Is Relativistic Mass Still Relevant in Modern Physics Discussions?

[JesseM]

No, the m in that equation is rest mass.

From this page

The rest mass (m) of a particle is the mass defined by the energy of the isolated (free) particle at rest, divided by c 2 . When particle physicists use the word ``mass,'' they always mean the ``rest mass'' (m) of the object in question

Yes, that's exactly what I just said, m is used to mean rest mass (although again, for a compound object the rest mass is defined to be the inertial mass in the compound object's rest frame, which is equal to its total energy divided by c^2 in that frame).

 

[Aer]

Yup, and of course this fits with my claim that the inertial mass of the deuteron is equal to its total energy divided by c^2, but it doesn't fit too well with your claim that its inertial mass is dependent only on the sum of the rest masses of its parts, and not on any other forms of energy that may be in the deuteron.

Why isn't this binding energy apart of the rest mass of the system? (I think your answer is rest mass is different from inertial mass - then what is the difference, explicitly)

 

[Aer]

Yes, that's exactly what I just said, m is used to mean rest mass (although again, for a compound object the rest mass is defined to be the inertial mass in the compound object's rest frame, which is equal to its total energy divided by c^2 in that frame).

But by that definition, you have to separate each particle separately and measure it's -rest mass-

What part of isolated particle is not clear?

 

[JesseM]

From this page
the energy that holds a nucleus together; the difference between the sum of the masses of the individual nucleons and the actual mass of the nucleus.

So is the energy not a part of the combined system?

Of course it's part of the combined system, but the energy doesn't have any rest mass of its own. So this contradicts your claim that the rest mass of the combined system is just the sum of the rest masses of its parts, and supports my claim that the rest mass of the combined system is the total energy of the system divided by c^2.

 

[Aer]

Of course it's part of the combined system, but the energy doesn't have any rest mass of its own. So this contradicts your claim that the rest mass of the combined system is just the sum of the rest masses of its parts, and supports my claim that the rest mass of the combined system is the total energy of the system divided by c^2.

It doesn't support your claim! BTW - I never made any claim regarding combining masses in chemistry!

If it supported your claim, then the mass of the deutron should be LARGER than the mass of the proton and neutron combined. Because it not only contains your proton and neutron, but also binding energy - which adds to the total energy.

 

[JesseM]

Why isn't this binding energy apart of the rest mass of the system? (I think your answer is rest mass is different from inertial mass - then what is the difference, explicitly)

It is! But that's assuming you use my definition that the "rest mass" of a composite system is the total energy in the system's rest frame divided by c^2. If you want to define the rest mass of a composite system as just the sum of the rest masses of each of its parts, then instead of looking at each components energy/c^2 from the point of view of the combined system's rest frame, you have to consider each part's energy/c^2 in that part's own rest frame (ie each part's rest mass), ignoring the rest of the system. But what is the rest mass of the binding energy on its own, ignoring all the particles? That doesn't seem to make any sense, binding energy is just a difference in potential energies, how can a difference in potential energies have a rest frame?

 

[JesseM]

Yes, that's exactly what I just said, m is used to mean rest mass (although again, for a compound object the rest mass is defined to be the inertial mass in the compound object's rest frame, which is equal to its total energy divided by c^2 in that frame).

But by that definition, you have to separate each particle separately and measure it's -rest mass-

It's your definition that the rest mass of a compound system is the sum of the rest mass of it's parts, which means you have to measure each part's rest mass separately. My definition is that the compound system's rest mass is the total energy divided by c^2, and I claim that the theory of relativity predicts this is equal to its inertial mass, which you can measure just by looking at the system's resistance to acceleration in its own rest frame.

 

[Aer]

If you want to define the rest mass of a composite system as just the sum of the rest masses of each of its parts, then instead of looking at each components energy/c^2 from the point of view of the combined system's rest frame, you have to consider each part's energy/c^2 in that part's own rest frame (ie each part's rest mass), ignoring the rest of the system. But what is the rest mass of the binding energy on its own, ignoring all the particles? That doesn't seem to make any sense, binding energy is just a difference in potential energies, how can a difference in potential energies have a rest frame?

I only said this about adding together particles that were separated but contained (as in a box), NOT regarding binding particles together - which LOSE mass.

 

[Aer]

It's your definition that the rest mass of a compound system is the sum of the rest mass of it's parts, which means you have to measure each part's rest mass separately. My definition is that the compound system's rest mass is the total energy divided by c^2, and I claim that the theory of relativity predicts this is equal to its inertial mass, which you can measure just by looking at the system's resistance to acceleration in its own rest frame.

Doesn't fit too well with your binding energy argument.

 

[Aer]

It doesn't support your claim! BTW - I never made any claim regarding combining masses in chemistry!

If it supported your claim, then the mass of the deutron should be LARGER than the mass of the proton and neutron combined. Because it not only contains your proton and neutron, but also binding energy - which adds to the total energy.

Which goes back to what I said, the binding energy comes from the proton and neutron's rest mass but is still apart of the system.

 

[JesseM]

If it supported your claim, then the mass of the deutron should be LARGER than the mass of the proton and neutron combined. Because it not only contains your proton and neutron, but also binding energy - which adds to the total energy.

No, the potential energy is greater when you pull the proton and neutron apart than when they are bound together--that's why they naturally tend to stick together! As it says in the wikipedia entry on binding energy, "A bound system has a lower potential energy than its constituent parts; this is what keeps the system together; it corresponds to a positive binding energy." In other words, binding energy is defined in a funny way, so that more positive binding energy is equivalent to less potential energy, and it's potential energy that you must use when calculating the total energy of different states.

 

[Aer]

No, the potential energy is greater when you pull the proton and neutron apart than when they are bound together--that's why they naturally tend to stick together! As it says in the wikipedia entry on binding energy, "A bound system has a lower potential energy than its constituent parts; this is what keeps the system together; it corresponds to a positive binding energy." In other words, binding energy is defined in a funny way, so that more positive binding energy is equivalent to less potential energy, and it's potential energy that you must use when calculating the total energy of different states.

I looked up the definition of binding energy, got:
"the energy that holds a nucleus together; the difference between the sum of the masses of the individual nucleons and the actual mass of the nucleus."

is this definition wrong?

 

[Aer]

Is the binding energy apart of the deutron or not?

 

[JesseM]

I looked up the definition of binding energy, got:
"the energy that holds a nucleus together; the difference between the sum of the masses of the individual nucleons and the actual mass of the nucleus."

is this definition wrong?

It's not wrong, but it could be misleading if you interpreted it to mean that there was some form of energy that increased in the bound state rather than decreased. It is the decrease in potential energy that holds a bound system together, and the fact that you have to climb a potential hill to separate the parts that makes it difficult to do so--do you disagree with this?

 

[Aer]

It's not wrong, but it could be misleading if you interpreted it to mean that there was some form of energy that increased in the bound state rather than decreased. It is the decrease in potential energy that holds a bound system together, and the fact that you have to climb a potential hill to separate the parts that makes it difficult to do so--do you disagree with this?

All this says is that the proton and neutron lose rest mass when they are bound together. This rest mass is referred to as the potential energy since all mass is essentially a form of energy. But this doesn't imply that kinetic energy or gravitational potential energy will become the potential energy that is considered mass. At least, there is nothing to assume that. What is needed is experimental evidence, not this endlessly pointless discussion.

 

[Aer]

Can we at least agree to disagree for now?

 

[Aer]

I guess at least I'll have to agree to disagree - going to bed, goodnight!

 

[JesseM]

Is the binding energy apart of the deutron or not?

I'm not sure what you mean by "a part of it". The potential energy of the bound and unbound state (which the binding energy is based on) must be taken into account when calculating the total energy of each state, but then the kinetic energy of each particle must be taken into account as well. The inertial mass of a compound object is proportional to the sum of potential, kinetic and rest mass energies of all its parts.

 

[JesseM]

All this says is that the proton and neutron lose rest mass when they are bound together.

No they don't! The rest mass of a proton in a deuteron nucleus is the same as the rest mass of a free proton, the rest mass of a given type of particle never changes, it's a constant of nature.

This rest mass is referred to as the potential energy since all mass is essentially a form of energy.

Not by any physicists, no.

But this doesn't imply that kinetic energy or gravitational potential energy will become the potential energy that is considered mass.

So electromagnetic potential energy (responsible for chemical binding between atoms) and strong-force potential energy (responsible for nuclear binding between protons and neutrons) can contribute to inertial mass, but somehow gravitational potential energy can't? And again, there is no existing theory of physics that explains changing potential energy between particles in terms of the particle's rest mass changing when the distance between them changes, you're just making stuff up off the top of your head now.

 

[JesseM]

I guess at least I'll have to agree to disagree - going to bed, goodnight!

OK, goodnight...

 

[Aer]

No they don't! The rest mass of a proton in a deuteron nucleus is the same as the rest mass of a free proton, the rest mass of a given type of particle never changes, it's a constant of nature. Not by any physicists, no. So electromagnetic potential energy (responsible for chemical binding between atoms) and strong-force potential energy (responsible for nuclear binding between protons and neutrons) can contribute to inertial mass, but somehow gravitational potential energy can't? And again, there is no existing theory of physics that explains changing potential energy between particles in terms of the particle's rest mass changing when the distance between them changes, you're just making stuff up off the top of your head now.

I'm making things up? I don't think so, the least I've done is inquire. You are the one making things up saying that kinetic and gravitational potential energy can be considered the same as the energy form of mass, consider this answer from Dr. Ken Mellendorf"

The mass of an atom is not the sum of the
masses of its individual parts. The mass of an atom is in fact less than
the mass of its parts.
The mass of an atom is the sum of the masses of its parts, minus (binding
energy)/c^2. Each proton and each neutron still have their original masses.
The loss of energy to the outside world results in a decrease of atomic
mass. At the level of particles and atoms, mass is NOT conserved. After an
event, you may end up with more or less mass than you started with. Total
energy, including E=mc^2, is conserved. Mass behaves like just another location
of energy. A negative potential energy can make the total energy less than
the sum of the other energies. At the atomic level, a negative potential
energy can make the total mass less than the sum of the individual masses.

Dr. Ken Mellendorf
Physics Professor
Illinois Central College

He uses total energy as the rest energy equation, E=mc^2.

Total energy in your situation (taking into account kinetic energies) is E=γ mc^2

 

[JesseM]

I'm making things up? I don't think so, the least I've done is inquire.

You're making stuff up when you say changes in potential energy are "really" changes in the rest masses of the particles (except in the case of gravitational potential, for some reason). There is no theory of physics that says this.

You are the one making things up saying that kinetic and gravitational potential energy can be considered the same as the energy form of mass

What does "the energy form of mass" mean? Do you mean inertial mass?

consider this answer from Dr. Ken Mellendorf"

The mass of an atom is not the sum of the
masses of its individual parts. The mass of an atom is in fact less than
the mass of its parts.
The mass of an atom is the sum of the masses of its parts, minus (binding
energy)/c^2. Each proton and each neutron still have their original masses.
The loss of energy to the outside world results in a decrease of atomic
mass. At the level of particles and atoms, mass is NOT conserved. After an
event, you may end up with more or less mass than you started with. Total
energy, including E=mc^2, is conserved. Mass behaves like just another location
of energy. A negative potential energy can make the total energy less than
the sum of the other energies. At the atomic level, a negative potential
energy can make the total mass less than the sum of the individual masses.

Dr. Ken Mellendorf
Physics Professor
Illinois Central College

He uses total energy as the rest energy equation, E=mc^2.

That sentence is ambiguous--when he says that total energy including E=mc^2 is conserved that could mean that "total energy" includes other things beyond E=mc^2 for each part--for example, the potential energy. Or, the "m" there may refer to the rest mass of the whole system, and as I've been saying, the rest mass of a composite system is defined to be equal to the total energy (which includes potential energy) divided by c^2. He also says that "the mass of an atom is in fact less than the mass of its parts", because you have to include the potential energy to get the total mass, and as he says, the potential energy is negative in the bound state (when compared to the unbound state). This is exactly what I've been saying! And it contradicts your claim that the mass of the atom is still equal to the sum of the mass of its parts, but that the mass of the proton and neutron have somehow decreased.

 

[Ich]

I don´t want to interfere, but I want to comment Mellendorf´s sentence "At the level of particles and atoms, mass is NOT conserved."
Mass is always conserved, as is Energy. After an an event (like n+p -> np + hf) the mass of the system still is the same. The photon contributes to the mass of the system, even though it has no mass itself.
Mass will change only when you change the system you´re considering, e.g. by neglecting the photon in the example.

 

[Aer]

He also says that "the mass of an atom is in fact less than the mass of its parts", because you have to include the potential energy to get the total mass, and as he says, the potential energy is negative in the bound state (when compared to the unbound state). This is exactly what I've been saying! And it contradicts your claim that the mass of the atom is still equal to the sum of the mass of its parts, but that the mass of the proton and neutron have somehow decreased.

That is not my claim when dealing with masses at the quantum level! If I said anything similar to that, it was because you were confusing the issue of whether we are talking about the quantum level or macroscopic level.

Just to be clear - this example is on the quantum level, in which energy and mass -do- lose distinction. Taking this to the next level - that is, putting macroscropic objects in a box with relative velocity to the box and claiming the kinetic energy -adds- to the mass at the macroscopic level, just like a negative energy -subtracts- from the mass at the microscoptic level is not sufficient.

You must show that this kinetic energy -adds- to the mass at the macroscopic level and not just state it to be so. THIS, and only this is the only point I am contending. Whether you believe physics is the same at the microscopic level and the macroscopic level is your prerogative. However - I know there is a difference as there is a thing called quantum physics! So unless you are willing to talk about your macroscopic level example, then you'll have to excuse me if I do not respond to your BS!

 

[learningphysics]

Just to be clear - this example is on the quantum level, in which energy and mass -do- lose distinction. Taking this to the next level - that is, putting macroscropic objects in a box with relative velocity to the box and claiming the kinetic energy -adds- to the mass at the macroscopic level, just like a negative energy -subtracts- from the mass at the microscoptic level is not sufficient.

The special theory of relativity predicts a change in mass whether or not the changes are quantum or macroscopic. Read Einstein's paper on mass-energy equivalence here:

http://www.ams.org/bull/2000-37-01/S0273-0979-99-00805-8/S0273-0979-99-00805-8.pdf

Note: "(6) The rest-energy changes, therefore, in an inelastic collision (additively) like the
mass. "

By mass, Einstein's referring to rest-mass.

His example uses a simple inelastic collision of two bodies. The lost kinetic energy goes into the rest energy of the two bodies and therefore their rest masses... he says nothing about the form of the energy... it could be heat or it could be nuclear binding energy... whatever. The case is general for any inelastic collision.

That's what the theory predicts. If two identical macroscopic baseballs collided in a symmetric inelastic collision losing some of their kinetic energy to heat, then each baseball would increase its rest energy, and therefore change its "rest mass". The increased "rest mass" is due to heat (which is the kinetic energy of the constituent particles that form the baseball).

I got this quote of Einstein's from this website:
http://www.cox-internet.com/hermital/book/holoprt3-1.htm

"In his 1938 book, The Evolution of Physics, 1 Einstein writes:

Energy, at any rate kinetic energy, resists motion in the same way as ponderable masses. Is this also true of all kinds of energy?
The theory of [special] relativity deduces, from its fundamental assumption, a clear and convincing answer to this question, an answer again of a quantitative character: all energy resists change of motion; all energy behaves like matter; a piece of iron weighs more when red-hot than when cool; radiation traveling through space and emitted from the sun contains energy and therefore has mass, the sun and all radiating stars lose mass by emitting radiation. This conclusion, quite general in character, is an important achievement of the theory of relativity and fits all facts upon which it has been tested.
Classical physics introduced two substances: matter and energy. The first had weight, but the second was weightless. In classical physics we had two conservation laws: one for matter, the other for energy.7 "

I cannot verify the accuracy of the quote as I don't have this book.

Note what he says... all energy resists change in motion. Therefore all energy has inertia... heat, kinetic energy, potential energy etc...

Also note that he says that a piece of iron weighs more red-hot... No nuclear changes need be involved.

 

[Aer]

Note: "(6) The rest-energy changes, therefore, in an inelastic collision (additively) like the
mass. "

By mass, Einstein's referring to rest-mass.

His example uses a simple inelastic collision of two bodies. The lost kinetic energy goes into the rest energy of the two bodies and therefore their rest masses... he says nothing about the form of the energy... it could be heat or it could be nuclear binding energy... whatever. The case is general for any inelastic collision.

You may have wanted to give the entire quote:

"(6) E0_bar - E0 = m_bar - m:
The rest-energy changes, therefore, in an inelastic collision (additively) like the
mass. As the former, from the nature of the concept, is determined only to within
an additive constant, one can stipulate that E0 should vanish together with m.
Then we have simply
E0 = m;"

That's what the theory predicts. If two identical macroscopic baseballs collided in a symmetric inelastic collision losing some of their kinetic energy to heat, then each baseball would increase its rest energy, and therefore change its "rest mass". The increased "rest mass" is due to heat (which is the kinetic energy of the constituent particles that form the baseball).

I would love to see you try to make two baseballs collide to become "one" - what you are referring to only happens on the quantum level, not the macroscopic level. All the kinetic energy will be given off as energy in another form in actuality.

I cannot verify the accuracy of the quote as I don't have this book.

You might want to verify it as it is in direct contradiction to the quote by Albert I gave.

 

[jtbell]

I don´t want to interfere, but I want to comment Mellendorf´s sentence "At the level of particles and atoms, mass is NOT conserved."
Mass is always conserved, as is Energy. After an an event (like n+p -> np + hf) the mass of the system still is the same. The photon contributes to the mass of the system, even though it has no mass itself.

Mellendorf's statement would have been phrased better as follows: "At the level of particles and atoms, (invariant) mass is not additive." The (invariant) mass of a system does not equal the sum of the (invariant) masses of the particles that it is composed of."

I put (invariant) in parentheses because many physicists (the ones who don't use the concept of "relativistic mass") would omit it. In this context, since we're discussing both kinds of mass, we need to be explicit about which one we're talking about.

 

[Aer]

Mellendorf's statement would have been phrased better as follows: "At the level of particles and atoms, (invariant) mass is not additive." The (invariant) mass of a system does not equal the sum of the (invariant) masses of the particles that it is composed of."

I put (invariant) in parentheses because many physicists (the ones who don't use the concept of "relativistic mass") would omit it. In this context, since we're discussing both kinds of mass, we need to be explicit about which one we're talking about.

Yes, and it is true only at "the level of particles and atoms" (i.e. quantum physics).

 

[pervect]

Here's another link addressing the topic - I'm not sure whether I've posted it to this particular thread before or not.

http://arxiv.org/abs/gr-qc/9909014

From the abstract

According to the general theory of relativity, kinetic energy contributes
to gravitational mass. Surprisingly, the observational evidence for this
prediction does not seem to be discussed in the literature. I reanalyze
existing experimental data to test the equivalence principle for the
kinetic energy of atomic electrons, and show that fairly strong limits
on possible violations can be obtained. I discuss the relationship
of this result to the occasional claim that “light falls with twice the
acceleration of ordinary matter.”
email: carlip@dirac.ucdavis.edu

and the introduction to the paper

The principle of equivalence—the exact equality of inertial and gravitational
mass—is a cornerstone of general relativity, and experimental tests of the universality
of free fall provide a large set of data that must be explained by any theory
of gravitation. But the implication that energy contributes to gravitational mass
can be rather counterintuitive. Students are often willing to accept the idea that
potential energy has weight—after all, potential energy is a rather mysterious
quantity to begin with—but many balk at the application to kinetic energy. Can
it really be true that a hot brick weighs more than a cold brick?
General relativity offers a definite answer to this question, but the matter is
ultimately one for experiment. Surprisingly, while observational evidence for the
equivalence principle has been discussed for a variety of potential energies, the
literature appears to contain no analysis of kinetic energy. The purpose of this
paper is to rectify this omission, by reanalyzing existing experimental data to look
for the “weight” of the kinetic energy of electrons in atoms.

 

[Aer]

Here's another link addressing the topic - I'm not sure whether I've posted it to this particular thread before or not.

http://arxiv.org/abs/gr-qc/9909014

From the abstract


and the introduction to the paper

So all you are saying is what I've said - the evidence is inconclusive. Or do you wish to offer some other analysis.