"Mass is that portion of energy which cannot be transformed away in any reference frame"

In this thread (post #18), Chronos said:

I have never heard ‘mass’ described/defined so succinctly and clearly. I sense that an epiphany could be coming my way and so would love to hear more along those lines.

Indeed, I understand that to delve into it in detail would take a textbook sized post. But, can somebody (?Chronos) elaborate even just a bit on that statement; to unpack it so to speak?

Many thanks!

Probably not quite what you’re looking for, but a similar statement is:
Heat is that portion of kinetic energy which cannot be transformed away in any reference frame

(a hot object is of course more massive than an otherwise-identical cold object)

That is indeed a provocative statement.

I believe what Chronos was referring to is that, since I had brought up the subject of different reference frames in the context of my proposed relationship between mass and gravitational potential energy, he was asserting that mass cannot be transformed away. Basically, if I understand his statement properly, he was claiming that I was completely wrong.

I had basically said that if a piece of mass enters into a gravitational well, gravity acts on it and it releases energy as a result (while it is falling), then some of the mass would be lost to account for the appearance of energy. I had also stated that this might only be true from an outside frame of reference (i.e. if you were to somehow measure this mass from a stable point further away from the gravitational well).

Of course, we all know that there are situations where mass can be transformed away to energy (matter + antimatter) but there clearly does not appear to be anywhere near enough antimatter in the universe to be able to annihilate with all the matter, so apparently it does not seem possible to transform the majority of the rest mass that exists in the universe into energy (the baryogenesis problem). This of course poses some fundamental philosophical questions since Einstein’s equations treat matter as equivalent to energy. Are the two really equivalent if there appears no obvious mechanism to inter-convert them?

“Epiphany may be coming into view, men. Stay frosty.”

Brief version:
The amount of energy a system has depends on the reference frame you view it from. If you pick a reference frame that is moving at high speed relative to a system, then you’ll judge the system as having fairly high energy, with much of that energy being kinetic energy (since you and the system have a high relative velocity). There is one reference frame in which the total energy of the system is the smallest possible from your point of view. You can’t make the energy you measure for the system any smaller by "transforming"sup[/sup] to another referencing frame. And, in that special reference frame, the energy you measure for the system is the mass of the system (E=mc[sup]2[/sup]). And, that special frame is the center of momentum frame of the system, where the (vector) sum of all the momenta of all the constituent parts is zero.

This is all true, but it is a consequence of a more general set of statements.

sup[/sup] “Transforming” here means “converting measured (or measurable) quantities (like energy) into what they would be if they were measured by an observer in a different reference frame (i.e., moving with a different velocity)”.

More details:
First, an analogy. Consider a long, thin rod. Let’s say one end of the rod is at location (0,0,0) in 3D space and the other end is at some (x,y,z) coordinate. If we calculate sqrt(x[sup]2[/sup]+y[sup]2[/sup]+z[sup]2[/sup]), we’ll get the length of the rod, thanks to Pythagoras.

If we rotate the rod around in space, keeping one end pinned at (0,0,0), we’ll end up with a different set of values for the other end’s (x,y,z) position. We can even tag the variables with “prime” symbols – i.e., (x’, y’, z’) – to remind us that we’ve rotated the stick around so we shouldn’t expect x and x’ (or y and y’…) to be the same numerical values. But we can still take sqrt(x’[sup]2[/sup]+y’[sup]2[/sup]+z’[sup]2[/sup]) to get the length of the rod.

What we have is a vector – the (x,y,z) position of the end of the rod – where the individual numerical values in the vector depend on how we’ve rotated the rod in space, yet regardless of the rotation we can still calculate the length of the rod through a fixed calculation.

Instead of actually rotating the rod we can rotate ourselves, rotating our chosen coordinate system with us. The change from (x,y,z) to (x’,y’,z’) is a transformation due to our picking a different reference frame to view the rod from. But the length of the rod is the same regardless of our choice of reference frame, and we can calculate it using the same calculation in any reference frame, always getting the same answer. The jargon of all this: the length of the rod is an “invariant” with respect to this particular transformation.

In relativity the transformations of interest are not rotations but rather changes of velocity. And the special vectors of interest have four components, not three. One such special vector is the “energy-momentum” four-vector of an object (we’ll generalize to multiple objects shortly). This four-vector’s components are (E, p[sub]x[/sub], p[sub]y[/sub], p[sub]z[/sub]). The first component is the energy of the object, and the next three components are the three components of momentum.

These are all measurable quantities, and you can write them down into this four-vector notation. The invariant quantity you can calculate with such four-vectors is analogous to a length. For the energy-momentum four-vector the quantity sqrt(E[sup]2[/sup] - p[sub]x[/sub][sup]2[/sup] - p[sub]y[/sub][sup]2[/sup] - p[sub]z[/sub][sup]2[/sup]) is an invariant. Note the minus signs, making this thing not actually a Euclidean length.

That invariant quantity has the same value regardless of the reference frame you measure (E, p[sub]x[/sub], p[sub]y[/sub], p[sub]z[/sub]) in. Changing reference frames means changing velocity (by definition), so the energy and the momenta all change when you change reference frames. But you will always get the same answer for sqrt(E[sup]2[/sup] - p[sub]x[/sub][sup]2[/sup] - p[sub]y[/sub][sup]2[/sup] - p[sub]z[/sub][sup]2[/sup]).

And the answer you always get is the mass of the object.

This is a good time to remind you that energy in relativity counts both rest mass and kinetic energy. If we pick a reference frame where our object is at rest, the the momentum parts of our four-vector are zero and the energy is just the rest mass. In any other reference frame, the energy would be higher.

So, the mass of an object is indeed the energy it has when viewed by a (relatively) stationary observer. But more generally, the mass of an object is also the invariant quantity calculable from the energy-momentum four-vector of an object as measured in any reference frame at all.

Multiple objects:
If our object splits into two or more objects, the total energy and total momentum of the system doesn’t change. Both energy and momentum are conserved in all processes, so no matter the process that busted the object apart, the total energy and total momentum will be unchanged. Thus, you can talk about the mass of the whole system as being the same invariant sqrt() quantity, where E=(total E of all the pieces), p[sub]x[/sub]=(total p[sub]x[/sub] of all the pieces), etc. And this mass will be the same as the mass of the original object. And it will be the mass of the system forever, as long as you don’t lose track of part of it.

Semantic caveat:
I like to point out in these threads that “mass” has multiple practical definitions used in science and even in physics specifically. If anyone tries to slap you down for saying “mass is converted into energy” or “the mass changed”, send them my way. Like with most English words, the meaning of “mass” is context dependent and it isn’t always the strict relativistic version given above.

“Transformed away” is a poor choice of words given the context of the thread. If you are talking about the mass of the system as a whole, then matter+antimatter leads to a system that has the same mass that it started with. If an electron and a positron annihilate to two photons, the total mass of the two-photon system is the same as the total mass of the electron/positron system. The constituent parts of the final system (the individual photons) are each massless, however.

It’s a minefield if one doesn’t make clear what parts of the system are being considered. To be sure: the gravitational pull of an electron/positron pair is the same as the gravitational pull of the two photons that result from their annihilation.

(Aside: the context in which mass is most often talked about as being “lost” is nuclear processes where energy or particles are streaming out never to be considered again. “Transformed away” is not standard usage for this, though.)

Hey, Pasta! Thanks for posting!

Just to add a little to what Pasta said:

As long as m, E are real and p has real components then:

m = sqrt(E[sup]2[/sup] - p[sub]x[/sub][sup]2[/sup] - p[sub]y[/sub][sup]2[/sup] - p[sub]z[/sub][sup]2[/sup])

implies that

m ≤ E

and for system of n objects of masses m[sub]1[/sub], m[sub]2[/sub],…,m[sub]n[/sub], the total mass m[sub]TOT[/sub] derived from the sum of the energy and the sum of the momentum:

m[sub]TOT[/sub] ≥ m[sub]1[/sub] + m[sub]2[/sub] + … + m[sub]n[/sub]

Thanks All!
It is a good day-I learned something today. At least, I understand something more clearly now. :slight_smile:

It may be easier to think in terms of an example:

Suppose you have a free electron at rest (in a specific reference frame). It has mass m[sub]0[/sub], and no kinetic energy, so the total energy it has is m[sub]0[/sub]c[sup]2[/sup].

Now suppose you have an electron moving at 0.5c in some direction (or alternatively, you are looking at the orginal electron in a reference frame moving at 0.5c in the opposite direction). Now the electron has a lot more energy, but not all of it is mass.

To determine which portion of the energy is the mass of the electron, note that there is a reference frame moving along with the electron (the electron’s rest frame) and in that reference frame, the electron has no kinetic energy, so the total energy of the electron is m[sub]0[/sub]c[sup]2[/sup]. In this frame the electron has less energy than it does in any other reference frame, so my Chronos’ definition, that energy is the mass of the electron – the only portion of the energy that we can’t get rid of by picking another reference frame.

Now suppose there are two electrons, e[sub]1[/sub] and e[sub]2[/sub] both moving an 0.5c but in opposite directions. We want to ask, “what is the mass of both electrons together?”.

Now both electrons have some kinetic energy, but since they are moving relative to each other there is no single reference frame that gives them zero kinetic energy. If we pick a frame in which e[sub]1[/sub] is at rest, e[sub]2[/sub] has a lot of kinetic energy and vice versa. The frame in which the total energy of the system is minimized is the one we are currently in where we see the electrons as having equal speed in opposite directions.

In particular this means that the mass of the two electron system is bigger than the sum of the masses of the individual electrons. This is because some of the kinetic energy that we can transform away if we are just looking at one of the electrons, we can no longer transform away if we are looking at both of them. I.e. some of the energy that was kinetic energy when looking at the individual electrons becomes mass, just by considering both electrons together.

Outstanding! Thank you for your answers. The meaning of the term ‘invariant mass’ has never been as clear to me (whether I appreciate* it as much tomorrow, remains to be seen).

*as in grok it

A couple of nitpicks to Pasta’s excellent answer:

1: There isn’t always a reference frame where the energy is at a minimum. If, for instance, your system of interest is a single photon, for any given reference frame, you can always find some other reference frame where the energy is lower. You can, in fact, get the energy to be any nonzero value you like, no matter how small, but you can’t get it completely to zero. This is what we mean when we say that a photon has no mass.

2: It’s not quite correct to say that the transformations of interest in relativity are not rotations. Rotations are a subset of the transformations of interest in relativity (collectively, these transformations form what’s called the Lorentz group). They’re just not the entire set of transformations any more.

Two other things that should be emphasized: First, everyone talks about how various quantities change in relativity, when you’re traveling at high speed. In practice, however, physicists doing relativity hardly ever work with those quantities. It’s much easier to instead work with what are called invariants, the quantities which are the same in all reference frames. Mass is one of these. And it turns out that almost all calculations of interest can in fact be done entirely in terms of invariants.

Second, mass is not additive. If I have one thing with one mass, and another thing with another mass, and put them together into a larger system, the mass of the larger system will not be equal to the sums of the masses of the two components. This is counterintuitive, because in most situations humans are familiar with, it’s almost equal, such that the difference is extremely difficult to detect. But once you get to velocities comparable to the speed of light, this breaks down. Hence, for instance, you can have that pair of photons with mass, even though neither individual photon does.

What is this “reference frame”?

For the record, I concur with these nitpicks. (I thought about pre-picking them both, but I figured it would obfuscate more than help. Thanks for adding the information.)

“Mission assessment. Has epiphany by means of grokking been determined?”