The electric scalar potential ϕ and the magnetic vector potential A are prime examples.
We work in 1+1 dimensions.
The Lorentz transformation (ignoring the gamma parameter, we assume v is small) is
ϕ = ϕ - v A
A = A - v ϕ.
Our new energy-momentum relation is
E^2 = p^2 + (m + V)^2,
where V is the scalar potential of the particle. When we Lorentz transform the relation, should we Lorentz transform m and V, too?
Traditionally, m is considered a scalar which stays the same in a Lorentz transformation. If we identify V with m, then V should not be transformed either. The energy-momentum relation transforms then like:
(E - v p)^2 = (p - v E)^2 + (m + V)^2.
If we try to Lorentz transform also m and V, we end up with a relation like:
(E - v p)^2 = (p - v E - v m - v V)^2
+ (m + V)^2.
If we have p = 0, that relation would indicate that the inertial mass of the particle is double of what we are used to. It may be more logical to think that the transformation of E in the momentum term already includes the transformation of m and V.
We are left with the question in which frame should we measure m or V. The simplest case is when we measure in the rest frame of the object which produces the electric potential V. We assume there is no magnetic field.
Conjecture 1. The correct energy-momentum relation is
E^2 = p^2 + (m + V)^2,
and we must not Lorentz transform m or V when we move to a new frame. V is the work we need to do to bring the particle from infinity to its current position, in the rest frame of the object that produces the field. We assume V does not change with time and that we move the particle under a constant potential V.
Suppose that we have an electron at rest. Then
E = m + V.
If we Lorentz transform, we get
E^2 = (-mv - v V)^2 + (m + V)^2.
We can interpret -v V either as the vector potential of V or as the momentum of the rest mass V.
How does Conjecture 1 differ from a traditional way to handle the potential in the energy-momentum relation?
Some people write
(E - V)^2 = (p + A)^2 + m^2,
where A is the vector potential associated with the scalar potential V. The value E' = E - V is considered the energy of the particle.
If A = 0 and p = 0, and we Lorentz transform the above, we get
(E - V)^2 = (-v (E - V) - v V)^2 + m^2
= (-v E)^2 + m^2,
where we Lorentz transformed V to get a vector potential A = -v V.
The Lorentz transformation of our new relation is
E^2 = (-v E)^2 + (m + V)^2.
The kinetic term -v E is the same as in the traditional case. Our new energy-momentum relation gives roughly the same results as a traditional one when |V| and |p| are small.
Suppose that we have an electron at rest. Then
E = m + V.
If we Lorentz transform, we get
E^2 = (-mv - v V)^2 + (m + V)^2.
We can interpret -v V either as the vector potential of V or as the momentum of the rest mass V.
How does Conjecture 1 differ from a traditional way to handle the potential in the energy-momentum relation?
Some people write
(E - V)^2 = (p + A)^2 + m^2,
where A is the vector potential associated with the scalar potential V. The value E' = E - V is considered the energy of the particle.
If A = 0 and p = 0, and we Lorentz transform the above, we get
(E - V)^2 = (-v (E - V) - v V)^2 + m^2
= (-v E)^2 + m^2,
where we Lorentz transformed V to get a vector potential A = -v V.
The Lorentz transformation of our new relation is
E^2 = (-v E)^2 + (m + V)^2.
The kinetic term -v E is the same as in the traditional case. Our new energy-momentum relation gives roughly the same results as a traditional one when |V| and |p| are small.
All rest mass is potential energy?
Can we consider also the rest mass as potential energy? We can bring an electron from nonexistence by creating a pair. What is the role of the antiparticle in this?
e+
e-
\O
|
E /\ ----------> E
=========================
- - - - - - - - - - -
x y
Let us introduce once again the man standing on a finite charged plane, who we had in our Nov 3, 2018 blog post. This time, the man uses an energy store E to create the electron far away from the plane by "detaching" it from the corresponding positron. That is, the man uses energy to create a pair.
We claim that both particles have an inertial mass 511 keV before the man brings the electron close to the plane, where the inertial mass of the electron is 511 keV plus the potential V.
The man moves the positron and the electron horizontally from x to y. After that, he fills an energy store at y by letting the electron move away and annihilate.
The inertial mass of the positron was 511 keV, that we know. The inertial mass of the electron had to be 511 keV plus V, otherwise the center of mass would have moved.
In this example, it makes sense to say that the rest mass of the electron is potential energy.
If the universe contains electrons that do not have an antiparticle, is it still reasonable to say that their rest mass is potential energy?
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