strong field E' + E
\ | /
\ | /
----------------------------------
| weak field E'
q • ---> v
| weak field E'
---------------------------------- Q
/ | \
/ | \ --> pW
strong field E' + E
"extra" field energy W
moves with the charge q
We have recently analyzed the experiment in more detail, looking at the Poynting theorem and the Poynting vector.
As the small charge q moves, it will carry a large amount of field energy W outside the tube. Inside the tube, the strong field E of the pipe is negligible. Only the weak field E' of the charge q is present.
From where does the momentum come to the extra energy W? The inertia of q is not changed
The momentum pW may be much larger than the momentum of q itself.
Let us accelerate q. The field lines of q bend to the left. The Poynting theorem suggests that W gains its momentum from the bent field lines of q which push the charge Q in the pipe to the left.
The inertia of q does not increase from the extra energy W. This solves the question which we have been investigating in this blog for many years! The inertia of a charge q does not depend on its potential, assuming that the charge Q which is causing the potential is held static.
Does W have any kinetic energy as it moves? No
The charge q does not feel the force of Q at all. We do not need to do any extra work as we accelerate q to the right, compared to the case where the pipe Q would not exist.
Thus, even though the extra energy W is accelerated to the right, it does not gain any kinetic energy of its own.
The bent field lines of q do not do any work on Q.
How to prevent a perpetuum mobile in the presence of retardation
Let q and Q have the same sign and q = Q. In this blog we have been perplexed about retardation which makes the repulsion of two accelerated charges q and Q to differ from
F = 1 / (4 π ε₀) * q Q / r²
in laboratory coordinates. If we accelerate q and Q toward each other, they will feel less repulsion. Does this allow us to construct a perpetuum mobile?
field lines field lines
bent to left bent to right
\ | | /
---- ● --> a a <-- ● ----
q Q
Since the field lines are sparser on the line from q to Q, the electric field and the repulsion is weaker.
But this does not allow a perpetuum mobile. Squeezing the field lines to the left and to the right requires energy. When q and Q come close, the left side of their joint field gets its energy from the field of q, and the right side from Q. We had to use enough energy to build the joint field of q and Q. We could not save any energy.
The energy which we save in the repulsion is consumed in squeezing and bending the electric lines of force. This is, of course, also a consequence of Poynting's theorem.
If q and Q would have opposite signs, then the attraction between q and Q would be surpfisingly weak, and we could harvest "too little" energy by moving q and Q close to each other. In this case, the surplus energy is radiated away as radio waves, when the joint field of q and Q settles to its low-energy state. No energy is lost.
Use gravity to construct a perpetuum mobile?
We try to prove that the "extra energy" W below does possess kinetic energy if it moves to the right.
In our earlier blog posts we realized that one can "grab" the mass-energy of W directly through gravity. In electromagnetism alone, we cannot "grab" W because it has no electric charge.
● M
\ ≈ c
v
W
\ | /
\ | /
----------------------------------
|
q • ---> v
|
---------------------------------- Q
/ | \
/ | \
Let us accelerate and move a large mass M at almost the speed of light to the left of the extra energy W. The mass M is stopped then.
When W is aware of the gravity of M, but before q knows anything about M, we move q to the right.
We make q static before the field of M arrives. The electric field of q will force the mass-energy of W to move farther from M. That requires some energy W'. Where does W' come from?
The electric field lines of q will "lift" W up in the gravity field, as the experiment settles down. The Poynting vector is not aware of gravity, and cannot describe the flow of energy to lift W up?
Could the squeezed gravity field of M supply the energy to lift W up? Why would it do that? If W were an ordinary lump of matter, there is no reason why the field of M would lift it up.
The electromagnetic lagrangian density
Can we combine the gravity potential to the lagrangian?
The potential would affect the term
What happens if we simply add the gravity potential to the term? Our analysis in the preceding section suggests that then energy is not conserved in the process which we described. But Noether's theorem says that energy is conserved in a stationary point of the action
S = ∫ L(x) dt.
Simulating gravity with acceleration
We can simulate gravity with acceleration. Can we construct a pathological process without gravity?
^ E
| pipe
Q -------------------------- ---> a
× B
<--- E
○ B'
q • ---> a
× B'
<--- E
○ B
-------------------------- ---> a
|
v E
Initially, the charged pipe Q is static. Its electric field E point outward from the pipe and is essentially zero inside the very long pipe. Also, the small charge q is static.
We suddenly start to accelerate the pipe to the right. The field E bends to the left, and a horizontal field component appears inside the pipe.
Simultaneously, we move q to the right and stop it soon. The electric field of q, E', becomes like the one in the Edward M. Purcell diagram.
field E' of q
The "tangential" part in the diagram is the electric field induced by a pulse of a magnetic field B' which we created as we moved q.
When the pulse B' meets the forming field E, the Poynting vector
S = 1 / μ₀ * E × B'
indicates that energy is flowing outward from the center of the pipe. The electric field
E + E'
on the circle has a larger energy density because it is stronger. The Poynting vector S describes that extra energy moving outward.
But where did that extra energy come from? We did not need to do much work moving the small charge q. Energy should be conserved locally, according to Poynting's theorem.
The extra energy may come from the magnetic field B associated with the forming field E?
The magnetic fields B and B' point to opposite directions. The extra energy probably comes from the magnetic field
B + B'.
The Poynting vector
1 / μ₀ * E' × B
describes the missing energy in the magnetic field moving outward. The total extra energy traveling with the pulse in the field of q would be zero.
Let us then wait until the field E' of q is updated also outside the pipe. In the comoving frame of the pipe, the extra energy W outside the pipe in the field E + E' will be moved to the right, "up", if we think of the acceleration a as gravity. From where does the energy to lift W "up" come?
<--- "gravity"
W ---> movement "up"
\ | / E + E'
\ | /
F ---> Q ---------------------------------- ---> a
| E'
q • ---> a
| E'
---------------------------------- ---> a
/ | \
/ | \ E + E'
The strongly bent pulse field E' of q pushes the charge Q in the pipe to the left. The force F accelerating the system has to do extra work when the pulse of E' passes the surface of the pipe, if the pipe is moving in the laboratory frame at that moment.
Some process has to use energy to push W up in the accelerating frame of the pipe. The force propelling the pipe will probably give the rest of the kinetic energy to W. It is like a person standing in an accelerating rocket and pushing a weight W up. The person has to do a little work. The motor propelling the rocket supplies the rest of the large kinetic energy increase of W.
Now we face a problem: the energy to push W up may be considerable. Where does it come from?
When we started accelerating the pipe, a radio wave carried quite a lot of energy away from the system, since the field of Q was deformed. Could it be that some of that energy still lingers and helps to lift W up?
Let us consider the case where q is attached to the pipe. Then the energy to propel W to the right clearly comes from the force pushing Q against the bent electric field of q.
If we suddenly move q up, the field is bent even more, as we accelerate q. When we decelerate q at the end, the field is less bent. Is this enough to give the energy to lift W up?
The kinetic/potential energy of W
Our thought experiments try to find out if behaves like any object which contains the mass-energy worth W.
Does W possess kinetic energy when it moves? Does W have a gravity potential, or an equivalent "gravity" potential in an accelerating frame?
Let us analyze the gravity example further.
W
| E + E'
------------------------- +Q
| E'
● --> • +q -->
M | E'
------------------------ +Q
| E + E'
W
We move the mass M very quickly close to the left end of the pipe. As M moves to its position, we can harvest from M the energy that M gains as it descends in the gravity potential of W.
Wikipedia states:
"the decrease in electromagnetic energy is the work done by the electromagnetic field on the gravity field plus the work on matter".
In this blog we have previously shown that the Einstein equations do not have a solution for any dynamic problem. We will forget about that for now.
Static configuration. Let us assume that M is close to the pipe, and the system has settled down. We move q upward. We have to do the work W' against the bent lines of force of E. How does this energy W' end up being gravity potential energy of W? What does the Poynting vector say?
Q
●
| E ^
| |
/ ○ B' pulse
v
• --> move W' = work done
q
If we move q suddenly to the right in the diagram, a magnetic pulse B' is born. The Poynting vector says that the horizontal component of E, and B' transports energy away from q in the diagram:
1 / μ₀ * E × B'.
The electromagnetic wave which is born from the movement, transports the work W' to the far parts of the field of Q. It is just like in the case where we would simply push q directly toward Q.
The tangential part of the field of q points to the same direction as the horizontal component of E. The sum of these two electric fields contains most of the energy W'. The energy of the magnetic pulse B' is small.
Dynamic configuration. Let us then assume that we move q before the field E has had time to bend to the left. Then we have to do much less work. The energy W' is much smaller.
However, when the wave reaches Q, the field E is already settled down. The wave must now be carrying as much energy as in the static case. Where did the energy come from?
The obvious source of the energy is the energy which is deforming E and bending the field lines. The existence of the field of q at the same location means that some of the deformation energy is spent to increase the value of W'.
But how can the wave of q steal energy from the wave associated with E? It is a linear theory. The waveform of E does not change in any way if it encounters the wave of q. Does the wave of q cause a destructive interference to the wave of E?
There probably is no paradox at all in this. If a wave hits any static electric field, then the energy of the electric field in the wave is locally increased, and that is compensated with a decreasing energy at some other location.
Further analysis of the Poynting vector
---> accelerate
Q
● -------------- pipe wall
| E
|
/
| ^ ○ B' pulse
| \ E'
v |
|
|
q • --> move
Let us first forget about q. As the field E inside the pipe settles down to a non-zero value E ≠ 0 , the Poynting vector must indicate energy flowing either from the walls of the pipe, or from the end of the pipe into the pipe.
Let us then assume that E has not yet settled down.
We use an inertial frame which is comoving with the pipe at the moment that the wave of E' meets the wave of E'. Then the only significant magnetic field is B'.
The Poynting vector
1 / μ₀ * (E + E') × B'
shows a significant energy flow radially outward from the center of the pipe when the waves of E and E' overlap. Where does that energy come from?
Conclusions
Let us close this long blog post. We need to study in more detail what is the electromagnetic field of a charge q like inside a strong gravity field.
An equivalence principle suggests that the field of a static charge q should look similar to the one when q is being uniformly accelerated. Then there probably is both a magnetic field and an electric field.
Under a uniform acceleration, q must continuously send energy to the far parts of its electric field, to supply them kinetic energy. How is this compatible with the fact that q cannot send any energy when it is static in a gravity field? Is this a contradiction?
Let us imagine an elastic sheet of rubber which is under a constant acceleration because of a rocket pushing it in the middle. There is energy flowing from the middle to the edges of the sheet. But if the rubber sheet is supported by a pole on Earth, there is no energy flow. Does the Poynting vector understand the difference? The "field" looks locally the same in both cases, but in just one there is an energy flow.
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