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

[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:

$$M=\Sigma{\gamma'_i m_i}$$

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:

$$M=\Sigma{\gamma'_i m_i}-U$$

 

[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:

$$M=\Sigma{\gamma'_i m_i}-U$$

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.