UPDATE October 6, 2023: This blog post is erroneous. We claim that the "inertia" of q is γ m, but that is not true if the velocity v and the acceleration a are parallel. In that case, the "inertia" is γ³ m.
The energy of the test charge q is γ m, but its "inertia" is different.
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UPDATE September 12, 2023: If, in the laboratory frame, Q moves radially relative to an initially static q, then the acceleration of q depends on the speed of of Q.
The Lorentz transformation probably is
E_x' = γ² E_x.
But if Q moves radially back and forth relative to q, then q probably sees an average electric field of
E_x' = γ E_x.
The difference comes from what happens when Q changes the direction of its velocity. The inertia shared by q and Q changes its direction, too.
That is, the "kinetic charge" (like kinetic energy for M) increases the Coulomb force, on the average, also in the radial direction. This would be analogous to gravity. We will check this.
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Maybe no one ever uses the Lorentz transformation formula of E_x in practical applications? It is conceivable that a purely theoretical formula could contain an error for 100 years.
The configuration
acceleration
a v
• -----> ● --->
q charge Q charge
m mass
^ y r = distance (q, Q) in
| laboratory frame
-------> x
Let us have the configuration in the diagram. The test charge q is initially static in the laboratory frame. The charge Q has the opposite sign to q.
The charge Q is carrying a very long ruler which extends all the way to the test charge q.
An observer stands on the ruler and measures the acceleration a' of q in the comoving frame of Q. A laboratory observer at the same moment measures the acceleration a of q in the laboratory frame.
We use the Lorentz transformation of an acceleration to transform between the laboratory frame and the comoving frame of Q. In the above formulae,
γ = 1 / sqrt(1 - v² / c²),
and u_x = 0, because q is initially static in the laboratory frame. Then
a' = a / γ³.
Let us then calculate the acceleration a' using the Lorentz transformation of the electric field E_x measured by the laboratory observer and the comoving observer at that moment at the position of q. We have
a = F / m
= q E_x / m,
and
a' = q E_x' / (γ m)
= q E_x / m * 1 / γ
= a / γ,
a different acceleration! The difference is large: if v = 0.1 C, then γ = 1.005 and the difference is 1%.
The denominator γ m comes from the fact that in the moving frame, q has kinetic energy and its inertia is larger.
Let us also present an alternative calculation, using the relativistic Lorentz force formula (with the green background) above. We calculate in the comoving frame.
We denote by V the velocity of q measured in the comoving frame. The 4-momentum of q in the comoving frame has the x component:
p'^x = γ m V,
where V = v, initially.
We have
dt' / dτ = γ,
where τ is the proper time of q. The 4-velocity components of q in the comoving frame are
U_x = γ V,
U_t = γ c.
The components of the electromagnetic tensor:
F^xx = 0,
F^xt = E_x' / c.
Then
dp^x / dτ = q E_x' / c * γ c
<=>
dp^x / dt' * dt' / dτ = γ q E_x'
<=>
d(γ m V) / dt' * γ = γ q E_x'
<=>
dV / dt' * m γ² = γ q E_x'
<=>
a' = 1 / γ * q E_x' / m.
<=>
a' = a / γ.
Let us still check with yet another formula. We have
d (γ m V) / dt' = q E_x' / c * c dt' / dt'
<=>
dV / dt' = 1 / γ * q E_x / m
<=>
a' = a / γ.
Checking the Lorentz transformation of accelerations
Who is right? Is the Lorentz transformation formula for accelerations incorrect?
Let us use the most basic Lorentz transformation formulae:
• - - - - - - - - - - - > •
t₀ = 0 t₁
x₀ = 0 x₁
Let a laboratory observer measure that an object is initially static at (t, x) = (0, 0), and later it is located at (t₁, x₁). He uses the familiar formula
x₁ = 1/2 a t₁²
to determine the acceleration. We assume that the final speed is slow relative to c.
The transformation gives
t₀' = 0,
x₀' = 0,
t₁' = γ (t₁ - v x₁ / c²),
x₁' = γ (x₁ - v t₁).
In the moving frame, the speed of the object is initially -v.
Let us determine a':
x₁' = 1/2 a' * t₁'² - v t₁'
<=>
x₁' = 1/2 a' * γ² (t₁ - v x₁ / c²)²
- γ v t₁ + γ v² x₁ / c²
<=>
γ x₁ = 1/2 a' γ² * (t₁ - v x₁ / c²)²
+ γ v² x₁ / c²
<=>
1/2 a t₁² = 1/2 a' γ (t₁ - v * 1/2 a t₁² / c²)²
+ v² * 1/2 a t₁² / c².
If t₁ is a very short time, we can discard t₁³ and t₁⁴ terms:
1/2 a t₁² = 1/2 a' γ t₁² + v² * 1/2 a t₁² / c²
<=>
(1 - v² / c²) a t₁² = γ a' t₁²
<=>
a / γ³ = a'.
The Lorentz transformation for an acceleration is correct.
Where is the error in the Richard Feynman (1964) derivation of the Lorentz transformation for E_x?
There is no error. We made an error in this blog post!
The Lorentz force formula is not Lorentz covariant
Above we showed that if q is initially static in the laboratory frame, and its acceleration is
a,
then its acceleration measured in the comoving frame is
a' = a / γ³.
An observer in the comoving frame sees q to approach Q at a velocity v. He may interpret that the inertia of q is γ m, because of v, and the electric field
E_x' = E_x / γ².
Let the observer stop q in the comoving frame and let it then fall freely toward Q. If the observer believes the Lorentz force formula, the acceleration now is
a'' = a / γ²,
because q no longer has the inertia from its kinetic energy.
Let us transform a'' to the laboratory frame:
a''' = a / γ⁵.
The laboratory observer believes the Lorentz force formula, and concludes:
a = a / γ⁴,
a contradiction!
The Lorentz force formula seems to be unaware of the fact that also the velocity v of a test charge greatly affects the Lorentz transformation of an acceleration.
Moreover, a radial movement of q in the field of Q changes the inertia of q, as we showed on August 29, 2023. The Lorentz force probably is oblivious of the changes in inertia.
Conclusions
Is there a calculation error above?
It is incredible if the problems with the Lorentz force formula and the Lorentz transformation of the electromagnetic field have gone undetected for over a century.
We will next calculate the accelerations of the test charge with our own methods which we introduced on August 29, 2023. It may be that we have to abandon the concept of one common, shared electric field E. We suspect that the concept of E does not store enough information about the movement of charges, and consequently, cannot calculate the path of a test charge precisely.
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