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Both the Photon and the Electron are point-like particles. Most massive particles can be thought of "bound collections of massless particles", but the electron assumed to be a "point particle with a point charge and no spatial extent."

They both contain energy. Energy and mass are equivalent, however, only the Electron has a mass from the Higgs field that prevents it from reaching the speed of light. If an electron and positron meet, the energy will be released in Photons that do travel the speed of light.

So how exactly DOES the Higgs field prevent the electron from going the speed of light?

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    $\begingroup$ Asking for an exact answer in non-mathematical terms is not realistic. Popular metaphors sometimes describe the Higgs field as acting like molasses for certain particles. I doubt that that satisfies you, but unfortunately I don’t know what to suggest beyond “learn quantum field theory”. Some things just aren’t easy to explain without math. $\endgroup$ – G. Smith Jul 17 at 18:21
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    $\begingroup$ Most physicists do not think of massive particles such as electrons or quarks as being bound collections of massless particles. (They certainly are not in the Standard Model.) The video was probably talking about the proton and neutron mass, which is mostly due to gluons. You seem to have drawn an overly broad conclusion. $\endgroup$ – G. Smith Jul 17 at 18:25
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/17944/2451 , physics.stackexchange.com/q/6450/2451 and links therein. $\endgroup$ – Qmechanic Jul 17 at 19:03
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    $\begingroup$ Good question, but at the end an electron could be accelerated only by the interactions with photons. Bombarding an object with another objects with velocity v never lead to an velocity higher v. QM and all its derivatives are developed for subatomic processes and the generalization to an electron as a disturbance of a self-sufficient electric field is to be taken with caution outside atomic physics. $\endgroup$ – HolgerFiedler Jul 18 at 4:35
  • $\begingroup$ @G.Smith I really have to stop asking these questions on the physics stackexchange. - Everytime I get a bunch of confusing technical answers (with people arguing in the comments) and end up no less confused than I did at the start. I have literally no idea which answer to choose as the correct one. ¯_(ツ)_/¯ $\endgroup$ – Benjamin Jul 20 at 17:41
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If it helps, the math is a close analogy to “Why does light slow down in glass?”

The Higgs boson, like glass, fills space with some density.

When an electron wave moves through that space (like light in glass), it couples to that Higgs density and excites it, like the light wave excites electrons in glass.

That excitation, through that same interaction, reradiated an electron (light) wave at a slightly later phase.

When that reradiated wave combines with the original, the result is later than the original. The more space you go through, the later you get: this is the same as having a slower speed.

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    $\begingroup$ It's a decent analogy, but far from a perfect one. For example, glass defines a preferred frame, but the Higgs field doesn't. $\endgroup$ – knzhou Jul 17 at 19:24
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    $\begingroup$ @knzhou True. The math of how it retards the wave is the same, but the (relavitistic) Higgs energy density is quite different from glass and atoms. Why that is might make a good Question: why is the Higgs density frame-independent? $\endgroup$ – Bob Jacobsen Jul 17 at 20:04
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    $\begingroup$ That's just because, being a scalar field vev, it behaves formally like dark energy. It has a pressure just so the energy density doesn't transform between frames. All nonrelativistic matter like glass has negligible pressure. $\endgroup$ – knzhou Jul 17 at 20:26
  • $\begingroup$ @BobJacobsen This sounds like a particle is slowed by traveling without a static spatial field, but I thought Einstein showed objects in space have no preferred reference frame? $\endgroup$ – Benjamin Jul 20 at 18:28
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For an electron, the energy and 3-momentum for a 4-vector:

$$ p_{\mu} = (E/c, \vec p)$$

that satisfies (in all reference frames):

$$ p^{\mu}p_{\mu} = E^2/c^2-p^2c^2 = m^2c^2$$

and in the rest frame ($\vec p = 0$) reduces to:

$$ E = mc^2 $$

If we look at the De Broglie relationship, we can describe a free electron with a 4-wave-vector via:

$$ p_{\mu} = \hbar k_{\mu} =(\hbar \omega/c, \hbar \vec k)$$

So that:

$$ k^{\mu}k_{\mu} = \omega^2/c^2 -||k^2|| = \frac{m^2c^2}{\hbar^2}$$

which gives a dispersion relation:

$$ \omega = \sqrt{(ck)^2 + \frac{m^2c^4}{\hbar^2}}$$

A photon satisfies the same conditions, with $m=0$: $$ E = pc$$ $$ \omega = ck $$ the latter being the well known relation describing dispersionless waves propagating at the speed of light, $c$.

Now if we consider light propagating in a waveguide, or say an O-wave in a plasma with plasma frequency $\omega_p$, the dispersion relation becomes:

$$ \omega = \sqrt{c^2k^2 + \omega_p^2}$$

Which mean EM waves with $\omega < \omega_p$ don't exist, because their is finite energy $(\hbar \omega_p)$ at zero wavenumber, and this is because the wave couples to the electrons in the plasma.

Note that this form is the same as the dispersion relation for an electron in free space. Because of coupling to the Higgs boson, there is non-zero frequency even at zero wavenumber:

$$ \omega_0 = \frac{mc^2}{\hbar}$$

which corresponds to finite energy at zero momentum, aka mass:

$$ E = \hbar\omega_0 = mc^2$$

So the Higgs acts more like a universal plasma or waveguide, than "molasses".

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The SM defines elementary particles as point like, with no spatial extent and no substructure:

  1. massless, like gluons and photons

  2. particles with rest mass, like the electron and quarks

Photons do travel at speed c when measured locally in a vacuum.

Now when in a medium, EM waves do travel slower than c, but that is because the classical EM wave is built up bu a herd of photons, that move in a zig-zag way through the medium as they interact with the atoms in the medium. Now photons still travel at speed c in between atoms in a vacuum. But since their path is zig-zag, longer then the path you calculate with, the wavefront's speed will be slower than $c$.

Now electrons do move slower than $c$, even in a vacuum, when measured locally. Why? As per SR:

  1. massless particles travel at speed c in a vacuum when measured locally
  2. particles with rest mass travel slower than c in a vacuum when measured locally

Now as per the SM, electrons, and quarks do have rest mass. What is rest mass?

Nobody has ever measured an electron at rest. Nor a quark, which is confined too. The rest mass is a mathematical theory, that fits the experimental data.

Now it is very important to understand that these are speeds for vacuum. What is a vacuum?

In quantum mechanics and quantum field theory, the vacuum is defined as the state (that is, the solution to the equations of the theory) with the lowest possible energy (the ground state of the Hilbert space).

https://en.wikipedia.org/wiki/Vacuum

In quantum field theory the vacuum expectation value (also called condensate or simply VEV) of an operator is its average, expected value in the vacuum. The vacuum expectation value of an operator O is usually denoted by {\displaystyle \langle O\rangle .} {\displaystyle \langle O\rangle .} One of the most widely used examples of an observable physical effect that results from the vacuum expectation value of an operator is the Casimir effect. This concept is important for working with correlation functions in quantum field theory. It is also important in spontaneous symmetry breaking. Examples are: 1. The Higgs field has a vacuum expectation value of 246 GeV [1] This nonzero value underlies the Higgs mechanism of the Standard Model. 2. The chiral condensate in Quantum chromodynamics, about a factor of a thousand smaller than the above, gives a large effective mass to quarks and distinguishes between phases of quark matter.

In a superconductor, photons do gain rest mass (travel slower than c) because of their interaction with the field.

How come a photon acts like it has mass in a superconducting field?

Now in a vacuum, photons do not couple to the Higgs field. They do not interact with it, so photons travel at speed c.

The Higgs field exists everywhere in space. Through the Yukawa interaction, the electron will interact with the Higgs field. This way the electron in a vacuum will not be able to travel at speed c, since these interactions with the Higgs field will slow the electron's propagation.

https://en.wikipedia.org/wiki/Higgs_mechanism

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    $\begingroup$ Photons dont zig-zag in a clear medium. If they did, coherent light wouldn’t stay coherent. $\endgroup$ – Bob Jacobsen Jul 17 at 19:49
  • $\begingroup$ @BobJacobsen In glass, photons do zig zag, but they do it so that their relative phase, energy and angle is kept. It is called inelastic scattering. $\endgroup$ – Árpád Szendrei Jul 18 at 10:12
  • $\begingroup$ @BobJacobsen It is the only way to keep a mirror image. Imagine the glass is made of layers of glass (where the thickness is one molecule thick). Imagine the whole image moving through glass, and at every atomic layer, the photons that build up the image, interact (inelastically scatter) with the atoms of that layer. The photons keeps the image, by keeping their relative phase, energy and relative angle. This way the image keeps zig zaging through the layers. $\endgroup$ – Árpád Szendrei Jul 18 at 10:20
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    $\begingroup$ Perhaps I don’t understand what you mean by zig zag. That carries a connotation of changed of direction that lengthen the path. Light does no such thing: it’s Poynting vector is always in the same direction. Maybe some other word than zig zag? $\endgroup$ – Bob Jacobsen Jul 18 at 13:35
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    $\begingroup$ The EM field propagates as a field. There are no individual physical photons during that process. $\endgroup$ – Bob Jacobsen Jul 18 at 13:55

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