The momentum of the field is
4/3 m v,
if m is the mass-energy of the field, and v is its linear velocity.
The stress-energy tensor of the field of a static charge Q
r
● • ----> Ex
Q observer
^ y
|
------> x
Do the pressures in T make the electric field "stable", in the sense that if it would be a solid matter object, it would not collapse or explode?
Q __-----
o r | | 1
----___| ring section,
1 points distance 1
out from screen
cylinder "o" points
out of the screen
Let us look at a very long uniformly charged cylinder, and the electric field pressure forces on a section of a ring drawn around it.
Let the radial electric field at a distance r be
E = 1 / r.
Let the negative radial pressure be
pr = -1 / r²,
and the tangential positive pressure be
pt = 1 / r².
The radial component of the force from the positive pressure is
Fr = 2 * 1/2 / r * 1 / r²
= 1 / r³.
The negative pressure difference between the left and the right side of the section is 2 / r³. That adds a force
Fr' = -2 / r³.
The area of the right side is larger by a factor 1 + 1 / r. That adds a force
Fr'' = 1 / r³.
We conclude that the pressures balance each other and make the electric field stable, if it were a solid matter object.
Poincaré stresses
We have not yet found a precise formulation of the Poincaré stresses (1906). Henri Poincaré proposed that non-electromagnetic forces must hold the "charge elements" of the electron together, to make the electron stable. In literature, it is claimed that the momentum in these force fields resets the factor 4/3 back to 1.
• • •
\ | /
• -----×----- • dQ string keeps dQ in place
/ | \
• • •
Q
We can keep the "charge elements" dQ of Q stable by adding strings which attach them to the center of Q.
The 4/3 problem for a solid matter sphere
Above we showed that we can imitate the stress-energy tensor of the static electric field with a solid matter object where the radial negative pressure is balanced by a positive tangential pressure.
But the negative pressure tends to infinity at r = 0. We cannot realize this in a solid object?
If we move such an object, does it exhibit the 4/3 problem of the electric field?
The repulsion on the charge elements dQ comes from the fact that the energy in the integral of T⁰⁰ over space is reduced if they move farther from each other.
In the case of a solid object, there is no such repulsion.
If we take an almost cubical section of a spherical shell, then there is a negative pressure -p in the radial direction, and a positive pressure p in the tangential directions. If we integrate over the entire sphere, then positive pressures win?
If we Lorentz transform for a velocity v to the x direction, that "mixes" the time coordinate t and the spatial coordinate x. The pressure T¹¹ in the original T will contribute momentum to T⁰¹' in the Lorentz transformed stress-energy tensor T'.
Maybe we forgot shear stresses inside the sphere?
-----> push shear
-------
Q ● pull <-- | | cube
-------
<-- push shear
If we look at a perfect cube, the radial negative pressure would pull it toward the charge Q. But shear forces on its sides must balance the pull from the negative pressure.
The shear forces arise from the components -σxy, etc., in the stress-energy tensor T. They are not zero on the sides of the cube.
The integral of T¹¹ over the electric field of a charge Q
Let us look at a spherical shell of a radius r around a charge Q. Then
•
/
/ r
/ θ
Q ● --------------> x
T¹¹ = ε₀ Ex² - 1/2 ε₀ E²
= ε₀ E² (cos²(θ) - 1/2),
where θ is the angle from the positive x axis. On a shell, |E| is constant.
The integral of cos²(θ) - 1/2 over a half shell is
π / 2
∫ cos²(θ) * 2 π sin(θ)
0
- 1/2 * 2 π sin(θ) dθ
= 2 π (1/3 - 1/2)
= -1/3 π.
We see that the integral of T¹¹ is not zero over the whole sphere. The positive pressure wins.
We calculated in a frame where Q is static. Let us Lorentz transform to a frame which moves to the x direction at a speed v. Then both T⁰⁰ and T¹¹ contribute to T⁰¹' in the new stress-energy tensor T'. No other non-zero elements of T contribute.
Lorentz transforming a pressure to a moving frame
pebbles
• • • -->
-a <-- ● ● --> a
M <-- • • • M
• ---> v
observer
-------> x
We implement a positive pressure between the two masses M in the way that both throw pebbles to each other.
Let us have an observer moving at a speed v to the right. What does he see at his own time t' = 0? In his coordinates, the time t' = 0 if
t' = 0
= 1 / sqrt(1 - v² / c²) * (t - v x / c²).
The observer sees the rightmost M at a later t time than the leftmost M.
The observer would think that momentum conservation is broken if he only looks at the masses M. There is too much momentum to the right.
^ t pebbles flying to the right
| • • •
| / / /
| / / /
| _____-------- t' = 0
-----------------------------------> x, t = 0
Since the line t' = 0 is skewed up, the density of pebbles flying to the right is reduced on the line t' = 0, and there are more pebbles flying to the left. The observer sees that the "pebble field" holds some momentum to the left. This is from length contraction.
We conclude that the "pressure field" of pebbles does hold momentum, if viewed in the moving frame of the observer. Special relativity is right in this.
The solution to the paradox of the solid sphere: the negative pressure singularity at the center
Henri Poincaré found the right explanation for the apparent 4/3 paradox in 1906.
In the case of a solid spherical object, the positive pressure does win the negative pressure in the outer layers of the sphere. But at the center we must have a core where there is just a negative pressure. Otherwise, we would have infinite pressures at the center.
If we think of the electron as a pointlike object, then we may imagine that the core of a negative pressure is a limit, an infinitesimal volume with an infinite negative pressure.
Discussion of the "paradox"
Is there any paradox in the pressurized solid sphere?
Let us have a moving sphere. Then we add there a negative radial pressure and the positive tangential pressures to the outer layers of the sphere. After that, the outer layers possess more momentum, and the core less momentum.
We can add the pressures extremely quickly, essentially instantaneously. The momentum moves in an instant into a new location in the sphere. Is this paradoxical?
It could not happen if the momentum was bound to moving mass-energy. But pressure can, in principle, contain zero energy.
One cannot "grab" the momentum contained in a pressure field, like one can grab moving particles.
The apparent "paradox" is in the nonintuitive behavior of special relativity. The momentum contained in pressure does not behave in the same way as momentum in moving particles.
In the case of the electron, the problem or the paradox is in the renormalization, or regularization, of the infinite energy of its field. The 4/3 problem is subject to that greater problem.
Conclusions
We conclude that Henri Poincaré solved the 4/3 problem already in 1906. The problem is equivalent to bookkeeping of the momentum in a moving solid sphere where there is a negative radial pressure and positive tangential pressures. In special relativity, pressure can carry momentum, and that momentum can behave in a surprising way.
The 4/3 problem shows that a significant portion of the momentum in a moving electric field comes from the positive pressure in the field. But in many cases we can ignore this pressure momentum because there always exists a negative pressure which counteracts it. If there is no negative pressure, then the system of charges is "exploding". Then the configuration is like in the Lorentz pebble example above. The role of the negative pressure is replaced by the change of simultaneousness in a moving frame.
The 4/3 problem in the case of an electron, or a pointlike charge, is a renormalization or regularization problem of the infinite energy in its electric field. It is not clear if 4/3 makes the renormalization problem any harder.
Now that we understand how the 4/3 problem is solved, let us look at the kinetic energy of the field paradox that we formulated in our August 24, 2024 blog post.
No comments:
Post a Comment