Hello all. I`m new here, and still very much a novice in physics. I was reading the "Does Light Have Mass?" thread, and saw that in the E=mc^2 equation, the "m" refers to rest mass. It was my understanding that there are no objects in the universe which are truly at rest, but that all objects are in constant motion relative to one another, so I am curious what exactly is meant by "rest mass?" I take it it must not mean the mass of a stationary object, but something more along the lines of "the mass of a body which moves only due to gravitational influence, and not through the expenditure of any energy of its own." Would anyone be willing to clear this up for me?
Thanks.

Welcome to the forum!

If so, then the E must refer only to rest energy too.

Rest mass, like all concepts in physics, is an idealization. Physics is about taking the real world and mentally replacing it with idealized situations that bring insight, understanding, and the ability to predict experiments to a useful degree of precision. It has never been, nor ever will be, about reaching exact characterizations of how things really are, so it does not present a problem that there is indeed no such thing as a particle that is truly at rest. Indeed, is there such a thing as a particle, at the exact level of precision of a particle at perfect rest? These are all experimentally verifiable concepts, no more and no less, and experiments never achieve perfect precision.

The important issue is not why it's moving, but how it's moving, relative to the observer. That's what is meant by a "stationary" object-- it is not an absolute state of being, it depends on who is observing it.

The above statement seems a little extreme.

An object under an acceleration is receiving energy whereas an object traveling with an inertial velocity is not receiving energy - there is a difference.

Particles in accelerators appear to gain mass because of their acceleration whereas that receding galaxy on the other side of the universe exhibits a constant mass ... or has the mass-energy of acceleration converted to an inertial mass?

Here is a probing question:

If two objects originally with the same rest mass are traveling with a differential velocity with respect to each other how can we determine which object was accelerated more and is; therefore, the fastest?

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Particles in particle accelerators gain mass because of their high velocity, not the amount of acceleration acting upon it. The mass relative to the observer is the rest mass multiplied by the Lorentz factor. See This Wikipedia article on relativistic mass

Thanks for the link.

Acording to what you say that "Particles in particle accelerators gain mass because of their high velocity" then the mass of a receding galaxy should appear to us to be greater than its actual rest mass.

"To see that kinetic energy of pairs of particles is an invariant property of a system itself (but not any one particle), consider a model system of two particles, A and B, moving away from each other, each with the same kinetic energy. In this system, the kinetic energy is equally shared between particles A and B. However, in an inertial frame centered on particle A, particle A is at rest and all of the kinetic energy in the system is found in particle B. From the viewpoint of particle B, however, all of the kinetic energy in the system is present in A. It is apparent that kinetic energy in this system is invariant for all observers, but different observers will disagree as to how it is distributed or "located" (the location is relative)."

The above quote from the Wiki site is somewhat reminiscent of a discussion about special relativity post #114 from here: The curvature of spacetime

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My invisible elf ??? Why he is made of dark matter and lives off of dark energy !!!
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No. There is a common misconception that energy is an absolute thing, but in fact it is also entirely reference frame dependent. To an observer accelerating with an object, the object gains no energy. You cannot specify the energy without also specifying the reference frame, and if the reference frame changes, so does the energy.

Mass-energy, in the sense you are using it, is also frame dependent.
We can't in any absolute way. The history of acceleration has no bearing on its current motion, it moves as we observe to move, right now, in our own reference frame.

To resolve this very subtle issue, you have to recognize that relativity is a measurement theory, and measurements are conducted locally. Hence relativity is a local description of physics. To do nonlocal conceptualizations, such as talking about what "distant galaxies" are doing, you are doing just that-- conceptualizing. It's fine to conceptualize, but you no longer have unique descriptions-- the only unique results are the measurements, and they are always local. Thus how you treat the motion of distant galaxies will also affect how you conceptualize their mass-energy, and their other attributes. This is related to my other point, that energy depends on the chosen reference frame. I think there may be ways to conceptualize the motion of a distant galaxy such that it has a greater relativistic mass. The standard approach, however, is to treat its motion only relative to a local reference frame that moves with the average motion of the matter, the comoving frame. In that coordinatization of space, even distant galaxies are essentially at rest, even though their distance to us is increasing rapidly. In that description, their mass-energy would still be essentially their rest mass.

By the way, this statement:
Is patently false. Just consider an observer whizzing by both partices at a very high speed! What is Wiki thinking here? That entire passage is complete garbage. All you can say is that the kinetic energy in the center of mass frame is added to the kinetic energy of the center of mass to get the total kinetic energy. The first piece identifies a frame, so it is the second piece that is frame dependent. Yet, the total kinetic energy is very much frame dependent.

In the Wiki article it seems to equate kinetic energy to rest mass and so when an object is accelerating its kinetic energy is increasing and; therefore, its mass is increasing.

Another quote: "The rest mass or invariant mass is an observer-independent quantity."

The above quote seems to indicate that every observer should be able to determine the rest mass of an object but according to variable reference frames the mass of the object is variable.

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My invisible elf ??? Why he is made of dark matter and lives off of dark energy !!!
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If it says that, it's totally wrong again-- rest mass does not change with acceleration, by definition.

That's true, if one means "relativistic mass", which is frame dependent. Usually, mass just means rest mass, by convention.

Okay, in the particle accelerator I see that the particle has gained mass and so it takes more energy to get a similar additional increase in velocity (which could, technically, be a mass measurement).

To calculate the rest mass of the particle (if I was not present at the start of the experiment) I would just subtract the apparent kinetic energy of the particle from the relativistic mass to receive the invariant rest mass?

If I was riding the test particle, yeehaw!!!, I would see the earth zipping along at a relativistic speed and so to calculate the rest mass of the earth I would have to calculate the kinetic energy of the earth and subtract that from the relativistic mass?

What I am trying to figure out is how to know the rest mass of a moving object - which I obviously can not stop and measure in my earth perspective - if what I am measuring is the relativistic mass.

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You already answered this yourself — you account for the kinetic energy, and the rest energy is what's left over.

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That is what I thought but in my example the earth was never accelerated and so subtracting the "apparent" kinetic energy of the earth would yield a value for the earth's mass that is incorrect.

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If you were riding the particle, you certainly would see the Earth accelerate. Otherwise, how would it go from not moving at all relative to you to moving very fast relative to you?

In general, you can determine what portion of the total energy is kinetic energy and what portion is rest energy from just the velocity (relative to you). If you also know the total energy (relative to you), you can work out the rest energy, and it will always be the same, regardless of how the object is moving relative to you.

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Conserve energy. Commute with the Hamiltonian.

Yes, that's right. The answers by swansont and Grey are also correct.

According to Einstein's relativity either perspective is valid: that of the particle or that of the earth.

Going back to this post:

Here is what I find disturbing about Einstein's relativity and rest mass:

For the sake of this "thought experiment" the accelerator is turned off and now the particle on which I am riding is now traveling with inertial velocity close to the speed of light.

I can see two sides:

A.) The particle on which I am riding is the moving entity or;
B.) The earth is the moving entity.

Examining the situation from perspective "A" I could calculate the correct rest mass for the earth: the measured mass minus the kinetic energy of the earth (kinetic energy which is zero since the earth is stationary from this perspective).

Examining the situation from perspective "B" I would calculate an incorrect rest mass for the earth: the measured mass minus the kinetic energy of the earth.

The result of this thought experiment is inconsistent with the stated definition of rest mass: "The rest mass or invariant mass is an observer-independent quantity."

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It seems like there are always two perspectives from which to view events but only one yields the correct interpretation.

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I think what is confusing you is that the kinetic energy must be consistent with the reference frame. Remember, these quantities are measureable or inferrable from measurements, you can't just "decide" that the Earth is stationary like it was a philosophical choice. For the Earth to have zero kinetic energy, it must look stationary, which it won't if you are riding the particle. No matter how you do it, if you observe both the relativistic mass and the kinetic energy of the Earth, and subtract, you'll get the same result. Relativity is a theory that applies to measurements, not arbitrary designations.

Okay let me restate the problem. With the particle accelerator we can "measure" the mass of the particle by attempting to accelerate it with a set amount of force ... and the same would apply to the particle's perspective.

So if the particle attempts to accelerate the earth the result will cause an equal but opposite reaction (as would the other perspective) and so the particle velocity would be affected from which the particle could infer a mass ratio which would, of course, be correlated with the particle's own perceived rest mass.

In our thought experiment the particle would receive the bigger acceleration velocity, since it is the least massive body, but could the particle consider the velocity imparted to be that of the earth's? - the only way the particle would "know" that it is itself being accelerated is by the "feel" of the force of acceleration of the attempt at massing the earth.

So the determination requires a perception of accelerations in order to make the correct assessment.

Since the earth is so much more massive than the particle we "feel" no acceleration when we attempt to accelerate the particle - but, in reality, the earth is oppositely accelerated, albeit, minutely.

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My invisible elf ??? Why he is made of dark matter and lives off of dark energy !!!
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Well, from the particle's perspective, the amount of energy going into the Earth is vastly more than what the particle gets from the Earth's perspective. There is not "reciprocity" in the amount of energy, if that is what is confusing you.

Equal and opposite reactions apply to real forces. If you enter an accelerating frame, you encounter ficticious forces that have no equal and opposite reaction.

You only need to "perceive" acceleration if you wish to attribute the cause to a ficticious force. If you just accept the existence of the ficticious force, as you do with gravity, then you do not need to perceive any acceleration for the particle.