E=mc^2 and rest mass?

[ratchetmouth]

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.

[Ken G]

Hello all. I`m new here, and still very much a novice in physics.

Welcome to the forum!

I was reading the "Does Light Have Mass?" thread, and saw that in the E=mc^2 equation, the "m" refers to rest mass.

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

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?"

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.

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." .

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.

[Squashed]

The important issue is not why it's moving, but how it's moving, relative to the observer.

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?

[Matthew]

Particles in accelerators appear to gain mass because of their acceleration

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

[Squashed]

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: http://www.bautforum.com/showthread.php?t=41631&page=3

[Ken G]

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

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.

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?

Mass-energy, in the sense you are using it, is also frame dependent.

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?

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.

[Ken G]

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 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:

"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.... 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)."

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.

[Squashed]

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.

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.

[Ken G]

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.

If it says that, it's totally wrong again-- rest mass does not change with acceleration, by definition.

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.

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

[Squashed]

...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.

[swansont]

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

[Squashed]

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

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.

[Grey]

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.

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.

[Ken G]

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?
.

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

[Squashed]

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.

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

Going back to this post:

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?

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."

- - - - - - - - - - - -

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

[Ken G]

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.

[Squashed]

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.

[Ken G]

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.

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.

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.

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.

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

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.

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.

That much is completely true, but yields no contradiction.

[Squashed]

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...

Okay, if the accelerations are fictious, according to relativity, then accelerated frames are applied equally to both reference frames: the particle and the earth.

For the particle, since it considers itself at rest and therefore "feels" the fictious force, the earth, which is in freefall motion with respect to the fictious force and so does not "feel" the force, is also under the influence of the fictious force and results in the earth's motion - correct? (I think Grey alluded to this in his post here.)

Note to readers: To properly parse the above sentence remove the discriptive parenthetical segments to read the actual sentence: "For the particle the earth is also under the influence of the fictious force and results in the earth's motion - correct?"

So from the particle's reference frame the particle can measure the fictious acceleration force and then measure the change in velocity (acceleration) of the earth and then calculate the apparent relativistic mass and then subtract the apparent kinetic energy to calculate the rest mass.

Therefore; to properly apply relativistic accelerated reference frames both the subject and the experimenter are considered in the same accelerated reference frame but under different circumstances: one at rest and one in freefall; and so one "feels" the force and the other does not "feel" the force (I can be so dense sometimes that I wonder why I do not implode into my own blackhole).

[Ken G]

For the particle the earth is also under the influence of the fictious force and results in the earth's motion - correct?

Yes, I'd say this and the other things you said are in line with my own limited understanding of doing physics in accelerated frames. I think you pretty much have to be a GR expert to be comfortable with translating GR into meaningful descriptions of what is happening in accelerated frames, however! It's much easier in classical mechanics, ironically, even though classical mechanics basically happens at a level inside the coordinate system, while general relativity happens in a coordinate-independent way (which is why it is both more "real" and more profound).

[publius]

Ken,

I don't want to get another convoluted discussion going again -- makes my head hurt -- but this may be another instance of my confusion between coordinate systems and reference frames.

You say that accelerating frame classical mechanics happens "inside the coordinate system". Consider a simple linear translating coordinate system accelerating with respect to some inertial frame at an acceleration 'a'. We can write Newton's Second Law in that frame as simply:

m*dv'/dt = F - m*a

where v' is the velocity in the accelerating coordinate system, and F is the "real force". Noting that F = m*dv/dt where dv/dt is the "real acceleration" in the inertial frame, we can note:

dv'/dt + a = dv/dt = F/m, with the expression on the left being simply the transform of dv/dt into the accelerating coordinate system.

That looks fairly coordinate system independent to me at least, with 'a' and the 'v's being vectors. Now, if we go into a rotating frame, our 'a' would break down into w x v' and w x w x r' terms.

When omega and v are vectors, that is fairly coordinate independent, save for the cross product being a "pseudo vector" (I forget about all that) that would flip between right and left handed coordinate systems. If we switched to a left-handed system, the coriolis term would flip signs. The centrifugal terms might not because of the double cross -- I'd have to think about it carefully and that makes my head hurt.

-Richard

[Ken G]

My point about being inside the coordinate system is in your equation, in the form of the "a". Where did that come from, physically? It came from the coordinate system, only. It is a "ficticious acceleration", which has no place in classical mechanics other than as a coordinate transformation from inertial frames where physics "makes sense" (has causes). But the equations of general relativity don't change form in other reference frames, they are the same equations, with the same meaning, in all reference frames that are achievable by an observer, or all coordinate systems that smoothly connect a bunch of achievable reference frames. (I'm not sure what requirements must be placed on coordinate systems such that they locally conform to reference frames, but the point is, a coordinate system is global and a reference frame is local. The equations of general relativity are also local, so that's why they apply to reference frames, and coordinate systems just appear as mathematical connections.) Maybe what I'm really saying is that classical mechanics counts gravity as a real force, so has to treat ficticious forces in a separate package, whereas general relativity accepts that ficticious forces are part of any description of reality because there is always gravity around. General relativity lives at a deeper level, and ficticious forces appear as a kind of "projection" onto local reference frames.

[Squashed]

Okay, if the accelerations are fictious, according to relativity, then accelerated frames are applied equally to both reference frames: the particle and the earth.

Therefore; to properly apply relativistic accelerated reference frames both the subject and the experimenter are considered in the same accelerated reference frame but under different circumstances: one at rest and one in freefall; and so one "feels" the force and the other does not "feel" the force.

The above sounds logical to me except when I reverse the process and use it from the "earth at rest" reference frame then I am in a quandry again.

For discussion's sake I will specify that the particle is undergoing an acceleration of 500g in the accelerator which works out fine from the particle's perspective because the particle "feels" the 500g force and the earth "appears" to be accelerating in a 500g environment but things fall apart in the earth perspective.

In the earth at rest perspective the earth "feels" only 1g of force but witnesses 500g of acceleration for the particle - so the accelerations do not match for the two perspectives.

I can make sense of the two by resorting to a momentum applied approach: since the particle is a lot less massive than the earth then imparting a set amount of momentum to the particle yields a high acceleration to the particle whereas the same amount of momentum applied to the earth leaves the earth essentially unmoved.

But now the question arises: when do I use acceleration to understand the observation and when do I use momentum to understand the observation?

[Grey]

In the earth at rest perspective the earth "feels" only 1g of force but witnesses 500g of acceleration for the particle - so the accelerations do not match for the two perspectives.

I can make sense of the two by resorting to a momentum applied approach: since the particle is a lot less massive than the earth then imparting a set amount of momentum to the particle yields a high acceleration to the particle whereas the same amount of momentum applied to the earth leaves the earth essentially unmoved.

But now the question arises: when do I use acceleration to understand the observation and when do I use momentum to understand the observation?

No need to use the momentum. Remember that the reason the particle is accelerating is that you're applying an external electromagnetic force. This is a real force, and it's still present in both reference frames. If we're considering the Earth at rest, then it's just an unopposed force and it produces a large acceleration. If we're considering the particle to be stationary, then that large force is opposing the huge gravitational field that suddenly appears, and keeps the particle from falling like everything else in the universe. That makes sense, just like the reason that you're stationary even though Earth is exerting a downward force on you is that the ground is exerting a force of equal magnitude in the opposite direction.

[Ken G]

And to follow that up, where did you get the idea that the Earth would experience "1g of force" if the particle experienced 500g? (And note, these are accelerations, not forces). If you correctly find the Earth acceleration, you will get the same answer as the momentum argument. But Grey's way of looking at it is the real point as far as general relativity goes.

[Squashed]

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.

Remember that the reason the particle is accelerating is that you're applying an external electromagnetic force. This is a real force, and it's still present in both reference frames.

You chaps are starting to confuse me.


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...where did you get the idea that the Earth would experience "1g of force"...

After I posted, I figured the 1g of force would cause confusion but from my earth perspective I only "feel" 1g of force which, by the way, is not a result of the particle accelerator. My point being that from the earth perspective there is no force being felt (although, technically, there is the equal but opposite reactionary force) but an object, the particle, is being accelerated.

It seems easier to just use the set amount of force to justify the two circumstances, that of the particle and that of the earth, as I did in this post#17:

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.

[Ken G]

You chaps are starting to confuse me.

Those statements you quoted are not in contradiction. Classically speaking, there are real forces and ficticious forces, the former have a real source and come in action/reaction pairs (like the EM forces accelerating the particle, which produce an equal and opposite force on the Earth, ultimately, but only a minute acceleration), whereas the latter are only there because your reference frame is accelerating, they have no source, and don't come in action reaction pairs. In either frame, the real force on the Earth has no significant effects. The ficticious force on the Earth only appears in the frame of the particle, and it is a staggeringly huge force, sufficient to accelerate the entire Earth virtually the speed of light. The energy involved is mind boggling, the relativistic mass of the Earth becomes spectacular. All that due to the reference frame, none due to any kind of real source, classically. But in general relativity, it's all just what gravity is doing, and lives at the level of spacetime curvature that underlies the arbitrary choice of reference frame.
That's my limited understanding, at least.

[publius]

Ken,

Just to be stinker because I can't help it, and you'll probably want to shoot me, but the mechanical EM forces will only be equal and opposite if we can ignore the light-time propagation delay. If we can't, they will miss somewhat, and radiate. (actually, I would assume we get significant radiation in particle accelerators when we accelerate things to near light speed in a short time) Of course, when we add the momentum of the radiation, then momentum is conserved, but the forces between the two "mechanical" parts of the system will not add to zero, as they will both be "pushing" against their radiation as well as each other........

It always struck me as sort of funny to speak of "mechanical momentum" with EM, but that's the way it is usually described. The vector point-function description (which comes from the Lorentz force expression) of the EM mechanical mometum is

dP_m/dt = rho*E + J x B

Where P_m is the "volume density" of mechanical momentum. IOW, this is the rate at which mechanical momentum (ie of moving charges) is being transferred by the fields.

Now, the fields themselves can "absord or release" momentum, and the momentum density of the field itself is D x B.

Add those together and that is the total rate at which momentum is flowing "at a point". Integrate that vector over a volume and that's the force acting on that volume. And you can pull a 2nd rank tensor expression out of that which can be integrated over the surface bounding that volume to get the force as well.

That is the Maxwell Stress Tensor, the most general and elegant (but very complex) way to calculate EM forces.

The 4-vector form of the Stress Tensor is a Stress-energy tensor, with the other 4 being the 4-Poynting vector. That is what you would plug in the Einstein Field equation to get the EM contribution to the metric equation.

Sorry for rambling on, but I just love the way Classical EM works. It's like poetry........

-Richard

PS: If any students of classical EM are lurking, 50 points to the one who can derive the form of the Maxwell Stress Tensor from the above mechanical momentum expression:

rho*E + J x B. Start out with the Lorentz force expression, F = q(E + v x B) for a single charged particle. Now, go to a 3D volume description, and get a "force per volume" expression, and you'll get (rho*E + J x B) as being integrated over a volume. Make the argument that is the point function form of the mechanical momentum density by F = dP/dt.

Now, plug in Maxwell into this. (ie rho = div D; J = curl H -dD/dt). Have some vector calculus fun. You'll find that "D x B" just falls right out of that with a time derivative in front, and if you move it over to the side with the above, you argue that has to be the total momentum, and so D x B has to be the momentum density of the field itself.

Now, you can take what's left on the other side and show it is a rank two tensor........

[publius]

Squashed,

Accelerating reference frames can get confusing. First, there is the concept of the "fictitous forces" in those frames. These "forces" are just there to make Newton's second law "work right" in that accelerating frame.

Consider a bunch of billiard balls floating in free space, all at rest with respect to each other. We define our inertial reference frame as this frame. Now imagine an observer who starts accelerating with respect to this frame. From his POV, his frame, those billiard balls are all now accelerating away from him at the same rate (note that magic phrase "accelerating at the same rate"). Now, he is the one who is "really accelerating" (at least in classical mechanics, as he is the one who "feels the force").

Now, to make F = ma work in that accelerating frame, we can introduce a force, proportional to mass, to "explain" why all those balls are accelerating away from us. If the observer is accelerating at a rate 'g', then every object experiences a fictitious force of "mg".

Now, if we apply a *real force* of mg in the opposite direction to one of those billiard balls, then that cancels out, and the object stops accelerating in our frame.

But in the inertial frame, of course, that billiard ball is now really accelerating at a rate 'g', coaccelerating with the accelerating observer.


So far so good. But real forces have to come in pairs; to accelerate something, we have to push against something else, throwing mass/energy off. In the inertial frame, say two of the billiard balls push against each to accelerate one at rate 'g'. The other billiard ball will accelerate in the opposite direction.

From the POV of the accelerating observer, that second billiard will accelerate at greater than 'g'. The real force adds to the fictitious force in his frame.

-Richard

[Ken G]

Just to be stinker because I can't help it, and you'll probably want to shoot me, but the mechanical EM forces will only be equal and opposite if we can ignore the light-time propagation delay. If we can't, they will miss somewhat, and radiate.

You're right-- we must look at all the forces to get the actual acceleration that occurs. We can still identify the part of the real force that is coming from things attached to the Earth and get the action/reaction on the Earth, but we also have to include the force of interaction with the radiation to get the full acceleration (and that second force is also real and also comes in action/reaction pairs).