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When I asked this question (probably in a less neutral form) to physicists, their answer was something along the lines that it's not gravity (i.e. unrelated to gravitons) but inertial mass. (So I wondered whether this is an analogous mechanism to gravitons, only that it explains inertia.) Now after some weeks of thinking (and reading) about this, I think I finally figured out what they were trying to tell me. This is related to the following comment for a similar question:

Have you made up your mind on what "mass" of a particle means to you in that question? Maybe that will help.

For me, the obvious candidates what "mass" might mean are

  • gravitational mass
  • inertial mass
  • rest mass

My current guess is that the Higgs mechanism explains why "other particles" (only fermions, or also other bosons?) have a non-zero rest mass. (I imagine it's some form of explanation for potential energy related to the mere presence of the particle, even in the absence of "interactions" with other particles.) However, at least some of the (popular science) explanations really seem to try to explain something related to motion and inertia, and I got the answer "inertial mass" so often that I wonder whether it's actually really the "inertial mass" (of fermions) that is "directly explained" by the Higgs mechanism (this doesn't preclude that this explanation might be "translated" into something equivalent to rest mass).

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    $\begingroup$ There isn't really a difference between inertial and rest mass. The term "rest mass" is used because, if you define inertial mass to be the ratio of force to rate of change of momentum, for large speeds relativity dictates that this "mass" increase without bound. The term rest mass is used to specify that this constant is taken in the particle's rest frame. $\endgroup$
    – Emilio Pisanty
    Jul 29, 2012 at 21:55
  • $\begingroup$ @EmilioPisanty When I think about rest mass, I'm mostly thinking about the potential energy related to the presence of the particle, including all interactions with other particles. So my question basically assumes that the Higgs mechanism only explains the part of the rest mass which isn't explained by other interactions. Even so rest mass and inertial mass are physically closely related, there is a difference between inertial and rest mass with respect to an explanation. (But I guess that explanations in terms inertial mass will be translatable into explanations in terms of rest mass.) $\endgroup$
    – Thomas Klimpel
    Jul 29, 2012 at 22:42
  • $\begingroup$ @EmilioPisanty: You said it wrong--- the ratio of force to time rate of change of momentum is 1, the ratio of force to time rate of change of velocity is the longitudinal/transverse mass. The relativistic mass is the ratio of momentum to velocity. $\endgroup$
    – Ron Maimon
    Jul 30, 2012 at 15:24

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Think of the Higgs mechanism as affecting rest-mass. This is the mass that a particle has when it is sitting still (you can weigh it to figure it out).

Think of gravity as affecting energy. More energy = more gravitational force. So an electron that is moving very quickly has a total energy of its rest mass (E = mc^2) + its kinetic energy.

Consider an electron and a positron. These both have rest-mass. When they collide and turn into two photons, all rest-mass is gone. Energy is conserved, so the system still weighs the same at all times. But the Higgs mechanism only affects the electron and positron, not the photons.

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  • $\begingroup$ Actually, I'm not sure about my answer anymore. Some other posts around here have me wondering if the Higgs field interacts in other ways... $\endgroup$
    – Nick
    Nov 23, 2014 at 8:37
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General relativity doesn't care about the difference between mass and energy. In the stress-energy tensor T$_{00}$ is the energy density and mass is just treated as energy divided by $c^2$. GR doesn't care what the Higgs' mechanism does, and will work just as well above the electroweak transition where the particles (well, the vector bosons at least) are all massless.

What the Higgs' mechanism does is explain why everything doesn't travel at the speed of light. It's the constant of proportionality between velocity and momentum. I guess in your terms it's the inertial mass/rest mass.

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  • $\begingroup$ The constant of proportionality between velocity and momentum is the inertial mass in my terms. So for me, what you are hinting at is an explanation in terms of the inertial mass, and not in terms of the rest mass. An explanation in terms of the rest mass would have to be more related to the energy involved in creating/destroying the particle. I know that special relativity tells us that inertial mass and rest mass are closely related, this is what I meant by "(this doesn't preclude that this explanation might be "translated" into something equivalent to rest mass)" in my question. $\endgroup$
    – Thomas Klimpel
    Jul 29, 2012 at 22:25
  • $\begingroup$ Rest mass is just the inertial mass in a frame where the velocity of the particle is zero. For example it predicts the recoil when your stationary particle is hit by another particle. Your reply to Emilio's comment says "potential energy related to the presence of the particle, including all interactions with other particles" but the only such mass related interaction I can think of is gravity, and this doesn't distinguish between mass and energy. I think you are attaching a special significance to "rest mass" that doesn't exist. $\endgroup$
    – John Rennie
    Jul 30, 2012 at 6:25
  • $\begingroup$ I guess you have a valid point that the special significance I attached to "rest mass" doesn't really exist. However, in order to confirm this (and also to better explain the special significance I used to attach to "rest mass"), I asked a question specifically about whether this special significance made any sense at all. $\endgroup$
    – Thomas Klimpel
    Jul 30, 2012 at 8:04
  • $\begingroup$ I was a bit surprised to see that general relativity is used in particle physics, because I heard that we still have no theory which satisfactorily unifies QM and GE. $\endgroup$
    – Thomas Klimpel
    Aug 2, 2012 at 22:03

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