Is Mass Dependent on Temperature According to Mass-Energy Equivalence?

[Caledon]

I'm a high school chem and physics teacher, participating in a recreational online chem course hosted by MIT.

One of the practice questions in this course asked how many atoms a certain mass would contain at 300K. This got me thinking about mass-energy equivalence, and whether the atomic mass units listed on the periodic table are defined according to a specific temperature, analogous to the way other values are defined at STP. This would mean that the number of atoms in a sample of mass X would vary inversely with the temperature of X.

My research led me to these conclusions: (and I know this is an incomplete understanding)...temperature is kinetic energy. Kinetic energy can alter the apparent mass of a system because gravity couples to momentum and energy, not just mass. Thus the mass may appear to change, but this is simply the aggregate effect of the (unchanged) mass and the kinetic energy. I found several sources that support this view, or at the very least, the statement that non-relativistic changes in temperature/kinetic energy will alter mass, although the values may be too small for us to measure with existing tools.

However, the response from the class's forum, including an MIT student administrating the forum, is that "mass is NOT temperature dependent" and will in no way affect it. I'm not sure if I'm completely misinterpreting mass-energy equivalence, but this doesn't make sense to me.

 

[WannabeNewton]

Energy certainly is temperature dependent but rest mass is not temperature dependent. Don't confuse rest mass with the energy of a particle as the two are only equal in the rest frame of the particle.

 

[PeterDonis]

whether the atomic mass units listed on the periodic table are defined according to a specific temperature

To follow up what WannabeNewton said, atomic mass units are units of rest mass, so they are independent of temperature. However, there is another wrinkle that's worth mentioning. You say:

Kinetic energy can alter the apparent mass of a system because gravity couples to momentum and energy, not just mass.

More precisely, the internal motions of the parts of a system can increase the *rest* mass of the system. For example, consider a solid object made up of lots of atoms in a crystal lattice. These atoms are not completely stationary; they are constantly undergoing small oscillations around their equilibrium positions in the lattice. If we increase the temperature of the object, the kinetic energy in these oscillations increases, and this shows up as a (very small) increase in the total rest mass of the object. This is probably what was being referred to by the sources you mentioned that seemed to support the view that changes in kinetic energy will change mass.

Note that this change in the rest mass of the solid object does *not* change the number of atoms in the object, nor does it change the rest masses of the individual atoms. It only changes the kinetic energy of the atom's oscillations about their equilibrium positions in the crystal lattice. But those oscillations do not produce any net motion of the object as a whole, so their kinetic energy contributes to the object's rest mass.

 

[Caledon]

Note that this change in the rest mass of the solid object does *not* change the number of atoms in the object, nor does it change the rest masses of the individual atoms. It only changes the kinetic energy of the atom's oscillations about their equilibrium positions in the crystal lattice. But those oscillations do not produce any net motion of the object as a whole, so their kinetic energy contributes to the object's rest mass.

Thank you, it sounds like we share the same understanding; the atoms retain a fixed individual rest mass, while the rest mass of the solid object increases with energy. And, yes, this does not change the number of atoms in the object, but (and this was the crux of my original inquiry) it would change the number of atoms necessary to produce a solid object with mass X. Or at least, I'm hoping that's a sensible conclusion. It seems that others are arguing that the very small change in rest mass is, in fact, nonexistant.

 

[xox]

I'm a high school chem and physics teacher, participating in a recreational online chem course hosted by MIT.

One of the practice questions in this course asked how many atoms a certain mass would contain at 300K. This got me thinking about mass-energy equivalence, and whether the atomic mass units listed on the periodic table are defined according to a specific temperature, analogous to the way other values are defined at STP. This would mean that the number of atoms in a sample of mass X would vary inversely with the temperature of X.

My research led me to these conclusions: (and I know this is an incomplete understanding)...temperature is kinetic energy. Kinetic energy can alter the apparent mass of a system because gravity couples to momentum and energy, not just mass. Thus the mass may appear to change, but this is simply the aggregate effect of the (unchanged) mass and the kinetic energy. I found several sources that support this view, or at the very least, the statement that non-relativistic changes in temperature/kinetic energy will alter mass, although the values may be too small for us to measure with existing tools.

However, the response from the class's forum, including an MIT student administrating the forum, is that "mass is NOT temperature dependent" and will in no way affect it. I'm not sure if I'm completely misinterpreting mass-energy equivalence, but this doesn't make sense to me.

The total energy of a system of particles is:

E=c^2\Sigma {\gamma_i m_i}

The total momentum of a system of particles is:

\vec{p}=\Sigma{\gamma_i m_i \vec{v_i}}

The total rest mass M of a system of particles satisfies the relationship:

E^2-(\vec{p}c)^2=M^2c^4

We can always find \vec{V} such that:

E=\gamma(V) M c^2
\vec{p}=\gamma(V) M \vec{V}

Obviously, from the above:

\gamma(V)M=\Sigma {\gamma_i m_i}

\vec{V}=\frac{\Sigma {\gamma_i m_i \vec{v_i}}}{\Sigma {\gamma_i m_i }}

and

M=\frac{\Sigma{\gamma_i m_i}}{\gamma(V)}

What Peter Donis has been telling you is that we can prove that:

M \ge \Sigma{m_i}

You can try to prove the above for the case i=2, it is pretty easy.

 

[Astronuc]

Thank you, it sounds like we share the same understanding; the atoms retain a fixed individual rest mass, while the rest mass of the solid object increases with energy. And, yes, this does not change the number of atoms in the object, but (and this was the crux of my original inquiry) it would change the number of atoms necessary to produce a solid object with mass X. Or at least, I'm hoping that's a sensible conclusion.

This interpretation is not quite sensible.

It seems that others are arguing that the very small change in rest mass is, in fact, nonexistant.

Or rather, negligible.

To put the matter into perspective, consider the energy equivalence of 300K with respect to electron binding energies and nuclear rest mass.

1 eV is equivalent to a temperature of ~11604.52 K. (from NIST)
http://physics.nist.gov/cuu/Constants/energy.html

300 K is equivalent to 300 K/(11604.52 K/eV) = 0.02585 eV.

Compare that to an electron binding energy, which is on the order to eVs. For example, the ionization energy of the electron in a hydrogen atom is ~13.6 eV.

The rest energy of an electron is ~ 511 keV.

The rest energy equivalence of an atomic mass unit is 931.494 MeV, so the contribution of 300 K to an atomic mass unit would be 1 part in 36 billion (3.6 E10), so it would be negligible.

But associating kinetic energy of a mass with a change in it's rest mass is not correct.

 

[Bill_K]

The rest energy equivalence of an atomic mass unit is 931.494 MeV, so the contribution of 300 K to an atomic mass unit would be 1 part in 36 billion (3.6 E10), so it would be negligible.

But since a typical sample contains ~10²³ atoms, a change of one part in 36 billion means that ~10¹³ fewer atoms will be necessary. Not so negligible!

 

[Caledon]

But since a typical sample contains ~10²³ atoms, a change of one part in 36 billion means that ~10¹³ fewer atoms will be necessary. Not so negligible!

Yes, it seems like some of the answers I've been reading elsewhere do not distinguish between "there is a negligible effect" and "there is NO effect". The class's original question referred to 13g of gold, so the result, according to these values, would have been trivial on the macroscopic scale but nontrivial on the atomic.

But associating kinetic energy of a mass with a change in it's rest mass is not correct.

Ok, this is where I assumed my understanding was breaking down. From my understanding of the equations that xox posted (thank you, xox), and what PeterDonis wrote, momentum and velocity are included in the consideration of total rest mass M, which is an aggregate of mass and energy and does not alter the original rest mass m_i of the constituent particles. This appears to be the divergent point: is there something that forbids kinetic energy from contributing to M?

 

[xox]

Ok, this is where I assumed my understanding was breaking down. From my understanding of the equations that xox posted (thank you, xox), and what PeterDonis wrote, momentum and velocity are included in the consideration of total rest mass M, which is an aggregate of mass and energy and does not alter the original rest mass m_i of the constituent particles. This appears to be the divergent point: is there something that forbids kinetic energy from contributing to M?

Your understanding is correct. The mass of a system of particles is NOT equal to the sum of the masses of the particles composing the system. It is always larger than that sum. By how much it is larger is a function of the energy of the particles. You can try to prove the above for the case i=2, it is pretty easy.
In general , I shy away from "literary" descriptions of physics and I ted to mathematical descriptions. "Literature" can be very imprecise, math, on the other hand, is very precise. (see my answer to your question).

 

[jtbell]

The mass of a system of particles is NOT equal to the sum of the masses of the particles composing the system. It is always larger than that sum.

If the particles are all at rest with respect to each other, that is, if there is an inertial reference frame in which all the particles are at rest; and the system does not have any binding energy; then the mass of the system does equal the sum of the masses of the particles.

 

[xox]

If the particles are all at rest with respect to each other, that is, if there is an inertial reference frame in which all the particles are at rest, then the mass of the system does equal the sum of the masses of the particles.

Yes, this is the ONLY case when M=\Sigma{m_i}. It is very improbable for such a frame to exist since it would require that all particles have the same velocity.
In ALL other cases M>\Sigma{m_i}.

 

[PeterDonis]

associating kinetic energy of a mass with a change in it's rest mass is not correct.

For a single object, no. But associating kinetic energy of *parts* of a system with the rest mass of the *system*, which is what I was describing, *is* correct; the rest mass of the system as a whole is *not* the sum of the rest masses of all the parts, if the parts are in relative motion (i.e., if the parts are moving in the center of mass frame of the system as a whole).

 

[xox]

The total energy of a system of particles is:

E=c^2\Sigma {\gamma_i m_i}

The total momentum of a system of particles is:

\vec{p}=\Sigma{\gamma_i m_i \vec{v_i}}

The total rest mass M of a system of particles satisfies the relationship:

E^2-(\vec{p}c)^2=M^2c^4

We can always find \vec{V} such that:

E=\gamma(V) M c^2
\vec{p}=\gamma(V) M \vec{V}

Obviously, from the above:

\vec{V}=\frac{\Sigma {\gamma_i m_i \vec{v_i}}}{\Sigma {\gamma_i m_i }}

and

M=\frac{\Sigma{\gamma_i m_i}}{\gamma(V)}

What Peter Donis has been telling you is that we can prove that:

M \ge \Sigma{m_i}

You can try to prove the above for the case i=2, it is pretty easy.

In the improbable (but not totally impossible) case when all particle velocities are equal:


\vec{V}=\frac{\Sigma {\gamma_i m_i \vec{v_i}}}{\Sigma {\gamma_i m_i }}=\vec{v}

resulting into:

M=\frac{\Sigma{\gamma_i m_i}}{\gamma(V)}=\Sigma{m_i}

 

[jtbell]

After I wrote my last post I went out for a walk, during which I realized that I hadn't fully appreciated the impact of my comment about binding energy, because I was focusing on the ∑m_i = M case.

Binding energy actually subtracts from the sum of the masses of the individual particles in a bound system! The most prominent example is an atomic nucleus, whose mass is less than the sum of the masses of the protons and neutrons that comprise it. In order to "rip" a nucleus apart completely into protons and neutrons, you have to supply at least enough energy to make up that difference in mass.

 

[xox]

After I wrote my last post I went out for a walk, during which I realized that I hadn't fully appreciated the impact of my comment about binding energy, because I was focusing on the ∑m_i = M case.

Binding energy actually subtracts from the sum of the masses of the individual particles in a bound system! The most prominent example is an atomic nucleus, whose mass is less than the sum of the masses of the protons and neutrons that comprise it. In order to "rip" a nucleus apart completely into protons and neutrons, you have to supply at least enough energy to make up that difference in mass.

While it is true that binding energy is the difference between the sum of the energies of the particles composing a nucleus and the total energy of the nucleus, the OP is about a system of atoms, binding energy operates only at the level of the nucleus. Besides, the part that you posted about all components having the same velocity indicated that you knew that the discussion was about a system of atoms, you added the part about the binding energy later.

 

[PeterDonis]

The OP is about a system of atoms, binding energy operates only at the level of the nucleus.

No, it doesn't. The electrons in the atom have a negative binding energy when bound to the nucleus, compared to free electrons. So the rest mass of the atom is smaller than the sum of the rest masses of the nucleus and the same number of free electrons. It's true that the difference is much smaller than the difference in the case of a nucleus (nuclear binding energies are in the millions of eV, whereas atomic binding energies are a few eV, or about a million times smaller); but it's still not zero.

Also, atoms in a bound object like a solid have negative binding energy relative to identical atoms that are free. So the rest mass of the solid object is smaller than the sum of the rest masses of its constituent atoms. These binding energies (i.e., chemical bond energies) are typically an eV or less, so somewhat smaller than atomic binding energies; but again, that's still not zero.

 

[xox]

No, it doesn't. The electrons in the atom have a negative binding energy when bound to the nucleus, compared to free electrons.

I stand corrected, nevertheless the point is that the OP is about the system of atoms, not about the energies inside the atoms.

So the rest mass of the atom is smaller than the sum of the rest masses of the nucleus and the same number of free electrons.

I would take exception from the above language, since a better way is to express the above in terms of energy, not "rest mass".

 

[PeterDonis]

the point is that the OP is about the system of atoms

Binding energy applies there too; see the edit I just made to my post.

 

[PeterDonis]

I would take exception from the above language, since a better way is to express the above in terms of energy, not "rest mass".

Rest mass *is* energy; which term is more appropriate is a matter of taste and convention, not physics. I would say that if we are making measurements of mass, such as weighing something on a balance, the term "mass" is more appropriate. Given that convention, the statement I made is correct as it stands; it can be confirmed directly by sufficiently accurate measurements of the respective rest masses, in the same way we normally measure mass (not the way we measure energy).

 

[xox]

Rest mass *is* energy;

"rest mass" is rest energy (m_0c^2). Energy is \gamma m_0 c^2. The particles constituting an atom are never at rest.

 

[PeterDonis]

The particles constituting an atom are never at rest.

Yes, this is true, and it means that the rest mass of the atom includes the kinetic energy of the particles that constitute it, *in addition to* the binding energy between those particles. So, for example, in the simplest case of a hydrogen atom, we would have

$$
M_0 = m_p + m_e + k_p + k_e - U
$$

where $M_0$ is the rest mass of the atom, $m_p$ is the rest mass of a proton, $m_e$ is the rest mass of an electron, $k_p$ and $k_e$ are the kinetic energies of the proton and electron (in the center of mass frame of the atom), and $U$ is the binding energy. In general, $k_p + k_e - U$ will not be zero, so the rest mass of the atom won't be equal to the sum of the rest masses of its constituents.

(In fact, the usually quoted "binding energy" of the electron in the hydrogen atom, 13.6 eV, includes *both* of the above effects: the binding energy due to the Coulomb attraction between electron and proton, *and* the kinetic energy of the electron in its orbit about the proton--the proton is assumed to be a rest, which is not precisely true in the center of mass frame but is a good enough approximation for this case.)

 

[xox]

Yes, this is true, and it means that the rest mass of the atom includes the kinetic energy of the particles that constitute it, *in addition to* the binding energy between those particles. So, for example, in the simplest case of a hydrogen atom, we would have

$$
M_0 = m_p + m_e + k_p + k_e - U
$$

where $M_0$ is the rest mass of the atom, $m_p$ is the rest mass of a proton, $m_e$ is the rest mass of an electron, $k_p$ and $k_e$ are the kinetic energies of the proton and electron (in the center of mass frame of the atom), and $U$ is the binding energy. In general, $k_p + k_e - U$ will not be zero, so the rest mass of the atom won't be equal to the sum of the rest masses of its constituents.

(In fact, the usually quoted "binding energy" of the electron in the hydrogen atom, 13.6 eV, includes *both* of the above effects: the binding energy due to the Coulomb attraction between electron and proton, *and* the kinetic energy of the electron in its orbit about the proton--the proton is assumed to be a rest, which is not precisely true in the center of mass frame but is a good enough approximation for this case.)

Thank you, this is interesting, how would you generalize the above formalism into answering the OP question? That is, what is the mathematical formalism that calculates the (rest) mass of a system of atoms? I would love to see something along the lines of the mathematical formalism that I posted. How would you work binding energy in the formulas?

 

[PeterDonis]

Thank you, this is interesting, how would you generalize the above formalism into answering the OP question?

It works the same for $N$ particles: just add up the $N$ rest masses and kinetic energies (in the center of mass frame of the system), and all of the applicable binding energies. In general there could be $N \left( N - 1 \right) / 2$ of the latter, since each pair of particles could have an interaction; but in many cases we can restrict attention to a much smaller number.

 

[Caledon]

Thank you both for your insights. Xox, I know exactly what you mean about linguistics being a relatively (haha) inferior means of describing these properties; it seems like conflicting definitions of mass and energy are the source of many of the contrary opinions I've been reading. Unfortunately I have to rely heavily on linguistics in my classroom because our district prefers to make math terrifying.

On this subject, let me extend another line of inquiry; this is what I meant in my original post by "mass defined at a particular temperature". If atoms are always in motion, i.e. they have some amount of kinetic energy, I assume that energy cannot be easily quantified, per Uncertainty, nor can it be removed, per thermodynamics. Thus there will always be some baseline rest mass M that is in fact the aggregate of particle mass m and baseline energies k and U, perhaps analogous to something like Planck's constant.

Is this the state at which an atom's atomic mass is calculated? Meaning, is it extrapolated as opposed to directly observed? Or are we just working with empirical measurements that include some ambient temperature, such as room temp, and the nuanced differences are too small for us to really care about in terms of anything except, say, the LHC? The SI definition of the mole says nothing about energy or temperature considerations.

 

[Bill_K]

Is this the state at which an atom's atomic mass is calculated? Meaning, is it extrapolated as opposed to directly observed? Or are we just working with empirical measurements that include some ambient temperature, such as room temp, and the nuanced differences are too small for us to really care about in terms of anything except, say, the LHC? The SI definition of the mole says nothing about energy or temperature considerations.

To determine a particle's rest mass, it is not necessary to bring it to rest. For example when they determined the rest mass of the Higgs boson, they did not bring it to rest. They measured its energy and its momentum, and calculated the rest mass from that. Generally the rest mass of an atom is determined using a mass spectrometer.

 

[Caledon]

To determine a particle's rest mass, it is not necessary to bring it to rest.

Yes, precisely what I'm trying to work around. It would in fact be impossible to bring it to true rest i.e. a state of zero motion. However, with that established, there are a variety of rest masses that could be measured: say, one at 300K, or as a plasma in the mass spectrometer, or one extrapolated down to a conjectured baseline near 0K, where the variable component of kinetic energy is at a minimum. Mind that I know we're talking about very small mass differences, but if the atoms in the spectrometer are being ionized at different energies, this would produce a non-standardized definition of mass.

I'll read the link you provided and see if I can gain a better understanding, thank you.

 

[PeterDonis]

If atoms are always in motion, i.e. they have some amount of kinetic energy

Don't confuse the kinetic energy of the *constituents* of the atom with the kinetic energy of the atom itself. The rest mass of the atom itself assumes that the atom has zero kinetic energy; note that I carefully specified that the kinetic energies of the constituents of the atom are in the center of mass frame of the atom, i.e., the frame in which the atom itself, as a whole system, is at rest.

The mass numbers of particular isotopes of elements (which is what I think you mean by "atomic mass"--the atomic weights for elements that appear in the periodic table are weighted averages of the mass numbers of the isotopes, according to how common each isotope is in nature) are given in the center of mass frame, as above. So they are "pure" rest mass, as it were; i.e., they don't include any kinetic energy of the atoms as a whole. (They do, as above, include the kinetic energies of the *constituents* of the atoms.)

The SI definition of the mole says nothing about energy or temperature considerations.

That's because a mole is just an Avogadro's number of atoms (or molecules, or whatever you're working with); there's nothing else required to define it except that number. In order to assign a mass to a mole, you have to know what kind of atom (or molecule) you have a mole of--and then, if you want to be really precise, you also have to know what state the mole of atoms (or molecules) is in, i.e., is it a gas at a certain temperature and pressure, or a liquid, or a solid, etc. But those things don't enter into the definition of a mole.

 

[Caledon]

Don't confuse the kinetic energy of the *constituents* of the atom with the kinetic energy of the atom itself.

I appreciate your help. I follow everything you're saying, and I'll try to be as precise in my language as I can. Yes, I was considering the kinetic energy of the constituents of the atom, but I do recognize that they are conflated into the atom's frame of reference, and at rest with zero kinetic energy in relation to this state. But what if we're talking about, say, a box of gas, where the atoms themselves are in motion as well? This is more of what I had in mind in terms of ascribing kinetic energy to the system. The temperature of the system will affect both the atoms and the atomic constituents.

Yes, by atomic mass I was actually referring to any particular isotope one happens to choose, not the averaged value on the table. So, as you said, they don't include any kinetic energy of the atoms as a whole. Ok. The way this energy-less value is obtained is one focus of my interest.

That's because a mole is just an Avogadro's number of atoms (or molecules, or whatever you're working with); there's nothing else required to define it except that number.

Actually (and perhaps this is the source of another miscommunication) the SI defines it in relationship to carbon.
From http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf
"The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12".
If an Avogadro's number of atoms is tied directly to a mass value, and (rest)masses can change, then the number of atoms necessary to achieve that mass value (which is an evaluation of rest mass) can also change. A mole will always be a mole (just a number) but the correlation between those numbers and mass will not be constant, correct?

 

[PeterDonis]

what if we're talking about, say, a box of gas, where the atoms themselves are in motion as well?

Then the rest mass of the gas will be the sum of the rest masses of all the atoms (which will include the kinetic energies and binding energies of the constituents of the atoms) plus the kinetic energies of all the atoms (in the center of mass frame of the gas as a whole) plus the binding energies due to inter-atomic forces like Van der Waals forces (which for a gas are much smaller than atomic or chemical binding energies, so they are often neglected in approximations like the derivation of the ideal gas law).

The temperature of the system will affect both the atoms

Yes; the temperature of the gas *is* the average kinetic energy of its atoms in the center of mass frame of the gas.

and the atomic constituents.

No; the atoms themselves are bound states of nuclei and electrons that are unaffected by the temperature of the gas (at least, as long as that temperature is less than the ionization energy of the atoms, so collisions between atoms don't ionize the atoms).

The way this energy-less value is obtained is one focus of my interest.

As I understand it, mass numbers of isotopes are usually measured by a mass spectrometer, which puts the atoms through a magnetic field and measures how they respond to it. The response depends on the ratio of charge to rest mass of the atoms; since the charge can be measured independently, the rest mass (mass number) can be determined.

"The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12".

Yes, this is the modern way of defining Avogadro's number. The "amount of substance" here just means "number of elementary entities". See below.

If an Avogadro's number of atoms is tied directly to a mass value

It isn't. It's tied directly to a number of atoms (or more correctly "elementary entities", as above, which may be molecules, or in more exotic cases may be ions, or even electrons or protons or some other type of particle). The mass value associated with that number of atoms depends, as I said before, on what kind of atoms (or other elementary entities) they are. The only function the mass value "0.012 kilogram of carbon 12" has is to pin down a particular number of elementary entities.

Note that the definition as given by SI is not entirely precise; it should really read "0.012 kilogram of carbon 12 at rest", since the intent of the definition is that you take 0.012 kilogram and divide by the mass number of carbon 12 to get Avogadro's number. That calculation gives a constant, since 0.012 kilogram and the mass number of carbon 12 are both constants. That this is the intent of the definition can be seen by looking just above the passage you quoted: "Amount of substance is defined to be proportional to the number of specified elementary entities in a sample, the proportionality constant being a universal constant which is the same for all samples".

the number of atoms necessary to achieve that mass value (which is an evaluation of rest mass) can also change.

No, this is not correct. A mole is always exactly the same number of elementary entities. See above.

 

[DrStupid]

Yes; the temperature of the gas *is* the average kinetic energy of its atoms in the center of mass frame of the gas.

The average kinetic energy is a measure of the temperature but not identical with temperature.

No; the atoms themselves are bound states of nuclei and electrons that are unaffected by the temperature of the gas (at least, as long as that temperature is less than the ionization energy of the atoms, so collisions between atoms don't ionize the atoms).

There are a lot of energy levels below ionization and in thermal equilibrium the population of these levels depends on temperature according to the Boltzmann distribution. Therefore the atoms actually are affected by temperature.

 

[PeterDonis]

The average kinetic energy is a measure of the temperature but not identical with temperature.

What's the difference, other than units? (Yes, the units are different, but that's not a matter of physics, it's a matter of convention. And the convention is not always the same: many physicists measure temperature in energy units.)

There are a lot of energy levels below ionization and in thermal equilibrium the population of these levels depends on temperature according to the Boltzmann distribution. Therefore the atoms actually are affected by temperature.

Hm, yes, good point; I was assuming that all the atoms would be in their ground states, but that's not in general going to be the case.

 

[DrStupid]

What's the difference, other than units?

In contrast to kinetic energy temperature is defined for thermodynamic systems in thermal equilibrium only and the extension of the thermodynamic temperature to non equilibrium states may lead to results that are completely different from the "kinetic temperature" (such as the negative temperature of a pumped laser).

 

[PeterDonis]

temperature is defined for thermodynamic systems in thermal equilibrium only

I think this is a bit narrow. My body is certainly not in thermodynamic equilibrium, but it has a temperature. (More precisely, it has a kinetic temperature; see below.)

the extension of the thermodynamic temperature to non equilibrium states may lead to results that are completely different from the "kinetic temperature" (such as the negative temperature of a pumped laser).

Yes, but this really means the word "temperature" can refer to at least two distinct concepts, "kinetic temperature" and "thermodynamic temperature" (which is defined as $1 / \beta$, where $\beta = dS / dE$, the rate of change of entropy with energy). To be precise, the one I was talking about was kinetic temperature, since that's the one that contributes to the rest mass of a system. The rest mass of a pumped laser *increases* because of the energy being pumped in, even though the photons inside it have a negative thermodynamic temperature.

 

[Caledon]

Alright. So a mole is just a number, and that number coincides with the atoms in 12g of C12 at rest, in the same way that the number 5 coincides with the fingers on my hand and doesn't change if I chop one off. The connection I'm having trouble making is, how can this be used to stoichiometrically calculate the atoms in a given sample that is not at rest, since we established that the rest mass M can change? I guess I'm attempting to make a connection between molar mass and rest mass, perhaps this is invalid.

 

[xox]

It works the same for $N$ particles: just add up the $N$ rest masses and kinetic energies (in the center of mass frame of the system), and all of the applicable binding energies. In general there could be $N \left( N - 1 \right) / 2$ of the latter, since each pair of particles could have an interaction; but in many cases we can restrict attention to a much smaller number.

We know that rest mass isn't additive (nor is its equivalent, rest energy), so why would the above be correct?

 

[xox]

Thank you both for your insights. Xox, I know exactly what you mean about linguistics being a relatively (haha) inferior means of describing these properties; it seems like conflicting definitions of mass and energy are the source of many of the contrary opinions I've been reading. Unfortunately I have to rely heavily on linguistics in my classroom because our district prefers to make math terrifying.

On this subject, let me extend another line of inquiry; this is what I meant in my original post by "mass defined at a particular temperature". If atoms are always in motion, i.e. they have some amount of kinetic energy, I assume that energy cannot be easily quantified, per Uncertainty, nor can it be removed, per thermodynamics. Thus there will always be some baseline rest mass M that is in fact the aggregate of particle mass m and baseline energies k and U, perhaps analogous to something like Planck's constant.

Is this the state at which an atom's atomic mass is calculated? Meaning, is it extrapolated as opposed to directly observed? Or are we just working with empirical measurements that include some ambient temperature, such as room temp, and the nuanced differences are too small for us to really care about in terms of anything except, say, the LHC? The SI definition of the mole says nothing about energy or temperature considerations.

Measure the energy and the magnitude of the momentum:

E=\gamma m_0 c^2
p=|\vec{p}|=\gamma m_0 v

From the above, the speed is :

v=\frac{E}{pc^2}

Armed with this, we calculate :

\gamma=\frac{1}{\sqrt{1-(v/c)^2}}

and finally

m_0=\frac{E}{\gamma c^2}

or

m_0=\frac{p}{\gamma v}

 

[PeterDonis]

We know that rest mass isn't additive (nor is its equivalent, rest energy), so why would the above be correct?

Um, because it doesn't just include rest masses, but also kinetic energies in the center of mass frame, and binding energies?

 

[xox]

Um, because it doesn't just include rest masses, but also kinetic energies in the center of mass frame, and binding energies?

Yet, according to your earlier post, the LHS is the rest mass, M_0.

 

[PeterDonis]

So a mole is just a number, and that number coincides with the atoms in 12g of C12 at rest

More precisely, it is the number of atoms in 12 g of C12 assuming that all of the atoms are at rest. (And also, to be precise, assuming that the binding energies between the atoms are insignificant.) In other words, it is 12 g divided by the mass number of the carbon-12 nucleus.

The connection I'm having trouble making is, how can this be used to stoichiometrically calculate the atoms in a given sample that is not at rest

First of all, once again, don't confuse the individual atoms being in motion with the sample being in motion. Do you do stoichiometry on samples that are flying around your lab? I'm betting the samples are sitting at rest on a table in your lab.

The fact that the individual atoms in the sample are not at rest does introduce a (very small) inaccuracy into the calculation of the number of atoms in the sample, yes. (So do the binding energies between the atoms.) But for practical purposes this inaccuracy is much too small to matter.

For example, suppose I have a lump of pure coal that is exactly 12 grams of carbon-12 atoms, as measured while it's sitting at rest in the lab. How many atoms will there be in the lump? It won't be exactly Avogadro's number (12 grams divided by the mass number of carbon-12); it will be a bit smaller, because the sample is at some finite temperature and there are binding energies associated with the covalent bonds between the carbon atoms. How much smaller? (That is, how much error am I making if I assume there are exactly one Avogadro's number of atoms in the sample?) Well, we just write down an equation similar to what I wrote above:

$$
M_0 = N \left( m_{12} + T \right) - U
$$

where $M_0 = 12$ is the rest mass of the sample (note that I'm using units in which $c = 1$, so mass and energy have the same units), $N$ is the actual number of atoms in the sample, $m_{12}$ is the mass number of carbon-12, $T$ is the absolute temperature of the sample (measured in energy units), and $U$ is the total of all the chemical bond energies. Rearrange this as follows:

$$
N = \frac{M_0 + U}{m_{12} + T} = \frac{M_0}{m_{12}} \left( \frac{1 + U / M_0}{1 + T / m_{12}} \right)
$$

So the size of the error is just the factor in parentheses on the right. How big is that? Let's plug in some numbers.

$U / M_0$ can be estimated from the heat of combustion of coal, which varies with the type of coal; the highest value I've found online is 32 kilojoules per gram, which equates to 32,000 times 10,000,000 ergs per Joule, divided by $c^2$ to get "grams per gram" of binding energy. This is a very small number, about $3.6 \times 10^{-10}$.

$T$ is just the absolute temperature in mass units, which for $T$ in Kelvins is just $k_B T / c^2$, where $k_B = 1.38 \times 10^{-23}$ is Boltzmann's constant (again in SI units, so we have to correct for that as well to get $T$ in grams). For an ordinary temperature of 300 K, this gives $T = 4.6 \times 10^{-35}$ in grams.

$m_{12}$ is exactly 12 atomic mass units, by definition, and an a.m.u is $1.66 \times 10^{-24}$ grams. This gives another very small number for $T / m_{12}$, about $2.3 \times 10^{-12}$. This is less than 1 percent of $U / M_0$, so it will, at best, reduce the error slightly (since it appears in the denominator) compared to the effect of $U / M_0$. So the total error involved in not counting the kinetic temperature or binding energy when estimating the number of atoms in the sample is at most a few parts in ten billion, i.e., way too small to matter in practical terms.

 

[PeterDonis]

Yet, according to your earlier post, the LHS is the rest mass, M_0.

The rest mass of the *system*, yes. And the formula therefore explicitly says that the rest mass of the system is *not* the sum of the rest masses of its constituents, since additional terms appear for the kinetic energies and binding energies.

 

[xox]

The rest mass of the *system*, yes.

No, what you state is that it is "the rest mass of the atom", I am interested in the extension of the formula from one atom to a system of atoms:

Yes, this is true, and it means that the rest mass of the atom includes the kinetic energy of the particles that constitute it, *in addition to* the binding energy between those particles. So, for example, in the simplest case of a hydrogen atom, we would have

$$
M_0 = m_p + m_e + k_p + k_e - U
$$

where $M_0$ is the rest mass of the atom, $m_p$ is the rest mass of a proton, $m_e$ is the rest mass of an electron, $k_p$ and $k_e$ are the kinetic energies of the proton and electron (in the center of mass frame of the atom), and $U$ is the binding energy. In general, $k_p + k_e - U$ will not be zero, so the rest mass of the atom won't be equal to the sum of the rest masses of its constituents.

And the formula therefore explicitly says that the rest mass of the system is *not* the sum of the rest masses of its constituents, since additional terms appear for the kinetic energies and binding energies.

I have no argument with that, what I find confusing is the way you plan to extend the formula from one atom to a system of atoms moving wrt. each other. To fix the ideas, could you please write the total (rest) mass of two atoms moving wrt. each other with speed v?

 

[PeterDonis]

No, what you state is that it is "the rest mass of the atom"

For the case where the "system" is one atom, which was the case under discussion when I defined the formula. For the case where the "system" contains many atoms, the LHS is the rest mass of the system. Sorry if that wasn't clear.

To fix the ideas, could you please write the total (rest) mass of two atoms moving wrt. each other with speed v?

In the center of mass frame, atom #1 is moving to the left at speed $u$ and atom #2 is moving to the right at $u$, where

$$
v = \frac{2 u}{1 + u^2}
$$

implicitly gives $u$ in terms of $v$. Then, assuming both atoms have rest mass $m$ and that $k$ is the kinetic energy of an object with rest mass $m$ moving at speed $u$, we have

$$
M_0 = 2 \left( m + k \right) = 2 \gamma m
$$

where $\gamma = 1 / \sqrt{1 - u^2}$ (assuming there is no significant interaction between the atoms and therefore no binding energy involved).

 

[xox]

$$
M_0 = 2 \left( m + k \right) = 2 \gamma m
$$

where $\gamma = 1 / \sqrt{1 - u^2}$ (assuming there is no significant interaction between the atoms and therefore no binding energy involved).

I am sorry if I were unclear, I was asking for the generalization (for two atoms) of the case you explained in post 21, that is for the case when binding energy IS significant. I know very well how to treat the case in the absence of binding energy. As an aside, I disagree with what you wrote above, 2 \gamma mc^2 is not the rest mass, it is the total energy. We seem to keep having this disagreement. The moment \gamma appears in the formula, you cannot longer claim to be talking about "rest" mass.

 

[jtbell]

Forget about the term "rest mass". All it does is confuse people. Do what Real Physicsists (TM) do, and call it just "mass", or if you want to make a point of distinguishing it from the so-called "relativistic mass", call it "invariant mass."

The ("rest") mass of a system equals the total energy of the system, divided by c², in the reference frame in which the total momentum of the system is zero. It is in that sense that we use the word "rest" here, even though in that frame the individual particles are still moving.

Or more generally, calculate the mass of the system using
$$mc^2 = \sqrt{(\Sigma E_i)^2 - (\Sigma \vec p_i c)^2}$$
which works in any inertial reference frame.

 

[xox]

Forget about the term "rest mass". All it does is confuse people. Do what Real Physicsists (TM) do, and call it just "mass", or if you want to make a point of distinguishing it from the so-called "relativistic mass", call it "invariant mass."

The ("rest") mass of a system equals the total energy of the system, divided by c², in the reference frame in which the total momentum of the system is zero. It is in that sense that we use the word "rest" here, even though in that frame the individual particles are still moving.

Or more generally, calculate the mass of the system using
$$mc^2 = \sqrt{(\Sigma E_i)^2 - (\Sigma \vec p_i c)^2}$$
which works in any inertial reference frame.

I already showed how this is done, early in the thread.
Peter and I are discussing a different situation, how to extend the calculations in the case of presence of binding energy. Can you shed some light?

 

[PeterDonis]

I am sorry if I were unclear, I was asking for the generalization (for two atoms) of the case you explained in post 21, that is for the case when binding energy IS significant.

Ok, then it would be $M_0 = 2 \gamma m - U$, where $U$ is the binding energy due to the interaction between the atoms.

2 \gamma mc^2 is not the rest mass, it is the total energy.

Since the system as a whole is at rest (because we are in the center of mass frame of the system), its total energy *is* its rest mass. Don't confuse the system with its constituents. The system as a whole is at rest even though its constituents are not.

The moment \gamma appears in the formula, you cannot longer claim to be talking about "rest" mass.

Yes, you can, as long as you're clear about *what* is at rest and what isn't. See above.

 

[PeterDonis]

how to extend the calculations in the case of presence of binding energy.

Doing this in the general case, i.e., in an arbitrary reference frame, raises some issues, because the definition of "binding energy" is the energy that would be required to break the system up and make each of its constituents a free object "at infinity" relative to all the others. This definition presupposes that there is in fact an "infinity" available, i.e., that the system is isolated; and if the system is isolated, it must pick out a definite center of mass frame, in which the binding energy is calculated.

Once the calculation is done, and a total 4-momentum for the system is obtained, that 4-momentum can be treated as an ordinary Lorentz-covariant 4-vector. But if binding energy is involved, there is no longer any Lorentz-invariant way to obtain that 4-vector by adding up 4-vector contributions from each constituent of the system, which is what jtbell's formula is equivalent to.

Another way of putting this is that binding energy works like potential energy, i.e., it's a function of position; but "position" is frame-dependent. In any system where a well-defined potential energy exists, it is only well-defined as a function of position in a particular frame, the system's center of mass frame. In other words, potential energy is not a Lorentz scalar, nor can it be modeled in general as a 4-vector or any other Lorentz-covariant mathematical object.

 

[xox]

Ok, then it would be $M_0 = 2 \gamma m - U$, where $U$ is the binding energy due to the interaction between the atoms.

Do you have a reference for this?

Since the system as a whole is at rest (because we are in the center of mass frame of the system), its total energy *is* its rest mass.

Then you cannot have any \gamma in the expression of energy for the mere reason that...\gamma=1.

Don't confuse the system with its constituents.

I am not confusing anything but I am sorry to say that your presentation is very poor.

 

[xox]

Doing this in the general case, i.e., in an arbitrary reference frame, raises some issues, because the definition of "binding energy" is the energy that would be required to break the system up and make each of its constituents a free object "at infinity" relative to all the others. This definition presupposes that there is in fact an "infinity" available, i.e., that the system is isolated; and if the system is isolated, it must pick out a definite center of mass frame, in which the binding energy is calculated.

In other words, the previous post:

Ok, then it would be $M_0 = 2 \gamma m - U$, where $U$ is the binding energy due to the interaction between the atoms.

has no justification.

 

[PeterDonis]

Do you have a reference for this?

I see that I should have used the term "potential energy" to describe $U$ rather than binding energy. Does that make it clearer? All I'm saying is that the system as a whole is described by a 4-momentum vector whose invariant length is $M_0$; and that invariant length is determined by summing up the rest masses, kinetic energies, and potential energies of the constituents, all evaluated in the center of mass frame of the system. You have to do it in that particular frame because potential energy as a function of position is only well-defined in that frame, as I said in my last post.

Then you cannot have any \gamma in the expression of energy for the mere reason that...\gamma=1.

Sigh. I specifically said that $\gamma = 1 / \sqrt{1 - u^2}$, and $u$ is not zero in the center of mass frame of the system (I defined what it was in an earlier post).

I am not confusing anything.

Well, you certainly seem to be, since you are evidently misinterpreting what $\gamma$ means. See above.

your presentation is very poor.

That may be; feel free to re-state what you believe to be correct in terms that seem to you to be appropriate. But nothing I've said contradicts the point you seemed most concerned about, that rest mass is not additive.