Well, lethe, let us resolve the problem “mass and rest mass”.

1. Everybody knows the famous Einstein’s formula

E = Mc² = [(Mo c²)² + p² c²]^½…………………(1)

Let me remind us that
E here is a total energy of body (a material point) in some , “Laboratory”, inertial reference frame;
Mo is its rest mass – the proper mass, i. e. mass measured in the inertial reference frame co-moving with body, where v = 0;
p = M v is a momentum of this body in “Laboratory”, where v is body’s velocity;
M is so called mass (inertial mass, dynamic mass, relativistic mass) of this body in “Laboratory”.

The term

Eo = Mo c² …………………………………….(2)

is called as “the rest energy” of this body (its proper total energy, measured in the inertial reference frame co-moving with body, where v = 0).

2. One very important notice: equation (1) is applicable only and only to a free body. If considered body (material point) interacts with some other body, expression (1) becomes invalid.
My question to everybody (and especially to lethe) is: “How (1) will transform in case of the interacting body?”
I will wait for your answers, before I continue further my argumentation…

Does it really?
I'm very much an amateur, so perhaps it's not surprising that I am surprised by this assertion.

I'm guessing that you mean that (eletrical, weak, and strong) field interactions alter a body's energy?

Pete,
your post proves how important this thread is - it was a major argument why I started it, although I did not wanted: too complex issue, as you will see later... But lethe very successfully ... forced me to do that.

If the argument was with that assertion, then why have you made it a premise to an argument?

Is this thread about whether expression (1) is invalid for interacting bodies? (this is my concern)
Or is it about the consequences of expression (1) being invalid for interacting bodies (a moot point, if the premise is false)?

Yuriy, I am highly dubious that the answer to this question will prove that there exists a particle with half the mass of an electron.

But anyway, I'll play along. Yes, for example a particle in an external potential will satisfy instead (E-V)²=p²+m²

and?

Pete,
This thread is about the mass and the rest mass; difference between them and why the mass is much more important in SRT them the rest mass and can not be eliminated from SRT by substitution by the rest mass without losing of a huge variety of physical interpretation, description and understanding of natural phenomena in SRT.

lethe,
Electron and its half rest mass analog in the spectrum of the elementary particles were left in the another thread, so, leave your doubts aside…
What is important here is to follow some uniform notifications at any expressions; including at your assertion that the main equation (1) at the presence of interaction of the given body (material point, particle) with potential energy U will be transformed to the following one:

(E - U)² = p² c² +(Mo c² )² …………………………(3)

(I do not think that we should in this Forum use the scientific system of units with c = 1; and following your slogan: “never, never, never use term mass anyhow as the rest mass”, I assumed that m in your notation was a rest mass, so, in mine it should be denoted as Mo; I also changed V on U to avoid possible mixing of it with a velocity)
If you do not have any objections, let us to left all as it is now and wait a little for possible other responses. If such responses will not be submitted, I will continue. OK?

So, because no other answers were posted, we can conclude with one that is expressed by equation (3).
But before we will go further, we should clarify one issue concerning to (3). And lethe that gave us this formula “owes” us this clarification.
And this clarification should answer the following questions:
Where from we get this equation (3)? Why we should be sure that it is a right one? Can one prove it validity?
Lethe, the next move is yours. We are waiting for proof of formula (3).

OK, so you admit that I was never in error, when I made that statement?
OK. If you don't mind, I will continue to use my preferred notations. Since I know what your notations mean, and you know what mine mean, we should have no further difficulties.

The equation follows from the equation of motion. It's really rather trivial. Do I really have to type up the calculations?

Please, be so kind... We are not alone in this thread and I am sure that many people will appreciate your efforts...

*sigh* OK, but next time you decide the readers of your thread would like to see a derivation, you do it yourself.

The lagrangian for a free particle is L₀=-m√(1-v²), so to find the motion for a particle in a potential, I subtract a potential: L=-m√(1-v²)-V. The canonical momentum is dL/dv=mγv=p. Inverting gives v=p/√(p²+m²)

Then the Legendre transformation gives H=pv-L = p²/√(p²+m²)+m²/√(p²+m²) + V = √(p²+m²)+V from which

(E-V)²=m²+p²

lethe,
for sake of economy of time only one question that stays unanswered:
what are the Lorentz transformation properties of quantities L, V and √(1-v²)
in your equation
L=-m√(1-v²)-V ?
Please, be very careful answering this question...

Last day of the semester has me very busy for the next couple of hours. I'm pretty sure I know the answer, but have no time to think carefully on it just now. I will be free in a few hours.

In order to understand the Lorentz transformation properties that you asked, we should restart from the begining.
In relativistic physics, the action that gives the equation of motion is : S = -m∫√dx^μdx_μ. This action is a Lorentz scalar. When you go to a specific reference frame, you can divide the integrand by dt and multiply it by dt in order to get: S = -m∫√(1-v²)dt, from which you can see that the Lagrangian is indeed L = m√(1-v²). When you write this expression, it does not have a good Lorentz transformation properties (the action, even written this way, is still a Lorentz scalar). When you go to the Hamiltonian, you do it with the Legendre transformation, as Lethe has shown. First you write the canonical momentum as p = ∂L/∂v in order to find that p = γmv. Then, you define the Hamiltonian as H = p.v - L in order to find that H = √(p² + m²). From this you can see that H² - p² = m², that is, (H,p) are a 4-vector whose "length" squared is given by m² (let us call this 4-vector P^μ).
Suppose that you have a potential energy Φ, you may be tempted to write that in this case H = √(p² + m²) + Φ or (H-Φ)² - p² = m². Obviously, by writting something like that, you get something that is wrong from the Lorentz transformation point of view, unless we say that Φ is the zeroth component of a 4-vector let's say A^μ = (Φ,A) which is the source of the force acting on the particle, and that in general, the relation between H, p, Φ and A will be that (H-Φ)² = (p - A)² + m², or (P^μ - A^μ)(P_μ - A_μ) = m².
What was given by Lethe was the special case where, in a specific reference frame, A = 0. In this case, you will indeed have H = √(p² + m²), however, if you go to another frame, you must Lorentz transform the 4-vector (Φ,0).
By the way, if you now take the Hamiltonian as H = √((p-A)² + m²) + Φ and do the inverse Legendre transformation to get the Lagrangian (you do it by writting that v = ∂H/∂p, solve p as a function of v) and write that L = p(v).v - H, you will get that L = -m√(1-v²)-V + A.v. Again, this is no good from the Lorentz transformation point of view, however the action will be now: S = ∫(-m√(1-v²)-V + A.v )dt which can be written as S = -m ∫ √dx^μdx_μ - A^μdx_μ which is of course a Lorentz scalar.

1100f,
after reading your post ... what are the Lorentz-transformation ptoperties of L, V and √(1-v²)
in equation
L=-m√(1-v²)-V ?
Please, no more stories, give a direct straigth and clear answer... if you can...

As I told you, L and V are not covariant, however ∫Ldt or ∫√(1-v²)dt are Lorentz scalar. V is the zeroth (or fourth) component of a 4-vector

Dear 1100f,
I heard it ... from your first post.
The problem is that evident form of Lorentz covariant features of L, V and
√(1-v²) in equation L=-m√(1-v²)-V is very important for a lot things I want to discuss in this thread.So, we have to bring them on...
Your constructive answer for now concerns only to V and it is:
V is 4th component of some 4D-Lorents-vector.
Good, but how about L and √(1-v²)?
And BTW, we should wait for lethe's answer. May be he will bring on light some another answer?
Until he will answer, you can think about such not too simple question:
If V is 4th component of some 4D-Lorentz vector, why we describe potential of the electromagnetic interaction as
ieΨ٭γμΨAμ,
i. e. as a Lorentz scalar term in the scalar Lagrangian density? (٭ is unitary conjugation) In other words, why we treat potential energy of the EM field as it is a Lorentz scalar? If you noticed in QED the Hamiltonian of interaction also is treated as a Lorentz scalar. How it corresponds to your statement that V is 4th component of 4D-Lorentz vector?
So, I can say that I keep you busy, is not it true?

First, I would like to mention that in quantum field theory, the action is a Lorentz scalar. The Lagrangian itself is not a Lorentz scalar, however, the Lagrangian is the integral over volume of the Lagrangian density so (if we denote the lagrangian density as L) we have S = ∫Ldtd³x = ∫Ld⁴x. d⁴x being a Lorentz scalar, this implies that the Lagrangian density is also a Lorentz scalar.

If you look at the interraction term in the QED lagrangian density, it is (you forgot a γ⁵, I don't know how to write here ψ bar = ψ*γ⁵). Any way, the interraction term is ψ*γ⁵γ^μψAγ_μ, which is a Lorentz scalar. If you decompose this sum you will see that you will have a part which is ψ*γ⁵γ⁰ψAγ₀ and a part which is ψ*γ⁵γ^kψAγ_k (where k are spatial indices from 1 to 3. A₀ is the potential that you mentioned, and it is not a Lorentz scalar.
Now you have two possibilities: the first one is to add the kinetic term of the EM field to the Lagrangian, the second one is to treat the EM field as an external given field. In case of an electrostatic field, the spatial part is zero and you are left only with the electric potential.

In fact this is equivalent for point particles (not fields) to what I wrote.
Remember what I wrote, in the presence of an external "force", the Hamiltonian is given by H = √((p-A)² + m²) + Φ. Since here we are dealing with a given "force", if A = 0, you get exactly what Lethe wrote. However, if not, you must use the formula that I wrote. which leads btw to the action given by S = ∫-m√(dx^μdx_μ) - A_μdx_μ.
Any way, you see that both in the point particle and both in QED, V is not a Lorentz scalar, but the zeroth component of a vector

1100f,
1. A small notice: we need separate gamma-5 matrix only to have of parity-pure psi-operators (then we use psi-bar); if we do not emphasize that feature the use of psi-unitary conjugated is enough and allows to write density of interaction potential as I did it.
2. Your analysis will in grate use in future development of this thread, but now we need the transformation properties of L and √(1-v²) in equation
L=-m√(1-v²)-V.
To recognize the importance of that, we can cogitate a little about case of the simple Mechanics of bodies (not elementary particles, but macroscopic bodies) in SRT. For instance, how looks the Newton’s 2 law at velocities close to c? What dynamics it pictures in sense of energy, mass, potential energy of body? How to get relativistic forces from relativistic potential energy? etc.
QED and QFT are a good theories, but what if pupil will ask us about simple ball attracted by fast moving meteorite somewhere in Cosmos? QED and QFT will not help us then: we have to have relativistic generalization of the Classic Dynamics too. Do you agree?

Because further development of the main issue of this thread in the regime of “dialog with lethe” does not go further, I have to do it in regime of monologue… I am sorry for that…
So, I continue my first post on this thread.

3. The main goal for now is to show what is the generalization of the Newtonian mechanics for velocities close to c. I. e. what is the Mechanics of SRT?

Before we do that, let me one more time attract your attention to basic formula (1).
In his famous work on SRT Einstein, considering transformation of whole body into an energy, had proven that

E = Mc² = [(Mo c²)² + p² c²]^½…………………(1)

where M was considered as the mass of moving body and E was considered as the total energy of this moving body.

Actually we have here two fundamental formulas:

E = Mc² …………………………………….……(4)
and
M = Mo / (1 - v²/c²)½……………………….……(5)

Formula (5) is very important by many reasons, but here we will mention the first one: because namely M keeps unchanged the definition of such an important characteristic of body as its momentum at the same meaning of notion “the velocity of body”:

p = Mv = Mo v/(1 - v²/c²)½……………………….(6)

First formula here, (4), actually can be considered as the relativistic generalization of classic notion “the kinetic energy of the inertial moving (including a macroscopic one) body”, which absolutely unexpectedly revealed the existence of the rest energy

Eo = Mo c² ……………………………………….(2)

as an energy associated with a resting body. So that if we want to keep valid separately the classic notion of the kinetic energy, we have to consider it as the following:

K = Mc² - Mo c² = Mo c² [1/(1 - v²/c²)½ - 1]………(7)

the classic limit of which (at c rushing to infinity) is a familiar K = Mo v²/2.

So, the relativistic generalization of the kinetic energy is (7), the relativistic generalization of mass is (5), the relativistic generalization of momentum is (6) and the relativistic generalization of the total energy of a free body is

E = Mc² = Eo + K……………………………….……(8)

at introducing of the new notion (purely relativistic one) of “the rest energy of body” defined by (2).

And now we are ready to generalize all those formula and notions for the case of an interacting body.