Momentum Energy and Higgs

So, as an object accelerates it gains energy. And energy is mass. So an object becomes more massive as it approaches the speed of light.

But, if mass is ONLY due to an object's interaction with the Higgs field (which I don't fully understand and have another question open about which you're free to answer as well =P ) then how does adding more kinetic energy make it interact more with the Higgs field if you're not adding more particles that interact with the higgs, only energy?????

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Your confusion arises because you are mixing an old concept (energy-dependent mass) with a relatively new one (particle masses from the Higgs mechanism).

The modern understanding is that (at least in the context of relativity and particle physics) there is only one kind of mass $m$, which used to be called the rest mass.

This mass is an invariant - it is the same in all reference frames - and is intrinsic to the particle under consideration. When you add energy, you are not increasing its mass - but you are making it more difficult to accelerate the particle (i.e. increasing its inertia) - and again, this is not because you are increasing the mass, but simply because mass and energy are related by a new formula in comparison to pre-relativistic physics. $$E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$$

So, no, an object doesn't become more massive as it approaches the speed of light. $m$ is always the same.

Now the mass $m$ for any given particle is a consequence of its interaction with the Higgs field. To unpack that statement - you start with a Lagrangian (an expression from which one can derive equations of motion - it's kind of the starting point) that doesn't explicitly look like it gives a mass to any of the particles. But in this Lagrangian, you also have a term that couples the Higgs field to all the other particle fields. The Higgs field is a special kind of field that induces something called spontaneous symmetry breaking, and when you rewrite your Lagrangian in such a way that everything is expanded around a stable vacuum of the Higgs field (as it should be), voila, you have new terms in your Lagrangian that look like they give particles masses. These masses are the $m$'s - they don't have anything to do with how fast the particles are moving - just with how the particle fields couple to the Higgs field.

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So the distinction is between Rest Mass and Inertia? So Einstein understood his equations to be pointing to a change in inertia as an object approaches C and not a change in some type of mass? I thought the point of this equivalence of Rest Mass and the acquired mass due to the acceleration was that you can't distinguish between an object at rest in a gravitational field and an accelerating object in open space? – John Dec 13 '11 at 14:35
Correct, inertia is related to energy, not mass. Note that rest mass is not equivalent to the "acquired mass" (which is actually kinetic energy) that arises due to an object's velocity (not acceleration). – David Z Dec 13 '11 at 15:30
So what of e=mc^2? Doesn't the M there refer to rest mass? And if rest mass is just how much a particle sticks to the Higgs field, then what explains the increase in mass of an accelerated object? – John Dec 13 '11 at 21:59
@John: I don't understand your first comment because you are still using terms like "acquired mass" which unnecessarily complicate things. The mass of an object doesn't increase when it is accelerated. The mass alone is not a measure of its inertia. The mass is intrinsic to the object. If you stare at the second equation here en.wikipedia.org/wiki/Special_relativity#Force for a while, you will realize that the force required for the same change in velocity is higher at higher velocities. That is what is meant by inertia increasing at higher velocities. The mass is always the same. – dbrane Dec 13 '11 at 23:24

An alternative view from an experimentalist to @dbrane 's answer here, (if one wants to keep defining mass as resistance to change as is the garden variety definition), is to consider what you mean with:

then how does adding more kinetic energy make it interact more with the Higgs field if you're not adding more particles that interact with the higgs, only energy?????

By considering how kinetic energy is "added" in the microscopic hbar dominated particle world, where the Higgs reigns:

You are adding more particles, i.e. more virtual exchanges of all type of particles, which are the carriers of the extra kinetic energy: There can be no interaction without virtual exchanges of particles ( according to the rules of the Lagrangian describing the system). These excess particles which are the impulse carriers and transfer the energy, are also interacting with the Higgs field . The interactions are not simply linear or vectorial additions but whenever one wants to change the kinetic energy by delta(E) of a moving particle they appear, due to the fields which will effect the change; the result is the resistance to change according to the relativistic mass at that energy, and not the rest mass.

It is simpler though when dealing with microscopic particle interactions to keep the point of view given by @dbrane and call particle mass the rest mass, which happens to be the "length" of the four vector describing the particle and is an invariant.

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"You are adding more particles, i.e. more virtual exchanges of all type of particles, which are the carriers of the extra kinetic energy." Can you explain this further? – John Dec 13 '11 at 14:31