UPDATE October 6, 2023: We corrected the error in our Lorentz transformation of E_x. The electric field in a frame has to be measured with a test charge q which is static in that frame.
Our error was to assume that the "inertia" of the test charge is the same as its mass-energy γ m.
----
The Feynman Lectures on Physics (1964) contains the same formulae as Wikipedia.
A lemma: the acceleration of a moving mass m under a force F
Let us have a moving mass m with a velocity v and a force F which accelerates it to the direction of v. What is the acceleration of m?
---> F
• ---> v
m
The total energy of m is
W = m c² / sqrt(1 - v² / c²).
When m moves a time dt, it gains the energy
dW = F v dt.
We have
dW / dv = m c² * -1/2 * (1 - v² / c²)^-3/2
* -2 v / c²
= m v γ³.
Then
dv = dW / (m v γ³)
= 1 / γ³ * F / m * dt
The acceleration is like the newtonian one, but must be corrected by the factor 1 / γ³.
The correct transformation is E_x' = E_x
a_x
• ---> ●
q Q
m
^ y
|
------> x
^ y'
|
------> x'
---> v
Let us have the following configuration in the laboratory frame. The test charge q is initially static and its mass is m. The charge Q is static.
Let us switch to a frame which moves to the positive x direction at a speed v. We denote
γ = 1 / sqrt(1 - v² / c²).
We denote parameters in the moving frame with the prime '.
We have
a_x = q E_x / m.
The test charge q does not feel a magnetic field in the laboratory frame or the moving frame, only an electric field.
The Lorentz transformation of the acceleration is:
a_x' = a_x / γ³.
Our test charge q is not static in the moving frame, but has a velocity v to the negative x' direction.
--> A_x
• --> v ●
q' Q
m
Let us use a test charge q' which is static in the moving frame. Its acceleration in the laboratory frame is
A_x = a_x / γ³.
Let A_x' be the acceleration of q' in the moving frame. In that frame, q' is static. Then
A_x = A_x' / γ³.
We obtain
A_x' = a_x.
That is, the electric field measured in the moving frame is
E_x' = E_x.
The transformation E_y' = γ E_y is approximately correct if Q and q are static
● Q
• q
m
^ y
|
-----> x
Let us switch to a moving frame, the speed is v to the positive x direction. We have:
a_y' = q / (γ m) * (v × B_z' + E_y'),
a_y = q / m * E_y,
and
a_y' = a_y / γ².
Let us assume that v is slow. Then
B_z' = -v C / c² * E_y
where C is very close to 1.
We have
q / (γ m) * (-v² C / c² * E_y + E_y')
= q / m * 1 / γ² * E_y
<=>
E_y' = 1 / γ * (1 + γ C v² / c²) E_y.
There γ C is close to 1, and approximately,
E_y' = 1 / γ * (1 + v² / c²) E_y
= γ E_y.
A moving Q and a static q
Q
● ---> v
• q
m
^ y
|
-----> x
In the laboratory frame, approximately,
B_z = v / c² * E_y,
a_y = q / m * E_y.
In the comoving frame of Q,
a_y' = q / (γ m) * E_y',
and
a_y' = a_y / γ².
Then
E_y' = E_y / γ.
Let us check the what Wikipedia says that E_y' should be:
E_y' = γ (E_y - v B_z)
= γ (E_y - v² / c² E_y)
= E_y γ * 1 / γ²
= E_y / γ.
Transformation of the magnetic field B_z if Q and q comove
Q
v <--- ●
. . . B_z'
v <--- •
q
m
^ y
|
-----> x
In the diagram we have a moving frame. In the laboratory frame Q and q are static.
Let us assume that Q is positive and q negative. The magnetic field B' lines point out of the screen (marked with . ). The magnetic force on the test charge,
F_y' = q v × B_z'
points down in the diagram. The acceleration is
a_y' = q / (γ m) * (v × B_z' + E_y').
Let us switch to the laboratory frame. There
a_y / γ² = a_y'
= q / m * 1 / γ * (v × B_z' + E_y')
= q / m * γ E_y'
<=>
(γ² - 1) E_y' = v × B_z'
<=>
B_z' = -(γ² - 1) / v * γ E_y.
Approximately,
-(γ² - 1) / v = -v² / c² * 1 / v
= -v / c².
The Lorentz transformation in Wikipedia is approximately correct in this particular case. Wikipedia claims that the formula is accurate, while it is not.
Transformation of B_z if Q and q have different velocities
Let us treat a case where B_z is non-zero in the laboratory frame.
Q
● ---> v
× × × B_z
• -------> V
q
m
The charge Q is positive and q negative. The magnetic field B_z points into the screen (marked with ×). We have, approximately:
B_z = v / c * E_y.
We denote
γ_V = 1 / sqrt(1 - V² / c²).
The acceleration of q is
a_y = q / (γ_V m) * (V × B_z + E_y).
We switch to the comoving frame of Q.
The velocity of q there is
V' = (V - v) / (1 - v V / c²).
The acceleration of q in the comoving frame is
a_y' = q / (γ_V' m) * E_y'.
The transformation of the acceleration is
a_y' = a_y / ( γ² (1 - v V / c²)² ).
We obtain
E_y' = γ_V' / γ_V * 1 / ( γ² (1 - v V / c²)² )
* (V × B_z + E_y).
It is a complicated formula. If v and V are slow, then, approximately:
γ_V' / γ_V * 1 / (γ² (1 - v V / c²)²)
= 1 + 1 / c²
* (1/2 (V - v)² * (1 + 2 v V / c²)
- 1/2 V² - v² + 2 v V)
= 1 + 1 / c² * (-1/2 v² + v V)
= 1 / γ + v V / c²,
where we dropped terms in which the velocities v or V appear in the third power. The result is approximately:
E_y' = (1 / γ + v V / c²) (E_y - V B_z)
= (1 - 1/2 v² / c² + v V / c²)
* (E_y - V v / c * E_y).
We drop terms with 1 / c³ and obtain
E_y' = E_y / γ.
This is also what Wikipedia predicts, as we saw in the section "A moving Q and a static q".
Transformation of B_x
Let us then check if the Wikipedia formula B_x' = B_x is correct.
------> B_x
• q moves up from the
screen at velocity V
^ y
|
------> x
In the laboratory frame there is no electric field, just a magnetic field B_x.
The acceleration of q is
a_y = q / (γ_V m) * V × B_x.
The acceleration in a frame moving at a velocity v to the positive x direction is
a_y' = a_y / γ².
If we would believe the Wikipedia formulae, then E_y' = 0. Then
a_y' = q / (γ_V' m) * V' × B_x',
where V' is the velocity of q in the moving frame. We obtain
1 / (γ_V γ²) * V × B_x
= 1 / γ_V' * V' × B_x'.
= 1 / γ_V' * 1 / γ * V × B_x'
<=>
B_x' = γ_V' / (γ_V γ) * B_x.
The formula B_x' = B_x is approximately correct.
The change in the inertia of q in the field of Q has a separate effect
In the calculations above, we ignored the change in the inertia of q in the electric field of Q. This is justified if the kinetic energy of q in various frames is much larger than the its potential energy in the field of Q.
If these energies are of the same order of magnitude, then it is easiest to do the calculations in the comoving frame of Q, and Lorentz transform the results to the desired frame. The concept of a Lorentz transformed electromagnetic field is not very useful then.
Conclusions
The Lorentz transformations of the electromagnetic field seem to be approximately correct in Wikipedia.
In practice, it is often better to use the Lorentz transformation of an acceleration, rather than transform E and B.
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