μξ)² + ½(∂_μη)² + μ²η² + constant + cubic and quartic terms in η, ξ​

That's what they claim, anyway. Ignoring the use of μ to mean two distinct things in that equation, my eyeball substitution said there should be terms linear in η and quadratic in ξ. When I worked through the algebra, however, those terms cancelled.

That's because you're expanding around a solution - a linear term would mean the action isn't stationary; i.e. you aren't expanding around a solution. Another way to say it is that a linear term in the energy means there's a force, which means the system will begin to accelerate, which means you weren't expanding around a stationary solution.

What's less trivial is the presence of a mass term for η, and the lack of one for ξ. The latter is a consequence of Goldstone's theorem.

 

[W.D.Clinger]

What's less trivial is the presence of a mass term for η, and the lack of one for ξ. The latter is a consequence of Goldstone's theorem.

Thank you. That was helpful.

 

[Farsight]

I can't stop long guys. So in brief:

You're falling for Farsight's argument by equivocation. He uses massless photons to argue that energy is the same as momentum. He then assumes energy is the same as momentum when talking about cannonballs.

No I don't. I refer to energy-momentum, and say energy and momentum are two aspects of energy-momentum. I've made this clear by describing how you can't remove the cannonball's kinetic energy without also removing its momentum.

So Farsight was quite wrong to say energy is the same thing as momentum, and Farsight was quite wrong when he wrote "You always divide by c to go from one to the other".

Straw man argument from a guy with a sincerity bypass. Enough.

 

[Farsight]

Farsight... you're not even a physicist, let alone a theoretical physicist.

Not professionally, but honestly, I talk to professional physicists, and they're forever calling me Professor Duffield. I guess that's because I obviously know so much physics. I always correct them of course, and tell them that I'm a well-read physics amateur with a Computer Science degree who's spent decades being analytical and logical and empirical.

So you'd like to commit to the statement that all transverse waves carry angular momentum now, rather than that all waves carry angular momentum?

No, go and look at what I said to sol again: Yes no problem. I made a mistake. I was thinking of electromagnetic waves and other transverse waves such as wind waves. A longitudinal wave such as a sound wave exhibits only a back-and-forth motion. In mitigation: we aren't talking about sound waves here. We're talking about light waves here. They convey angular momentum. Stop trying to waste everybody's time with trivia because you can't fault my argument.

Field, with negative mass squared not negative mass, in a hypothetical situation that doesn't apply to reality, and you might want to go away and check on the distinction between a tachyon and a tachyonic field.

No thanks. And I won't be checking up on negative mass squared either.

 

[TjW]

So, if momentum is the same thing as energy, and particles are made of energy, it stands to reason one could as easily say that particles are made of momentum.
When a particle's momentum is measured to be zero, what is it made of then?

 

[Farsight]

Referring to that equation in a previous post doesn't change the fact that it contradicts what you said in a later post

It doesn't contradict it. I gave the full equation and said how only the momentum term applied for photons, and later talked about cannonball momentum.

What is "the book's potential energy" if not it's interaction with the gravitational field? If the field were not present, there'd be no potential energy, and the potential energy is dependent upon the book's position in the field

It's how much of the book's mass-energy the gravitational field will convert into the book's kinetic energy. A deeper gravitational field will convert more.

It gains kinetic energy from it's interaction with the field, of course, energy is conserved so the total energy remains the same, but so what? The book isn't moving, as it falls it accelerates, and it's kinetic energy increases What causes that acceleration? The interaction with the gravitational field.

Which converts a portion of the book's mass-energy that we label potential energy into the book's macrosopic kineitc energy. When you lift a book you do work on it. You give it potential energy. You increase its mass-energy just as surely as you do when you warm it up. Hurl it straight up at 11km/s and it takes that potential energy away with it. That energy doesn't go into the gravitational field. Conservation of energy applies.

Position a detector 100 meters above your photon emiter, and you'll find that it's frequency is lower at the detector than when emitted, move the detector up another 100 meters and you'll find the frequency lower still What causes the frequency to change? Gravity

No. It doesn't change. Your measurement devices run faster when they're higher, that's all. Conservation of energy applies again.

Why? You said the book has more kinetic energy on the shelf than on the floor Can you support that in the case of a book, or not?

I said the book on the shelf has more potential energy than the book on the floor, and that this is hidden kinetic energy inside the book that is converted into the falling book's macroscopic kinetic energy. I illustrated this using a standing wave in a box. I can support this in the case of the book because in atomic orbitals "electrons exist as standing waves". It's akin to a whole collection of standing waves in boxes.

Sorry guys, I have to go.

 

[edd]

Not professionally, but honestly, I talk to professional physicists, and they're forever calling me Professor Duffield. I guess that's because I obviously know so much physics. I always correct them of course, and tell them that I'm a well-read physics amateur with a Computer Science degree who's spent decades being analytical and logical and empirical.

You're talking to at least three at a minimum professional physicists here who do not hold that opinion of you and very probably a maximum of zero here that do (if its any more than zero they're staying extremely improbably quiet).

No, go and look at what I said to sol again: Yes no problem. I made a mistake. I was thinking of electromagnetic waves and other transverse waves such as wind waves. A longitudinal wave such as a sound wave exhibits only a back-and-forth motion. In mitigation: we aren't talking about sound waves here. We're talking about light waves here. They convey angular momentum. Stop trying to waste everybody's time with trivia because you can't fault my argument.

It's not trivia. Your clarification has clarified little. You've just requoted exactly what I sought clarification upon. Do all transverse waves in your opinion carry angular momentum? Yes or no?

 

[ben m]

In previous posts I've referred to this as "hidden kinetic energy", but it's hidden momentum too. The thing that's hiding is energy-momentum, and it makes the cannonball's mass increase a little. In similar vein its mass increases a little when you heat it up.

No one else caught this, but it's worth jumping in to say *how thoroughly contra-Einstein this is*.

You just said that, if you fire a cannonball upwards, its rest mass will vary along with its distance from Earth. Imagine an observer in a sealed capsule who comes along and finds that a cannonball has punctured their hull. "Either we just flew very fast past a stationary cannonball, or we're at rest and someone fired a cannonball at us," he says. "Although, since the capsule has no rockets, the only reason it would be moving fast would be if we're deep in a gravity well."

The fundamental principle of GR is that they can't tell the difference. All of the laws of physics are invariant in all free-falling reference frames. That's why it's a problem when Farsigh beams aboard, saying, "No, I can tell you quite a lot about your reference frame. Using this specially-designed spring scale for moving objects, let's measure the mass of the cannonball as it flies by. If the mass is large, we're deep in a gravity well. If the mass is small, we must be far from the well." Thus Farsight contradicts Einstein on the indistinguishability of free-falling reference frames.

 

[Farsight]

You know what, Farsight? After reading pages and pages of this thread, I have no idea what you think it is about the Higgs mechanism that contradicts E-mc^2. Literally no idea. And I'm pretty sure the same goes for everyone else here.

Very droll, sol. Nice try. But it won't work because the mass of of body is a measure of its energy-content will never square with the mass of a body is a measure of its interaction with the Higgs field. It's one or the other. You can't have your cake and eat it, not when for a linear wave, momentum is a measure of resistance to change of motion, and for a standing wave, mass is a measure of resistance to change of motion. Not when you can create an electron (and a positron) from a photon in pair production, and when you can diffract an electron, and when in atomic orbitals "electrons exist as standing waves". What do you think a free electron consists of? You know, that thing that has magnetic moment and spin angular momentum? A point particle? No, it's a wave, and it isn't going past you at c. It's just sitting there right there in front of you. So it's a standing wave.

It's very specific. Among many others, one prediction is that the Higgs couples to fermions with a strength that's exactly proportional to their mass. That's quite easy to falsify - if it's wrong.

Oh how so very convenient when mass is a measure of energy content. Here, try disproving this: mass is a measure of the interaction with fairies and it's always proportional to energy content. You can't disprove it when mass is a measure of energy content. And you know it. So that little bit of sophistry is a busted flush, n'est pas?

Oh really? Light waves always carry angular momentum, do they?

Yep. Take a look at Susskind's lecture. Two minutes fifty seconds in. He refers to Planck's constant of action, the h in E=hf. Action has the same dimensionality as angular momentum. And Susskind said angular momentum is quantized. Come on sol, this is kid's stuff.

The mass is imaginary, not negative.

Mass is just a measure of energy content. It's never negative, and it certainly isn't imaginary.

Not negative, imaginary. The mass SQUARED is negative. Of course, you haven't a clue how to interpret that or what it could possibly mean. Which is a shame, because it's actually quite interesting.

I know all about imaginary numbers. In physics, they're asscociated with rotation. Funnily enough, so is mass. When the energy-momentum is locked into rotational motion instead of linear motion at c, the result is standing-wave mass instead of linear-wave momentum.

It's fun to actually understand things, Farsight. That's what makes physics so fascinating - it's rarely what you expect. It's a shame you'll never get to experience that feeling.

I experience it all the time sol. One day, you will too.

 

[Perpetual Student]

Disclaimer: I am not a physicist, and I know nothing about particle physics.

In an earlier post, I quoted the first sentence of the Higgs paper and asked for help in understanding it.

Lorentz-covariant field theories, symmetry, Lie groups? I was okay with that.

The Goldstone theorem, spontaneous breakdown of symmetry, gauge fields? I hadn't a clue.

Perpetual Student and edd offered some helpful suggestions, which I read. Those readings reminded me of a book ben m mentioned in another thread, which has been sitting on my bookshelf for over a year, unread:

Francis Halzen and Alan D Martin. Quarks & Leptons: An Introductory Course in Modern Particle Physics. John Wiley and Sons, 1984.​

The Higgs mechanism is covered in chapter 14, and the Higgs particle in chapter 15. Sections 14.6 (on spontaneous symmetry breaking) and 14.7 (spontaneous breaking of a global gauge symmetry) were most helpful to me.

Since we've been speaking of algebra, here's a personal anecdote. Section 14.7 starts with the following Lagrangian:

ℒ = (∂_μφ)*(∂^μφ) - μ²φ*φ - λ(φ*φ)²​

That's equation (14.48). If you write the complex scalar field as φ = (φ₁ + iφ₂)/(√2), with φ₁ and φ₂ real, then the potential energy part of that Lagrangian is minimal on the circle with

(φ₁² + φ₂²) = v² = - μ²/λ.​

That's equation (14.49). Picking φ₁ = v and φ₂ = 0 as the values of those real fields at some convenient representative point on that circle, we can examine the Lagrangian in the neighborhood of that point by substituting

φ(x) = (1/√2) (v + η(x) + iξ(x))​

into the Lagrangian, where η(x) and ξ(x) are infinitesimal real fields that model the variation in φ as you move away from the representative point. The result of that substitution is equation (14.51):

ℒ' = ½(∂_μξ)² + ½(∂_μη)² + μ²η² + constant + cubic and quartic terms in η, ξ​

That's what they claim, anyway. Ignoring the use of μ to mean two distinct things in that equation, my eyeball substitution said there should be terms linear in η and quadratic in ξ. When I worked through the algebra, however, those terms cancelled.

The authors immediately explain the geometric reason, shown in Figure 14.5, but I hadn't read that far when I did the algebra.

Section 14.8 (the Higgs mechanism) is basically the same calculation for a local gauge symmetry, so the partial derivatives of the Lagrangian in section 14.7 are replaced by covariant derivatives, which introduces a vector gauge field. To keep the algebraic manipulations from creating the appearance of an unphysical real field, a different substitution is used. We end up with a massive vector boson and a Higgs particle.

That, at least, is what I understood from skimming chapter 14 last night. I'm going to have to read most of the book before I can do justice to chapters 14 and 15. That will take me a while. (The authors' preface suggests chapters 3 through 6 could be the basis for an undergraduate course on QED.)

On the other hand, I am no longer stuck on Higgs's first sentence. I can now read the entire paper, noting the details I still don't understand.

Judging by the rest of your post, the second statement at least is false.



The Goldstone theorem has a fairly simple mathematical proof that will probably make sense to you. But let me try to give an intuitive explanation.

Suppose you have a symmetry of the laws of physics, like translation invariance. Translation invariance means the laws of physics are the same everywhere in space.

Now suppose we consider a state containing some object that is localized at some position in space. Physicists say that such a state "spontaneously breaks" translation invariance, because the state itself is not invariant under translations even though the laws of physics are. But translation invariance still tells us something - it tells us that the energy (and charge and momentum and every other property) of that object cannot depend on where we put it.

Consider the a degree of freedom that corresponds to the location of that object, let's call it x. What is the energy of associated with x? Clearly, if the object is in motion, there will be extra energy (the kinetic energy of the object, proportional to the time derivative of x, squared). But there cannot be any energy if the object is at rest, no matter where it is. Therefore, there cannot be a term like x^2 or x^4 in the Lagrangian or Hamiltonian that describes this system. Therefore, x is "massless" - in field theory, x would be a field, the symmetry would be a symmetry of the field space, and the absence of x^2 terms would mean x is a massless field.

And that's the basic idea behind Goldstone's theorem - broken symmetries give rise to massless particles, or so-called Goldstone bosons.



That's because you're expanding around a solution - a linear term would mean the action isn't stationary; i.e. you aren't expanding around a solution. Another way to say it is that a linear term in the energy means there's a force, which means the system will begin to accelerate, which means you weren't expanding around a stationary solution.

What's less trivial is the presence of a mass term for η, and the lack of one for ξ. The latter is a consequence of Goldstone's theorem.

Thank you both for the above posts. As someone who has only recently been making a serious attempt to get a handle on quantum field theory, the above discussion is challenging to the extreme for me. My pursuit of QFT is quite genuine, so I hope to have a better understanding of this discussion in the coming months. In contrast, it's quite ludicrous that Farside, with demonstrably no understanding of QFT -- and making no attempt to correct that ignorance -- has the unmitigated gall to continue to babble about his pretend physics. Considering the significance of the Higgs mechanism, these discussions (excluding Farside) are very much appreciated.

 

[Farsight]

No, I'm just arguing that the formulae were mathematically valid as stated.

If E = hf and p = hf/c are valid, then p = E/c follows, unequivocally!

if A = xy and B = xy/n then B = A/n

These are in agreement. I would be interested to see any argument that state they aren't.

I literally cannot comment on the physics being argued. You are all miles over my head!!

You are right smartcooky. Stick to your guns, and don't think this is all miles over your head. You can comment, and you can understand it. However there are people here who are trying to persuade you that you can never understand it, and that you should just roll over and believe what they say, even though they can't explain anything at all. Don't fall for it. Be skeptical instead.

 

[Stimpson J. Cat]

I wonder if Farsight realizes that the mechanism by which the Higgs field gives mass to particles in the Standard Model is essentially the same as the mechanism by which photons acquire a non-zero rest mass in a superconductor? Or that the tachyonic field issue he dismissed out of hand also applies there?

I wonder if he also thinks that photons acquiring mass in a superconductor also violates E=mc^2? Or if he accepts the empirically verifiable fact that photons do acquire rest mass in a superconductor, does he insist that the Standard Model's explanation for how it does so must also be wrong? After all, a photon with non-zero rest mass is a "body", and a body's mass is due to its energy content, so it can't be due to interaction with a Cooper pair condensate.

 

[jhunter1163]

Just FYI, disclosure of personal information without consent is a breach of Rule 8. Please do not attempt to get others to disclose personal information in threads. Thank you for your anticipated cooperation.

Replying to this modbox in thread will be off topic  Posted By: jhunter1163

 

[Tubbythin]

I haven't made any errors, despite your wishful thinking.

That is simply not true. You said:

And like space and time, energy and momentum are merely two aspects of the same thing called energy-momentum. You always divide by c to go from one to the other, for example for a photon energy E = hf whilst momentum p = hf/c. Hence momentum p=E/c. Forget about the vector aspect of momentum in all this, it's a distraction from energy-momentum. Anyway, Einstein's E=mc² paper is similar to this in that the upshot is you divide by c again for inertia aka mass, so m=E/c². It has to be like that because p=mv, so m=p/v so m = (E/c)/v, then replace v with c.

First off you don't always divide by c to go from energy to momentum. But that only works for massless particles. Hence, you were wrong. Then you claim that "It has to be like this" and then use the equation for momentum for a non-relativistic massive particle. Hence you were wrong again. That's two errors in that one paragraph alone.

 

[W.D.Clinger]

Farsight... you're not even a physicist, let alone a theoretical physicist.

Not professionally, but honestly, I talk to professional physicists, and they're forever calling me Professor Duffield.

When I talk to professional physicists, they usually call me "Will".

I guess that's because I obviously know so much physics. I always correct them of course, and tell them that I'm a well-read physics amateur with a Computer Science degree who's spent decades being analytical and logical and empirical.

You don't tell them about your A-level maths and tutoring experience?

Edited by LashL: 

Removed quote of moderated content.

Speaking of dishonesty and creepy disregard for evidence, let's look at this example:

For example, look at this:

Nobody apart from straw-man-mongers have suggested that energy and mass are the same thing. Or energy and momentum.

In the following quotation, Farsight appears to suggest that energy and momentum aren't different things:

Energy and momentum aren't two different things.

When someone goes out of his way to argue that two things aren't different, classical logic tells us he's suggesting they're the same.

Combining that inference with what Farsight wrote above, we might conclude that Farsight thinks of himself as a "straw-man-monger".

That chain of reasoning fails because the final step implicitly assumes a fact not in evidence: that Farsight is capable of applying logic or other objective criteria to his own words and arguments.

Edited by LashL: 

Removed quote of moderated content.

 

[Perpetual Student]

Originally Posted by Farsight

Nobody apart from straw-man-mongers have suggested that energy and mass are the same thing. Or energy and momentum.

Originally Posted by Farsight
Energy and momentum aren't two different things.

Yes, Mr. Clinger, this is indeed quite revealing.