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

[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.

 

[DrStupid]

He's not adding energy, he's adding mass.

According to his original statement you would be right because he defined the kinetic energy of the particles "(in the center of mass frame of the atom)". But if I understand the following discussion correctly he changed that to the rest frame of the total assembly of all atoms. In that case he is actually adding energies. Maybe the term "rest mass" for "rest energy" is confusing at this place but that's the current convention.

Moreover, not any energy is additive, total energy is additive.

If the energies are related to the same frame of reference they should be additive.

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

And what about the potential energy related to interactions between different atoms (e.g. the binding energy of H₂)?

 

[xox]

According to his original statement you would be right because he defined the kinetic energy of the particles "(in the center of mass frame of the atom)". But if I understand the following discussion correctly he changed that to the rest frame of the total assembly of all atoms. In that case he is actually adding energies.

No, he's not adding energies. Let me explain this to you in more detail: in post 5 I have shown that the mass of a system of particles can be calculated, from base principles, as:

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

where:

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

Later on, I showed that in the "center of momentum" (not "center of mass") the mass of the system is:


M=\Sigma{\gamma&#039;_i m_i}

by virtue of the fact that \vec{V}=0 in the center of momentum frame. In the above:

\gamma&#039;_i=\frac{1}{\sqrt{1-(v&#039;_i/c)^2}}

where v&#039;_i represent the speeds of the particles transformed into the "center of momentum" frame.

There is NO justification to simply write :

M=\Sigma{\gamma&#039;_i m_i}-U

in the case of a system with non-null binding energy. One cannot slap the binding energy onto the formula of mass derived (from base principles) for a system with null binding energy and declare that the correct answer. The approach is unacceptable on many levels:

1. There is no derivation from base principles
2. There is no reason to combine mass with binding energy (the formula isn't even dimensionally correct, U should appear divided by c^2)
3. There is no clear understanding of what is U, (the total binding energy? )
4. Is U frame invariant? Or is it frame variant?

And the list can go on. This is not a proper derivation, it is just some stuff carelessly slapped together.

If the energies are related to the same frame of reference they should be additive.

This is not true.

And what about the potential energy related to interactions between different atoms (e.g. the binding energy of H₂)?

No idea, this is why I am objecting to the whole approach.

 

[PeterDonis]

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

If that happens to be the case, we would just write $U = \Sigma U_i$. I didn't say anything about the form of $U$; it doesn't have to be a single number or a single function.

In the case where a well-defined potential energy as a function of position exists, the form of $U_i$ will be $U_i = \Phi(x_i)$, where $\Phi$ is the potential energy function. So the total invariant mass of the system will be $M_0 = \Sigma \left( \gamma'_i m_i - \Phi ( x_i ) \right)$.

But, as I noted in a previous post, you won't be able to transform this expression into an arbitrary frame, because the potential energy function won't transform correctly, since its argument is a 3-vector in a particular frame, not a 4-vector. You can define a total 4-momentum vector for the system whose norm is $M_0$, but that's all.

To give a concrete example, consider a hydrogen-1 atom, i.e., a proton and an electron in a bound state. For concreteness, assume it's the ground state. As an approximation to the actual CoM frame, we'll work in a frame in which the proton is at rest, and we'll define the potential energy function $\Phi$ in this frame using the proton's field only, i.e., it's only nonzero for the electron. (This is a pretty good approximation because the proton's rest mass is so much higher than the electron's.) So we have $\Phi = k e^2 / r$, where $k$ is the Coulomb constant, $e$ is the magnitude of the charge on the electron (and the proton), and $r$ is the distance between them. That gives

$$
M_0 = m_p + \gamma_e m_e - k e^2 / r
$$

If we take $r$ to be the Bohr radius and calculate $\gamma_e$ using the classical formula for the orbital velocity at radius $r$ in an inverse square field, we get $\gamma_e \approx 1 + 2.66 \times 10^{-5}$, i.e., the kinetic energy of the electron is about $2.66 \times 10^{-5}$ times its rest mass, or 13.6 eV. For $k e^2 / r$, with $r$ the Bohr radius, we get 27.2 eV, so the overall result is M_0 = m_p + m_e + 13.6 eV - 27.2 eV = m_p + m_e - 13.6 eV. 13.6 eV is, of course, the usually quoted binding energy of the hydrogen atom.

I'll defer discussion of how to derive all this from "first principles" to a separate post.

I see no reason why the binding energies of the constituents can be represented by a single variable, U, as you did it above.

It's not a "single variable"; it's just a placeholder for whatever the total of the binding energies in the system turns out to be. Again, I made no assumptions whatever about the form of $U$. Its specific form will depend on the system.

if I understand the following discussion correctly he changed that to the rest frame of the total assembly of all atoms.

My intent all along was to work in the rest frame of the *system*. If the system is a single atom, that frame is the rest frame of the atom. If the system is composed of multiple atoms, that frame is the rest frame of the total assembly of all the atoms. There's no fundamental difference between these cases, because the atom itself is composed of multiple particles in relative motion, i.e., it's a "system", not just a single particle.

In that case he is actually adding energies.

That's what it amounts to, yes. But note that, as I said above, in the case where binding energy is present, you can only add energies this way in a particular frame, the frame in which potential energy is a well-defined function of position. The potential energy function won't transform correctly into an arbitrary frame.

If the energies are related to the same frame of reference they should be additive.

Yes, that's why I was careful to specify that I was doing all this in one particular frame.

 

[PeterDonis]

2. There is no reason to combine mass with binding energy (the formula isn't even dimensionally correct, U should appear divided by c^2)

I said I was using units in which $c = 1$. This is standard procedure in relativity physics: mass and energy have the same units, since they're just different forms of the same thing.

3. There is no clear understanding of what is U, (the total binding energy? )
4. Is U frame invariant? Or is it frame variant?

I tried to clarify this in my previous post.

This is not true.

Sure it is: energies that are all relative to the same frame are additive. They may not all transform correctly into another frame, but that's a different issue.

1. There is no derivation from base principles

Deferring that to a separate post since it raises further issues (one of which is the issue of how to do all this in a Lorentz invariant way when binding energy is present).

 

[xox]

I said I was using units in which $c = 1$. This is standard procedure in relativity physics: mass and energy have the same units, since they're just different forms of the same thing.

Doesn't matter, there is still no justification for your formula. More correctly, there is no derivation, you just put it by hand.

If that happens to be the case, we would just write $U = \Sigma U_i$.

What justifies that? How do you know that binding energy is additive?

In the case where a well-defined potential energy as a function of position exists, the form of $U_i$ will be $U_i = \Phi(x_i)$, where $\Phi$ is the potential energy function. So the total invariant mass of the system will be $M_0 = \Sigma \left( \gamma'_i m_i - \Phi ( x_i ) \right)$.

I see absolutely no reason why the above would be true since I see no derivation, you are just putting a formula by hand. It might be wrong, it might be right, you are simply tacking the binding energy to the formula of mass that I have derived in a rigorous fashion from base principles. What entitles you to do that? How do you even know that binding energy is additive? The fact that you can calculate the mass of an atom by adding (subtracting) its binding energy doesn't mean that you can do the same thing to a system of atoms.

Sure it is: energies that are all relative to the same frame are additive. They may not all transform correctly into another frame, but that's a different issue.

Total energy is the one that is additive (by virtue of being a component of the energy-momentum four-vector). The other forms of energy are not. You know very well that rest energy, for example, is NOT additive. I have no idea about the case of binding energy but I see no reason why it should be additive.

Deferring that to a separate post since it raises further issues (one of which is the issue of how to do all this in a Lorentz invariant way when binding energy is present).

Ok, I'll wait...

 

[DrStupid]

No, he's not adding energies.

That's not, how I understand it. In #21 he wrote

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.

and than in #23

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(N−1)/2 of the latter, since each pair of particles could have an interaction

with an additional explanation in #42

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 my understanding this means replacing "atom" by "system" in the entire text resulting in

M = \sum\limits_{i = 1}^N {m_i } + \sum\limits_{i = 1}^N {k_i } - U

where $M$ is the rest mass of the system, $m_i$ are the rest masses of the particles, $k_i$ the kinetic energies the particles (in the rest frame of the system), and $U$ the sum of the up to N(N−1)/2 binding energies for each particle pair. With c=1 this is identical to

M = \sum\limits_{i = 1}^N {m_i } + \sum\limits_{i = 1}^N {\left( {\gamma _i - 1} \right) \cdot m_i } - U = \sum\limits_{i = 1}^N {\gamma _i \cdot m_i } - U

All summands in this equation are energies even though the current convention requires the term mass for the rest energy. I cannot see why you disagree in this point, the more so as you already agreed with the result without binding energy.

No idea, this is why I am objecting to the whole approach.

If you have not idea, how can you make the proposal above? In PeterDonis' equation U includes the sum of all binding energies (see his explanation above) whereas you include the inner binding energies of the particles only. What was your intention behind this approach?

 

[xox]

To my understanding this means replacing "atom" by "system" in the entire text resulting in

M = \sum\limits_{i = 1}^N {m_i } + \sum\limits_{i = 1}^N {k_i } - U

There is no reason to believe that this is correct, since you are putting it together by hand, just the same way as he's been doing it for quite a while.

If you have not idea, how can you make the proposal above?

I made no proposal, I made a rigorous derivation for the case of absent binding energy, see post5. I am objecting to the way he (and now you) are trying to extend it to the case of non-null binding energy. Let me make it clearer:

There is no argument that:

M=\Sigma{\gamma&#039;_i m_i}

and

m_{0i}=m_p+\gamma_{ei}(v_e) m_e-u_i (for ONE atom)

From the above, it DOES NOT follow that, for a system of atoms:

M=\Sigma{\gamma&#039;_i m_i}-U

 

[DrStupid]

There is no reason to believe that this is correct

As I have no reason not do believe that this is correct, please teach me what I might have miss here.

 

[xox]

As I have no reason not do believe that this is correct, please teach me what I might have miss here.

Read the whole post, until the end. You are "calculating" the rest mass of a system (rest energy) as the sum of the rest masses of the components. That is known to be false. Not to mention that you are using the wrong kinetic energy ...

 

[DrStupid]

I made no proposal, I made a rigorous derivation for the case of absent binding energy, see post5.

I was talking about this:

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

To my taste that sounds like a proposal including binding energy.

From the above, it DOES NOT follow that, for a system of atoms:

M=\Sigma{\gamma&#039;_i m_i}-U

If you agree with this equation for a single atom why not for any system? What makes an atom so special compared to other systems?

 

[DrStupid]

You are "calculating" the rest mass of a system (rest energy) as the sum of the rest masses of the components.

No, I don't. Did you miss the γ in my equation?

 

[PeterDonis]

There is no argument that:

<...>

m_{0i}=m_p+\gamma_{ei}(v_e) m_e-u_i (for ONE atom)

I'm confused; this is exactly the formula (including binding energy) that you said I had not derived rigorously, because I put in the binding energy "by hand". If my formula is not justified, how is the one in the quote above justified? The atom is a "system" containing multiple particles.

 

[PeterDonis]

I guess you mean m_0=\gamma m_p+\gamma m_e-u

I assumed he was using the same approximation I did in a previous post, using the frame in which the proton is at rest as an approximation to the CoM frame.

 

[xox]

No, I don't. Did you miss the γ in my equation?

No, I didn't miss it, your formula is still wrong , you need to learn how to find your own mistakes. Hint: you applied \gamma to the whole atom. You need to apply it only to the electron.

 

[xox]

I guess you mean m_0=\gamma m_p+\gamma m_e-u

Nope, m_0=m_p+\gamma m_e-u.

If you agree with this equation for a single atom why not for any system? What makes an atom so special compared to other systems?

Because there is absolutely no justification to do the addition.
Because your addition contains an error.

 

[xox]

I'm confused; this is exactly the formula (including binding energy) that you said I had not derived rigorously,

One more time, I am not objecting to your formula for a single atom, I am objecting to you attempt at generalizing it for a system of atoms:

There is no argument that:

M=\Sigma{\gamma&#039;_i m_i}

and

m_{0i}=m_p+\gamma_{ei}(v_e) m_e-u_i (for ONE atom)

From the above, it DOES NOT follow that, for a system of atoms:

M=\Sigma{\gamma&#039;_i m_i}-U

 

[DrStupid]

Because there is absolutely no justification to do the addition.

If there is absolutely no justification why do you accept it for a single atom?

you applied \gamma to the whole atom. You need to apply it only to the electron.

No, I applied it to every individual particle no matter what kind of particle it is.

 

[PeterDonis]

I am not objecting to your formula for a single atom, I am objecting to you attempt at generalizing it for a system of atoms:

And my question is, *why* are you not objecting to the formula for a single atom? A single atom is a system containing multiple particles, so if you object to the formula for a system of atoms, you should also object to it for the system of particles that is a single atom. I don't understand what you think the difference is between these two cases.

Let me put this another way: if you think the formula is justified for a single atom, then presumably you could give a derivation of that formula for a single atom from first principles. If you do so, I will be able to take that derivation and use it to produce a derivation of the formula for a system containing multiple atoms. So if you believe the formula is justified for a single atom, you should already have the answer to the question that is still pending, namely, how to derive the formula for a system of multiple atoms.

 

[xox]

And my question is, *why* are you not objecting to the formula for a single atom?

Good point, I find your formula for a single atom just as objectionable. So, now you have an even tougher task: derive the formula for a single atom from base principles and then generalize it for multiple atoms. I have already listed the objections against your attempt at this generalization in my earlier posts (posts 58,60 and 62).

 

[xox]

No, I applied it to every individual particle no matter what kind of particle it is.

The individual particles have different "gammas". I already pointed out this mistake. Peter Donis pointed it out as well. You need to start paying attention.

 

[PeterDonis]

I find your formula for a single atom just as objectionable.

Ok, at least that's consistent.

So, just as a question of physics, independently of how the physics is represented in math, do you think that, for example, the rest mass of a hydrogen-1 atom is (1) less than, (2) equal to, or (3) greater than the rest mass of a proton plus the rest mass of an electron? In other words, if we measured all three rest masses very, very accurately, what do you think the results of those measurements would show?

 

[PeterDonis]

derive the formula for a single atom from base principles

I did that in an earlier post for the hydrogen-1 atom; that's how I got the result $M_0 = m_p + \gamma_e m_e - k e^2 / r$ (in the frame in which the proton is at rest). I can fill in more details since I only really sketched the derivation in that post, but first I'd like to know whether that's worth doing. Would that count as a derivation "from base principles"? If not, why not?

 

[xox]

I did that in an earlier post for the hydrogen-1 atom; that's how I got the result $M_0 = m_p + \gamma_e m_e - k e^2 / r$ (in the frame in which the proton is at rest). I can fill in more details since I only really sketched the derivation in that post, but first I'd like to know whether that's worth doing. Would that count as a derivation "from base principles"? If not, why not?

Putting results by hand doesn't count as derivation from base principles.
What we know for a fact (as I demonstrated in post 5) is that in the frame of the center of momentum:

m_0=\gamma_p(v_p) m_p+\gamma_e(v_e) m_e

Questions:

1. What makes you think that in the CoM frame v_p=0 as you chose it?
2. What makes you think that there is a CoM frame given the fact that the electron describes (complex) orbitals?
3. What makes you think that in the CoM frame the binding energy is \frac{ke}{r}
4. How do you generalize the MASS formula above to the case of the mass of multiple atoms, all moving at different speeds, with various makeups in terms of the number of protons, neutrons and electrons?

 

[PeterDonis]

Putting results by hand doesn't count as derivation from base principles.

Using the standard Coulomb potential for a two-body system of charges is not "putting results by hand".

What we know for a fact (as I demonstrated in post 5) is that in the frame of the center of momentum:

m_0=\gamma_p(v_p) m_p+\gamma_e(v_e) m_e

No, we do not know that, because you derived that formula under the assumption that no binding energy was present. In other words, the proton and electron cannot be in a bound state if this formula is true. So the hydrogen-1 atom could not exist.

1. What makes you think that in the CoM frame v_p=0 as you chose it?

It isn't exactly zero in the CoM frame, but it's very close to zero because $m_p$ is so much larger than $m_e$. I already explained that I was using this approximation in the previous post. I already said that that post was only a sketch, and asked you whether it was worth expanding on it. It would really be nice to get a plain yes or no answer to a simple question instead of a bunch of further questions that don't tell me whether you consider this whole line of discussion to even be moving towards your desired goal.

2. What makes you think that there is a CoM frame given the fact that the slectron describes orbitals?

If we're going to bring in QM, then please start a separate thread in the Quantum Physics forums. I've been talking about an idealized "classical" hydrogen-1 atom in which, in the proton's rest frame, the electron is following a classical circular orbit at the Bohr radius. I entirely agree that that's not what a real hydrogen-1 atom is like, but remember that this thread is in the SR forum and is supposed to be about how the rest mass of a composite system is determined from the properties of its constituents, in the classical (non-quantum) case.

3. What makes you think that in the CoM frame the binding energy is \frac{ke}{r}

Same answer as #1; the formula I gave is for the proton's rest frame, which is not exactly the same as the CoM frame but is very close.

4. How do you generalize the MASS formula above to the case of the mass of multiple atoms, all moving at different speeds, with various makeups in terms of the number of protons, neutrons and electrons?

I don't even want to consider this case until we have the case of a single atom worked out. You still haven't told me whether, if I gave a more detailed derivation of the formula I gave for the hydrogen-1 atom, that would even count as a "derivation from base principles". If it wouldn't, this whole subthread is pointless.

Also, I would really like a simple, straightforward answer to the physics question I posed in post #81.

 

[xox]

Using the standard Coulomb potential for a two-body system of charges is not "putting results by hand".
No, we do not know that, because you derived that formula under the assumption that no binding energy was present. In other words, the proton and electron cannot be in a bound state if this formula is true. So the hydrogen-1 atom could not exist.

I thought I was quite clear that the formula was derived for the case of absent binding energy. I was just trying to point out the origins of your derivation. Come to think of it, your formula becomes less and less viable since you are simply shoehorning a macroscopic theory into a particle theory. This makes your approach even more objectionable.

It isn't exactly zero in the CoM frame, but it's very close to zero because $m_p$ is so much larger than $m_e$. I already explained that I was using this approximation in the previous post. I already said that that post was only a sketch, and asked you whether it was worth expanding on it. It would really be nice to get a plain yes or no answer to a simple question instead of a bunch of further questions that don't tell me whether you consider this whole line of discussion to even be moving towards your desired goal.

No, it is not worth expanding. The more I think about it, the more objectionable it becomes, you cannot apply classical, macroscopic SR to a problem that really belongs in the realm of QED.

If we're going to bring in QM, then please start a separate thread in the Quantum Physics forums.I've been talking about an idealized "classical" hydrogen-1 atom in which, in the proton's rest frame, the electron is following a classical circular orbit at the Bohr radius.

True, I do not think that your "classical" approach is valid.

I entirely agree that that's not what a real hydrogen-1 atom is like, but remember that this thread is in the SR forum and is supposed to be about how the rest mass of a composite system is determined from the properties of its constituents, in the classical (non-quantum) case.

I think that at this point we should abandon the issue since it is clearly one of those case where forcing a "classical" approach is rendering unusable answers.

I don't even want to consider this case until we have the case of a single atom worked out. You still haven't told me whether, if I gave a more detailed derivation of the formula I gave for the hydrogen-1 atom, that would even count as a "derivation from base principles". If it wouldn't, this whole subthread is pointless.

Yes, it is pointless, time to drop it.

Also, I would really like a simple, straightforward answer to the physics question I posed in post #81.

Less than.

 

[PeterDonis]

you cannot apply classical, macroscopic SR to a problem that really belongs in the realm of QED.

No argument. But what if we made the problem legitimately classical by considering macroscopic objects bound together by some interaction?

Less than.

Ok, good, at least we agree on the actual physical observable.

 

[xox]

No argument. But what if we made the problem legitimately classical by considering macroscopic objects bound together by some interaction?

We could, this is why I was going to accept the formula for one atom, just to see where things are going. But things got totally off the tracks with the attempt at generalizing the formula to multiple atoms. Reminiscent of other attempts at shoehorning theories in places where they do not fit.

Ok, good, at least we agree on the actual physical observable.

Yes :-)
I am really sorry for being so tough but I do not like derivations that lack rigor. I appreciate all your efforts and your perseverance.

 

[PeterDonis]

We could

Ok, then we would need to find a macroscopic example to consider. Unfortunately the only ones that I can think of involve gravitational binding, things like planets orbiting the Sun, and using gravity raises other issues since we can't really model gravity with SR.

I am really sorry for being so tough but I do not like derivations that lack rigor. I appreciate all your efforts and your perseverance.

Thanks! I agree rigor is highly desirable in derivations (although physicists and mathematicians often disagree on what constitutes rigor ).

 

[xox]

Ok, then we would need to find a macroscopic example to consider.

There must be some literature on this subject, why don't we try finding it and returning to the subject when we can find a rigorous treatment.

Unfortunately the only ones that I can think of involve gravitational binding, things like planets orbiting the Sun, and using gravity raises other issues since we can't really model gravity with SR.

I agree , this would be a bad idea.

Thanks! I agree rigor is highly desirable in derivations (although physicists and mathematicians often disagree on what constitutes rigor ).

I am an applied mathematician :-)

 

[PeterDonis]

There must be some literature on this subject, why don't we try finding it and returning to the subject when we can find a rigorous treatment.

I've been looking and haven't found any good macroscopic examples so far; everything I find is concerned with binding energy at the microscopic level (chemical, atomic, and nuclear) or the gravitational case.

I am an applied mathematician :-)

I'm not a physicist (at least, not in terms of my actual day job), but my standards of rigor are a lot more like a physicist's than a mathematician's. A good thing to be aware of, both ways.

 

[DrStupid]

The individual particles have different "gammas". I already pointed out this mistake.

The particles have different gammas because they have different velocities. I do not see the problem.

 

[PAllen]

m_{0i}=m_p+\gamma_{ei}(v_e) m_e-u_i (for ONE atom)

From the above, it DOES NOT follow that, for a system of atoms:

M=\Sigma{\gamma&#039;_i m_i}-U

I think part of this is simply definition of U, independent of any pairwise model, such that it can even apply to non-linear interactions. You have a system of particles 'at infinity'. As they come together and bind, radiation is released. The mass of the system is reduced by the radiation released/c^2 (else conservation violated). We call this released energy = mass deficit * c^2 = binding energy = U by convention. U is generically a function of the system as a whole, with a maximum value defining the ground state of the system.