Moving charge Q and an approaching test charge q
Q ● ---> v
W = q E Δy
^ u ^ q E Lorentz force
| |
• q -----> q v × B
/
\
/ spring
| F = -q E
v
^ y
|
------> x
Let Q be positive and q negative. The charge Q creates an electric field E, and a magnetic field B which steers q to the right, according to the Lorentz force.
The spring and the force F on it cancels out the electric attraction q E between q and Q.
There is no retardation in the field of Q. If Q is to the y direction from q at the time t = 0, then the electric field E of Q points directly to the y direction at the time t = 0.
Switch to a comoving frame of Q
● Q
W = q E Δy
^ u ^ q E Lorentz force
| |
-v <--- • q
/
\
/ spring
| F = -q E
v
^ y'
|
-----> x'
The electric field E' points to the y' direction, and there is no magnetic field B. The momentum of q does not change with time.
The Lorentz force says that q does not have an acceleration in the x' direction in this case.
This seems to break Lorentz covariance. The problem is that the Lorentz force does not take into account the momentum held by the energy
W = q E Δy,
which q receives from the field of Q. By Δy we denote the distance which q moves to the y direction.
In the first diagram, the energy W holds momentum to the positive x direction (and this also explains the magnetic field B: we do not really need to consider B at all).
Is the Poynting vector aware of the momentum in W?
Maybe the Lorentz force simply is an imprecise description of the electromagnetic action? Let us check if the action understands the momentum held by W.
Does the Poynting vector
1 / μ₀ * E × B
understand that in the first diagram, the energy W flowing to q contains momentum to the positive x direction? And that in the second diagram, W does not contain momentum to the x direction?
The energy W flows further to the spring. We assume that in the first diagram, that flow does not contain x momentum. The momentum is retained by q.
Second diagram. The magnetic field of the negative charge q moving to the y direction makes the Poynting vector 1 / μ₀ * E × B to point toward q on both sides of q:
● Q
|
| E
v
^ u
|
-v <--- • q
B × ○ B
---> <---
E × B E × B
Above, × denotes a magnetic field B line produced by q coming up from the screen, and ○ is a field line going into the screen. We ignore the magnetic field associated with the -v velocity vector.
The Poynting vector 1 / μ₀ * E × B brings field energy to q. Is the diagram symmetric versus the energy flows, or does q receive more energy from the left, as q moves to the left at the speed v?
First diagram.
Q ● ---> v
|
| E
v
^ u
|
• q
B × ○ B
---> <---
E × B E × B
The first diagram looks more symmetric. The Poynting vector seems to bring the same energy flux from the left and from the right. But the charge q should accelerate to the right. There is something wrong.
Does the magnetic field of Q change this somehow?
Replace Q with a part of a wire with an electric current: analysis of the Poynting vector to q
Let us check if the Poynting vector handles a pure magnetic field correctly. We put an opposite static charge -Q close to Q, so that the electric field E of them is essentially zero at q.
-Q ●
Q ● ---> v
^ u
|
• q
Bq × ○ Bq
---> <---
E × Bq E × Bq E = 0
^ y
|
------> x
The fields close to q are now:
1. The electric field produced by Q and -Q is EQ = 0.
2. The magnetic field BQ produced by Q is significant.
3. The fields Eq and Bq produced by q are significant.
The Poynting vector Eq × Bq cannot bring any momentum to q. Only
1 / μ₀ * Eq × BQ
might possibly bring momentum. But Eq is almost radial to q, and Eq × BQ is orthogonal to Eq. If q accelerates to the right, is the electric field deformed enough, so that the Poynting vector could give momentum to q?
(Richard Fitzpatrick, 2014)
Conservation of the electromagnetic 4-momentum is this formula:
The term j × B is the force which pushes q to the right, to the positive x direction. We are interested in the coordinate i = 1, which is the x coordinate.
Is -∂ (ε₀ E × B) / ∂t then non-zero?
Make q very heavy and push it suddenly toward Q and -Q: momentum conservation works correctly
We suddenly push q to the positive y direction, so that it acquires a velocity u. The x coordinate of Q, -Q and q is the same after the push operation.
Can q somehow acquire momentum to the right, from the electromagnetic field? The speed of light is finite. The momentum must come from a close location to q.
-Q ●
Q ● ---> v
|
| line L of cylindrical symmetry
|
|
^ u
|
•
q
^ y
|
○ ----> x
z points out
of the screen
Let us assume that q is extremely heavy, so that its acceleration to the right is very small, even though it gets a lot of momentum to the right from the magnetic field BQ. Since q moves almost exactly to the positive y direction, then the fields Eq and Bq of q are almost exactly cylindrically symmetric relative to the line L which contains q and runs to parallel to the y axis.
Let us assume that the distance from q to Q and -Q is very large, and that Q and -Q are very close to each other.
Then the electric field EQ-Q of Q and -Q is almost exactly zero close to q. The magnetic field BQ of Q is almost exactly parallel to the z axis, and is almost exactly uniform.
The field momentum is
The field momentum close to q in Eq × Bq points to the y direction, on the average.
Let us study the field Eq of q at some distance r.
Daniel V. Schroeder (1999) drew the electric field lines if a charge q is initially static, but is pushed and acquires a speed u upward. The diagram shows the field lines after a little while.
What is the vector Eq × BQ like?
Below q in the picture, that vector points to the left, and above q, the vector points to the right. As q moves up, the field will contain more momentum to the left. This may explain how q itself gains momentum to the right?
The Poynting vector does not bring any new energy to q, but the movement of q changes the amount of x momentum in the integrated Poynting vector.
If we stop the y movement of q, then Q and q absorb the x momentum stored in the field.
The electromagnetic momentum conservation probably works correctly in this case.
The Lorentz force does work correctly
Let us look again at the case where the electric attraction of Q is counteracted with a spring, rather than putting a charge -Q close to Q.
We may assume that the spring is attached to the laboratory table.
● Q
W = q E Δy
^ u ^ q E Lorentz force
| |
-v <--- • q
/
\
/ spring
-v <--- × attachment
| F = -q E
v
^ y'
|
-----> x'
In the moving frame, the spring moves to the left along with q. Now we see that the movement of q to the left will slow down as time passes. That is because q receives an energy packet W which is static relative to Q. But q has to pass W to the spring which is moving to the left. Then q has to give up some of its own momentum to the left. We conclude that q accelerates to the right, just as it should do.
Does the energy W which q gains in the field of Q have a zero momentum relative to Q?
● Q
| EQ
v
W = F Δy
^ F
^ u | Coulomb's force q EQ
|
-v <--- • q
^ y
|
------> x
We still have to check if the energy packet W really is static relative to Q. Or does the velocity -v of q affect this?
If the Lorentz force formula, or Coulomb's force formula is correct, then q does not receive any momentum in the x direction. Then W really has a zero momentum relative to Q.
We can take as an axiom that Coulomb's force points exactly toward a static charge Q. If Q is accelerating, then this gets more complicated.
Conclusions
Classical electromagnetism fared well in this test of ours. Lorentz covariance and momentum conservation were worked ok.
We can take as an axiom that Coulomb's force points directly toward a static charge, and that it gives an energy packet W which is static relative to Q.
Then we can derive the Biot-Savart law for a wire segment. The derivation of Zile and Overduin (2014) can be corrected this way. We will write a new blog post about Biot-Savart. We will also look at the August 28, 2024 overlapping electric fields again, now that we understand 4-momentum conservation better.
No comments:
Post a Comment