How can the Higgs boson have mass if it creates mass?

[Eli137]

As you may know, the mass of the Higgs Boson is 125 GeV. My question is, "How can the particles themselves that create mass by interacting with others have mass?"

 

[Nugatory]

https://today.duke.edu/2012/11/higgsmisconceptions

 

[Eli137]

If

https://today.duke.edu/2012/11/higgsmisconceptions

If the Higgs Bosons are just the product of an energy field, are they really necessary for the interactions with particles that gives them mass?

 

[Drakkith]

If the Higgs Bosons are just the product of an energy field, are they really necessary for the interactions with particles that gives them mass?

What do you mean by 'are they really necessary'? You certainly don't need to have Higgs Bosons lying about everywhere just so particles have mass, the Higgs field does this by itself.

 

[Eli137]

M

What do you mean by 'are they really necessary'? You certainly don't need to have Higgs Bosons lying about everywhere just so particles have mass, the Higgs field does this by itself.

Because of that, my question is: What do Higgs bosons do if they aren't the reason particles have mass?

 

[Drakkith]

Well, they don't 'do' anything except exist and obey the laws of physics, just like every other particle. I'm still not quite sure what you're trying to get at. Are you asking what kinds of laws they obey and how they work? Are you asking if they interact with other particles?

Edit: I'm leaving school right now. It will be at least an hour before I can get back to this.

 

[Eli137]

I apologize for the wording. How do they interact with other particles?

 

[Drakkith]

I apologize for the wording. How do they interact with other particles?

Unfortunately I can't answer that. I'll have to let other here on PF that are more knowledgeable than myself answer that.
@mfb, can you answer this?

 

[Orodruin]

The Higgs bosons are a necessary consequence of the Higgs mechanism. If the Higgs field exists, so do the Higgs bosons. The Higgs field must exist if it is to take a non zero vacuum expectation value, which is what gives other particles their mass and therefore the Higgs boson must exist because it does if the field does. (You may think of the field as the surface of a pond and the particle as ripples on the surface - if the surface is there, you can make ripples.)

Since the Higgs boson and particle masses are both predictions of the same theory, we can deduce some of the properties the Higgs boson must have (such as its interactions) from knowledge of particle masses.

 

[Eli137]

T

The Higgs bosons are a necessary consequence of the Higgs mechanism. If the Higgs field exists, so do the Higgs bosons. The Higgs field must exist if it is to take a non zero vacuum expectation value, which is what gives other particles their mass and therefore the Higgs boson must exist because it does if the field does. (You may think of the field as the surface of a pond and the particle as ripples on the surface - if the surface is there, you can make ripples.)

Since the Higgs boson and particle masses are both predictions of the same theory, we can deduce some of the properties the Higgs boson must have (such as its interactions) from knowledge of particle masses.

Thank you

 

[Drakkith]

Orodruin, do you know which of the fundamental forces the higgs boson interacts via?

 

[mfb]

It doesn't fit in that framework of two or three "forces". You can consider its interactions as a separate "force" if you like.

 

[ChrisVer]

the Higgs boson can interact with fermions (via Yukawa terms)
the Higgs boson can interact with the weak bosons W,Z.
You can get additional effective interactions (if I can call them such) via its couplings to fermions, where the Higgs can interact with gluons or photons (eg H->gg or H->γγ with a top-quark triangle) something that the pion was also known to be able to do.
Finally the Higgs boson (as a scalar field) should be able to interact with other higgs bosons too.

 

[Orodruin]

It doesn't fit in that framework of two or three "forces". You can consider its interactions as a separate "force" if you like.

That being said, the Higgs is part of a complex SU(2) doublet with non-zero hypercharge. As such, it has gauge interactions with the weak gauge bosons and it just so happens that the component which becomes the physical higgs is electrically neutral after SSB (that may be a poor choice of words, it must be because of the nature of SSB). The remaining three real degrees of freedom of the original Higgs doublet are eaten by the Ws and the Z and therefor become an integral part of what we call weak interactions.

Apart from the gauge interactions, you have the Yukawas, but those are not gauge interactions and have a lot of free parameters (essentially the particle masses and mixings) and the quartic self-interaction term required to make the Hamiltonian bounded from below (since the quadratic term comes with a negative sign - hence leading to symmetry breaking).

 

[vanhees71]

It's difficult to answer this question, because it involves pretty advanced math to understand, why the Higgs mechanism (as a theoretical concept) is necessary to give the particles mass. The reason is that the mass of particles is doubly "problematic" in the quantum-field theoretical model that describe electroweak interaction. First of all it is a socalled chiral theory, and it must be so, because the weak interaction is violating the symmetry of the physical laws under space reflections, i.e., if you look at certain processes that involve the interaction of particles through the weak interaction, and somebody takes a movie of this process but films the image in a mirro rather than directly the experiment, you can immediately realize this. For a process involving only the strong or electromagnetic interaction, this is not possible, because any process that happens implies that also that the same process looked at in a mirror could be as well a really possible process.

So we have to understand, what chirality ("handedness") means an what it has to do with mass. The answer is again a bit difficult without the mathematical machinery of QFT, but let's try. The point is that particles in addition to mass and various charges can also have a angular-momentum like quantity associated with it, the spin. This usually implies that it has a magnetic moment, i.e., it is like a little permanent magnet. Now an angular momentum is a funny kind of vector, a socaled axial vector. It has to do with rotation rather than translation as the usual vectors (sometimes called polar vectors in this context). If we take the usual orbital angular momentum in classical mechanics, it's defined as $\vec{L}=\vec{r} \times \vec{p}$, where $\vec{r}$ is the position vector of a particle and $\vec{p}$ its momentum. Now a space reflection is defined by the substitution $t \rightarrow t$, $\vec{r} \rightarrow -\vec{r}$. Since $\vec{p}=m \mathrm{d} \vec{x}/\mathrm{d} \vec{t}$ and since we assume that $m \rightarrow m$ under space reflections also $\vec{p} \rightarrow -\vec{p}$, but $\vec{L} \Rightarrow (-\vec{r}) \times (-\vec{p})=\vec{r} \times \vec{p}=\vec{L}$. So $\vec{L}$ doesn't flip its sign under space reflections. This is what's called an axial vector.

Now let's look at chirality. The helicity of a particle is the projection of its (total) angular momentum to the direction of its momentum, i.e., $h=\vec{p} \cdot \vec{J}/|\vec{p}|$. Now since $\vec{p}$ is a polar and $\vec{J}$ a axial vector and if $h \neq 0$ under space reflections $h \rightarrow -h$, i.e., the helicity is a socalled pseudoscalar, because it doesn't change under rotations as a usual scalar but flips sign under space reflections. In this sense it defines a "handedness", because for $h>0$ the rotation described by the angular momentum is in the sense of the right-hand rule when considering the direction of the particle's momentum. For $h<0$ it's like a "left-hand rule" relative to the momentum. Thus in some sense you can call a particle in a right- or left-handed state if it has positive or negative helicity, respectively.

Now there's a little subtlety here: If you want to define an intrinsic property of a particle it should not depend on the particle's motion. For a massive particle the sign of the helicity is not such an intrinsic property (although it can make a lot of sense to describe particles in terms of helicity in a fixed reference frame). Since a massive particle goes always with a speed less than that of light, I can choose a another reference frame, which is going faster than the particle in the original frame. In the new frame the helicity has the other sign.

However, if you now have a massless particle, it always goes with the speed of light with respect to any frame, and you cannot overtake it by moving faster, because you cannot move faster than the speed of light. This implies that for massless particles the sign of the helicity is an invariant and thus a proper definition for an intrinsic quantity. In this case you call this sign its chirality.

Now you can define chirality also for massive particles, but this is hard to explain without the math. It boils however down to the idea that you can define eigenstates of chirality and any particle wave function can be written in terms of these eigenstates. In other words you can project to left- and right-handed parts of the wave function. For massless particles, the chirality is conserved. Formally this implies that you can rotate the phase of the left- and the phase of the right-handed part of the wave function independently from each other, if you consider a model where the Hamiltonian does not mix left- and right-handed components of the wave function. For massless particles (for the standard model most important for massless spin-1/2 particles) it is easy to build such chirally symmetric models. However, if you just introduce another term to the Hamiltonian that describes the mass of these particles, you destroy the chiral symmetry, because this term mixes left- and right-handed parts of the wave function. Thus to have a chirally symmetric model you better choose models with massless particles.

Now, in the standard model there are no massless fermions (we know nowadays that even the neutrinos have mass!). On the other hand the weak interaction should be somehow chiral, because empirical evidence shows that only the left-handed part of the neutrinos participate in the weak interaction. You may argue that one could simply forget about chiral symmetry and introduce mass terms by hand, but that's forbidden for another reason I come to in a moment. For the moment we just think about ways to describe massive particles in a chirally symmetric model. This brings another very important idea into the game, namely that you can have models, where the dynamics, i.e., the equations of motion, are symmetric under some transformations, but the ground state of the system does not need to be necessarily symmetric. An example is a ferromagnet. The laws of electromagnetism are invariant under rotations, but for a ferromagnetic material at not too high temperatures, it is energetically favorable to spontaneously direct a lot of magnetic moments into one direction (which is due to a quantum effect, the exchange forces, but that's not so important here), i.e., at 0 temperature, where the material is in its ground state, you have a preferred direction, because you have a non-zero magnetic moment. Of course, because the dynamical laws of electromagnetism are symmetric under rotations, which specific direction this magnetic moment has, isn't imporant. All directions have the same energy. One says the ground state is degenerate, i.e., there are many different states with the same lowest possible energy. The theory behind this was worked out in the 1960ies by Nambu and Goldstone for Quantum Field theory. The result was a famous theorem, according to which you can give mass to the particles by spontaneously breaking chiral symmetry, without making the dynamics unsymmetric under chiral transformations. However, it was inevitable that then one also had a massless scalar field for each broken symmetry, and of course there should be massless particles as excitations of this field. In electroweak theory, nobody needs such a massless scalar particle, because it's not seen in nature.

Now the old models (going back to Fermi's theory of beta decay of 1934) had anyway the flaw to be not renormalizable. So there was the idea to use a socalled gauge theory to describe the weak interaction analogous to the electromagnetic interaction. So the idea was to introduce vector bosons as "force carriers" similar to the photons which are the "force carriers" of the electromagnetic interaction. The only problem with that was that in gauge theories the vector bosons should be massless, this time because of local gauge symmetry. If you introduce a mass term by hand, the gauge invariance is broken, and this has far more severe consequences as in the case that a global symmetry is broken. In the latter case then this global symmetry is just not there or it is valid only approximately (as is the case for chiral symmetry in the strong sector of the Standard Model, QCD, where it is an important theoretical tool to build effective models of hadrons). If you break a gauge symmetry, the whole model is useless, because you violate basic principles as the unitarity of the S-matrix, so that you don't get physically meaningful predictions for cross sections.

Now the trick is to combine the idea of spontaneous symmetry breaking with gauging that symmetry. In the electroweak standard model you gauge the chiral SU(2) of the weak interaction, i.e., you make the global symmetry local by introducing vector bosons, which then are gauge bosons of the gauge theory. In fact you use the larger symmetry $\mathrm{SU}(2) \times \mathrm{U}(1)$ of the weak isospin and the weak hypercharge. The gauge bosons are now massless to begin with, but you can make some massive by attempting to spontaneously break the local gauge symmetry. As was found out a bit earlier already by the solid-state physicists in the context of the theory of superconductivity, doing so has the funny consequence, that no massless Nambu-Goldstone bosons occur in that case, but the corresponding scalar field-degrees of freedom get "eaten up" by the gauge bosons, which at the same time become massive, and still the local gauge symmetry is not broken! That you need additional field-degrees of freedom becomes clear from the fact that a massless vector boson has only two polarization-degrees of freedom. This is known from the electromagnetic field, which has only two polarization degrees of freedom. You can choose these two degrees of freedom as the helicity states, and for a massless vector boson you only have helicities $\pm 1$ but not $0$. A massive vector boson, however has, as one would expect in the first place, three spin states, namely $\sigma_z \in \{-1,0,1 \}$, and the additional spin-degree of freedom comes in the above described Higgs mechanism from the absorbed "would-be Goldstone boson modes".

So now we have solved two problems: (a) we have massive gauge bosons by "spontaneously breaking" the electroweak $\mathrm{SU}(2)_{\text{wiso}} \times \mathrm{U}(1)_{Y}$ local gauge theory to the electromagnetic $\mathrm{U}(1)$ symmetry. So you get 3 massive gauge bosons (the W and Z bosons) and one massless gauge boson, which is the photon. So you have absorbed 3 would-be-Goldstone bosons to get the 3 massive gauge bosons. If you want to keep the model renormalizable, you have to introduce at least a Higgs wiso doublet, which has 4 real field-degrees of freedom. Three are absorbed, providing the mass to the gauge bosons. Then one of the real field degrees of freedom gets a constant vacuum expectation value, which via the Yukawa Couplings of the quarks and leptons to the Higgs doublet provides also mass for the quarks and leptons, again without violating the chiral gauge symmetry). Now there is still one massive scalar boson left, and this is the famous Higgs boson which motivated the physicists to build the Large Hadron Collider at which this last building block of the Standard Model indeed has been found in 2012.

 

[Drakkith]

Now a space reflection is defined by the substitution t→tt \rightarrow t, r⃗ →−r⃗ \vec{r} \rightarrow -\vec{r}. Since p⃗ =mdx⃗ /dt⃗ \vec{p}=m \mathrm{d} \vec{x}/\mathrm{d} \vec{t} and since we assume that m→mm \rightarrow m under space reflections also p⃗ →−p⃗ \vec{p} \rightarrow -\vec{p}, but L⃗ ⇒(−r⃗ )×(−p⃗ )=r⃗ ×p⃗ =L⃗ \vec{L} \Rightarrow (-\vec{r}) \times (-\vec{p})=\vec{r} \times \vec{p}=\vec{L}. So L⃗ \vec{L} doesn't flip its sign under space reflections. This is what's called an axial vector.

Interesting. It's like the right-hand rule for angular momentum becomes a left-hand rule in a reflection. Is that right?

 

[vanhees71]

Interesting. It's like the right-hand rule for angular momentum becomes a left-hand rule in a reflection. Is that right?

Yes, if you have a usual right-handed skrew and you look at it in a mirror, it looks left-handed. So everything that has some sense of winding flips the relative orientation of the corresponding direction and the associated rotation.

One should note, however that a mirror flips only one component of axial directions. If the mirror is in the $xy$ plane all polar vectors undergo the transformation $(x,y,z) \rightarrow (x,y,-z)$. Usually the space reflection is a reflection at a point (usually the origin of your coordinate system). So the mirror reflection is a space reflections followed by a rotation by an angle $\pi$ around the normal direction of the mirror.

 

[Drakkith]

One should note, however that a mirror flips only one component of axial directions. If the mirror is in the xy plane all polar vectors undergo the transformation (x,y,z)→(x,y,−z)

So you're saying an object moving in the +z direction looks like its moving in the -z direction when viewed in the mirror, but any x and y motions are unaffected?

Usually the space reflection is a reflection at a point (usually the origin of your coordinate system). So the mirror reflection is a space reflections followed by a rotation by an angle π around the normal direction of the mirror.

Okay, it took me a little while (and some wiki browsing) to understand, but I think I get it. A point reflection swaps every point (x,y,z) with (-x,-y,-z), and a rotation of angle π about an axis running perpendicular to the mirror surface turns (-x,-y,-z) into (x,y,-z)?

 

[Vanadium 50]

Vahhees, that's quite interesting - but given the level of the OP, do you think he got any of that?

 

[vanhees71]

Who knows? I hope, he or she will ask further questions if not!

 

[Orodruin]

Who knows? I hope, he or she will ask further questions if not!

The "B" tag in the thread title means beginner level (i.e., the OP is at high-school level or below). The OP is likely not even familiar with the cross product. The BIA tags were introduced specifically for people to be able to gauge the level of answer required for the OP.

 

[vanhees71]

Well, I had not the impression that with the other answers the question was even addressed. If this is what a "B" tag mandatorily means, I'd rather not post to B-tagged questions anymore!

 

[Vanadium 50]

Who knows? I hope, he or she will ask further questions if not!

Yes, but this is a "B" question. If you have a rank beginner, would they even know what to ask? If he's a freshman in high school (the sort of person who might ask a B), your answer is, what 10-12 years beyond him? If I were presented with something 10-12 years beyond where I was, I don't think I'd even know what question to ask.

 

[Orodruin]

Well, I had not the impression that with the other answers the question was even addressed. If this is what a "B" tag mandatorily means, I'd rather not post to B-tagged questions anymore!

Just to say that this was the entire idea behind the introduction of the BIA tags. For people who are not interested in replying to B level threads to know that they can save the time it would take to open the thread, for people who are at B level not having to open A level threads, and for responders who are happy responding to all levels to understand the appropriate level of response which may be understood by the OP.

 

[vanhees71]

Very complicated, and I thought that this posting is ok for a high-school student. Sorry for that. I'll refrain from answering to B threads (to be honest, however, I don't look too much at these tags; so if I should answer a B-thread in the future, I apologize in advance).

 

[ChrisVer]

Very complicated, and I thought that this posting is ok for a high-school student. Sorry for that. I'll refrain from answering to B threads (to be honest, however, I don't look too much at these tags; so if I should answer a B-thread in the future, I apologize in advance).

That's the spirit vanshees!
Never give up the closest to right answer only for the B-tagging system.
(but oh well, you could always simplify an answer)

 

[vanhees71]

Well, sometimes the hardest questions are tagged with a B-level sign ;-)). Anyway, there's a another warning from Einstein concerning physics in general and popularization in particular: "One should explain things as simple as possible but not simpler."

 

[Buzz Bloom]

It's difficult to answer this question, because it involves pretty advanced math to understand, why the Higgs mechanism (as a theoretical concept) is necessary to give the particles mass. The reason is that the mass of particles is doubly "problematic" in the quantum-field theoretical model that describe electroweak interaction.

Hi Vanhees:

Although I will need some further education to understand some of your presentation, I found it very illuminating.

One general aspect about which I have failed to achieve any intuitive grasp is the relationship between a particle and it's corresponding field, in this case the Higgs particle and the Higgs field. The following interpretation is my best effort.

Fields are continuous and particles are discrete. A particle is the quantized equivalent of it's field. Fields interact with fields, and particles interact with fields, but particles do not directly interact with particles. When a particle interacts with a field, or a field interacts with a field, some quantities are exchanged, for example: charge, spin, energy, and momentum. Charge and spin are of course quantized, but energy and momentum in classical physics are continuous. However, for quantum fields, the specific values of energy and momentum exchanged are limited to a set of discrete values.

Is this interpretation reasonable? If not, can you suggest any improvements?

Regards,
Buzz

 

[nikkkom]

Fields are continuous and particles are discrete. A particle is the quantized equivalent of it's field. Fields interact with fields, and particles interact with fields, but particles do not directly interact with particles. When a particle interacts with a field, or a field interacts with a field, some quantities are exchanged, for example: charge, spin, energy, and momentum. Charge and spin are of course quantized, but energy and momentum in classical physics are continuous. However, for quantum fields, the specific values of energy and momentum exchanged are limited to a set of discrete values.

Particles are excitations of fields.

There is a very, very good explanation of all this with a very small amount of math:

http://profmattstrassler.com/articl...-basics/fields-and-their-particles-with-math/

 

[Buzz Bloom]

Particles are excitations of fields.
There is a very, very good explanation of all this with a very small amount of math

Hi nikkkom:

Thanks very much for your post. I took a quick look at the Stassler "conversation" you cited, and I think I will be able to understand it and learn some interesting physics.

I feel my current state of learning difficulty is more related to presentations that have too many physics concepts that are new to me rather than too much math. I am still able to handle most of the undergraduate and graduate level math I learned many years ago. As I age I find that I cannot keep simulataneously in my head as many new concepts at one time as I did when I was younger.

Regards,
Buzz