But we did not then analyze the extra inertia of q inside the electric field of Q. Let us do it now.
A static Q and a moving frame
r = distance (q, Q)
---> F
• --> a ●
q Q
m
The test charge q is negative and the charge Q positive.
The acceleration a is from the familiar Coulomb force
a = 1 / m * 1 / (4 π ε₀) * q Q / r²
= F / m.
Let us then have a test charge q moving to the right at the speed v.
r
--> A
• ---> v ●
q Q
m
What is the acceleration A? Let us switch to comoving frame of q, which we denote with the prime'. The electric field in that frame at q is
E' = E,
that is, the same as in the laboratory frame. Then
a' = a,
and
A = a / γ³,
where
γ = 1 / sqrt(1 - v² / c²).
However, this ignores the deceleration which is due to the extra inertia of q close to Q. If the absolute value of potential energy of q inside the field of Q is
W > 0,
we have claimed that the extra inertia in a tangential motion relative to Q is
W / c²,
and in a radial motion
2 W / c².
Let us assume that v is not very large and W is neither. Let q move toward Q for a short time t. How large is the deceleration correction because of the increased inertia? The potential energy of q changes by
W = v t F.
Since the inertia grows by 2 W / c², the velocity v decelarates by
v * 2 W / (m c²)
= 2 t v² / c² * F / m.
The correction to the acceleration in the laboratory frame is
-2 v² / c² * a.
We have
A = a (1 - 3/2 v² / c²).
The corrected A is
Ac = a (1 - 7/2 v² / c²)
= a / γ⁷.
The corrected acceleration in the comoving frame of q is
ac' = a / γ⁴,
quite different from the value a which we got from the Lorentz transformation of the electric field E!
v <--- • ●
q Q
What if q moves to the left at a speed v? Let us consider this in the comoving frame of q.
The Lorentz transformation of the electric field E gives the same result as in the previous case where v was to the right.
Since the inertia of q inside the electric field of Q is decreases to the left, q gets an additional acceleration to the left. The result for ac' is the same as for when v is to the right.
The case where both q and Q are negative
A <-- • ---> v ●
q (neg) Q (neg)
m
Let W be the absolute value of the potential of q in the electric field of Q, and W is small. We have claimed that the inertia of q is diminished by
W / c²
in a tangential motion relative to Q, and diminished by
2 W / c²
in a radial motion.
As q moves closer to Q, the inertia of q decreases, which speeds up its velocity v. The acceleration A has to be corrected to have a smaller absolute value. We conclude that also in this case,
ac' = a / γ⁴.
Transformation of the electric field Eₓ
Theorem. If we define the electric field E in a frame through the acceleration of an initially static test charge q in that frame, then the transformation of the electric field is
Eₓ' = Eₓ / γ⁴,
where Eₓ is the electric field in the frame where the charge Q is static.
Lorentz transformations of the electric and magnetic fields
v
<---
\ | /
\|/
----- ● Q ---
/|\
/ | \ field lines
^ y'
|
-----> x
In the configuration above, the Lorentz transformation of Ey is
Ey' = γ (Ey - v Bz).
In the formula, a non-zero value of Bz reveals that the frame is moving relative to Q.
We could define
Ey' = γ Ey,
where Ey is the electric field in the frame where Q is static. This definition would be analogous to the transformation for Eₓ which we introduced in the previous section. In this sense, our transformation is kind of a Lorentz transformation for the electric field.
Conclusion
We have a hypothesis that a test charge q feels an increased or decreased inertia in the field of another charge Q. Though this hypothetical change is not present in quantum mechanical systems like the hydrogen atom.
If the inertia hypothesis is correct, it affects significantly the radial acceleration of q relative to Q. The "Lorentz transformation" of Eₓ becomes
Eₓ' = Eₓ / γ⁴,
where
γ = 1 / sqrt(1 - v² / c²),
v is the radial velocity of q relative to Q, the x axis is parallel to (q, Q), and Eₓ is the electric field in the frame where Q is static.
We could define a "magnetic field" which would allow us to drop the dependency of the transformation formula to a frame where Q is static. The "magnetic field" would code the speed of Q in a frame.
In our September 9, 2023 blog post we aimed to reveal the effect of inertia on the Lorentz transformation of Eₓ, but we miscalculated and got confused.
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