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If an electron would otherwise be moving at the speed of light if it weren't constantly interacting with the Higgs Field, how is conservation of momentum preserved if it's constantly bouncing off of virtual Higgs Bosons? Why does this not lead to particles like electrons experiencing Brownian motion from all the random collisions?

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The cartoon of particles getting their mass by repeatedly bumping into Higgs field is truly misleading--to the extent that it has led the OP to ask this question (which to most high energy physicists sounds "silly").

Here, I can offer another slightly better cartoon that addresses the conceptual issue:

Imagine the Higgs field as a large collection of closely-spaced stumps jutting out of the surface of a vast body of water. Water waves (representing the quantum massless particles in this analogy) attempting to traverse the area with stumps will be multiply scattered, causing them to be "slowed down" which looks like these waves corresponding to particles have acquired a mass. Since we have have replaced the massless particles with waves, this cartoon offers a "smoother" picture which makes it easier to discard the notion that any Brownian motion would occur.

Please keep in mind that this is just another cartoon: every analogy that highlights a particular feature (in this case, the lack of any Brownian motion in the Higgs mechanism) always falls short in something else (the loss of wave packet localization due to the numerous haphazard scatterings, and the apparent loss of momentum conservation as the waves are scattered).

In reality, the waves undergo only forward-coherent scattering with the Higgs field. The "coherent" aspect of scattering holds the wave packet together. The "forward" scattering aspect means that the wave number (i.e. the momentum) remains unchanged. It works just like how glass changes the dispersion relation of light waves due to forward coherent scattering with the glass molecules.

None of this is particularly tied to the fact that it's a left/right chirality flipping interaction.

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  • $\begingroup$ I like your analogy with the forward scattering, it is actually a very good one, especially the comment on the forward scattering are nice (with a bit of stretch one could think of the 1->1 scattering amplitude as the corrections to the propagator and m). But I just slightly disagree with the very last comment: it seems precisely because it is a left-right chiral flipping interaction that the dispersion relation for the wave is changed more dramatically than, to follow your analogy, just changing the refraction index. To go from $\omega=k$ to $\omega=\sqrt{k^2+m^2}$ one really needs those. $\endgroup$ – TwoBs Oct 15 '16 at 8:01
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    $\begingroup$ I honestly don't think it's a silly question, that's quite unfair to the OP $\endgroup$ – innisfree Oct 15 '16 at 14:41
  • $\begingroup$ @innisfree did you mean unfair to the cartoon? $\endgroup$ – QuantumDot Oct 15 '16 at 14:53
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    $\begingroup$ no, unfair to the OP $\endgroup$ – innisfree Oct 16 '16 at 0:07
  • $\begingroup$ @innisfree I don't know how calling the question silly is unfair to the OP. The problem lies with the cartoon. Not with the OP. $\endgroup$ – QuantumDot Oct 16 '16 at 1:38
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That's an interesting question. As @JgL says, particles that interact with the Higgs field (e.g. electrons) experience a drag, which makes them massive. You wonder whether these interactions could impart momentum to the electron.

Let's extend the analogy to a ball being thrown through water. If the water is sloshing about, as well as providing drag, the waves could impart net momenta on the ball, making it deviate from a straight line. This is only an analogy; there are aspects of drag (e.g. energy loss as ball slows down) that is quite unlike the Higgs field.

However, the part of the Higgs field that provides drag, denoted by $v$, resulting in masses, is homogeneous (i.e. unchanging) in space and time: $$ \text{Higgs field}(x) = h(x) + v $$ That means there is no sloshing or waves that impart momentum on the electron. The homogeneous part of the Higgs field is like perfectly still water that provides drag and nothing else. The other part of the Higgs field, $h(x$), results in physical Higgs bosons.

The physical Higgs bosons, $h(x)$, could indeed impart momenta; however, this would be nothing but collisions (elastic scattering) between Higgs bosons and electrons. As there is no big source of background Higgs bosons, the probability of any interactions is negligible, and it doesn't cause the electron to deviate from a straight line.

Finally, concerning the effects virtual Higgs bosons (as opposed to physical Higgs bosons considered above) on the motion of an electron. Virtual Higgs bosons cannot be present in the initial or final states. We begin with an electron and end with a electron. By conservation of momentum, it cannot have changed momentum, i.e. it does not deviate from a straight line due to virtual Higgs bosons.

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  • $\begingroup$ The drag analogy makes no sense to me. To me drag is friction which would stall an electron to a stop. I'd tend to think it'd be more like a photon in glass, except that, as I understand it, the reason the photon is slower is because it spends a fractions of a second absorbed by an electron before it is reemitted. But in the case of the Higgs field, an electron is clearly not absorbed, only interacting with the field. What I need is a better conceptual description of the interaction with the field. $\endgroup$ – Shufflepants Oct 13 '16 at 14:37
  • $\begingroup$ All I can get are people trying give poor analogies and that the real interaction is waves in fields. Can no one give me a good picture of what the interaction between an electron and the Higgs field is? We have such good representations for double slit interference images.slideplayer.com/13/3866150/slides/slide_2.jpg or for refraction file2.answcdn.com/answ-cld/image/upload/… What's conceptually going on with electron-higgs field scattering? $\endgroup$ – Shufflepants Oct 13 '16 at 14:44
  • $\begingroup$ @Shufflepants : a photon in glass has a slight effective mass because it's moving at less than c. Because it's interacting with the glass. When you catch a photon in a mirror-box, then whilst it's going round and round inside the box, it's effectively at rest in aggregate with respect to you. Then all of its energy-momentum is effective as mass. The mass of the system is increased. When you open the box it's a radiating body that loses mass. $\endgroup$ – John Duffield Oct 15 '16 at 14:47
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The Higgs field is a field that permeates all of space (very much like the electromagnetic field), Higgs bosons are 'excitations' of this field. In some contexts, you can roughly think of these excitations as ripples or waves in the field, though this is a bit of a sloppy analogy. I would say these 'excitations' have a much richer structure than just being simple waves, which allows them to also behave as what we normally think of as particles.

When the Higgs field interacts with other elementary fields, the excitations of these other elementary fields (which you can think of as other elementary particles) experience what you can think of as 'drag', because due to their interaction with the Higgs field they have to 'plow through' this Higgs field while they are moving through space. This is the typical way the Higgs field gives what we normally think of as 'mass' to e.g. electrons (the excitations fo the electron field). If this interaction would not be there, the excitations of the electron field would be free to move through space at the speed of light.

(Sometimes people prefer to explain Quantum Field Theory in terms of particles to avoid having to mention 'excitations' or 'fields'. When they do this they often talk about 'virtual particles' where they are actually talking about interactions between fields. This, unfortunately is a problem that arises often when you try to explain things by analogy. Your explanation will only seem to make sense as far as your analogy holds. )

Edit: If you know some electromagnetism you can maybe try to play around with two classical fields $\phi$ and $\chi$ who's interaction is described by the Lagrangian $$ \mathcal{L} = \partial_\mu \phi \partial^\mu \phi + \partial_\mu \chi \partial^\mu \chi + \phi^2 \chi^2 $$ If the coupling term $\phi^2 \chi^2$ wasn't there, waves in the two fields could freely propagate at the speed of light. The behaviour of both fields would be described by a differential equation of the form $$ \frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2} = \nabla^2 \phi $$ The coupling $\phi^2 \chi^2$ causes waves in the $\phi$ field to interact with the $\chi$ field. Waves in the $\phi$ field will as a result experience some form of 'drag' (not in terms of friction, but as a consequence of the way the fields interact with each other). When we turn this coupling on, the above differential equation that describes the behaviour of the $\phi$ field will be changed, an be of the form $$ \frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2} = \nabla^2 \phi - \chi^2 \phi $$ The amplitude of the $\chi$ field at a point $x_i$ now behaves as an effective mass that is experienced by the waves in the $\phi$ field at that point. This is analogous to the way the Yukawa coupling of the Higgs field gives an effective mass to many other fundamental fields.

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  • $\begingroup$ "Your explanation will only seem to make sense as far as your analogy holds." And I'm looking for more detailed answer than the "drag" analogy provides. My physics intuition reads "drag" as friction that would slow a massive particle to a stop. See the comments on the main post by @QuantumDot for something closer to what I'm looking for. I'd compile it into an answer myself, but I've only barely got a conceptual grasp on it, and I'm sure I'd butcher it. And it probably needs some details filled in. $\endgroup$ – Shufflepants Oct 12 '16 at 20:49
  • $\begingroup$ I've read a ton of the super layman's descriptions and have tried to read some that go into some of the math, and that's what's lead to my confusion. I've read a lot of physics, but my math background for quantum mechanics is lacking. Looking for something in between that explains the concepts accurately without needing to fully understand all the math involved. $\endgroup$ – Shufflepants Oct 12 '16 at 20:55
  • $\begingroup$ The 'drag' is a fundamental interaction between the Higgs field and the other fields involved. You can compare it in that sense with electric charge. Why are some particles electrically charged? Because they interact with the electromagnetic field. It is one of the basic assumptions of the theory in order to fit our observations. The different fields and their interactions are the assumptions - carefully chosen to fit our observations - on which the theory is based. Particles are less fundamental within the framework of the Standard Model, they are excitations of these fields. $\endgroup$ – JgL Oct 13 '16 at 9:53
  • $\begingroup$ @Shufflepants - ' Looking for something in between that explains the concepts accurately without needing to fully understand all the math involved.' - This may be a problem. One theme in the answers is that analogies don't really give an accurate picture. There is no royal road to physics. $\endgroup$ – mmesser314 Oct 15 '16 at 15:33
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If an electron would otherwise be moving at the speed of light if it weren't constantly interacting with the Higgs Field,

Here is the first misunderstanding , the electron is not "interacting" with the Higgs field in terms of exchanging energy or at least momentum, which is what "interaction" means classically and quantum mechanically.

Quantum mechanical interactions are represented by interaction vertices in Feynman diagrams which have a definite coupling constant of interaction and describe the integral that has to be evaluated to get crossections and lifetimes. The acquisition of mass happens at the wave function level, before probabilities for interactions are calculable.

The acquisition of mass for the massless particles happens below a specific energy when the electroweak symmetry breaks and from then on all electrons acquire a fixed mass. In the present standard model of the universe that happened at 10^-10 seconds after the Big Bang., and the average particle energy was of the order of 100GeV.

how is conservation of momentum preserved if it's constantly bouncing off of virtual Higgs Bosons?

It is not bouncing off Higgs bosons, it is just the the units of reference of the vacuum have changed, there is a non zero vacuum expectation value for the Higgs field, whereas the electron field and all the other particle fields have zero expectation value if no particle is there, the Higgs field has a value and that imposes constraints on all the other fields, not interactions.

Why does this not lead to particles like electrons experiencing Brownian motion from all the random collisions?

There are no collisions, no interactions of the classical type. In terms of wavefunctions, the wavefunction of the electron is superposed on the Higgs field. An electron path generated with creation and annihilation operators on the electron field gets a fixed non zero mass when the Higgs field has the vacuum expectation value it has after symmetry breaking (246GeV), than before symmetry breaking when the electron mass was zero and the Higgs vev was zero.

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If particles get mass from the Higgs field, why do we not see Brownian motion?

Because they don't.

If an electron would otherwise be moving at the speed of light if it weren't constantly interacting with the Higgs Field

It isn't. We make electrons and positrons in gamma-gamma pair production. We start with photons moving at the speed of light. These photons interact with each other, such that each changes direction and starts interacting with itself. Then it continually changes direction, in a chiral spin ½ fashion. Only then we don't call it a photon any more. We call it an electron. Or a positron if it has the opposite chirality. And as Einstein said, the mass of a body is a measure of its energy-content. Not the measure of its interaction with some field. See https://arxiv.org/abs/1508.06478 by van der Mark and (not the Nobel) 't Hooft. If you catch a massless photon in a mirror-box, you increase the mass of that system. When you open the box it's a radiating body that loses mass.

how is conservation of momentum preserved if it's constantly bouncing off of virtual Higgs Bosons?

Momentum is preserved because it isn't bouncing off virtual Higgs bosons. See this answer. Virtual particles only exist in the mathematics of the model.

Why does this not lead to particles like electrons experiencing Brownian motion from all the random collisions?

Because it's a fairy tale. See page 174 of A Zeptospace Odyssey where CERN physicist Gian Guidice says 98% of proton mass results from E=mc², whilst electromagnetic effects and the Higgs mechanism account for 1% each.

enter image description here

As for whether that 1% is correct, well. Don't forget that we've never ever seen a free quark. However we have seen low-energy proton-antiproton annihilation to gamma photons.

enter image description here

Image credit CSIRO, see The Big Bang & the Standard Model of the Universe

I think the best way to think of this is that it's like opening one box with another. Then each is a radiating body losing mass. All of it. And then that box isn't there any more.

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