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

[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.

 

[PeterDonis]

In other words, the previous post:

has no justification.

Sure it does; I specified that I was doing that calculation in the center of mass frame, where it *is* valid.

 

[xox]

I see that I should have used the term "potential energy" to describe $U$ rather than binding energy. Does that make it clearer?

No, it doesn't since all you are doing is waving your arms without being able to present a complete , coherent argument. If you were to do all the calculations accompanying your prose, then we could have a discussion.

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.

Once again, could you please put this prose into math? Two atoms would be sufficient. If you have a reference, that is even better, I can go and read on my own.

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.

This is a strawman, I never said that rest mass is additive, quite the opposite. On the other hand, the manipulations that you have posted are all over the map, so I am trying to see if you could write one coherent, non-contradictory, complete , mathematical treatment on the subject. I wrote one for the case of absent binding energy, can you write one for the case of binding energy being present? I am interested in the answer , if you cannot produce it , just say so and we are done.

 

[xox]

Sure it does; I specified that I was doing that calculation in the center of mass frame, where it *is* valid.

Let me explain this to you:

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

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

There is a frame, called "center of momentum" (not "center of mass") where \vec{V}=0.
In that frame the mass of the system is :

M=\Sigma{\gamma'_i m_i}

where

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

and v'_i are the speeds of the system constituents in the "center of momentum" frame.

 

[PeterDonis]

There is a frame, called "center of momentum" (not "center of mass") where \vec{V}=0.

I'll agree that "center of momentum" frame is a more accurate term. I've encountered both plenty of times, though.

In that frame the mass of the system is :

M=\Sigma{\gamma'_i m_i}

where

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

and v'_i are the speeds of the system constituents in the "center of momentum" frame.

Which is exactly what I wrote down for the 2-particle case with zero binding energy; I specified $| v_1 | = | v_2 | = u$ and $m_1 = m_2 = m$. (I used units where $c = 1$.)

Now please tell me how you would modify what you wrote for the case of nonzero binding energy.

 

[PeterDonis]

I never said that rest mass is additive

And I never said you said that. You keep on appearing to read me as saying exactly the opposite of what I said. I am quite aware that you said that rest mass is *not* additive, and I agree with you on that point.

 

[xox]

I'll agree that "center of momentum" frame is a more accurate term. I've encountered both plenty of times, though.

"Center of mass" doesn't make any sense, "center of momentum" obviously does.

Which is exactly what I wrote down for the 2-particle case with zero binding energy; I specified $| v_1 | = | v_2 | = u$ and $m_1 = m_2 = m$. (I used units where $c = 1$.)

It was your use of "center of mass" and your use of "rest mass" that threw me off, now we are in clear.

Now please tell me how you would modify what you wrote for the case of nonzero binding energy.

I have no idea, this is precisely I was hoping that you would clarify. Once again, can you write down an explanation (math, please) or give a pertinent reference?

 

[PeterDonis]

"Center of mass" doesn't make any sense

Apparently it does to a lot of people who write physics textbooks and articles.

I have no idea, this is precisely I was hoping that you would clarify. Once again, can you write down an explanation (math, please) or give a pertinent reference?

Then I'm confused, because you appeared to think there was something wrong with what I already wrote down, whereas to me it seems obvious. If, in the absence of binding energy (i.e., if the system as a whole is not in a bound state--there are no interactions between the constituents), we have

$$
M_0 = \Sigma \gamma'_i m_i
$$

then isn't it obvious that, in the presence of binding energy (i.e., if the system as a whole is in a bound state), we must have

$$
M_0 = \Sigma \gamma'_i m_i - U
$$

where $U$ is some positive number? All this is saying is that it would take some input of energy to convert the system from its current bound state to a state in which all the constituents were free.

We could also define a "constituent mass" $M_c = \Sigma m_i$, i.e., the simple sum of the constituent rest masses, and say that, for any real bound system, we must have $M_0 < M_c$, i.e., $M_0 = M_c - V$, where $V$ is some positive number--i.e., we are now saying it would take some input of energy to convert the system from its current bound state (i which some constituents may be moving) to a state in which all the constituents were free *and at rest*. (For any real bound state, we would expect this to be true.) This is obviously a more stringent condition on $M_0$ than the above; it requires that the $U$ in the formula I gave above is larger than $V$, possibly significantly larger (if the constituents of the system in the CoM frame are moving at significant velocities). Is this what you were trying to get at?

If you're looking for a more specific form for $U$ or $V$, that's going to depend on the specific system. I wasn't trying to go into that kind of detail since it didn't seem like the OP needed it. Is that what you're interested in?

 

[xox]

Apparently it does to a lot of people who write physics textbooks and articles.



Then I'm confused, because you appeared to think there was something wrong with what I already wrote down, whereas to me it seems obvious. If, in the absence of binding energy (i.e., if the system as a whole is not in a bound state--there are no interactions between the constituents), we have

$$
M_0 = \Sigma \gamma'_i m_i
$$

then isn't it obvious that, in the presence of binding energy (i.e., if the system as a whole is in a bound state), we must have

$$
M_0 = \Sigma \gamma'_i m_i - U
$$

where $U$ is some positive number? All this is saying is that it would take some input of energy to convert the system from its current bound state to a state in which all the constituents were free.

No, it isn't obvious at all. While M= \Sigma \gamma&#039;_i m_i was derived from base principles, you put by hand M= \Sigma \gamma&#039;_i m_i-U. What allows you to do that? Why not M= \Sigma \gamma&#039;_i m_i-U_i? Do you have a reference to either formula?

We could also define a "constituent mass" $M_c = \Sigma m_i$, i.e., the simple sum of the constituent rest masses, and say that, for any real bound system, we must have $M_0 < M_c$, i.e., $M_0 = M_c - V$, where $V$ is some positive number--i.e., we are now saying it would take some input of energy to convert the system from its current bound state (i which some constituents may be moving) to a state in which all the constituents were free *and at rest*. (For any real bound state, we would expect this to be true.) This is obviously a more stringent condition on $M_0$ than the above; it requires that the $U$ in the formula I gave above is larger than $V$, possibly significantly larger (if the constituents of the system in the CoM frame are moving at significant velocities). Is this what you were trying to get at?

No, not at all, I am trying to find a derivation of the mass of a system of particles from base principles, like I did it in post 5, for the case when binding energy is not zero. I see no reason why the binding energies of the constituents can be represented by a single variable, U, as you did it above.

If you're looking for a more specific form for $U$ or $V$, that's going to depend on the specific system. I wasn't trying to go into that kind of detail since it didn't seem like the OP needed it. Is that what you're interested in?

No, I am not looking at this level of detail, I am simply looking at a consistent, coherent methodology for driving the mass of the system in the presence of binding energies.

 

[DrStupid]

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

Because energy is additive.

Why not M= \Sigma \gamma&#039;_i m_i-U_i?

What do you mean with U_i?

 

[xox]

Because energy is additive.

He's not adding energy, he's adding mass (with binding energy divided by c^2). I have no idea what allows him to do that. Moreover, not any energy is additive, total energy is additive.

What do you mean with U_i?

Each atom has its own binding energy, hence the U_i.