Nope. The equation e=sqrt(p^2 + m^2) is a statement of kinematics. It is a description of how the electron's energy depends on its motion
Correct, but only when the electron is not bound within an atom or a material. It is also valid when the free electron is not moving and then it gives for a SOLITARY electron within its own inertial reference frame that e=m . This is the energy that a free electron has at the vacuum level. Thus, equation cannot give the energy when the electron has an energy that is less than its free energy: i.e. when e<m.
It does NOT include other forms of energy, like electrostatic energy, or gravitational potential, or spring constants, or what-have-you.
Exactly! But when you include these energies and calculate the energy of the electron to be e=m
/ where m
/ is smaller than m, then according to this equation the electron cannot have kinetic energy.
You are incorrect (or, rather, Farisght is incorrect) to try to squeeze non-kinematic energies into the kinematic term.
No Farsight is not doing this: This is what you people are doing:
You are claiming that for an electron energy e less than m, the kinematic term p
2 is still valid. We are saying that when the energy of an electron e is less than m, there can be no kinetic energy whatsoever: Just as Einstein has proved. The equation e=sqrt(p
2+m
2) does then not apply.
Never mind, though, Johan! All of the experiments that disagree with you are fraudulent, right?
You are again projecting yourself, Sol and Ziggurat.
It is really you who ignore experimental results that we all agree are correct. The fact remains that for a hydrogen atom with rest mass M
H, formed from a proton with rest mass m
P and an electron with rest mass m
e one has that:
(m
P+m
e)*c
2-M
H*c
2=h(nu)=13.6 eV.
Do you deny that this equation is correct? Do you deny that this equation clearly proves that the rest masses of the proton and the electron must have decreased during bonding? So e=sqrt(p
2+m
2) cannot apply anymore since it would require that p
2 must be negative.
Now of course you want to wriggle out of this dilemma by postulating that e consists of both potential energy as well as kinetic energy, and that for this reason it only seems that the rest mass is decreasing. Why do you claim kinetic energy and potential energy? Because the Hamilton operator that is used can be written in opertor format as H
op=-p
op2/2m +V
op(r)=T
op+V
op.
When you solve the Schroedinger equation you obtain a spectrum of possible eigen-energies of which the lowest for a hydrogen atom is the s-state, e
s. Next you claim that this energy must consist of the sum of kinetic energy e
T and potential energy e
V. But this is quantum mechanically impossible, since neither the operator for kinetic energy T
op nor the operator for the potential energy V
op commute with H
op. This means that these two energies cannot manifest simultaneously with the total energy e
s. Thus there are no kinetic and potential components for e
s.
To wriggle yourselves out of this dilemma you tried to bully me to calculate <T
op> and <V
op> in order to argue that since these expectation values are not zero, there must be kinetic energy and potential energy. But this is absurd since these average values can only be obtained when one measures the kinetic energy of an ensemble of s-states and with each measurement you detroy the s-state and generate a new quantum state that is an eigen-state of the kinetic energy. This is further proof that the energy of the s-state does not have a kinetic energy component. The same for the potential energy.
I can generate at randon many operators which do not commute with H
op and calculate their expectation values. Say I call an operator a cat operator CAT
op, which does not commute with H
op and find that <CAT
op> is not zero: Does this mean that the s-state of the hydrogen atom has a cat within it? Stop being absurd!