UPDATE Nov 15, 2018: There is something wrong with either classical electromagnetism or our proof below. Our post on Nov 15, 2018 studies that problem.
Theorem 1. The potential energy of a particle adds to its inertial mass, in other words, to the rest mass. The exact amount of inertial mass which the potential energy adds to each particle depends on the geometry of the setup.
Note that if there are unknown forces between the electron and the platform, then the inertia of the electron can be arbitrarily high. The proof below gives the minimum possible inertia. See our Nov 15, 2018 post.
Theorem 1. The potential energy of a particle adds to its inertial mass, in other words, to the rest mass. The exact amount of inertial mass which the potential energy adds to each particle depends on the geometry of the setup.
Note that if there are unknown forces between the electron and the platform, then the inertia of the electron can be arbitrarily high. The proof below gives the minimum possible inertia. See our Nov 15, 2018 post.
Proof. Let us give a more detailed proof where the potential energy of a particle has to be added to its rest mass, if we believe the conservation law of the center of mass.
z r
e- <--- F
O/
|\_______________________o m + V
/\ ---> F
====================
- - - - - - - -
x y
-V +V
Let the electron have an inertial mass m when it is static at position z.
A weightless man living on a not infinite uniform negatively charged plane uses an energy store V to pull an electron from position z closer to the plane at x. The plane must not be infinite because then the potential energy of the electron would be infinite.
We assume that the plane is very massive, so that the round trip of the electron only changes the tilt of the plane a negligible amount.
Since the electron comes closer at x, the plane will be tilted by a small angle, leaning down on the right.
We assume that the plane is very massive, so that the round trip of the electron only changes the tilt of the plane a negligible amount.
Since the electron comes closer at x, the plane will be tilted by a small angle, leaning down on the right.
The magnetic force between the electron and the plane is proportional to v^2 where v is the relative speed of the electron and the plane. If the man moves the electron slowly, we can ignore the magnetic force.
The man moves the electron from position x to y. Since the electric field is not perfectly uniform close to the plane (force F), he moves the electron along a slightly curved path where the electron stays at a constant electric potential relative to the plane. He then does not need to use any energy to win horizontal forces on the electron.
The man has to win the inertia of the electron to move it. He pulls on a long weightless rod which is attached to an electrically neutral mass of m + V. We claim that m + V is the inertial mass of the electron at x.
Once the man is at y, he moves the electron to r and stores the released potential energy V in an energy store at y.
Then he moves the electron back to y. This time he uses a rod attached to a weight m to win the inertia of the electron. He moves the electron at a constant electric potential so that he does not need to win horizontal forces.
The end result of the process is that the electron is back at its original horizontal position. Horizontal forces affecting the plane were negligible. Thus, the center of mass of the plane is at the original horizontal position, though the plane did tilt slightly when the electron completed its loop.
The energy V has moved a distance s = y - x to the right.
A weight V + m moved the distance s to the left and then the weight m moved back the same distance.
If the energy stores do not have any interaction with the plane, then we can treat them as separate objects that have an inertial mass V. The center of mass of the plane stayed still. The center of mass of energy V moved the distance s to the right and it was balanced by a weight V moving s to the left.
We conclude that the center of mass of the whole system the plane & the electron & energy stores & weights stayed still.
If the inertial mass of the electron would have been some m + V' != m + V at x, then the center of mass would have moved. QED.
If we have a fully symmetric setup of two charges, then obviously we cannot add the full potential energy to the inertial mass of both particles. That would breach conservation of momentum.
The energy of an electric field is stored in a density |E|^2 where E is the electric field strength. In our plane example above, the extra energy in the electric field is mostly stored just upwards from the electron. That is the obvious reason why the inertial mass of the electron appears to grow by the full amount V, and the plane's inertial mass remains constant.
Theorem 2. The inertial mass of a particle is the same to each direction at a location x. The mass may change when the particle moves, because the geometry of the setup changes.
UPDATE Nov 18, 2018: This theorem is wrong. See our post about inertial mass Nov 17, 2018.
Proof. We only prove this for the special setup of Theorem 1. The general proof is left for future.
Theorem 1 essentially handles the case where the particle moves perpendicular to the force field.
____________________
V x ________________ e-z|
ds | | pipe
_________________| |
dV y ___________________ r|
\O
|
/\
==============
- - - - - - - -
Let us repeat the consideration when the energy stores are vertical relative to each other. The stores are rigidly attached to the plane. There is a pipe as in the diagram. The pipe extends very far from the charged plane and is rigidly attached to the plane. The plane is very massive, so that the round trip of the electron does not tilt it significantly.
The distance between x and y is small, ds.
Let the inertial mass of the electron be m at z.
The man uses the energy V to pull the electron to x.
Then he lets the (small) energy dV pull on the electron so that its force exactly cancels the electric repulsion. The man uses a rod attached to a weight m + V to win the inertia of the electron an pushes the electron down to y.
Then he lets the electron to slide away in the pipe back to r. A weightless string runs from y to the electron. The string stores the energy at y.
In the final trip r -> z the man lets the energy store pull on the string so that it exactly cancels the electric repulsion. The man uses a rod attached to a weight m to win the inertia of the electron and pushes the electron back to z.
The energy V moved from x to y. A weight m + V moved up ds and a weight m moved down ds. There was no net vertical force on the plane in any phase, and its center of mass did not move. The plane tilted a negligible angle.
We see that the center of mass of the whole system was conserved. If the inertial mass of the electron would have been different from m + V at x, the center of mass would have moved. QED.
An ages-old problem is where is the potential energy stored in a bound system. Suppose that we have a bound system where the particles are static and the forces between them do not change significantly if we move a particle a short distance dr to an arbitrary direction.
How is the energy divided between the particles? We cannot take the system apart and weigh each particle individually. But we can measure the inertia of each particle if we use a rod to move it a distance dr. The sum of these measurements must be equal to the inertia of the whole system when all particles are moved simultaneously. The breakdown of inertial masses is a measure of how much potential energy each individual particle possesses.
The full energy-momentum relation should obviously contain all the particles in a system as well as the energy of their fields. When the particles move, the field energy moves. For any move dr of an individual particle, the field energy is moved some dr'.
https://en.wikibooks.org/wiki/Modern_Physics/Potential_Momentum
In the link, the energy-momentum relation is given as
(E - V)^2 = (p - q)^2 + m^2.
We assume c = 1 and just one spatial dimension. A potential momentum q is subtracted from the momentum p to make the equation Lorentz-invariant. Can this work?
p E
^ ^
| /
| /
|/
------> m
If we draw m and p as orthogonal vectors, the mass-energy relation (without potential) states that E is the length of their vector sum. The formula above means that we reduce the length of the p vector by amount q (potential momentum) to make the sum vector E shorter by amount V.
p - q E - V
^ ^
| /
| /
------> m
The procedure is equivalent to some "agent" drawing part V of the kinetic energy of the particle and storing the energy to some energy storage at location x.
Suppose that a man far away uses a rod and a weight m to move the particle to location y in the potential field.
When the particle leaves the potential, it gets the energy back from the storage.
But how can the potential field store energy at location x and give back the same amount of energy at location y? The field after the particle left is the same as before it arrived. The energy stored at x must have flowed from x to y.
What force moved the energy V from x to y? It must have been the man who used the rod. He had to overcome the inertia of energy V.
The potential momentum concept forgets that someone has to move the energy V, too. That error leads to thinking that rest mass (= inertial mass) of the particle stays the same.
Location of the extra inertial mass
If we have a fully symmetric setup of two charges, then obviously we cannot add the full potential energy to the inertial mass of both particles. That would breach conservation of momentum.
The energy of an electric field is stored in a density |E|^2 where E is the electric field strength. In our plane example above, the extra energy in the electric field is mostly stored just upwards from the electron. That is the obvious reason why the inertial mass of the electron appears to grow by the full amount V, and the plane's inertial mass remains constant.
Theorem 2. The inertial mass of a particle is the same to each direction at a location x. The mass may change when the particle moves, because the geometry of the setup changes.
UPDATE Nov 18, 2018: This theorem is wrong. See our post about inertial mass Nov 17, 2018.
Proof. We only prove this for the special setup of Theorem 1. The general proof is left for future.
Theorem 1 essentially handles the case where the particle moves perpendicular to the force field.
____________________
V x ________________ e-z|
ds | | pipe
_________________| |
dV y ___________________ r|
\O
|
/\
==============
- - - - - - - -
Let us repeat the consideration when the energy stores are vertical relative to each other. The stores are rigidly attached to the plane. There is a pipe as in the diagram. The pipe extends very far from the charged plane and is rigidly attached to the plane. The plane is very massive, so that the round trip of the electron does not tilt it significantly.
The distance between x and y is small, ds.
Let the inertial mass of the electron be m at z.
The man uses the energy V to pull the electron to x.
Then he lets the (small) energy dV pull on the electron so that its force exactly cancels the electric repulsion. The man uses a rod attached to a weight m + V to win the inertia of the electron an pushes the electron down to y.
Then he lets the electron to slide away in the pipe back to r. A weightless string runs from y to the electron. The string stores the energy at y.
In the final trip r -> z the man lets the energy store pull on the string so that it exactly cancels the electric repulsion. The man uses a rod attached to a weight m to win the inertia of the electron and pushes the electron back to z.
The energy V moved from x to y. A weight m + V moved up ds and a weight m moved down ds. There was no net vertical force on the plane in any phase, and its center of mass did not move. The plane tilted a negligible angle.
We see that the center of mass of the whole system was conserved. If the inertial mass of the electron would have been different from m + V at x, the center of mass would have moved. QED.
An ages-old problem is where is the potential energy stored in a bound system. Suppose that we have a bound system where the particles are static and the forces between them do not change significantly if we move a particle a short distance dr to an arbitrary direction.
How is the energy divided between the particles? We cannot take the system apart and weigh each particle individually. But we can measure the inertia of each particle if we use a rod to move it a distance dr. The sum of these measurements must be equal to the inertia of the whole system when all particles are moved simultaneously. The breakdown of inertial masses is a measure of how much potential energy each individual particle possesses.
The full energy-momentum relation should obviously contain all the particles in a system as well as the energy of their fields. When the particles move, the field energy moves. For any move dr of an individual particle, the field energy is moved some dr'.
Potential momentum
https://en.wikibooks.org/wiki/Modern_Physics/Potential_Momentum
In the link, the energy-momentum relation is given as
(E - V)^2 = (p - q)^2 + m^2.
We assume c = 1 and just one spatial dimension. A potential momentum q is subtracted from the momentum p to make the equation Lorentz-invariant. Can this work?
p E
^ ^
| /
| /
|/
------> m
If we draw m and p as orthogonal vectors, the mass-energy relation (without potential) states that E is the length of their vector sum. The formula above means that we reduce the length of the p vector by amount q (potential momentum) to make the sum vector E shorter by amount V.
p - q E - V
^ ^
| /
| /
------> m
The procedure is equivalent to some "agent" drawing part V of the kinetic energy of the particle and storing the energy to some energy storage at location x.
Suppose that a man far away uses a rod and a weight m to move the particle to location y in the potential field.
When the particle leaves the potential, it gets the energy back from the storage.
But how can the potential field store energy at location x and give back the same amount of energy at location y? The field after the particle left is the same as before it arrived. The energy stored at x must have flowed from x to y.
What force moved the energy V from x to y? It must have been the man who used the rod. He had to overcome the inertia of energy V.
The potential momentum concept forgets that someone has to move the energy V, too. That error leads to thinking that rest mass (= inertial mass) of the particle stays the same.
The Aharonov-Bohm effect
Yakir Aharonov and David Bohm noted, among many other physicists, that a potential affects the phase of a wave function, and that happens also in a constant potential where the particle does not meet "field lines" of the force field.
This fact is, of course, evident in the Schrödinger equation. The interference pattern of a particle depends on the choice of the zero level of the potential V.
The effect has been measured for a magnetic vector potential but not for a static electric potential.
All rest mass is potential energy?
The rest mass of the electron can be considered potential energy which the particle acquired in pair creation.
The same might hold for all particles with a non-zero rest mass.
Let us look at the Schrödinger equation:
-1 / (2m) * d^2Ψ/dx^2 + V Ψ = i dΨ/dt.
What is the role of the rest mass m versus the role of V?
The hamiltonian operator
H = p^2 / (2m) + V
does not include the (potential) energy in the rest mass. Why does the equation then give accurate predictions?
A more correct hamiltonian would have m + V in the place of V.
The free Schrödinger particle is
exp(-i (E t - p x)),
where E only includes the kinetic energy. Why it does not contain the rest mass energy?
If we include the rest mass energy, the free particle would be
exp(-i ((m + E) t - p x))
where E only contains the kinetic energy.
A "full-energy" Schrödinger equation is
-1 / (2(m + V)) d^2Ψ / dx^2 + (m + V) Ψ
= i dΨ / dt.
If we are looking for a solution of type
Ψ(x) * exp(-i (m + E) t)
for the full-energy equation, we get on the right side of the equation
(m + E) Ψ(x) exp(-i (m + E) t)
and on the left side
-1/(2 (m + V)) d^2Ψ(x)/dx^2 * exp(-i(m+E)t)
+ (m + V) Ψ(x) exp(-i (m + E)).
We can divide by exp(...) to remove a common factor. We remove the term m Ψ... on both sides. In 1 / (2(m + V)), V is negligible. The end result is the familiar time-independent Schrödinger equation.
We conclude that in the Schrödinger equation we can treat m as potential energy.
The hamiltonian operator
H = p^2 / (2m) + V
does not include the (potential) energy in the rest mass. Why does the equation then give accurate predictions?
A more correct hamiltonian would have m + V in the place of V.
The free Schrödinger particle is
exp(-i (E t - p x)),
where E only includes the kinetic energy. Why it does not contain the rest mass energy?
If we include the rest mass energy, the free particle would be
exp(-i ((m + E) t - p x))
where E only contains the kinetic energy.
A "full-energy" Schrödinger equation is
-1 / (2(m + V)) d^2Ψ / dx^2 + (m + V) Ψ
= i dΨ / dt.
If we are looking for a solution of type
Ψ(x) * exp(-i (m + E) t)
for the full-energy equation, we get on the right side of the equation
(m + E) Ψ(x) exp(-i (m + E) t)
and on the left side
-1/(2 (m + V)) d^2Ψ(x)/dx^2 * exp(-i(m+E)t)
+ (m + V) Ψ(x) exp(-i (m + E)).
We can divide by exp(...) to remove a common factor. We remove the term m Ψ... on both sides. In 1 / (2(m + V)), V is negligible. The end result is the familiar time-independent Schrödinger equation.
We conclude that in the Schrödinger equation we can treat m as potential energy.
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